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Other approaches to simplify $\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}$ I want to simplify the trigonometric expression $\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}$.
My approach,
Here I used the abbreviation $s,c,t$ for $\sin x$ and $\cos x$ and $\tan x$ respectively,
Numerator is, $$\frac{s^2}{c^2}-s^2=\frac{s^2-s^2c^2}{c^2}=\frac{s^4}{c^2}=s^2t^2.$$
And denominator is,
$$\frac{2t}{1-t^2}-2t=\frac{2t^3}{1-t^2}.$$
So $$\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}=s^2t^2\times\frac{1-t^2}{2t^3}=\sin^2x\times \frac1{\tan 2x}=\frac{1-\cos 2x}{2\tan 2x}.$$
I'm looking for alternative approaches to simplify the expression.
| My method uses Euler's formula and Computer Algebra Systems.
Let $\,X := e^{i\,x}.\,$ Then
$$ s := \sin(x) = \frac{X-X^{-1}}{2i}, \;\;
t := \tan(x) = \frac{X-X^{-1}}{i(X+X^{-1})}, \\
c := \cos(x) = \frac{X+X^{-1}}{2}, \;\;
t_2 := \tan(2x) = \frac{X^2-X^{-2}}{i(X^2+X^{-2})}. $$
Simplify a little to get
$$ s = \frac{1-X^2}{2iX}, \;\;
c = \frac{1+X^2}{2X}, \;\;
t = \frac{1-X^2}{i(1+X^2)}, \;\;
t_2 = \frac{1-X^4}{i(1+X^4)}. $$
Now get with Computer Algebra System factoring
$$ f := \frac{t^2 - s^2}{t_2-2t} =
\frac{(1 - X^2)(1 + X^4)}{4i(1 + X^2)X^2}. $$
Rearrange the factors of the expression as
$$ f = \frac{(1-X^2)}{i(1+X^2)}\frac{(1+X^4)}{4X^2}
= \frac{t\,c_2}2 = \frac{\tan(x)\cos(2x)}2 . $$
Alternatively rewrite it as
$$ f = \frac{(1-X^2)^2}{4i^2X^2}
\frac{i(1+X^4)}{(1-X^2)(1+X^2)} = \frac{s^2}{t_2}
= \frac{\sin(x)^2}{\tan(2x)}.$$
Yet another alternative is
$$ \frac18\frac{(1-X^2)(1+X^2)(1+X^4)}{2iX^4}
\frac{4X^2}{(1+X^2)^2} = \frac{s_4}{8c^2} =
\frac{\sin(4x)}{8\cos(x)^2}.
$$
Which one of these alternatives is simpler than the
others is not clear unless a precise definition of
simplicity is given in this context.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Which number is bigger: $e^{\sqrt{5}} + e^{\sqrt{7}}$ or $2e^{\sqrt{6}}$? Let $f:(0, + \infty) \rightarrow \mathbb{R}$ be $f(x) = e^{\sqrt{x}}$.
Which number is bigger: $e^{\sqrt{5}} + e^{\sqrt{7}}$ or $2e^{\sqrt{6}}$?
I would like to ask for a hint on how to approach this question, I don't have an idea.
| We have
\begin{align*}
e^{\sqrt{5}} > 2^{\sqrt{5}} > 2^1 \implies \frac{1}{e^\sqrt{5}} < \frac{1}{2} \tag{$*$}\\
e^{\sqrt{7}} > 2^{\sqrt{7}} > 2^1 \implies \frac{1}{e^\sqrt{7}} < \frac{1}{2} \tag{$**$}
\end{align*}
Now because $\ln{x}$ is an increasing function, we can write:
\begin{align*}
\ln(e^{\sqrt{5}} + e^{\sqrt{7}}) &= \ln\left(\frac{e^{\sqrt{5}} + e^{\sqrt{7}}}{e^{\sqrt{5}}e^{\sqrt{7}}}\right) + \ln(e^{\sqrt{5}}) + \ln(e^{\sqrt{7}})\\
&= \ln\left(\frac{1}{e^{\sqrt{5}}} + \frac{1}{e^\sqrt{7}}\right) + \sqrt{5} + \sqrt{7}\\
(*), (**) &< \ln\left(\frac{1}{2} + \frac{1}{2}\right) + \sqrt{5} + \sqrt{7}\\
&= \sqrt{5} + \sqrt{7}\\
(?) &< 2\sqrt{6}\\
&= \ln(e^{2\sqrt{6}})
\end{align*}
Notice that for $(?)$ we can write:
\begin{align*}
\sqrt{5} + \sqrt{7} < 2\sqrt{6} &\iff (\sqrt{5} + \sqrt{7})^2 < (2\sqrt{6})^2\\
&\iff 12 + 2\sqrt{35} < 24\\
&\iff \sqrt{35} < 6
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $x^2+2(\alpha-1)x-\alpha+7=0$ has distinct negative solutions... Let $\displaystyle{ \alpha }$ be real such that the equation $\displaystyle{ x^2+2(\alpha-1)x-\alpha+7=0 }$ has two different real negative solutions. Then
*
*$ \ \displaystyle{ \alpha<-2 }$ ;
*$ \ \displaystyle{ 3<\alpha<7 }$ ;
*it is impossible ;
*none of (a)-(c).
$$$$
I have done the following :
The value $\alpha$ is real and such that the equation $x^2+2(\alpha-1)x-\alpha+7=0$ has two different real negative solutions.
The solutions of the equation are given from the quadratic formula \begin{align*}x_{1,2}&=\frac{-2(\alpha-1)\pm \sqrt{[2(\alpha-1)]^2-4\cdot 1\cdot (-\alpha+7)}}{2}\\ & =\frac{-2(\alpha-1)\pm \sqrt{4(\alpha^2-2\alpha-1)-4\cdot (-\alpha+7)}}{2}\\ & =-(\alpha-1)\pm \sqrt{2(\alpha^2-2\alpha-1)-2\cdot (-\alpha+7)} \\ & =-(\alpha-1)\pm \sqrt{2\alpha^2-4\alpha-2+2\alpha-14} \\ & =-(\alpha-1)\pm \sqrt{2\alpha^2-2\alpha-16}\end{align*}
So that we have two different solutions the discriminant must be non-zero.
So that we have two negative solutions, it must hold $-(\alpha-1)\pm \sqrt{2\alpha^2-2\alpha-16}<0$.
So that we have real solutions the expression under the square root must be non negative.
The expression under the square root has the sign of the coefficient of $x^2$, i.e. positive, outside the roots.
We have that \begin{equation*}2\alpha^2-2\alpha-16=0 \Rightarrow \alpha_{1,2}=\frac{1}{2}\pm \frac{\sqrt{33}}{2}\end{equation*}
So we have that the expression under the square root if $\alpha<\frac{1}{2}- \frac{\sqrt{33}}{2}$ and if $\alpha>\frac{1}{2}+ \frac{\sqrt{33}}{2}$.
Is my attempt correct so far? Now do we check if the first two intervals of $\alpha$ can hold for all these conditions? Or how do we continue? Or is there a better way to solve that exercise ?
| Sum of roots$=-2(\alpha-1)<0\implies \alpha>1$
Product of roots$=(7-\alpha)>0\implies\alpha<7$
with the condition that the discriminant, $(2\alpha-2)^2-4(7-\alpha)>0\\\implies (\alpha-3)(\alpha+2)>0\\\implies \alpha>3\ or\ \alpha<-2\\
\therefore 3<\alpha<7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Prove if $\sqrt{x+1}+\sqrt{y+2}+\sqrt{z+3}=\sqrt{y+1}+\sqrt{z+2}+\sqrt{x+3}=\sqrt{z+1}+\sqrt{x+2}+\sqrt{y+3}$, then $x=y=z$. Let $x$, $y$, $z$ be real numbers satisfying $$
\begin{align}
&\sqrt{x+1}+\sqrt{y+2}+\sqrt{z+3}\\
=&\sqrt{y+1}+\sqrt{z+2}+\sqrt{x+3}\\
=&\sqrt{z+1}+\sqrt{x+2}+\sqrt{y+3}.
\end{align}$$
Prove that $x=y=z$.
I tried assuming $x>y>z$, $x>y=z$,$x<y<z$, etc., but none of the directions work. Please help me solve this problem.
| Not sure if this is a good approach, but:
Let $f(t) = \sqrt{t+2}-\sqrt{t+1}$. Observe $f$ is strictly decreasing. Then subtract $\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}$ from each part of the equation to get: $$\begin{align}
&f(y)+f(z)+f(z+1)\\
=&f(z)+f(x)+f(x+1)\\
=&f(x)+f(y)+f(y+1)
\end{align}$$
This reduces to two variables in each part of the equation. To simplify further, I will subtract each term from $f(x)+f(y)+f(z)$: $$\begin{align}
&f(x)-f(z+1)\\
=&f(y)-f(x+1)\\
=&f(z)-f(y+1)
\end{align}$$
Suppose without loss of generality that $x$ is the largest of $(x, y, z)$. Because $f$ is strictly decreasing, $f(x) \leq f(y)$ and $f(x + 1) \leq f(z + 1)$, so $f(y) - f(x + 1) \geq f(x) - f(x + 1) \geq f(x) - f(z + 1)$ with equality holding only when $x=y=z$. But, equality holds by the equation system above, so we have our result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 2,
"answer_id": 1
} |
Why this transformation matrix $A$ has $\begin{pmatrix}0 \\ 1\end{pmatrix}$ as Eigenvector? I have the following transformation matrix:
$$
A=\begin{pmatrix}
1 & 0 \\
-1 & 4
\end{pmatrix}
$$
If I resolve to find the eigenvalues I get:
$$
\begin{vmatrix}
A-\lambda I
\end{vmatrix} = 0
$$
which leads to:
$$
\lambda_1 = 1;
\lambda_2 = 4
$$
Now if I try to calculate the eigenvectors
For $\lambda_1$ I get:
$$
(A - 1 I)\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix}0 \\ -x_1+3 x_2 \end{pmatrix}
$$
and for $\lambda_2$:
$$
(A - 1 I)\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix}-3 x_1 \\ -x_1 \end{pmatrix}
$$
I see (and can compute using a symbolic calculation program) that there are two eigenvectors:
$$
e1 = \begin{pmatrix}0 \\ 1\end{pmatrix}
$$
$$
e2 = \begin{pmatrix}3 \\ 1\end{pmatrix}
$$
I can easily see why $\begin{pmatrix}3 \\ 1\end{pmatrix}$ is an eigenvector for the eigenvalue $\lambda_2$.
But I am struggling to understand why $\begin{pmatrix}0 \\ 1\end{pmatrix}$ is an eigenvector.
Could someone help me to understand why?
| Straightforward calculation shows that:
$$A\begin{pmatrix}0 \\ 1 \end{pmatrix}=\begin{pmatrix}0 \\ 4 \end{pmatrix}=4\begin{pmatrix}0 \\ 1 \end{pmatrix}$$
So $\begin{pmatrix}0 \\ 1 \end{pmatrix}$ is an eigenvector corresponding to the eigenvalue $\lambda=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$x_2 + x_3 + x_4 + x_5 = n$ where $x_i$ is not divisible by $i$? How can I find the generating function for the number of solutions of non-negative integers for the equation $x_2$ + $x_3$ + $x_4$ + $x_5 = n$ where $x_i$ is not divisible by $i$?
Attempt:
I know $x_2$ is a variable for odd numbers and its generating function is $\frac{x}{1-x^2}$.
But how do I find the other three?
| You have to multiply the series
$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 2} } x^i \right)$$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 3} } x^i \right)$$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 4} } x^i \right)$$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 5} } x^i \right)$.
We have
$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 2} } x^i \right) = \left(\sum_{i \geq 0} x^i\right) - \left(\sum_{i \geq 0} x^{2i}\right) = \frac{1}{1-x} - \frac{1}{1-x^2} = \frac{x}{1-x^2}$.
In a similar way, we also have
$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 3} } x^i \right) = \left(\sum_{i \geq 0} x^i\right) - \left(\sum_{i \geq 0} x^{3i}\right) = \frac{1}{1-x} - \frac{1}{1-x^3} = \frac{x + x^2}{1-x^3}$,
$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 4} } x^i \right) = \left(\sum_{i \geq 0} x^i\right) - \left(\sum_{i \geq 0} x^{4i}\right) = \frac{1}{1-x} - \frac{1}{1-x^4} = \frac{x+x^2+x^3}{1-x^4}$,
and
$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 5} } x^i \right) = \left(\sum_{i \geq 0} x^i\right) - \left(\sum_{i \geq 0} x^{5i}\right) = \frac{1}{1-x} - \frac{1}{1-x^5} = \frac{x +x^2+x^3+x^4}{1-x^5}$.
Multiplying the results and having some simplification yields the desired answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac {x \csc x + y \csc y}{2} < \sec \frac {x + y}{2}$
If $0 < x,y < \frac {\pi}{2}$, prove that:
$$
\frac {x \csc x + y \csc y}{2} < \sec \frac {x + y}{2}
$$
My attempt. First, I tried to change this inequality:
$$
\frac {\frac{x}{\sin x} + \frac{y}{\sin y}}{2} < \frac{1}{\cos \frac{x+y}{2}}
$$
Then,it's easy to know:
$$
LHS = \frac{1}{2} \left ( \frac {x}{2\sin \frac {x}{2} \cos \frac{x}{2}} + \frac{y}{2 \sin \frac{y}{2} \cos \frac{y}{2}} \right ) < \frac{1}{2} \left ( \frac{1}{\cos^{2} \frac{x}{2}} +\frac{1}{\cos^{2} \frac{y}{2}} \right )
$$
$$
RHS > \frac{1}{\cos \frac{x}{2} \cos \frac{y}{2}}
$$
How to solve it next? It seems this way is wrong.
| We have
$$\frac{x}{\sin x} = \frac{x}{2\sin \frac{x}{2} \cos \frac{x}{2}} = \frac{x}{4\sin \frac{x}{4}\cos\frac{x}{4} \cos \frac{x}{2}} = \frac{\frac{x}{4}}{\tan \frac{x}{4} \cos^2\frac{x}{4}\cos\frac{x}{2}} \le \frac{1}{\cos^2\frac{x}{4}\cos\frac{x}{2}}$$
where we have used $\frac{x}{4} \le \tan \frac{x}{4}$.
Also, we have
$$\cos \frac{x+y}{2} = \cos \frac{x}{2} \cos \frac{y}{2} - \sin\frac{x}{2}\sin \frac{y}{2} \le \cos \frac{x}{2} \cos \frac{y}{2}.$$
It suffices to prove that
$$\frac{1}{\cos^2\frac{x}{4}\cos\frac{x}{2}} + \frac{1}{\cos^2\frac{y}{4}\cos\frac{y}{2}} \le \frac{2}{\cos \frac{x}{2} \cos \frac{y}{2}}$$
or
$$\frac{2}{(1 + \cos\frac{x}{2})\cos\frac{x}{2}} + \frac{2}{(1 + \cos \frac{y}{2})\cos\frac{y}{2}} \le \frac{2}{\cos \frac{x}{2} \cos \frac{y}{2}}$$
which is true since
$$\mathrm{LHS} \le \frac{2}{(\cos \frac{y}{2} + \cos\frac{x}{2})\cos\frac{x}{2}} + \frac{2}{(\cos\frac{x}{2} + \cos \frac{y}{2})\cos\frac{y}{2}} = \mathrm{RHS}.$$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve $(x^3+1)=2\sqrt[3]{2x-1}$ algebraically? I'm trying to solve the said equation in the thread title algebraically.
$$(x^3+1)=2\sqrt[3]{2x-1}$$
Cubing both sides and simplifying:
$$x^9+3x^6+3x^3-16x+9 = 0$$
Not sure if this can be solved algebraically?
Edit: WA gives $3$ solutions $x=1,\frac{1}{2}(-1-\sqrt{5}),\frac{1}{2}(-1+\sqrt{5})$
|
$$f(x)=x^9+3x^6+3x^3-16x+9$$
$$f(x)=0\Rightarrow (x-1)(x^2+x-1)(x^6+2x^4+2x^3+4x^2+2x+9)=0$$
where $$x^6+2x^4+2x^3+4x^2+2x+9=x^6+x^4+x^2(x+1)^2+2x^2+(x+1)^2+8>0$$ for $\forall x\in \mathbb{R}$, so we get:
$$(x-1)(x^2+x-1)=0$$
There are three real roots:
$$x_1=1, x_2=\frac{-1-\sqrt{5}}2, x_3=\frac{-1+\sqrt{5}}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find projection point on elipse Given the ellipse
$$
E(x, \ y) = ax^2 + bxy + cy^2 + dx + ey + f = 0
$$
With $4ac - b^2 = 1$
How can I compute the projection $(x_p, \ y_p)$ of the point $(x_0, y_0)$ in this ellipse?
$$
E(x_p, \ y_p) = 0
$$
Motivation: There is already a direct algorithm to fit ellipses from a set of datapoints as shown here where I can get the coefficients.
In linear interpolation, there's the R-Squared indicator bellow that tells us how good the model is. With $R_2 = 1$, all the points are in the line.
$$
R_2 = \dfrac{\left[n\sum x_iy_i - \left(\sum x_i\right)\left(\sum y_i\right) \right]^2}{\left[n\sum x_i^2 - \left(\sum x_i\right)^2\right]\left[n\sum y_i^2 - \left(\sum y_i\right)^2\right]}
$$
But for the ellipse problem I don't know the formula of $R_2$.
Then I want to compute the mean distance between the points and the ellipse to know how well my datapoints fits an ellipse.
To compute the distance of each point I need the projection point.
| This answer is not the one that I expected but it solves my problem in a non-elegant way.
I will divide this answer in three parts:
*
*Transform the ellipse with center $(x_c, \ y_c)$ and rotated counter clockwise with angle $\varphi$
$$E(x, y) = ax^2+bxy+cy^2 + dx + ey+f = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
into another ellipse
$$F(u , v) = \left(\dfrac{u}{A}\right)^2 + \left(\dfrac{v}{B}\right)^2 - 1 =0 \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$
With $A > B$
*Scale into a new ellipse of major axis equal 1
$$G(a, b) = a^2 + \left(\dfrac{b}{H}\right)^2 - 1 \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$
*Compute the new point $(u_0, \ v_0)$ using the translation and rotations operations of point $(x_0, y_0)$
*Compute $(a_0, \ b_0)$ by scaling $(u_0, \ v_0)$
*Compute $(a, \ b)$ such we get the minimal distance from $(a_0, \ b_0)$ to $(a, \ b)$
*Scale, rotate and translate back
Part 1: Ellipse transformation
We can describe the variable $x$ and $y$ using
$$
x = u \cdot \cos \varphi + v \cdot \sin \varphi + x_c \ \ \ \ \ \ \ \ \ \ \ \ \ (4)
$$
$$
y = -u \cdot \sin \varphi + v \cdot \cos \varphi + y_c \ \ \ \ \ \ \ \ \ \ \ \ \ (5)
$$
With $(1)$, $(2)$, $(4)$ and $(5)$ we can solve it. The computations are a little bit hard but we get
The center
$$
x_c = be-2cd
$$
$$
y_c = bd-2ae
$$
The angle
$$
\varphi = \dfrac{1}{2}\arctan\left(\dfrac{b}{c-a}\right)
$$
The major and minor axis
$$
\kappa = \sqrt{(a-c)^2+b^2}
$$
$$
\eta = \left(f+bde\right) -\left(ae^2+ cd^2\right)
$$
$$
A = \sqrt{\dfrac{-\eta}{a+c-\kappa}}
$$
$$
B = \sqrt{\dfrac{-\eta}{a+c+\kappa}}
$$
Part 2: Scalling ellipse
$$
u = A \cdot a \ \ \ \ \ \ \ \ \ v = A \cdot b \ \ \ \ \ \ \ \ \ H = \dfrac{B}{A}
$$
Part 3: Point transformation
To make a translation we only subtract the center $(x_c, \ y_c)$ and to make a clockwise rotation we put
$$
\begin{bmatrix}
u_0 \\ v_0
\end{bmatrix} = \begin{bmatrix}
\cos \varphi & -\sin \varphi \\
\sin \varphi & \cos \varphi
\end{bmatrix}\begin{bmatrix}
x_0 - x_c \\ y_0 - y_c \end{bmatrix}
$$
Part 4: Scale point
Then we have
$$
a_0 = \dfrac{u_0}{A} \ \ \ \ \ \ \ \ \ \ \ \ \ b_0 = \dfrac{v_0}{A}
$$
Part 5: Compute distance
Let's find the distance of an arbitrary point $(a_0, b_0)$ to the ellipse $(3)$
The distance square is given by
$$
D^2 = (a-a_0)^2 + (b-b_0)^2
$$
Option 3.1: Describe $a$ and $b$ in terms of $\theta$
Let
$$ a = \cos \theta \ \ \ \ \ \ \ \ \ \ \ \ \ \ b = H \cdot \sin \theta $$
Then
$$ D^2 = (\cos \theta - a_0)^2 + (H\sin \theta-b_0)^2 $$
Getting the minimal of $D^2$ we get
$$ \sin \theta (a_0-\cos \theta) - H\cos \theta (b_0-H\sin \theta) = 0 $$
$$ - (1-H^2) \cdot \sin \theta \cos \theta+ a_0 \cdot \sin \theta - Hb_0 \cdot \cos \theta = 0 $$
Then solve for $\theta$ with a numerical method.
Option 3.2: Use Lagrange multiplier
We have the minimizing function $D^2$ with the constraint of $(3)$.
That means we have to find the solution for
$$ \nabla D^2 + \lambda \nabla G = \vec{0} $$ $$ \begin{bmatrix}
2(a-a_0) \\ 2(b-b_0) \end{bmatrix} + \lambda \begin{bmatrix} 2a \\
\dfrac{2b}{H^2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$
Gives us
$$ a = \dfrac{u_0}{1+\lambda} \ \ \ \ \ \ \ \ \ \ \ \ b = \dfrac{v_0}{1+\frac{\lambda}{H^2}} $$
And in $(2)$ gives us
$$ u_0^2 \left(H+\dfrac{\lambda}{H}\right)^2 + v_0^2 \left(1+\lambda\right)^2 - \left(1+\lambda\right)^2
\left(H+\dfrac{\lambda}{H}\right)^2 = 0 $$
Then solve for $\lambda$ with a numerical method
Part 6: Scale back
Once we have $\theta$ or $\lambda$, we can compute the point $a$ and $b$
$$
a = \cos \theta \ \ \ \ \ \ \ \ \ \ \ \ \ \
b = H \sin \theta
$$
or
$$ a = \dfrac{a_0}{1+\lambda} \ \ \ \ \ \ \ \ \ \ \ \ \ \ b = \dfrac{b_0}{1+\dfrac{\lambda}{H^2}}$$
To get the projection point we have to do
$$
\begin{bmatrix}
x \\ y
\end{bmatrix} =
\begin{bmatrix}
\cos \varphi & \sin \varphi \\
-\sin \varphi & \cos \varphi
\end{bmatrix}
\begin{bmatrix}
A \cdot a \\
A \cdot b
\end{bmatrix}+
\begin{bmatrix}
x_c \\
y_c
\end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Bounding a simple ratio of bivariate functions For any $x,y \ge 0$, define $f(x,y) := x^2+y^2 + 2c xy$, where $c := \sqrt{2/\pi}$. It is clear that
$$
\sqrt{2/\pi}\cdot (x+y)^2 \le f(x,y) \le (x+y)^2.
\tag{1}
$$
This is an immediate consequence of the fact that $c \le 1 \le 1/c$.
Question. What is a good upper-bound for $\alpha := \sup_{x,y}\dfrac{(x+y)^2}{f(x,y)}$ ?
Note that (1) gives the trivial upper-bound $\alpha \le \sqrt{\pi/2}$.
| Since $c<1$ and $f(x,y) =(x+y)^2-2xy(1-c)$ we have
$$\frac1\alpha := \inf_{x,y}\dfrac{f(x,y)}{(x+y)^2}= 1-(1-c)\sup_{x,y}\frac{2xy}{(x+y)^2}$$
It remains to compute
$$\sup_{x,y}\frac{2xy}{(x+y)^2}= \sup_{x,y}\frac12\frac{(x+y)^2-(x-y)^2}{(x+y)^2}= \frac12-\frac12\inf_{x,y}\frac{(x-y)^2}{(x+y)^2}=\frac12$$
Indeed since $x,y\geq 0$, taking $x=y=1$ we find that
$$\inf_{x,y}\frac{(x-y)^2}{(x+y)^2}=\min_{x,y}\frac{(x-y)^2}{(x+y)^2}=0$$
Thus it follows that
$$\frac1\alpha := \inf_{x,y}\dfrac{f(x,y)}{(x+y)^2}= 1-(1-c)\sup_{x,y}\frac{2xy}{(x+y)^2}= \frac{1+c}{2}$$
that is taking $x=y=1$ we find
$$\alpha=\sup_{x,y}\dfrac{(x+y)^2}{f(x,y)}=\max_{x,y}\dfrac{(x+y)^2}{f(x,y)}= \frac{2}{1+c}\leq \frac1c.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is $f(x)=\frac{x}{e^x-1}$ continuous? My exercise says:
$$f(x)=\frac{x}{e^x-1},\quad x\neq0$$
$$f(0)=1$$
Can someone explain whatever this means? And why does the graph not cut off at $x=0$?
Edit : Apologies ,it's not $f(0)=0$ but $f(0)=1$.
And to clarify the exercise asks to confirm that $f'(0)$ exists (But that's not what puzzles me right now ).
| $\frac{x}{e^x- 1}$ is NOT continuous. In order for a function, f(x), to be continuous at x= a, three things must be true.
*
*$\lim_{x\to a} f(x)$ exist.
*f(a) exist.
*$\lim_{x\to a} f(x)= f(a)$.
(For 3 to make sense, 1 and 2 must be true so often only
3 is stated.)
$\lim_{x\to 0} \frac{x}{e^x- 1}$ exists and is 0 (see below) so 1 is true. But since $e^0- 1= 0$ and we cannot divide by 0, the function is not defined at 0. 2 is NOT true. In order to have a function that is continuous at x= 0, we need to redefine it at x= 0 to be that limit, 1.
The MacLaurin series (Taylor series at 0) for e^x is
$1+ x+ x^2/2+ x^3/3!+ \cdot\cdot\cdot+ x^n/n!+ \cdot\cdot\cdot$.
$e^x- 1= x+ x^2/2+ x^3/3!+ \cdot\cdot\cdot+ x^n/n!+ \cdot\cdot\cdot$ so $\frac{x}{e^x- 1}= \frac{x}{x+ x^2/2+ x^3/3!+ \cdot\cdot\cdot+ x^n/n!+ \cdot\cdot\cdot}= \frac{x}{x(1+ x/2+ x^2/3!+ \cdot\cdot\cdot+ x^{n-1}/n!+ \cdot\cdot\cdot)}= \frac{1}{1+ x/2+ x^2/3!+ \cdot\cdot\cdot+ x^{n-1}/n!+ \cdot\cdot\cdot}$
The limit as x goes to 0 is $\frac{1}{1}= 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving $\frac{dx}{dt}=\frac{xt}{x^2+t^2},\ x(0)=1$ I have started self-studying differential equations and I have come across the following initial value problem
$$\frac{dx}{dt}=\frac{xt}{x^2+t^2}, \quad x(0)=1$$
Now, since $f(t,x)=\frac{xt}{x^2+t^2}$ is such that $f(rt,rx)=f(t,x)$ for every $r\in\mathbb{R}\setminus\{0\}$, we can use the change of variables $y=\frac{x}{t}$ to rewrite it in the form
\begin{align}
y+t\frac{dy}{dt} &=\frac{t^2 y}{t^2(1+y^2)} \\
&=\frac{y}{1+y^2} \\
\implies t\frac{dy}{dt} &=\frac{y}{1+y^2}-y \\
&=-\frac{y^3}{1+y^2} \\
\implies \frac{dy}{dt} &= \left(-\frac{y^3}{1+y^2}\right)\cdot\frac{1}{t}
\end{align}
which is separable, and becomes:
\begin{align}
\left(\frac{1+y^2}{y^3}\right)dy &= -\frac{dt}{t} \\
\implies \int_{y_1}^{y_2} \left(\frac{1+y^2}{y^3}\right) dy &= -\int_{t_1}^{t_2} \frac{dt}{t} \\
\implies -\frac{1}{2y_2^2}+\ln \left\lvert \frac{y_2}{y_1} \right\rvert + \frac{1}{2y_1^2} &= -\ln \left\lvert \frac{t_2}{t_1} \right\rvert
\end{align}
but now I don't see how to go forward and find $y(t)$. Also, I integrated from a generic time $t_1$ to a generic time $t_2$ because the right hand side wouldn't have converged otherwise.
So, I would appreciate any hint about how to go forward in solving this IVP.
Thanks
| $$\frac{dx}{dt}=\frac{xt}{x^2+t^2},\ x(0)=1$$
Multiply by $2x$:
$$2x{dx}=\frac{2x^2t}{x^2+t^2}dt$$
$${dx^2}=\frac{x^2}{x^2+t^2}dt^2$$
Substitute $u=x^2,v=t^2$:
$${du}=\frac{udv}{u+v}$$
This is a first order linear DE $uv'=u+v$ that you can also solv as this:
$$u{du}={udv}-vdu$$
$$\dfrac {du}u=\dfrac {udv-vdu}{u^2}$$
$$\dfrac {du}u=d \left (\dfrac {v}{u}\right)$$
Integrate:
$$\ln u=\dfrac {v}{u}+C$$
$$\ln x^2= \dfrac{t^2}{x^2}+C$$
Apply initial condition.
$$x(0)=1 \implies C=0$$
$$\ln x^2= \dfrac{t^2}{x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4520385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Given $3x+4y=15$, $\min(\sqrt{x^2+y^2})=?$ (looking for other approaches)
Given, $(x,y)$ follow $3x+4y=15$. Minimize $\sqrt{x^2+y^2}$.
I solved this problem as follows,
We have $y=\dfrac{15-3x}{4}$,
$$\sqrt{x^2+y^2}=\sqrt{x^2+\frac{(3x-15)^2}{16}}=\frac{\sqrt{25x^2-90x+225}}4=\frac{\sqrt{(5x-9)^2+144}}{4}$$Hence $\min(\sqrt{x^2+y^2})=3$.
I'm wondering is it possible to solve this problem differently?
| By Cauchy-Schwarz inequality,
$\sqrt{x^2+y^2}\ge\frac{15}{\sqrt{3^2+4^2}}=3$
and by the case of Cauchy-Schwarz equality, this value is attained, hence it is the min you look for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Show that $\frac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\frac{3\pi}{4}+\alpha\right)$ Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$
I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For the LHS:
$$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\dfrac{1-2\sin\alpha\cos\alpha}{1+2\sin\alpha\cos\alpha}=\dfrac{\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha}{\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha}=\dfrac{\left(\sin\alpha-\cos\alpha\right)^2}{\left(\sin\alpha+\cos\alpha\right)^2}$$ I don't know if this is somehow useful as I can't get a feel of the problem and what we are supposed to notice to solve it.
| $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha} = \left(\frac{\tan \alpha-1}{\tan \alpha+1}\right)^2= \left(\frac{1-\tan \alpha}{1+\tan \alpha}\right)^2$$
$$=\tan^2\left(\dfrac{\pi}{4}-\alpha\right) =\tan^2\left(\alpha-\dfrac{\pi}{4}\right)$$
Inverse tangent function satisfies two angles in range $0, 2 \pi.\;$We are allowed to add $k\cdot\pi$ to any angle for tangent function value to be same. So for $k=1,$
$$ =\tan^2\left(\alpha-\dfrac{\pi}{4}+\pi \right) =\tan^2\left(\alpha+\dfrac{3\pi}{4}\right)= RHS $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 2
} |
Find the range of $f(x)=\frac1{1-2\sin x}$ Question:
Find the range for $f(x)= 1/(1-2\sin x)$
Answer :
$ 1-2\sin x \ne 0 $
$ \sin x \ne 1/2 $
My approach:
For range :
$ -1 ≤ \sin x ≤ 1 $
$ -1 ≤ \sin x < 1/2$ and $1/2<\sin x≤1 $ , because $\sin x≠1/2$
$ -2 ≤ 2\sin x <1 $ and $ 1< 2\sin x≤2 $
$ -1 < -2\sin x ≤2 $ and $ -2≤ -2\sin x<-1 $
$ 0 < 1-2\sin x ≤3 $ and $ -1≤ 1-2\sin x<0 $
I am stuck at the last step. If I take reciprocal i.e., $\frac 1{1-2\sin x}$ I get :
$ \frac10 < \frac1{1-2\sin x} ≤\frac13 $ and $ -1≤ \frac1{1-2\sin x} < \frac 10 $
$1/0$ can be interpreted as infinity so the second equation gives range of $f(x)$ as $[-1,∞)$. But the correct answer is : $(−∞,−1]∪[\frac13,∞)$
Any alternative solutions are welcome :)
P.S. : Also $\frac1{1-2\cos x}$ will also have the same range, right? As $\sin x$ and $\cos x$ both lie between $[-1,1]$. So the answer will proceed in similar fashion to the given one.
| You made an error when you took reciprocals. When $a, b$ have the same sign, $$a < b \implies \frac{1}{a} > \frac{1}{b}$$
You failed to reverse the direction of the inequalities when you took reciprocals.
Since the denominator cannot be zero, you correctly concluded that the function is defined for those real numbers $x$ such that $\sin x \neq \dfrac{1}{2}$.
Since $-1 \leq \sin x \leq 1$ and $\sin x \neq \dfrac{1}{2}$, $-1 \leq \sin x < 1/2$ or $1/2 \leq \sin x \leq 1$. Note the use of the word or since those two inequalities cannot simultaneously be true.
Let's consider the inequality $-1 \leq \sin x < \dfrac{1}{2}$.
\begin{align*}
-1 \leq & \sin x < \frac{1}{2}\\
-2 \leq & 2\sin x < 1\\
2 \geq & -2\sin x > -1\\
3 \geq & 1 - 2\sin x > 0
\end{align*}
If $a, b > 0$, then $a < b \implies \dfrac{1}{a} > \dfrac{1}{b}$. To see this, observe that
$$\frac{1}{a} - \frac{1}{b} = \frac{b - a}{ab} > 0$$
since $b - a > 0$ by definition of $a < b$ and $ab > 0$ since the product of positive numbers is positive. Thus, when you took reciprocals, you should have obtained
$$\frac{1}{3} \leq \frac{1}{1 - 2\sin x}$$
Now, let's consider the inequality $\dfrac{1}{2} < \sin x \leq 1$.
\begin{align*}
\frac{1}{2} & < \sin x \leq 1\\
1 & < 2\sin x \leq 2\\
-1 & > -2\sin x \geq -2\\
0 & > 1 - 2\sin x \geq -1
\end{align*}
If $a, b < 0$, then $a < b \implies \dfrac{1}{a} > \dfrac{1}{b}$. The proof is identical to that given above for positive numbers except that $ab > 0$ since the product of two negative numbers is positive. Hence, when you took reciprocals, you should have obtained
$$\frac{1}{1 - 2\sin x} \leq -1$$
Since $\dfrac{1}{3} \leq \dfrac{1}{1 - 2\sin x}$ or $\dfrac{1}{1 - 2\sin x} \leq -1$, the range of the function $f(x) = \dfrac{1}{1 - 2\sin x}$ defined on all real numbers except those for which $\sin x = 1/2$ is
$$(-\infty, -1] \cup \left[\frac{1}{3}, \infty\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find all values of a so that the circle $x^2 - ax + y^2 + 2y = a$ has the radius 2 My goal is to find all values of "a" so that the circle $x^2 - ax + y^2 + 2y = a$ has the radius 2
The correct answer is: $a = -6$ and $a = 2$
I tried solving it by doing this:
$x^2 - ax + y^2 +2y=a$
$x^2 - ax + (y+1)^2-1=a$
$(x - \frac a2)^2 - (\frac a2)^2 + (y+1)^2-1=a$
$(x - \frac a2)^2 - {a^2\over 4} + (y+1)^2-1=a$
$(x - \frac a2)^2 + (y+1)^2=a + {a^2\over 4} + 1$
$(x - \frac a2)^2 + (y+1)^2={a^2+4a + 4\over 4}$
We want the radius to be 2 so set this ${a^2+4a + 4\over 4}$ equal to 2
${a^2+4a + 4\over 4}=2$
$a^2+4a + 4=8$
$a^2+4a -4=0$
Solve for a:
$a=-2 \pm \sqrt{4+4}$
$a=-2 \pm \sqrt{8}$
This is not correct as you can see. I don't understand what I do wrong, I'm not sure if there is one of those tiny mistakes somewhere in my solving process or if I'm completely wrong from the beginning. Thanks in advance.
| $\frac{a^2+4a+4}{4}$ should equal $2^2=4$, not $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned}
$$
Then, we have
$$
\begin{aligned}
0<x<2\\
\implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned}
$$
Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$.
This leads,
$$
\begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$.
Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$.
We have:
$$
\begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$.
Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$.
This case implies, $\left(2x-\frac a2\right)^2≥0$. This means, $f(x)≥\frac {3a^2}{4}-5a+4$
Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$.
Finally, we have to combine all the solution sets we get.
I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?
| Might be calculative but this works
Find the roots using quadratic formula then if the condition satisfies then there are 3 cases
$$\text{Case1}:-$$ $$\text{first the smaller of the 2 roots (the one with the negative sign) is greater than 2 this gives the quadratic}$$ $a^2-13a-68\ge0$ and $3a^2-20a+16\le0$
$$\text{Case2}:-$$
Again the greater root is less than or equal to 0 $$a^2-5a+4\le0$$ and $3a^2-20a+16\le0$
$$\text{Case3}:-$$
Discriminant is negative ,ie $3a^2-20a+16\ge0$
Solving case by case we get all solutions
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 11,
"answer_id": 7
} |
How to calculate the gradient of $\mathbf{x}^T\mathbf{W}^2(\mathbf{W}^2)^T\mathbf{x}$ w.r.t. $\mathbf{W}$? I need to calculate the gradient of $\mathbf{x}^T\mathbf{W}^2(\mathbf{W}^2)^T\mathbf{x}$ w.r.t. $\mathbf{W}$. Here is what I have tried. Let $A=W^2$, then the form reduces to
\begin{align*}
\frac{\partial \mathbf{x}^T\mathbf{AA}^T\mathbf{x}}{\partial \mathbf{A}}
=&\frac{\partial \mathbf{x}^T\mathbf{B}^T\mathbf{B}\mathbf{x}}{\partial \mathbf{B}^T} \quad\quad (\text{where }\mathbf{A}=\mathbf{B}^T) \\
=&\left(\frac{\partial \mathbf{x}^T\mathbf{B}^T\mathbf{B}\mathbf{x}}{\partial \mathbf{B}}\right)^T \\
=&\left(\mathbf{B}(\mathbf{x}\mathbf{x}^T+\mathbf{x}\mathbf{x}^T)\right)^T\\
=&2\mathbf{x}\mathbf{x}^T\mathbf{B}^T\\
=&2\mathbf{x}\mathbf{x}^T\mathbf{A}
\end{align*}
which follows from the formula (77) in Matrix Cookbook, specifically,
$$
\frac{\partial \mathbf{b}^T\mathbf{X}^T\mathbf{X}\mathbf{c}}{\partial\mathbf{X}}=\mathbf{X}(\mathbf{b}\mathbf{c}^T+\mathbf{c}\mathbf{b}^T).
$$
I was trying to use the chain rule since we already know
$$
\frac{\partial \mathbf{x}^T\mathbf{W}^2(\mathbf{W}^2)^T\mathbf{x}}{\partial \mathbf{W}^2}=2\mathbf{x}\mathbf{x}^T\mathbf{W}^2.
$$
The next step is supposed to be
$$
\frac{\partial \mathbf{x}^T\mathbf{W}^2(\mathbf{W}^2)^T\mathbf{x}}{\partial \mathbf{W}}=\frac{\partial\mathbf{W}^2}{\partial\mathbf{W}}\frac{\partial x^T\mathbf{W}^2(\mathbf{W}^2)^Tx}{\partial \mathbf{W}^2}
$$
where we use the denominator layout. However, the dimension of
$
\frac{\partial\mathbf{W}^2}{\partial\mathbf{W}}
$
does not fit the dimension of $\frac{\partial \mathbf{x}^T\mathbf{W}^2(\mathbf{W}^2)^T\mathbf{x}}{\partial \mathbf{W}^2}$, namely $2\mathbf{x}\mathbf{x}^T\mathbf{W}^2$. Does anyone give me any clues? I appreciate it.
Update: I tried to use Frobenius inner product to do it as follows. Let $z=\mathbf{x}^T\mathbf{W}^2(\mathbf{W}^2)^T\mathbf{x}$, then we have
\begin{align*}
\mathrm{d}z&=2\mathbf{x}\mathbf{x}^T\mathbf{A}:\mathrm{d}\mathbf{A} \\
&=2\mathbf{x}\mathbf{x}^T\mathbf{A}:\mathrm{d}\mathbf{W}^2\\
&=2\mathbf{x}\mathbf{x}^T\mathbf{A}:(\mathrm{d}\mathbf{W}\mathbf{W}+\mathbf{W}\mathrm{d}\mathbf{W})\\
&=2\mathbf{x}\mathbf{x}^T\mathbf{A}:\mathrm{d}\mathbf{W}\mathbf{W}+2\mathbf{x}\mathbf{x}^T\mathbf{A}:\mathbf{W}\mathrm{d}\mathbf{W}\\
&=2\mathbf{x}\mathbf{x}^T\mathbf{A}\mathbf{W}^T:\mathrm{d}\mathbf{W}+2\mathbf{W}^T\mathbf{x}\mathbf{x}^T\mathbf{A}:\mathrm{d}\mathbf{W}\\
&=2(\mathbf{x}\mathbf{x}^T\mathbf{A}\mathbf{W}^T+\mathbf{W}^T\mathbf{x}\mathbf{x}^T\mathbf{A}):\mathrm{d}\mathbf{W}
\end{align*}
which gives the solution
$$
\frac{\partial \mathbf{x}^T\mathbf{W}^2(\mathbf{W}^2)^T\mathbf{x}}{\partial \mathbf{W}}=2(\mathbf{x}\mathbf{x}^T\mathbf{A}\mathbf{W}^T+\mathbf{W}^T\mathbf{x}\mathbf{x}^T\mathbf{A}).
$$
I tested this result with a $2\times 2$ $\mathbf{W}$ using the auto-differentiation tool by PyTorch which gives an identical result. This implies that the above derivations are correct.
| Let
$\mathbf{y}=(\mathbf{W}^2)^T \mathbf{x}$.
It holds
\begin{eqnarray*}
d\phi
&=& 2 \mathbf{y}:d\mathbf{y} \\
&=& 2 \mathbf{y}\mathbf{x}^T:d(\mathbf{W}^2)^T \\
&=& 2 \mathbf{B}:d(\mathbf{W}^2) \\
&=& 2
\left[\mathbf{W}^T\mathbf{B}+
\mathbf{B} \mathbf{W}^T \right]
:d\mathbf{W}
\end{eqnarray*}
where
$\mathbf{B}
=\mathbf{x}\mathbf{y}^T
=\mathbf{x}\mathbf{x}^T\mathbf{W}^2$.
The LHS term is the gradient you found by yourself.
| {
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"url": "https://math.stackexchange.com/questions/4529245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\left|\int_a^bf(x)dx \right|\le \frac{(b-a)^3}{24} \max_{a\le x\le b}|f''(x)|$ Suppose that $f\in C^2 [a, b]$ and that $f(\frac{a+b}{2}) =0$ then prove that \begin{equation} \bigg|\int_a^bf(x)dx \bigg|\le \frac{(b-a)^3}{24} \max_{a\le x\le b}|f''(x)| \end{equation}.
I know that since $f\in C^2$, by the Taylor expansion of $f$, in a neighborhood of $x_0$ up to the 2nd derivative I can express $f$ as
$$f(x) = f(x_0) +\frac{(x-x_0)}{1!}f'(x_0)+\frac{(x-x_0)^2}{2!}f''(x_0)$$
I'm not sure if this is right, but I go ahead to write the integral as
$$\bigg| \int_a^b f(x) dx\bigg|\le \max_{a\le x\le b}|f''(x) | \bigg| \int_a^b \frac{(x-x_0)^2} {2!} dx\bigg|. $$
But I don't seem to be getting what is needed because I have to use the condition $f(\frac{a+b} {2})$. What is wrong here with my approach?
| Solution:
The Taylor expansion (with Lagrange remainder) of $f(x)$ at the point of $x=\dfrac{a+b}{2}$ is
\begin{align}
f(x)&=\color{blue}{\underbrace{f\left(\frac{a+b}{2}\right)}_{0}}+f^\prime\left(\frac{a+b}{2}\right)\left(x-\frac{a+b}{2}\right)+\frac{1}{2}f^{\prime\prime}\left(\xi\right)\left(x-\frac{a+b}{2}\right)^2\\
&=f^\prime\left(\frac{a+b}{2}\right)\left(x-\frac{a+b}{2}\right)+\frac{1}{2}f^{\prime\prime}\left(\xi\right)\left(x-\frac{a+b}{2}\right)^2,~~x\in[a,b],
\end{align}
where $\xi\in[a,b]$.
Integrate both sides, then we have
\begin{align}
\int_a^bf(x)dx&=f^\prime\left(\frac{a+b}{2}\right)\color{blue}{\underbrace{\int_a^b\left(x-\frac{a+b}{2}\right)dx}_{0}}+\frac{1}{2}\int_a^bf^{\prime\prime}\left(\xi\right)\left(x-\frac{a+b}{2}\right)^2dx\\
&=\frac{1}{2}\int_a^bf^{\prime\prime}\left(\xi\right)\left(x-\frac{a+b}{2}\right)^2dx.
\end{align}
Finally, there is
\begin{align}
\left|\int_a^bf(x)dx\right|&\le \frac{1}{2}\int_a^b\left|f^{\prime\prime}\left(\xi\right)\left(x-\frac{a+b}{2}\right)^2\right|dx\\
&\le\frac{A}{2}\int_a^b\left(x-\frac{a+b}{2}\right)^2dx=\frac{(b-a)^3}{24}A,
\end{align}
where $A=\underset{a\le x\le b}{\max}\left|f^{\prime\prime}(x)\right|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4535917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $\frac{\sqrt[7]{x-\sqrt2}}{2}-\frac{\sqrt[7]{x-\sqrt2}}{x^2}=\frac{x}{2}\sqrt[7]{\frac{x^3}{x+\sqrt2}}$ Solve the equation $$\dfrac{\sqrt[7]{x-\sqrt2}}{2}-\dfrac{\sqrt[7]{x-\sqrt2}}{x^2}=\dfrac{x}{2}\sqrt[7]{\dfrac{x^3}{x+\sqrt2}}$$ We have $x\ne0;-\sqrt2$.
Let's multiply both sides of the equation by $2x^2\ne0$ to get $$x^2\sqrt[7]{x-\sqrt2}-2\sqrt[7]{x-\sqrt2}=x^3\sqrt[7]{\dfrac{x^3}{x+\sqrt2}}$$$$(x^2-2)\sqrt[7]{x-\sqrt2}=x^3\sqrt[7]{\dfrac{x^3}{x+\sqrt2}}$$ Let's multiply both sides of the equation by$\sqrt[7]{x+\sqrt2}\ne0$ to get $$(x^2-2)\sqrt[7]{x^2-2}=x^3\sqrt[7]{x^3}$$ $$(x^2-2)^8=x^{24}$$
| Starting where you left off, at $(x^2-2)^8=x^{24}$, rearrange the equation to:
$$x^{24} - (x^2-2)^8 = 0$$
You could use the Binomial Theorem to expand the $(x^2-2)^8$ term, but for now I won't. Just note that the polynomial's constant term (and the product of all its roots) is $-(-2)^8 = -256$.
If you just want real solutions
By the Rational Root Theorem, any rational roots of the polynomial must be factors of 256, i.e., $x \in \pm\{ 1, 2, 4, 8, 16, 32, 64, 128, 256 \}$. It turns out that $x = 1$ and $x = -1$ are the only ones that work, and then there are no other real solutions.
If you want complex solutions
The exponents are even, so the 24th-degree polynomial can be factored as a difference of two squares:
$$(x^{12} + (x^2-2)^4)(x^{12} - (x^2-2)^4) = 0$$
And we can do this two more times:
$$(x^{12} + (x^2-2)^4)(x^6 + (x^2-2)^2)(x^6 - (x^2-2)^2) = 0$$
$$(x^{12} + (x^2-2)^4)(x^6 + (x^2-2)^2)(x^3 + (x^2-2))(x^3 - (x^2-2)) = 0$$
Or, un-nesting the parentheses in the cubics:
$$(x^{12} + (x^2-2)^4)(x^6 + (x^2-2)^2)(x^3 + x^2 - 2)(x^3 - x^2 + 2) = 0$$
You can use the Rational Root Theorem to find one factor of each of the two cubics, and then the quadratic formula to find the other two:
*
*$x^3 + x^2 - 2 = 0 \implies (x-1)(x^2 + 2x + 2) = 0 \implies x \in \{ 1, -1 + i, -1 - i \}$
*$x^3 - x^2 + 2 = 0 \implies (x+1)(x^2 - 2x + 2) = 0 \implies x \in \{ -1, 1+i, 1-i \}$
This gives us 6 of the original polynomial's 24 roots. To find others, we can use $a^2+b^2 = a^2-i^2b^2 = (a+ib)(a-ib)$.
$$(x^{12} + (x^2-2)^4)(x^6 + (x^2-2)^2) = 0$$
$$(x^{6} + i(x^2-2)^2)(x^{6} - i(x^2-2)^2)(x^3 + i(x^2-2))(x^3 - i(x^2-2)) = 0$$
This gives us two more cubics. There is a cubic formula that can find the roots, but they'll be messy nested-radical things.
Finally, for the sextic polynomials, use $\sqrt{i} = \pm \frac{1 + i}{\sqrt 2}$ to do the sum/difference of squares thing one more time. Again, I'm not going to even try to write the radical forms.
| {
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"answer_count": 2,
"answer_id": 1
} |
Find $\cos\frac{\pi}{12}$ given $\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$
Find $\cos\frac{\pi}{12}$ given $\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$
From a question I asked before this, I have trouble actually with the numbers manipulating part.
Using trigo identity, $\sin^2 \frac{\pi}{12} + \cos^2 \frac{\pi}{12} = 1$ so , $\cos^2 \frac{\pi}{12} = 1- \sin^2 \frac{\pi}{12}$
To find $\cos \frac{\pi}{12} = \sqrt{1- \sin^2 \frac{\pi}{12}}$
$\sin^2 \frac{\pi}{12} = (\frac{\sqrt{3} -1}{2 \sqrt{2}})^2 = \frac{(\sqrt{3}-1)^2}{(2\sqrt{2})^2} = \frac{2- \sqrt{3}}{4}$
$\cos \frac{\pi}{12} = \sqrt{1-(\frac{\sqrt{3} -1}{2 \sqrt{2}})^2} $
$\cos \frac{\pi}{12} = \sqrt{1- \frac{2-\sqrt{3}}{4}}$
$\cos \frac{\pi}{12} = \frac{\sqrt{2+\sqrt{3}}}{2}$
What is wrong with my steps?
| Nothing wrong.
If you prefer, we can do some simplification.
Let me focus on $\sqrt{2+\sqrt3}$.
Let $$\sqrt{2+\sqrt3}+\sqrt{2-\sqrt3}=x$$
$$2+\sqrt3+2-\sqrt3+2=x^2$$
Hence, we have $$\sqrt{2+\sqrt3}+\sqrt{2-\sqrt3}=\sqrt6.$$
Similarly, we have $$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$
Hence $$\sqrt{2+\sqrt3}=\frac{\sqrt6+\sqrt2}{2}=\frac{\sqrt3+1}{\sqrt2}.$$
$$\cos \frac{\pi}{12}=\frac{\sqrt3+1}{2\sqrt2}$$
| {
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"answer_id": 2
} |
Arithmetic-Geometric limit $\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} y_n$
For $x,y>0$, define two sequences $(x_n)$ and $(y_n)$ by $x_1=x,y_1=y$ and $x_{n+1}=(x_n+y_n)/2$ and $y_{n+1}=\sqrt{x_ny_n}$. Prove that $\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} y_n= \dfrac{\pi}{\int_0^\pi \dfrac{d\theta}{\sqrt{x^2 \cos^2\theta + y^2\sin^2\theta}}}.$
I think it might be easier to prove $\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} y_n.$ Let the LHS of this equation be denoted $L$ and let the RHS be denoted $M$. By the AM-GM inequality and induction, $x_{n}\leq y_n$ for all $n\ge 2$. It could be useful to define a new sequence with a limit that's easier to evaluate. We have $L\leq M$ by limit properties, so we just need to show $L\ge M$ to get $L=M$. Suppose for a contradiction that $L < M.$ Then by definition, there exists $N$ so that for all $n\ge N, x_n < \frac{L+M}2$ and $y_n > \dfrac{L+M}2.$ How can I proceed from here?
As for showing it equals an expression involving a given integral, I think the integral is actually fairly hard to compute explicitly, so one should use some properties of the sequences to show the desired equality.
| I will handle the integral at the end using the transformation given by Gauss. An alternative transformation is available on my blog (linked in comments to question).
Let us write $$I(x, y) =\int_0^{\pi/2}\frac{dt}{\sqrt{x^2\cos^2t+y^2\sin^2t}}\tag{1}$$ for $x, y>0$ and we prove $$I(x, y) =I\left(\frac{x+y} {2},\sqrt {xy} \right) \tag{2}$$ Gauss used the substitution $$\sin t=\frac{2x\sin u} {x+y+(x-y) \sin^2u}\tag{3}$$ to establish $(2)$. The substitution maps the interval $[0,\pi/2]$ to $[0,\pi/2]$.
Let us first observe that $$\cos t=\frac{\sqrt{(x+y)^2+(x-y)^2\sin^4u-2(x^2+y^2)\sin^2u}} {x+y+(x-y) \sin^2u} $$ which can be rewritten as $$\frac{\sqrt{4a^2+4(a^2-b^2)\sin^4u-4(2a^2-b^2)\sin^2u}} {x+y+(x-y) \sin^2u} $$ where $2a=x+y,b^2=xy$. The above can be further simplified as $$\cos t=\frac{2\cos u\sqrt{a^2\cos^2u+b^2\sin^2u}}{x+y+(x-y)\sin^2u}\tag{4}$$ Next we have $$x^2\cos^2t+y^2\sin^2t=\frac{4x^2\cos^2u(a^2\cos^2u+b^2\sin^2u)+4x^2y^2\sin^2u}{(x+y+(x-y)\sin^2u)^2}$$ which can be rewritten as $$\frac{x^2[(x+y)^2\cos^4u+4xy\sin^2u\cos^2u+4y^2\sin^2u]}{(x+y+(x-y)\sin^2u)^2} $$ Replacing $\cos^2u$ with $1-\sin^2u$ we get $$\sqrt{x^2\cos^2t+y^2\sin^2t}=\frac{x(x+y-(x-y)\sin^2u)}{x+y+(x-y)\sin^2u}\tag{5}$$ Differentiating equation $(3)$ with respect to $u$ we get $$\cos t\cdot \frac{dt}{du} =\frac{2x\cos u(x+y-(x-y) \sin^2u)} {(x+y+(x-y)\sin^2u)^2} \tag{6}$$ Using $(4),(5),(6)$ we get $$\int_{0}^{\pi/2}\frac{dt}{\sqrt{x^2\cos^2t+y^2\sin^2t}}=\int_{0}^{\pi/2}\frac{du}{\sqrt{a^2\cos^2u+b^2\sin^2u}} $$ or $I(x,y) =I(a,b) $.
Thus we get $I(x, y) =I(x_n, y_n) $ and taking limits as $n\to\infty$ we get $$I(x, y) =\lim_{n\to\infty} I(x_n, y_n) =I(\lim_{n\to\infty} x_n, \lim_{n\to\infty} y_n) =I(L, L) $$ We have used the fact that $I(x, y) $ is a continuous function of $x, y$ and that the sequences $x_n, y_n$ tend to a common limit, say $L$.
Now $I(L, L) =\pi/(2L)$ it follows that $$L=\frac{\pi} {2I(x,y)}=\dfrac{\pi}{\displaystyle \int_0^{\pi}\dfrac{dt}{\sqrt{x^2\cos^2t+y^2\sin^2t}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Show that discretization matrix is positive-definite
Given the following coefficient matrix $A^h$, resulting from the finite difference approximation of the biharmonic equation on a mesh with mesh size $h$:
\begin{equation*}
A^h = \frac{1}{h^4}
\begin{pmatrix}
5 & -4 & 1 & & \\
-4 & 6 & \ddots & \ddots & \\
1 & \ddots & \ddots & \ddots & 1 \\
& \ddots & \ddots & 6 & -4 \\
& & 1 & -4 & 5
\end{pmatrix}.
\end{equation*}
Show that $A^h$ is symmetric, positive-definite.
Of course it is symmetric, since $(A^h)^T = A^h$. In order to show that it is also positive-definite, we need to let $\mathbf{u}\in \mathbb{R}^n\setminus\{\mathbf{0}\}$ and have that $\mathbf{u}^TA\mathbf{u} >0$. I wrote it out:
\begin{align}
\mathbf{u}^TA\mathbf{u} &= \frac{1}{h^4}
\begin{pmatrix}
u_1 & \cdots & u_n
\end{pmatrix}
\begin{pmatrix}
5 & -4 & 1 & & \\
-4 & 6 & \ddots & \ddots & \\
1 & \ddots & \ddots & \ddots & 1 \\
& \ddots & \ddots & 6 & -4 \\
& & 1 & -4 & 5
\end{pmatrix}
\begin{pmatrix}
u_1\\ \\ \vdots\\ \\ u_n
\end{pmatrix} \\[2ex]
&= \frac{1}{h^4}
\begin{pmatrix}
u_1 & \cdots & u_n
\end{pmatrix}
\begin{pmatrix}
5u_1 - 4u_2 + u_3 \\
-4u_1 + 6u_2 -4u_3 + u_4 \\
u_1 - 4u_2 + 6 u_3 - 4u_4 + u5 \\
\vdots \\
u_{i-2} - 4u_{i-1} + 6u_i - 4u_{i+1} + u_{i+2} \\
\vdots
\end{pmatrix} \\[2ex]
&= \frac{1}{h^4}\left[5u_1^2 - 8u_1u_2 + 2u_1u_3 + 6u_2^2 - 8u_2u_3 + 2u_2u_4 + 6u_3^2 -8u_3u_4 + 2 u_3u_5 + \dots\right]
\end{align}
This where I get stuck. I know that I have to show that above can be expressed as the sum of squares, wich means it is positive. I've tried rewriting it using terms like $(u_1 - u_2)^2$ and $(u_1 + u_2 + u_3)^2$ etc, but I cannot see how it all fits together. Help would be greatly appreciated.
| Every symmetric, positive-definite matrix has a square root.
In particular, the root may be asked to be symmetric and positive-definite as well, and then it is uniquely determined.
For the given discretisation matrix $A^h$, which is highly structured, (t-)his square root
$$\sqrt{A^h} \;=\; \frac1{h^2}
\begin{pmatrix}
2 & -1 & 0 & &\\ -1 & 2 & -1 & & & \\
0 & -1 & \ddots & \ddots &\\ & & \ddots & \ddots & -1 & 0\\
& & & -1 & 2 & -1\\ & & & 0 & -1 & 2
\end{pmatrix}.$$
is not so far away, let's say by an educated guess which extends
$$\begin{pmatrix}
5 & -4\\ -4 & 5
\end{pmatrix} \;=\;
\begin{pmatrix}
2 & -1\\ -1 & 2
\end{pmatrix}^2\:.$$
And $\sqrt{A^h}$ is positive-definite:
Cf$\,$
https://en.wikipedia.org/wiki/Definite_matrix#Examples for the $3\times 3$ matrix. The method should generalise to higher dimensions.
And see here on this site:
The full statement including proof of $(7.4.7)$ Theorem starts on page 537 in https://zhilin.math.ncsu.edu/TEACHING/MA580/Stoer_Bulirsch.pdf .
An explicit Sum of squares expression as looked for in the OP is given by
$$(2u_1-u_2)^2 \,+\, \sum_{j=2}^{n-1}\,(-u_{j-1}+2u_j-u_{j+1})^2 \,+\, (2u_n-u_{n-1})^2\tag{SoS}$$
Having the knowledge of the square root $\sqrt{A^h}\,$ it results from expanding
\begin{align}
h^4\cdot\mathbf{u}^TA^h\,\mathbf{u} & \:=\:
\left(h^2\cdot\mathbf{u}^T\sqrt{A^h}\right) \left(h^2\cdot\sqrt{A^h}\,\mathbf{u}\right) \\[2ex]
& \:=\:
\big(2u_1-u_2, \dots, -u_{n-1}+2u_n\big)
\begin{pmatrix}
2u_1 -u_2\\ -u_1 +2u_2 -u_3\\
\vdots \\
-u_{n-2} +2u_{n-1} -u_n \\ -u_{n-1} +2u_n
\end{pmatrix}
\end{align}
The expression $(\text{SoS})$ gets zero only if $\mathbf{u}= \mathbf{0}$.
This is equivalent to the linear system $\sqrt{A^h}\,\mathbf{u} = \mathbf{0}$ having only the trivial solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4546583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
easy way to calculate the limit $\lim_{x \to 0 } \frac{\cos{x}- (\cos{x})^{\cos{x}}}{1-\cos{x}+\log{\cos{x}}}$ I have been trying to use L'Hôpital over this, but its getting too long, is there a short and elegant solution for this?
The Limit approaches 2 according to wolfram.
| Start with $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6) \\
\log(\cos(x))=-\frac{x^2}{2}-\frac{x^4}{12}+O(x^6) \\
\cos(x)\log(\cos(x))=-\frac{x^2}{2}+\frac{x^4}{6}+O(x^6)$$
Then $$\cos(x)-e^{\cos(x)\log(\cos(x))}=1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)-\left\{1+\cos(x)\log(\cos(x))+\frac{(\cos(x)\log(\cos(x)))^2}{2}+O\left((\cos(x)\log(\cos(x)))^3\right)\right\}\\
=-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)-\left\{-\frac{x^2}{2}+\frac{x^4}{6}+O(x^6)+\frac{x^4}{8}+O(x^6)+O(x^6)\right\} \\
=-\frac{x^4}{4}+O(x^6) \,.$$
Similarly,
$$1-\cos(x)+\log(\cos(x))=1-\left\{1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)\right\}+\left\{ -\frac{x^2}{2} - \frac{x^4}{12} + O(x^6) \right\}\\
=-\frac{x^4}{8}+O(x^6) \, .$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
The convergence speed of $ \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x $? I have already known how to prove
\begin{equation*}
\lim _{n \rightarrow \infty} \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x = \sqrt{\frac{\pi}{2n}}
\end{equation*}
with Wallis's formula
\begin{equation*}
\quad \frac{\pi}{2}=\frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdot 8 \cdot 8 \cdots}{1 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot 7 \cdot 9 \cdots}
\end{equation*}
But the method I used was considered not to be universal.
How to prove that
\begin{equation*}
\lim _{n \rightarrow \infty} \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x = \frac{\sqrt{2 \pi}}{2} \cdot \frac{1}{n^{\frac{1}{2}}}-\frac{\sqrt{2 \pi}}{8} \cdot \frac{1}{n^{\frac{3}{2}}}+\frac{\sqrt{2 \pi}}{64} \cdot \frac{1}{n^{\frac{5}{2}}}
\end{equation*}
And is
\begin{equation*}
\lim _{n \rightarrow \infty} \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x = \frac{\sqrt{2 \pi}}{2} \cdot \frac{1}{n^{\frac{1}{2}}}-\frac{\sqrt{2 \pi}}{8} \cdot \frac{1}{n^{\frac{3}{2}}}+ \dots +
(-1)^{k}\cdot\frac{\sqrt{2 \pi}}{2^{\frac {k(k+1)}{2}}} \cdot \frac{1}{n^{\frac{2k+1}{2}}}
\end{equation*}
true? Are there any more powerful tools, like numerical methods to calculate the integration?
|
Are there any more powerful tools, like numerical methods to calculate the intergration?
Yes. There are.
Interval Integration
e.g.
We can take the sine to the n power of x intervals and then integrate those intervals.
This would probably be the easiest way to solve it:
$$
\begin{align*}
z &= \int_{0}^{\frac{\pi}{2}} \lim_{{n} \to {\infty}} \sin(x)^{n} ~\operatorname{d}x\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= \lim_{{n} \to {\infty}} \mathrm{e}^{\ln(\sin(x)^{n})}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= \lim_{{n} \to {\infty}} \mathrm{e}^{n \cdot \ln(\sin(x))}\\
\ln(\sin(x)) &= \begin{cases}
\ln(\sin(x)) = 0, \text{ if } x = \frac{\pi}{2} + 2 \cdot k \cdot \pi\\ \ln(\sin(x)) < 1, \text{ if } x \ne \frac{\pi}{2} + 2 \cdot k \cdot \pi
\end{cases}\\
\ln(\sin(x)) &= \begin{cases}
\ln(\sin(x)) \leq 0, \text{ if } x \in \mathbb{R}\\
\end{cases}\\
\ln(\sin(x)) &\leq \begin{cases}
0, \text{ if } x \in \mathbb{R}\\
\end{cases}\\
\lim_{{n} \to {\infty}} n \cdot \ln(\sin(x)) &= \begin{cases}
\lim_{{n} \to {\infty}} n \cdot 0, \text{ if } x \in \mathbb{R}\\ \end{cases}\\
\lim_{{n} \to {\infty}} n \cdot \ln(\sin(x)) &= \begin{cases}
0, \text{ if } x \in \mathbb{R}\\
\end{cases}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= \begin{cases}
\lim_{{n} \to {\infty}} \mathrm{e}^{n \cdot \ln(\sin(x))}, \text{ if } x \in \mathbb{R}\\
\end{cases}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &\leq \begin{cases}
\mathrm{e}^{0}, \text{ if } x \in \mathbb{R}\\
\end{cases}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &\leq
\mathrm{e}^{0}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &\leq
1\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &=
0\\
F(x) = \int_{0}^{x} \lim_{{n} \to {\infty}} \sin(x)^{n} ~\operatorname{d}x &= \int_{0}^{x} \lim_{{n} \to {\infty}} 0 ~\operatorname{d}x \\
F(x) = \int_{0}^{x} \lim_{{n} \to {\infty}} \sin(x)^{n} ~\operatorname{d}x &= 0\\
\\
\int_{0}^{\frac{\pi}{2}} \lim_{{n} \to {\infty}} \sin(x)^{n} &= F(\frac{\pi}{2}) - F(0) = 0 - 0 = 0\\
z &= 0\\
\end{align*}
$$
Complex Numbers Way
$$\begin{align*}
z &= \int_{0}^{\frac{\pi}{2}} \lim_{{n} \to {\infty}} \sin(x)^{n} ~\operatorname{d}x\\
\sin(x) &= \frac{\mathrm{e}^{x \cdot \mathrm{i}} - \mathrm{e}^{-x \cdot \mathrm{i}}}{2 \cdot \mathrm{i}}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= \lim_{{n} \to {\infty}} (\frac{\mathrm{e}^{x \cdot \mathrm{i}} - \mathrm{e}^{-x \cdot \mathrm{i}}}{2 \cdot \mathrm{i}})^{n}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= \lim_{{n} \to {\infty}} (\frac{\mathrm{e}^{x \cdot \mathrm{i}} - \mathrm{e}^{-x \cdot \mathrm{i}}}{2 \cdot \mathrm{i}})^{n}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= \lim_{{n} \to {\infty}} (\frac{\mathrm{e}^{x \cdot \mathrm{i}} - \mathrm{e}^{-x \cdot \mathrm{i}}}{2} \cdot -\mathrm{i})^{n}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= \lim_{{n} \to {\infty}} \frac{(\mathrm{e}^{x \cdot \mathrm{i}} - \mathrm{e}^{-x \cdot \mathrm{i}})^{n}}{2^{n}} \cdot (-\mathrm{i})^{n}\\
&= \dots
\end{align*}
$$
This way is method is way harder...
It may be some nice taskt to train solving integrals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Computing the eigenvalues and eigenvectors of a $ 3 \times 3$ with a trick The matrix is: $
\begin{pmatrix}
1 & 2 & 3\\
1 & 2 & 3\\
1 & 2 & 3
\end{pmatrix}
$
The solution says that
$ B\cdot
\begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}
=
\begin{pmatrix}
6 \\
6 \\
6\end{pmatrix}$
$ B\cdot
\begin{pmatrix}
1 \\
1 \\
-1
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$
$ B\cdot
\begin{pmatrix}
3 \\
0 \\
-1
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$
Thus the eigenvalues are $ \lambda_{1}=6,\lambda_{2}=0 $
My question is, how can I easily find
$\begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$ and $\begin{pmatrix} 6 \\
6 \\
6\end{pmatrix}$?
Is there any way to see it "quickly"?
| There are many ways of finding eigenvectors and eigenvalues, but they all come down to solving a linear system of equations. For example, in this case you can use the fact that $\lambda_1=6$ is an eigenvalue (which you can easily find by solving the characteristic equation $-\lambda^3 + 6\lambda^2 - 9\lambda = 0$), to solve for the eigenvector corresponding to that eigenvalue. The eigenvector is the solution to the system of equations
$$
\begin{bmatrix}
-6 & 2 & 3 \\
1 & -6 & 3 \\
1 & 2 & -6
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}.
$$
Note that the coefficient matrix of this system is the identity matrix, except for the $-6$ in each diagonal entry, which is exactly what you would expect to see for an eigenvalue of 6.
This linear system is actually very easy to solve, even without using software, because the equations are all multiples of the first equation:
$$
\begin{align*}
-6x + 2y + 3z &= 0 \\
-6x + 2y + 3z &= 0 \\
-6x + 2y + 3z &= 0.
\end{align*}
$$
That means that the solution can only have $x=y=z$, so we can simplify to
$$
\begin{align*}
-6x + 2x + 3x &= 0 \\
-6x + 2x + 3x &= 0 \\
-6x + 2x + 3x &= 0,
\end{align*}
$$
which is just
$$
\begin{align*}
x &= 0 \\
x &= 0 \\
x &= 0.
\end{align*}
$$
Thus, the only solution to the linear system is the trivial solution $x=y=z=0$, which means that the only eigenvector corresponding to $\lambda_1=6$ is the trivial vector $(0,0,0)^T$.
Now that we have found one eigenvector, we can use it to find the other eigenvector. We know that $\lambda_2=0$ is an eigenvalue (which you can easily find by solving the characteristic equation), so we can solve for the eigenvector corresponding to that eigenvalue. The eigenvector is the solution to the system of equations
$$
\begin{bmatrix}
0 & 2 & 3 \\
1 & 0 & 3 \\
1 & 2 & 0
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}.
$$
Again, the linear system is very easy to solve. The first equation implies that $z=-\frac{2}{3}y$, so we can substitute that value of $z$ into the second and third equations to get
$$
\begin{align*}
2y &= 3x \\
-\frac{4}{3}y &= 3x,
\end{align*}
$$
which is
$$
3x = \frac{4}{3}y.
$$
That means that we can choose $x$ and $y$ arbitrarily, as long as $x=\frac{4}{9}y$. So the solution to the linear system is
$$
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
=
\begin{bmatrix}
\frac{4}{9}y \\ y \\ -\frac{2}{3}y
\end{bmatrix}
=
y
\begin{bmatrix}
\frac{4}{9} \\ 1 \\ -\frac{2}{3}
\end{bmatrix}
=
y\begin{bmatrix}
\frac{4}{9} \\ 1 \\ -\frac{2}{3}
\end{bmatrix}.
$$
Note that this is a non-trivial eigenvector, which means that $\lambda_2=0$ is a non-zero eigenvalue.
Thus, the eigenvectors of the matrix are $\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}$ and $\begin{bmatrix}
\frac{4}{9} \\ 1 \\ -\frac{2}{3}
\end{bmatrix}$, and the eigenvalues are $\lambda_1=6$ and $\lambda_2=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4551846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Spivak, Ch. 22, "Infinite Sequences", Problem 1(iii): How do we show $\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right ]=0$? The following is a problem from Chapter 22 "Infinite Sequences" from Spivak's Calculus
*
*Verify the following limits
(iii) $\lim\limits_{n\to \infty} \left
[\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right ]=0$
The solution manual says
$$\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right
]$$
$$=\lim\limits_{n\to \infty} \left [\left
(\sqrt[8]{n^2+1}-\sqrt[8]{n^2}\right )+\left
(\sqrt[4]{n}-\sqrt[4]{n+1}\right )\right ]$$
$$=0+0=0$$
(Each of these two limits can be proved in the same way that
$\lim\limits_{n\to \infty} (\sqrt{n+1}-\sqrt{n})=0$ was proved in the
text)
How do we show $\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[8]{n^2}\right ]=0$?
Note that in the main text, $\lim\limits_{n\to \infty} (\sqrt{n+1}-\sqrt{n})=0$ was solved by multiplying and dividing by $(\sqrt{n+1}+\sqrt{n})$ to reach
$$0<\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2\sqrt{n}}<\epsilon$$
$$\implies n>\frac{1}{4\epsilon^2}$$
| Consider a different approach as seen by the following.
\begin{align}
\sqrt[m]{n^p + 1} - \sqrt[m]{n^p} &= \sqrt[m]{n^p} \, \left( \sqrt[m]{1 + \frac{1}{n^p}} - 1 \right) \\
&= \sqrt[m]{n^p} \, \left( e^{\frac{1}{m} \, \ln\left(1 + \frac{1}{n^p}\right)} - 1 \right) \\
&= \sqrt[m]{n^p} \, \left( \frac{1}{m} \, \ln\left(1 + \frac{1}{n^p}\right) + \frac{1}{2 \, m^2} \, \ln^{2}\left(1 + \frac{1}{n^p}\right) + \mathcal{O}\left(\frac{1}{m^3}\right) \right) \\
&= \sqrt[m]{n^p} \, \frac{1}{m} \, \ln\left(1 + \frac{1}{n^p}\right) \, \left( 1 + \frac{1}{2 \, m} \, \ln\left(1 + \frac{1}{n^p}\right) + \mathcal{O}\left(\frac{1}{m^2}\right) \right)
\end{align}
which gives the limit as
\begin{align}
\lim_{n \to \infty} \left( \sqrt[m]{n^p + 1} - \sqrt[m]{n^p} \right) &= \lim_{n \to \infty} \, \sqrt[m]{n^p} \, \frac{1}{m} \, \ln\left(1 + \frac{1}{n^p}\right) \, \left( 1 + \frac{1}{2 \, m} \, \ln\left(1 + \frac{1}{n^p}\right) + \mathcal{O}\left(\frac{1}{m^2}\right) \right) \\
&= \frac{1}{m} \, \ln(1 + 0) \, \left(1 + \frac{\ln(1)}{2 \, m} + \cdots \right) \times \left(\lim_{n \to \infty} \, \sqrt[m]{n^p} \right) \ \\
&= 0.
\end{align}
Now, for the main problem:
\begin{align}
\sqrt[8]{n^2 + 1} - \sqrt[4]{n+1} &= \sqrt[8]{n^2 + 1} - n^{1/4} + n^{1/4} - \sqrt[4]{n+1} \\
&= ( \sqrt[8]{n^2 + 1} - \sqrt[8]{n^2}) - (\sqrt[4]{n+1} - \sqrt[4]{n})
\end{align}
such that
\begin{align}
\lim_{n \to \infty} \left( \sqrt[8]{n^2 + 1} - \sqrt[4]{n+1} \right) &= \lim_{n \to \infty} \left( \sqrt[8]{n^2 + 1} - \sqrt[8]{n^2} \right) - \lim_{n \to \infty} \left( \sqrt[4]{n + 1} - \sqrt[4]{n} \right) \\
&= (0) - (0) = 0.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the total number of roots of $(x^2+x+1)^2+2=(x^2+x+1)(x^2-2x-6)$, belonging to $(-2,4)$.
Find the total number of roots of $(x^2+x+1)^2+2=(x^2+x+1)(x^2-2x-6)$, belonging to $(-2,4)$.
My Attempt:
On rearranging, I get, $(x^2+x+1)(3x+7)+2=0$
Or, $3x^3+10x^2+10x+9=0$
Derivative of the cubic is $9x^2+20x+10$
It is zero at minus zero point something and minus one point something.
So, even at local minima, the cubic is positive. It means it would cross x-axis only once.
At $x=-2$, cubic is positive and at $-3$, it is negative.
It means the only root is minus two point something.
Is there any other way to solve this question? Something that doesn't involve calculator? Or maybe something that doesn't involve calculus?
| Expanding on Anne Bauval's comment.
Rearranged equation is $(x^2+x+1)(3x+7)+2=0$
Discriminant of $x^2+x+1$ is negative. Thus, this quadratic is always positive.
$-2\lt x\lt4\implies-6\lt3x\lt12\implies1\lt3x+7\lt19$
It means the rearranged equation, on the given interval is (positive)(positive)+positive. Thus, never zero.
So, the answer to the question is zero zeros in the interval.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Closed Form Formula for Nonlinear Recurrence $a_{n+1}=\frac{a_{n}}{2} + \frac{5}{a_{n}}$ I'm trying to find a closed form solution to the sequence
$a_{n+1}=\frac{a_{n}}{2} + \frac{5}{a_{n}}$
I tried using a generating function approach in the following way:
Let $$f(x) = \sum_{n=1}^\infty a_n x^n$$
Then multiplying the original equation with $x^n$ and summing over all $n$, we get:
$$\sum_{n=1}^\infty a_{n+1}x^n= \sum_{n=1}^\infty \frac{a_{n}}{2} x^n + \sum_{n=1}^\infty\frac{5}{a_{n}} x^n$$
$$= \frac{f(x) - a_1}{x} = \frac{f(x)}{2} + 5 \sum_{n=1}^\infty \frac{x^n}{a_n}$$
But now I don't know how to simplify the last term and I get stuck. I put it on Wolframalpha and
found the general solution to the recurrence relation as:
$$a_n = -i \sqrt{10} \cot\left(c_1 2^n\right)$$
| Let's make a change of variable $x_n = \frac{a_n}{\sqrt{10}}$, then
$$x_{n+1} = \frac{x_n^2 +1}{2x_n}$$
We have
$$x_{n+1} -1 = \frac{(x_n -1)^2}{2x_n}$$
$$x_{n+1} + 1 = \frac{(x_n +1)^2}{2x_n}$$
Then
$$\frac{x_{n+1} -1}{x_{n+1} + 1} = \left(\frac{x_n -1}{x_n +1}\right)^2$$
By reccurence, we obtain
$$\frac{x_{n} -1}{x_{n} + 1} = \left(\frac{x_0 -1}{x_0 +1}\right)^{2^n}$$
and we can deduce the closed form solution of $x_n$ and then $a_n$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Given any $a,b,c \geq 1$, prove that $a^2 + b^2 + c^2 \geq 2a\sqrt{b-1} + 2b\sqrt{c-1} + 2c\sqrt{a-1}$ Given any $a,b,c \geq 1$, prove that:
$a^2 + b^2 + c^2 \geq 2a\sqrt{b-1} + 2b\sqrt{c-1} + 2c\sqrt{a-1}$
I tried using most of the popular inequalities and I didn't end up anywhere. Can anyone guide me through this problem?
| $RHS^2 \leq (4a^2+4b^2+4c^2)(a+b+c-3)$, so it suffices to show $a+b+c -3 \leq \frac{1}{4}(a^2+b^2+c^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I prove this limit result associating with an infinite nested radical Let $a_n=\sqrt{1+\sqrt{2+\sqrt{3+...+\sqrt{n}}}}$
I can show that $\lim\limits_{n\to∞}a_n$ converges,let $l=\lim\limits_{n\to∞}a_n$
Now what puzzles me is that how to prove $\lim\limits_{n\to∞}\sqrt{n}\sqrt[n]{l-a_n}=\frac{\sqrt{e}}{2}$
I have already figured out that $\lim\limits_{n\to∞}\sqrt{n}\sqrt[n]{l-a_n}\le\frac{\sqrt{e}}{2}$
How about the other half?
P.S. Here's how I work the upper bound out:
Since $a_n$ converges, we have
$\begin{aligned}
l-a_n=& \sum\limits_{i=n}^∞(a_{i+1}-a_i)\\
=& \sum\limits_{i=n}^∞(\sqrt{1+\sqrt{2+\sqrt{3+...+\sqrt{i+1}}}}-\sqrt{1+\sqrt{2+\sqrt{3+...+\sqrt{i}}}})\\
=&\sum\limits_{i=n}^∞\frac{\sqrt{2+\sqrt{3+...+\sqrt{i+1}}}-\sqrt{2+\sqrt{3+...+\sqrt{i}}}}{\sqrt{1+\sqrt{2+\sqrt{3+...+\sqrt{i+1}}}}+\sqrt{1+\sqrt{2+\sqrt{3+...+\sqrt{i}}}}}\\
\le&\sum\limits_{i=n}^∞\frac{\sqrt{2+\sqrt{3+...+\sqrt{i+1}}}-\sqrt{2+\sqrt{3+...+\sqrt{i}}}}{2\sqrt{1}}\\
\le&...(\text{repeat the process n times})\\
\le&\sum\limits_{i=n}^∞\frac{\sqrt{i+1}}{2^i\sqrt{i!}}\\
<&\sum_{i=n}^\infty \frac{\sqrt{i+1}}{2^i\sqrt{i!}}\\
\le&\frac{1}{2^n\sqrt{n!}}\sum_{i=n}^\infty\frac{\sqrt{i+1}}{2^{i-n}}\\
=&\frac{1}{2^n\sqrt{n!}}\sum_{i=0}^\infty\frac{\sqrt{n+i+1}}{2^i}
=\frac{\sqrt{n}}{2^n\sqrt{n!}}\sum_{i=0}^\infty\frac{1}{2^i}+\frac{1}{2^n\sqrt{n!}}\sum_{i=0}^\infty\frac{\sqrt{n+i+1}-\sqrt{n}}{2^i}\\
=&\frac{2\sqrt{n}}{2^n\sqrt{n!}}+\frac{1}{2^n\sqrt{n!}}\sum_{i=0}^\infty\frac{1}{2^i}\cdot\frac{i+1}{\sqrt{n+i+1}+\sqrt{n}}<\frac{2\sqrt{n}+C}{2^n\sqrt{n!}}<\frac{2\sqrt{n}+n}{2^n\sqrt{n!}}
\end{aligned}
$
Now by this estimation of the upper bound, we can show $\lim\limits_{n\to∞}\sqrt{n}\sqrt[n]{l-a_n}\le\frac{\sqrt{e}}{2}$ by calculation(with Stirling's approximation)
| This is more like long observation with I think it fits better in this section than in the comment section.
It seems that your bounding came a little too soon (as comedians would say).
For each $n$ and $1\leq k\leq n$ set
\begin{align}
x_n&:=\sqrt{1+\sqrt{2+\ldots +\sqrt{n}}}\\
x_{k,n}&:=\sqrt{k+\sqrt{(k+1)+\ldots +\sqrt{n}}}
\end{align}
The sequence $x_n$ is monotone increasing and bounded (with a little effort, $x_n<2$).
Taking conjugates one gets
\begin{align}
x_{n+1}-x_n &=\sqrt{1+\sqrt{2+\ldots +\sqrt{n+\sqrt{n+1}}}}-\sqrt{1+\sqrt{2+\ldots +\sqrt{n}}}\\
&=\frac{\sqrt{2+\sqrt{3+\ldots +\sqrt{n+\sqrt{n+1}}}}-\sqrt{2+\sqrt{3+\ldots +\sqrt{n}}}}{x_{1,n+1}+x_{1,n}}\\
&=\frac{\sqrt{3+\ldots +\sqrt{n+\sqrt{n+1}}} - \sqrt{3+\ldots +\sqrt{n}}}{(x_{1,n+1}+x_{1,n})(x_{2,n+1}+x_{2,n})}\\
&=\ldots\\
&=\frac{\sqrt{n+1}}{\prod^n_{k=1}(x_{k,n+1}+x_{k,n})}
\end{align}
The key part then is finding good asymptotics for the product $\prod^n_{k=1}(x_{k,n+1}+x_{k,n})$.
We have the obvious relations:
*
*$x_{k,n}=\sqrt{k+x_{k+1,n}}$
*and by induction $\sqrt{k}\leq x_{k,n}<\sqrt{k}+1$:
The right-hand side is obvious; as for the second, notice that $x_{n, n}<\sqrt{n}+1$. Suppose statement holds for all $k+1,k+2,\ldots, n$. Then
\begin{align}
x_{k,n}=\sqrt{k+x_{k+1,n}}\leq \sqrt{k+\sqrt{k+1}+1}\leq \sqrt{k}+1
\end{align}
since $\sqrt{k+1}\leq 2\sqrt{k}$ for all $k\in\mathbb{N}$.\
*Let $\phi(x)=\sqrt{1+x}$. Expanding around $0$ we have
$$\phi(1+x)=\phi(0)+\phi'(0)x+\frac12\phi''(\xi)x^2=1+\frac12 x - \frac14(1+\xi)^{-3/2}x^2$$
where $\xi$ is between $0$ and $x$.
Hence, by (3)
\begin{align}x_{k, n}&=\sqrt{k}\sqrt{1+\tfrac{x_{k+1,n}}{k}}=\sqrt{k}\Big(1+\frac12\frac{x_{k+1,n}}{k}-\frac{1}{4}\frac{1}{(1+\xi_{k,n})^{3/2}}\frac{x^2_{k+1,n}}{k^2}\Big)\\
&=\sqrt{k}+\frac12+O\big(\tfrac{1}{\sqrt{k}}\big)
\end{align}
I think this asymptote is not enough yet to solve the OP. But I think that is a good start.
I will leave to his as Community wiki in the hope that others, including the OP, may be able to contribute and have a complete and clear solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding the big-$O$ estimate of the solutions to $\cot x = x$ So a problem for my class I am asked to find the big-$O$ estimate (up to 3 terms) of the solutions to $\cot x = x$. We label the solutions to this equation in increasing order $0<x_1<x_2<\cdots$ and we wish to find the expansion of $x_n$ as $n \rightarrow\infty$.
I've computed the expansion
$$\cot x = x^{-1} - \frac{1}{3}x - \frac{1}{45}x^3 + O(x^5)$$
I observe graphically that for integer $m\ge0$ there exists exactly one solution between $m\pi$ and $(m+1/2)\pi$ ($x_{m+1}$ by the previous labelling). So we know that the leading order of $x_n$ should be $(n-1)\pi$. I'm not sure where to proceed from here. I tried iterating $x$ in the series expansion, but the increasing powers of $x$ gives larger errors with each iteration.
| You already find the first term $n\pi$ for $x_n$ (with $0=x_0<x_1<...$). So, let's denote $x_n=n\pi+y_n$ with $y_n=\mathcal{o}(n)$ and $0<y_n<\frac{\pi}{2}$.
$$\begin{align}
&\Longrightarrow \cot(y_n)=n\pi+y_n \xrightarrow{n\to+\infty}+\infty \\
&\Longrightarrow y_n \xrightarrow{n\to+\infty} 0
\end{align}$$
So $y_n=\mathcal{o}(1)$ and we can compute the series expansion of $\cot(y_n)$ when $y_n \to 0$, we have
$$\begin{align}
n\pi+y_n=\cot(y_n) = \frac{1}{y_n}+\mathcal{O}(y_n) &\Longleftrightarrow n\pi = \frac{1}{y_n}+\mathcal{O}(y_n)\\
&\Longleftrightarrow y_n = \frac{1}{n\pi + \mathcal{O}(y_n)}\\\
&\Longleftrightarrow y_n = \frac{1}{n\pi } \left(1 + \mathcal{O}(\frac{y_n}{n}) \right)\\\
&\Longleftrightarrow y_n = \frac{1}{\pi}\cdot\frac{1}{n}+\mathcal{o}\left(\frac{1}{n} \right)\\
\end{align}$$
We obtain then the second term $\frac{1}{\pi}\cdot\frac{1}{n}$.
Let's denote $x_n=n\pi+ \frac{1}{\pi}\cdot\frac{1}{n}+z_n$ with $z_n=\mathcal{o}\left(\frac{1}{n} \right)$. We have
$$\begin{align}
n\pi+\frac{1}{\pi}\cdot\frac{1}{n}+z_n &=\cot\left(\frac{1}{\pi}\cdot\frac{1}{n}+z_n \right) \\&=\frac{1}{\frac{1}{\pi}\cdot\frac{1}{n}+z_n} -\frac{1}{3}\left( \frac{1}{\pi}\cdot\frac{1}{n}+z_n \right)+\mathcal{o}\left( \frac{1}{n} \right)\\
&=\pi n \left(1+ \pi n z_n \right)^{-1}-\frac{1}{3}\left( \frac{1}{\pi}\cdot\frac{1}{n}+z_n \right)+\mathcal{o}\left( \frac{1}{n} \right)\\
&=\pi n \left(1- \pi n z_n +\mathcal{o}(n^2z_n^2) \right)-\frac{1}{3\pi}\cdot\frac{1}{n}+\mathcal{o}\left( \frac{1}{n} \right)\\
&=\pi n - \pi^2 n^2 z_n +\mathcal{o}(n^2z_n)-\frac{1}{3\pi}\cdot\frac{1}{n}+\mathcal{o}\left( \frac{1}{n} \right)\tag{1}\\
\end{align}$$
Then
$$\begin{align}
(1) &\Longleftrightarrow \frac{4}{3\pi}\cdot\frac{1}{n} = -\pi^2 n^2 z_n++\mathcal{o}(n^2z_n)+\mathcal{o}\left( \frac{1}{n} \right)\\
&\Longleftrightarrow z_n = -\frac{4}{3\pi^3}\cdot\frac{1}{n^3} +\mathcal{o}\left( \frac{1}{n^3} \right)\\\
\end{align}$$
Conclusion:
$$x_n = n\pi+\frac{1}{\pi}\cdot\frac{1}{n}-\frac{4}{3\pi^3}\cdot\frac{1}{n^3} +\mathcal{o}\left( \frac{1}{n^3} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Determine the values of c and d so that E[Y] = c and Var(Y ) = 1 Question
Supppose $X$ is a random PDF $f(x)=\frac{2x}{c^{2}d^{2}}, \quad 0<x<cd$ where $c > 0, \; d > 0$.
Determine the values of $c$ and $d$ so that $\mathbb{E}[X]= c $ and $\mathrm{Var}(x) = 1$
My approach
My approach has been to try and isolate the variables using the given information. I have used $\mathrm{Var}(X) = 1$ as a reference point to start. My work has been as follows:
\begin{align}
1 &= \mathbb{E}(X^2) - \mathbb{E}[X]^2 \\
1 &= \int_{0}^{cd} x^{2} \cdot \frac{2x}{c^{2}d^{2}} - \left(\int_{0}^{cd} x \cdot \frac{2x}{c^{2}d^{2}}\right)^2
\end{align}
Following this, I have approached this result:
\begin{align}
1 = \frac{(cd)^4}{2c^2d^2} - \frac{(cd)^4}{c^4d^4}
\end{align}
Which yields the following:
\begin{align}
1 &= \frac{c^2d^2}{2} \\
2 &= c^2d^2 \\
\end{align}
I have then derived the possible inter solutions of $c$ and $d$.
\begin{align*}
\therefore c &= \pm 1\; ,\pm 2 \\
d &= \pm 1\; , \pm 2
\end{align*}
I have then tried to prove the $\mathbb{E}[X] = c$ through the following case. Suppose $c = 1$ and $d = 2$ thus $0 < x < 2$.
\begin{align}
\mathbb{E}[X] &= \int_{0}^{2} x\cdot \frac{2y}{c^{2}d^{2}} \; \mathrm{d}y \\
&= \int_{0}^{2} \frac{2x^2}{4} \; \mathrm{d}x \\
&= \frac{1}{4}\int_{0}^{2} 2x^2 \; \mathrm{d}x \\
&= \frac{1}{4} \cdot \frac{2x^3}{3} \Biggr|_{0}^{2}
\end{align}
However, the answer does not evalute to the case. I have taken the wrong approach and am lost on how to approach the problem. Any help with the solution would be really appreciated.
Update
Thank you for the suggestions. Errors were made in the integration. The error was found within the 2nd integral.
$$
\frac{(cd)^4}{c^4d^4} \rightarrow \frac{2(cd)^3}{3(c^4d^4)}
$$
Squaring the results would give the following $\frac{4c^2d^2}{9}$
The derivation of the expected value for $c$ is as follows
\begin{align}
c &= \int_{0}^{ac} x \cdot \frac{2x}{c^2d^2} \; \mathrm{d}x \\
c &= \frac{2}{c^2d^2} \int_{0}^{cd} x^2 \\
c &= \frac{2}{c^2d^2} \cdot \frac{x^3}{3} \Biggr|_{0}^{cd} \\
c &= \frac{2x^3}{3c^2d^2} \xrightarrow[]{} \frac{2(cd)^3}{3c^2d^2} \\
c &= \frac{2}{3}cd \\
&\therefore d = \frac{3}{2}
\end{align}
Following that, the answer would be simply substituting it into the corrected equation for variance.
\begin{align}
1 &= \frac{1}{2}c^2\cdot \frac{9}{4} - \frac{4}{9}c^2\cdot\frac{9}{4} \\
1 &= \frac{9}{4}c^2\left(\frac{1}{2}-\frac{4}{9}\right) \\
18 &= \frac{9}{4}c^2 \\
8 &= c^2 \\
c &= 2\sqrt{2}
\end{align}.
| Here's a quick way to solve this. Brief hints are provided for the path to follow. Reveal the spoilers if you get stuck in computations.
You have $\mathbb E[X]=c$. Then put this into the Variance equation.
\begin{align*}1=\mathrm Var[X]&=\mathbb E[X^2]-\mathbb E[X]^2\\&=\int_{0}^{cd}x^2\frac{2x}{c^2d^2}\ \mathrm dx-c^2\\&=\frac12c^2d^2-c^2\end{align*}
For the expectation, you have
$$c=\int_{0}^{cd}x\frac{2x}{c^2d^2}\ \mathrm dx=\frac{2}{3}cd\Rightarrow d=\frac32$$
Plugging this in, we get
$$1=\frac12c^2\frac94-c^2=\frac{c^2}{8}\Rightarrow c=2\sqrt 2$$
and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can we generalize factorisation of $(a+b)^n-(a^n+b^n)$
How can we generalize factorisation of
$$(a+b)^n-(a^n+b^n)\,?$$
where $n$ is an odd positive integer.
I found the following cases:
$$(a+b)^3-a^3-b^3=3ab(a+b)$$
$$(a+b)^5-a^5-b^5=5 a b (a + b) (a^2 + a b + b^2)$$
$$(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$$
Exapansions and factoring seem terrible. Maybe there is a good way for generalization.
The question is, can we generalize these factorisations?
$$(a+b)^n-(a^n
+b^n)$$
for all odd integer $n$?
| For $n=9$ we have
$$
(a+b)^9-(a^9+b^9)=3(3a^6 + 9a^5b + 19a^4b^2 + 23a^3b^3 + 19a^2b^4 + 9ab^5 + 3b^6)(a + b)ab,
$$
where the first part is irreducible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solve the equation $\sqrt{x^2+x+1}+\sqrt{x^2+\frac{3x}{4}}=\sqrt{4x^2+3x}$ Solve the equation $$\sqrt{x^2+x+1}+\sqrt{x^2+\dfrac{3x}{4}}=\sqrt{4x^2+3x}$$
The domain is $$x^2+\dfrac{3x}{4}\ge0,4x^2+3x\ge0$$ as $x^2+x+1>0$ for every $x$. Let's raise both sides to the power of 2: $$x^2+x+1+x^2+\dfrac{3x}{4}+2\sqrt{(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)}=4x^2+3x\\2\sqrt{(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)}=2x^2+\dfrac{5x}{4}$$ Let's raise both sides to the power of 2 again but this time the roots should also satisfy $A:2x^2+\dfrac54x\ge0$:$$4(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)=(2x^2+\dfrac54x)^2$$ I came at $$x(2x^2+\dfrac{87}{16}x+3)=0$$ I obviously made a mistake as the answer is $x=-4$, but is there an easier approach?
| Let me try. One can rewrite equation as below.
$$\sqrt{x^2+x+1} + \frac{1}{2}\sqrt{4x^2+3x} = \sqrt{4x^2+3x},$$
$$\sqrt{x^2+x+1} = \frac{1}{2}\sqrt{4x^2+3x},$$
$$x^2+x+1 = \frac{1}{4}(4x^2+3x)$$
$$4x^2+4x+4 = 4x^2+3x,$$
$$x = -4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4566074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve a first order PDE derived from exponential generating function The PDE comes from a counting problem:
Suppose there is a bag of $m$ red, $n$ blue balls, and each time one randomly removes one ball until there are only balls of one color in the bag. What is the expected number of balls left?
Let $f(m,n)$ be the expectation. We have
\begin{equation*}
f(m,n)=\frac{m}{m+n}f(m-1,n)+\frac{n}{m+n}f(m,n-1),\quad f(m,0)=m,\; f(0,n)=n.
\end{equation*}
Multiply by $x^my^n/(m!n!)$ and sum over $m,n\geq 1$ to obtain the PDE:
\begin{equation*}
xF_x+yF_y-(x+y)F=xe^x+ye^y,\quad F(x,y)=\sum_{m,n\geq 0}f(m,n)\frac{x^my^n}{m!n!}.
\end{equation*}
My question is how to solve it?
Remark. Of course, if you know or can guess out the formula $f(m,n)=\frac{m}{n+1}+\frac{n}{m+1}$, the original problem can be solved easily by induction. Knowing the formula, you can also find
\begin{equation*}
F(x,y)=\left(\frac{x}{y}+\frac{y}{x}\right)e^{x+y}-\frac{x}{y}e^x-\frac{y}{x}e^y.
\end{equation*}
It does satisfy the PDE.
| You have
$$x F_{x} + y F_{y} = (x + y) F + x e^{x} + y e^{y}$$
Using the method of characteristics in the parameterisation invariant form, we have
$$\frac{dx}{x} = \frac{dy}{y} = \frac{dF}{(x + y) F + x e^{x} + y e^{y}}$$
Solving across the first equality gives
$$\frac{x}{y} = C_{1}$$
Using this result, we can rewrite the last ratio as
\begin{align}
\frac{dF}{(x + y) F + x e^{x} + y e^{y}} &= \frac{dF}{y (x/y + 1) F + y (x/y) e^{y (x/y)} + y e^{y}} \\
&= \frac{dF}{y (C_{1} + 1) F + C_{1} y e^{C_{1} y} + y e^{y}}
\end{align}
and solving across the second equality gives
\begin{align}
\frac{dy}{y} &= \frac{dF}{y (C_{1} + 1) F + C_{1} y e^{C_{1} y} + y e^{y}} \\
\implies F' - (C_{1} + 1) F &= C_{1} e^{C_{1} y} + e^{y} \\
\implies (e^{-(C_{1} + 1) y} F)' &= C_{1} e^{- y} + e^{- C_{1} y} \\
\implies e^{-(C_{1} + 1) y} F &= - C_{1} e^{- y} - C_{1}^{-1} e^{- C_{1} y} + C_{2} \\
\implies F &= - C_{1} e^{C_{1} y} - C_{1}^{-1} e^{y} + C_{2} e^{(C_{1} + 1) y}
\end{align}
Replacing the constant $C_{1} = x/y$ and noting $C_{2} = g(C_{1})$ for some arbitrary differentiable function $g$, we have
$$F = - \frac{x}{y} e^{x} - \frac{y}{x} e^{y} + g \left( \frac{x}{y} \right) e^{x + y}$$
which is the result you found, with $g(x/y) = x/y + y/x$. Note that $g$ is determined by the boundary conditions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4566287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing vector to equal vector forces/magnitudes I have this problem with its respective solution:
Assuming that the superior vector is $\overrightarrow{A}$, the middle vector is $\overrightarrow{B}$, and the inferior vector is $\overrightarrow{F}$
My computation was:
$$\overrightarrow{A} \cdot \overrightarrow{B}=\overrightarrow{B} \cdot \overrightarrow{F}\\
|\overrightarrow{A}||\overrightarrow{B}|cos(40)=|\overrightarrow{B}||\overrightarrow{F}|cos(35)\\
|\overrightarrow{F}| = \frac{20000*cos(40)}{cos(35)} \space\space\space\space \text{(given} |\overrightarrow{A}|=20000 \text{)} \\
|\overrightarrow{F}| = 18703
$$
Where am I wrong? I'm looking for a vector solution.
| Although I don't recommend it for this application, one can get a vector solution using $$\vec{A}\times \vec{B}=S\vec{F}\times \vec{B}$$
Where variable $S$ is the unknown force.
$$
\vec{A} \, = \, \left( \begin{align}20000 \; \cos \left( 2 \cdot \frac{\pi }{9} \right) \\ 20000 \; \sin \left( 2 \cdot \frac{\pi }{9} \right) \\ 0 \end{align} \right)$$
$$
\vec{B} \, = \, \left( \begin{align}1 \\ 0 \\ 0 \end{align} \right)
$$
$$
\vec{F} \, = \, \left( \begin{align}S \; \operatorname{cos} \left( 7 \cdot \frac{\pi }{36} \right) \\ S \; \operatorname{sin} \left( 7 \cdot \frac{\pi }{36} \right) \\ 0 \end{align} \right)
$$
Now $$\vec{A}\times\vec{B}=\left( \begin{align}0 \\ 0 \\ -20000 \; \sin \left( \frac{2}{9} \; \pi \right) \end{align} \right)
$$
$$\vec{F}\times\vec{B}=\left( \begin{align}0 \\ 0 \\ -S \; \sin \left( \frac{7}{36} \; \pi \right) \end{align} \right)$$
Solving for $S$
$$ \left\{ S = 20000 \cdot \frac{\sin \left( 2 \cdot \frac{\pi }{9} \right)}{\sin \left( 7 \cdot \frac{\pi }{36} \right)} \right\}=22413.32 $$
Just because you can, doesn't mean you should.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4568190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Stuck on Liebniz Integration Technique This is a problem from the textbook Advanced Calculus by Edwin Wilson from Chapter 11 (page 281-288). The problem asks,
Evaluate by any means the following: $$\int_0^{\frac{\pi}{2}} \frac{ln(1 + \cos{a}\cos{x})}{\cos{x}} \, dx$$
I think I'm on the right track... but I'm getting stuck. I have the following:
\begin{align*}
\phi(a) &= \int_0^\frac{\pi}{2}\, \, \, \frac{\ln(1 + \cos{a}\cos{x})}{\cos{x}} \, dx\\
\\
\\
\implies \phi'(a) &= \int_0^\frac{\pi}{2}\, \, \,\frac{1}{\cos{x}} \cdot \frac{\partial}{\partial a}\ln(1 + \cos{a}\cos{x}) \, dx\\
&= \int_0^\frac{\pi}{2}\, \, \, \frac{1}{\cos{x}} \cdot \left [ \frac{1}{1 + \cos{a}\cos{x}} \cdot \cos{x}(-\sin{a}) \right ] \, dx\\
&= \int_0^\frac{\pi}{2}\, \, \,\frac{-\sin{a}}{1+\cos{a}\cos{x}} \, dx\\
\\
\\
\implies \phi'(a) &= -\sin{a} \cdot \int_0^\frac{\pi}{2}\, \, \,\frac{1}{1 + \cos{a}\cos{x}} \, dx
\\
\\
&\\
\end{align*}
Here, I've seen the suggestion to use the substitution $t = \tan\left({\frac{x}{2}}\right)$, and I make some headway but then I get stuck... I see that with this substitution we get $\cos{x} = \dfrac{1 - t^2}{1+t^2}$ and $dx = \dfrac{2}{1 + t^2} dt$, and the bounds change to $t = 0$ and $t = 1$. Substituting some things in..
\begin{align*}
\phi'(a) &= -\sin{a} \int_0^\frac{\pi}{2} \frac{1}{1 + \cos{a}\cos{x}} \, dx\\
\\
&= -\sin{a} \int_{t=0}^{t=1} \frac{1}{1 + \cos{a} \cdot \left(\frac{1 - t^2}{1 + t^2} \right )} \cdot \left (\frac{2}{1 + t^2} \right ) dt\\
\\
&=-2\sin{a} \int_0^1 \frac{1}{\left(1 + t^2 \right) \cdot \left (1 + \frac{1-t^2}{1+t^2} \cdot \cos{a} \right )} \, \, dt\\
\\
&= -2\sin{a} \int_0^1 \frac{1}{(1 + t^2) + \left[(1 - t^2) \cos{a}\right]} \, dt
\end{align*}
And this is where I get stuck. Any assistance would be great!
| Let $\alpha := \cos(a)$ and $\beta := \sqrt{1-\alpha}$ and $\gamma := \sqrt{1+\alpha}$. (The latter two are well defined since $\cos(a) \in [-1,1]$.) Then the integral of concern is, just with some factorizations,
$$\mathcal{I}:= \int_0^1 \frac{1}{1+t^2 +\alpha - \alpha t^2} \, \mathrm{d}t
= \int_0^1 \frac{1}{t^2 \beta^2 + \gamma^2} \, \mathrm{d}t$$
Factor out a $\gamma^2$:
$$\mathcal{I}:= \frac{1}{\gamma^2} \int_0^1 \frac{1}{1+(\beta t/\gamma)^2} \, \mathrm{d}t$$
From here, the solution is simply a further, obvious $u$-substitution and then the arctangent derivative. You may also have to do a little extra casework in the event $\gamma$ or $\beta$ or both are $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4570259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
I can't find the solution of $\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}}$ I can't find the solution of
$$\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}}$$
Computing for $x$ goes to $0$ it gives a $1^\infty$ type of indeterminate form. I tried to solve it by making it similar to $0/0$ type of indeterminate form by taking log of both sides and writing as
$$\frac{\ln\left(1+\frac{x}{(x-1)^2}\right)}{\sqrt{1+x}-1}$$
Then I applied L'Hôpital's rule to find the limit and find it infinity for $0^+$ and $0$ for $0^-$ but my solution did not match with the answer.
Thanks for help!
| Given
$$\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}}$$ we have with $u= e ^{ \ln u}$,
$$ \lim_{x \to 0} \left(\frac{x}{\left(x - 1\right)^{2}} + 1\right)^{\frac{1}{\sqrt{x + 1} - 1}} = \lim_{x \to 0} e^{\ln{\left(\left(\frac{x}{\left(x - 1\right)^{2}} + 1\right)^{\frac{1}{\sqrt{x + 1} - 1}} \right)}} =$$
$$= \lim_{x \to 0} e^{\ln{\left(\left(\frac{x}{\left(x - 1\right)^{2}} + 1\right)^{\frac{1}{\sqrt{x + 1} - 1}} \right)}} = \lim_{x \to 0} e^{\frac{\ln{\left(\frac{x}{\left(x - 1\right)^{2}} + 1 \right)}}{\sqrt{x + 1} - 1}} $$
Since we have an indeterminate form of type $\left(\frac 00\right)$, we can apply the l'Hopital's rule:
$$e^{ \lim_{x \to 0} \frac{\frac{d}{dx}\left(\ln{\left(\frac{x}{\left(x - 1\right)^{2}} + 1 \right)}\right)}{\frac{d}{dx}\left(\sqrt{x + 1} - 1\right)}} = e^{ \lim_{x \to 0} \frac{2 \sqrt{x + 1} \left(- \frac{2 x}{\left(x - 1\right)^{3}} + \frac{1}{\left(x - 1\right)^{2}}\right)}{\frac{x}{\left(x - 1\right)^{2}} + 1}}= $$
$$=e^{ \lim_{x \to 0} \frac{2 \sqrt{x + 1} \left(- \frac{2 x}{\left(x - 1\right)^{3}} + \frac{1}{\left(x - 1\right)^{2}}\right)}{\frac{x}{\left(x - 1\right)^{2}} + 1}} = e^{ \lim_{x \to 0}\left(- \frac{2 \left(x + 1\right)^{\frac{3}{2}}}{\left(x - 1\right) \left(x + \left(x - 1\right)^{2}\right)}\right)}= e^{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Seeking for other methods to evaluate $\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx$ for $n\geq 2$. Inspired by my post, I go further to investigate the general integral and find a formula for
$$
I_n=\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx =-\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left[\gamma+\psi\left(1-\frac{1}{n}\right)\right] \tag*{}
$$
Let’s start with its partner integral
$$
I(a)=\int_0^{\infty}\left(x^n+1\right)^a d x
$$
and transform $I(a)$, by putting $y=\frac{1}{x^n+1}$, into a beta function
$$
\begin{aligned}
I(a) &=\frac{1}{n} \int_0^1 y^{-a-\frac{1}{n}-1}(1-y)^{-\frac{1}{n}-1} d y \\
&=\frac{1}{n} B\left(-a-\frac{1}{n}, \frac{1}{n}\right)
\end{aligned}
$$
Differentiating $I(a)$ w.r.t. $a$ yields
$$
I^{\prime}(a)=\frac{1}{n} B\left(-a-\frac{1}{n}, \frac{1}{n}\right)\left(\psi(-a)-\psi\left(-a-\frac{1}{n}\right)\right)
$$
Then putting $a=-1$ gives our integral$$
\begin{aligned}
I_n&=I^{\prime}(-1) \\&=\frac{1}{n} B\left(1-\frac{1}{n}, \frac{1}{n}\right)\left[\psi(1)-\psi\left(1-\frac{1}{n}\right)\right] \\
&=-\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left[\gamma+\psi\left(1-\frac{1}{n}\right)\right]\end{aligned}
$$
For examples,
$$
\begin{aligned}& I_2=-\frac{\pi}{2} \csc \frac{\pi}{2}\left[\gamma+\psi\left(1-\frac{1}{2}\right)\right]=\pi \ln 2,\\ & I_3=-\frac{\pi}{3} \csc \left(\frac{\pi}{3}\right)\left[\gamma+\psi\left(\frac{2}{3}\right)\right]=\frac{\pi \ln 3}{\sqrt{3}}-\frac{\pi^2}{9} ,\\ &I_4=-\frac{\pi}{4} \csc \left(\frac{\pi}{4}\right)\left[\gamma+\psi\left(\frac{3}{4}\right)\right]=\frac{3 \pi}{2 \sqrt{2}}\ln 2-\frac{\pi^2}{4 \sqrt{2}},\\
& I_5=-\frac{\pi}{5} \csc \left(\frac{\pi}{5}\right)\left[\gamma+\psi\left(\frac{4}{5}\right)\right]=-\frac{2 \sqrt{2} \pi}{5 \sqrt{5-\sqrt{5}}}\left[\gamma+\psi\left(\frac{4}{5}\right)\right], \\
& I_6=-\frac{\pi}{6} \csc \left(\frac{\pi}{6}\right)\left[\gamma+\psi\left(\frac{5}{6}\right)\right]=\frac{2 \pi}{3} \ln 2+\frac{\pi}{2} \ln 3-\frac{\pi^2}{2 \sqrt{3}},
\end{aligned}
$$
Furthermore, putting $a=-m$, gives
$$\boxed{I(m,n)=\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{(x^n+1)^m} dx = \frac{1}{n} B\left(m-\frac{1}{n}, \frac{1}{n}\right)\left[\psi(m)-\psi\left(m-\frac{1}{n}\right)\right] }$$
For example,
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(x^6+1\right)}{(x^6+1)^5} dx & =\frac{1}{6} B\left(\frac{29}{6}, \frac{1}{6}\right)\left[\psi(5)-\psi\left(\frac{29}{6}\right)\right] \\
& =\frac{1}{6} \cdot \frac{21505 \pi}{15552} \cdot\left(-\frac{71207}{258060}-\frac{\sqrt{3} \pi}{2}+\frac{3 \ln 3}{2}+2 \ln 2\right) \\
& =\frac{21505 \pi}{93312}\left(-\frac{71207}{258060}-\frac{\sqrt{3} \pi}{2}+\frac{3 \ln 3}{2}+2 \ln 2\right)
\end{aligned}
$$
Are there any other methods?
Your comments and alternative methods are highly appreciated.
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\newcommand{\ic}{\mathrm{i}}
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\begin{align}
\color{#44f}{{\tt I}_{n}} & \equiv
\color{#44f}{\int_{0}^{\infty}{\ln\pars{x^{n} + 1} \over x^{n} + 1}\dd x} \sr{x^{n}\ \mapsto\ x}{=}
\int_{0}^{\infty}{\ln\pars{1 + x} \over 1 + x}\
{1 \over n}\, x^{1/n - 1}\,\,\dd x
\\[5mm] & =
\left.{1 \over n}\partiald{}{\nu}\int_{0}^{\infty}x^{\color{red}{1/n} - 1}\,\,
\pars{1 + x}^{\nu - 1}\,\dd x\right\vert_{\nu\ =\ 0}.
\\[5mm] & \mbox{Note that}\ \pars{1 + x}^{\nu - 1} =
\sum_{k = 0}^{\infty}{\nu - 1 \choose k}x^{k} =
\sum_{k = 0}^{\infty}\bracks{{k - \nu \choose k}\pars{-1}^{k}}x^{k}
\\[5mm] & =
\sum_{k = 0}^{\infty}{\Gamma\pars{k - \nu + 1} \over \Gamma\pars{1 - \nu}}{\pars{-x}^{k} \over k!}\quad \mbox{such that}
\\[5mm] \color{#44f}{{\tt I}_{n}} & \equiv
\color{#44f}{\int_{0}^{\infty}{\ln\pars{x^{n} + 1} \over x^{n} + 1}\dd x} =
{1 \over n}\partiald{}{\nu}\overbrace{\bracks{\Gamma\pars{\color{red}{1 \over n}}
{\Gamma\pars{-\color{red}{1/n} - \nu + 1} \over \Gamma\pars{1 - \nu}}}}^{\ds{Ramanujan's\ Master\ Theorem}}\hspace{-1mm}_{\substack{\\[2mm]\nu\ =\ 0}}
\\[5mm] & = \bbx{\color{#44f}{%
-\,{\pi\csc\pars{\pi/n}H_{-1/n}\,\, \over n}}} \\ &
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find All Complex solutions for $z^3+3i\bar{z}=0$ Find All Complex solutions for $z^3+3i \bar z =0$.
I tried substituting $z=a+bi$ and simplifying as much as I can
and this is what I ended up with: $a^3-2b^2a+3b+i(3ba^2-b^3+3a)=0$.
I just did not understand how do I get the values of $z$ from this equation
| You let $z=a+bi$ and the system of equations to solve is: $a^3-3ab^2+3b=0$ and $3ba^2-b^3+3a=0$. If you multiply the first equation by $-a$ and the second by $b$ and then add the resulting equations you get $a^4+b^4=6a^2b^2$. Let $m=\frac{b}{a}$. Then, $$m^4-6m^2+1=0.$$ The solution of this equation is: $m=\tan(\frac{3\pi}{8}+\frac{k\pi}{2})$ where $k=0,1,2,3$. Hence, we found the Arguments (angles) of the 4 solutions. The other 4 Arguments can be found by formula $\pi-\theta$ from these Arguments, since $\tan(\pi-\theta)=-\tan\theta$ satisfies the equation too.
To find the magnitudes of them, multiply the first equation by $a$ and the second by $b$ and then add the resulting equations you get $a^2+b^2=\frac{6ab}{b^2-a}$. Then, since $b=ma$, we have $|z|=\sqrt\frac{6m}{m^2-1}$. We find out that $\sqrt\frac{6m}{m^2-1}=\sqrt{3}$ for all $m$ values we found above. I checked by WolframAlpha trying to prove by hand...
The solution $z=0$ could be found by polar method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4578068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How far can we go with the integral $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$ Inspired by my post, I decided to investigate the integral in general
$$
I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$$
by the powerful substitution $x=\frac{1-t}{1+t} .$
where $n$ is a natural number greater $1$.
Let’s start with easy one
\begin{aligned}
I_1 &=\int_0^1 \frac{\ln \left(\frac{2 t}{1+t}\right)}{1+t^2} d t \\ &=\int_0^1 \frac{\ln 2+\ln t-\ln (1+t)}{1+t^2} d t \\&=\frac{\pi}{4} \ln 2+\int_0^1 \frac{\ln t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2} d t \\&=\frac{\pi}{4} \ln 2-G-\frac{\pi}{8} \ln 2 \\
&=\frac{\pi}{8} \ln 2-G\end{aligned}
By my post
$$I_2= \frac{\pi}{4} \ln 2-G $$
and
$$\begin{aligned}I_4 &=\frac{3 \pi}{4} \ln 2-2 G
\end{aligned}
$$
$$
\begin{aligned}
I_3=& \int_0^1 \frac{\ln (1-x)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1+x+x^2\right)}{1+x^2} d x \\=& \frac{\pi}{8} \ln 2-G+\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+x+x^2\right)}{1+x^2} d x-G
\\ =& \frac{\pi}{8} \ln 2-\frac{4G}{3} +\frac{\pi}{6} \ln (2+\sqrt{3})
\end{aligned}
$$
where the last integral refers to my post.
Let’s skip $I_5$ now.
$$
I_6=\int_0^1 \frac{\ln \left(1-x^6\right)}{1+x^2} d x=\int_0^1 \frac{\ln \left(1+x^3\right)}{1+x^2} d x+I_3\\
$$
$$\int_0^1 \frac{\ln \left(1+x^3\right)}{1+x^2} d x = \int_0^1 \frac{\ln (1+x)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1-x+x^2\right)}{1+x^2} d x\\=\frac{\pi}{8}\ln 2+ \frac{1}{2} \int_0^{\infty} \frac{\ln \left(1-x+x^2\right)}{1+x^2} d x-G \\= \frac{\pi}{8}\ln 2+ \frac{1}{2}\left( \frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G \right)- G \\= \frac{\pi}{8} \ln 2+\frac{\pi}{3} \ln (2+\sqrt{3})-\frac{5}{3} G $$
Hence $$I_6= \frac{\pi}{4} \ln 2+\frac{\pi}{2} \ln (2+\sqrt{3})-3 G$$
How far can we go with the integral $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x?$
| As @FShrike’s solution, I want to express the integral in terms of an infinite series of Diagamma functions.
Using $\ln \left(1-x^n\right)=-\sum_{k=1}^{\infty} \frac{x^{n k}}{k}$ for $|x|<1$, we have
$$
\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x=-\sum_{k=1}^{\infty} \frac{1}{k} \underbrace{\int_0^1 \frac{x^{n k}}{1+x^2} d x}_{I_k}
$$
$$
\begin{aligned}
I_k &=\int_0^1 \frac{y^{\frac{n k}{2}}}{1+y} \cdot \frac{1}{2 \sqrt{y}} d y \quad \textrm{, where }y=x^2\\
&=\frac{1}{2} \int_0^1 \frac{y \frac{n k-1}{2}}{1+y} d y
\end{aligned}
$$
As $\int_0^1 \frac{x^a}{1+x} d x=\frac{1}{2}\left[\psi\left(\frac{a}{2}+1\right)-\psi\left(\frac{a}{2}+\frac{1}{2}\right)\right]$, we can conclude that
$$
\boxed{
\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x =\frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k}\left[\psi\left(\frac{n k+1}{4}\right)-\psi\left(\frac{n k+3}{4}\right)\right]}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
} |
What is the solution for $x+22=-6\sqrt{2x+9}$ So, I want to solve for x in the radical equation:
$x + 22 = -6\sqrt{2x+9}$
By Squaring each expression we get:
$(x + 22)^2 = (-6\sqrt{2x+9})^2$
$ x^2 + 44x + 484 = 36\cdot(2x+ 9) $
$ x^2 + 44x + 484 = 72x + 324 $
Now by solving the quadratic equation:
$ x^2 - 28x + 160 = 0 $
$x^2 - 20x - 8x + 160 = 0$
$x\cdot(x-20) -8\cdot(x-20) = 0 $
$ (x-20)\cdot(x-8) = 0 $
$ x = 20 $ or $ x = 8 $
But, none of the values of x satisfies the equation $x + 22 = -6\sqrt{2x+9}$, they satisfy the equation $x + 22 = 6\sqrt{2x+9}$. There should be an extraneous root that satisfies the equation $x + 22 = 6\sqrt{2x+9}$, but boot the roots satisfies this equation and none of them satisfies $x + 22 = -6\sqrt{2x+9}$.
Why is that so? And is there any complex/imaginary solution to the equation?
| Sometimes it can be useful to find the domain of the equation before solving equations involving radicals.
By definition of the principal square root, if $\sqrt {f(x)}=g(x)$, then $f(x)\geqslant0\wedge g(x)\geqslant 0$ holds, where $f(x)$ and $g(x)$ are some algebraic expressions.
Thus, you have:
$$
\begin{align}&\begin{cases}\sqrt {2x+9}=\frac {x+22}{-6}\geqslant 0\\ 2x+9\geqslant0\end{cases}\\\\
\implies &\begin{cases}x\leqslant-22\\ x\geqslant-\frac 92\end{cases}\\\\
\implies &x\in\color{#c00}{\emptyset.}\end{align}
$$
Therefore, we don't need to solve the equation. Because, the domain of the equation is the empty set. This means, the real solution, indeed, doesn't exist.
In this part of the answer, we want to prove that the equation $x+22=-6\sqrt{2x+9}$ has no complex roots in general.
Let $\sqrt {2x+9}=z, z\in\mathbb C\setminus \mathbb R$. Then we have:
$$
\begin{align}x+22=\frac {z^2}{2}+\frac {35}{2}=\frac {z^2+35}{2}\end{align}
$$
This leads to:
$$
\begin{align}&\frac {z^2+35}{2}=-6z\\
\implies &z^2+12z+35=0\\
\implies &\Delta_z=36-35=1\geqslant 0.\end{align}
$$
This implies that $z\in\mathbb R$, which gives a contradiction. This completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Literature containing the process of solving $ f''(x)+ae^{bx}f(x)=0 $ I tried to solve the following ODE using infinite summs, but failed:
$$ f''(x)+ae^{bx}f(x)=0 \tag1$$
From a comment in this post, I found out that the solution of the above equation contains a linear combination of the Bessel functions. That means that there is a good probability that this equation is solved in detail in some literature.
My question is, does equation $(1)$ have a specific name? I want to google it so I can go through the process of solving it. If not, is there any other way I can find some literature that goes through this problem?
| Consider the differential equation:
$$ \frac{d^2 y}{d x^2} + a^2 \, e^{b x} \, y = 0. $$
First note that if $a = 0$ then $y(x) = c_{0} + c_{1} \, x$. Second note that if $b = 0$ then $f(x) = c_{0} \, \cos(a x) + c_{1} \, \sin(a x)$. Now consider the case when $a \neq 0$ and $b \ne 0$, which is as follows.
Let $t = e^{b x/2}$ to obtain $\frac{d t}{d x} = \frac{b}{2} \, t$ which is required in
\begin{align}
\frac{d}{dx} &= \frac{d t}{d x} \, \frac{d}{dt} = \frac{b}{2} \, t \, \frac{d}{d t} \\
\frac{d^2}{d x^2} &= \frac{b^2}{4} \, t \, \frac{d}{d t} \left( t \, \frac{d}{dt} \right) = \frac{b^2}{4} \, \left( t^2 \, \frac{d^2}{d t^2} + t \, \frac{d}{dt} \right)
\end{align}
and leads to
$$ t^2 \, f^{''} + t \, f^{'} + \frac{4 a^2}{b^2} \, t^2 \, f = 0. $$
In order to remove the parameter in the last term let $t = p \, u$ and repeat the change of variables process to obtain, which is
$$ \frac{d}{dt} = \frac{1}{p} \, \frac{d}{du} \hspace{6mm} \text{and} \hspace{6mm} \frac{d^2}{dt^2} = \frac{1}{p^2} \, \frac{d^2}{du^2},
$$
and gives
$$ u^2 \, f^{''} + u \, f^{'} + \frac{4 \, a^2 \, p^2}{b^2} \, f = 0. $$
Let
$$ p^2 = \frac{b^2}{4 \, a^2} \hspace{5mm} \text{which gives} \quad p = \pm \frac{b}{2 a}$$
and the equation $ u^2 \, f^{''} + u \, f^{'} + u^2 \, f = 0$. This equation is the Bessel differential equation of zeroth order and provides the solution as
$$ f(u) = A \, J_{0}(u) + B \, Y_{0}(u).$$
Reverting one step yields
$$ f(t) = A \, J_{0}\left( \pm \frac{2 a}{b} \, t \right) + B \, Y_{0}\left( \pm \frac{2 a}{b} \, t \right). $$
Note that since $J_{n}(-x)$ is related to $J_{n}(x)$ for $n$ being an integer then choosing the $+$ sign leads to no other solutions to determine. Now reverting the last step then
$$ f(x) = A \, J_{0}\left( \frac{2 a}{b} \, e^{b x/2} \right) + B \, Y_{0}\left( \frac{2 a}{b} \, e^{b x/2} \right). $$
From this, the solutions to $y^{''} + a^2 \, e^{b x} \, y = 0$ are
$$ y(x) = \begin{cases} c_{0} + c_{1} \, x & a = 0 \\ c_{0} \, \cos(a x) + c_{1} \, \sin(a x) & b = 0 \\ A \, J_{0}\left( \frac{2 a}{b} \, e^{b x/2} \right) + B \, Y_{0}\left( \frac{2 a}{b} \, e^{b x/2} \right) & a \neq 0 \, \text{and} \, b \neq 0 \end{cases}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4582183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve $\cos^{40}x-\sin^{40}x=1$ for real and imaginary solutions Solve : $$\cos^{40}x-\sin^{40}x=1$$
I found by induction that $\sin(x)=0$ satisfy the equation so $$x=n\pi$$ must be the solution but there should be more solutions real or imaginary , how to find them ?
| Notice that, with $t = \cos x$ and $y = \sin x$, the equation factors into the form
\begin{align}
1 &= (t - y) (t + y) (t^2 + y^2) (t^4 + y^4) (t^4 - t^3 y + t^2 y^2 - t y^3 + y^4) \\
& \hspace{5mm} \cdot (t^4 + t^3 y + t^2 y^2 + t y^3 + y^4) (t^8 - t^6 y^2 + t^4 y^4 - t^2 y^6 + y^8) \\
& \hspace{5mm} \cdot (t^{16} - t^{12} y^4 + t^8 y^8 - t^4 y^{12} + y^{16}).
\end{align}
By using $t^2 + y^2 = 1$ then this reduces by a factor. There are seven remaining factors. Each polynomial has roots and finding these roots will help lead to the corresponding $x$ values. Otherwise it can be determined that $x = 2 \, n \, \pi$, $n = \pm \, \text{integer}$, and solutions will be of the form $x = 2 \, n \, \pi - \tan^{-1}(\text{some value})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4583530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $(x^2−1) \bmod 8$ $\in \{ 0 , 3 , 7 \}, \forall x \in \mathbb{Z}$. It must be verified that for all $x \in \mathbb{Z}$ it holds that $x^2 - 1 \bmod{8} \in \{0, 3, 7\}$. First some definitions. Using the following theorem a definition for $\bmod$ is provided:
Theorem. For all $a \in \mathbb{Z}$ and $d \in \mathbb{N}$ unique integers q, r exist satisfying $a = q \cdot d + r \wedge 0 \leq r <d$
The integers $q$, $r$ correspond to the quotient and remainder, respectively, of $a$, these being defined as:
$a = (a / d) \cdot d + a \bmod d \wedge 0 \leq a \bmod d < d$
With $q = (a/d)$ and $r = a \bmod d$. To verify the claim I use induction. I use a brute force approach where I search for all numbers $x$ such that $x^2 - 1 \bmod 8 = 0$, $x^2 - 1 \bmod 8 = 3$ or $x^2 - 1 \bmod 8 = 7$. First I search for all $x$ such that $x^2 - 1 \bmod 8 = 0$. If $x^2 - 1 \bmod 8 = 0$ then 8 is a divisor of $x^2 - 1$, i.e.: $8 \mid x^2 - 1$. Now I search for some other $x$ such that 8 is a divisor of $x^2 - 1$. Let $f(x) = x^2 - 1$ and $8 \mid f(x)$ then $8 \mid f(x + a)$ for some $a \in \mathbb{N}$. I compute this a:
$8 \mid f(x) = 8 \mid x^2 - 1 \implies 8 \mid (x + a)^2 - 1 = 8 \mid x^2 + a^2 + 2ax - 1 = 8 \mid x^2 - 1 + a^2 + 2ax$
The implication holds for e.g.: $a = 4$ since:
$8 \mid x^2 - 1 + a^2 + 2ax \implies 8 \mid x^2 - 1 + 16 + 8x = 8 \mid f(x) \wedge 8 \mid 16 + 8x = true$
So it can be concluded that if $8 \mid f(x)$ then $8 \mid f(x + 4)$. In fact it may be concluded that if $8 \mid f(x)$ then $8 \mid f(x + 4k), \forall x \in \mathbb{Z} \wedge k \in \mathbb{N}$, since:
$8 \mid (x + 4k)^2 - 1 = 8 \mid x^2 - 1 + 16k^2 + 8kx = 8 \mid f(x) \wedge 8 \mid 16k^2 + 8kx$
So it holds that $8 \mid f(x) \implies 8 \mid f(x + 4k)$. By inspection one finds that $8 \mid f(1) = 8 \mid 1^2 - 1 = 8 \mid 0 = true$ and so $f(x) \bmod 8 = 0$ for all $x \in \{1, 5, 9, 13, ...\}$. Since $f(x)$ is symmetric it holds that $f(-x) = f(x)$ and so it follows that $f(x) \bmod 8 = 0$ for all $x \in \{-1, -5, -9, -13, ...\}$. Upon further inspection one finds that $8 \mid f(3) = 8 \mid 3^2 - 1 = 8 \mid 8 = true$ and so it follows that $x^2 - 1 \bmod 8 = 0$ for all $x \in \{3, 7, 11, 15, ...\}$ and all $x \in \{-3, -7, -11, -15, ...\}$
Next I search for all x such that $x^2 - 1 \bmod 8 = 3$, or equivalently $x^2 - 4 \bmod 8 = 0$. Using the same approach I find that $x^2 - 1 \bmod 8 = 3$ for all $x \in \{2, 6, 10, 14, ...\}$ and all $x \in \{-2, -6, -10, -14, ...\}$. Finally $x^2 - 1 \bmod 8 = 7$ for all $x \in \{0, 4, 8, 12, ...\}$ and $x \in \{0, -4, -8, -12, ...\}$. And so one concludes that for all $x \in \mathbb{Z}$ it holds that $x^2 - 1 \bmod{8} \in \{0, 3, 7\}$.
This question comes from a course in discrete mathematics in computer science. I feel that my approach is overkill and that I'm doing something wrong and that there must be some cleverer way of solving the problem. If anyone can help with a better, cleaner, approach, or point out errors, it will be greatly appreciated :).
| $$x^2-1=(x-1)(x+1)$$
If $x$ is even, $x-1$ and $x+1$ are both odd. The possibilities are then $7\times1$, $1\times3$, $3\times5$ and $5\times7$, which gives you $3$ or $7$.
If $x$ is odd, $x-1$ and $x+1$ are both even, and one of them is divisible by $4$, so the remainder is always $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\operatorname{lcm}$ of [$\binom{n}{1}$, $\binom{n}{2}$, ... ,$\binom{n}{n}$] = $\operatorname{lcm}(1, 2, ...,n+1)/(n+1)$ How to prove that:
$$\operatorname{lcm}\left(\binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n}\right) = \frac{\operatorname{lcm}(1, 2, \ldots, n+1)}{n+1}$$
There is a hint:
$p$ is a prime and consider the highest power of $p$ in $\binom{n}{k}$, that is $p^e \:||\: \binom{n}{k}$, e is the highest power of p.
| Lemma: $[\frac{n}{p}]-[\frac{n-k}{p}]-[\frac{k}{p}]\le 1$
Note that $$v_p(\binom{n}{k})=\sum_{i=1}^{\infty} [\frac{n}{p^i}]-[\frac{n-k}{p^i}]-[\frac{k}{p^i}]$$
Suppose $p^e\le n+1 <p^{e+1}$ for some $p<n$, then $v_p(\operatorname{lcm}(1, ..., n+1))=e$.
We need to prove that
$$(n+1)v_p(\operatorname{lcm}(\binom{n}{1}, ..., \binom{n}{n}) = v_p((n+1)\operatorname{lcm}(\binom{n}{1}, ..., \binom{n}{n}) = e$$
Now $(n+1)\binom{n}{k} = (k+1)\binom{n+1}{k+1}$, so $$v_p((n+1)\operatorname{lcm}(\binom{n}{1}, ..., \binom{n}{n}) = \max\limits_{1\le k \le n}v_p((k+1)\binom{n+1}{k+1})=\max\limits_{1\le k\le n}(v_p(k+1)+v_p\binom{n+1}{k+1})$$
Let $v_p(k+1)=s$ then by the lemma $v_p(\binom{n+1}{k+1}) \le e-s$, therefore $v_p((k+1)\binom{n+1}{k+1}) \le e$ for all $k=1, 2, ..., n$.
Also if $k=p^e-1$, $v_p((k+1)\binom{n+1}{k+1}) \ge e$. So we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $2^\frac{2x-1}{x-1}+2^\frac{3x-2}{x-1}=24$, find all values of $x$ that satisfy this As title suggests, the problem is as follows:
Given that $$2^\frac{2x-1}{x-1}+2^\frac{3x-2}{x-1}=24$$ find all values of $x$ that satisfy this.
This question was shared in an Instagram post a few months ago that I came across today. Examining it at first, it seems there are many ways to solve this. I'll show my own approach here, please let me know if there are any issues in my solution and please share your own solution too!
Here's my approach for the problem:
Let $a=2^\frac{2x-1}{x-1}$ and $b=2^\frac{3x-2}{x-1}$
We then get $a+b=24$.
Now notice that:
$$(3x-2)-(2x-1)=x-1$$
That gives us a motivation to perform division with $a$ and $b$ (as the denominator and numerator of the exponent will be equal, hence reducing the exponent) thus:
$$\frac{b}{a}=\frac{2^\frac{3x-2}{x-1}}{2^\frac{2x-1}{x-1}}$$
$$\frac{b}{a}=2^\frac{x-1}{x-1}=2$$
$$b=2a$$
Therefore:
$$2a+a=24$$
$$3a=24$$
$$a=8$$
$$2^\frac{2x-1}{x-1}=8$$
$$2^\frac{2x-1}{x-1}=2^3$$
$$\frac{2x-1}{x-1}=3$$
$$2x-1=3x-3$$
Thus, $x=2$
| First of all, your solution looks great.
Here's a slightly different way to discover that $x=2$. Noticing the $x-1$ denominators in the exponents, I would start by making the substitution $u = x - 1$, so $x = u + 1$. Thus,
$$
\frac{2x-1}{x-1} = \frac{2u+1}{u} = 2 + \frac{1}{u}
$$
and
$$
\frac{3x-2}{x-1} = \frac{3u+1}{u} = 3 + \frac{1}{u}.
$$
Now the left-hand side of the equation becomes
$$
2^{2 + \frac{1}{u}} + 2^{3 + \frac{1}{u}}
= 2^{2 + \frac{1}{u}} \bigl( 1 + 2 \bigr)
= 3 \cdot 2^{2 + \frac{1}{u}},
$$
so the equation reduces to
$$
2^{2 + \frac{1}{u}} = 8.
$$
Since the exponential function $t \mapsto 2^t$ is one-to-one on the real numbers$^\dagger$ and $8 = 2^3$, it follows that
$$
2 + \frac{1}{u} = 3,
$$
hence
$$
\frac{1}{u} = 1.
$$
From there, it's clear that $u = 1$, so $x = 2$ is the only solution (which you can verify by plugging in!)
Note: the detour into the variable $u$ is not necessary, but it makes the relationship between the two exponents more obvious.
Challenge: Looking at the graph, it appears that the expression on the left-hand side of the equation is always decreasing on its domain. If you know differential calculus, prove this! This is another way to show that your solution is unique. Also, what is the domain? What is the range?
$^\dagger$This is no longer true if we are solving the equation in the complex numbers, where there will be an infinite family of solutions, stemming from the fact that $\exp(z + 2\pi k i) = \exp z$ for all integers $k$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why i calculate this limit like this? Calculate $ \lim_{x \to ∞} f(x) = ($$\frac{x^2-2x+7}{x^2+3x+1})^\frac{1}{\sin(1/x)}$
The solution involves using $ \lim_{t \to 0} (1+t)^\frac{1}{t}=e $
But can i say that if $ \lim_{x \to ∞} ($$\frac{x^2-2x+7}{x^2+3x+1})$ = 1
and $ \lim_{x \to ∞}\frac{1}{\sin(1/x)} = 1 $
then $ \lim_{x \to ∞} f(x) = ($$\frac{x^2-2x+7}{x^2+3x+1})^\frac{1}{\sin(1/x)} = 1$?
I guess it's wrong but I don't really know why
| Assuming the limit exists, let
$\displaystyle L:= \lim_{x \to +\infty} \left(\frac{x^2-2x+7}{x^2+3x+1}\right)^\frac{1}{\sin(1/x)}$
Then we have:
$$\log L=\lim_{x \to +\infty} \frac{1}{\sin(1/x)} \log \left(\frac{x^2-2x+7}{x^2+3x+1}\right)$$
Since
$$\sin \left(\frac{1}{x}\right) \sim \frac{1}{x} \quad \text{as}\quad x\to+\infty\\
\log \left(\frac{x^2-2x+7}{x^2+3x+1}\right)=\log \left(1+\frac{-5x+6}{x^2+3x+1}\right)\sim \frac{-5x+6}{x^2+3x+1}
$$
It follows that:
$$\frac{1}{\sin(1/x)} \log \left(\frac{x^2-2x+7}{x^2+3x+1}\right)\sim \frac{6x-5x^2}{x^2+3x+1} \to -5 \quad \text{as}\quad x\to+\infty$$
Finally $$\boxed{L=\text{e}^{-5}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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I got a different value with wolfram of evaluating $~\int_{0}^{2\pi}\left|\sin\theta+\cos\theta\right|\mathrm d \theta$ $$\begin{align}
A&:=\int_{0}^{2\pi}\left|\sin\theta+\cos\theta\right|\mathrm{d}\theta\\
\end{align}$$
$$\begin{align}
\sin\theta+\cos\theta&<0\leftrightsquigarrow\theta\neq{\pi\over2},{3\pi\over 2}\\
\cos\theta&<-\sin\theta\\
1&<-\tan\theta\\\iff-1&>\tan\theta
\\\equiv\tan\theta&<-1\\
f(\theta)&:=\sin\theta+\cos\theta
\\
A&=\int_{0}^{{\pi\over 2}}f(\theta)\mathrm d\theta-\int_{{\pi\over 2}}^{{\pi\over 2}+{\pi\over 4}}f(\theta)\mathrm d\theta+\int_{{\pi\over 2}+{\pi\over 4}}^{3\pi\over 2}f(\theta)\mathrm d\theta\\~~~&-\int_{{3\pi\over 2}}^{{3\pi\over2}+{\pi\over 4}}f(\theta)\mathrm d\theta+\int_{{3\pi\over 2}+{\pi\over 4}}^{2\pi}f(\theta)\mathrm d\theta
\end{align}$$
And I got $~ A=0 ~$ but wolfram shows me the value of $~ A ~$ should be $~ 4\sqrt{2} ~$
Where I've made (a) mistake(s)?
| I would like to evaluate the integral from definition of absolute value by splitting the integration interval into 3 intervals.
$$
\begin{aligned}
\int_0^{2 \pi}|\sin \theta+\cos \theta| d \theta
\stackrel{\theta\mapsto\theta-\frac{\pi}{4}}{=}& \sqrt{2} \int_0^{2 \pi}\left|\cos \left(\theta-\frac{\pi}{4}\right)\right| d \theta \\
= &\sqrt{2} \int_{-\frac{\pi}{4}}^{\frac{7 \pi}{4}}|\cos \theta| d \theta \\
= & \sqrt{2}\left(\int_{-\frac{\pi}{4}}^{\frac{\pi}{2}}|\cos \theta| d \theta+\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}|\cos \theta| d \theta+\int_{\frac{2 \pi}{2}}^{\frac{7 \pi}{4}} \cos \theta d \theta\right) \\
= & \sqrt{2}\left(\int_{-\frac{\pi}{4}}^{\frac{\pi}{2}} \cos \theta d \theta-\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \cos \theta d \theta+\int_{\frac{3 \pi}{2}}^{\frac{\pi \pi}{4}} \cos \theta d \theta\right) \\
= & \sqrt{2}\left([\sin \theta]_{-\frac{\pi}{4}}^{\frac{\pi}{2}}-[\sin \theta]_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}+[\sin \theta]_{\frac{3 \pi}{2}}^{\frac{7\pi}{4}}\right) \\
= & \sqrt{2}\left(1+\frac{1}{\sqrt{2}}+2-\frac{1}{\sqrt{2}}+1\right) \\
= & \ 4 \sqrt{2}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find all three complex solutions of the equation $z^3=-10+5i$ Let $z\in \mathbb{C}$. I want to calculate the three solutions of the equation $z^3=-10+5i$. Give the result in cartesian and in exponential representation.
Let $z=x+yi $.
Then we have $$z^2=(x+yi)^2 =x^2+2xyi-y^2=(x^2-y^2)+2xyi$$
And then $$z^3=z^2\cdot z=[(x^2-y^2)+2xyi]\cdot [x+yi ] =(x^3-xy^2)+2x^2yi+(x^2y-y^3)i-2xy^2=(x^3-3xy^2)+(3x^2y-y^3)i$$
So we get
$$z^3=-10+5i \Rightarrow (x^3-3xy^2)+(3x^2y-y^3)i=-10+5i \\ \begin{cases}x^3-3xy^2=-10 \\ 3x^2y-y^3=5\end{cases} \Rightarrow \begin{cases}x(x^2-3y^2)=-10 \\ y(3x^2-y^2)=5\end{cases}$$
Is everything correct so far? How can we calculate $x$ and $y$ ? Or should we do that in an other way?
| Use a primitive third root of unity, say $\zeta_3=e^{\frac {2\pi i}3}.$
Take $\omega $ with $w^3=-10+5i$, say $\omega =-\sqrt {125}^{\frac 13}e^{\frac{i\arctan -\frac 12}3}=-\sqrt 5e^{\frac{i\arctan -\frac 12}3}.$
Then the roots are $\{\omega, \zeta_3 \omega, \zeta_3 ^2\omega \}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Draw an area on the complex plane On the complex plane draw the area:
$$
\begin{equation}
\begin{cases}
|z+4i| < 3 \\
|\arg(z-5-5i)|<\frac{\pi}{3}
\end{cases}
\end{equation}
$$
Where $ \arg(z) \in (-\pi, \pi ]$
I can draw $|z+4i| < 3$:
$|x + iy + 4i|<3 \Rightarrow \sqrt{x^2 + (y + 4)^2}<3 \Rightarrow x^2 + (y + 4)^2<9$:
However, I have no idea how to draw and intersect with $|arg(z-5-5i)|<\frac{\pi}{3}$
| You must intersect the circle with:
$$|arg(z-5-5i)|<\frac{\pi}{3}$$
which means:
$$\left| \arctan \left(\frac{\Im(z - 5 - 5i)}{\Re(z - 5 - 5i)} \right) \right | < \frac{\pi}{3}$$
$$\Longleftrightarrow \Re \left\{\arctan \left(\frac{\Im(z) - 5}{\Re(z) - 5} \right)^2 \right\} + \Im \left\{\arctan \left(\frac{\Im(z) - 5}{\Re(z) - 5} \right) ^2\right\} < \frac{\pi^2}{9}$$
You know the argument is an angle in the complex plane. It is thus always real:
$$\arctan \left(\frac{\Im(z) - 5}{\Re(z) - 5} \right) < \pm \frac{\pi}{3}$$
$$\Longrightarrow
\frac{\Im(z) - 5}{\Re(z) - 5} < \pm \tan\frac{\pi}{3} = \pm \sqrt 3$$
$$\Longleftrightarrow \Im (z) < \pm \sqrt 3(\Re (z)-5) + 5$$
You can now draw in the complex plane:
$$y = \pm \sqrt 3 (x - 5) + 5$$
Finally, considering $<$, you obtain your domain:
There is thus no connection at all with your circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4602374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Does $\sum\limits_{n = 1}^{\infty} \frac{3^n + 4^n}{2^n + 5^n}$ converge? My Attempt
First, check the limit
$$\lim_{n \to \infty} \frac{3^n + 4^n}{2^n + 5^n} =
\lim_{n \to \infty} \frac{\left(\frac{3}{5}\right)^n + \left(\frac{4}{5}\right)^n}{\left(\frac{2}{5}\right)^n + 1} = 0.$$
So, we cannot conclude anything.
I used Comparison Test and Ratio Test.
Consider that
$$\frac{3^n + 4^n}{2^n + 5^n} < 2\frac{5^n}{2^n + 5^n}.$$
If I can proof the convergence of the series on the right-side, then it's done by comparison test.
I used ratio test, in order to proof the convergence of
$$\sum_{n= 1}^{\infty} \frac{5^n}{2^n + 5^n}.$$
$$\lim_{n \to \infty} \frac{55^n}{2(2^n) + 5(5^n)} \frac{2^n + 5^n}{5^n} = 5 \lim_{n \to \infty} \frac{2^n + 5^n}{2(2^n) + 5(5^n)} = 1.$$
I didn't find the way to proof.
Any suggestion? Thanks in advanced.
Solution (@abiessu & @Thomas Andrew)
Consider that
$$\frac{3^n + 4^n}{2^n + 5^n} < 2\frac{4^n}{2^n + 5^n}.$$
Proof this series converge.
$$\sum_{n= 1}^{\infty} \frac{4^n}{2^n + 5^n}.$$
Proof (Ratio Test)
$$\lim_{n \to \infty} \frac{44^n}{2(2^n) + 5(5^n)} \frac{2^n + 5^n}{4^n} = 4 \lim_{n \to \infty} \frac{2^n + 5^n}{2(2^n) + 5(5^n)} = \frac{4}{5}.$$
The series converge.
Hence
$$\sum_{n = 1}^{\infty} \frac{3^n + 4^n}{2^n + 5^n}$$
converge.
| $$\left(\frac{2}{5}\right)^n + 1 > 1 $$
Hence
$$\frac{\left(\frac{3}{5}\right)^n + \left(\frac{4}{5}\right)^n}{\left(\frac{2}{5}\right)^n + 1} < \left(\frac{3}{5}\right)^n + \left(\frac{4}{5}\right)^n $$
And both $\left(\frac{3}{5}\right)^n$ and $\left(\frac{4}{5}\right)^n$ are general terms of a convergent geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Sum of two subspaces: representing it with equations I found the following excercise:
Let $W_1 = \{(x_1, ..., x_6) : x_1 + x_2 + x_3 = 0, x_4 + x_5 + x_6 = 0 \}$. Let $W_2$ be the span of $S := \{(1, -1, 1, -1, 1, -1), (1, 0, 2, 1, 0, 0), (1, 0, -1, -1, 0, 1), (2, 1, 0, 0, 0, 0)\}$.
Give a base, a dimension and an equation representation of $W_1 + W_2$
I'm new to the concept of sum of subspaces. But as I understand it, the first step would be to note that any $\textbf{x} = (x_1, ..., x_6) \in W_1$ satisfies
\begin{equation*}
\begin{cases}
x_1 = -x_2 - x_3 \\
x_4 = -x_5 - x_6
\end{cases}
\end{equation*}
so that its general form is
\begin{equation*} \textbf{x} =
\begin{pmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5 \\
x_6
\end{pmatrix} = \begin{pmatrix}
-x_2 - x_3 \\
x_2 \\
x_3 \\
-x_5 - x_6 \\
x_5 \\
x_6
\end{pmatrix}
\end{equation*}
We also know any $\textbf{y} \in W_1$ is of the general form
\begin{align*} \textbf{y} =
\begin{pmatrix}
y_1 \\
y_2 \\
y_3 \\
y_4 \\
y_5 \\
y_6
\end{pmatrix} = \begin{pmatrix}x + y +z + 2w \\
-x + w \\
x + 2y - z \\
-x + y - z \\
x\\
-x + z\end{pmatrix}
\end{align*}
Then for generals $\mathbf{x}, \mathbf{y}$ we have
\begin{align*}
\textbf{x} + \textbf{y} &= \begin{pmatrix}
x + y +z + 2w + (-x_2 - x_3)\\
-x + w + x_2\\
x + 2y - z + x_3\\
-x + y - z + (-x_5 - x_6)\\
x + x_5\\
-x + z + x_6
\end{pmatrix}
\end{align*}
One can then conclude
$$W_1 + W_2 = \Big\{\big(x + y + z + 2w - x_2 - x_3\big), \big(-x + w + x_2 \big), \big(x +2y - z + x_3 \big), \big(-x + y - z - x_5 - x_6 \big), \big(x + x_5 \big), \big(-x + z + x_6 \big) \mid x, y, z, x_2, x_3 \in \mathbb{R} \Big\}$$
But what would be an representation via equations of this system? I'm very new to linear algebra so go easy on me!
| You already found that a vector $(a,b,c,d,e,f)$ belongs to $W_1+W_2$ if and only if there exist real numbers $x_2,x_3,x_5,x_6,x,y,z,w$ such that
$$\begin{cases}a&=x+y+z+2w-x_2-x_3\\b&=-x+w-x_2\\c&=x + 2y - z + x_3\\
d&=-x + y - z-x_5 - x_6\\
e&=x + x_5\\
f&=-x + z + x_6.
\end{cases}$$
Using for instance the 2nd, 3rd, 5th, and 6th equations to first eliminate the $x_i$'s, this is equivalent to the existence of real numbers $x,y,z,w$ such that
$$\begin{cases}a&=3x+3y+w+b-c\\
d&=-x + y-e -f
\end{cases}$$
For any $(a,b,c,d,e,f)\in\Bbb R^6,$ such numbers $x,y,z,w$ always exist (e.g. choose $x,z$ arbitrarily and take $y=d+e+f+x,$ and $w=a-b+c-3x-3(d+e+f+x)$) hence
$$W_1+W_2=\Bbb R^6.$$
You could see it more rapidly since you already knew that $\dim(W_1\cap W_2)=2$: $\dim(W_1)$ is $6-2=4$ (since its 2 equations are clearly independent), $\dim(W_2)=4$ (since its 4 generating vectors are - less clearly - independent) hence by Grassmann's formula,
$$\dim(W_1+W_2)=4+4-2=6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$3\times3$ determinant using standard basis I am trying to get from a $2\times2$ determinant to a $3\times3$ determinant.
$$\left|\begin{array}{c1 c2 c3}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|
$$
How does one get to
$$ \det(A)=\sum_{j=1}^3 a_{j1} \; \det(e_j, a_2, a_3) $$
and then end up with
$$ \det(A)=\sum_{i,j,k=1}^3 a_{i1}a_{j2}a_{k3} \; \det(e_i, e_j, e_k) $$
| Consider the given determinant as $\begin{vmatrix}\mathbf{a}_1 & \mathbf{a}_2 &\mathbf{a}_3 \end{vmatrix}$, where $\mathbf{a}_j$ is the $j-$th column.
By multi-linearity of the determinants (in particular keeping the second and third columns the same), we get
$$\begin{vmatrix}\color{red}{\mathbf{u} +\lambda\mathbf{b}} & \mathbf{v} &\mathbf{w} \end{vmatrix}=\begin{vmatrix}\color{red}{\mathbf{u}} & \mathbf{v} &\mathbf{w} \end{vmatrix}+\begin{vmatrix}\color{red}{\lambda\mathbf{b}} & \mathbf{v} &\mathbf{w} \end{vmatrix}=\begin{vmatrix}\color{red}{\mathbf{u}} & \mathbf{v} &\mathbf{w} \end{vmatrix}+\color{red}{\lambda}\begin{vmatrix}\color{red}{\mathbf{b}} & \mathbf{v} &\mathbf{w} \end{vmatrix}$$
Observe that
$$\mathbf{a}_1=a_{11}\mathbf{e}_1+a_{21}\mathbf{e}_2+a_{31}\mathbf{e}_3=
\sum_{j=1}^3a_{j1}\mathbf{e}_j$$
Using multi-linearity of the determinants, we get
\begin{align*}
\begin{vmatrix}\mathbf{a}_1 & \mathbf{a}_2 &\mathbf{a}_3 \end{vmatrix}&=\begin{vmatrix}\sum_{j=1}^3a_{j1}\mathbf{e}_j & \mathbf{a}_2 &\mathbf{a}_3 \end{vmatrix}\\
&=\sum_{j=1}^3\begin{vmatrix}a_{j1}\mathbf{e}_j & \mathbf{a}_2 &\mathbf{a}_3 \end{vmatrix}\\
&=\sum_{j=1}^3a_{j1}\begin{vmatrix}\mathbf{e}_j & \mathbf{a}_2 &\mathbf{a}_3 \end{vmatrix}.
\end{align*}
Hope you can now proceed from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Recurrence Relation in Reduction formula for Integral $\sin^n(x)$ For the integral $\displaystyle \int\sin^n(x) dx$ there exists the following reduction formula, that is a recurrence relation:
$\displaystyle I_n = \frac{n-1}{n} \cdot I_{n-2}-\frac{\sin^{n-1}(x) \cdot\cos(x)}{n}$
I have now been trying to solve this recurrence relation and was able to find a solution for the homogeneous problem:
$I_n = \frac{n-1}{n} \cdot I_{n-2}$
I arrived at the following:
$\displaystyle I_n=\frac{\displaystyle \prod_{i=1}^{\frac{n}{2}} (2i-1)}{\displaystyle \prod_{i=1}^{\frac{n}{2}} 2i} \cdot C_1$, if $n$ is even
and
$\displaystyle I_n=\frac{\displaystyle \prod_{i=1}^{\frac{n-1}{2}} 2i}{\displaystyle \prod_{i=1}^{\frac{n-1}{2}} (2i+1)} \cdot C_2$, if $n$ is odd
which I then simplified further using some identities for products:
$\displaystyle I_n=\frac{n!}{2^n \cdot \left( \left(\displaystyle \frac{n}{2} \right)! \right)^2} \cdot C_1$, if $n$ is even
and
$\displaystyle I_n=\frac{2^{n-1} \cdot \left( \left(\displaystyle \frac{n-1}{2} \right)! \right)^2}{n!} \cdot C_2$, if $n$ is odd
which combined results in:
$I_n=\displaystyle \frac{(-1)^n+1}{2} \cdot \displaystyle \frac{n!}{2^n \cdot \left( \left(\displaystyle \frac{n}{2} \right)! \right)^2} \cdot C_1 + \displaystyle \frac{(-1)^{n+1}+1}{2} \cdot \frac{2^{n-1} \cdot \left( \left(\displaystyle \frac{n-1}{2} \right)! \right)^2}{n!} \cdot C_2$
But now I'm struggling with finding a particular solution to the non-homogeneous problem:
$\displaystyle I_n = \frac{n-1}{n} \cdot I_{n-2}-\frac{\sin^{n-1}(x) \cdot\cos(x)}{n}$
| After separating even and odd terms of the sequence, i.e.
$$
\begin{cases}
\displaystyle
a_n := I_{2n} = \frac{2n-1}{2n}I_{2n-2} - \frac{\sin^{2n-1}(x)\cos(x)}{2n} = \frac{2n-1}{2n}a_{n-1} - \frac{\sin^{2n-1}(x)\cos(x)}{2n} \\
\displaystyle
b_n := I_{2n+1} = \frac{2n}{2n+1}I_{2n-1} - \frac{\sin^{2n}(x)\cos(x)}{2n+1} \,= \frac{2n}{2n+1}b_{n-1} - \frac{\sin^{2n}(x)\cos(x)}{2n+1}
\end{cases}
$$
you get two first-order inhomogeneous linear recurrence relations with non-constant coefficients, which are always solvable. Each can be rewritten in the following manner :
$$
c_n = \alpha_nc_{n-1} + \beta_n
$$
with $\alpha_n = \frac{2n-1}{2n}$ and $\beta_n = - \frac{\sin^{2n-1}(x)\cos(x)}{2n}$ for the first sequence for instance. It is possible to get rid of the non-constant coefficient $\alpha_n$ by dividing by ${\prod_{k=1}^n\alpha_k}$, such that
$$
d_n = d_{n-1} + \gamma_n \verb+ +\mathrm{where}\verb+ + d_n := \frac{c_n}{\prod_{k=1}^n\alpha_k} \verb+ +\mathrm{and}\verb+ + \gamma_n := \frac{\beta_n}{\prod_{k=1}^n\alpha_k}
$$
Then, $d_n$ is found with the help of a telescoping series :
$$
d_n-d_0 = \sum_{k=1}^n (d_k-d_{k-1}) = \sum_{k=1}^n \gamma_k
$$
hence finally, after switching back to the initial parametrization,
$$
c_n = \prod_{k=1}^n\alpha_k \cdot \left(A + \sum_{k=1}^n \frac{\beta_k}{\prod_{j=1}^k\alpha_j}\right)
$$
where $A = c_0$ is a constant to be found thanks to the initial condition of the considered sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How do we prove $x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0$? Question
How do we prove the following for all $x \in \mathbb{R}$ :
$$x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0 $$
My Progress
We can factorise the left hand side of the desired inequality as follows:
$$x^6+x^5+4x^4-12x^3+4x^2+x+1=(x-1)^2(x^4+3x^3+9x^2+3x+1)$$
However, after this I was unable to make any further progress in deducing the desired inequality.
I appreciate your help
| $$
x^4+3x^3+9x^2+3x+1 = x^2\left(x+\frac 32\right)^2 + \frac{27}{4}\left(x+\frac 2 9\right)^2 + \frac 23
$$
is strictly positive for all real $x$.
How did I come up with that? I started by completing the square in
$$
x^4+3x^3+9x^2 = x^2(x^2+3x+9) = x^2\left( (x+\frac 32)^2 + \frac{27}{4}\right)
$$
so that
$$
x^4+3x^3+9x^2 +3x+1= x^2\left( x+\frac 32 \right)^2 + \frac{27}{4}\left( x^2 +\frac 4 9 x + \frac{4}{27}\right)
$$
and then completed the square in $x^2 +\frac 4 9 x + \frac{4}{27}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
} |
If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$, show that $a\sqrt{bc} + b\sqrt{ac} + c\sqrt{ab} \leq abc$. If $a$, $b$, $c$ are nonzero natural numbers and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$, show that $a\sqrt{bc} + b\sqrt{ac} + c\sqrt{ab} \leq abc$.
I basically have no idea how I can start this problem.
I thought of demonstrating that $a\sqrt{bc}\leq abc$ but I just got lost in all the equations.
Also, I thought to give a common factor but I don't really find a good point of starting the problem with.
If you have any other idea, please write in the comments down below. Any idea is welcome!
Hope one of you can help me! Thank you so much!
| Multiply $abc$ in $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ to get:$$bc+ac+ab=abc$$Now, by AM-GM: $$a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\le\frac{1}{2}(a(b+c)+b(a+c)+c(a+b)=\frac{1}{2}(ab+ac+ab+bc+ac+bc)=ab+bc+ac=abc$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
determine the product of all possible values of $|a+b+c|$
Let $a,b,c$ be complex numbers such that $|a|=|b|=|c|=1$. If $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}=1$ as well, then find the product of all possible values of $|a+b+c|$.
Suppose we know that $s^3 = abc(3|s|^2-2),$ where $s = a+b+c.$ Then $|s|$ is a nonnegative root of $x^3 \pm (-3x^2+2)$. For $x^3-3x^2+2=(x-1)(x^2-2x-2),$ the positive roots are $1$ and $1+\sqrt{3}.$ Note that $x^3+3x^2-2 = (x+1)(x^2+2x-2)$ and so it has the single positive real root $-1+\sqrt{3}.$ So the desired product is indeed 2, provided all of the above values are obtainable.
Question: why are all the above values obtainable? I'd prefer an answer that at least provides some motivation as to how to show that both $\pm 1+\sqrt{3}$ are possible values for s.
To show that $s^3 = abc(3|s|^2-2)$, we have the following series of equalities:
$\begin{align}
s^3 &= (a+b+c)^3\\
&= (a+b+c)(a^2 + b^2 + c^2 + 2(ab+ac+bc))\\
&= (a+b+c)(a^2 +b^2 + c^2 - (ab+ac+bc)) + 3(a+b+c)(ab+ac+bc)\\
&= a^3 + b^3 + c^3 - 3abc + 3(a^2 b + ab^2 + abc + a^2 c + abc + ac^2 + abc+b^2 c + bc^2)\\
&= abc-3abc +3abc(a+b+c)(1/a+1/b+1/c)\\
&= abc(-2 + 3|s|^2),
\end{align}$
where the last step follows from the fact that $\overline{s} = 1/a+1/b+1/c$, which in turns follows from the fact that $|a|=|b|=|c|=1$.
| As a different way which shows why the values are obtainable:
Let's assume $a=e^{i \theta_1}, b=e^{i \theta_2}$, and $c=e^{i \theta_3}$. Then, because of $\frac{a^2}{bc}+ \frac{b^2}{ac}+\frac{c^2}{ab}=1$, we must have:
$$\cos (2\theta_1-\theta_2-\theta_3)+\cos (2\theta_2-\theta_1-\theta_3)+\cos (2\theta_3-\theta_2-\theta_1)=1 \\
\sin (2\theta_1-\theta_2-\theta_3)+\sin (2\theta_2-\theta_1-\theta_3)+\sin (2\theta_3-\theta_2-\theta_1)=0.$$
Now, set $x=2\theta_1-\theta_2-\theta_3$, $y=2\theta_2-\theta_1-\theta_3$, and $z=2\theta_3-\theta_2-\theta_1$, and WLOG, we may assume $0\leq y\leq x\leq 2\pi.$
Note that $x+y=-z$. Therefore:
$$\sin x +\sin y=\sin (x+y)\\ \cos x+ \cos y=1- \cos (x+y)\\ \implies \\ 2\sin x \sin y+\sin^2 x+\sin^2y=\sin^2(x+y) \\ 2\cos x \cos y+\cos^2 x+\cos^2y=\ 1+cos^2(x+y)-2 \cos (x+y) \\ \implies \\ 2+2\cos (x-y)=2-2\cos (x+y) \\ \implies \\ \cos (x-y)=- \cos (x+y).$$
Hence, Two cases occure.
Case $1$: $\cos (x-y)= \cos (x+y)=0$.
In this case, regarding $0\leq y\leq x\leq 2\pi$, we conclude that $(x,y)=(\frac{\pi}{2}, 0), (2\pi, \frac{\pi}{2}),$ or $(\frac{3\pi}{2}, 0)$; note that not every solution to $\cos (x-y)=- \cos (x+y)$ is acceptable because the major equations are $\sin x +\sin y=\sin (x+y)$ and $\cos x+ \cos y=1- \cos (x+y)$.
Now, if $(x,y)=(\frac{\pi}{2}, 0)$, then $x-y=3(\theta_1-\theta_2)=\frac{\pi}{2}\implies \theta_1-\theta_2 =\frac{\pi}{6}.$
Moreover,
$$|a+b+c|^2=(\sin \theta_1+\sin \theta_2+\sin \theta_3)^2+(\cos \theta_1+\cos \theta_2+\cos \theta_3)^2\\
=3+2(cos (\theta_1-\theta_2)+cos (\theta_3-\theta_2)+cos (\theta_1-\theta_3))\\=3+2\cos(\theta_1-\theta_2)+4\cos(\frac{2\theta_3-\theta_2-\theta_1}{2}) \cos(\frac{\theta_1-\theta_2}{2});$$
but $\theta_1-\theta_2=\frac{\pi}{6}$, and $\frac{z}{2}=\frac{2\theta_3-\theta_2-\theta_1}{2}=\frac{-\pi}{4}$. A simple calculation yields $|a+b+c|^2=4+2\sqrt 3.$
Similarly, if $(x,y)=(2\pi,\frac{\pi}{2}),$ it is easy to check that $|a+b+c|^2=1.$
Similarly, if $(x,y)=(\frac{3\pi}{2},0),$ it is easy to check that $|a+b+c|^2=1.$
Case $2$: $x+y=\pi+x-y.$
In this case $y=\frac{\pi}{2}$, and it is not hard to show that $(x,y)=(\frac{3\pi}{2}, \frac{\pi}{2})$ (or $(2\pi, \frac{\pi}{2})$, which is a repeated case).
Similarly, if $(x,y)=(\frac{3\pi}{2}, \frac{\pi}{2})$, then $\theta_1-\theta_2=\frac{\pi}{3}$ and $\frac{z}{2}=\frac{2\theta_3-\theta_2-\theta_1}{2}=-\pi$. So, $|a+b+c|^2=4-2\sqrt 3.$
Notice that $\pi -(x-y)=x+y$ is possible which leads to a repeated case.
Finally, observe that $x+y=(2k+1)\pi+ x-y$ where $k=1,2,...$ doesn't give valid values for $x$ and $y$ because of the assumption $0\leq y\leq x\leq 2\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2$ Solve the equation $$\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2$$
We have $$D_x:\begin{cases}1-2x>0\\6x^2-5x+1>0\\1-3x>0\\1-3x\ne1\\4x^2-4x+1>0\iff(2x-1)^2>0\iff x\ne\dfrac12\end{cases}\iff x\in(-\infty;0)\cup(0;\dfrac{1}{3})$$
Also the quadratic $6x^2-5x+1$ factors as $(2x-1)(3x-1)$. The equation then becomes $$\log_{1-2x}(2x-1)(3x-1)-\log_{1-3x}(2x-1)^2=2\\\log_{1-2x}(2x-1)(3x-1)-2\log_{1-3x}(1-2x)=2,$$ as $\log_{1-3x}(2x-1)^2=2\log_{1-3x}|2x-1|,$ but we know from $D_x$ that $2x-1<0$,
$$\log_{1-2x}(2x-1)+\log_{1-2x}(3x-1)-\dfrac{2}{\log_{1-2x}(1-3x)}=2$$ I don't know what to do next.
| The key here is to use the change of base formula for logs: $\log_c(x)=\frac{\log_b(x)}{\log_b(c)}$. Using this allows us to express everything in terms of a common base, which then lets us use other log rules.
$$\frac{\log (1-2x)(1-3x)}{\log (1-2x)}-\frac{\log[(1-2x)^2]}{\log(1-3x)}=2.$$
Using log properties, this simplifies to
$$1+\frac{\log (1-3x)}{\log (1-2x)}-2\frac{\log[(1-2x)]}{\log(1-3x)}=2.$$
For ease of notation, let us write $u=\log(1-2x), v=\log(1-3x)$. Then $\frac{v}{u}-2\frac{u}{v}=1$. The easiest approach is to notice that we reciprocals, so that if $t=u/v$, then $1/t - 2t=1$. Multiplying through by $t$ yields a quadratic to solve for $t$, and then for $x$. However, if we do not make that realization, we may proceed as follows.
Multiplying by $uv$ to clear denominators, and moving everything to one side, we get
$$v^2-uv-2u^2=0$$
This factors as $$(v+u)(v-2u)=0$$
and so either $v+u=0$, or $v-2u=0$, i.e., either $\log(1-2x)(1-3x)=0$ so $(1-2x)(1-3x)=1$, or $\log\left(\frac{1-3x}{(1-2x)^2}\right)=0$, so $\frac{1-3x}{(1-2x)^2}=1$. This gives a linear and a quadratic equation to try to solve for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4615148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Simplifying $\frac{2x^3-9x^2+27}{3x^3-81x+162} $ Simplify $$\frac{2x^3-9x^2+27}{3x^3-81x+162} $$
All I can see is thus far is the factor 3 in the denominator can be taken out. Then I am stuck because I don't recognize any of the usual patterns in simplification.
The answer is $\;\dfrac{2x+3}{3(x+6)}\;,\;$ so clearly I am missing something here. Thanks.
| By the rational roots theorem, the rational roots of the numerator are $-\frac32$ and $3$ and, if you divide the numerator by $\left(x+\frac32\right)(x-3)$, you will get $2x-6$. Therefore, the numerator is equal to $(2x+3)(x-3)^2$. On the other hand, the rational roots of the denominator are $-6$ and $3$; in fact, the denominator is equal to $3(x+6)(x-3)^2$. Therefore, the quotient is indeed $\frac{2x+3}{3(x+6)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4617155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Showing $x_n-x_nx_1+\sum_{k=1}^{n-1} (x_k-x_kx_{k+1})\leq\left\lfloor\frac{n}{2}\right\rfloor$, with $x_i\in[0,1]$
Let $x_1, x_2,\ldots, x_n$ be arbitrary numbers from the interval $[0,1]$ with $n>1$.
Show that $$x_n-x_nx_1+\sum_{k=1}^{n-1} (x_k-x_kx_{k+1})\leq\left\lfloor\frac{n}{2}\right\rfloor$$
I tried to factor out the $x_k$ from each term to show that if the coefficient $x_k$ of $x_k(1-x_{k+1})$ is larger than $\dfrac{1}{2}$, then the term $x_{k-1}(1-x_k)$ must be smaller than $\dfrac12$, but I don't know where to go from here or if it is even the right approach.
| When $n$ is even, we have
\begin{align*}
x_1(1-x_2) + x_2(1-x_3)
= 1 - (1 - x_1)(1 - x_2) - x_2 x_3 &\le 1,\\
x_3(1-x_4) + x_4(1 - x_5) = 1 - (1-x_3)(1-x_4) - x_4x_5 &\le 1,\\
\cdots \cdots \cdots \cdots &\qquad\\
x_{n-1}(1 - x_n) + x_n(1 - x_1)
= 1 - (1 - x_{n-1})(1 - x_n) - x_nx_1 &\le 1.
\end{align*}
Adding them up, the desired result follows.
When $n$ is odd, we split into two cases:
If $1 - x_1 - x_{n-1} \ge 0$, then
\begin{align*}
x_2(1 - x_3) + x_3(1-x_4) = 1 - (1-x_2)(1-x_3) - x_3x_4 &\le 1, \\
x_4(1-x_5) + x_5(1-x_6) = 1 - (1-x_4)(1-x_5) - x_5x_6 &\le 1,\\
\cdots \cdots \cdots \cdots &\qquad\\
x_{n-3}(1-x_{n-2}) + x_{n-2}(1-x_{n-1}) = 1 - (1-x_{n-3})(1-x_{n-2}) - x_{n-2}x_{n-1} &\le 1,
\end{align*}
and
\begin{align*}
&x_{n-1}(1 - x_n) + x_n(1-x_1) + x_1(1-x_2)\\
={}& (1 - x_1 - x_{n-1})x_n + x_{n-1} + x_1(1-x_2)\\
\le{}& (1-x_1-x_{n-1}) + x_{n-1} + x_1(1-x_2)\\
={}&1 - x_1x_2\\
\le{}& 1.
\end{align*}
Adding them up, the desired result follows.
If $1 - x_1 - x_{n-1} < 0$, then
\begin{align*}
x_1(1-x_2) + x_2(1-x_3)
= 1 - (1 - x_1)(1 - x_2) - x_2 x_3 &\le 1,\\
x_3(1-x_4) + x_4(1 - x_5) = 1 - (1-x_3)(1-x_4) - x_4x_5 &\le 1,\\
\cdots \cdots \cdots \cdots &\qquad\\
x_{n-4}(1 - x_{n-3}) + x_{n-3}(1 - x_{n-2})
= 1 - (1 - x_{n-4})(1 - x_{n-3}) - x_{n-3}x_{n-2} &\le 1,
\end{align*}
and
\begin{align*}
&x_{n-2}(1 - x_{n-1}) + x_{n-1}(1 - x_n) + x_n(1 - x_1)\\
={}& x_{n-2}(1 - x_{n-1}) + x_{n-1} + x_n( 1 - x_1 - x_{n-1})\\
\le{}& x_{n-2}(1 - x_{n-1}) + x_{n-1} \\
\le{}& (1 - x_{n-1}) + x_{n-1}\\
={}& 1.
\end{align*}
Adding them up, the desired result follows.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How to derive a closed form of a recursion (maybe using generating functions) Let $a_0=9$ and consider the following recurrence relation: $$a_n=36(n+1)2^{n-2}+2a_{n-1},$$ I'm looking for the closed form of $\{a_n\}.$
I have tried using generator functions:
\begin{align*}
f(x)&=\sum_{n=0}^\infty a_nx^n\\
&=9 +\sum_{n=1}^\infty \left(4.5(n+1)2^{n+1}+2a_{n-1}\right)x^n\\
&=9 +4.5\sum_{n=1}^\infty (n+1)2^{n+1}x^n+2\sum_{n=1}^\infty a_{n-1}x^n\\
&=9 +\frac{4.5}{x}\sum_{n=1}^\infty (n+1)2^{n+1}x^{n+1} + 2x\sum_{n=1}^\infty a_{n-1}x^{n-1}\\
&=9 +\frac{4.5}{x}\sum_{n=2}^\infty n2^{n}x^{n} + 2xf(x)
\end{align*}
What should I do next? Is there a quicker way to solve this?
Thanks in advance
| An easy way to solve this recurrence.
As the recurrence is linear, we have that
$$
a_n = a_n^h + a_n^p
$$
such that
$$
\cases{a_n^h= 2 a_{n-1}^h\\ a_n^p = 2 a_{n-1}^p+36(n+1)2^{n-2}}
$$
so $a_n^h = c_0 2^n$. Now assuming $a_n^p = c_0(n)2^n$ after substitution into the complete equation, we have
$$
c_0(n)2^n = 2c_0(n-1)2^{n-1}+36(n+1)2^{n-2}
$$
or
$$
c_0(n) = c_0(n-1)+9(n+1)
$$
hence
$$
c_0(n) = 9\frac{(n+1)(n+2)}{2}
$$
then
$$
a_n^p = 9\frac{(n+1)(n+2)}{2}2^n
$$
and finally
$$
a_n = \left(9\frac{(n+1)(n+2)}{2}+c_0\right)2^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve the equation $8^x+3\cdot2^{2-x}=1+2^{3-3x}+3\cdot2^{x+1}$ Solve the equation $$8^x+3\cdot2^{2-x}=1+2^{3-3x}+3\cdot2^{x+1}$$ The given equation is equivalent to $$2^{3x}+\dfrac{12}{2^x}=1+\dfrac{8}{2^{3x}}+6\cdot2^x$$ If we put $a:=2^x>0$, the equation becomes $$a^3+\dfrac{12}{a}=1+\dfrac{8}{a^3}+6a$$ which is $$a^6-6a^4-a^3+12a^2-8=0$$ The LHS factors as $(a+1)(a-2)(a^4+a^3-3a^2-2a+4)$, which is in no case obvious. Let's say that we find the roots $1$ and $-2$, then how do we show that $(a^4+a^3-3a^2-2a+4)$ does not factor any more? Taking these into consideration, I believe there is an another approach. Any ideas would be appreciated.
| Letting $y=2^{x-\frac12}>0$
$$\begin{align}8^x+3\cdot2^{2-x}=1+2^{3-3x}+3\cdot2^{x+1}&\iff y^3+\frac3y=\frac1{2\sqrt2}+\frac1{y^3}+3y
\\&\iff\left(y-\frac1y\right)^3=\left(\frac1{\sqrt2}\right)^3\\&\iff\dots\\&\iff y=\sqrt2\\&\iff x=1.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Compute $\int_a^b e^x dx$ as a Riemann Sum I tried computing the integral $$\int_a^b e^x dx$$ as a Riemann sum. Therefore split the interval in to $n$ parts of the length $$\frac{b-a}{n}$$
and then took the limit of the Riemann sum.
$$\lim _{n \rightarrow \infty} \frac{b-a}{n} \sum_{k=0}^n e^{\frac{k(b-a)}{n}}$$.
When I computed this sum I got a limit, but not the right one.
$$
\begin{aligned}
& \int_a^b e^x d x=\lim _{n \rightarrow \infty} \frac{b-a}{n} \sum_{k=0}^n e^{\frac{k(b-a)}{n}} \\
& =\lim _{n \rightarrow \infty} \frac{b-a}{n} \sum_{k=0}^n\left(e^{\frac{b-a}{n}}\right)^k \\
& =\lim _{n \rightarrow \infty} \frac{b-a}{n}\left(\frac{\left(e^{\frac{b-a}{n}}\right)^{n+1}-1}{e ^{\frac{b-a}{n}}-1}\right) \\
& =\lim _{n \rightarrow \infty} \frac{(b-a)\left(\left(e^\frac{b-a}{n}\right)^{n+i}-1\right)}{n\left(e ^{\frac{b-a}{n}}-1\right)} \\
& =\lim _{n \rightarrow \infty} \frac{(b-a)\left(e^{\frac{(b-a)(n+1)}{n}}-1\right)}{n\left(\left(1+\frac{1}{n}\right)^{b-a}-1\right)} \\
& =\lim _{n \rightarrow \infty} \frac{(b-a)\left(e^{\frac{(b-a)(n+1)}{n}}-1\right)}{n\left(1+\frac{b-a}{n}-1\right)} \text { after Taylor Expansion } \lim _{x \to 0}(1+x)^a=1+x \cdot a\\
& =\lim _{n \rightarrow \infty} e^{\frac{(b-a)(n+1)}{n}}-1 \\
& =\lim _{n \rightarrow \infty} e^{b-a}-1 \\
&
\end{aligned}
$$
Does somebody spot my mistake?
| $$
\begin{aligned}
\int_a^b e^x d x & =\lim _{n \rightarrow \infty} \frac{b-a}{n} \sum_{k=0}^n e^{a+\frac{k(b-a)}{n}} \\
& =e^a \lim _{n \rightarrow \infty} \frac{b-a}{n} \sum_{k=1}^n\left(e^{\frac{b-a}{n}}\right)^k \\
& =e^a \lim _{n \rightarrow \infty} \frac{b-a}{n} \cdot \frac{\left(e^{\frac{b-a}{n}}\right)^n-1}{e^{\frac{b-a}{n}}-1} \\
& =e^a\left(e^{b-a}-1\right) \frac{1}{\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)} \textrm{, where }x=\frac{b-a}{n} \\
& =e^b-e^a
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving $\left\lfloor(\frac{1+\sqrt{5}}{2})^{4n+2}\right\rfloor-1$ is a perfect square for $n=0,1,2,\ldots$ Let $$S_n = \left \lfloor\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\rfloor-1$$ ($n=0, 1, 2, \ldots$).
Prove that $S_n$ is a perfect square.
In Art of Problem Solving website, there is a hint
$$
\begin{align}
\left\lfloor\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\rfloor-1 & =\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}+\left(\frac{1-\sqrt{5}}{2}\right)^{4n+2}-2\\
&=\left(\left(\frac{1+\sqrt{5}}{2}\right)^{2n+1}+\left(\frac{1-\sqrt{5}}{2}\right)^{2n+1}\right)^2
\end{align}
$$
I don't know how to get the first equal sign, is it mean
$$
\left\lfloor\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\rfloor = \left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}-\left\{\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\} $$
But how to prove the decimal part of $\phi^{4n+2}$
$$ \left\{\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2} \right\} =1-\left(\frac{1-\sqrt{5}}{2}\right)^{4n+2}$$
| Let $\varphi:=\frac{1+\sqrt5}2$ and $\bar\varphi:=\frac{1-\sqrt5}2.$ The hint you wonder about,
$$\left\lfloor\varphi^{2m}\right\rfloor=\varphi^{2m}+\bar\varphi^{2m}-1,$$
is due to the fact that $\varphi^k+\bar\varphi^k$ is an integer (the $k$-th Lucas number) and
$$\varphi^{2m}+\bar\varphi^{2m}-1\le\varphi^{2m}<\varphi^{2m}+\bar\varphi^{2m},$$
i.e.
$$0<\bar\varphi^{2m}\le1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Inverse of difference of two digamma functions I recently encountered the expression below for which I was interested in solving for $x$:
\begin{equation}
\psi(x+n+1) - \psi(x+1) =y
\end{equation}
$\psi$ is the digamma function, $n$ is a positive integer and $x,y>0$. Initially, I solved this numerically but I was interested to see if there was the possibility to solve this, either in closed form or as a series. I tried to find some way by using the relationship to harmonic numbers, e.g. rewriting as
\begin{equation}
\psi(x+n+1) - \psi(x+1) = \sum_{k=1}^{n} \frac{1}{k+x} = \int_0^1 \frac{t^x(1-t^n)}{1-t}dt
\end{equation}
to see if there was something which I could do, but I didn't notice anything apparent. The only thing I've come up with is that for large values of $x$, $y$ is small and using $\psi(x) \sim\ln(x)$, that $x \approx \dfrac{n}{e^y - 1}$.
I'm happy for any suggestions on how to proceed!
| $$\psi(x+n+1) - \psi(x+1) =y$$
For large values of $x$, the lhs write
$$\frac{n}{x}-\frac{n (n+1)}{2 x^2}+\frac{n (n+1) (2 n+1)}{6
x^3}-\frac{n^2 (n+1)^2}{4
x^4}+O\left(\frac{1}{x^5}\right)$$ Using series reversion
$$x_{(1)}=\frac{n}{y}-\frac{n+1}{2}+\frac{\left(n^2-1\right)}{12
n}y+O\left(y^3\right)$$
Comparing with your approximation
$$x_{(2)}= \dfrac{n}{e^y - 1}=\frac{n}{y}-\frac{n}{2}+\frac{n }{12}y+O\left(y^3\right)$$ It is really good since
$$x_{(2)}-x_{(1)}=-\frac{1}{2}-\frac{y}{12 n}+O\left(y^3\right)$$
Using the above for $n=1000$ and $y=1$, the exact solution is $581.477$ while $x_{(2)}=582.833$ and $\frac{1000}{e-1}=581.977$.
Edit
We could do better using more terms in the series expansion, then series reversion and transformation of the result into a $[n,n+1]$ Padé approximant.
This would give
$$x_{(3)}=\frac{60 (n-1) n^3 -24 n^2 \left(n^2-4\right)y+3 n \left(n^3-3 n^2-9 n-13\right)y^2} {60 (n-1) n^2 y+6 n \left(n^2+11\right)y^2+(n-1) \left(n^2+11\right)y^3 }$$
For the worked example, this gives $x_{(3)}=581.590$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $7 \mid 11^n - 4^n$ with mathematical induction I want to prove that $7 \mid 11^n - 4^n$ with mathematical induction. This is what I wrote:
*
*For $n = 1$, we have $7 \mid 11^1 - 4^1 \Rightarrow 7 \mid 7$ which is obviously true. $\checkmark$
*Assume that the statement is true for $n = k$. So: $7 \mid 11^k - 4^k$ and therefore there is an integer such that $m$ such that: $11^k - 4^k = 7m$. Thus: $4^k = 11^k - 7m$.
*Now we prove its truth for $n = k + 1$. So we want to prove that $7 \mid 11^{k + 1} - 4^{k + 1}$. We can write this like this: $7 \mid \big(11\cdot11^k\big) - \big(4\cdot4^k\big)$ and since $4^k = 11^k - 7m$, so in fact we have to prove: $7 \mid \big(11 \cdot11^k\big) - \big(4\cdot(11^k - 7m)\big) \Longrightarrow 7 \mid 11\cdot11^k - 4\cdot11^k + 7(4m) \Longrightarrow 7 \mid 7\cdot11^k + 7(4m) \Longrightarrow 7 \mid 7(11^k + 4m)$.
But from here you can clearly see that $7(11^k + 4m)$ is divisible by 7. So according to the principle of mathematical induction, the statement is proved. $\blacksquare$
Now my question is, is this proof correct? And is there another way to prove this statement by mathematical induction?
| Let's modify your induction proof so that we directly use the induction hypothesis that $7 \mid 11^k - 4^k$ for some positive integer $k$.
Let $P(n)$ be the the statement that $7 \mid 11^n - 4^n$.
You handled the $n = 1$ case correctly.
Since the statement $7 \mid 11^n - 4^n$ holds for $n = 1$, we may assume $P(k)$ holds for some positive integer $k$. That is, $7 \mid 11^k - 4^k$ holds for some positive integer $k$. Then there exists an integer $m$ such that $7m = 11^k - 4^k$.
Thus far, I have essentially reformulated what you said in your proof. Now, we will express $11^{k + 1} - 4^{k + 1}$ in terms of $11^k - 4^k$ so that we can use the induction hypothesis in the induction step.
Let $n = k + 1$. Then
\begin{align*}
11^{k + 1} - 4^{k + 1} & = 11 \cdot 11^k - 4 \cdot 4^{k}\\
& = (7 + 4)11^k - 4 \cdot 4^k\\
& = 7 \cdot 11^k + 4 \cdot 11^k - 4 \cdot 4^k\\
& = 7 \cdot 11^k + 4(11^k - 4^k)\\
& = 7 \cdot 11^k + 4 \cdot 7m && \text{by the induction hypothesis}\\
& = 7(11^k + 4m)
\end{align*}
Since $11^k + 4m$ is an integer, $7 \mid 11^{k + 1} - 4^{k + 1}$. Thus, $P(k) \implies P(k + 1)$ for each positive integer $k$. Since $P(1)$ holds and $P(k) \implies P(k + 1)$ for each positive integer $k$, $P(n)$ holds for each positive integer $n$ by the Principle of Mathematical Induction.
| {
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"url": "https://math.stackexchange.com/questions/4621513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Nth Derivative of $\frac{x}{x-1}$ So starting off, rewrite $x*\dfrac1{x-1}$. $d/dx=\dfrac{d}{dx}(x)*\dfrac{1}{x-1}+\dfrac{d}{dx}\dfrac{1}{x-1}*x$.
$\dfrac{d}{dx}(x)=1$. $x^{n}=nx^{n-1}$.
$\dfrac{1}{x-1}=(x-1)^{-1}$.
$\dfrac{d}{dx}(\dfrac{1}{x-1})=(-1)(x-1)^{-2}=\dfrac{-1}{(x-1)^{2}}$.
Combining them both we get $\dfrac{1}{x-1}+\dfrac{-1}{(x-1)^{2}}*x$.
$\dfrac{d}{dx}=\dfrac{1}{x-1}+\dfrac{-x}{(x-1)^{2}}$.
Now to find $\dfrac{d^2y}{d^2x}$, using product rule and sum rule
$\dfrac{d}{dx}\dfrac{1}{x-1}+\dfrac{d}{dx}-x\dfrac{1}{(x-1)^2}+\dfrac{d}{dx}\dfrac{1}{(x-1)^{2}}x$, so anyway doing it we get
$\dfrac{-1}{(x-1)^2}+\dfrac{-1}{(x-1)^2}+\dfrac{-2x}{(x-1)^{3}}$.
Simplifying $\dfrac{d^2y}{d^2x}=\dfrac{-2}{(x-1)^2}+\dfrac{-2x{(x-1)^{3}}$; and finally $\dfrac{d^3y}{d^3x}$. $\dfrac{d}{dx}\dfrac{-2}{(x-1)^2}+\dfrac{d}{dx}-2x\dfrac{1}{(x-1)^{3}}+\dfrac{d}{dx}\dfrac{1} {(x-1)^{3}}*(-2x)$,which gives $\dfrac{4}{(x-1)^3}+\dfrac{-2}{(x-1)^3}+\dfrac{6x{(x-1)^4}$; simplyfing we get
$\dfrac{2}{(x-1)^3}+\dfrac{6x}{(x-1)^4}$; now there is supposed to be a pattern here but i don't see it.
is there something wrong with my answer?
| How about
$$f(x)=\frac x{x-1}=\frac1{x-1}+1$$
Therefore:
$$\frac d{dx}f(x)=-1\frac1{(x-1)^2}\\\frac{d^2}{dx^2}f(x)=-1\cdot-2\frac1{(x-1)^3}\\\frac {d^3}{dx^3}f(x)=-1\cdot -2\cdot -3\frac1{(x-1)^4}\\\vdots\\\frac{d^n}{dx^n}f(x)=\frac{(-1)^n n!}{(x-1)^{n+1}}$$
One may use factorial power $u^{(v)}$ to get:
$$\boxed{\frac{d^n}{dx^n}\frac x{x-1}=\frac{(-1)^{(n)}}{(x-1)^{n+1}},n\ge 1}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Integration $\int_0^{\pi/2} \frac{dx}{(3 + 5 \cos x)^2}$ I had tried to solve this integral; using the substitution $\tan(x/2) =t$, and $\cos x= \frac{1-t^2}{1+t^2}$. But after making terms in $t$, I am not able to integrate further as numerator contains quadratic and denominator contains biquadratic.
$\int\limits_0^{\pi/2} \frac{1}{(3 + 5 \cos x)^2}\ dx$.
| Well let's use Weierstrass and partial fractions ig and evaluate the definite with FTC II.
Using Weierstrass with the tangent half angle substitution, our integral becomes
$$\int \frac{1}{(3 + 5 \cos x)^2}\ dx = \int \frac{1}{\left(3+5\left(\frac{1-t^2}{1+t^2}\right)\right)^2}\cdot \frac{2dt}{1+t^2} = \frac12\int {t^2 + 1\over(t - 2)^2 (t + 2)^2}dt$$
Perform partial fraction decomposition by setting it up like this
$$ {t^2 + 1\over2(t - 2)^2 (t + 2)^2}dt = \frac{A}{2(t-2)} + \frac{B}{2(t-2)^2} + \frac{C}{2(t+2)} + \frac{D}{2(t+2)^2}$$
Then just multiply by the denominator, match powers, and solve the system of equations for the unknown. We get
$$ {t^2 + 1\over2(t - 2)^2 (t + 2)^2}dt = \frac{3}{64(t-2)} + \frac{5}{32(t-2)^2} + \frac{-3}{64(t+2)} + \frac{5}{32(t+2)^2}$$
Now integrate termwise (log for single power, 1/(whatever) for double power) and simplify to get
$$\int \frac{3}{64(t-2)} + \frac{5}{32(t-2)^2} + \frac{-3}{64(t+2)} + \frac{5}{32(t+2)^2} dt $$$$= {-5t\over 16(t^2-4)} + {3\over 64}\ln(2-t) -\frac3{64}\ln(t+2)+C$$
The bounds transform as follows $\tan\left({\frac\pi2\over2}\right)=1$ and the lower one remains $0$ so
$${-5t\over 16(t^2-4)} + {3\over 64}\ln(2-t) -\frac3{64}\ln(t+2)\Big|^1_0 = \frac5{48} - {3\ln(3)\over64}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculate the limit of a function involving greatest integer function
If $f:(0,\infty)\rightarrow\mathbb{N}$ and
$$f(x)=\left[\frac{x^2+x+1}{x^2+1}\right]+\left[\frac{4x^2+x+2}{2x^2+1}\right]+\left[\frac{9x^2+x+3}{3x^2+1}\right]+\cdots+\left[\frac{(nx)^2+x+n}{nx^2+1}\right]$$
Find the value of $$\lim_{n\rightarrow\infty}\left[\frac{f(x)-n}{(f(x))^2-\frac{n^3(n+2)}{4}}\right]$$
where $[.]$ denotes the greatest integer function.
I am not able to calculate $f(x)$. Like if I'll put summation on the $nth$ term, how will I calculate it$?$ Moreover, how will I handle the greatest integer function.
Any help is greatly appreciated.
| $f(x)$ is easy to calculate once you simplify the algebraic expressions under it.
$f(x)=\begin{equation}
\sum_{k=1}^{n}\left[\frac{(nx)^2+x+n}{nx^2+1}\right]
\end{equation}$
$\left[\frac{(nx)^2+x+n}{nx^2+1}\right] = \left[\frac{(nx)^2+n}{nx^2+1}+\frac{x}{nx^2+1}\right]$
$=\left[\frac{n(nx^2+1)}{nx^2+1}+\frac{x}{nx^2+1}\right]$
$=\left[{n}+\frac{x}{nx^2+1}\right]$
Since $x < nx^2+1$ , the expression can be evaluated to n.
Hence, $f(x) = \sum_{k=1}^{n}n = \frac{n(n+1)}{2}$
$\lim_{n\rightarrow\infty}\left[\frac{f(x)-n}{(f(x))^2-\frac{n^3(n+2)}{4}}\right]$
Substituting the value of $f(x)$ , the limit simplifies to,
$\lim_{n\rightarrow\infty}\left[\frac{2(n-1)}{n}\right]$
Hope this is enough to evaluate the limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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All natural number solutions for the equation $a^2+b^2=2c^2$ $a$, $b$ and $c$ of all Pythagorean triplets can be written in the form
$$
\begin{split}
a &= 2mn\\
b &= m^2-n^2 \\
c &= m^2+n^2
\end{split}
$$
where $m$ and $n$ are natural numbers. For any natural number $m$ and $n$, this set of equations will give a Pythagorean triplet. And all Pythagorean triplets satisfy this set of equations.
Can $a$, $b$ and $c$ of all triplets satisfying the equation $$a^2+b^2=2c^2$$ where $a$, $b$ and $c$ are natural numbers, be written as a set of equations as for the Pythagorean triplets?
So, I need a set of equations that generates triplets that satisfy the equation $a^2+b^2=2c^2$ for any natural numbers I plug into the set of equations. Also, every natural number triplets satisfying the equation $a^2+b^2=2c^2$ must satisfy the set of equations.
I tried to derive the set of equations myself, no attempts have been successful yet.
I would like to have the proof of the set of equations, (otherwise I won't know if every triple will satisfy the set of equations)
Any comments that helps to give an insight into solving the problem are really appreciated.
| Noting that
$a^2+b^2=2c^2 \Rightarrow a $ and $b$ are of same parity. Hence there exists natural numbers $u$ and $v$ such that
$$a+b=2u \textrm{ and } a-b=2v$$
Then $$a=u+v \textrm{ and }b=u-v$$
$$
\begin{gathered}
(u+v)^2+(u-v)^2=2 c^2 \Rightarrow
u^2+v^2=c^2
\end{gathered}
$$
There exists Pythagorean triple such that
$$
\begin{aligned}
& \left\{\begin{array}{l}
u=2k m n \\
v=k(m^2-n^2) \\
c=k(m^2+n^2)
\end{array}\right. \Rightarrow \left\{\begin{array}{l}
a=k(2 m n+m^2-n^2 )\\
b=k(2 m n-m^2+n^2) \\
c=k(m^2+n^2)
\end{array}\right. ,\\
&
\end{aligned}
$$
where $k,m,n \in N$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find the radical centre of the four spheres :$x^2+y^2+z^2+2x+2y+2z+2=0,x^2+y^2+z^2+4y=0,x^2+y^2+z^2+3x-2y+8z+6=0,x^2+y^2+z^2-x+4y-6z-2=0$ Determine the radical centre of the spheres $x^2+y^2+z^2+2x+2y+2z+2=0,x^2+y^2+z^2+4y=0,x^2+y^2+z^2+3x-2y+8z+6=0,x^2+y^2+z^2-x+4y-6z-2=0.$
I tried solving the problem by assuming $S_1=x^2+y^2+z^2+2x+2y+2z+2,S_2=x^2+y^2+z^2+4y,S_3=x^2+y^2+z^2+3x-2y+8z+6,S_4=x^2+y^2+z^2-x+4y-6z-2.$ Now, the radical plane of the $3$ spheres $S_1,S_2,S_3$ are $S_1-S_2=0,S_2-S_3=0,S_3-S_1=0.$ Now, these radical planes, intersect at the straight line $S_1=S_2=S_3.$ Thus, the radical axis of $S_1,S_2,S_3$ is $S_1=S_2=S_3.$ Similarly, the other possible radical axis of the spheres taken three at a time are: $S_1=S_2=S_4,S_1=S_3=S_4,S_2=S_3=S_4.$ Thus, the $4$ radical axis are $S_1=S_2=S_3,$$S_1=S_2=S_4,S_1=S_3=S_4,S_2=S_3=S_4.$ Now, the point where these $4$ radical lines intersects is called the radical centre. But how to calculate $C$ ? If we try to solve these $4$ equations, then I think we should require a fast processing and computing software or otherwise it becomes a tremendously huge calculation. Is there any shorter way to determine the radical centre $C$ using only elementary analytic geometry of spheres ?
| Here is the computation.
$\quad\quad$ $(x,y,z)$ is the radical centre
$\iff S_1(x,y,z)=S_2(x,y,z)=S_3(x,y,z)=S_4(x,y,z)$
$\iff 2x+2y+2z+2=4y=3x-2y+8z+6=-x+4y-6z-2$
$\iff x=y-z-1\text{ and }x=2y-\frac83z-2=0\text{ and } x=-6z-2$
$\iff x=y-z-1\text{ and }y-\frac53z-1=0\text{ and } y+5z+1=0$
$\iff x=-\frac15, y=\frac12, z=-\frac3{10}$
I can hardly see why "we should require a fast processing and computing software or otherwise it becomes a tremendously huge calculation".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the limit $\lim_{x\to2}(x^3-2x-4)\tan\frac{\pi x}{4}$ Find the limit $$\lim_{x\to2}(x^3-2x-4)\tan\dfrac{\pi x}{4}$$
I can't even determine what type of indeterminate form we have. It's $0\times\text{undefined}$. We can see that $2$ is a root of the polynomial $x^3-2x-4$ and it factors as $(x-2)(x^2+2x+2)$. Additionally, when $x\to2$, $\tan\dfrac{\pi x}{4}\to\tan\dfrac{\pi}{2},$ which is not defined. How do we approach the problem and what's the intuition? I suppose somehow the limit $\lim_{f(x)\to0}\dfrac{\sin f(x)}{f(x)}=1$ may come in handy.
| We have:
$$ (x^3-2x-4) \tan(x\pi/4) = \frac{\sin(x\pi/4)(x-2)(x^2+2x+2)}{\cos(x\pi/4)}$$
Let $u = x-2$, then:
$$ = \frac{u\sin((u+2)\pi/4)((u+2)^2+2(u+2)+2)}{\cos((u+2)\pi/4)} $$
Where $\sin((u+2)\pi/4) = \cos(u\pi/4)$ and $\cos((u+2)\pi/4) = -\sin(u\pi/4)$ using the addition formulas for sine and cosine. So:
$$ \frac{u\cos(u\pi/4)((u+2)^2+2(u+2)+2)}{-\sin(u\pi/4)} = \frac{u\pi/4 \cos(u\pi/4)((u+2)^2+2(u+2)+2)}{-\pi/4\sin(u\pi/4)} =$$
$$ = -4/\pi \cdot \frac{u\pi/4}{\sin(u\pi/4)} \cdot \cos(u\pi/4) \cdot ((u+2)^2+2(u+2)+2)$$
We let $u \rightarrow 0$. First factor tends to $1$, second factor to $1$, last factor tends to $10$, the limit is therefore $-\frac{40}{\pi}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I find individual values of $\sin(\varphi)$ and $\cos(\varphi)$ from $x = a\sin^{2}(\varphi) + b\cos^{2}(\varphi)$? If $x = a\sin^2\phi+ b\cos^2\phi$, express $\sin$ and $\cos$ in terms of $x$ ( $a$ and $b$ are real constants)
I know how to find values of $T$ ratios in equations like $x = a\sin^2t$ or $x = b\cos^2t$ but how do I find the values of $\sin(t)$ and $\cos(t)$
in expressions like $x = \sin^2t + 5\cos^2t$ or $x = a\sin t + b\cos t$?
| 1. $$x=a\sin^2\phi + b\cos^2\phi$$
2. $$x=a\sin^2\phi+b-b\sin^2\phi$$
3. $$x=b+(a-b)\sin^2\phi$$
4. $$\sin\phi=\pm\sqrt\frac{x-b}{a-b}$$
$$\cos\phi=\pm\sqrt\frac{a-x}{a-b}$$
Hope this helps!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631766",
"timestamp": "2023-03-29T00:00:00",
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Solve $(x^2+1)(x-1)^2=ax^2, \ a\in\mathbb{R}$
Solve $(x^2+1)(x-1)^2=ax^2, \ a\in\mathbb{R}$. Discuss.
This is the entire text of the problem. I am really unsure what the authors meant by discuss here? Is it to find $x$ based on $a$ and then separate by cases? What I tried doing is to raise to power and multiply the parentheses, but I'm not sure what to do next:
$$(x^2+1)(x-1)^2-ax^2 =0 \iff (x^2+1)(x^2-2x+1)-ax^2 = 0 \iff \\ \iff x^4 -2x^3+(2-a)x^2-2x+1 = 0$$
What I noticed is that this looks like a symmetric polynomial, would that help? And, again, how would I "discuss" that?
| $(x^2+1)(x-1)^2=ax^2\quad\color{blue}{(*)}$
Since $\,x=0\,$ is not a solution of the equation $(*)\,,\,$ we can divide by $\,x^2$ both sides of $(*)$ and get the following equivalent equation:
$\left(x+\dfrac1x\right)\left(x+\dfrac1x-2\right)=a\;.$
By letting $\;t=x+\dfrac1x\,,\,$ we get the following equation:
$t\,(t-2)=a\quad\color{blue}{(**)}$
which does not have any real solution if $\,a<-1\,.$
Whereas, if $\,a\geqslant-1\,,\,$ $(**)$ has the following real solutions :
$t=1\pm\sqrt{a+1}$.
Moreover, since $\;t=x+\dfrac1x\,,\,$ it follows that
$x+\dfrac1x=1\pm\sqrt{a+1}\;\;,$
$x^2-\left(1\pm\sqrt{a+1}\right)x+1=0\quad\color{blue}{(*\!*\!*)}$
which is equivalent to the equation $(*)$.
If $\;a<0\,,\,$ the equation $(*\!*\!*)$ does not have any real solution.
If $\;0\leqslant a<8\,,\,$ the equation $(*\!*\!*)$ has the following two real solutions:
$x=\dfrac{1+\sqrt{a+1}\pm\sqrt{a-2+2\sqrt{a+1}}}2\,.$
Whereas, if $\,a\geqslant8\,,\,$ the equation $(*\!*\!*)$ also has other two real solutions:
$x=\dfrac{1-\sqrt{a+1}\pm\sqrt{a-2-2\sqrt{a+1}}}2\,.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4632973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Without Calculator find $\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$
Without Calculator find $$\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$$
Where $\left \lfloor x \right \rfloor $ represents floor function.
My Try:
Let $x=2\cos(50^{\circ})+\sqrt{3}$. We have
$$\begin{aligned}
& \cos \left(50^{\circ}\right)<\cos \left(45^{\circ}\right) \\
\Rightarrow \quad & 2 \cos \left(50^{\circ}\right)<\sqrt{2} \\
\Rightarrow \quad & x<\sqrt{3}+\sqrt{2}<3.14
\end{aligned}$$
Now I am struggling hard to prove that $x>3$
| 50° is between 45° and 60° ( $50°=45°+(60°-45°)/3$ ), and in this range, function $cos$ is concave. So $\cos(50°) > \cos 45°+(\cos60°-\cos45°)/3$
$\cos 50° > \sqrt{2}/2 + (1/2- \sqrt{2}/2)/3$.
Approximate $\sqrt{2}$ with $1.414$ and $\sqrt{3}$ with $1.732$, and you are in.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
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Find the roots of the equation $(1+\tan^2x)\sin x-\tan^2x+1=0$ which satisfy the inequality $\tan x<0$ Find the roots of the equation $$(1+\tan^2x)\sin x-\tan^2x+1=0$$ which satisfy the inequality $$\tan x<0$$
Shold I solve the equation first and then try to find which of the roots satisfy the inequality? Should I use $\tan x$ in the solution itself and obtain only the needed roots?
I wasn't able to see how the inequality can be used beforehand, so I solved the equation. For $x\ne \dfrac{\pi}{2}(2k+1),k\in\mathbb{Z}$, the equation is $$\dfrac{\sin^2x+\cos^2x}{\cos^2x}\sin x-\left(\dfrac{\sin^2x}{\cos^2x}-\dfrac{\cos^2x}{\cos^2x}\right)=0$$ which is equivalent to $$\sin x+\cos2x=0\\\sin x+1-2\sin^2x=0\\2\sin^2x-\sin x-1=0$$ which gives for the sine $-\dfrac12$ or $1$. Then the solutions are $$\begin{cases}x=-\dfrac{\pi}{6}+2k\pi\\x=\dfrac{7\pi}{6}+2k\pi\end{cases}\cup x=\dfrac{\pi}{2}+2k\pi$$ How do I use $\tan x<0$?
| $\frac\pi2+2k\pi$ should never appear because the initial equation is not defined when $\cos x=0.$
You forgot a $-$ sign in front of $\frac{5\pi}6+2k\pi.$
As for your final question: simply discard $-\frac{5\pi}6+2k\pi$ (whose $\tan$ is $>0$) and retain $-\frac\pi6+2k\pi$ (whose $\tan$ is $<0$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4639727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Keeping track of basis changes when computing the smith normal form I don't understand how to keep track of the basis change as you compute smith normal form. We did an example in lecture where:
$A = \begin{pmatrix}
[1] & 2 & 3 & 4 \\
5 & 6 & 7 & 8 \\
9 & 10 & 11 & 12 \\
\end{pmatrix}$
Where [1] is the pivot position. The first operations performed were:
$R_2\mapsto R_2 - 5R_1$
$R_3\mapsto R_3 - 9R_1$.
Which then gives:
$\begin{pmatrix}
1 & 2 & 3 & 4 \\
0 & -4 & -8 & -12 \\
0 & -8 & -16 & -24 \\
\end{pmatrix}$.
He then writes: $U\mapsto V$: Basis of V:
$\begin{pmatrix}
1\\
5\\
9 \\
\end{pmatrix}$
$\begin{pmatrix}
0\\
1\\
0 \\
\end{pmatrix}$
$\begin{pmatrix}
0\\
0\\
1 \\
\end{pmatrix}$.
Afterwards, he column transformations:
$C_2\mapsto C_2 - 5C_1$
$C_3\mapsto C_3 - 9C_1$
$C_4\mapsto C_4 - 4C_1$
Getting
$\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & -4 & -8 & -12 \\
0 & -8 & -16 & -24 \\
\end{pmatrix}$ and basis of U:
$\begin{pmatrix}
1\\
0\\
0 \\
0\\
\end{pmatrix}$
$\begin{pmatrix}
-2\\
-1\\
0 \\
0 \\
\end{pmatrix}$
$\begin{pmatrix}
-3\\
0\\
1 \\
0 \\
\end{pmatrix}$
$\begin{pmatrix}
-4\\
0\\
0 \\
1 \\ \end{pmatrix}$
and then so on, updating the basis after each step. What I don't understand is how these basis are found after each step.
| You can think of the decomposition process as follows: starting with $A_0 = A$, $P_0 = I_3$, and $Q_0 = I_4$, we go from $(A_k,P_k,Q_k)$ to $(A_{k+1},P_{k+1},Q_{k+1})$ in such a way that at all points in the process, we have $A = P_k A_k Q_k$. Note that $P_k$ is a change of basis matrix over the codomain and $Q_k$ is a change of basis matrix over the domain.
On the other hand, note that applying a row-operation to $A$ amounts to computing $RA$ for some invertible matrix $R$. Similarly, applying a column-operation to $A$ amounts to computing $C^{-1}A$ for some invertible matrix $C$.
With all that established, consider the following. We begin with
$$
A = \overbrace{\pmatrix{1&0&0\\0&1&0\\0&0&1}}^{P_0}
\overbrace{\begin{pmatrix}
1 & 2 & 3 & 4 \\
5 & 6 & 7 & 8 \\
9 & 10 & 11 & 12 \\
\end{pmatrix}}^{A_0}
\overbrace{\pmatrix{1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1}}^{Q_0}.
$$
Applying your row operations amounts to the multiplication $RA$, where
$$
R = \pmatrix{1&0&0\\-5&1&0\\-9&0&1}.
$$
Note: to find the matrix corresponding to a certain sequence of row operations, simply apply those operations to the identity matrix. Similarly, the matrix corresponding to a certain sequence of columns operations can be found by applying those operations to an identity matrix.
With that in mind, we have $A = (P_0R^{-1})(RA_0)Q_0$. That is, we can take $P_1 = P_0R^{-1} = R^{-1}$. Note that the inverse of $R$ will be the matrix corresponding to the reverse of the row operations associated with $R$.
Similarly, the column operations applied to $A$ are those corresponding to the matrix
$$
C = \pmatrix{1&-2&-3&-4\\0&1&0&0\\0&0&1&0\\0&0&0&1}.
$$
With that in mind, we have
$$
A = (P_0R^{-1})(RA_0C)(C^{-1}Q_0).
$$
That is, our work so far amounts to taking $P_1 = P_0R^{-1} = R^{-1}$, $A_1 = RA_0C$, and $Q_1 = C^{-1}Q_0 = C^{-1}$.
After continuing these process, we will eventually end up with $A_k$ equal to the Smith normal form $A_k = D$, so that $A = P_k D Q_k$ is a Smith normal form decomposition. We will have $P_k = R_1^{-1}R_2^{-1} \cdots R_k^{-1}$ for some sequence of row operations with corresponding matrices $R_1,\dots,R_k$. Similarly, we will have $Q_k = C_k^{-1} \cdots C_2^{-1}C_1^{-1}$.
$P_k$ is a change of basis matrix that takes us to the desired basis from the standard basis. Thus, the columns of $P_k$ are the elements of that basis. $Q_k$ is a change of basis that takes us from the desired basis to the standard basis. Thus, the columns of $Q_k^{-1} = C_1C_2 \cdots C_k$ will be the elements of that basis.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx$
Evaluate $$\large{\int} \small{\sqrt{\frac{1+x^2}{x^2-x^4}} \space {\large{dx}}}$$
Note that this is a Q&A post and if you have another way of solving this problem, please do present your solution.
| In order to evaluate the integral over all domain $ x\in (-1,0)\cup(0,1)$, substitute $t=x\sqrt{x^2}$. Then, $dt =2\sqrt{x^2}\ dx$ and
\begin{align}
\int \sqrt{\frac{1+x^2}{x^2-x^4}}\ dx
=& \int \frac{1+x^2}{\sqrt{x^2(1-x^4)}}\ dx
=\frac12\int \frac{1+\sqrt{t^2}}{\sqrt{t^2(1-t^2)}}\ dt \\
=& \ \frac12\int \frac{1}{\sqrt{1-t^2}}+ \frac{1}{\sqrt{t^2(1-t^2)}}\ dt\\
=& \ \frac12 \sin^{-1}t-\frac12\sinh^{-1}
\frac{\sqrt{1-t^2}}t \\
= & \ \frac12 \sin^{-1}\left( x\sqrt{x^2}\right)-\frac12\sinh^{-1}\bigg(
\frac1x \sqrt{\frac{1-x^4}{x^2}}\bigg)\\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given such $n$th degree polynomial $P(x)$ and positive numbers $a, b\in\mathbb{R}$, does $\sqrt[n]{P(a+b)} \leq \sqrt[n]{P(a)} + \sqrt[n]{P(b)}$ hold? Given any $n$th degree polynomial $P(x)$ with positive coefficients and positive numbers $a, b\in\mathbb{R}$, does $\sqrt[n]{P(a+b)} \leq \sqrt[n]{P(a)} + \sqrt[n]{P(b)}$ hold?
I've worked it out for $n=2$:
$$P(x) = Ax^2 + Bx + C$$
$$\sqrt{Aa^2 + Ba + C} + \sqrt{Ab^2 + Bb + C} = \sqrt{(\sqrt{Aa^2 + Ba + C} + \sqrt{Ab^2 + Bb + C})^2} $$
$$= \sqrt{(Aa^2 + Ba + C) + 2\sqrt{(Aa^2 + Ba + C)(Ab^2 + Bb + C)} + (Ab^2 + Bb + C)}$$
$$ \geq \sqrt{Aa^2 + Ba + C + 2\sqrt{(Aa^2)(Ab^2)} + Ab^2 + Bb + C}$$
$$ \geq \sqrt{Aa^2 + Ba + C + 2Aab + Ab^2 + Bb}$$
$$ = \sqrt{A(a^2 + b^2 + 2ab) + B(a+b) + C} = \sqrt{A(a+b)^2 + B(a+b) + C}$$
But I couldn't figure out the generalization from there.
I also pondered whether Jensen's inequality would be of help, but I don't think $\sqrt[n]{P(x))}$ is concave.
I have used various higher degree polynomials to plot $\sqrt[n]{P(x))}$ and the inequality always seems to hold.
| From https://artofproblemsolving.com/community/c6h1440215p8188094 on AoPS:
If $P(x) = \sum_{k=0}^n c_k x^k$ is a polynomial with nonnegative coefficients and $a, b \ge 0$ then
$$
\begin{align}
\sqrt[n]{P(a+b)} &= \left( \sum_{k=0}^n c_k (a+b)^k\right)^{1/n}\\
&\overset{(1)}{\le} \left(\sum_{k=0}^n c_k( a^{k/n} + b^{k/n})^n\right)^{1/n} \\
&\overset{(2)}{\le} \left(\sum_{k=0}^n c_k a^k\right)^{1/n} + \left(\sum_{k=0}^n c_k b^k\right)^{1/n}\\
&= \sqrt[n]{P(a)} + \sqrt[n]{P(b)}
\end{align}
$$
where
*
*$(1)$ follows from How to show that $(a+b)^p\leq a^p + b^p$ for $a,b\geqslant 0$ and $0<p<1$? with $p=k/n$, and
*$(2)$ is Minkowski's inequality.
Or, using the fact that $Q(x) = P(x)/x^n$ is decreasing:
$$
\begin{align}
\sqrt[n]{P(a+b)} &= a \sqrt[n]{Q(a+b)} + b \sqrt[n]{Q(a+b)} \\
&\le a \sqrt[n]{Q(a)} + b \sqrt[n]{Q(b)} \\
&= \sqrt[n]{P(a)} + \sqrt[n]{P(b)}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Convergence of infinite product and its limit I wanted to find $\prod_{n=2}^{\infty}(1+\frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+...)$ and ended up simplifying it as
$\prod_{n=2}^{\infty} \frac{n^2}{n^2-1}$. Now the partial product is $\frac{2n}{n+1}$ it converges and it's limit is 2.. am I correct? Kindly share your views
| We can check your results, in order to verify your answer.
*
*First, we can show that
$$1+\frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots =\sum_{k=0}^{+\infty}\frac{1}{n^{2k}}.$$
*If $n\geqslant 1$, we can show that
$$\sum_{k=0}^{+\infty}\frac{1}{n^{2k}}=\sum_{k=0}^{+\infty}\left(\frac{1}{n^2}\right)^k=\frac{n^2}{n^2-1}$$
*Then,
$$\prod_{n=2}^{+\infty}\left(1+\frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots \right)=\prod_{n=2 }^{+\infty}\sum_{k=0}^{+\infty}\left(\frac{1}{n^2}\right)^k=\prod_{n=2}^{+\infty}\left(\frac{n^2}{n^2-1}\right)$$
*If you showed that for all $N\in\mathbf{N}$ we can write
$$\prod_{n=2}^{N}\left(\frac{n^2}{n^2-1}\right)=\frac{2N}{N+1},$$
(this is indeed true, but you should have to justify it; a way it is using partial fraction), then taking $N\to +\infty$ both sides we find
$$\prod_{n=2}^{+\infty}\left(\frac{n^2}{n^2-1}\right)=\lim_{N\to +\infty}\prod_{n=2}^{N}\left(\frac{n^2}{n^2-1}\right)=\lim_{N\to +\infty}\left(\frac{2N}{N+1}\right)=2$$
(it follows directly by definition) and then we get the answer $2$ as you said.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the number of ways to arrange so that at least two are followed
Suppose there is a sentence containing only sequences of three characters and nothing more. The three characters are $X,Y,Z$ and it is given that $X$ has occurred $a$ times, $Y$ has occurred $b$ times and $Z$ has occurred $c$ times in the sentence. What is the probability that a $X$ will be followed by a $Y$ at least $2$ times$?$
This problem looks a bit complicated to me, so I decided to break it into some parts. At least $2$ times means, all cases$-$$($exactly $0$ time$+$exactly $1$ time$)$
The number of all cases are simply $$\binom{a+b+c}{a}\binom{b+c}{b}\binom{c}{c}$$
For the exactly $0$ times I'm able to think of a logic but it is hard to explain. I think that we should first select a $Z$ and fix it. Then we arrange $b$ $Y's$ and $(c-1)$ $Z's$ on one side of the fixed $Z$ and on other side put all the $X$. One more is to put all the $X's$ between two $Z's$ and then do the rest of the arrangement. So you see I'm not been able to think of all such cases. One more is put all $Y's$ and then arrange rest such that no $X$ goes behind any $Y$.
For exactly $1$ I have similar incomplete cases. How to find all possible cases in each sub problem$?$
Any help is greatly appreciated.
| As this answer has now been accepted, I should mention that it’s unnecessarily complicated and the stars-and-bars argument provided by Daniel Mathias in a comment below is much more elegant.
You can proceed as in this nice answer by A.J. to What is the probability of 2 named cards appearing sequentially in a randomly shuffled deck if suits are ignored?.
First, to find the probability that no $X$ is followed by a $Y$, arrange the $X$s and $Z$s in some way and then insert the $Y$s. For the first insertion, there are $a+c+1$ equiprobable slots where the $Y$ can go, and $a$ of them are behind an $X$. If we ever put a $Y$ behind an $X$, it’s over (since we can’t prevent that $X$ from being followed by a $Y$ by inserting further $Y$s behind it). The probability to survive the first insertion is $\frac{c+1}{a+c+1}$, to survive the second insertion $\frac{c+2}{a+c+2}$, and so on. Thus, the probability that no $X$ is followed by a $Y$ is
$$
\frac{c+1}{a+c+1}\cdot\frac{c+2}{a+c+2}\cdots\frac{c+b}{a+c+b}=\frac{(c+b)!(a+c)!}{c!(a+c+b)!}\;.
$$
For the probability that exactly one $X$ is followed by a $Y$, we need to replace one of the factors by its complement, and the numerator of all factors after that is increased by $1$ because we can now put $Y$s behind the $X$ that we put a $Y$ behind. Thus, we get an additional factor $c+b+1$ in the numerator, and if the $k$-th $Y$ is inserted after an $X$ the factors $c+k$ and $c+k+1$ are replaced by $a$. Thus the probability that exactly one $X$ is followed by a $Y$ is
$$
\frac{(c+b+1)!(a+c)!}{c!(a+c+b)!}\cdot a\left(\frac1{(c+1)(c+2)}+\frac1{(c+2)(c+3)}+\cdots+\frac1{(c+b)(c+b+1)}\right)\;.
$$
The sum in parentheses telescopes because
$$\frac1{(c+k)(c+k+1)}=\frac1{c+k}-\frac1{c+k+1}\;,$$
so the sum is $\frac1{c+1}-\frac1{c+b+1}=\frac b{(c+1)(c+b+1)}$, for a probability
$$
ab\cdot\frac{(c+b)!(a+c)!}{(c+1)!(a+c+b)!}\;.
$$
Subtracting these two from $1$ yields the probability that at least two $X$s are followed by a $Y$ as
$$
1-\frac{(c+b)!(a+c)!}{c!(a+c+b)!}\left(1+\frac{ab}{c+1}\right)\;.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Valid proof for integral of $1/(x^2+a^2)$ I'm trying to prove some integral table formulae and had a concern over my proof of the following formula:
$$\int\frac{1}{x^2+a^2}\;dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C$$
Claim:
$$\frac{1}{x^2+a^2}=\frac{1}{a^2}\sum_{k=1}^\infty(-1)^{k-1}\left(\frac{x}{a}\right)^{2k}\hspace{5mm}\forall\;x\in(-a,a)$$
Proof:
$$\frac{1}{x^2+a^2}=\frac{1}{a^2}-\frac{x^2}{a^4}+\frac{x^4}{a^6}-\frac{x^6}{a^8}+\dots$$
$$\implies 1=(x^2+a^2)\left(\frac{1}{a^2}-\frac{x^2}{a^4}+\frac{x^4}{a^6}-\frac{x^6}{a^8}+\dots\right)$$
$$\implies 1=\left(\frac{x^2}{a^2}-\frac{x^4}{a^4}+\frac{x^6}{a^6}-\frac{x^8}{a^8}+\dots\right)+\left(1-\frac{x^2}{a^2}+\frac{x^4}{a^4}-\frac{x^6}{a^6}+\dots\right)$$
$$\implies 1=1\hspace{5mm}\forall\;x\in(-a,a)$$
My proof then uses the following:
$$\int\frac{1}{x^2+a^2}\;dx=\frac{1}{a^2}\int\sum_{k=1}^\infty(-1)^{k-1}\left(\frac{x}{a}\right)^{2k}\;dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C$$
My concern is that this isn't a valid proof since the radius of convergence of arctangent's Taylor Series is finite. I'm 7 years removed from taking calculus so I'm admittedly forgetful of the fine details on this. Could someone explain if this is a valid approach or, if not, why?
| It’s valid for $|x|\leq 1$ due to the uniform convergence of the Taylor series you obtain after integrating to arctan, but not outside this interval and has to do with not being able to pull that sum outside the integral on this region. I can give you a simpler way to integrate this if you’d like?
| {
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"timestamp": "2023-03-29T00:00:00",
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Difficulties with estimation of epsilon-delta limit proof I have to proof $\lim_{x\to 5}\frac{4x-9}{3x-16}=-11$. I have hard time to evaluate $\frac{1}{|3x-16|}$.
So I start that let $|x-5|<$. I need to show that $|\frac{4x-9}{3x-16}-(-11)|<ε$.
$|\frac{4x-9}{3x-16}-(-11)|$ = $|\frac{37x-185}{3x-16}|$ = $\frac{5|x-5|}{|3x-16|}$ < $ε$.
I choose $:=1$, then $|x-5|<1$, where I get $-1 < x-5 < 1$, then multiply with 3 and subtract -1, so $-4 < 3x-16 < 2$. If I think about it as $\frac{-1}{4} > \frac{1}{3x-16} > \frac{1}{2}$ contradiction.
Is it wrong to continue the way that $ |\frac{1}{3x-16}| < \frac{1}{2}$ ? So I get that $\frac{5|x-5|}{|3x-16|}$< $5 * * \frac{1}{2}$ <= $ε $ if $ = \frac{2}{5}*ε$.
Finaly I got that := min {1, $\frac{2}{5}*ε$}
I would need some help with $\frac{1}{|3x-16|}$.
| A systematic advice in such a situation is to "translate the variable to $0$", i.e. here: let
$$x=5+h,$$
so that $x\to5\iff h\to0.$ If $x\ne\frac{16}3,$
$$\frac{4x-9}{3x-16}+11=\frac{37h}{-1+3h}.$$
If $h<\frac13$ then
$$\begin{align}\left|\frac{37h}{-1+3h}\right|<\varepsilon&\iff37|h|<\varepsilon(1-3h)\\&\Longleftarrow37|h|<\varepsilon(1-3|h|)\\&\iff|h|<\frac\varepsilon{37+3\varepsilon}.
\end{align}$$
A convenient $\delta$ is therefore $\min\left(\frac13,\frac\varepsilon{37+3\varepsilon}\right).$
Note that José's $\delta=\min\left(\frac\varepsilon{74},\frac16\right)$ is smaller, hence also convenient.
| {
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How can you prove that the square root of two is irrational? I have read a few proofs that $\sqrt{2}$ is irrational.
I have never, however, been able to really grasp what they were talking about.
Is there a simplified proof that $\sqrt{2}$ is irrational?
| Here's a short algebraic proof. It nowhere uses rules about primes or even numbers.
You need to first show that $1<\sqrt{2}<2,$ but that is obvious.
We first assume that $\sqrt{2}$ is rational. Then pick the smallest positive $q$ so that $p=q\sqrt{2}$ is an integer. Then $q<p<2q.$
Now compute:
$$\left(\frac{2q-p}{p-q}\right)^2 = \frac{4q^2-4pq+p^2}{p^2-2pq+q^2}=\frac{6q^2-4pq}{3q^2-2pq}=2$$
But $q<p<2q$ means $0<p-q<q$, and $\frac{2q-p}{p-q}=\sqrt{2},$ contradicting the assumption that $q$ was the least possible positive denominator.
More generally
We can prove, more generally, that if $n$ is an integer and $n^2<D<(n+1)^2$ then $\sqrt{D}$ is irrational. In the case $D=2$ we have $n=1$.
If $\sqrt{D}$ is rational, find the least positive $q$ such that there is a $p$ such that $\frac{p}{q}=\sqrt{D}$. So $p^2=Dq^2$ means $n^2q^2<p^2<(n+1)^2q^2$ and hence $nq<p<(n+1)q$.
Therefore $0<p-nq<q$.
But then:
$$\left(\frac{Dq-pn}{p-qn}\right)^2=\frac{D^2q^2-2Dpqn + p^2n^2}{p^2-2pqn+q^2n^2}=\frac{D^2q^2-2Dpqn + Dq^2n^2}{Dq^2-2pqn+q^2n^2}=D\tag{*}$$
contradicting the fact that $q$ was the smallest positive denominator for $\sqrt{D}$.
You can prove if $D\geq 0$ is an integer, then there is exactly one non-negative integer $n$ such that $n^2\leq D<(n+1)^2$. We first prove $n$ exists:
Since $(1+D)^2=D+(1+D+D^2)$, we know that $D<(1+D)^2$ and hence there is a least positive $m$ such that $D<m^2.$ We know $m\neq 0$ because $D\geq 0^2$, so $m\geq 1$. Let $n=m-1$. Then $n^2\leq D<(n+1)^2$.
Uniqueness follows from:
If $0\leq m<n$ then $1\leq m+1\leq n$ and thus $(m+1)^2\leq n^2$.
So if $n^2\leq D< (n+1)^2$ and $m^2\leq D< (m+1)^2$, then we can't have $m<n$ or we'd have $D<(m+1)^2\leq n^2\leq D$. Similarly, we can't have $m>n$. So we must have $m=n$.
Together, the above say that if $D\geq 0$ then $\sqrt{D}$ is rational if and only if $\sqrt{D}$ is an integer.
(*) The magic trick in the above computation is that If $\frac{np}{nq}=\sqrt{D}$ then $\frac{Dq}{p}=\sqrt{D}.$ And if $\frac{a}{b}=\frac{c}{d}$ then $d\neq b$ then $$\frac {a-c}{b-d}=\frac{a}{b}.$$
The expression is arrived by computing (using that $p=q\sqrt{D}):$
$$0<\left(p+q\sqrt{D}\right)\left(\sqrt{D}-n\right)=(qD-np)+(p-nq)\sqrt{D}$$
From this we see $(qD-np)^2-D(p-nq)^2=0.$
And $p-nq=q(\sqrt{D}-n)<q.$ since $0<\sqrt{D}-n<1.$
| {
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Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$
Does it converge? If so, what is its sum?
| Let's say you have a sequence of nonnegative numbers $a_1 \geq a_2 \geq \dots$ tending to zero. Then it is a theorem that the alternating sum $\sum (-1)^i a_i$ converges (not necessarily absolutely, of course).
This in particular applies to your series.
Incidentally, if you're curious why it converges to $\log(2)$ (which seems somewhat random), it's because of the Taylor series of $\log(1+x)$ while letting $x \to 1$.
| {
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Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!
Problem:
For any natural number $n , n^3 + 2n$ is divisible by $3.$
This makes sense
Proof:
Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
$2 \times 0 = 0.$ So it is divisible by $3.$
Induction: Assume that for an arbitrary natural number $n$,
$n^3+ 2n$ is divisible by $3.$
Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use
the induction hypothesis. Got it
$$( n + 1 )^3+ 2( n + 1 ) = ( n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}$$
$$ = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying
and regrouping}\}$$
$$ = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out
the 3}\}$$
which is divisible by $3$, because $(n^3 + 2n )$ is divisible by $3$
by the induction hypothesis. What?
Can someone explain that last part? I don't see how you can claim $(n^3+ 2n ) + 3( n^2 + n + 1 )$ is divisible by $3.$
| We can take three cases (it's worth reading up on this by the way, the idea is called 'modular arithmetic)
$n≡0$ mod $3$
$n≡1$ mod $3$
$n≡2$ mod $3$
In the first case we are immmediately done because $n^3+2n=n(n^2+2)$ and since n factors out, by assumption we are done.
In the second and thirdcase we will substitute $'n'$ by it's respective residue in the equation and hope that the outcome will equal $0 mod 3$
$n≡1$ mod $3$ gives us $1^3$ mod $3$ $+ 1*2$ mod $3$ $≡ 1+2$ mod $3$ $≡ 0$ mod $3$
$n≡2$ mod $3$ gives us $2^3$ mod $3$ $+ 2*2$ mod $3$ $≡ 8+4$ mod $3$ $≡ 12$ mod $3$ $≡ 0$ mod $3$
| {
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Find the coordinates in an isosceles triangle Given:
$A = (0,0)$
$B = (0,-10)$
$AB = AC$
Using the angle between $AB$ and $AC$, how are the coordinates at C calculated?
| Let $a,b$ and $c$ be the side lengths and $A,B$ and $C$ the angles.
$a^{2}=x^{2}+\left( y+10\right) ^{2}$
$b^{2}=x^{2}+y^{2}=10^{2}$
$b=c=10$
By the (Neper) theorem of tangents (corollary of the Law of tangents):
$\tan \frac{A-B}{2}=\frac{a-b}{a+b}\cot \frac{C}{2}$
On the other hand
$\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}\quad C<\pi $
and by the theorem of sinus
$c\sin A=a\sin C\iff \left( x^{2}+\left( y+10\right) ^{2}\right) \sin
C=10\sin A$
Compiling, we get:
$\frac{A-B}{2}=\arctan (\frac{\sqrt{x^{2}+\left( y+10\right) ^{2}}-10}{\sqrt{%
x^{2}+\left( y+10\right) ^{2}}+10}\cot \frac{C}{2})$
$\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}\quad C<\pi $
$(x^{2}+(\sqrt{100-x^{2}}+10)^{2})\sin C=10\sin A$
$y^{2}=10^{2}-x^{2}$
We have to solve this system of four equations and four unknowns $x,y,A,B$.
Edit: I started this approach before the question has been updated.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding the $N$-th derivative of $f(x)=\frac {x} {x^2-1}$ I'm practicing some problems from past exams and found this one:
Find the n-th derivative of this function:
$$f(x)=\frac {x} {x^2-1}$$
I have no idea how to start solving this problems. Is there any theorem for finding nth derivative?
| Maybe we can add some more help -just in case you didn't succeed to find the answer yourself yet. Let
$$
\frac{x}{x^2 - 1} = \frac{A}{x-1} + \frac{B}{x+1}
$$
be the splitting into partial fractions. (I'm too lazy to compute the coeffitients $A$ and $B$.) Then
$$
\frac{d}{dx} \frac{x}{x^2 - 1} = -\frac{A}{(x-1)^2} - \frac{B}{(x+1)^2} \ .
$$
Differentiating again,
$$
\frac{d^2}{dx^2}\frac{x}{x^2 - 1} = \frac{2A}{(x-1)^3} + \frac{2B}{(x+1)^3} \ .
$$
One more time:
$$
\frac{d^3}{dx^3} \frac{x}{x^2 - 1} = - \frac{3\cdot 2 A}{(x-1)^4} - \frac{3\cdot 2 B}{(x+1)^4} \ .
$$
And sure enough you can find the general pattern now, can't you? Then, use induction to prove your guess.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Simplification of expressions containing radicals As an example, consider the polynomial $f(x) = x^3 + x - 2 = (x - 1)(x^2 + x + 2)$ which clearly has a root $x = 1$.
But we can also find the roots using Cardano's method, which leads to
$$x = \sqrt[3]{\sqrt{28/27} + 1} - \sqrt[3]{\sqrt{28/27} - 1}$$
and two other roots.
It's easy to check numerically that this expression is really equal to $1$, but is there a way to derive it algebraically which isn't equivalent to showing that this expression satisfies $f(x) = 0$?
| Yes. The first thing to try is to guess that $\sqrt[3]{ \left( \sqrt{ \frac{28}{27} } \pm 1 \right) } = \pm \frac{1}{2} + \sqrt{a}$ for some $a$. Cubing both sides then gives
$$\frac{2}{9} \sqrt{21} \pm 1 = \pm \frac{1}{8} + \frac{3}{4} \sqrt{a} \pm \frac{3}{2} a + a \sqrt{a}.$$
Setting $1 = \frac{1}{8} + \frac{3a}{2}$ gives $a = \frac{7}{12}$, and we can verify that
$$\frac{3}{4} \sqrt{a} + a \sqrt{a} = \frac{1}{8} \sqrt{21} + \frac{7}{72} \sqrt{21} = \frac{2}{9} \sqrt{21}$$
as desired. If this method doesn't work then the problem becomes harder.
| {
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"url": "https://math.stackexchange.com/questions/4680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
| Another proof i have (re?)discovered.
I want to prove that,
$\displaystyle J:=\int_0^1 \frac{\ln(1+x)}{x}dx=\frac{\pi^2}{12}$
Let $f$, be a function, such that, for $s\in[0;1]$,
$\displaystyle f(s)=\int_0^{\frac{\pi}{2}} \arctan\left(\frac{\sin t}{\cos t+s}\right)\,dt$
Observe that,
$\begin{align} f(0)&=\int_0^{\frac{\pi}{2}}\arctan\left(\frac{\sin t}{\cos t}\right)\,dt\\
&=\int_0^{\frac{\pi}{2}} t\,dt\\
&=\left[\frac{t^2}{2}\right]_0^{\frac{\pi}{2}}\\
&=\frac{\pi^2}{8}
\end{align}$
For $t$ in $\left[0,\frac{\pi}{2}\right]$,
$\begin{align} \frac{\sin t}{\cos t+1}&=\frac{2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)}{\cos^2\left(\frac{t}{2}\right)-\sin^2\left(\frac{t}{2}\right)+1}\\
&=\frac{2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)}{2\cos^2\left(\frac{t}{2}\right)}\\
&=\tan\left(\frac{t}{2}\right)
\end{align}$
Therefore,
$\begin{align} f(1)&=\int_0^{\frac{\pi}{2}}\arctan\left(\frac{\sin t}{\cos t+1}\right)\,dt\\
&=\int_0^{\frac{\pi}{2}}\arctan\left(\tan\left(\frac{t}{2}\right)\right)\,dt\\
&=\int_0^{\frac{\pi}{2}} \frac{t}{2}\,dt\\
&=\left[\frac{t^2}{4}\right]_0^{\frac{\pi}{2}}\\
&=\frac{\pi^2}{16}
\end{align}$
For $s$ in $[0,1]$,
$\begin{align}
f^\prime(s)&=-\int_0^{\frac{\pi}{2}}\frac{\sin t}{1+2s\cos t+s^2}\,dt\\
&=\left[\frac{\ln(1+2s\cos t+s^2)}{2s}\right]_0^{\frac{\pi}{2}}\\
&=\frac{1}{2}\frac{\ln\left(1+s^2\right)}{s}-\frac{\ln\left(1+s\right)}{s}
\end{align}$
Therefore,
$\begin{align}
f(1)-f(0)&=\int_0^1 f^\prime(s)ds\\
&=\frac{1}{2}\int_0^1\frac{\ln\left(1+s^2\right)}{s}\,ds-\int_0^1 \frac{\ln\left(1+s\right)}{s}\,ds\\
\end{align}$
In the first integral perform the change of variable $y=s^2$, therefore,
$\displaystyle f(1)-f(0)=-\frac{3}{4}J$
But,
$\begin{align} f(1)-f(0)&=\frac{\pi^2}{16}-\frac{\pi^2}{8}\\
&=-\frac{\pi^2}{16}
\end{align}$
Therefore,
$\boxed{\displaystyle J=\frac{\pi^2}{12}}$
PS:
To obtain the value of $J$ knowing that $\displaystyle \zeta(2)=-\int_0^1 \frac{\ln(1-x)}{x}dx$
$\begin{align}
\int_0^1 \frac{\ln(1+t)}{t}\,dt+\int_0^1 \frac{\ln(1-t)}{t}\,dt=\int_0^1 \frac{\ln(1-t^2)}{t}\,dt
\end{align}$
Perform the change of variable $y=t^2$ in RHS integral,
$\begin{align}
\int_0^1 \frac{\ln(1+t)}{t}\,dt+\int_0^1 \frac{\ln(1-t)}{t}\,dt=\frac{1}{2}\int_0^1 \frac{\ln(1-t)}{t}\,dt
\end{align}$
Therefore,
$\begin{align}
\int_0^1 \frac{\ln(1+t)}{t}\,dt=-\frac{1}{2}\int_0^1 \frac{\ln(1-t)}{t}\,dt
\end{align}$
$\boxed{\displaystyle \int_0^1 \frac{\ln(1+t)}{t}\,dt=\frac{1}{2}\zeta(2)}$
| {
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"url": "https://math.stackexchange.com/questions/8337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "814",
"answer_count": 48,
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} |
Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However, doing the following gets a completely different answer:
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{\sin x}{\sin^2(x)\cos x} dx\\
&=&\int \frac{\sin x}{(1-\cos^2(x))\cos x} dx.
\end{eqnarray*}
let $u=\cos x, du=-\sin x dx$; then
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{-1}{(1-u^2)u} du\\
&=&\int \frac{-1}{(1+u)(1-u)u}du\\
&=&\int \left(\frac{-1}{u} - \frac{1}{2(1-u)} + \frac{1}{2(1+u)}\right) du\\
&=&-\ln|\cos x|+\frac{1}{2}\ln|1-\cos x|+\frac{1}{2}\ln|1+\cos x|+C
\end{eqnarray*}
I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?
| If I take the derivative of your second answer (call it $g(x)$), I get:
\begin{eqnarray*}
\frac{dg}{dx}
& = & -\frac{-\sin x}{\cos x} + \frac{\sin x}{2(1-\cos x)} + \frac{-\sin x}{2(1+\cos x)}\\
& = & \frac{\sin x\left(1-\cos^2 x + \frac{1}{2}\cos x(1+\cos x) - \frac{1}{2}\cos x(1-\cos x)\right)}{\cos x(1-\cos x)(1+\cos x)}\\
& = & \frac{\sin x\left( 1- \cos^2 x + \frac{1}{2}\cos x + \frac{1}{2}\cos^2 x - \frac{1}{2}\cos x + \frac{1}{2}\cos^2 x\right)}{\cos x(1-\cos^2 x)}\\
& = & \frac{\sin x}{\cos x\>\sin^2 x} = \frac{1}{\cos x\sin x}.
\end{eqnarray*}
So I'm not sure why Mathematica says the second method is not "the right answer".
| {
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"timestamp": "2023-03-29T00:00:00",
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function asymptotic where $f(x) = \frac{a + O(\frac{1}{\sqrt{x}})}{b + O(\frac{1}{\sqrt{x}})}$ If $a$ and $b$ are positive real numbers, and if $f(x)$ has the following asymptotic property
$f(x) = \frac{a + O(\frac{1}{\sqrt{x}})}{b + O(\frac{1}{\sqrt{x}})}$
then is the following true?
$f(x) = \frac{a}{b} + O(\frac{1}{\sqrt{x}})$
This might look like homework but it isn't.
| Yes. One way to see this is to actually do the long division (like the kind you learned in elementary school)! Unfortunately, typesetting that in full on this forum will overtax my LaTeX powers.
Anyway, dividing $b + O\left(\frac{1}{\sqrt{x}}\right)$ into $a + O\left(\frac{1}{\sqrt{x}}\right)$ yields $\frac{a}{b}$ with a remainder of $O\left(\frac{1}{\sqrt{x}}\right)$. So we have
$$\frac{a + O\left(\frac{1}{\sqrt{x}}\right)}{b + O\left(\frac{1}{\sqrt{x}}\right)} = \frac{a}{b} + \frac{O\left(\frac{1}{\sqrt{x}}\right)}{b + O\left(\frac{1}{\sqrt{x}}\right)} = \frac{a}{b} + O\left(\frac{1}{\sqrt{x}}\right),$$
since $b + O\left(\frac{1}{\sqrt{x}}\right) = O(1).$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Algebraic Identity $a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}$ Prove the following: $\displaystyle a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}$.
So one could use induction on $n$? Could one also use trichotomy or some type of combinatorial argument?
| Can we build a combinatorial argument along these lines?
Say if we have say $n$ students to be allotted in $a$ rooms with $b$ of the $a$ room being non air-conditioned. (Assume the students are distinguishable so we can order the student as $1,2,3,...,n$)
The total number of ways is $a^n$.
Suppose all students are allotted to the $b$ rooms, the number of ways is $b^n$.
If the first $n-1$ students are allotted to the $b$ rooms, and the final dude in some other room, the number of possible ways is $b^{n-1} \times (a-b)$.
Now if the first $n-2$ students are allotted to the $b$ rooms, and the remaining two students are now left. If the $(n-1)^{th}$ student chooses from the $b$ rooms then we are back to the earlier case. So the $(n-1)^{th}$ student needs to choose from the remaining $(a-b)$ rooms. Now the $n^{th}$ student can choose from any of the $a$ rooms. The number of possible ways is $b^{n-2} \times (a-b) \times a$.
In general, if the first $n-k$ students are allotted to the $b$ rooms, and the remaining $k$ students are now left. If the $(n-k+1)^{th}$ student chooses from the $b$ rooms then we are back to the previous case. So the $(n-k+1)^{th}$ student needs to choose from the remaining $(a-b)$ rooms. Now the students from $(n-k+2)$ to $n$ can choose from any of the $a$ rooms. The number of possible ways is $b^{n-k} \times (a-b) \times a^{k-1}$.
So, the total number of ways is $b^n + \displaystyle \sum_{k=1}^n b^{n-k} \times (a-b) \times a^{k-1}$.
Both the counting must add up and hence we get $a^n = b^n + \displaystyle \sum_{k=1}^n b^{n-k} \times (a-b) \times a^{k-1}$.
You could use geometric series to conclude the result as well.
The right hand side is $(a-b) b^{n-1} \displaystyle \sum_{k=0}^{n-1} (\frac{a}{b})^k = (a-b) b^{n-1} \frac{((\frac{a}{b})^n - 1)}{(\frac{a}{b})-1} = (a-b) \frac{a^n - b^n}{a-b} = a^n - b^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/11618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Limit of $S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$ - Part II This is a follow up of Limit of $S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$
More details can be found in the above thread.
Let $S(n) = \displaystyle \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$
Mike has proved that $S(n)$ in fact diverges at-least faster than $\log_2(\lfloor n-1 \rfloor)$.
Now based on what Mike has worked this conjectures arises:
$\displaystyle \lim_{n \rightarrow \infty} (2 \log_{2}(n) - S(n)) = \alpha$.
Also, can $\alpha$ be expressed in terms of other familiar constants. $\frac{\pi \gamma}{e}$ seems to be a close guess.
The numerical evidence seem to suggest they are true. For example, we have the following graph of $2 \log_2 n - S(n)$ for $n \leq 300$.
(More numerical evidence: The value of $2 \log_2 n - S(n)$, is, for $n = 1000$, $2000$, and $3000$, respectively, $0.667734$, $0.667494$, and $0.667413$.)
An alternative expression for $S(n)$ was worked out by Moron in the previously-mentioned question:
$$S(n) = - \sum_{k=1}^{n-1} \frac{s(n,k)}{2^{n-k}-1},$$
where $s(n,k)$ is a Stirling number of the first kind.
| Here's an argument that pushes the lower bound for $S(n)$ closer to a factor of $2$ times
$\log_2 n.$ More precisely, we show
$$ S(n) \ge \left( 2 - \frac{1}{e} \right) \left( \lfloor \log_2 n \rfloor – 1 \right). $$
Let $ a= \lfloor \log_2 n \rfloor $ and write
$ f(n,x) = 1 - \prod_{j=1}^{n-1} ( 1 – jx),$
and so
$$ S(n) = \sum_{k=1}^\infty \left( 1 - \prod_{j=1}^{n-1} \left( 1 - \frac{j}{2^k} \right) \right)$$
$$ = \sum_{k=1}^\infty f(n, \frac{1}{2^k} ) =
a-1 + \sum_{k=a}^\infty f(n, \frac{1}{2^k} ),$$
since for each $k \le a-1$ $\exists$ $j=2^k$ in $\prod_{j=1}^{n-1} ( 1 - j/2^k ),$ and thus this product is $0.$
So neglecting terms with $k \ge 2a-1,$ we have
$$S(n) \ge a-1 + \sum_{k=a}^{2a-2} f(n, \frac{1}{2^k} ). \qquad (1)$$
By the AM-GM for $x \le 1/(m-1)$ we have for $m \ge 2$ and $2/m(m-1) \le x \le 1/(m-1)$
$$(1-x)(1-2x) \cdots (1-(m-1)x) \le
\left( 1 - \frac{mx}{2} \right)^{m-1} \le
\left( 1 - \frac{1}{m-1} \right)^{m-1} < \frac{1}{e}. $$
Thus for $k=a, a+1,\ldots, 2a-2$ we have
$$ f(n, \frac{1}{2^k} ) \ge 1 - \frac{1}{e}$$
and substituting this into $(1)$ the result follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Suggest a tricky method for this problem Find the value of:
$$ 1+ \biggl(\frac{1}{10}\biggr)^2 + \frac{1 \cdot 3}{1 \cdot 2} \biggl(\frac{1}{10}\biggr)^4 + \frac{1\cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \biggl(\frac{1}{10}\biggr)^6 + \cdots $$
| Note that we can rewrite the series as
$$ \displaystyle\sum\limits_{n=0}^\infty \frac{(2n)!}{2^n(n!)^2} \cdot \bigg(\frac{1}{10}\bigg)^{2n}$$
which is exactly
$$ \displaystyle\sum\limits_{n=0}^\infty \frac{(2n)!}{2^n(n!)^2} \cdot x^{2n}$$
evaluated at $x = \frac{1}{10}$. It is easy to see that this sum has radius of convergence $\frac{1}{\sqrt{2}}$, and so converges absolutely for $x = \frac{1}{10}$.
Also note that this sum is exactly the Taylor series for $\frac{1}{\sqrt{1-2x^2}}$ centered at $x = 0$. Thus
$$ \displaystyle\sum\limits_{n=0}^\infty \frac{(2n)!}{2^n(n!)^2} \cdot \bigg(\frac{1}{10}\bigg)^{2n} = \frac{10}{7\sqrt{2}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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In every power of 3 the tens digit is an even number How to prove that in every power of $3$, with natural exponent, the tens digit
is an even number?
For example, $3 ^ 5 = 243$ and $4$ is even.
| A slightly different approach:
Let $n$ be the exponent of the variable power of $3$. Therefore, the power of $3$ takes the form $3^n$. Now, $n$ can either be even or odd. Hence two cases are formed:-
Case1: $n$ is even.
Since $n$ is even, let us say it is of the form $2k$.
Therefore, $3^n = 3^{2k} = (3^k)^2$. This means that is $n$ if even, $3^n$ is a square. Notice that the even powers of $3$ result in a number with the last digit $\in \{1,9\}$
And $(2m + 1)^2 \equiv 1(mod 4)$
$\Rightarrow 3^n \equiv 1(mod4)$
$\Rightarrow 3^n -1 \equiv 0(mod4)$
$\Rightarrow 4 |(3^n-1)$.
We know that, if $4$ divides a number, it must divide $(10($$the$ $second$ $last$ $digit$$) + $$the$ $last$ $digit$$)$
Therefore, if $4 |(3^n-1)$,$ $ $ $ $ $ $ $ $ $ $ $ $4|(10($$the$ $second$ $last$ $digit$$) + $$the$ $last$ $digit$$)-1$
$\Rightarrow 4|(10($$the$ $second$ $last$ $digit$$) + 1-1)$ and $4|(10($$the$ $second$ $last$ $digit$$) + 9-1)$.
$\Rightarrow 4|(10($$the$ $second$ $last$ $digit$$) + 0)$ and $4|(10($$the$ $second$ $last$ $digit$$) + 8)$.
Since, $4| 0$ and $4| 8$, $4|(10($$the$ $second$ $last$ $digit$$)$
This is true only if $the$ $second$ $last$ $digit$ is even.
Case2: $n$ is odd.
Since $n$ is odd, let us say it is of the form $2k+1$.
Therefore, $3^n = 3^{2k+1} = (3^k)^2.(3)$. This means that if $n$ is odd, $3^n$ is a $($square $* 3)$. Notice that the odd powers of $3$ result in a number with the last digit $\in \{3,7\}$
And $(2m + 1)^2 \equiv 1(mod 4)$
$\Rightarrow (2m + 1)^{2p+1} \equiv (2m + 1)(mod4)$
$\Rightarrow 3^n \equiv 3(mod4)$
$\Rightarrow 3^n -3 \equiv 0(mod4)$
$\Rightarrow 4 |(3^n-3)$.
We know that, if $4$ divides a number, it must divide $(10($$the$ $second$ $last$ $digit$$) + $$the$ $last$ $digit$$)$
Therefore, if $4 |(3^n-3)$,$ $ $ $ $ $ $ $ $ $ $ $ $4|(10($$the$ $second$ $last$ $digit$$) + $$the$ $last$ $digit$$)-3$
$\Rightarrow 4|(10($$the$ $second$ $last$ $digit$$) + 3-3)$ and $4|(10($$the$ $second$ $last$ $digit$$) + 7-3)$.
$\Rightarrow 4|(10($$the$ $second$ $last$ $digit$$) + 0)$ and $4|(10($$the$ $second$ $last$ $digit$$) + 4)$.
Since, $4| 0$ and $4| 4$, $4|(10($$the$ $second$ $last$ $digit$$)$
This is true only if $the$ $second$ $last$ $digit$ is even.
Therefore, the tens digit of any power of $3$ is always even.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
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Golden Number Theory The Gaussian $\mathbb{Z}[i]$ and Eisenstein $\mathbb{Z}[\omega]$ integers have been used to solve some diophantine equations. I have never seen any examples of the golden integers $\mathbb{Z}[\varphi]$ used in number theory though. If anyone happens to know some equations we can apply this in and how it's done I would greatly appreciate it!
| Formally the Riemann zeta-function can be expressed as
$$ \zeta(z)=\prod_{k=0}^{\infty}\;\;\prod_{p\in \mathbb{P}}\bigg\{\left( 1 -\varphi^{-1}\;p^{-5^{k}z} +p^{-2\cdot5^{k}z}\right)\left( 1 +\varphi\;p^{-5^{k}z} +p^{-2\cdot5^{k}z}\right)\bigg\} \;for\;z>1$$
where
$
\varphi=\frac{1+\sqrt{5}}{2}
$
is the Golden Ratio. This follows from the fact that the zeta function can be expressed as
$$ \zeta(z)=\prod_{p\in \mathbb{P}}\;\;\sum_{k=0}^{\infty}\frac{1}{p^{k\;z}}$$
and after some manipulations its easy to get the above representation.
To justify this, take the series
$$
f(x)=1+x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}+x^{7}+x^{8}+x^{9}+x^{10}+x^{11}+x^{12}+\cdots
$$
we can write this as
$$
f(x)=1+x+x^{2}+x^{3}+x^{4}+x^{5}(1+x+x^{2}+x^{3}+x^{4})+x^{10}(1+x+x^{2}+x^{3}+x^{4})+\cdots
$$
or equivalently
$$
f(x)=(1+x+x^{2}+x^{3}+x^{4})(1+x^{5}+x^{10}+x^{15}+x^{20}+x^{25}+x^{30}+x^{35}\cdots)
$$
and again
$$
f(x)=(1+x+x^{2}+x^{3}+x^{4})(1+x^{5}+x^{10}+x^{15}+x^{20}+x^{25}(1+x^{5}+x^{10}+x^{15}+x^{20})+x^{25}(1+x^{5}+x^{10}+x^{15}+x^{20})+\cdots)
$$
or
$$
f(x)=(1+x+x^{2}+x^{3}+x^{4})(1+x^{5}+x^{10}+x^{15}+x^{20})(1+x^{25}+x^{50}+x^{75}+x^{100}+\cdots)
$$
so the general pattern is
$$
f(x)=\prod_{k=0}^{\infty}(1+x^{1\cdot 5^{k}}+x^{2\cdot 5^{k}}+x^{3\cdot 5^{k}}+x^{4\cdot 5^{k}})
$$
now make $y=x^{5^{k}}$, one has that
$$
1+y+y^{2}+y^{3}+y^{4}=\left(y^{2}-\frac{\sqrt{5}-1}{2}y+1\right)\left(y^{2}+\frac{\sqrt{5}+1}{2}y+1\right)
$$
where $\varphi=\frac{\sqrt{5}+1}{2}$ and $\frac{1}{\varphi}=\frac{\sqrt{5}-1}{2}$
so
$$
1+y+y^{2}+y^{3}+y^{4}=\left(y^{2}-\frac{1}{\varphi}y+1\right)\left(y^{2}+\varphi y+1\right)
$$
now remember that
$$
\zeta(s)=\prod_{p \in \mathbb{P}}\left(1+\frac{1}{p^{s}}+\frac{1}{p^{2s}}+\frac{1}{p^{3s}}+\frac{1}{p^{4s}}+\cdots \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/18589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "61",
"answer_count": 4,
"answer_id": 2
} |
How to deduce trigonometric formulae like $2 \cos(\theta)^{2}=\cos(2\theta) +1$? Very important in integrating things like $\int \cos^{2}(\theta) d\theta$ but it is hard for me to remember them. So how do you deduce this type of formulae? If I can remember right, there was some $e^{\theta i}=\cos(\theta)+i \sin(\theta)$ trick where you took $e^{2 i \theta}$ and $e^{-2 i \theta}$. While I am drafting, I want your ways to remember/deduce things (hopefully fast).
[About replies]
*
*About TPV's suggestion, how do you attack it geometrically??
$\cos^{2}(x) - \sin^{2}(x)=\cos(2x)$
plus $2\sin^{2}(x)$, then
$\cos^{2}(x)+\sin^{2}(x)=\cos(2x)+2\sin^{2}(x)$
and now solve homogenous equations such that LHS=A and RHS=B, where $A\in\mathbb{R}$ and $B\in\mathbb{R}$. What can we deduce from their solutions?
| For Proving $\sin(\alpha+\beta)=\sin\alpha\cdot \cos\beta + \cos\alpha \cdot \sin\beta$ you can see this link:
*
*http://www.math.wisc.edu/~leili/teaching/math222s11/problems/quizzes/trig.pdf
By the above by substituting $\beta=\alpha$ you have
*
*$\sin{2\alpha} = \sin\alpha \cdot \cos\alpha + \cos\alpha \cdot \sin\alpha =2 \sin\alpha \cos\alpha$
*$\sin{2\alpha} = \frac{2 \tan\alpha}{1+\tan^{2}\alpha} = \displaystyle 2\tan{\alpha}\cdot\cos^{2}\alpha=2\sin\alpha \cdot \cos\alpha$
We have $\cos(\alpha + \beta) = \cos\alpha\cdot \cos\beta - \sin\alpha \cdot \sin\beta$. Using this you can obtain the following:
*
*$\cos{2 \alpha} = \cos\alpha\cdot \cos\alpha - \sin\alpha\cdot \sin \alpha =\cos^{2} \alpha - \sin^{2} \alpha$
*$\cos{2 \alpha}= 1 - 2 \sin^{2}\alpha \qquad \Bigl[ \because \ \sin^{2}\theta = 1 - \cos^{2}\theta \Bigr]$
*$ \cos{2 \alpha}=2 \cos^{2}\alpha - 1$
*$\cos{2 \alpha}=\displaystyle \cos^{2}\alpha \cdot \Bigl[1-\frac{\sin^{2}\alpha}{\cos^{2}\alpha}\Bigr]=\displaystyle \frac{1-\tan^{2}\alpha}{1+\tan^{2}\alpha} \ \Bigl[\because \ \cos^{2}\alpha = \frac{1}{\sec^{2}\alpha}\Bigr]$.
Using the above formulas for $\sin(\alpha + \beta)$ and $\cos(\alpha + \beta)$ we have $$\tan(\alpha+\beta) = \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} = \frac{\sin\alpha\cdot \cos\beta + \cos\alpha \cdot \sin\beta}{\cos\alpha\cdot \cos\beta - \sin\alpha\cdot \sin\beta}$$
Dividing the last term by $\cos\alpha \cdot \cos\beta$ gives, $$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan \beta}{1 - \tan\alpha \cdot \tan\beta}$$
*
*From this if you put $\beta = \alpha$ you get $$\tan{2\alpha} = \frac{\tan\alpha + \tan\alpha}{1 - \tan\alpha \cdot \tan\alpha} = \frac{2 \tan \alpha}{1 - \tan^{2}\alpha}$$
Now moving on to formulas for $3\alpha$.
\begin{align*}
\sin{3\alpha} &= \sin(2\alpha + \alpha) = \sin(2\alpha)\cdot\cos{\alpha} + \cos(2\alpha)\cdot\sin\alpha \\ &= 2\sin\alpha\cdot \cos^{2}\alpha + (\cos^{2}\alpha - \sin^{2}\alpha) \cdot \sin\alpha \\ &= 2\sin\alpha \cdot (1-\sin^{2}\alpha) + (1-\sin^{2}\alpha)\sin\alpha - \sin^{3}\alpha \\ &= 3\sin{\alpha} - 4 \sin^{3}\alpha
\end{align*}
Again proceed similarly for $\cos{3\alpha}$ and convert all the $\sin^{2}\alpha$'s which appear in the expression to $(1-\sin^{2}\alpha)$. Then you get the value as:
*
*$\cos(3\alpha) = 4\cos^{3}\alpha - 3 \cos\alpha$.
*$\displaystyle \tan{3\alpha} = \frac{3\tan\alpha - \tan^{2}\alpha}{1-3\tan^{2}\alpha}$. Look here
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/19876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
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