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Prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$ I'm trying to prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$
Suposse that $\sqrt{3}=a+b\sqrt{2}$
$\begin{align*}
\sqrt{3}&=a+b\sqrt{2}\\
3&=(a+b\sqrt{2})^2\\
3&=a^2+2\sqrt{2}ab+b^2\\
(3-a^2-12b^2)^2&=(2\sqrt{2}ab)^2\\
9-6a^2-12b^2+4a^2b^2+a^4+4b^4&=8a^2b^2
\end{align*}$
But I don't know what else I can do here.
| You can actually continue from your third line by using that $\sqrt 2$ is irrational so if $$a^2+b^2+2\sqrt 2ab=3,$$ then actually $$a^2+b^2=3 \text{
and } 2\sqrt 2ab=0$$ so that either $a$ is zero or $b$ is $0$. Since we can't solve $a^2=3$ or $b^2=3$ in the rationals, you can conclude by using that $\sqrt 3$ is irrational .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2489947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find local maximum, minimum and saddle points of $f(x,y) = x^4 + y^4 - 4xy + 2$
Find maximum, minimum and saddle points of $f(x,y) = x^4 + y^4 - 4xy + 2$.
For critical points, $f_{x} = f_{y} = 0$, $f_{x} = 3x³ - 4y$ and $f_{y} = 3y³ - 4x$. Therefore $f_{xx} = 9x²$, $f_{yy} = 9y²$, and $f_{xy} = f_{yx} = -4$.
Hessian matrix determinant is positive for P($\frac{2}{\sqrt3},\frac{2}{\sqrt3},-14)$ and Q($\frac{-2}{\sqrt3}$,$\frac{-2}{\sqrt3}$,-14), so they are minimum points.
And R(0,0,2) is a saddle point.
Is this correct?
| By AM-GM $$x^4+y^4+2\geq4\sqrt[4]{x^4y^4}=4|xy|\geq4xy.$$
The equality occurs for $x=y=1$, which gives that $0$ is a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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smallest possible sum of b and c
In the set of positive integers {$2, 6, 8, 9, 11, b, c$}, the mean is larger than the unique mode, which is larger than the median. What is the smallest possible sum of $b$ and $c$?
So I just started plugging in a few values to try to see what works, but values close to 6, 8 and 9 don't seem to work, and I think there would be a much more efficient and mathematical way to see this... Any suggestions?
| First off, either $b = c$ or not. First, let's assume $b = c$. Then $b$ is the unique mode, so we must make sure that it is smaller than the mean and larger than the median. If $b \leq 9$, then the mode is not larger than the median. At the same time, the mean is equal to $\frac{2+6+8+9+11+2b}{7} = \frac{36 + 2b}{7}$ and we want this to be larger than $b$, which gives
$$
b < \frac{36+2b}{7}\\
7b<36+2b\\
5b<36\\
b<\frac{36}7<6
$$
which clearly contradicts what we found above.
So, $b$ and $c$ must be two different numbers. Then exactly one of them (say $b$) must be equal to one of the given numbers, because we want the mode to be unique. Again, we check what happens to the mode vs. the median first. The smallest we can possibly make the median is $6$, so clearly $b>6$. That leaves three possibilities:
*
*$b = 8$. Even in this case, the median is at least $b$, so it's not feasible
*$b = 9$. In this case, the median is less than $9$ iff $c$ is less than $9$. However, the mean is $\frac{2+6+8+9+9+11+c}{7} = \frac{45 + c}{7}$, and if we want this to be larger than $9$, then we get $c>18$, so this is not feasible
*$b = 11$. In this case, the mode is larger than the median no matter what happens, so we just have to choose as low a $c$ as we can to make $\frac{2+6+8+9+11+11+c}{7}<11$, which gives $c>30$.
Thus the lowest $c$ we can pick when picking the only $b$ we can pick is $31$, so $b+c$ is at least $42$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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l'Hôpital vs Other Methods Consider the first example using repeated l'Hôpital:
$$\lim_{x \rightarrow 0} \frac{x^4}{x^4+x^2} = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x^4)}{\frac{d}{dx}(x^4+x^2)} = \lim_{x \rightarrow 0} \frac{4x^3}{4x^3+2x} = ... = \lim_{x \rightarrow 0}\frac{\frac{d}{dx}(24x)}{\frac{d}{dx}(24x)} = \frac{24}{24}=1 $$
Consider the following example using a different method:
$$ \lim_{x \rightarrow 0} \frac{x^4}{x^4+x^2} = \lim_{x \rightarrow 0}\frac{\frac{x^4}{x^4}}{\frac{x^4}{x^4}+\frac{x^2}{x^4}} = \lim_{x \rightarrow 0} \frac {1}{1 +\frac{1}{x^2}} = \frac {1}{1+\infty} = \frac{1}{\infty}=0 $$
The graph here clearly tells me the limit should be $0$, but why does l'Hôpital fail?
| $$\lim_{x \rightarrow 0} \frac{x^4}{x^4+x^2} = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x^4)}{\frac{d}{dx}(x^4+x^2)} = \lim_{x \rightarrow 0} \frac{4x^3}{4x^3+2x} = \lim_{x\to0} \frac{12x^2}{12x^2+2} = \frac{0}{0+2} = 0$$
There. You can't apply l'Hospital there because the denominator doesn't go to $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Modular arithmetic and probability $a$, $b$ and $c$ are chosen from $1$-$999$ with order and replacement, what is the probability that $a^2 + bc$ is divisible by $3$? (each choice is equally likely)
I split it up into 2 cases
$a^2$ has a remainder of $1$ while $bc$ has a remainder of $2$
$a^2$ has a remainder of $1$, $2/3$ of the time, $bc$ has a remainder of $2$ if $b$ and $c$ have $2$ and $1$. $(2/3 \times 1/3 \times 1/3)2$ (because $b$ and $c$ can be interchanged) = $4/27$
$a^2$ has a remainder of 0 while $bc$ has a remainder of $0$
$a^2$ has a remainder of $0$ $1/3$ of the time while $bc$ has a remainder of $0$ if $b$ or $c$ have remainders of $0$. probability $1/3 \times 1/3 \times 2 = 2/9$
$4/27 + 2/9 = 10/27$
Please tell me what i did wrong.
| $b$ or $c$ have remainders of $0$ with probability $\frac{5}{9}$, rather than your $\frac13 \times 2$, since you have to take account of double counting the possibility they both do: $\frac13 \times 2 -\left(\frac13\right)^2 = \frac59$
That will turn your final calculation into $\frac4{27}+\frac5{27}=\frac13$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $x^{x+y}=y^{y-x}$ in positive integers.
Let $x,y$ are positive integers. Solve the equation
$$x^{x+y}=y^{y-x}$$
smartplot(x^(x+y)=y^(y-x));
I used http://www.wolframalpha.com
| Here I will show that $y= x^{a/b}$ where a and be are two consecutive odd or even numbers:
Clearly x and y have common factors, lets suppose $y= k x$ we may write:
$y = x ^ {\frac{x+y}{x-y}}= x^{\frac{(k+1)x}{(k-1)x}}=x^{\frac{k+1}{k-1}}$
$k+1 $ and $k -1$ are two consecutive odd or even numbers:
Examples:
$k=3 ⇒ a=k+1 =4 , b =k -1= 2 ⇒ y =x^2$; solution
x=3 and y=9 is a result $
$ k=4 ⇒ a= 4+1=5, b=4-1=3 ⇒y=x^{5/3}$;solution
x=8 and y=32 is a result$
Hence equation $x^{x+y} =y^{y-x}$ is reduced to a simple form like $ y =x^{a/b}$ where a and b are two consecutive odd or even numbers. Clearly a and b can have certain values as showed in examples.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $a \sin\theta + b \cos\theta = c$ Could someone help me with the steps for solving the below equation $$a \sin\theta + b \cos\theta = c$$
I know that the solution is $$\theta = \tan^{-1} \frac{c}{^+_-\sqrt{a^2 + b^2 - c^2}} - \tan^{-1} \frac{a}{b} $$
I just can't figure out the right steps to arrive at this solution.
| Recall the sum formula for sine
$$A\sin(\theta+\alpha)=A\sin(\theta)\cos(\alpha)+A\cos(\theta)\sin(\alpha)$$
Equating this with the L.H.S gives
$$a\sin(\theta)+b\cos(\theta)=A\cos(\alpha)\sin(\theta)+A\sin(\alpha)\cos(\theta)$$
so we get the system
$$a=A\cos(\alpha)\\b=A\sin(\alpha)$$
and so $A=\sqrt{a^2+b^2}$.
Recall that we're defining some angle $\alpha$ such that both $\cos(\alpha)=\frac{a}{\sqrt{a^2+b^2}}$ and $\sin(\alpha)=\frac{b}{\sqrt{a^2+b^2}}$, so $\alpha$ is unique over one rotation and falls within a quadrant that depends on the signs of $a$ and $b$. You should be able to convince yourself that $\alpha$ takes the value
$$\alpha=\begin{cases}
\arctan(b/a), &a\gt0\\
\arctan(b/a)+\pi, &a\lt0
\end{cases}$$
So for $a>0$, we get
$$
a\sin(\theta)+b\cos(\theta)=\sqrt{a^2+b^2}\sin(\theta+\arctan(b/a))
$$
and from the identity $\sin(\theta+\pi)=-\sin(\theta)$, then for $a<0$ we get
$$
\sqrt{a^2+b^2}\sin(\theta+\arctan(b/a)+\pi)=-\sqrt{a^2+b^2}\sin(\theta+\arctan(b/a))
$$
so we can conclude that in general
$$
a\sin(\theta)+b\cos(\theta)=\frac{a}{|a|}\sqrt{a^2+b^2}\sin(\theta+\arctan(b/a))\quad (a\neq 0)
$$
For $-\sqrt{a^2+b^2}\leq c\leq\sqrt{a^2+b^2}$, we get
$$
\frac{a}{|a|}\sqrt{a^2+b^2}\sin(\theta+\arctan(b/a))=c\\
\implies\sin(\theta+\arctan(b/a))=\frac{|a|c}{a\sqrt{a^2+b^2}}\\
\implies\theta_1=\arcsin\left(\frac{|a|c}{a\sqrt{a^2+b^2}}\right)-\arctan\left(\frac{b}{a}\right),\\
\theta_2=\pi-\arcsin\left(\frac{|a|c}{a\sqrt{a^2+b^2}}\right)-\arctan\left(\frac{b}
{a}\right)
$$
Use the identity $\arcsin(x)=\arctan\left(\frac{x}{\sqrt{1-x^2}}\right)$ with $x=\frac{|a|c}{a\sqrt{a^2+b^2}}$ to get
$$
\arcsin\left(\frac{|a|c}{a\sqrt{a^2+b^2}}\right)=\arctan\left(\frac{|a|c}{a\sqrt{a^2+b^2}}\cdot\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2-c^2}}\right)\\
=\arctan\left(\frac{|a|c}{a\sqrt{a^2+b^2-c^2}}\right)
$$
and conclude that the general solution becomes
$$
\theta_1=\arctan\left(\frac{|a|c}{a\sqrt{a^2+b^2-c^2}}\right)-\arctan\left(\frac{b}{a}\right)+2\pi k_1\\
\theta_2=-\arctan\left(\frac{|a|c}{a\sqrt{a^2+b^2-c^2}}\right)-\arctan\left(\frac{b}{a}\right)+\pi(2k_2+1)
$$
where $k_1,k_2\in\mathbb{Z}$ and $a\neq 0$.
As desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Basis for the space of 2 by 3 matrices whose nullspace contains $(2,1,1)$ To find a basis for the space of $2$ by $3$ matrices whose nullspace contains $(2,1,1)$
My attempt:
Let A= \begin{bmatrix}
a & b & c \\
d & e & f \\
\end{bmatrix}
Then $(2,1,1)$ is a solution to $Ax=0$
So I get,
$$2a+b+c=0$$
$$2d+e+f=0$$
How will I proceed further?
| You can use the Gram–Schmidt process to do this. For a certain $x_1$ and $x_2$ (such that $x$, $x_1$ and $x_2$ are linearly independent) you can get
\begin{align}
y &=\begin{bmatrix}a \\ b \\ c\end{bmatrix}=x_1-x\cdot \frac{x\cdot x_1}{x \cdot x}, \\
z &=\begin{bmatrix}d \\ e \\ f\end{bmatrix}=x_2-x\cdot \frac{x\cdot x_1}{x \cdot x} -y\cdot \frac{y\cdot x_1}{y \cdot y}.
\end{align}
If you choose $x_1 = \begin{bmatrix} 0 & 6 & 0\end{bmatrix}^T$ and $x_2 = \begin{bmatrix} 0 & 0 & 30\end{bmatrix}^T$, you will end up with
$$
\begin{bmatrix} y^T \\ z^T\end{bmatrix} = \begin{bmatrix}a & b & c \\ d & e & f\end{bmatrix} = \begin{bmatrix} -2 & 5 & -1 \\ -12 & 0 & 24\end{bmatrix}.
$$
In order to find a basis for the space of $2$ by $3$ matrices whose nullspace contains $\begin{bmatrix}2&1&1\end{bmatrix}$, I write the linear equations you mentioned in matrix format:
$$
\begin{bmatrix}2 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 1\end{bmatrix}
\begin{bmatrix}a\\b\\c\\d\\e\\f\end{bmatrix}=
\begin{bmatrix}0\\0\end{bmatrix}.
$$
Now let
$$
B = \begin{bmatrix}2 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 1\end{bmatrix},
$$
then the solution to your problem is the nullspace of $B$. This can be easily found using the result above:
$$
\text{Null}(B) = \begin{bmatrix}-2 & -12 & 0 & 0 \\5 & 0 & 0 & 0\\-1 & 24 & 0 & 0 \\0 & 0 & -2 & -12 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & -1 & 24\end{bmatrix}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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find the value of ''a'' and ''b'' for which function is always increasing
Question Let f(x)= x$^{3}+ax^{2}+bx+5sin^{2}x$ be an increasing
function$\forall$ x $\in$$\mathbb{R}$$ Then$
(a) $a^2 -3b-15$<0
(b) $a^2 -3b-15$>0
(c)$a^2 -3b+15<$0
(d)a>0,b>0
My Approach i tried by making perfect square of the derivative
$\left\{ x+\frac{a}{3}\right\} ^{2}$+$\frac{b}{3}$-$\frac{a}{9}-5\geq$0
$\left\{ sin2x\geq-1\right\} $ $\Longrightarrow$a$^{2}$-3b+45$\leq$0
but it doesn't gives the required result
| Choice $(a)$ is correct.
For $a,b \in \mathbb{R},\;$let $f:\mathbb{R} \to \mathbb{R}\;$be defined by
$$f(x)= x^3+ax^2+bx+5\sin^2(x)$$
Suppose $a,b$ are such that
$\qquad{\small{\bullet}}\;\;f\;$is an increasing function.
$\qquad{\small{\bullet}}\;\;a^2-3b -15 \ge 0$.
Our goal is to derive a contradiction.
Since $f$ is increasing, it follows that $f'(x) \ge 0$, for all $x \in \mathbb{R}$.
Note that
\begin{align*}
f'(x) &= 3x^2 + 2ax + b + 10\sin(x)\cos(x)\\[4pt]
&= 3x^2 + 2ax + b + 5\sin(2x)\\[4pt]
&=p(x) + 5 + 5\sin(2x)\\[4pt]
\end{align*}
where $p(x) = 3x^2 + 2ax + b - 5$.
By the quadratic formula, $p$ has roots $r_1,r_2$ given by
\begin{align*}
r_1 &= \frac{-a - \sqrt{a^2 + 3b + 15}}{3}\\[4pt]
r_2 &= \frac{-a + \sqrt{a^2 + 3b + 15}}{3}\\[4pt]
\end{align*}
From $a^2 - 3b-15 \ge 0$, we get $a^2 - 3b + 15 \ge 30$.
It follows that $r_1,r_2$ are distinct real numbers, and $p(x) < 0$, for $x \in (r_1,r_2)$.
\begin{align*}
\text{Also,}\;\;r_2 - r_1 &=
2\left(
\frac
{\sqrt{a^2 + 3b + 15}}
{3}
\right)\\[5pt]
&\ge \frac{2\sqrt{30}}{3}\\[5pt]
& > \frac{2\sqrt{25}}{3}\\[5pt]
&=\frac{10}{3}\\[5pt]
& > \pi\\[4pt]
\end{align*}
Since $r_2-r_1 > \pi$, it follows that for some integer $n$, we have $r_1 < \frac{3\pi}{4} + n\pi < r_2$.
Then letting $t = {\large{\frac{3\pi}{4}}} + n\pi$, we get
\begin{align*}
f'(t) &= p(t) + 5 + 5\sin(2t)\\[4pt]
&=p(t) + 5 + 5\sin\left({\small{\frac{3\pi}{2}}} + 2n\pi\right)\\[4pt]
&=p(t) + 5 + 5\sin\left({\small{\frac{3\pi}{2}}}\right)\\[4pt]
&=p(t) + 5 + 5(-1)\\[4pt]
&=p(t)\\[4pt]
&< 0\\[4pt]
\end{align*}
contradiction.
It follows that choice $(a)$ is correct, as claimed.
If it's given that only one of the choices $(a),(b),(c),(d)$ is correct, then since $(a)$ is correct, we're done.
If we need to show that choice $(a)$ is the only correct choice, then we have more work to do . . .
To invalidate choices $(b)$ and $(d)$, use the values $a=0,\,b=6$.
Then for all $x \in \mathbb{R}$,
\begin{align*}
f'(x)&=3x^2 + 2ax + b + 5\sin(2x)\\[4pt]
&=3x^2 + 6 + 5\sin(2x)\\[4pt]
&\ge 3x^2 + 6 + 5(-1)\\[4pt]
&= 3x^2 +1\\[4pt]
& > 0\\[4pt]
\end{align*}
so $f$ is an increasing function.
But then
*
*Choice $(d)$ fails since$\;a,b\;$are not both positive.$\\[4pt]$
*Choice $(b)$ fails since$\;a^2 - 3b - 15 = (0)^2 - 3(6) - 15 = -33 < 0$.
To invalidate choice $(c)$, let
\begin{align*}
a=&{\small{\frac{\pi}{4}}}\\[4pt]
b=&{\small{\frac{a^2}{3}}}
+\left(
5 - {\small{\frac{1}{10}}}
\right)
\\[4pt]
\end{align*}
and verify that $f'(x) > 0\;$for all $x \in \mathbb{R}$.
The easiest verification is graphical (i.e., graph $f',\;$and see that the graph lies entirely above the $x$-axis).
It can also be verified algebraically, with a little more work.
I'll omit the algebraic verification for now, but I'll supply the (somewhat messy) details, if requested.
So for the values of $a,b,\;$as given above, $f$ is an increasing function, but
$$a^2 - 3b + 15 = {\small{\frac{3}{10}}} > 0$$
hence choice $(c)$ fails.
Therefore choice $(a)$ is the only correct choice.
| {
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"timestamp": "2023-03-29T00:00:00",
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$a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $
Let $a_n$ be a sequence satisfying $a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $
Attempt at solution using strong induction:
Base cases: $n = 1$ and $n = 2 \implies a_1 = a_2 = 1 \implies 1\le 1 \le 2$
Inductive assumption (strong induction): Assume that for all $m\in \mathbb{N}$ such that $1\le m \le k$, where $k\in \mathbb{N}$, the condition $1\le a_m \le 2$ holds True.
Show that $m = k+1$ holds true
$a_{k+1} = \frac{1}{2} \cdot (a_k + \frac{2}{a_{k-1}})$
I know that $a_k$ and $a_{k-1}$ are satisfying $ 1\le a_m\le2$ but I am not sure how to use that to prove that $a_{k+1}$ holds true for the condition.
| Hint: Try maximizing and minimizing $ (a_k + \frac{2}{a_{k-1}})$ using
(1) If $a \le c$ and $b \le d$ then $a + b \le c + d$
(2) If $a \ge c$ and $b \ge d$ then $a + b \ge c + d$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of an Exponential Equation My problem is the $\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{3x}$. I got $$\ln(B) = \ln\left(\lim_{x \rightarrow \infty} 3x \cdot \ln\left(1-\frac{4}{x}\right)\right)$$
which equates to $$\frac{\ln\left(1-\frac{4}{x}\right)}{\frac{1}{3x}} = \frac{0}{0}$$
so I can use L'Hôpital's rule
$$\frac{1}{1-\frac{4}{x}} \cdot \frac{1}{3x} - \ln\left(1-\frac{4}{x}\right) \cdot \left(-\frac{1}{3x^2}\right)$$
I'm stuck here. Am I even doing this correctly? If I am what are the next steps?
| Since $f(x)=x^{-12}$ is a continuous function in $x=e$, we obtain:
$$\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{3x}=\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{-\frac{x}{4}\cdot\left(-12\right)}=\left(\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{-\frac{x}{4}}\right)^{-12}=e^{-12}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $n(n^2-1)$ is divisible by 24, if $n$ is an odd integer greater than $2$. How can I show that $n(n^2-1)$ is divisible by 24, if $n$ is an odd integer greater than $2$?
I am thinking that since odd numbers have the form of $2n-1$ in which if it is to be more than $2$, it will be $2n-1+1 = 2n+1$. So would it be correct to use this and try solving through induction?
| Induction on odd numbers is easier when expressed as $P(n_0)$ is true, $P(n)
\implies P(n+2)$.
Let $f(n)=n(n^2-1)$. Then $f(1)=24$ and $f(n+2)-f(n)=6 (n + 1)^2$.
If $n$ is odd, then $n+1$ is even and so $(n+1)^2$ is a multiple of $4$.
Therefore, $f(n+2)-f(n)$ is a multiple of $24$ and so is $f(n+2)$ by induction.
For fun, here is a different approach using finite differences.
Write $n=2t+1$. Then
$$
n(n^2-1) = 8 t^3 + 12 t^2 + 4 t
= 24 \binom{t}{1} + 72 \binom{t}{2} + 48 \binom{t}{3}
$$
clearly a multiple of $24$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2508927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A curious way of generating series expansion for $\cos x$ If we take the approximation $\sin x \approx x$, then, using the trigonometric identity $1- \cos 2x = 2\sin^2 x$, and take $2\sin^2 x \approx 2x^2$, we get, after making the substitution $x \to x/2$, that
$$\cos x \approx 1 - \frac{x^2}{2}$$
Now, using the trigonometric identity
$$\cos(2x)=2\cos^2(x)-1$$
and using the last approximation $\cos x \approx 1 - \frac{x^2}{2}$, then we get a new approximation for $\cos x$, namely
$$\cos(2x)\approx2\left(1-\frac{x^2}2\right)^2-1=1-2x^2+\frac{x^4}2$$
Then let $x\to\frac x2$ to get
$$\cos(x)\approx1-\frac{x^2}2+\frac{x^4}{32}$$
So, we repeat:
$$\cos(2x)=2\cos^2(x)-1\approx2\left(1-\frac{x^2}2+\frac{x^4}{32}\right)^2-1$$
and so on. This seems to generate a series expansion for $\cos x$, similar to Taylor's series, but with greater denominators.
The question is: Does the iterative procedure described above generates better and better approximations to $\cos x$, that is, a Taylor-like series one, or this iterative procedure doesn't converge to $\cos x$ to arbitrary accuraty for real $x$?
| You get weird coefficients for terms beyond $x^2$ because you re using an inaccurate input for the double angle formula. If you have
$\cos x = 1-(x^2/2)+O(x^4)$,
you must accept
$\cos 2x = 1-(2x^2)+(x^4/2)+O(x^4)$;
and putting $x/2$ for $x$ then gives
$\cos x = 1-(x^2/2)+(x^4/32)+O(x^4)$
You did not make the error term any smaller in order of magnitude than the quartic term you added, so you cannot defend the quartic term.
How to get the right quartic term? Assume
$\cos x = 1-(x^2/2)+(ax^4)+O(x^6)$.
Apply the double angle formula; after simplifying you get
$\cos 2x = 1-(2x^2)+((4a+(1/2))x^4)+O(x^6)$,
where everything that's a multiple of $x^6$ or a higher power is lost in the "noise" of the error term. Put $x/2$ for $x$ to then get
$\cos x = 1-(x^2/2)+((a/4+(1/32))x^4)+O(x^6)$.
This matches your original assumption if $a=(a/4)+(1/32)$; thus, properly, $a=(1/24)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2509858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Can we split a function while doing differentiation? I have a silly question here. Assume a function $$f(x)= \frac{x}{(x-1)^2} \cdot (x+2)$$
Can I write
$$\frac{d}{dx} f(x) = \frac{d}{dx} x \cdot \frac{d}{dx} (x-1)^{-2} \cdot \frac{d}{dx} (x+2)^{-1}$$
If not, then why?
| \begin{align}
f(x) &= \frac{x}{(x-1)^2} \cdot (x+2)\\
&= \frac{x^2+2x}{(x-1)^2}\\
&= \frac{x^2}{(x-1)^2} + \frac{2x}{(x-1)^2}\\
\end{align}
Now
\begin{align}
f^\prime(x) &= \frac{x^2}{(x-1)^2}\,\frac{d}{dx} + \frac{2x}{(x-1)^2}\,\frac{d}{dx}\\
&= -\frac{2x}{(x-1)^3}\ - \frac{2(x+1)}{(x-1)^3}\\
&= -2\left(\frac{x}{(x-1)^3} + \frac{(x+1)}{(x-1)^3}\right)\\
\end{align}
$$f^\prime(x)= -2\frac{2x + 1}{(x-1)^3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a system of equations $(x+y-z=1), (x^2+y^2-z^2=5-2xy), (x^3+y^3-z^3=43-3xy)$
I have to solve this system of equations \begin{cases} x+y-z=1 \\ x^2+y^2-z^2=5-2xy \\x^3+y^3-z^3=43-3xy \end{cases}
From the third equation I have done this to manipulate it:
$$ x^3+y^3-z^3=43-3xy \implies (x+y)(x^2-xy+y^2)-z^3=43-3xy\implies (x+y)((x+y)^2-3xy) -z^3=43-3xy$$
Notice from the first equation $x+y=z+1$ and $(x+y)^2 = z^2+5$
So the above equation reduces to $z^2+5z-38-3xyz=0$
How do I get rid of the $xyz$?
| $$ x^3+y^3-z^3=43-3xy \implies ...\implies (x+y)((x+y)^2-3xy) -z^3=43-3xy$$
From the first equation: $\color{blue}{x+y}=\color{green}{z+1}$
$$(\color{blue}{x+y})^2=z^2+5$$
$$(\color{green}{z+1})^2=z^2+5$$
$$\vdots$$
It's easy from this point
| {
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"timestamp": "2023-03-29T00:00:00",
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Consider the following table representing the distribution of the discrete random variable x: $$\begin{array}{|c|c|c|c|c|}\hline X & -4 & 0 & 1 & 4 \\ \hline f(x)=P(X=x) & 0.1& 0.3 & 0.2 & 0.4 \\\hline\end{array}$$
Compute:
a) \begin{array}E(x)&=\sum_x xf(x)\\&=(-4\cdot 0.1)+0+(1\cdot 0.2)+(4\cdot0.4) \\
&=-0.4+0.2+1.6 \\
&=1.4 \end{array}
b)\begin{align}E(X^2-X+1)&=\sum_x (x^2-x+1)f(x)\\&=((-4)^2-(-4)+1)\cdot0.1+(0^2-0+1)\cdot 0.3+(1^2-1+1)\cdot 0.2+(4^2-4+1)\cdot 0.4 \\
\\&=2.1+0.3+0.2+5.2
\\&=7.8 \end{align}
c) \begin{align}V(X)&=\sum x^2p-\mu^2\\&=(1.6+0+0.2+6.4)-(1.4)^2
\\&=8.2-1.96
\\&=6.24 \end{align}
d) \begin{align}\sigma_X&=\sqrt{\operatorname{Var}(X)}\\&=\sqrt{6.24}\\&=\frac{2\sqrt{39}}5\\&=2.497999199\end{align}
This is the work I did I was wondering if I did someone could reassure me or give me any tips if I did something wrong. Thanks in advance! P.S. Feel free to fix/adjust any symbols in the formating.
| There is an error in part $c$, when you compute $\mathbb{E}[X^2]$.
$1.6+0+0.2+6.4=8.2 \neq 5$.
After obtaining $E(X^2)$ in part $c$, I will use it to check my answer in part $b$. Notice that $$E(X^2-X+1)=E(X^2)-E(X)+1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of the integral Taking $b,c \in \mathbb{R^+}$, the integral $$\int_{-\infty}^{\infty} \dfrac{\mathrm{d}x}{\sqrt{\left(a-x\right)^2+b}}$$
diverges, however the integral
$$\int_{-\infty}^{\infty} \left(\dfrac{1}{\sqrt{\left(a-x\right)^2+b}}-\dfrac{1}{\sqrt{\left(a-x\right)^2+c}}\right)\mathrm{d}x$$
converges to
$\ln\left(c\right)-\ln\left(b\right)$
The analitical result of the first and second integrals without limits of integration are respectively
$$\ln\left(\left|\sqrt{\left(x-a\right)^2+b}+x-a\right|\right)+C
$$
$$\ln\left(\left|\sqrt{\left(x-a\right)^2+b}+x-a\right|\right)-\ln\left(\left|\sqrt{\left(x-a\right)^2+c}+x-a\right|\right)+C
$$
I haven't succeeded on evaluating the limit of the second expression, is there any trick?
| As Jack D'Aurizio and Messney said in their comments, we can eliminate the constant $a$ by renaming our variable $x-a \rightarrow x$, leaving the evaluation of our integral as
$$\left. \ln\left(\left|\frac{\sqrt{\left(\frac{x}{\sqrt{b}}\right)^2+1}+\frac{x}{\sqrt{b}}}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2+1}+\frac{x}{\sqrt{c}}}\right|\right)\right|_{-\infty}^{+\infty}$$
wich equals
$$\lim_{x \to \infty} \ln\left(\left|\frac{\sqrt{\left(\frac{x}{\sqrt{b}}\right)^2+1}+\frac{x}{\sqrt{b}}}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2+1}+\frac{x}{\sqrt{c}}}\right|\right)-\ln\left(\left|\frac{\sqrt{\left(\frac{x}{\sqrt{b}}\right)^2+1}-\frac{x}{\sqrt{b}}}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2+1}-\frac{x}{\sqrt{c}}}\right|\right)$$
Dividing the numerator and denominator by $x$ on the first term, inserting the limit inside the logarith and multiplying and dividing by the conjugate of the numerator and denominator of the second we get
$$ \ln\left(\lim_{x \to \infty}\left|\frac{\sqrt{\left(\frac{1}{\sqrt{b}}\right)^2+\frac{1}{x^{2}}}+\frac{1}{\sqrt{b}}}{\sqrt{\left(\frac{1}{\sqrt{c}}\right)^2+\frac{1}{x^{2}}}+\frac{1}{\sqrt{c}}}\right|\right)-
\ln \left(\frac{1}{1}\right)+\lim_{x \to \infty} \ln\left(\left|\frac{\sqrt{\left(\frac{x}{\sqrt{b}}\right)^2+1}+\frac{x}{\sqrt{b}}}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2+1}+\frac{x}{\sqrt{c}}}\right|\right)$$
The second terrm is zero while the first and third terms are equal and both result in
$$\ln\left(\sqrt{\frac{c}{b}}\right) = \frac{1}{2}\ln\left(\frac{c}{b}\right) $$
So their sum is
$$\ln\left(\frac{c}{b}\right)=\ln(c)-\ln(b) $$
| {
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"timestamp": "2023-03-29T00:00:00",
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The number of integers $k$ for which the equation $x^3-27x+k$ has at least two distinct integer roots is? The number of integers $k$ for which the equation $x^3-27x+k$ has at least two distinct integer roots is?
My try:
Since imaginary numbers always form roots in pairs the question is the same as having two integer roots and one other real root.
If the roots are $A$,$B$ and $C$ of which $A$ and $B$ are two distinct integers and $C$ is a real number then we can write
1.
$$-k=ABC$$
2.
$$-27=AB+BC+CA$$
3.
$$A+B+C=0$$
Equation 3. Implies that even $C$ has to be an integer.
Now I don't know how I can find the number of values of k for which all roots are integers.
| Substituting $c = -\frac{k}{ab}$ into the third equation gives
$$a+b-\frac{k}{ab}=0\quad\Rightarrow\quad k = ab(a+b).$$
Using that in the second equation and simplifying gives
$$(a+b)^2 = ab+27,\text{ or }a^2+ab+b^2=27.$$
Thus
$$b = \frac{1}{2}(-a\pm\sqrt{3(36-a^2)}\,).$$
Since $a$ and $b$ must both be integers, one quickly gets the pairs
$$(a,b) = (-6,3), (-3,-3), (-3,6), (3,-6), (3,3), (6,-3),$$
each of which gives a solution. However, only two of those solutions have the linear term equal to $-27$: $x^2-27x\pm 54$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the series $\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$ converges
Show that the series $$\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$$ converges.
I showed it using Abel's theorem and limit comparison test. Any other simpler method?
| Write
$$\frac{(n^3+1)^{\frac{1}{3}}-n}{\log n} = \frac{n^3+1-n^3}{\log n((n^3+1)^{2/3}+(n^3+1)^{1/3}n+n^2)} = \frac{1}{\log n((n^3+1)^{2/3}+(n^3+1)^{1/3}n+n^2)}$$
and observe that the denominator goes at $0$ faster than $\frac{1}{n^2}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove using Mathematical Induction that $2^{3n}-3^n$ is divisible by $5$ for all $n≥1$. I did most of it but I stuck here I attached my working
tell me if I did correct or not thanks
My working:
EDITED: I wrote the notes as TEX
Prove using induction that $2^{3n} - 3^n \mod{5} = 0$.
Statement is true for $n = 1$:
$$2^{3 * 1} - 3^1 = 2^3 - 3 = 8 - 3 = 5$$
$$5 \mod{5} = 0$$
Now for $n = p$ and $n = p + 1$:
$$2^{3(k+1)} - 3{k + 1} = 2 * 2^p - 3$$
$$=2(5n + 3) - 3=10n + 6 - 3 = 10n+3$$
| We first show for the $n=1$ case:
$2^3-3 = 8-3 = 5$
which is clearly divisible by 5. Next is the inductive step where we assume
$2^{3n} - 3^n=5m $ for some integer $m$.
We can rewrite this as
$2^{3n}=5m+3^n$.
Now we want to prove that $2^{3(n+1)}-3^{n+1}$ is divisible by 5. We first can simplify this a bit:
$2^{3(n+1)}-3^{n+1} = 2^{3n+3}-3^{n+1}=2^3\cdot2^{3n}-3\cdot 3^n$.
We can now make the substitution from the previous statement to write this as
$2^3(5m+3^n)-3\cdot 3^n=5(8m)+8\cdot 3^n-3\cdot 3^n = 5(8m)+5\cdot3^n=5(8m+3^n)$
which is clearly divisible by 5 and proves the statement.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the probability of success in Hi-Lo? The way the game works is you are given natural numbers from 1 - 10, and one of these numbers is selected. You have to then guess whether the next randomly selected number will be higher or lower than the previous one, hence the name Hi-Lo.
For example, first number is 6. Choose "Lower". Next number is 3. Here, I am correct [because 3<6]. I choose "higher". Next number is 2. I am wrong. And so on.
My question is, given that I will always pick the more probable solution [aka if number is 8, I obviously will pick a number lower than 8 because my chance of being correct is higher], what will my success/fail ratio look like?
As an extension, what would success/fail ratio look like for numbers selected between 1 - n, where n is a natural number?
Note: I am a year 12 high school student so if possible please try to keep explanation simple.
| Notice that we have independence here so the "success to fail ratio", if we carry on this experiment in infinitum, is the same as in an individual trial. Assuming that we lose in the case of a tie, the chances of winning given that the first number is $1$ is $\frac{9}{10}$, $2$ is $\frac{8}{10}$, $3$ is $\frac{7}{10}$, $4$ is $\frac{6}{10}$, $5$ is $\frac{5}{10}$, $6$ is $\frac{5}{10}$, $7$ is $\frac{6}{10}$, $8$ is $\frac{7}{10}$, $9$ is $\frac{8}{10}$, and $10$ is $\frac{9}{10}$.
Each of these numbers are equally likely to be selected initially so we just average these probabilities:
$$\frac{\frac{9}{10}+\frac{8}{10}+\frac{7}{10}+\frac{6}{10}+\frac{5}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}}{10}=0.7$$
Or in terms of "success to fail ratio", we have $7$ successes for every $3$ losses giving $\frac{7}{3}$.
In general, for even $n$ the probability is
$$\frac{2\cdot\displaystyle\sum_{i=\frac{n}{2}}^{n-1}i}{n^2}$$
In general, for odd $n$ the probability is
$$\frac{\left\lfloor{\frac{n}{2}}\right\rfloor+2\cdot\displaystyle\sum_{i=\left\lceil\frac{n}{2}\right\rceil}^{n-1}i}{n^2}$$
or more simply
$$\frac{\frac{n-1}{2}+2\cdot\displaystyle\sum_{i=\frac{n+1}{2}}^{n-1}i}{n^2}$$
The success to fail ratio would just be $\frac{p}{1-p}$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $n(n+1)a_{n+1}=n(n-1)a_{n}-(n-2)a_{n-1}.$ >Then $\lim_{n\rightarrow \infty}a_{n}$
Let $a_{0}=1,a_{1}=2$ and for $n\geq 1\;, n(n+1)a_{n+1}=n(n-1)a_{n}-(n-2)a_{n-1}.$
Then $\lim_{n\rightarrow \infty}a_{n}$
$\bf{Attempt:}$ $2a_{2} = a_{0} = 1$ so $\displaystyle a_{2} = \frac{1}{2}$
$\displaystyle 6a_{3}=2a_{2} = 1$ so $\displaystyle a_{3} = \frac{1}{6}$
could some help me to solve it, thanks
| Working through the first few $n$ values it can be seen that, for $n \geq 2$,
$$a_{n} = \frac{a_{0}}{n!}.$$
This yields the limit
$$\lim_{n \to \infty} a_{n} = \lim_{n \to \infty} \frac{a_{0}}{n!} = 0.$$
For verification:
\begin{align}
\frac{n(n+1) a_{n+1} - (n-1) a_{n}}{(n+1)(n+2)} &= \frac{a_{0} \, (n - (n-1))}{(n+2)!} = \frac{a_{0}}{(n+2)!} = a_{n+2}.
\end{align}
| {
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For which value(s) of a and b do we have, : $\lim_{x\to 0}{\frac{e^x+a x \sin(2x)-b \cos(2x)-x}{x^2}}=\frac{9}{2}?$ I have a limit
$$\lim_{x\to 0}{\frac{e^x+a x \sin(2x)-b \cos(2x)-x}{x^2}}=\frac{9}{2},$$
where $a,b\in\mathbb{R}$, and I want to find $a$ and $b$.
In other cases, I evaluate the limits of numerator and denominator separately, which results in a relationship between $a$ and $b$. But now both the denominator and numerator are going to zero. How can I solve this problem?
| From Taylor expansion at $x=0$ we have, $$\sin(2x)\sim 2x-\frac{(2x)^3}{6}$$
$$\cos(2x)\sim 1-\frac{(2x)^2}{4}$$
$$e^x \sim 1+x+\frac{x^2}{2}$$
Hence,
$$ \frac{e^x+ax\sin2x-b \cos2x-x}{x^2}\sim \frac{1+x+\frac{x^2}{2} +2ax^2 -ax\frac{(2x)^3}{6} -b-b\frac{(2x)^2}{4}-x}{x^2}\\= \frac{1-b + (\frac{1}{2} -b+2a)x^2 -ax\frac{(2x)^3}{6} }{x^2} $$
Hence,
$$\lim_{x\to 0}\frac{e^x+ax\sin2x-b \cos2x-x}{x^2}= \frac{1}{2} -b+2a + \lim_{x\to 0}\frac{1-b }{x^2} =\begin{cases} \frac{1}{2} -b+2a & b=1\\
\infty& b<1\\-\infty& b>1
\end{cases}$$
Necessary we have, $b=1$ and, $\frac92= -frac12 +2a$ that is, $a=2$
We have $$\color{red}{a=2,~~~\text{and}~~~b=1}$$
| {
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Show recursion in closed form I've got following sequence formula:
$ a_{n}=2a_{n-1}-a_{n-2}+2^{n}+4$
where $ a_{0}=a_{1}=0$
I know what to do when I deal with sequence in form like this:
$ a_{n}=2a_{n-1}-a_{n-2}$
- when there's no other terms but previous terms of the sequence.
Can You tell me how to deal with this type of problems?
What's the general algorithm behind solving those?
| One approach that works to get to the final form is to take the formal power series
$$f(x) = \sum_{n=0}^{\infty} a_n x^n$$
and try and rewrite it in terms of itself. Applying the initial conditions where necessary:
\begin{align}
f(x) &= \sum_{n=0}^{\infty} a_n x^n\\
&= a_0 + a_1 x +\sum_{n=2}^{\infty} \left(2a_{n-1}-a_{n-2} + 2^n + 4\right) x^n\\
%
&= 2\sum_{n=2}^{\infty}a_{n-1}x^n - \sum_{n=2}^{\infty}a_{n-2}x^n + \sum_{n=2}^{\infty}2^nx^n + 4\sum_{n=2}^{\infty}x^n\\
%
&= 2\sum_{n=1}^{\infty}a_nx^{n+1} - \sum_{n=0}^{\infty}a_n x^{n+2} + \sum_{n=0}^{\infty}(2x)^{n+2} + 4\sum_{n=0}^{\infty}x^{n+2}\\
%
&= 2x\left(\sum_{n=0}^{\infty}a_nx^n - a_0\right) - x^2\sum_{n=0}^{\infty}a_n x^n + 4x^2\sum_{n=0}^{\infty}(2x)^n + 4x^2\sum_{n=0}^{\infty}x^n\\
%
&= 2xf(x) - x^2f(x) + 4x^2 \frac{1}{1-2x} + 4x^2 \frac{1}{1-x}\\
%
&= (2x-x^2)f(x) + 4x^2\frac{(1-x)+(1-2x)}{(1-2x)(1-x)}\\
%
&= (2x-x^2)f(x) + \frac{4x^2(2-3x)}{(1-2x)(1-x)}
\end{align}
Solving for $f(x)$:
$$f(x) = \frac{4x^2(2-3x)}{(1-2x)(1-x)(1-2x+x^2)} = \frac{4x^2(2-3x)}{(1-2x)(1-x)^3}$$
Applying partial fraction expansion and using the well-known result that
$$\sum_{n=0}^{\infty} (n+1)(n+2)\dotsb(n+m)x^n
= \frac{d^m}{dx^m}\left(\frac{1}{1-x}\right)
= \frac{m!}{(1-x)^{m+1}}$$
we get
\begin{align}
\sum_{n=0}^{\infty} a_n x^n &=
\frac{4x^2(2-3x)}{(1-2x)(1-x)^3}\\
&= \frac{4}{1-2x} + \frac{4}{1-x} - \frac{12}{(1-x)^2} + \frac{4}{(1-x)^3}\\
&= 4\frac{1}{1-2x} + 4\frac{1}{1-x} - 12\frac{1!}{(1-x)^{1+1}} + 2 \frac{2!}{(1-x)^{2+1}}\\
&= 4 \sum_{n=0}^{\infty} (2x)^n + 4 \sum_{n=0}^{\infty} x^n - 12 \sum_{n=0}^{\infty} (n+1) x^n + 2 \sum_{n=0}^{\infty} (n+1)(n+2) x^n\\
&= \sum_{n=0}^{\infty} \left(4\cdot 2^n + 4 - 12(n+1) + 2(n+1)(n+2)\right)x^n\\
&= \sum_{n=0}^{\infty} \left(2^{n+2} + 2n^2 - 6n -4\right)x^n
\end{align}
Equating coefficients, we find
$$a_n = 2^{n+2} + 2n^2 - 6n -4$$
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$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \cdots $ up to $n$ terms in terms of $x$ and $n$. There's a series which I can't seem to find a way to sum. Any help would be highly appreciated. It goes as follows $$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \cdots $$ up to $n+1$ terms. The sum is to be expressed in terms of $x$ and $n$. I tried setting a formula for the $n$-th term and setting it into a difference but ran into a dead end.
| Hint:
If $f(x)=\dfrac1{1+x}+\dfrac2{1+x^2}+\dfrac4{1+x^4}+\cdots+\text{ up to } n+1\text{ terms}$
$$\dfrac1{1-x}+f(x)=\dfrac1{1-x}+\dfrac1{1+x}+\left(\dfrac2{1+x^2}+\dfrac4{1+x^4}+\cdots+\text{ up to } n+1\text{ terms}\right)$$
$$=\dfrac2{1-x^2}+\dfrac2{1+x^2}+\left(\dfrac4{1+x^4}+\cdots+\text{ up to } n\text{ terms}\right)$$
$$=\dfrac4{1-x^4}+\dfrac4{1+x^4}+\left(\dfrac8{1+x^8}+\cdots+\text{ up to } n-1\text{ terms}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2528731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
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What's the value of $x$ if $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$ and... Given $x, y, z\in \mathbb{R}$, such that $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$ and $ \frac{1}{y}+\frac{1}{z+x}=\frac{1}{3}$ and $\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4}$. Find the value of $x$.
| Let $xy=c$, $xz=b$ and $yz=a$.
Thus, $$\frac{x+y+z}{c+b}=\frac{1}{2},$$
$$\frac{x+y+z}{c+a}=\frac{1}{3}$$ and
$$\frac{x+y+z}{a+b}=\frac{1}{4},$$
which gives
$$\frac{c+a}{c+b}=\frac{3}{2}$$ and
$$\frac{a+b}{c+b}=2.$$
From here we obtain $$b=\frac{3}{5}a$$ and
$$c=\frac{1}{5}a$$ or
$$xz=\frac{3}{5}yz$$ and
$$xy=\frac{1}{5}yz,$$ which gives
$$y=\frac{5}{3}x$$ and
$$z=5x.$$
Id est, $$\frac{1}{x}+\frac{1}{\frac{5}{3}x+5x}=\frac{1}{2},$$ which gives
$$x=2.3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2529442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the integral $\int_0^{\pi} \frac{dx}{a^2\sin^2{x}+b^2\cos^2{x}}$ where $ab\neq 0$ I’ve got some problems evaluating the integral $\int_0^{\pi} \frac{dx}{a^2\sin^2{x}+b^2\cos^2{x}}$. I’ve found a solution in my textbook as follows:
The integral
$\int \frac{dx}{a^2\sin^2{x}+b^2\cos^2{x}}=\frac{1}{ab}\arctan(\frac{a}{b}\tan{x})+C$
Hence by Newton-Leibniz Formulas, $\int_0^{\pi} \frac{dx}{a^2\sin^2{x}+b^2\cos^2{x}}= \frac{1}{ab}\arctan(\frac{a}{b}\tan{x})|_0^{\pi}=0$.
But it doesn’t make sense. $F(x)=\frac{1}{ab}\arctan(\frac{a}{b}\tan{x})$ is not continuous where $x=\frac{\pi}{2}$.
Am I right? Or how to evaluate this integral?
And a Cauchy’s example.
$\int_0^{\frac{3\pi}{4}} \frac{\sin{x}}{1+\cos^2{x}} dx=\arctan(\sec{x})|_0^{\frac{3\pi}{4}}=-\arctan\sqrt{2}-\frac{\pi}{4}.$
The second $=$ is wrong because $\arctan(\sec{x})$ is not continuous where $x=\frac{\pi}{2}$ so that we cannot use Newton-Leibniz formula.
Am I right???
| The integrand is $\pi$-periodic and even around $0$. Therefore,
$$\begin{align}
\int_0^\pi \frac{1}{a^2\sin^2(x)+b^2\cos^2(x)}\,dx&=2\int_0^{\pi/2}\frac{1}{a^2\sin^2(x)+b^2\cos^2(x)}\,dx\\\\
&=2\int_0^{\pi/2}\frac{\sec^2(x)}{a^2\tan^2(x)+b^2}\,dx\\\\
&=2\int_0^\infty \frac{1}{a^2u^2+b^2}\,du
\end{align}$$
Can you finish now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2530207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Simplify the expression: $(2x+1)×\left(x^2+(x+1)^2 \right)×\left( (x^4+(x+1)^4) \right)×...×\left(x^{64}+(x+1)^{64}\right)$ Simplify the expression:
$$(2x+1)×\left(x^2+(x+1)^2 \right)×\left( (x^4+(x+1)^4) \right)×...×\left(x^{64}+(x+1)^{64}\right)$$
I used the general method:
$(2x+1)(2x^2+2x+1)(2x^4+4x^3+6x^2+4x+1)×...$
But, But I'm stuck here.
| Write $2x + 1 = (x+1)^2 - x^2$ and use the factorization for $a^2 - b^2$ repeatedly.
| {
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"url": "https://math.stackexchange.com/questions/2531082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$T(n) = T(n-2) + log n$ Does anyone know how to solve this example?
I start with this :
$T(n) = T(n-2) + \log n$
$T(n-2) = T(n-4) + \log(n-2)$
$T(n-4) = T(n-6) + \log (n-4)$
Now i substitute back:
$T(n-2) = T(n-6) + \log(n-4) + \log(n-2)$
$T(n) = T(n-6) + \log(n-4) + \log(n-2) + \log n$
Now i see a pattern here so i substitute it again:
$T(n) = T(n-2\times k) +$ $\sum\nolimits_{k=2}^n$ $(\log 2\times k)$
then $n-2\times k = 1 \implies k = \frac{n-1}{2}$
Now i am stuck with this term:
$T(n) = T(1) +$ $\sum\nolimits_{k=2}^n$ $\log 2\times \frac{n-1}{2}$
How shall I go on?
| The correct recursion for even terms would be
$$T(2n) = T(2n-2) + \log(2n) = T(2n-2k) + \sum_{j=0}^{k-1}\log(2n-2j) $$
Substituting $k=n$ gives
\begin{align}
T(2n) &= T(0) + \sum_{j=0}^{n-1}\left[\log(2) + \log(n-j)\right]\\
&= T(0) + \log(2)\cdot n + \sum_{j=0}^{n-1} \log(n-j)\\
&= T(0) + \log(2)\cdot n + \log\left(\prod_{j=0}^{n-1} (n - j)\right)\\
&= T(0) + \log(2)\cdot n + \log(n!)\\
&\overset{\text{Stirling's}}{\sim} T(0) + \log(2)\cdot n + \ln\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n \right)
\in \Theta(n\log n)
\end{align}
For odd terms we ultimately find
\begin{align}
T(2n+1) &= T(1) + \sum_{j=0}^{n-1} \log(2n - 2j -1)\\
&= T(1) + \log\left(\prod_{j=0}^{n-1} (2n-2j-1)\right)
\end{align}
The product is just $$(1)\cdot (3)\cdot \dotsb\cdot(2n-1)$$
that we can deal with by writing it in terms of factorials:
\begin{align}(1)\cdot (3)\cdot \dotsb\cdot(2n-1) &= \frac{(1)(2)(3)(4)\dotsb(2n-1)(2n)}{(2)(4)\dotsb (2n)}\\
&= \frac{(1)(2)(3)(4)\dotsb(2n-1)(2n)}{(1)(2)\dotsb (n) 2^n}
= \frac{(2n)!}{n! 2^n} \end{align}
So using Stirling's again:
\begin{align}
T(2n-1) &= T(1) + \log \frac{(2n)!}{n!2^n}\\
&\sim T(1) + \log \frac{\sqrt{2\pi \cdot(2n)}\left(\frac{(2n)}{e}\right)^{(2n)} }{\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n} 2^n}\\
&= T(1) + \log\left(\sqrt{2}\cdot 2^{2n-1} n^{2n} e^{-n}\right)\\
&\sim \log(n^{2n}) = 2n \log (n) \in \Theta(n\log n)
\end{align}
Therefore $T(n)$ is $\Theta(n \log n)$.
Alternatively If we look at the computation of area under $\log(2x)$, we may approximate the sum of logs:
$$\int_1^n \log(2x)\;dx
\leq \sum_{k=1}^n \log\big(2(k+1)\big)
\leq \int_2^{n+1} \log(2x)\;dx$$
given that $\int \log(2x) \;dx = x\left[\log(2x) - 1\right] + C$, we find that both the upper and lower bound are $\Theta\big( n\log (n) \big)$. Hence
$$ \sum_{k=1}^n \log\big(2(k+1)\big) = T(2n) - T(0) + \log(2n+2) \in \Theta\big( n\log (n) \big)$$ so $T(2n) \in \Theta\big( n\log (n) \big)$.
A near identical argument for $T(2n+1)$ shows the same thing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Proving $\sum\limits_{r=1}^n \cot \frac{r\pi}{n+1}=0$ using complex numbers Let $x_1,x_2,...,x_n$ be the roots of the equation $x^n+x^{n-1}+...+x+1=0$.
The question is to compute the expression $$\frac{1}{x_1-1} + \frac{1}{x_2-1}+...+\frac{1}{x_n-1}$$
Hence to prove that $$\sum_{r=1}^n \cot \frac{r\pi}{n+1}=0$$
I tried rewriting the expression as $$\sum_{i=1}^n \frac{\bar{x_i}-1}{|x_i-1|^2}$$
I then used the fact that $$x^{n+1}-1=(x-1)(x^n+x^{n-1}+...+x+1=0$$ so $x_i$ are the complex nth roots of unity.Using cosine formula I found that $$|x_i-1|^2=2-2\cos(\frac{2i\pi}{n+1})=4(\sin \frac{\pi}{n+1})^2$$
After substituting this I couldn't simplify the resulting expression.Any ideas?Thanks.
| @Jaideep Khare already gave a nice answer to first question, namely he proved that,
$$\frac{1}{x_1-1} + \frac{1}{x_2-1}+...+\frac{1}{x_n-1}=- \frac{n}{2}$$
where $x_i$'s satisfy
$$\color{red}{\frac{x^{n+1}-1}{x-1}= }x^n+x^{n-1}+...+x+1=0$$
It is easy to check that, $\color{blue}{x_k = e^{\frac{2r\pi}{n+1}}}$ with $0\le k\le n $
is easy to see that, $x_k$'s are the roots of the equation
$$\color{red}{\frac{x^{n+1}-1}{x-1} } =x^n+x^{n-1}+...+x+1=0$$
Now using: $\sin x =\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos x =\frac{e^{ix}+e^{-ix}}{2} $ one can check that,
$$\cot x =i\frac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}}= i +i \frac{2}{e^{2ix}-1}$$
thus,
$$\color{blue}{\sum_{r=1}^n \cot \frac{r\pi}{n+1} = ni+ 2i\sum_{r=1}^n \frac{1}{\displaystyle{e^{\frac{2r\pi}{n+1}}}-1} = ni+ 2i\sum_{r=1}^n \frac{1}{x_r-1} =0}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Laplace transform $\mathcal{L}_x\left[\frac{1}{(x+9) \sqrt{x+8}}\right](s)$ How to find Laplace transform ?:
$$\mathcal{L}_x\left[\frac{1}{(x+9) \sqrt{x+8}}\right](s) $$
I have been stuck on this problem for quite a bit. I only found in tables:
$$\mathcal{L}_x\left[\frac{1}{(x+8) \sqrt{x+8}}\right](s)=\frac{1}{\sqrt{2}}-2 e^{8 s} \sqrt{\pi } \sqrt{s} \text{erfc}\left(2 \sqrt{2} \sqrt{s}\right)$$
My attempt:
$$\mathcal{L}_x\left[\frac{1}{(x+9) \sqrt{x+8}}\right](s)
=\mathcal{L}_x\left[\int_0^{\infty } \frac{\exp (-t (x+9))}{\sqrt{x+8}} \, dt\right](s)=\int_0^{\infty }
\mathcal{L}_x\left[\frac{\exp (-t (x+9))}{\sqrt{x+8}}\right](s) \, dt=\int_0^{\infty } \frac{e^{8s-t} \sqrt{\pi } \text{erfc}\left(2 \sqrt{2} \sqrt{s+t}\right)}{\sqrt{s+t}}
\, dt$$
or:
$$\mathcal{L}_x\left[\frac{1}{(x+9) \sqrt{x+8}}\right](s)=\int_0^{\infty } -\frac{2 e^{9 s+t^2} \text{Ei}\left(-9 \left(s+t^2\right)\right)}{\sqrt{\pi }} \, dt$$
| The integral you need to compute is
$$
I(s)=\int_0^\infty dx\frac{e^{-sx}}{(x+9)\sqrt{x+8}}\ .
$$
Call $\sqrt{x+8}=t$. Then $x=t^2-8$ and $dx=2t dt$.
So
$$
I(s)=\int_{2\sqrt{2}}^\infty 2t dt\frac{e^{-s(t^2-8)}}{t(t^2+1)}=2e^{8s}\int_{2\sqrt{2}}^\infty dt\frac{e^{-s t^2}}{t^2+1}\ .
$$
This can be rewritten as
$$
I(s)=2e^{8s}\left[\int_0^\infty dt\frac{e^{-s t^2}}{t^2+1}-\int_0^{2\sqrt{2}}dt\frac{e^{-s t^2}}{t^2+1}\right]=2e^{8s}\left[\frac{1}{2} \pi e^s \text{erfc}\left(\sqrt{s}\right)-2\pi e^s \frac{1}{2\pi}\int_0^{2\sqrt{2}}dt\frac{e^{-s (t^2+1)}}{t^2+1}\right]
=2e^{9s}\pi\left[\frac{1}{2} \text{erfc}\left(\sqrt{s}\right)-2 T(\sqrt{2s},2\sqrt{2})\right]\ ,
$$
where $T(x,a)$ is the Owen's T-function (https://reference.wolfram.com/legacy/v9/ref/OwenT.html).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $A+B+C=180^\circ, \cot A=2$ and $\cot B= \frac34$, what is $\sin C$?
If $A+B+C=180^\circ, \cot A=2$ and $\cot B= \frac34$, what is $\sin C$?
$$\cot A = 2\\
\frac{\cos A }{ \sin A }= 2\\
\cot B = \frac34\\
\frac{\cos B }{ \sin B} = \frac34$$
What should I do with $A+B+C=180^\circ$?
| $\tan C = -\tan (A+B)= \dfrac{\tan A+\tan B}{\tan A\tan B- 1}=...=-\dfrac{11}{2}\implies \cot C = -\dfrac{2}{11}\implies \cot^2C= \dfrac{4}{121}\implies \csc^2C= 1+\cot^2C= \dfrac{125}{121}\implies \sin^2C = \dfrac{121}{125}\implies \sin C = \dfrac{11}{5\sqrt{5}}$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2537575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove by induction with summation and factorials Sorry I'm not sure how to format the text for using sum. Could someone help me out with that as well. Much appreciated.
\begin{eqnarray*}
\sum_{r=1}^{n} (r^2+1)r! =n(n+1)!
\end{eqnarray*}
for all $n \geq 1 $.
| $$\sum_{r=1}^{n} (r^2+1)r! =n(n+1)!$$
$$\sum_{r=1}^{1} (r^2+1)r! = (1^2 + 1)1! = 2 * 1 = 2 = 1(1 + 1)$$
So it is true for $n=1$. Now let's see for $n=p$
$$\sum_{r=1}^p (r^2+1)r! =p(p+1)!$$
$$\sum_{r=1}^{p+1} (r^2+1)r! = (p+1)(p+2)!$$
$$\sum_{r=1}^{p} (r^2+1)r! + ((p+1)^2 + 1)(p+1)! = (p+1)(p+1)! *(p+2)$$
$$\sum_{r=1}^{p} (r^2+1)r! + (p^2 + 2p + 2)(p+1)! = (p+1)(p+2)(p+1)!$$
$$\sum_{r=1}^{p} (r^2+1)r! + (p^2 + 2p + 2)(p+1)! = (p^2 + 3p + 2)(p+1)!$$
$$\sum_{r=1}^{p} (r^2+1)r! + (p^2 + 2p + 2)(p+1)! = (p + 3 + 2 * \frac{1}{p})p(p+1)!$$
$$1 + ((p^2 + 2p + 2)(p+1)!)/(p(p+1)!) = (p + 3 + 2\frac{1}{p})$$
$$\frac{p^2 + 2p + 2}{p} + 1 = p + 3 + 2 * \frac{1}{p}$$
$$p + 2 + 2 * \frac{1}{p} + 1 = p + 3 + 2 * \frac{1}{p}$$
Which is true, so the statement is true, proven by induction
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{x \to 0} \frac{1-(x^2/2) -\cos (x/(1-x^2))}{x^4}$
find the limits with Using :$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\frac12$
$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}$$
My Try :
$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}=\lim_{x \to 0} \frac{(1-\cos (\dfrac{x}{1-x^2}))+(-\dfrac{x^2}2)}{x^4}$$
$$\lim_{x \to 0} \frac{(\dfrac{(1-\cos (\dfrac{x}{1-x^2}))}{(\dfrac{x}{1-x^2})^2})(\dfrac{x}{1-x^2})^2+(-\dfrac{x^2}2)}{x^4}$$
now what ?
| Letting $u = x/(1 - x)^2$, we can write this as:
$$\lim_{x \rightarrow 0} \frac{1 - \frac{1}{2}u^{2} - \cos(u) + \frac{1}{2}(u^{2} - x^{2})}{u^{4}(1 - x^{2})^{4}}$$
$$= \lim_{u \rightarrow 0} \frac{1 - \frac{1}{2}u^{2} - \cos(u)}{u^{4}} + \lim_{x \rightarrow 0}\frac{u^{2} - x^{2}}{2x^{4}}$$
The first limit is a bit more complicated than the limit it said to use- but it's easy with Taylor series or l'Hospital's rule, and the answer is $-\frac{1}{24}$. The second limit is just a rational function in $x$:
$$\lim_{x \rightarrow 0}\frac{u^2 - x^2}{2x^{4}} = \lim_{x \rightarrow 0} \frac{x^{2} - x^{2}(1 - x^{2})^{2}}{2x^{4}(1 - x^{2})^{2}} = \lim_{x \rightarrow 0}\frac{x^{2} - x^{2} + 2x^{4} - x^{6}}{2x^{4}(1 - x^{2})^{2}} = \lim_{x \rightarrow 0}\frac{2 - x^{2}}{2(1 - x^{2})^{2}} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2541078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Evaluating $\int \frac{x \sqrt{2\sin(x^2+1) - \sin2(x^2+1)}}{2 \sin(x^2+1)+\sin2(x^2+1)} dx$ The question is to evaluate $$\int \frac{x \sqrt{2\sin(x^2+1) - \sin2(x^2+1)}}{2 \sin(x^2+1)+\sin2(x^2+1)} dx$$ if $x^2 \neq (n \pi-1) \forall n\in N$
I tried to rewrite the integral as $$\int \frac{x\sqrt{(2\sin(x^2+1))(1-\cos(x^2+1))}}{(2\sin(x^2+1))(1+\cos(x^2+1))}$$
If there would have been a radical sign in denominator too the problem would be quite simple but I couldn't understand how to proceed from here.Any ideas?
| Hint:
Let $x^2+1=u$ then your integral simplifies to
$$\int\dfrac{\sqrt{8}}{8}\dfrac{du}{\left(\cos\frac{u}{2}\right)^\frac52}$$
now let $\sin\dfrac{u}{2}=w$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How can I calculate the limit $\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$ without L'Hospital's rule? I have a problem with calculation of the limit:
$$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$$
Is there a way to calculate it? How can I do it?
| $$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}=\lim\limits_{x \to 7} \frac{\sqrt{9+x-7} - \sqrt[3]{27+x-7}}{\sqrt[4]{16+x-7} - 2}=$$
$$=\lim\limits_{x \to 7} \frac{3\sqrt{1+\frac{x-7}{9}} - 3\sqrt[3]{1+\frac{x-7}{27}}}{2\sqrt[4]{1+\frac{x-7}{16}} - 2}=\frac{3}{2}\lim\limits_{x \to 7}\frac{\left(1+\frac{x-7}{9}\right)^{\frac{1}{2}}-\left(1+\frac{x-7}{27}\right)^{\frac{1}{3}}}{\left(1+\frac{x-7}{16}\right)^\frac{1}{4}-1}$$
Now recall that $\lim\limits_{t \to t_0}\left[1+\left(f(t)-f(t_0)\right)\right]^{\alpha}=1+\alpha\left[f(t)-f(t_0)\right]+o\left(f(t)-f(t_0)\right)$ if $f(t)\to f(t_0)$ for $t \to t_0$
$$\frac{3}{2}\lim\limits_{x \to 7}\frac{\left(1+\frac{x-7}{9}\right)^{\frac{1}{2}}-\left(1+\frac{x-7}{27}\right)^{\frac{1}{3}}}{\left(1+\frac{x-7}{16}\right)^\frac{1}{4}-1}=\frac{3}{2}\lim\limits_{x \to 7}\frac{1+\frac{1}{2}\frac{x-7}{9}-1-\frac{1}{3}\frac{x-7}{27}+o\left(x-7\right)}{1+\frac{1}{4}\frac{x-7}{16}-1+o\left(x-7\right)}=$$
$$=\frac{3}{2}\lim\limits_{x \to 7}\frac{\frac{x-7}{18}-\frac{x-7}{81}+o\left(x-7\right)}{\frac{x-7}{64}+o\left(x-7\right)}=96\lim\limits_{x \to 7}\frac{(x-7)\left(\frac{1}{18}-\frac{1}{81}\right)+o\left(x-7\right)}{(x-7)\left[1+o\left(x-7\right)\right]}=96\cdot \frac{28}{648}=\frac{112}{27}$$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Find the sum of the series $S = \sum_{k=1}^{n} \frac{k}{k^{4} + k^{2} + 1} $ $$ S = \sum_{k=1}^{n} \frac{k}{k^{4} + k^{2} + 1} $$
I started by factorizing the denominator as $k^2+k+1$ and $k^2-k+1$
The numerator leaves a quadratic with $k$ and $k-1$ or a constant with $k+1$ and $k-1.$
I tried writing the individual terms, ofcourse, it was useless.
How do I do this?
| So your term is equal to $$\frac{1}{2}\left(\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right)$$
Now note $(k+1)^2-(k+1)+1=k^2+k+1$, so your term is: $$\frac{1}{2}\left(\frac{1}{k^2-k+1}-\frac{1}{(k+1)^2-(k+1)+1}\right)$$
and you can apply a telescoping series technique to establish the sum to $n$ is just half of $\frac{1}{1^2 -1 +1}-\frac{1}{(n+1)^2-(n+1)+1}$. And if you are also looking for the limit, half of $\frac{1}{1^2 -1 +1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2544518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Proving the equivalence relationship within a relation $(a,b)R(c,d)$ To prove that for a relation $R$ on $\mathbb{R} \times \mathbb{R}$ such as $(a,b)R(c,d)$ exists if and only if $\exists k \in \mathbb{N}$ so that :
$k \le \sqrt{a^2+b^2} < k+1$
$k \le \sqrt{c^2+d^2} < k+1$
*
*I understand that to prove I need to show that it is reflexive, symetric and transitive. However, the proof just seems silly because the equation on both sides are the same. Would this be an acceptable proof?
Reflexive & Symmetric:
$(a,b)\, \mathscr{R}\, (a,b) \iff k \le \sqrt{a^2+b^2} < k+1 \iff k \le \sqrt{a^2+b^2} < k+1$
$(c,d)\, \mathscr{R}\, (a,b) \iff k \le \sqrt{c^2+d^2} < k+1 \iff k \le \sqrt{a^2+b^2} < k+1$
Because we can substitute $(a,b)$ for $(c,d)$ and obtain the exact same equation on both sides.
Alternatively I was also thinking that we could prove that if:
$\sqrt{a^2+b^2} = k$ and $\sqrt{c^2+d^2} = k$ then:
$\sqrt{a^2+b^2} - \sqrt{a^2+b^2} = 0$
$\sqrt{c^2+d^2} - \sqrt{a^2+b^2} = 0$
Transitive:
$(a,b)R(c,d) \rightarrow (c,d)R(e,f) \rightarrow (a,b)R(e,f)$
$k \le \sqrt{a^2+b^2} < k+1 \iff k \le \sqrt{c^2+d^2} < k+1 \iff k \le \sqrt{e^2+f^2} < k+1$
Because all three will reside between a value $k \le (x,y) \le k+1$
I am not sure how to go about proving this, I've seen other examples in the book but they really rely on one condition that links both $a,b$ and $c,d$. Given that in this example they are seperated I am confused as to how I should be tackling the problem. I feel as if what I wrote doesn't prove much, but rather just shows the definition of the given properties.
*Also how would you show the equivalence classes graphically?
Thanks,
| Sometimes proving equivalence relations IS trivial.
For example if the equivalence is based adding or multiplication of two numbers symmetry and reflexivity are automatic as addition and multiplication is commutative.
However this one, although very easy, is not as trivial as you think. Note $(a,b) R (c,d)$ if there exists some $k$ so that $k \le \sqrt{a^2+ b^2}, \sqrt{c^2 + d^2} < k+1$ but that if $(a,b) R (cd)$ and $(e,f)R (g,h)$ then the $k$ so that $k \le \sqrt{a^2+ b^2}, \sqrt{c^2 + d^2} < k+1$ need not be the same naturall number $j$ so that $j \le \sqrt{e^2+ f^2}, \sqrt{g^2 + h^2} < j+1$.
But it is still almost trivial.
Obviously as $\sqrt{a^2 + b^2} = \sqrt{a^2 + b^2}$ Reflexivity holds. You can spell it out if you want to (and you probably should if 1) you are teaching a student learning the stuff 2) you are a student learning the stuff or 3) you don't know how fussy you teacher is and s/he might assume you don't really get this stuff if you don't) by writing the absurdly obvious: $k \le \sqrt{a^2 + b^2} < k+1 \iff k \le \sqrt{a^2 + b^2} < k+1$.
Oh, we should also point out the obvious: that i) As $a^2 + b^2 \ge 0$, $\sqrt{a^2 + b^2}$ always exists and ii) by the archimedian principal there always exist a $k \in \mathbb Z$ so that $k \le \sqrt{a^2 + b^2} < k+1$ and iii) as $\sqrt{a^2 + b^2} \ge 0$ then the $k$ in ii) is a natural number (we're including zero I presume), then all ordered pairs are actually in an equivalence class.
And as $k = k$ and $k +1 = k+1$ Symmetry and Transitivity hold. I.E. If $k\le \sqrt{a^2 + b^2} < k+1$ and $k\le \sqrt{c^2 + d^2} < k+1$ and $k\le \sqrt{e^2 + f^2} < k+1$ then .... welll, $k\le \sqrt{a^2 + b^2} < k+1$ and $k\le \sqrt{c^2 + d^2} < k+1$ and $k\le \sqrt{e^2 + f^2} < k+1$, duh.
But we can spell them out specifically:
Symmetry:($k\le \sqrt{a^2 + b^2} < k+1$ and $k\le \sqrt{c^2 + d^2} < k+1$) $\iff$ ($k\le \sqrt{c^2 + d^2} < k+1$ and $k\le \sqrt{a^2 + b^2} < k+1$)
And Transitivity: ($k\le \sqrt{a^2 + b^2} < k+1$ and $k\le \sqrt{c^2 + d^2}<k+1)$ AND ($k\le \sqrt{c^2 + d^2} < k+1$ and $k\le \sqrt{e^2 + f^2})$\implies ($k\le \sqrt{a^2 + b^2} < k+1$ and $k\le \sqrt{e^2 + f^2})$
....
So, yes, the proof WAS that trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Proving trigonometric identity: $ \frac{\sin (y+x)}{\sin (y-x)} = \frac{\tan y + \tan x}{\tan y - \tan x} $
Prove that
$$ \frac{\sin (y+x)}{\sin (y-x)} = \frac{\tan y + \tan x}{\tan y - \tan x} $$
This is my working -
$$\text{LHS} = \frac{\sin (y+x)}{\sin (y-x)} = \frac{\sin y \cos x + \cos y \sin x}{\sin y \cos x - \cos y \sin x} $$
I used the compound angle relations formula to do this step.
However , now I'm stuck.
I checked out the answer and the answer basically carried on my step by dividing it by
$ \frac{\cos x \cos y}{\cos x \cos y} $
If I'm not wrong , we cannot add in our own expression right? Because the question is asking to prove left is equal to right. Adding in our own expression to the left hand side will be not answering the question, I feel .
| $$\frac{\tan x+\tan y}{\tan x-\tan y} =\frac{\frac{\sin y\cos x+\cos y\sin x}{\cos x\cos y}}{\frac{\sin y\cos x-\cos y\sin x}{\cos x\cos y}}= \frac{\sin y\cos x+\cos y\sin x}{\sin y\cos x-\cos y\sin x}=\frac{\sin (y+x)}{\sin (y-x)} $$
which gives the results since we know that
$$ \sin(x-y) =\sin y\cos x-\cos y\sin x $$and
$$ \sin(x+y) =\sin y\cos x+\cos y\sin x $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2547140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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} |
What is $\operatorname{ord}_{22}(5^6)$?
Find $\operatorname{ord}_{22}(5^6)$.
So, basically we want to find:$$\operatorname*{arg\,min}_k 5^{6k} \equiv 1\pmod {22}$$
I found that $5^5 \equiv 1\pmod {22}$ so I know that for $k=5$ we have $5^{6k} \equiv 1 \pmod {22}$. Therefore, $\text{ord}(5^6) \le 5$.
I guess that I could proceed with the somewhat tedious checking:
*
*$k=4$: $(5^6)^4 \equiv (5^4)^6 \equiv 5^4 \equiv 9 \pmod {22}$
*$k=3$ $(5^6)^3 \equiv (5^3)^6 \equiv 5^3 \equiv 15 \pmod {22}$
*$k=2$ $(5^6)^2 \equiv (5^2)^6 \equiv 5^2 \equiv 3 \pmod {22}$
*$k=1$ $(5^6)^1 \equiv (5^1)^6 \equiv 5 \pmod {22}$
So $\text{ord}_{22}(5^6) = 5$.
Questions:
*
*Is that what I'm expected to do? Or is there a simpler way? (I had to use a calculator or to tediously calculate it myself)
*Why can we get rid of the exponent when we do modular arithmetic? (I actually used it along the proof) i.e. $$x \equiv a \pmod m \iff x^b \equiv a \pmod m$$
| By the Chinese remainder theorem, it is enough to find the order of $5^6$ $\!\!\pmod{2}$ (which is obviously $1$) and the order of $5^6\equiv5$ $\!\!\pmod{11}$. Since $5$ is a quadratic residue $\!\!\pmod{11}$, due to
$$\left(\frac{5}{11}\right)=\left(\frac{11}{5}\right)=\left(\frac{1}{5}\right)=1$$
the order of $5\pmod{11}$ is five and that is also the order of $5^6\pmod{22}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Show that $\left|\int_{-n}^{n}e^{iy^2}dy\right|\le 2$ for $n\ge 5.$ Question is to show that
$$\left|\int_{-n}^{n}e^{iy^2}dy\right|\le 2$$
when $n\geq5$, $x \in \mathbb R $ and $i$ is an imaginary unit.
My effort:
$$|\int_{-n}^{n}e^{iy^2}dy|\leq \int_{-n}^{n}|e^{iy^2}|dy=\int_{-n}^{n}|\cos(y^2)+i\sin(y^2)|dy$$
$$ \leq \int_{-n}^{n}|\cos(y^2)|dy+\int_{-n}^{n}|i||\sin(y^2)|dy$$
$$\leq \int_{-n}^{n}|\cos(y^2)|dy+\int_{-n}^{n}|\sin(y^2)|dy$$
It is also known that $|\cos(x)|,|\sin(x)|\leq1$
but its leading nowhere since integral then evalutes to $0$..
Any tips?
| $$\left|\int_{-n}^{n}e^{iy^2}dy\right|^2 =\left|\int_{-n}^{n}\cos{y^2}dy +i\int_{-n}^{n}\sin{y^2}dy\right|^2 \\= 4\left(\int_{0}^{n}\cos{y^2}dy \right)^2 + 4\left(\int_{0}^{n}\sin{y^2}dy \right)^2\\= 4[I^2_n + J^2_n]$$
This is somehow related to this: Prove only by transformation that: $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $. Employing the change of variables $2u =x^2$ We get $$I_n=\int_0^n\cos(x^2) dx =\frac{1}{\sqrt{2}}\int^{n^2/2}_0\frac{\cos(2x)}{\sqrt{x}}\,dx$$ $$ J_n=\int_0^n \sin(x^2) dx=\frac{1}{\sqrt{2}}\int^{n^2/2}_0\frac{\sin(2x)}{\sqrt{x}}\,dx $$ Using the same change of variables like here one readily get
$$J_n = \frac{\sin^2 \frac{n^2}{2}}{n}+\frac{1}{2\sqrt{2}} \int^{\frac{n^2}{2}}_0\frac{\sin^2 x}{x^{3/2}}\,dx < \frac{1}{n}+\frac{1}{2\sqrt{2}} \int^{\infty}_0\frac{\sin^2 x}{x^{3/2}}\,dx $$
and
$$I_n =\frac{1}{2} \frac{\sin 2 n^2}{n} +\frac{1}{4 }\frac{\sin^2 \frac{n^2}{2}}{n} +\frac{3}{8\sqrt{2}} \int^{\frac{n^2}{2}}_0\frac{\sin^2 x}{x^{5/2}}\,dx \\< \frac{3}{4n} +\frac{3}{8\sqrt{2}} \int^{\infty}_0\frac{\sin^2 x}{x^{5/2}}\,dx $$
Since $0\le \sin^2 x\le 1$. From this we have,
$$\int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx = \frac{4\sqrt \pi}{3}$$
similarly, we have $$\int^{\infty}_0\frac{\sin^2 x}{x^{3/2}}\,dx =\sqrt\pi$$
Whence,
$$J_n < \frac{1}{n}+\frac{\sqrt\pi}{2\sqrt{2}}= \frac{1}{n}+ \sqrt{\frac{\pi}{8}} $$
and
$$I_n < \frac{3}{4n}+\frac{\sqrt\pi}{2\sqrt{2}} = \frac{3}{4n}+ \sqrt{\frac{\pi}{8}} $$
Since,
$$\color{red}{\lim_{n\to\infty}4 \left[ \left(\frac{3}{4n}+ \sqrt{\frac{\pi}{8}} \right)^2+ \left(\frac{1}{n}+ \sqrt{\frac{\pi}{8}} \right)^2\right] = \pi <4}$$
That is for very large $n$ we get,
$$\color{red}{\left|\int_{-n}^{n}e^{iy^2}dy\right| = 2[I^2_n + J^2_n]^{1/2} \le 2\left[ \left(\frac{3}{4n}+ \sqrt{\frac{\pi}{8}} \right)^2+ \left(\frac{1}{n}+ \sqrt{\frac{\pi}{8}} \right)^2\right]^{1/2}<2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2552190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Calculate $\lim\limits_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$ In preparation for finals, I am trying to calculate $$\lim_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$$ with proof.
Here is my approach/what I have done so far:
If we can find a dominating function, we have $$\lim_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x = \int\limits_0^\infty \lim_{n\rightarrow \infty} \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$$ by the Dominated Convergence Theorem. If we let $f_{n} = \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)}$, then $f_{n}(x)$ converges to $0$ for all $x > 0$, which implies the limit is equal to 0 because the Dominated Convergence Theorem only requires a.e. convergence (so not having convergence at $x = 0$ is no issue). Operating under the assumption that dominating function exists, is this correct?
As far as finding a dominating function is concerned, we have
$$
|f_{n}| = \left| \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \right| = \frac{|n^2 \sin(n/x)|}{n^3x + x(1 + x^3)} \leq \frac{n^2}{n^3x + x(1 + x^3)},
$$
which is where I get stuck. The two directions that seemed the most clear from here was to either
$$
\frac{n^2}{n^3x + x(1 + x^3)} \leq\frac{n^2}{n^3x} = \frac{1}{x} \quad \text{or} \quad \frac{n^2}{n^3x + x(1 + x^3)} \leq \frac{n^2}{x(1 + x^3)}.
$$
The former is not integrable and I cannot seem to grapple with the $n^2$ on the latter and sufficiently bound it. So my main question is how can I bound $|f_{n}|$?
| $$\int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x=\int\limits_0^\infty \frac{n^3 \sin(y)}{n^4y + n y(1 +n^3 y^3)} \, d y=\frac{1}{n} \int\limits_0^\infty \frac{\sin(y)}{y + y^3 +y/n^3} \, d y$$
Here we make the change of variables $y=n x$.
The integral $$ \int\limits_0^\infty \frac{\sin(y)}{y + y^4 +y/n^3}$$ is finite. Thus, the limit under discussion tends to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Coefficient of $x^3$ in $(1-2x+3x^2-4x^3)^{1/2}$ The question is to find out the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$
I tried using multinomial theorem but here the exponent is a fraction and I couldn't get how to proceed.Any ideas?
| It is quite practical to exploit the identity
$$ (1+x)^2 (1-2x+3x^2-4x^3) = 1-5x^4-4x^5 \tag{A}$$
from which
$$\begin{eqnarray*} \sqrt{1-2x+3x^2-4x^3} &=& \frac{\sqrt{1-5x^4-4x^3}}{1+x} \\&=&\left(1+O(x^4)\right)(1-x+x^2-x^3+O(x^4))\tag{B}\end{eqnarray*}$$
and the coefficient of $x^3$ in the RHS of $(B)$ is trivially $\color{red}{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2555399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Find the sum of the series of $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+...$ Find the sum of the series
$$\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots$$
My attempt solution:
$$\frac13\cdot\left(1+\frac15\right)+\frac17\cdot\left(\frac15+\frac19\right)+\frac1{11}\cdot\left(\frac19+\frac1{13}\right)+\cdots$$
$$=\frac13\cdot\left(\frac65\right)+\frac17\cdot \left(\frac{14}{45}\right)+\frac1{11}\cdot\left(\frac{22}{117}\right)+\cdots$$
$$=2\cdot\left(\left(\frac15\right)+\left(\frac1{45}\right)+\left(\frac1{117}\right)+\cdots\right)$$
$$=2\cdot\left(\left(\frac15\right)+\left(\frac1{5\cdot9}\right)+\left(\frac1{9\cdot13}\right)+\cdots\right)$$
It is here that I am stuck. The answer should be $\frac12$ but I don't see how to get it. Any suggestions?
Also, a bit more generally, are there good books (preferably with solutions) to sharpen my series skills?
| $$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\frac{1}{9\cdot 11}...=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}\right)+\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\cdots=\dfrac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Limit of $\lim_{n\to\infty}n\cdot \sqrt[n]{\sin\left(\frac{1}{1}\right)\cdot \sin\left(\frac{1}{2}\right)\cdots\sin\left(\frac{1}{n}\right)}$ $$\lim _{n\to \infty }\left(n\cdot \sqrt[n]{\sin\left(\frac{1}{1}\right)\cdot \sin\left(\frac{1}{2}\right)\cdot \cdot \cdot \sin\left(\frac{1}{n}\right)}\right)$$
I have noticed that the nth root is in fact the nth term in the series of geometric means of the series $ \sin\left(\frac{1}{n}\right)$.
Therefore i believe that the series behaves like $ \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}}$ and therefore approaches 1. but i cant find a way of proving this.
I see that for each n we have $ \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}} \le \left(n\cdot \sqrt[n]{\sin\left(\frac{1}{1}\right)\cdot \sin\left(\frac{1}{2}\right)\cdot \cdot \cdot \sin\left(\frac{1}{n}\right)}\right)$ and that this implies that the limit is greater than or equal to 1 but cant get much further than this.
This problem is meant to be solved without the notion of continuity and without certainly without taylor series expansion, lhospitals rule etc..
Any help would be appreciated. Thanks!
| $$\begin{align}\lim _{n\to \infty }\left(n\cdot \sqrt[n]{\sin\left(\frac{1}{1}\right)\cdot \sin\left(\frac{1}{2}\right)\cdot \cdot \cdot \sin\left(\frac{1}{n}\right)}\right)&=\lim _{n\to \infty }\left(n\cdot \sqrt[n]{\prod_{i=1}^n \sin\dfrac1i}\right) &\\=& \lim _{n\to \infty }\left(n\cdot \sqrt[n]{\prod_{i=1}^n \sin\dfrac1i}\right) &\\=& \lim _{n\to \infty }\left(\dfrac{n}{\sqrt[n]{n!}}\cdot \sqrt[n]{ \prod_{i=1}^n \dfrac{\sin (1/i)}{1/i}}\right)&\\=& \lim _{n\to \infty }\dfrac{n}{\sqrt[n]{n!}}\cdot\lim_{n\to \infty} \sqrt[n]{ \prod_{i=1}^n \dfrac{\sin (1/i)}{1/i}}&\\=& e\cdot\exp\left(\lim_{n\to\infty}\dfrac1n\sum^n_{i=1}\ln\left(\dfrac{\sin (1/i)}{1/i}\right)\right)&\\=& e\cdot\exp\left(0\right)=e \end{align}$$
I used Finding the limit of $\frac {n}{\sqrt[n]{n!}}$ and Convergence and value of infinite product $\prod^{\infty}_{n=1} n \sin \left( \frac{1}{n} \right)$?.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate the limit $\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}$ which involves rational functions?
Find $$\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}.$$
I have tried rationalizing but there is no pattern that I can observe.
Edit:
So we forget about the $x$ that is multiplied to both the functions and try to work with the expression $(x^2+1)^{1/2}-(x^3+1)^{1/3}.$ Thus we have, $$(x^2+1)^{1/2}-(x^3+1)^{1/3}=\frac{((x^2+1)^{1/2}-(x^3+1)^{1/3})((x^2+1)^{1/2}+(x^3+1)^{1/3})}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}$$
$$=\frac{x^2+1-(x^3+1)^{2/3}}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}=\frac{(x^2+1)^2-(x^3+1)^{4/3}}{(x^2+1)^{3/2}+(x^2+1)(x^3+1)^{1/3}+x^3+1+(x^3+1)^{2/3}(x^2+1)}=??$$
| I like the Taylor expansion approach, but wondered if there was a more "Cal I" approach.
Using
$$
A-B = (A^{1/6} - B^{1/6})(A^{5/6} + A^{4/6}B^{1/6} + A^{3/6}B^{2/6} + \cdots + B^{5/6})
$$
with $A = (x^2+1)^3$ and $B = (1+x^3)^2$ gives
$$
\begin{multline*}
(x^2+1)^{1/2} - (x^3+1)^{1/3}\\
=\frac{(x^2+1)^3 - (x^3+1)^2}{(x^2+1)^{5/2}+ (x^2+1)^{2}(1+x^3)^{1/3} + \cdots + (1+x^3)^{5/3}}\\
=\frac{3x^4+2x^3+3x^2}{(x^2+1)^{5/2}+ (x^2+1)^{2}(1+x^3)^{1/3} + \cdots + (1+x^3)^{5/3}}
\end{multline*}
$$
Now
$$
\begin{multline*}
x((x^2+1)^{1/2} - (x^3+1)^{1/3})\\
=\frac{x^5(3+2/x+3/x^2)}{x^5((1+1/x^2)^{5/2}+ (1+1/x^2)^{2}(1+1/x^3)^{1/3} + \cdots + (1+1/x^3)^{5/3})}\\
\end{multline*}
$$
which goes to $3/6$ as $x\to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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$\iiint (y^2+z) dxdydz$ under region $x+y+z=2, x=2, y=1, z=y$ I'm having trouble trying to find the interval.
I tried mapping it to $u=x+y+z$, $v=x$, $w=y$ to get
$$u=2$$
$$v=2$$
$$w=1$$
$$u-v-w=w$$
Is this in the right direction?
| Your region is the union of the two regions $\{(x,y,z)\in\mathbb{R}^3:2-y-z\leq x\leq 2, z\leq y\leq 1, 0\leq z\leq 1\}$ and $\{(x,y,z)\in\mathbb{R}^3:2-y-z\leq x\leq 2, -z\leq y\leq 1, -1\leq z\leq 0\}$.
This can be determined by sketching the region in the $(y,z)$ plane.
Your integral is
$$\int_0^1\int_z^1\int_{2-y-z}^2(y^2+z)\,dx\,dy\,dz + \int_{-1}^0\int_{-z}^1\int_{2-y-z}^2(y^2+z)\,dx\,dy\,dz$$
The first integral can be calculated as follows:
$$\begin{align}\int_0^1\int_z^1\int_{2-y-z}^2(y^2+z)\,dx\,dy\,dz &= \int_0^1\int_z^1x(y^2+z)|_{2-y-z}^2\,dy\,dz\\
&= \int_0^1\int_z^1(y^3+yz+zy^2+z^2)\,dy\,dz\\
&=\int_0^1 \left(\frac{y^4}{4}+\frac{y^2}{2}+\frac{zy^3}{3}+z^2y\right)\Bigg|_z^1\,dz\\
&=\int_0^1 \left(\frac{1}{4}+\frac{5}{6}z+z^2-\frac{3}{2}z^3-\frac{7}{12}z^4\right)\,dz\\
&=\left(\frac{1}{4}z+\frac{5}{12}z^2+\frac{1}{3}z^3-\frac{3}{8}z^4-\frac{7}{60}z^5\right)\Bigg|_0^1\\
&=\frac{61}{120}
\end{align}$$
The second integral can be calculated similarly, and is equal to $\frac{7}{120}$. Therefore, the answer is $\frac{68}{120}=\frac{17}{30}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Solution for this inequality $a^2 - 4b + 8 > 0$ I dont know where I messed up, this is what I have done:
This is the isolated x of a function
$$x = \frac{1}{2}\biggl(\pm\sqrt{a^2 - 4 b + 8} - a - 4\biggr)$$
The root above must be positive and greater than $0$ so a function could have inflection points.
So: $a^2 - 4 b + 8 > 0$
$b$ isolated:
$b<\frac{1}{4}\ (a^2 + 8)$
And $a$ isolated:
$a> \sqrt{4b-8}$
$b$ must be $\geq 2$ so that the root above could be real
With this in mind
if $a=1$
then $b<\frac{9}{4}\ $
And here is where I dont know what's wrong because It works even if I use $b$ values greater than $\frac{9}{4}\ $
| To guarantee that exist
$$\sqrt{a^2 - 4 b + 8} $$
you have to set
$$a^2 - 4 b + 8 \geq 0\implies a^2\geq4b-8$$
and consider 2 cases
*
*$$4b-8<0 \implies a^2\geq0>4b-8$$
*$$4b-8\geq0 \implies -\sqrt{4b-8}\leq a^2\leq \sqrt{4b-8}$$
To solve
$$x = \frac{1}{2}\biggl(\pm\sqrt{a^2 - 4 b + 8} - a - 4\biggr)>0 \iff\pm\sqrt{a^2 - 4 b + 8} - a - 4>0$$
consider two cases
CASE 1
$$\sqrt{a^2 - 4 b + 8} - a - 4>0 \implies\sqrt{a^2 - 4 b + 8}>a + 4$$
The system to solve is
$$\begin{cases}\sqrt{a^2 - 4 b + 8}>a + 4\\ a^2 - 4 b + 8 \geq 0\end{cases}$$
If:
$a+4<0$ you only have to check that $a^2 - 4 b + 8 \geq 0$
$a+4\geq0$ you can square and the system becomes
$$\begin{cases}a^2 - 4 b + 8>a^2+8a + 16\\ a^2 - 4 b + 8 \geq 0\end{cases}\implies \begin{cases}2a+b+2<0\\ a^2 - 4 b + 8 \geq 0\end{cases}$$
CASE 2
$$-\sqrt{a^2 - 4 b + 8} - a - 4>0 \implies a+4<-\sqrt{a^2 - 4 b + 8}$$
The system to solve is
$$\begin{cases}a+4<-\sqrt{a^2 - 4 b + 8}\\ a^2 - 4 b + 8 \geq 0\end{cases}$$
If:
$a+4\geq0$ no solutions
$a+4<0$ you can square reversing the sign and the system becomes
$$\begin{cases}a^2+8a + 16>a^2 - 4 b + 8\\ a^2 - 4 b + 8 \geq 0\end{cases}\implies \begin{cases}2a+b+2>0\\ a^2 - 4 b + 8 \geq 0\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Splitting the square root of complex function into real and imaginary parts I have these functions below: $$\sqrt{(x+iy)^2-a^2}$$
$$\frac{b(x+iy)}{\sqrt{(x+iy)^2-a^2}}$$
How do I split these to get the real and imaginary parts of these functions?
If anyone could help me out, that really would be helpful!!!!
It will help me with the method of manufactured solutions for my fracture problem that has the Westergaard solution (as the analytical solution)!!!
Thank you very much,
Mousumi
| Let $\sqrt{(x+iy)^2-a^2}=c+id$
$\implies (x+iy)^2-a^2=(c+id)^2$
Comparing the real & the imaginary parts
$\implies c^2-d^2=x^2-y^2-a^2\ \ \ \ (1)$
$2cd=2xy$
$(c^2+d^2)^2=(c^2-d^2)^2+(2cd)^2=(x^2-y^2-a^2)^2+(2xy)^2=(x^2+y^2)^2+a^4-2a^2(x^2-y^2)$
$c^2+d^2=\sqrt{(x^2+y^2)^2+a^4-2a^2(x^2-y^2)}\ \ \ \ (2)$
Use $(1),(2)$
Observe that $cd$ will have the same sign as $xy$
| {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd The question:
Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd.
Hence, prove $7 | 6^n + 8^n \iff n ~$ is odd
I realise that this is a proof by induction, and this is what I have so far:
\begin{align}
f(n) & = 6^n + 8^n \\
& = (2\cdot 3)^n + (2^3)^n \\
& = 2^n \cdot 3^n + 2^{3n} \\
\end{align}
\begin{align}
\text{Assume} ~ f(k) & = 6^k + 8^k \\
& = 2^k \cdot 3^k + 2^{3k} \\
f(k+1) & = 2^{k+1} \cdot 3^{k+1} + 2^{3(k+1)} \\
\end{align}
Where do I go from here?
| $6\equiv1\pmod7, ~8\equiv-1\pmod7$, hence $$6^n+8^n\equiv1^n+(-1)^n\pmod7$$ $1^n+(-1)^n=0~\iff~n$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2567658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the probability of getting at least one pair in Poker? Problem:
A 5-card hand is dealt from a perfectly shuffled deck of playing cards. What is the probability of each
of the hand has at least two cards with the same rank.
Answer:
By a rank, I mean a card like a $2$ or a king. The set of all poker hands is ${52 \choose 5}$. Let $p$ be the
probability we seek. I note that there are $13$ ranks of cards and for each rank there are $4$ cards.
\begin{eqnarray*}
p &=& \frac{ 13{ 4 \choose 2 }{ 50 \choose 3 } } { { 52 \choose 5 } } \\
{ 4 \choose 2} &=& \frac{4(3)(2)}{2} = 12 \\
{50 \choose 3} &=& \frac{50(49)(48)}{3(2)} = 25(49)(16) \\
%
{52 \choose 5} &=& \frac{52(51)(50)(49)(48)}{5(4)(3)(2)} = \frac{52(51)(10)(49)(48)}{4(3)(2)} \\
&=& \frac{52(51)(10)(49)(12)}{3(2)} = 52(51)(10)(49)(2) \\
p &=& \frac{ 13(12)(25)(49)(16) } { 52(51)(10)(49)(2) } = \frac{ 13(12)(25)(49)(8) } { 52(51)(10)(49) } \\ &=& \frac{ 13(6)(25)(8) } { 52(51)(5) } = \frac{ 13(3)(25)(8) } { 26(51)(5) } \\ &=& \frac{ 13(3)(5)(8) } { 26(51) } \\ &=& \frac{ 1560 } {1586 } \\
\end{eqnarray*}
I am fairly sure that my answer is wrong but I do not understand where I went wrong. I am hoping somebody can tell me where I went wrong.
Thanks,
Bob
| As Eric Fisher pointed out in his answer, the probability of obtaining at least one pair can be found by subtracting the probability of obtaining no pairs from $1$. If no pairs are obtained, we must select five different ranks. There are $\binom{13}{5}$ ways to select the five ranks. For each rank, we must select one of four suits. Hence, the number of hands that do not contain a pair is
$$\binom{13}{5}4^5$$
so the probability of not obtaining a pair is
$$\frac{\dbinom{13}{5}4^5}{\dbinom{52}{5}}$$
Hence, the probability of obtaining at least one pair is
$$1 - \frac{\dbinom{13}{5}4^5}{\dbinom{52}{5}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$? Consider the following:
$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$
$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$
$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$
In General is it true for further increase i.e.,
Is
$$\sum_{i=1}^n i^5+i^7=2\left( \sum_{i=1}^ni\right)^4$$ true $\forall $ $n \in \mathbb{N}$
| Here are some interesting observations, which are too long to be included as a comment.
A useful reference can be found here.
Denoting $\displaystyle\sum_{r=1}^n r^m=\sigma_m$, we note that
$$\begin{align}\sigma_3&=\sigma_1^2\tag{1}\\
\frac{\sigma_5}{\sigma_3}&=\frac {4\sigma_1-1}{3}\tag{2}\\
\frac {\sigma_7}{\sigma_3}&=\frac {6\sigma_1^2-4\sigma_1+1}3\tag{3}\end{align}$$
Adding $(2),(3)$ and using $(1)$ gives
$$\begin{align}
\frac {\sigma_5+\sigma_7}{\sigma_3}&=2\sigma_1^2=2\sigma_3\\
\sigma_5+\sigma_7&=2\sigma_3^2=2\sigma_1^4\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 3
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How can I calculate $\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$ without L'Hôpital's rule? How can I calculate following limit without L'Hôpital's rule
$$\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$$
I tried L'Hôpital's rule and I found the result $2$.
| $$\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}\sqrt[3] {\cos 3x}}{x^2}=$$
$$=\lim_{x\rightarrow0}\frac{\cos^6x-\cos^32x\cos^23x}{6x^2}=$$
$$=\lim_{x\rightarrow0}\frac{\cos^2x(1-\cos^2x)(128\cos^8x-256\cos^6x+200\cos^4x-69\cos^2x+9)}{6x^2}=$$
$$=\frac{128-256+200-69+9}{6}=2.$$
I used $$x^6-y^6=(x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^5+y^5),$$
$$\lim_{x\rightarrow0}\frac{\sin{x}}{x}=1$$ and
$$\lim_{x\rightarrow0}\cos{x}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Evaluate integral : How to evaluate integral
$$\int \frac{x^2}{\sqrt{x^2-25}}dx=?$$
My Try :
$$\int \frac{x^2+25-25}{\sqrt{x^2-25}}dx=\int \frac{x^2-25}{\sqrt{x^2-25}}+\frac{25}{\sqrt{x^2-25}}dx$$
$$\int \frac{x^2-25}{\sqrt{x^2-25}}+\int\frac{25}{\sqrt{x^2-25}}dx$$
Now what ?
| Continuing with your method, $$\int \frac{x^2}{\sqrt{x^2-25}}\,dx=\int \left(\sqrt{x^2-25} + \frac{25}{\sqrt{x^2-25}}\right)\,dx$$ Now $$\int \sqrt{x^2-25}\, dx=\frac{x\sqrt{x^2-25}}2-\frac{25}2\ln(x+\sqrt{x^2-25})+C_1 \tag{1}$$ using integration by parts, with $f(x)=x^2-25$ and $g'(x)=\dfrac1{\sqrt{x^2-25}}$.
Also, $$\int \frac{25}{\sqrt{x^2-25}}\,dx=25\ln(x+\sqrt{x^2-25})+C_2 \tag{2}$$ where $C_1$ and $C_2$ are constants.
Adding $(1)$ and $(2)$ together, we have $$\boxed{\int \frac{x^2}{\sqrt{x^2-25}}\,dx=\frac{x\sqrt{x^2-25}}2+\frac{25}2\ln(x+\sqrt{x^2-25})+C}$$ where $C=C_1+C_2$.
P.S. This works for all numbers, not just square numbers. In general, $$\int \frac{x^2}{\sqrt{x^2-a}}\,dx=\frac{x\sqrt{x^2-a}}2+\frac{a}2\ln(x+\sqrt{x^2-a})+C$$ for $a \in \mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find this binomial summation How to calculate this?$$\sum_{x = 0}^n \sum_{y = 0}^n a^x b^y\binom{x + y}{y}$$
I tried the following approach by replacing $x + y = t$ and then solving $\sum_{t\ =\ 0}^{2n}\sum_{r=max(0,\ t - n)}^{min(n,\ t)} a^r b^{t - r}\binom{t}{r}$ but this got me nowhere. Maybe we can reduce it to one summation instead of two which can be solved in $\mathcal{O(n)}$ time then?
| We can generate a recurrence like this.
Let's denote $S(t) = \sum_{r\ =\ max(0, t - n)}^{min(n, t)}a^rb^{t - r}\binom{t}{r}$. Now $\forall 1 \leq t \leq n,\ S(t) = (a+b)S(t - 1)$
So, for calculating $S(t),\ \forall t > n$ notice \begin{align*}
S(n + 1) = (a+b)^{n+1} - \binom{n + 1}{0}a^{n + 1} - \binom{n + 1}{0}b^{n + 1}\\
S(n + 1) = (a + b)S(n) - \binom{n + 1}{0}a^{n + 1} - \binom{n + 1}{0}b^{n + 1}
\end{align*}
Now, using the identity $$\binom{n}{r} = \binom{n-1}{r - 1} + \binom{n}{r - 1}$$. We can write $S(n + 2)$ as
\begin{align*}
S(n + 2) = (a + b) * S(n + 1) - \binom{n + 2}{1}a^{n+1}b^1 - \binom{n + 2}{1}a^1b^{n + 1}
\end{align*}
Similarly, we can calculate $$S(i) = (a+b)S(i-1) - \binom{i}{i - n - 1}a^{n + 1}b^{i - n - 1} - \binom{i}{i - n - 1}b^{n + 1}a^{i -n - 1}$$
Hence, each time we are decreasing two appropriate terms after multiplying it $(a + b)$. This all works because of the identity stated above. Hence, you can calculate it all $S(i)$ in $\mathcal{O(n)}$ time.
| {
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"timestamp": "2023-03-29T00:00:00",
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Feasibility and boundedness of non-linear programming In linear programming, we have the following results:
*
*primal problem infeasible --> dual problem infeasible or unbounded
*dual problem infeasible --> primal problem infeasible or unbounded
*primal problem unbounded --> dual problem infeasible
*dual problem unbounded --> primal problem infeasible
For a non-linear convex optimization problem, the last 2 items still hold, since the weak duality $d^* \le p^*$ ($d^*$ is dual optimal and $p^*$ is primal optimal).
Then, for a non-linear programming problem, do the first 2 items hold?
| Here's an example involving a primal-dual pair of semidefinite programming problems in which the primal has multiple optimal solutions but the dual is infeasible.
Our SDP primal-dual pair has the primal problem:
$\min \mbox{tr}(CX)$
subject to
$\mbox{tr}(A_{i}X)=b_{i}$, $i=1, 2, \ldots, m$
$X \succeq 0$
with the dual problem:
$\max b^{T}y$
subject to
$\sum_{i=1}^{m} y_{i}A_{i} + Z = C$
$Z \succeq 0$.
Consider the primal-dual pair with
$C=\left[
\begin{array}{cccc}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}
\right]
$
$A_{1}=\left[
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1
\end{array}
\right]
$
$b_{1}=0$
$A_{2}=\left[
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0
\end{array}
\right]
$
$b_{2}=0$
Since $X \succeq 0$, we must have $X_{i,i} \geq 0$, $i=1, 2, 3, 4$. From the second primal constraint, $X_{1,1}+X_{3,3}=0$, so $X_{1,1}=0$ and $X_{3,3}=0$. From the first primal constraint, $X_{2,2}=X_{4,4} \geq 0$. If $X_{i,i}=0$ and $X$ is positive semidefinite, then $X_{i,j}=X_{j,i}=0$. Thus the feasible solutions to the primal problem are of the form:
$X=\left[
\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & X_{2,2} & 0 & X_{2,4} \\
0 & 0 & 0 & 0 \\
0 & X_{4,2} & 0 & X_{4,4}
\end{array}
\right]
$
where $X_{2,2}=X_{4,4}\geq 0$.
Note that all of these primal feasible solutions are singular matrices. Thus we have no solutions that are strictly feasible with respect to the positive semidefiniteness constraint and Slater's condition is not satisfied.
Because $C$ has $0$ entries in the relevant positions, all solutions of this form have the primal optimal objective function value of $0$.
Next, consider the dual constraints
$y_{1}A_{1}+y_{2}A_{2}+Z=C$
$Z \succeq 0$.
Since the (1,2) and (2,1) entries in $A_{1}$ and $A_{2}$ are 0, we must have $Z_{1,2}=Z_{2,1}=1$. Since $Z$ is positive semidefinite, we must
have $Z_{1,1} > 0$ and $Z_{2,2}>0$ (in fact, $Z_{1,1}Z_{2,2} \geq 1$.) Note that these are strict inequalities! From the (2,2) position in the dual constraints, we have $y_{1}+Z_{2,2}=C_{2,2}=0$. Thus $y_{1}<0$.
From the (4,4) position in the dual constraint, we have
$-y_{1}+Z_{4,4}=C_{4,4}=0$. However, since
$Z$ is positive semidefinite we must also have $Z_{4,4} \geq 0$. We can't simultaneously have $-y_{1}>0$ and $Z_{4,4} \geq 0$ and $-y_{1}+Z_{4,4}=0$. Thus the dual problem is infeasible.
| {
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Aczel's inequality Can anyone here please explain how to notice Aczel's identity in a certain question. I mean in questions involving inequalities, on understanding the inequality one gets a hint to apply a certain inequality e. g. Cauchy Schwarz, Jensen's inequality, Chebyshev's inequality, etc. So how does a question involving Aczel's inequality be identified.
Eg. In this question how does one get to know that this question uses Aczel's identity.
Suppose $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are real numbers such that $$(a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1)(b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 - 1) > (a_1 b_1 + a_2 b_2 + \cdots + a_n b_n - 1)^2.$$Prove that $a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 > 1$ and $b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 > 1$
Note: If someone wishes he/she can post the solution to this problem.
| Suppose $$(a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1)(b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 - 1) > (a_1 b_1 + a_2 b_2 + \cdots + a_n b_n - 1)^2 \tag{A}$$
The quantity on the RHS is nonnegative so for signs to work out there are just two cases:
$$(a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1) > 0 \text{ and } (b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 - 1) > 0\tag{1} $$
$$(a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1) < 0 \text{ and } (b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 - 1) < 0\tag{2} $$
In case (1), we have immediately $a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 > 1$ and $b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 > 1$.
In case (2) we have that $1^2 > a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2$ (and $1^2 > b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2$) so by Aczel's inequality
$$(1-a_1 ^ 2 - a_2 ^ 2 - \cdots - a_n ^ 2)(1-b_1 ^ 2 - b_2 ^ 2 - \cdots - b_n ^ 2) \leq (1 - a_1 b_1 + a_2 b_2 + \cdots + a_n b_n)^2$$ which contradicts (A).
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that this polynomial has three different real roots There is polynomial $f(x)=x^3 +ax^2+bx+c$ and $ab=9c$ and $b<0$. Like in the title. I've only come up so far with $9x^3 + 9ax^2 + 9bx+ab=0$ and factoring here does nothing.
| I have expanded Eric Towers' comments into an answer. Hope this can help OP. The second method is more efficient and error-prone doesn't work well, unlike the first one.
Method 1: Discriminant
Given that $ab=9c$ and $b<0$, the discriminant is
\begin{align}
\Delta &= 18(1)abc-4a^3c+a^2b^2-4(1)b^3-27(1)^2c^2 \\
&= 162c^2-4a^3c+81c^2-4b^3-27c^2 \\
&= 216c^2-4a^3c-4b^3 \\
&= 4\left(54c^2-a^3c-b^3\right) \\
&= 4\left[54c^2-\left(\frac{9c}{b}\right)^3c-b^3\right] \\
&= 4\left[54c^2+\frac{3^6c^4}{(-b)^3}+(-b)^3\right] \\
&= \frac{4}{(-b)^3} \left[\left(3^3c^2\right)^2+2\left(3^3c^2\right)(-b)^3+(-b)^6\right] \\
&= \frac{4}{\underbrace{(-b)^3}_{b<0}} (\underbrace{27c^2}_{\ge0}+\underbrace{(-b)^3}_{b<0})^2 > 0
\end{align}
Hence $f(x)=x^3 +ax^2+bx+c$ has three distinct real roots.
Method 2: Descartes rule of signs
Edit: As pointed out in the comments, this argument is incomplete. I tried proof by contradiction, and dividing by $x-r$ ($r$ is the positive root) in the case $a>0$, but I can't figure out how this works.
I'm leaving this, hoping some motivated user complete this
Refer to Wiki's example to know how to apply this result.
Rewrite $f(x)=x^3 +ax^2+bx+c$ into an array of $\pm$: $(+,?,-,?)$. Since $ab=9c$ with $b<0$, the first and the third $?$'s are opposite to each other.
\begin{array}{c|c|c}
\text{Case} & \text{+ve roots} & \text{-ve roots} \\ \hline
a>0 & +,+,-,- & -,+,+,- \\
& \text{1 sign change} & \text{2 sign changes} \\ \hline
a<0 & +,-,-,+ & -,-,+,+ \\
& \text{2 sign changes} & \text{1 sign change} \\ \hline
a=0 & +,\phantom{+},-,\phantom{+} & -,\phantom{+},+,\phantom{+} \\
& \text{1 sign change} & \text{1 sign change}
\end{array}
*
*Divide the situation into three cases according to the sign of $a$.
*Extract the signs from $f(x)$ and write them down in the first column.
*Count the number of sign changes ($+,-$ or $-,+$) in each cell. This represents the parity of the number of positive root(s) in each case.
*Write down the signs for $f(-x)$ in the second column.
*
*Copy the signs representing even powers of $x$.
*Invert the signs representing odd powers of $x$.
*Repeat (3) in the second column. This represents the parity of the number of negative root(s) in each case.
*
*If $a\ne0$, we can't conclude anything from it.
*If $a=0$, the condition $ab=9c$ gives $c=0$, so $x=0$ is a root not counted by this method. We have three distinct real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Solving the quadratic congruence (using some lemma)
Find all $x$'s such that $x^2 \equiv 17 \pmod{ 128}$
Now, I've read one solution here (fixed URL). What I don't understand is how in the first place one can find a particular solution, $a$? (in order to find the all four).
| $$x^2 \equiv 17 \pmod{128}$$
$$x^2 \equiv 1 \pmod 4 \implies x \in \{1,3\}$$
\begin{align}
(1+4n)^2 &\equiv 1 \pmod{16} \\
1 + 8n &\equiv 1 \pmod{16} \\
8n &\equiv 15 \pmod{16} \\
n &\in \{ \ \}
\end{align}
\begin{align}
(3+4n)^2 &\equiv 1 \pmod{16} \\
9 + 8n &\equiv 1 \pmod{16} \\
8n &\equiv 8 \pmod{16} \\
n &\equiv 1 \pmod{2} \\
x &\in \{ 7,15 \} \\
x &\in \{ 1, 9 \} &\text{(That's $-15$ and $-7$)}
\end{align}
\begin{align}
(1+16n)^2 &\equiv 17 \pmod{128} \\
1 + 32n &\equiv 17 \pmod{128} \\
32n &\equiv 16 \pmod{128}
\end{align}
\begin{align}
(7+16n)^2 &\equiv 17 \pmod{128} \\
49 + 96n &\equiv 17 \pmod{128} \\
96n &\equiv 96 \pmod{128} \\
3n &\equiv 3 \pmod{4} \\
n &\equiv 1 \pmod{4} \\
x &\in \{23, 87\} \\
x &\in \{41, 105\} &\text{(That's $-87$ and $-23$)}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2579196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Relation between integral, gamma function, elliptic integral, and AGM The integral $\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}}$ is equal to $\displaystyle \frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi }}$.
It is calculated or verified with a computer algebra system that $\displaystyle \frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi }} = K\left(\frac{1}{2}\right)$ , where $K(m)$ is the complete elliptic integral of the first kind. This is in relation to what is called the elliptic integral singular value.
It is also known or verified that
$\displaystyle K\left(\frac{1}{2}\right) =\displaystyle \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{1-\frac{\sin ^2(t)}{2}}} \, dt= \frac{1}{2} \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{\sin (t) \cos (t)}} \, dt$.
Can one prove directly or analytically that
$\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}} =\frac{1}{2} \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{\sin (t) \cos (t)}} \, dt =\displaystyle \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{1-\frac{\sin ^2(t)}{2}}} \, dt = K\left(\frac{1}{2}\right) $ ?
| This one is a cakewalk. Just use the definition $$K(1/2)=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-(1/2)\sin^{2}x}}=\frac{\sqrt{2}}{2}\int_{0}^{\pi}\frac{dx}{\sqrt{2-\sin^{2}x}}$$ and then put $t=\tan (x/2)$ so that $dx=2/(1+t^2)\,dt$ and $$2-\sin^{2}x=2-\frac{4t^2}{(1+t^2)^{2}}=2\cdot\frac{1+t^{4}}{(1+t^{2})^{2}}$$ and hence we have $$K(1/2)=\frac{1}{\sqrt{2}}\int_{0}^{\infty}\frac{1+t^{2}}{\sqrt{2}\sqrt{1+t^{4}}}\cdot\frac{2}{1+t^2}\,dt=\int_{0}^{\infty}\frac{dt}{\sqrt{1+t^{4}}}$$ Similarly put $\tan t=x^2$ to get $$\frac{1}{2}\int_{0}^{\pi/2}\frac{dt}{\sqrt{\sin t\cos t}} =\int_{0}^{\infty} \frac{dx} {\sqrt{1+x^{4}}}$$ This avoids the more complicated approaches from the theory of elliptic and theta functions and AGM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2579911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Probability for a Poisson Process I have a problem with a solution. I find it hard to understand some of the steps in the solving process. Can someone help me understand whats going on here?
"Calculate the probability P(X(1)=1, X(4)=4|X(2)=2, X(3)=3, X(5)=5) for a Poisson process X(t) with arrival rate λ > 0."
Solution:
\begin{align}
(i) \ \ P(X(1)=1, X(4)=4|X(2)=2, X(3)=3, X(5)=5) \\
\
\end{align}
\begin{align}
(ii) = \ \frac{P(X(1)=1, X(4)=4, X(2)=2, X(3)=3, X(5)=5)}{ P(X(2)=2, X(3)=3, X(5)=5)} \\
\
\end{align}
\begin{align}
(iii) = \ \frac{P(X(1)=1, X(2)−X(1) = 1, X(3)−X(2) = 1, X(4)−X(3) = 1, X(5)−X(4) = 1)}{ P(X(2)=2, X(3)−X(2) = 1, X(5)−X(3) = 2)} \\
\
\end{align}
\begin{align}
(iiii) = \ \frac{[P(X(1)=1)]^5}{P(X(2)=2)P(X(2)=1)P(X(2)=2)} \\
\
\end{align}
\begin{align}
= \ \frac{[[λ^1/((1!) · e^λ)]^4}{[(2λ)2/((2!) · e^2λ)]^2} \\
\
\end{align}
\begin{align}
= \ \frac{1}{4} \\
\
\end{align}
I don´t understand what happens between (ii)->(iii)
P(X(1)=1, X(4)=4, X(2)=2, X(3)=3, X(5)=5) -> P(X(1)=1,X(2)−X(1)=1,X(3)−X(2)=1,X(4)−X(3)=1,X(5)−X(4)=1)
Is this the independent increments property?
I don´t understand why we applying this property here.
Hope someone can explain this solution to me.
| No, it's not using the independent increments property, but it's kind of a preparation to use it… let's consider a simpler example.
You want to calculate: $$P(X(1) = 1, X(2) = 3)$$ unfortunately $X(1)$ and $X(2)$ aren't independent so we cannot split this into two probabilities… but we can rewrite this a bit to get increments for which we know they are independent so we can split it!
So let's do an equivalent transformation to get an increment independent of $X(1)$ to use independence. Obviously if holds:
$$\begin{align*} && X(2) &= 3 \\ \iff && X(2) - X(1) &= 3 - X(1)\end{align*}$$
we just removed $X(1)$ on both sides… so we get:
$$\begin{align*} &\;P(X(1) = 1, X(2) = 3) \\= &\;P(X(1) = 1, X(2) - X(1) = 3 - X(1))\end{align*}$$
But now we have the additional information given, that $X(1) = 1$ so we can replace $X(1)$ on the the far right by $1$ and get:
$$\begin{align*} = &\;P(X(1) = 1, X(2) - X(1) = 3 - 1) \\= &\;P(X(1) = 1, X(2) - X(1) = 2)\end{align*}$$
And now we have the probability of independent increments and can split this and use that $X(2) - X(1) \sim X(1)$ to get
$$\begin{align*} = &\;P(X(1) = 1)\cdot P(X(2) - X(1) = 2) \\= &\;P(X(1) = 1)\cdot P(X(1) = 2)\end{align*}$$
And nothing else was done in the step from (ii) to (iii) just 4 times more to get 5 independent increments.
So actually it's nothing else then transform a system of equations, e.g. for the nominators:
$$X(1)=1, X(4)=4, X(2)=2, X(3)=3, X(5)=5$$ is equivalent to $$X(1)=1, X(2)−X(1) = 1, X(3)−X(2) = 1, X(4)−X(3) = 1, X(5)−X(4) = 1$$ and so it follows both probabilities are the same, hence
$$\begin{align*}
&\; P(X(1)=1, X(4)=4, X(2)=2, X(3)=3, X(5)=5 ) \\ =&\; P(X(1)=1, X(2)−X(1) = 1, X(3)−X(2) = 1, X(4)−X(3) = 1, X(5)−X(4) = 1)\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2582931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Elementary Inequality from a local olympiad test The problem I'm stuck with is the following :
Let $a$, $b$, $c$, $x$, $y$ and $z$ be real numbers such that
$a+x \ge b+y \ge c+z \ge 0 $ and $ a+ b+c = x+y+z$
Prove that $ay+bx \ge ac+xz$
As easy as it sounds, I didn't achieve any significant progress after several tries.
Any advice, hint would be very appreciated. Thanks.
| \begin{align}
(ay+bx) - (ac+xz) &= a(\color{red}{y-c}) + x(b-z) \\
&= a(\color{red}{a+b - x -z}) + x(b-z)\\
&= a(a-x) + (a+x)(b-z)\\
&= \tfrac12(a-x)^2 + \tfrac12(a^2-x^2) + (a+x)(b-z)\\
&= \tfrac12(a-x)^2 + \tfrac12(a+x)(a-x+2b-2z)\\
&= \tfrac12(a-x)^2 + \tfrac12(a+x)(\color{red}{a+b-x-z}+b-z)\\
&= \tfrac12(a-x)^2 + \tfrac12(a+x)(\color{red}{y-c} + b-z)\\
&= \tfrac12(a-x)^2 + \tfrac12(a+x)(b+y-c-z)\\
&\geqslant 0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How many tries on average before I see the same value $N$ times in a row? If I repeatedly roll a fair, $X$-sided die, on average how many rolls should I expect to make before I happen to roll the same value $N$ times in a row?
I've found questions on here with answers when $X=2$, or where $N=2$, or where you're looking for a specific result $N$ times in a row; I'm interested in the situation where I don't care what the specific value is, I just want it to be $N$ times in a row. The closest I can find is Suppose we roll a fair $6$ sided die repeatedly. Find the expected number of rolls required to see $3$ of the same number in succession., but I can't quite figure out how to generalize it beyond $N=3$.
| Let $E_k$ denote the expected number of rolls that are still needed if $k$ rolls have passed with equal result. We write recurrence relations for $E_k,0\leq k\leq N-1$ and calculate the wanted expectation $E=E_0$ from them.
We obtain
\begin{align*}
E_0&=E_1+1\tag{1}\\
E_1&=\frac{1}{X}(E_2+1)+\frac{X-1}{X}(E_1+1)\\
E_2&=\frac{1}{X}(E_3+1)+\frac{X-1}{X}(E_1+1)\\
E_3&=\frac{1}{X}(E_4+1)+\frac{X-1}{X}(E_1+1)\tag{2}\\
&\ \ \ \vdots\\
E_{N-3}&=\frac{1}{X}(E_{N-2}+1)+\frac{X-1}{X}(E_1+1)\\
E_{N-2}&=\frac{1}{X}(E_{N-1}+1)+\frac{X-1}{X}(E_1+1)\\
E_{N-1}&=\frac{1}{X}+\frac{X-1}{X}(E_1+1)\tag{3}\\
\end{align*}
Comment:
*
*In (1) we note that any roll of the $X$-sided die will do.
*In (2) we are in the situation that a run of length $3$ is given. With probability $\frac{1}{X}$ we get a run of length $4$ and with probability $\frac{X}{X-1}$ we get a different result and start with a run of length $1$.
*In (3) we get a run of length $n$ with probability $\frac{1}{X}$ and can stop or we are back at the beginning and start again with a run of length $1$.
This system can be easily solved. We iteratively obtain be respecting all equations from above besides the last one (tagged with (3)).
\begin{align*}
E_1&=E_0-1\\
E_2&=E_0-(1+X)\\
E_3&=E_0-(1+X+X^2)\\
E_4&=E_0-(1+X+X^2+X^3)\\
&\ \ \ \vdots\\
E_{N-1}&=E_0-(1+X+X^2+\cdots+X^{N-2})\\
&=E_0-\frac{X^{N-1}-1}{X-1}\tag{4}\\
\end{align*}
From equation (4) and (3) we finally conclude
\begin{align*}
E_{N-1}&=E_0-\frac{X^{N-1}-1}{X-1}\\
(1-X)E_0+XE_{N-1}&=1
\end{align*}
from which
\begin{align*}
\color{blue}{E=E_0=\frac{X^N-1}{X-1}}
\end{align*}
follows.
Note: This approach is a generalisation of @lulu's answer of this MSE question. In fact we get as small plausibility check with $N=3$ and $X=6$ the result
\begin{align*}
E=\frac{6^3-1}{6-1}=\frac{215}{5}=43.
\end{align*}
in accordance with @lulu's result.
[Add-on] The strongly related problem: Rolling a fair $X$-sided die until we get a run of length $N$ of a specific value can be solved by the system above with the first equation (1) replaced with
\begin{align*}
E_0&=\frac{1}{X}(E_1+1)+\frac{X-1}{X}(E_0+1)
\end{align*}
which gives the result
\begin{align*}
\color{blue}{E=E_0=\frac{X\left(X^N-1\right)}{X-1}}
\end{align*}
This is also plausible compared with the result above, since the probability to roll a specific value is $\frac{1}{X}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find orthonormal vectors for given vectors using Gram-Schmidt Doing an exercise after completing the G.S. Lecture 17 and stuck.
Suppose we have three vectors $\begin{pmatrix}1 \\ 1 \\0 \end{pmatrix}$ , $\begin{pmatrix}1 \\ 0 \\1 \end{pmatrix}$, $\begin{pmatrix}0 \\ 1 \\1 \end{pmatrix}$ and the aim is to transform these vectors into orthonormal vectors $(A,B,C)$ using the Gram-Schmidt process.
My working:
Let $u_1$ = $\begin{pmatrix}1 \\ 1 \\0 \end{pmatrix}$ , $u_2=$ $\begin{pmatrix}1 \\ 0 \\1 \end{pmatrix}$ and $u_3$ $\begin{pmatrix}0 \\ 1 \\1 \end{pmatrix}$
Fix $A=\frac{u_1}{||u_1||}= \frac{1}{\sqrt{2}} \begin{pmatrix}1 \\ 1 \\0 \end{pmatrix}= \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\0 \end{pmatrix}$
Then we can find $ B = \frac{b}{||b||} $ and to find $b$ we use the projection property: $b=b- \frac{u_1^{T}u_2}{u_1^{T}u_1}u_1= \begin{pmatrix}1 \\ 0 \\1 \end{pmatrix} - \frac{1}{2}\begin{pmatrix}1 \\ 1 \\0 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \\ \frac{-1}{2} \\ 1 \end{pmatrix}$
Which means $ B = \frac{\sqrt{6}}{3}\begin{pmatrix} \frac{1}{2} \\ \frac{-1}{2} \\ 1 \end{pmatrix} =\begin{pmatrix} \frac{\sqrt{6}}{6} \\ \frac{-\sqrt{6}}{6} \\ \frac{\sqrt{6}}{3} \end{pmatrix}$
Lastly, we need to find $C=\frac{c}{||c||}$ and to find $c$ we use projection once again:
$ c = u_3 -\frac{u_1^{T}u_3}{u_2^{T}u_1}u_1 - \frac{u_2^{T}u_3}{u_2^{T}u_1}u_2 = \begin{pmatrix}0 \\ 1 \\1 \end{pmatrix} - \frac{1}{2} \begin{pmatrix}1 \\ 1 \\0 \end{pmatrix} - \frac{1}{2} \begin{pmatrix}1 \\ 0 \\1 \end{pmatrix} = \begin{pmatrix} -1 \\ \frac{1}{2} \\ \frac{1}{2}\end{pmatrix}$
But apparently this is wrong, it should be: $\begin{pmatrix} \frac{-1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{pmatrix} $, where did I go wrong?
| HINT
$u_2-u_3$ is orthogonal to $u_1$
in this way you can simplify the calculation assuming $u_1*=u_1$, $u_2*=u_2-u_3$
NOTE
When you calculate c you should use $u_3$, $u_1$ and $b$.
Remember the meaning of the procedure: you are subtracting to $u_3$ the components of $u_3$ on orthogonal vectors $u_1$ and $b$, to do this you use dot product by the unitaru vectors, for example for $u_1$:
$$\left(\frac{u_1}{|u_1|}\cdot u_3\right)\frac{u_1}{|u_1|}=\frac{u_1^{T}u_3}{|u_1|^2}u_1=\frac{u_1^{T}u_3}{u_1^{T}u_1}u_1$$
thus
$$c = u_3 -\frac{u_1^{T}u_3}{u_1^{T}u_1}u_1 - \frac{b^{T}u_3}{b^{T}b}b = \begin{pmatrix}0 \\ 1 \\1 \end{pmatrix} - \frac{1}{2} \begin{pmatrix}1 \\ 1 \\0 \end{pmatrix} - \frac{1}{3} \begin{pmatrix}\frac12 \\ -\frac12 \\1 \end{pmatrix} = \begin{pmatrix} -\frac23 \\ \frac23 \\ \frac23\end{pmatrix}$$
$$|c|=\sqrt{\frac43}=\frac{2}{\sqrt3}\implies C= \begin{pmatrix} -\frac{\sqrt3}{3} \\ \frac{\sqrt3}{3} \\ \frac{\sqrt3}{3}\end{pmatrix}= \begin{pmatrix} -\frac{1}{\sqrt3} \\ \frac{1}{\sqrt3} \\ \frac{1}{\sqrt3}\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculate the determinant of $A-5I$ Question
Let $ A =
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
6 & 7 & 8 & 9 & 10 \\
11 & 12 & 13 & 14 & 15 \\
16 & 17 & 18 & 19 & 20 \\
21& 22 & 23 & 24 & 25
\end{bmatrix}
$.
Calculate the determinant of $A-5I$.
My approach
the nullity of $A$ is $3$, so the algebraic multiplicity of $\lambda = 0$ is $3$, i.e. $\lambda_1 = \lambda_2 = \lambda_3 = 0.$
Now trace($A$) = $\lambda_4 + \lambda_5 = 1+6+13+19+25 = 65$
Then det($A-5I$) = $(\lambda_1-5)(\lambda_2-5)(\lambda_3-5)(\lambda_4-5)(\lambda_5-5)=(-5)^3(\lambda_4\lambda_5 - 5 \times 65 + 25)$
We need to calculate the value of $\lambda_4 \lambda_5$, which includes sum of lots of determinant of $2 \times 2$ matrices.
Is there any fast way to calculate the determinant?
| Row-reducing the matrix $A-5I$ is quicker.
$$
\begin{gather}
\begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
6 & 2 & 8 & 9 & 10 \\
11 & 12 & 8 & 14 & 15 \\
16 & 17 & 18 & 14 & 20 \\
21 & 22 & 23 & 24 & 20
\end{pmatrix} \longrightarrow
\begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
14 & -2 & 2 & 1 & 0 \\
23 & 6 & -1 & 2 & 0 \\
32 & 9 & 6 & -2 & 0 \\
37 & 14 & 11 & 8 & 0
\end{pmatrix} \\
\longrightarrow \begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
14 & -2 & 2 & 1 & 0 \\
-5 & 10 & -5 & 0 & 0 \\
60 & 5 & 10 & 0 & 0 \\
-75 & 30 & -5 & 0 & 0
\end{pmatrix} \longrightarrow
\begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
14 & -2 & 2 & 1 & 0 \\
-5 & 10 & -5 & 0 & 0 \\
50 & 25 & 0 & 0 & 0 \\
-70 & 20 & 0 & 0 & 0
\end{pmatrix} \\
\longrightarrow \begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
14 & -2 & 2 & 1 & 0 \\
-5 & 10 & -5 & 0 & 0 \\
50 & 25 & 0 & 0 & 0 \\
-110 & 0 & 0 & 0 & 0
\end{pmatrix}
\end{gather}
$$
This takes not more than $30$ multiplications in all. The determinant is clearly $$5\cdot (-1) \cdot (-5) \cdot (-25) \cdot (-110) = 68750.$$
I chose to row-reduce the matrix in this fashion to make use of the fact that every entry in the last column is a multiple of $5$. It made the calculations easier when doing it by hand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the value of $n$. Write the value of $n$ if the sum of n terms of the series $1+3+5+7...n =n^2$.
I'm not getting the right value if I proceed with the general formula for finding sum of n terms of a arithmetic series. The general summation formula for arithmetic series is $\frac{n(2a+(n-1)d)}{2}$, where $a$ is the first term, $d$ is the common difference and $n$ is the number of terms.
| If we add $2+4+...+(n-1) = 2(1+2+3+...+\frac{n-1}{2})$ to both sides of the equation, we get
$$1+2+3+...+n = n^2+2\bigg(1+2+3+...+\frac{n-1}{2}\bigg)$$
which can be written as:
$$\frac{n(n+1)}{2} = n^2+2\bigg[\frac{\big(\frac{n-1}{2}\big)\big(\frac{n+1}{2}\big)}{2} \bigg]$$
that is
$$\frac{n^2+n}{2} = n^2+\frac{n^2-1}{4} \implies n^2+n = \frac{5n^2-1}{2} \implies 3n^2-2n-1 = 0 \implies n = 1$$
because $n$ is not a negative number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find max value of $a$ Let $a, x, y, z \in \mathbb{R}_{>0}$ such that $xyz=1$.
Find maximum value of $a$ such that is satisfied the inequality
$$ \sum_{cyc} \frac{x}{(x+1)(y+a)} \ge \frac{3}{2(1+a)} $$
After doing some calculation the max value of $a$ seems to me to be $2$ but I can’t prove it.
| Let $y=z$ and $x\rightarrow+\infty.$
Thus, $$\frac{1}{a}\geq\frac{3}{2(1+a)},$$
which gives $a\leq2$.
We'll prove that $a=2$ it's a maximal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{x}{(x+1)(y+2)}\geq\frac{1}{2}$$ or
$$\sum_{cyc}(2x^2y+xy-3x)\geq0.$$
Now, let $x=\frac{p}{q}$ and $y=\frac{q}{r}$, where $p$, $q$ and $r$ are positives.
Thus, $z=\frac{r}{p}$ and we need to prove that
$$\sum_{cyc}\left(\frac{2p^2}{qr}+\frac{p}{r}-\frac{3p}{q}\right)\geq0$$ or
$$\sum_{cyc}(2p^3+p^2q-3p^2r)\geq0$$ or
$$\sum_{cyc}(2p^3+p^2q-3q^2p)\geq0$$ or
$$\sum_{cyc}\left(p(p-q)(2p+3q)-\frac{5}{3}(p^3-q^3)\right)\geq0$$ or
$$\sum_{cyc}(p-q)^2(p+5q)\geq0.$$
Done!
A bit of easier way.
Let $x=\frac{b}{a}$ and $y=\frac{c}{b},$ where $a,$ $b$ and $c$ are positives.
Thus, by C-S we obtain:
$$\sum_{cyc}\frac{x}{(x+1)(y+2)}=\sum_{cyc}\frac{\frac{b}{a}}{\left(\frac{b}{a}+1\right)\left(\frac{c}{b}+2\right)}=$$
$$=\sum_{cyc}\frac{b^2}{(a+b)(c+2b)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a+b)(c+2b)}=\frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given a fair coin, on average what is the number of tosses needed to get 3 heads in a row? My attempt:
Let $H$ be the outcome heads, and $T$ be tails. Let $X$ be the random variable "number of tosses required to get 3 consecutive heads". The minimum number of trials (tosses) is 3. Thus we have infinite sample space:
$$\{HHH, THHH, TTHHH, HTHHH, TTTHHH, THTHHH, HTTHHH, HHTHHH,....\}$$
For the first 2 elements of sample space, we have the probabilities $(1/2)^{3}$ and $(1/2)^{4}$ respectively. Then we have 2 ways to get 3 consecutive heads out of 5 tosses, hence the probability is $2(1/2)^{5}$. Similar for 4th, 5th and 6th tosses. For $n = 7, 8, 9,... \infty$ tosses, we have
$$2^{n-4} - \frac{(n-5)(n-6)}{2}$$
ways of getting 3 consecutive heads.
Thus the expectation is
$$E(X) = 3(1/2)^{3} + 4(1/2)^{4} + 5 \cdot 2 (1/2)^{5} + 6 \cdot 4(1/2)^{6} + \sum_{n=7}^{\infty} \frac{n}{2^{n}}\left(2^{n-4} - \frac{(n-5)(n-6)}{2}\right) $$
However the infinite sum above doesn't converge, so obviously the approach is wrong.
I have looked at related posts (i.e. Expected Number of Coin Tosses to Get Five Consecutive Heads) but I don't see how my approach fails.
Edit: computation of $$2^{n-4} - \frac{(n-5)(n-6)}{2}, \qquad n \geq 7.$$
I need to compute the number of ways that X -values can be next to each other for a given number of tosses $n \geq 7$. For this $n$, I have a sample space element of the form $...TXXX$, consisting of $n$ symbols $T$ and $H$. Thus, I have a choice of $2^{n-4}$ arrangements for the first $n-4$ symbols. Out of those, I need to exclude the sequences of symbols of the form $XXX...X$, for 3 or more values of $X$. There are $(n-6)$ such arrangements for 3 consecutive $X$ values, $(n-7)$ arrangements for 4 consecutive $X$s, $(n-8)$ arrangements for 5 $X$s, and so on. Hence for the total of such arrangements, we have the arithmetic progression
$$n-6, n-7, n-8,...0.$$
For a given $n$, the sum turns out to be $$\frac{(n-5)(n-6)}{2}$$
Hence, we have the result $2^{n-4} - \frac{(n-5)(n-6)}{2}$ as stated above...
| The more general case of asking for the average number $E$ of rolls of a fair $X$-sided die ending with a run of $N$ consecutive heads is
answered in this MSE post as
\begin{align*}
E=\frac{X(X^N-1)}{X-1}\tag{1}
\end{align*}
Here we apply (1) by considering a coin as two-sided die ($X=2$) and tosses ending in heads with run-length $N=3$. We conclude the average number of trials is
\begin{align*}
\color{blue}{E(X=2,N=3)=\frac{2\left(2^3-1\right)}{2-1}=14}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
For a primitive pythagorean triple $(a, b, c)$, prove $(c+b), (c-b)$ are both squares. Problem
For a primitive pythagorean triple $(a, b, c)$ (with $a$ odd, $b$ even), prove $(c+b), (c-b)$ are both squares.
My thoughts
Having $a$ be odd, gives $a^2$ as odd also.
From $a^2 + b^2 = c^2$ we have $a^2 = (c-b)(c+b)$.
I have figured out that $\gcd(c-b, c+b) = 1$, which seems relevant.
From the fundamental theorem of arithmetic, we now know that the prime factorizations of $c-b, c+b$ will have no common factors. This also seems relevant.
But I'm having a hard time tying all this together.
Any help would be appreciated!
| Given $A=m^2-n^2\quad B=2mn\quad C=m^2+n^2$
$$C-B=m^2+n^2-2mn=(m-n)^2$$
$$C+B=m^2+n^2+2mn=(m+n)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Error while integrating reciprocal of irreducible quadratic $\int{\frac{1}{ax^2 + bx + c}}{dx} = ?$ I am trying to derive a formula for indefinite integral of reciprocal of irreducible real quadratic.
$$\int{\frac{1}{ax^2 + bx + c}}{dx} = ?$$, $b^2 - 4ac \lt 0$. According to WolframAlpa it should come out as: $$\frac{2\arctan{\frac{2ax+b}{\sqrt{4ac - b^2}}}}{\sqrt{4ac - b^2}} + C$$ Instead I get $$\frac{\arctan{\frac{x + \frac{b}{2a}}{\sqrt{\frac{c}{a} - (\frac{b}{2a})^2}}}}{a(\frac{c}{a} - (\frac{b}{2a})^2)^\frac{3}{2}} + C$$ which, no matter how I try to simplify it, doesn't give the right answer. Here are my steps, what am I doing wrong?
1) Reduce to a monic quadratic and complete the square: $$\int{\frac{1}{ax^2 + bx + c}}{dx} = \frac{1}{a}\int{\frac{1}{x^2 + \frac{b}{a}x + \frac{c}{a}}}{dx} = \frac{1}{a}\int{\frac{1}{(x + \frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2}}{dx}$$
3) Substitute $x = u - \frac{b}{2a}$, $u = x + \frac{b}{2a}$, $k = \frac{c}{a} - (\frac{b}{2a})^2$: $$\frac{1}{a}\int{\frac{1}{(x + \frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2}}{dx} = \frac{1}{a}\int{\frac{1}{u^2 + k}}{du}$$
4) Substitute $u = v\sqrt{k}$, $v = \frac{u}{\sqrt{k}}$, $du = \frac{1}{\sqrt{k}}\hphantom ddv$:
$$\frac{1}{a}\int{\frac{1}{u^2 + k}}{du} = \frac{1}{a}\int{\frac{\frac{1}{\sqrt{k}}}{kv^2 + k}}{dv} = \frac{1}{a}\int{\frac{1}{k\sqrt{k}(v^2 + 1)}}{dv} = \frac{1}{ak^\frac{3}{2}}\int{\frac{1}{v^2 + 1}}{dv}$$
5) Use common integral $$\frac{1}{ak^\frac{3}{2}}\int{\frac{1}{v^2 + 1}}{dv} = \frac{\arctan{v}}{ak^\frac{3}{2}} + C$$
6) Substitute back $$\frac{\arctan{v}}{ak^\frac{3}{2}} + C = \frac{\arctan{\frac{u}{\sqrt{k}}}}{ak^\frac{3}{2}} + C = \frac{\arctan{\frac{x + \frac{b}{2a}}{\sqrt{\frac{c}{a} - (\frac{b}{2a})^2}}}}{a(\frac{c}{a} - (\frac{b}{2a})^2)^\frac{3}{2}} + C$$
| In my opinion you can save many hassles with a slightly different method. Set $\sqrt{4ac-b^2}=D$ and do
\begin{align}
\int\frac{1}{ax^2+bx+c}\,dx
&=\int\frac{4a}{4a^2x^2+4abx+4ac}\,dx \\[6px]
&=\int\frac{4a}{(2ax+b)^2+(4ac-b^2)}\,dx \\[6px]
&=\int\frac{4a}{D^2t^2+D^2}\frac{D}{2a}\,dt
&& 2ax+b=Dt,\quad dx=\frac{D}{2a}\,dt \\[6px]
&=\frac{2}{D}\int\frac{1}{t^2+1}\,dt \\[6px]
&=\frac{2}{D}\arctan t+C \\[6px]
&=\frac{2}{\sqrt{4ac-b^2}}\arctan\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)
\end{align}
Note that if you set $u=Kv$, then $du=K\,dv$, not $du=\frac{1}{K}\,dv$. For instance, in the substitution above,
$$
2a\,dx=D\,dt
$$
so
$$
dx=\frac{D}{2a}\,dt
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to prove :$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} <3$ In fact initially I wanted to prove that
$$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} <2$$
Which by the accepted answer here fails to be true.
@Barry Cipra advised me to ask this question in a different post:
How can I now prove that:
$$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} <3$$
Note that the answers here do not provide an estimate of this sequence
Can anyone have any idea?
| For $n\geq 2$,
\begin{align*}
\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} & = 2^{1/4+1/8+1/16+\ldots+1/2^n} \cdot 3^{1/8+1/16+1/32+\ldots+1/2^n}\cdot \ldots \cdot n^{1/2^n} \\
& < 2^{1/2}\cdot 3^{1/4} \cdot 4^{1/8} \cdot \ldots \cdot n^{1/2^{n-1}} \\
& = \sqrt{2\sqrt{3\sqrt{\cdots\sqrt{n}}}} \\
&<3\end{align*}
The last inequality was proved here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 3
} |
Solving $\frac{1}{|x+1|} < \frac{1}{2x}$ Solving $\frac{1}{|x+1|} < \frac{1}{2x}$
I'm having trouble with this inequality. If it was $\frac{1}{|x+1|} < \frac{1}{2}$, then:
If $x+1>0, x\neq0$, then
$\frac{1}{(x+1)} < \frac{1}{2} \Rightarrow x+1 > 2 \Rightarrow x>1$
If $x+1<0$, then
$\frac{1}{-(x+1)} < \frac{1}{2} \Rightarrow -(x+1) > 2 \Rightarrow x+1<-2 \Rightarrow x<-3$
So the solution is $ x \in (-\infty,-3) \cup (1,\infty)$
But when solving $\frac{1}{|x+1|} < \frac{1}{2x}$,
If $x+1>0, x\neq0$, then
$\frac{1}{x+1} < \frac{1}{2x} \Rightarrow x+1 > 2x \Rightarrow x<1 $
If $x+1<0$, then
$\frac{1}{-(x+1)} < \frac{1}{2x} \Rightarrow -(x+1) > 2x \Rightarrow x+1 < -2x \Rightarrow x<-\frac{1}{3}$
But the solution should be $x \in (0,1)$.
I can see that there can't be negative values of $x$ in the inequality, because the left side would be positive and it can't be less than a negative number. But shouldn't this appear on my calculations?
| The left hand side of $$\frac{1}{|x+1|} < \frac{1}{2x}$$ is positive so the right hand side must be positive. Therefore $x>0$ which implies $x+1>0$ and $|x+1|=x+1.$
Upon substitution, the inequality becomes $$\frac{1}{x+1} < \frac{1}{2x}$$ with the condition $x>0.$ The solution is $0<x<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Factoring the polynomial $3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$ I'm trying to factor the following polynomial:
$$3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$$
What I've done:
$$3(x^2 -1)^3 + 7(x^2-1)^2 + 4(x^2 -1)$$
Then I set $p=x^2 -1$ so the polynomial is:
$$3p^3 + 7p^2 + 4p$$
Therefore: $$p(3p^2 + 7p + 4)$$
I apply Cross Multiplication Method: $$p(p+3)(p+4)$$
I substitute $p$ with $x^2-1$:
$$(x^2-1)(x^2-1+3)(x^2-1+4)$$
$$(x-1)(x+1)(x^2-2)(x^2-3)$$
I don't know if I've done something wrong or if I have to proceed further and how. The result has to be: $x^2(3x^2+1)(x+1)(x-1)$. Can you help me? Thanks.
| it is $$(x^2-1)(3(x^2-1)^2+7(x^2-1)+4)=(x^2-1)x^2(3x^2+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How do I integrate these functions? I deleted my last question, since some of you wanted me to rewrite the question properly. I feel sorry for inconvenience, but please understand that this is the first time I use $\texttt{MathJax}$.
Up to now, I tried every method I know in integration, like substituition, partial fractions, uv-method, etc. But seems like nothing works. I would appreciate to have your help. Thanks.
$$
\int_{0}^{1}\frac{\mathrm{d}x}{\left(x + 1\right)
\left[x^{2}\left(1 - x\right)\right]^{1/3}}\,,
\qquad\qquad\int_{0}^{1}\frac{\mathrm{d}x}{\left(x^{2} + 1\right)
\left[x^{2}\left(1 - x\right)\right]^{1/3}}
$$
| Here is an alternative evaluation for the second integral.
\begin{align*}
\int_0^1 \frac{dx}{(1 + x^2) [x^2 (1 - x)]^{1/3}} \, dx
&= \int_0^1 \sum_{n = 0}^\infty (-1)^n x^{2n} \frac{1}{x^{2/3} (1 - x)^{1/3}} \, dx\\
&= \sum_{n = 0}^\infty (-1)^n \int_0^1 x^{2n - 2/3} (1 - x)^{-1/3} \, dx\\
&= \sum_{n = 0}^\infty (-1)^n \text{B} \left (2n + \frac{1}{3}, \frac{2}{3} \right )\\
&= \sum_{n = 0}^\infty (-1)^n \frac{\Gamma (2n + 1/3) \Gamma (2/3)}{\Gamma (2n + 1)}\\
&= \Gamma \left (\frac{2}{3} \right ) \sum_{n = 0}^\infty \frac{(-1)^n \Gamma (2n + 1/3)}{(2n)!} \tag1
\end{align*}
Here $\text{B}(x,y)$ is Euler's Beta function while $\Gamma (x)$ is the familiar Gamma function. Now we just need to find a closed form for the sum appearing in (1).
From the integral definition for the Gamma function, namely
$$\Gamma (z) = \int_0^\infty t^{z - 1} e^{-t} \, dt,$$
we can write
$$\Gamma \left (2n + \frac{1}{3} \right ) = \int_0^\infty t^{2n - 2/3} e^{-t} \, dt,$$
so the sum appearing in (1) can be rewritten as
\begin{align*}
\sum_{n = 0}^\infty \frac{(-1)^n \Gamma (2n + 1/3)}{(2n)!} &= \sum_{n = 0}^\infty \frac{(-1)^n}{(2n)!} \int_0^\infty \frac{t^{2n} e^{-t}}{t^{2/3}} \, dt\\
&= \int_0^\infty t^{-2/3} e^{-t} \left [\sum_{n = 0}^\infty \frac{(-1)^n t^{2n}}{(2n)!} \right ] \, dt\\
&= \int_0^\infty t^{-2/3} e^{-t} \cos t \, dt\\
&= \mathfrak{R} \int_0^\infty t^{-2/3} e^{-(1 - i) t} \, dt\\
&= \mathfrak{R} \left (\frac{1}{\sqrt[3]{1 - i}} \right ) \cdot \int_0^\infty u^{-2/3} e^{-u} \, du\\
&= \Gamma \left (\frac{1}{3} \right ) \mathfrak{R} \left (\frac{1}{\sqrt[3]{1 - i}} \right ).
\end{align*}
Now it can be readily shown that
$$\frac{1}{\sqrt[3]{1 - i}} = \frac{1}{2^{1/6}} \exp \left (\frac{i \pi}{12} \right ).$$
Thus
$$\mathfrak{R} \left (\frac{1}{\sqrt[3]{1 - i}} \right ) = \frac{1}{2^{1/6}} \cos \left (\frac{\pi}{12} \right ) = \frac{1 + \sqrt{3}}{2 \cdot 2^{2/3}},$$
since $\cos (\pi/12) = (1 + \sqrt{3})/(2\sqrt{2})$, and we have
$$\sum_{n = 0}^\infty \frac{(-1)^n \Gamma (2n + 1/3)}{(2n)!} = \Gamma \left (\frac{1}{3} \right ) \frac{1 + \sqrt{3}}{2 \cdot 2^{2/3}}.$$
Substituting this result into (1), on taking advantage of the duplication formula for the Gamma function, namely
$$\Gamma \left (\frac{1}{3} \right ) \Gamma \left (\frac{2}{3} \right ) = \Gamma \left (\frac{1}{3} \right ) \Gamma \left (1 - \frac{1}{3} \right ) = \frac{\pi}{\sin (\pi/3)} = \frac{2\pi}{\sqrt{3}},$$
the value for the integral will be
$$\int_0^1 \frac{dx}{(1 + x^2) x^{2/3} \sqrt[3]{1 - x}} = \frac{\pi (1 + \sqrt{3})}{2^{2/3} \sqrt{3}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 5
} |
Recurrent equation system How to solve this recurrent equation system?
$$\begin{cases}a_{n+1}=a_n+2b_n\\
b_{n+1}=2a_n+b_n\end{cases}$$
The possible solutions should be
\begin{cases}a_n=\frac{1}{2}3^n+\frac{1}{2}(-1)^n\\
b_n=\frac{1}{2}3^n-\frac{1}{2}(-1)^n\end{cases}
| Assume we know $a_1$ and $b_1$.
First, $a_{n+1}+b_{n+1}=3(a_n+b_n) = \cdots = 3^n(a_1+b_1)$.
And secondly, $a_{n+1}-b_{n+1}=-(a_n - b_n) = \cdots = (-1)^n (a_1 - b_1)$.
And
$$ a_{n+1} = \frac{(a_{n+1}+b_{n+1})+(a_{n+1}-b_{n+1})}{2}$$
$$ b_{n+1} = \frac{(a_{n+1}+b_{n+1})-(a_{n+1}-b_{n+1})}{2}$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
If $\cos3A + \cos3B + \cos3C = 1$ in a triangle, find one of its length I would like to solve the following problem.
In $\triangle ABC, AC = 10, BC = 13$. If $\cos3A + \cos3B + \cos3C = 1$, compute the length of $AB$.
I thought that I could apply the Law of Cosines. Using the fact that $A+B+C=\pi$, I attempted to build the equation up from there.
What I got was that $$\cos3A+\cos3B-\cos(3A+3B)=1$$
Expanding, I got $$\cos3A+\cos3B-\cos3A\cos3B-\sin3A\sin3B=1$$
Now, it's possible that I'd be able to factor it somehow by rewriting it all in terms of cosine and arrive at the answer, but is there a better way to solve the problem? Thanks!
| Using the identity
\begin{align}
\cos3\alpha+\cos3\beta+\cos3\gamma
&=
1-\frac{r\,(3\,\rho^2-(r+3\,R)^2)}{R^3}
\tag{1}\label{1}
,
\end{align}
where $r$ is the inradius, $R$ is the circumradius
and $\rho$ is the semiperimeter
of the triangle,
condition
\begin{align}
\cos3\alpha+\cos3\beta+\cos3\gamma&=1
\end{align}
results in
\begin{align}
3\,\rho^2-(r+3\,R)^2&=0
,\\
3\,\rho^2-\left(\frac{S}{\rho}+3\,\frac{abc}{4S}\right)^2&=0
,\\
3\,\rho^2
-\frac{S^2}{\rho^2}
-\frac{3abc}{2\rho}
-\tfrac{9}{16}\frac{(abc)^2}{S^2}&=0
\tag{2}\label{2}
,\\
\end{align}
where $S$ is the area of triangle.
With the help of substitution
\begin{align}
\rho&=\tfrac12(a+b+c)
,\\
S^2&=\tfrac1{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c)
,\\
a&=13
,\quad
b=10
\end{align}
equation \eqref{2} becomes
\begin{align}
\frac{(c^2-399)(c^2+13c+69)(c^2+10c-69)}{(c^2-9)(c^2-23^2)}
&=0
,
\end{align}
with just two suitable roots,
\begin{align}
c_1&=\sqrt{399}
,\\
c_2&=\sqrt{94}-5
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
$2^{x-3} + \frac {15}{2^{3-x}} = 256$ $$2^{x-3} + \frac {15}{2^{3-x}} = 256$$
*
*Find the unknown $x$.
My attempt:
We know that $x^y . x^b = x^{y+b}$.
$$2^x . 2^{-3} + 15. 2^{-3+x} = 2^8$$
and
$$2^x . 2^{-3} + 15. 2^{-3} . 2^x = 2^8$$
From here, we get
$$2^x + 15 = 2^8$$
However, I'm stuck at here and waiting for your kindest helps.
Thank you.
| "$2^x . 2^{-3} + 15. 2^{-3} . 2^x = 2^8"$
"From here, we get"
"$2^x + 15 = 2^8$"
Uh....What? No, you don't.
You get
$2^x . 2^{-3}*2^3 + 15. 2^{-3}2^3 . 2^x = 2^8*2^3"$
$2^x + 15*2^x = 2^{8+3}$
So $16 *2^x = 2^{11}$
$2^{x+4} = 2^{11}$
$x+4 = 11$
$x = 7$.
====
I'd find it easier to do. Let $y = x-3$ so
$2^y + \frac 15{2^{-y}} = 256$
$2^y + 15*2^y = 256$
$2^y(1 + 15) = 256$
$2^y(16)= 256$
$2^y = \frac {256}{16} = \frac {2^8}{2^4} = 2^4$
$y = x-3 = 4$
$x = 3$.
=====
Someone else suggested: Let $t = 2^{x-3}$ so
$t + \frac {15}{\frac 1t} = 256$ so
$t + 15t = 256$ etc....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
limit with summation and product
Given $L=\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\bigg(\frac{1}{r!}\prod^{r}_{i=1}\left(\frac{i}{2}+\frac{1}{3}\right)\bigg)$. then $\lfloor L \rfloor$ is
Try: $$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{1}{r!}\bigg[\left(\frac{1}{2}+\frac{1}{3}\right)\cdot \left(\frac{2}{2}+\frac{1}{3}\right)\cdots \cdots \left(\frac{r}{2}+\frac{1}{3}\right)\bigg]$$
could some help me to solve it, thanks
| Comment too long,
so entered as an answer:
Wolfy also says that
$\sum_{n=1}^∞ \prod_{j=1}^n(1/2 + 1/(b j))
= 2^{1+2/b} - 1
$.
Wolfy also says that
(the first line is what it returns,
the remaining lines
are my attempt
to properly simplify)
$\begin{array}\\
\sum_{n=1}^∞ \prod_{j=1}^n(\frac1{a} + \frac1{b j})
&= -\dfrac{(\frac{a - 1}{a})^{-a/b} (a (\frac{a - 1}{a})^{a/b} -
(\frac{a - 1}{a})^{a/b} - a)}{a - 1}\\
&= -\dfrac{a - 1 - a(\frac{a - 1}{a})^{-a/b}}{a - 1}\\
&= \dfrac{ a(\frac{a - 1}{a})^{-a/b}}{a - 1}-1\\
&= \dfrac{ a^{1+a/b}}{(a - 1)^{1+a/b}}-1\\
&= \left(\dfrac{ a}{a - 1}\right)^{1+a/b}-1\\
\end{array}
$
when $|a|>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction: $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ The whole problem has been translated from German, so apologies if I made any mistakes. Thank you for taking the time to help!
So this is a problem from my math book
Prove that $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ for all $n\ \varepsilon \ N$
and I have come as far as this:
*
*let $n_0 = 1$. Then $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = (-1)^{1+1}1^2 = 1 = \frac{(-1)^{1+1}1(1+1)}{2} = 1 $
*For one $n\ \varepsilon \ N$ $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ is true.
*Now we have to show that $\sum\limits_{k=1}^{(n+1)} (-1)^{k+1}k^2 = \frac{(-1)^{(n+1)+1}(n+1)((n+1)+1)}{2} = \sum\limits_{k=1}^{(n+1)} (-1)^{k+1}k^2 = \frac{(-1)^{(n+2)}(n+1)((n+2)}{2}$ for all $n\ \varepsilon \ N$
From here, I did this: $\sum\limits_{k=1}^{(n+1)} (-1)^{k+1}k^2 = \sum\limits_{k=1}^{(n)} (-1)^{k+1}k^2 + (-1)^{(n+2)}(n+1)^2 = \frac{(-1)^{n+1}n(n+1)}{2} + (-1)^{(n+2)}(n+1)^2 = \frac{(-1)^{n+1}n(n+1)}{2} + \frac{2((-1)^{(n+2)}(n+1)^2)}2$
at which point I got stuck.
A look at the solutions at the back of the book told me that the next step is to factor (n+1) and $(-1)^{(n+2)}$, which resulted into this:
$\frac{(-1)^{n+2}(n+1)((-1)n+2(n+1))}{2} = \frac{(-1)^{n+2}(n+1)(-n+2n+2)}{2} = \frac{(-1)^{n+2}(n+1)(n+2)}{2}$
Which makes sense, but here is my question: How on earth do I get from $\frac{(-1)^{n+1}n(n+1)}{2} + \frac{2((-1)^{(n+2)}(n+1)^2)}2$ to $\frac{(-1)^{n+2}(n+1)((-1)n+2(n+1))}{2}$? I have been trying for quite a while and I just can't seem to figure it out. Any help is greatly appreciated!
| Because if $$1^2-2^2+...+(-1)^n(n-1)^2=\frac{(-1)^nn(n-1)}{2}$$ then
$$1^2-2^2+...+(-1)^n(n-1)^2+(-1)^{n+1}n^2=$$
$$=(-1)^{n+1}\left(n^2-\frac{n(n-1)}{2}\right)=(-1)^{n+1}\frac{n(n+1)}{2}.$$
The base is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $n \in \mathbb{N}\setminus\{1\}$ then $\gcd(n^2-1,3n+1)=1$ if and only if $n$ is even If $n\in\mathbb N\setminus\{1\}$ then $\gcd(n^2-1,3n+1)=1$ if and only if $n$ is even.
Can somebody help me prove this problem?
| Let $a=n^2-1$, and let $b = 3n+1$.
Let $d = \text{gcd}(a,b)$.
First, suppose $n$ is odd.
Then $a,b$ are both even, so $d > 1$.
Next, suppose $n$ is even.
Then $a,b$ are both odd, so $d$ must be odd.
Now simply note that
\begin{align*}
9a-(3n-1)b &= 9(n^2-1)-(3n-1)(3n+1)\\[4pt]
&=(9n^2-9)-(9n^2-1)\\[4pt]
&=-8\\[4pt]
\end{align*}
hence $d$ must divide $8$.
Since $d$ is odd, it follows that $d=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\sum\limits_{cyc}\frac{a}{a^{11}+1}\leq\frac{3}{2}$ for $a, b, c > 0$ with $abc = 1$
Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leq\frac{3}{2}.$$
I tried homogenization and the BW (https://artofproblemsolving.com/community/c6h522084),
but it does not work.
Indeed, let $a=\frac{x}{y}$, $b=\frac{y}{z}$, where $x$, $y$ and $z$ are positives.
Hence, $c=\frac{z}{x}$ and we need to prove that $$\sum_{cyc}\frac{xy^{10}}{x^{11}+y^{11}}\leq\frac{3}{2},$$
which has a problem around $(x,y,z)=(7,5,6)$.
For these values $$\frac{3}{2}-\sum_{cyc}\frac{xy^{10}}{x^{11}+y^{11}}=0.0075...$$
I tried also TL, uvw, C-S, Lagrange multipliers and more, but without success.
Also, Vasc's Theorems don't help.
Also, the following method does not help here. Find the maximum of the expression
Because the inequality $\frac{x}{x^{11}+1}\leq\frac{3(a^9+1)}{4(a^{18}+a^9+1)}$ is wrong.
| In principle it is, if not always, almost very often that such a problem can be solved using techniques from optimization. For instance one can consider the following maximization problem:
\begin{equation}
\max_{a,b,c\in C} f(a,b,c)
\end{equation}
where the constraint set $C:=\{a,b,c\in\mathbb{R}_+:abc=1\}$ and $f(a,b,c):=\sum_{cyc}a/(a^{11}+1)$. If one shows that $3/2$ is the maximum value $f(a,b,c)$ attains in $C$ then this solves the inequality problem. We will follow the same idea however first we transform the given inequality into an equivalent form together with an appropriate constraint which makes it easier to solve it as a maximization problem. The original inequality is given by:
\begin{equation}
\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leqslant \frac{3}{2}
\end{equation}
and $abc=1$. After proper rearrangements this last inequality is equivalent to:
$$2(a(b^{11}+1)(c^{11}+1)+b(a^{11}+1)(c^{11}+1)+c(a^{11}+1)(b^{11}+1))\leqslant 3(a^{11}+1)(b^{11}+1)(c^{11}+1)$$
or equivalently:
$$2\Big((b^{10}+\frac{1}{b})(c^{10}+\frac{1}{c})+(a^{10}+\frac{1}{a})(c^{10}+\frac{1}{c})+(a^{10}+\frac{1}{a})(b^{10}+\frac{1}{b})\Big)\leqslant 3(a^{10}+\frac{1}{a})(b^{10}+\frac{1}{b})(c^{10}+\frac{1}{c})$$
Let $f(x):=x^{10}+1/x$ then the last inequality is the same as:
$$\frac{1}{f(a)}+\frac{1}{f(b)}+\frac{1}{f(c)}\leqslant\frac{3}{2}$$
It is sufficient to look at the problem:
$$\max_{a,b,c}F(a,b,c):=\frac{1}{f(a)}+\frac{1}{f(b)}+\frac{1}{f(c)}$$
subject to $abc=1$. The Lagrangian for this problem is:
$$L(a,b,c,\lambda):=F(a,b,c)-\lambda(1-abc)$$
From the first order conditions we get the following equations:
$$\frac{f'(a)}{f^2(a)}=\lambda bc\Leftrightarrow a\frac{f'(a)}{f^2(a)}=\lambda \\
\frac{f'(b)}{f^2(b)}=\lambda ac\Leftrightarrow b\frac{f'(b)}{f^2(b)}=\lambda \\
\frac{f'(c)}{f^2(c)}=\lambda ab\Leftrightarrow c\frac{f'(c)}{f^2(c)}=\lambda $$
A possible obvious solution to this system is $a=b=c=1$ and $\lambda=9/4$. If one calculates the Hessian of $L(a,b,c,\lambda)$ (bordered Hessian) we get:
$$\text{Hess}_L(a,b,c,\lambda)=
\begin{bmatrix}
0 & bc & ac & ab\\
bc & F_{aa} & \lambda c& \lambda b \\
ac &\lambda c & F_{bb} & \lambda a\\
ab &\lambda b & \lambda a & F_{cc}
\end{bmatrix}
$$
where $F_{xx}:=-\Big[(f''(x)f^2(x)-2(f'(x))^2f(x))/f^4(x)\Big]$. Evaluating at $(1,1,1,9/4)$ gives:
$$\text{Hess}_L(1,1,1,9/4)=
\begin{bmatrix}
0 & 1 & 1 & 1\\
1 & -11/4 & 9/4& 9/4 \\
1 &9/4 & -11/4 & 9/4\\
1 &9/4 & 9/4 & -11/4
\end{bmatrix}
$$
From this follows $\det \text{Hess}_L(1,1,1,9/4)=-75$ and $\det M_L(1,1,1,9/4)=10$ where $M$ is the submatrix
$$M_L:=\begin{bmatrix}
0 & 1 & 1 \\
1 & -11/4 & 9/4 \\
1 &9/4 & -11/4
\end{bmatrix}
$$
These satisfy the second order conditions for our bordered Hessian (the alternating sign condition) for a local maximum at $(1,1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
} |
Solve the initial value problem $x^2y'+y(x-y)=0$ $x^2y'+y(x-y)=0$
$y(1)=-1$
Can someone point me in the right direction on this? I started doing this:
$y'+p(x)y=q(x)y^n$
So
$p(x)=x-y, q(x)=1/y^2, n=2$
$x^2y'+y(x-y)=(1/y^2)y^2=1$
Then ran out of ideas...
(I don't want you give give me the answer, just steer me the right way please!)
| $$y'+\frac{y}{x}-\frac{y^2}{x^2}=0$$
$$xy'+y=\frac{y^2}{x}$$
$$(xy)'=\frac{(xy)^2}{x^3}$$
Substitute $z=xy$:
$$z'=\frac{z^2}{x^3}$$
integrate:
$$\int \frac {dz}{z^2}=\int\frac{dx}{x^3}=\frac{-1}{2x^2}+K$$
$$ \frac {1}{z}=\frac{1}{2x^2}+K$$
$$ \frac {1}{xy}=\frac{1}{2x^2}+K$$
$$ x=y(\frac 1 2 +Kx^2)$$
$$ y=\frac x {(\frac 1 2 +Kx^2)}=\frac {2x} {(1 +Kx^2)}$$
$$....$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2603855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Fourier series expansion of $\frac{\pi^4}{96}$ and $\frac{\pi^4}{90}$ I have to prove that:
\begin{equation}
\frac{\pi^4}{96}=\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^4} \qquad \mbox{and} \qquad \frac{\pi^4}{90}=\sum_{n=1}^{+\infty}\frac{1}{n^4}
\end{equation}
My idea:
I've expanded $f(x)=|x|$ in a real Fourier series $\forall x\in Q:=[-\pi,\pi)$:
\begin{equation}
f(x)=|x|=\frac{\pi}{2}-\sum_{n=0}^{+\infty} \frac{1}{(2n+1)^2\pi}\cos(nx)
\end{equation}
Now, using Parseval's Identity, I get:
\begin{equation}
\frac{2\pi^2}{3}=\sum_{n=0}^{+\infty}\frac{16}{n^4\pi^2}+\frac{\pi^2}{2}
\end{equation}
hence:
\begin{equation}
\frac{\pi^4}{96}=\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^4}
\end{equation}
Now, how can I prove the second claim?
| From what you've already proved:
$$\sum_{n=1}^\infty\frac 1{n^4}=\sum_{n=0}^\infty\frac 1{(2n+1)^4}=\sum_{n=1}^\infty\frac 1{(2n)^4},$$
whence
$$\sum_{n=1}^\infty\frac 1{n^4}-\sum_{n=1}^\infty\frac 1{(2n)^4}=\frac{15}{16}\sum_{n=1}^\infty\frac 1{n^4}=\frac{\pi^4}{96}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2604850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the integral $\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx.$ The question is $$\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx.$$
I have tried to multiply both numerator and denominator by $1-\sqrt{x}$ but can't proceed any further, help!
| We have
$\mathrm{\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\ dx}$
Put $\mathrm{\sqrt{x} = \cos t}$
Differentiate Both Sides w.r.t. $\mathrm{x}$
$\mathrm{\frac{dx}{2\sqrt{x}}= - sint \ dt}$
$\mathrm{dx =- 2 \ cost\ sint \ dt}$
So, we get now:
$\mathrm{-\int \sqrt{\frac{1- cost}{1+ cost}}\ \times 2\ cost \ sint \ dt}$
= $\mathrm{-\int \sqrt{\frac{2 \ sin^2 \frac{t}{2}}{2 \ cos^2 \frac{t}{2}}}\ \times 2\ cost \ sint \ dt}$
= $\mathrm{-\int tan\frac{t}{2} \times 2\ cost \ sint \ dt}$
= $\mathrm{-\int \frac{sin\frac{t}{2}}{cos\frac{t}{2}} \times 2\ \left( 2cos^2\frac{t}{2} -1 \right) \ sint \ dt}$
= $\mathrm{-\int 4 \ sin\frac{t}{2}cos\frac{t}{2}sint \ dt + 2\int \frac{sin\frac{t}{2}}{cos\frac{t}{2}} sint \ dt}$
= $\mathrm{-2 \int sin^2 t dt + 4 \int sin^2 \frac{t}{2}dt}$
= $\mathrm{-2 \int \frac{1}{2}\ dt + 2 \int cos2t\ dt + 4 \int \frac{1}{2}dt - 4\int cost \ dt}$
=$\mathrm{-t + sin2t + 2t - 4 \ sint}$
= $\mathrm{t + sin2t - 4 \ sint}$
= $\mathrm{cos^{-1}\sqrt{x}+ 2\sqrt{x} \ sin(cos^{-1}\sqrt{x}) - 4 sin(cos^{-1}\sqrt{x})}$
= $\mathrm{cos^{-1}\sqrt{x} + 2\sqrt{x}(\sqrt{1-x}) - 4(\sqrt{1-x}) }$
= $\mathrm{cos^{-1}\sqrt{x} + 2(\sqrt{x(1-x)}) - 4(\sqrt{1-x}) + C}$
And, This is your final answer!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2605444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
limit of: $\begin{align} \lim_{n\to \infty}\sqrt{n-2\sqrt{n}}-\sqrt{n} \end{align}$ How to get the limit of:
$\begin{align}
\lim_{n\to \infty}\sqrt{n-2\sqrt{n}}-\sqrt{n}
\end{align}$
I have formed it to:
$\begin{align}
\lim_{n\to \infty}\frac{n(\sqrt{n-2\sqrt{n}}-\sqrt{n})}{n}
\end{align}$
And tried to use L'Hopital but unfortunately I am not able to get a result with that:
$f'(x)=\frac{d}{dn}(n(\sqrt{n-2\sqrt{n}}-\sqrt{n}))= (\sqrt{n-2\sqrt{n}}-\sqrt{n})+n\left( \frac{1-\frac{1}{\sqrt{n}}}{2\sqrt{n-2\sqrt{n}}} - \frac{1}{2\sqrt{n}}\right)$
$g'(x)=\frac{d}{dn} n = 1$
$\begin{align}
\lim_{n \to \infty}\frac{f'(x)}{g'(x)} &=
\lim_{n \to \infty}{\frac{(\sqrt{n-2\sqrt{n}}-\sqrt{n})+n\left( \frac{1-\frac{1}{\sqrt{n}}}{2\sqrt{n-2\sqrt{n}}} - \frac{1}{2\sqrt{n}}\right)}{1}} \\
&= \lim_{n \to \infty}{\frac{(\sqrt{n-2\sqrt{n}}-\sqrt{n})+n\left( 0\right)}{1}} \\
&= \lim_{n \to \infty}{\frac{(\sqrt{n-2\sqrt{n}}-\sqrt{n})}{1}} \\
&= \lim_{n \to \infty}{\frac{(\sqrt{\frac{1}{n}-2\sqrt{\frac{1}{n}}}-\sqrt{\frac{1}{n}})}{\frac{1}{n}}} \\
&= 0\\
\end{align}$
Which is wrong. Wolfram says that the limit is $-1$
I appreciate every help :)
| $$\lim_{n\to \infty}\sqrt{n-2\sqrt{n}}-\sqrt{n}=\lim_{n\to \infty}\frac{-2\sqrt{n}}{\sqrt{n-2\sqrt{n}}+\sqrt{n}}=\lim_{n\to \infty}\frac{-2\sqrt{n}}{\sqrt{n}-1+\sqrt{n}}=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove the next cyclic inequality Let $a,b,c>0$ such that $$a+b+c=1$$
prove that
$$cyclic\sum \frac {ab}{\sqrt {c+ab}}\le\frac {1}{2} $$
By symetrie, i proved it by assuming $a=b=c$ but i cannot justify this hypothesis.
| By C-S $$\left(\sum_{cyc}\frac{ab}{\sqrt{c+ab}}\right)^2=\left(\sum_{cyc}\frac{ab}{\sqrt{c(a+b+c)+ab}}\right)^2=\left(\sum_{cyc}\frac{ab}{\sqrt{(a+c)(b+c)}}\right)^2\leq$$
$$\leq\sum_{cyc}ab\sum_{cyc}\frac{ab}{(a+c)(b+c)}.$$
Thus, it's enough to prove that
$$\sum_{cyc}ab\sum_{cyc}\frac{ab}{(a+c)(b+c)}\leq\frac{(a+b+c)^2}{4},$$ which is easy after full expanding.
Indeed, we need to prove that
$$\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2-2a^3bc+2a^2b^2c)\geq0$$ or
$$\sum_{cyc}(a-b)^2ab(a+b-c)\geq0.$$
Let $a\geq b\geq c$.
Thus,
$$\sum_{cyc}(a-b)^2ab(a+b-c)\geq$$
$$\geq(a-c)^2ac(a+c-b)+(b-c)^2bc(b+c-a)\geq$$
$$\geq(b-c)^2ac(a-b)+(b-c)^2bc(b-a)=(b-c)^2(a-b)^2c\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Convergence of the sequence $ \sqrt {n-2\sqrt n} - \sqrt n $ Here's my attempt at proving it:
Given the sequence $$ a_n =\left( \sqrt {n-2\sqrt n} - \sqrt n\right)_{n\geq1} $$
To get rid of the square root in the numerator:
\begin{align}
\frac {\sqrt {n-2\sqrt n} - \sqrt n} 1 \cdot \frac {\sqrt {n-2\sqrt n} + \sqrt n}{\sqrt {n-2\sqrt n} + \sqrt n} &= \frac { {n-2\sqrt n} - \ n}{\sqrt {n-2\sqrt n} + \sqrt n} = \frac { {-2\sqrt n}}{\sqrt {n-2\sqrt n} + \sqrt n} \\&= \frac { {-2}}{\frac {\sqrt {n-2\sqrt n}} {\sqrt n} + 1}
\end{align}
By using the limit laws it should converge against:
$$
\frac { \lim_{x \to \infty} -2 } { \lim_{x \to \infty} \frac {\sqrt {n-2\sqrt n}}{\sqrt n} ~~+~~\lim_{x \to \infty} 1}
$$
So now we have to figure out what $\frac {\sqrt {n-2\sqrt n}}{\sqrt n}$ converges against:
$$
\frac {\sqrt {n-2\sqrt n}}{\sqrt n} \leftrightarrow \frac { {n-2\sqrt n}}{ n} = \frac {1-\frac{2\sqrt n}{n}}{1}
$$
${\frac{2\sqrt n}{n}}$ converges to $0$ since:
$$
2\sqrt n = \sqrt n + \sqrt n \leq \sqrt n ~\cdot ~ \sqrt n = n
$$
Therefore $~\lim_{n\to \infty} a_n = -1$
Is this correct and sufficient enough?
| There's a small mistake. You seem to think that to study the convergence of the sequence$$\left(\frac{\sqrt{n-2\sqrt n}}{\sqrt n}\right)_{n\in\mathbb N}$$is equivalent to the convergence of its square. It is true in this case because it is a sequence of real numbers greater than $0$; but it is false in general.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Evaluate limit containing $\sum{n^6}$ Evaluate:
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6+\ldots+n^6}{(1^2+2^2+3^2+\ldots+n^2)(1^3+2^3+3^3+\ldots+n^3)}}$$
I can solve the denominator as:
$$\frac{n(n+1)(2n+1)}{6}\cdot\frac{n^2(n+1)^2}{4}$$
$$n^7\cdot\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}\cdot\frac{(1+\frac{1}{n})}{4}$$
$$=\frac{n^7}{12}$$
How can I reduce the numerator?
| Here is yet another answer, which does not require any of the full, explicit formulae for sums of powers.
Consider the sum
$$
S(n, p) = \sum_{m = 1}^{n} m^p\, .
$$
I can rewrite this as
$$
S(n, p) = n^{p+1}\sum_{m = 1}^{n} \frac{1}{n}\;\left(\frac{m}{n}\right)^p\, .
$$
The sum in the above expression is a Riemann sum which approximates the integral
$$
\sum_{m = 1}^{n}\frac{1}{n}\;\left(\frac{m}{n}\right)^p
\approx \int_0^1 x^p\, dx \;=\; \frac{1}{p+1}\, .
$$
Thus, for large $n$
$$
S(n, p) \approx \frac{n^{p+1}}{p+1}\, .
$$
We can use this expression to replace all three sums in your expression:
$$
\lim_{n\rightarrow\infty} \frac{1^6 + \;...\; + n^6}{\left(1^2 +\;...\;+ n^2\right) \left(1^3 +\;...\;+ n^3\right)}
\;=\;
\lim_{n\rightarrow\infty} \frac{n^7 / 7}{\left(n^3 / 3\right) \left(n^4 / 4\right)}
\;=\; \frac{12}{7}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
The complex equation $x^3 = 9 + 46i$ has a solution of the form $a + bi$ where $a,b\in \mathbb Z$. Find the value of $a^3 + b^3$ The complex equation $x^3 = 9 + 46i$ has a solution of the form $a + bi$ where $a,b\in \mathbb Z$. Find the value of $a^3 + b^3$ .
| Let $x=a+bi$, where $\{a,b\}\subset\mathbb R$.
Thus, $$(a+bi)^3=9+46i$$ or
$$a^3-3ab^2=9$$ and
$$3a^2b-b^3=46,$$
which gives
$$46(a^3-3ab^2)=9(3a^2b-b^3)$$ or
$$(2a+3b)(23a^2-48ab+3b^2)=0,$$ which gives
$$2a+3b=0.$$
Thus, $$3\cdot\frac{9}{4}b^3-b^3=46,$$ which gives $b=2$, $a=-3$ and $a^3+b^3=-19.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Geometry Problem: Find the area of $ABCD$ square
My Problem is:
The $E-$ point on the $AC$ diagonal is marked on the $ABCD$ square.If $BE= 13$, $CE=17$ find the square area.
My way:
$$a=\frac{k+17}{\sqrt2}$$
$$13^2=k^2+\frac {(k+17)^2}{2}-\frac {2k(k+17)}{\sqrt2}×\frac{\sqrt2}{2} \Rightarrow k^2+289=338 \Rightarrow k^2=49 \Rightarrow k=7$$
$$S_{ABCD}=\frac {(k+17)^2}{2}=\frac {(7+17)^2}{2}=288 $$
It looks ugly solution.
Is this solution correct?
| Let $ABE_1E$ be a parallelogram.
Thus, $BECE_1$ is isosceles trapezoid ($BE=E_1C=13$) and since $EE_1\perp BC$,
we obtain:
$$BE_1^2+EC^2=BE^2+E_1C^2$$ or
$$BE_1^2+17^2=13^2+13^2,$$ which gives $$BE_1=7.$$
Id est, $$AC=AE+EC=7+17=24$$ and
$$S_{ABCD}=\frac{1}{2}\cdot24^2=288.$$
| {
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Find the area of underneath the curve $ \ f(x)=4-3x \ $ over the interval $ \ [2,4] \ $ by dividing the interval into $ \ n \ $ Find the area of underneath the curve $ \ f(x)=4-3x \ $ over the interval $ \ [2,4] \ $ by dividing the interval into $ \ n \ $ equal subintervals and taking limit.
Answer:
The length of the interval $ \ [2,4] \ $ is
$ \ 4-2=2 \ $ ,
Thus subinterval size $ =\Delta x=\frac{2}{n} , \ $
Then the required area is given by
$ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x \\ = \lim_{n \to \infty} \sum_{i=1}^{n} [4-3 (2+i \frac{2}{n} )] \frac{2}{n}
\\ = -\lim_{n \to \infty} \sum_{i=1}^{n} \left[2+i \frac{6}{n}\right] \frac{2}{n} \\ \\ = -\large \lim_{n \to \infty} \left(\frac{4}{n}+12 \{\frac{1}{n^2} + \frac{2}{n^2} +..........+\frac{n}{n^2}\} \right) \\ =\large 0- \lim_{n \to \infty} \frac{12}{n^2} (1+2+3+.....+n) \\ = -\large \lim_{n \to \infty} \frac{12}{n^2} \cdot \frac{n(n+1)}{2} \\ =-6 $
Thus the required area $ \ =6 \ $
But $ \ \int_{2}^{4} (4-3x) dx=-10 \ $ i.e., the area is $ \ 10 \ $
Why such difference in approximation above.
I thought I have done something wrong in calculation .
Kindly examine my work and help me out
| $$\begin{align} \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x& = \lim_{n \to \infty} \sum_{i=1}^{n} [4-3 (2+i \frac{2}{n} )] \frac{2}{n}
\\&= -\lim_{n \to \infty} \sum_{i=1}^{n} \left[2+i \frac{6}{n}\right] \frac{2}{n}\\ &= -\large \lim_{n \to \infty} \left(\frac{4}{n}+12 \{\frac{1}{n^2} + \frac{2}{n^2} +\cdots+\frac{n}{n^2}\} \right) \end{align}$$
The first row is OK, the second one is as well, but the second one is wrong.
What you should get is
$$\sum_{i=1}^n \frac{2}{n}\left(2+i\frac6n\right) = \sum_{i=1}^n\left(\frac4n + i\frac{12}{n^2}\right) \\=\left(\frac4n+1\cdot\frac{12}{n^2}\right) + \left(\frac4n+2\cdot\frac{12}{n^2}\right)+\cdots +\left(\frac{4}n+n\cdot\frac{12}{n^2}\right)$$
which we can reorder into
$$\frac4n + \frac4n + \cdots + \frac4n + 12\left(\frac1{n^2}+\frac12{n^2}+\cdots+\frac{n}{n^2}\right)$$
and therefore equals
$$\color{red}{n\cdot} \frac{4}{n}+12 \{\frac{1}{n^2} + \frac{2}{n^2} +\cdots+\frac{n}{n^2}\}$$
whic is close to what you got, i.e.
$$\frac{4}{n}+12 \{\frac{1}{n^2} + \frac{2}{n^2} +\cdots+\frac{n}{n^2}\}$$
(note the $n$ in front of $n\cdot \frac4n$)
| {
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is there away to give $= \sqrt[3]{5 + 10i } + \sqrt[3]{5- 10i}$ question is there away to give
$$= \sqrt[3]{5 + 10i }
+ \sqrt[3]{5- 10i}$$
in just reals?
I am thinkting maybe use some clever algebra raise it to like 3 and magically imaginary terms dissapear. Or expressed the complex numbers in polar form but idk
Background (might delete if this is not really the question)
Let $$ m(x)=x^3-15x-10 $$
Find the roots of $m(x)$ using the cubic formula and show that they are all real
Using Howell complex approach
Lets change variables $$ \begin{aligned}
&z^3-15z-10
\\=&z^3+0z^2-15z-10
\end{aligned}$$
so, general form is denoted as
$$ z^3+a_2z^2+a_1z+a_0=0$$
so $$\begin{aligned}
a_2&=0
\\a_1&=-15
\\a_0&=-10
\end{aligned}$$
we get by substituting that $z=x-a_2/3$. $a_2=0$ its already depressed cubic but so theses steps are unnecesary for this one problem but will use them for other problems
$$ x^3+bx+c$$
where $$\begin{aligned}
b&=a_1-\frac{a_2^2}{3}
=-15-\frac{0^2}{3}=-15
\\c&=\frac{-a_1a_2}{3}+\frac{2}{27}a_2^3+a_0
=\frac{-(-15)(0)}{3}+\frac{2}{27}(0)^3+-10=-10
\end{aligned} $$
our depressed cubic is of $$ x^3-15x-10 $$
Ferro-Tartaglia Formula
$$ x=\sqrt[3]{\frac{-c}{2} + \sqrt{\frac{c^2}{4} +\frac{b^3}{27}}}\
+ \sqrt[3]{\frac{-c}{2} - \sqrt{\frac{c^2}{4} -\frac{b^3}{27}}}\ $$
in our case $$
\begin{aligned}
x&=\sqrt[3]{\frac{10}{2} + \sqrt{\frac{100}{4} +\frac{(-15)^3}{27}}}\
+ \sqrt[3]{\frac{10}{2} - \sqrt{\frac{100}{4} +\frac{(-15)^3}{27}}}\
\\&= \sqrt[3]{5 + \sqrt{25 -125}}
+ \sqrt[3]{5- \sqrt{25 -125}}
\\ &= \sqrt[3]{5 + 10i }
+ \sqrt[3]{5- 10i}
\end{aligned}
$$
| No there isnt. This is a well known phenomenon with a latin name, Casus Irreducibilus.
If a third degree equation with real coefficients is irreducible, and has all real solutions then these solutions cannot be expressed as roots of real numbers.
| {
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Solving $\int(x^2 + 1)^7x^3{\mathrm d}x$ without expansion What is $\int(x^2 + 1)^7x^3{\mathrm d}x$?
I wasn't able to come up with a substitution so I attempted integration by parts:
$$u = x^2 + 1, u' = 2x, v' = x^3, v = \frac{x^4}{4}$$
$$(x^2+1)\frac{x^4}{4} - \frac{x^6}{12}$$
The derivative clearly shows that this is wrong. How can I solve this without using the binomial theorem?
| By definition of integral without substitutions:
$$(x^2+1)^7x^3=(x^2+1)^7(x^2+1-1)x=(x^2+1)^8x-(x^2+1)^7x.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of $(x_n)_n$ if $2018^2=x_{n+1}\sqrt[n+1]{x_1\cdot x_2\cdots x_n}$ $\forall~n$ I became interested in solving a question in a recently deleted post concerning the relation $$2018^2=x_{n+1}\sqrt[n+1]{x_1\cdot x_2\cdot\cdots\cdot x_n}$$ for $n\in\mathbb{N}$.
The question was whether it converges and if so what the limit is.
My thoughts:
We have $$\begin{array}{c|c}n&x_{n+1}\\\hline1&2018^{2/1}x_1^{-1/2}\\2&2018^{4/3}x_1^{-1/6}\\3&2018^{7/6}x_1^{-1/12}\\...&...\end{array}$$ The pattern is that for $n\in\mathbb{N}$, $$x_{n+1}=2018^{1+[2/n(n-1)]}x_1^{-1/n(n-1)}.$$ Note that we suppose $x_1$ is fixed. This means that as $n\to\infty$, $$x_\infty=2018^{1+0}x_1^{0}=2018$$ so the sequence converges.
Is this correct?
|
Your answer is correct in fact we have the following relations
$$\color{blue}{x_{n+2} =2018^{\frac{2}{n+2}}x_{n+1}^{\frac{n}{n+2}}}$$and
$$\color{red}{x_{n+1}=2018^{1+\frac{2}{n(n+1)}}x_1^{-\frac{1}{n(n+1)}}.} \to 2018$$
See the prove below
Let us establish a recursive formula3. we have $$2018^2=x_{n+1}\sqrt[n+1]{x_1\cdot x_2\cdot\cdots\cdot x_n} \implies {x_1\cdot x_2\cdot\cdots\cdot x_n} = \left(\frac{2018^2}{x_{n+1}}\right)^{n+1}$$
then $$x_{n+2} = 2018^2\left(x_1\cdot x_2\cdot\cdots\cdot x_nx_{n+1}\right)^{-\frac{1}{n+2}} = 2018^2\left(\left(\frac{2018^2}{x_{n+1}}\right)^{n+1}x_{n+1}\right)^{-\frac{1}{n+2}} \\= \left(2018^2\right)^{1-\frac{n+1}{n+2}}\left(\left(x_{n+1}\right)^{-n-1}x_{n+1}\right)^{-\frac{1}{n+2}} =2018^{\frac{2}{n+2}}\left(x_{n+1}\right)^{\frac{n}{n+2}} $$
That is we have the recurrence relation, $$\color{blue}{x_{n+2} =2018^{\frac{2}{n+2}}x_{n+1}^{\frac{n}{n+2}}}$$
Now assume that $x_n$ has the form $$x_n =\ell^{a_n}x_1^{b_n}~~~~with~~~~\ell=2018.$$
Obviously $a_2 = 2$ and $b_2= -\frac{1}{2}$
Then by the aforementioned recursive relation we have
$$\ell^{a_{n+1}}x_1^{b_{n+1}} = x_{n+1}= \ell^{\frac{2}{n+1}}x_{n}^{\frac{n-1}{n+1}}=\ell^{\frac{2}{n+1}}\left(\ell^{a_n}x_1^{b_n}\right)^{\frac{n-1}{n+1}}$$
after identification one remains with
$$a_{n+1} =a_n\frac{n-1}{n+1}+\frac{2}{n+1}~~~and~~~~b_{n+1} =b_n\frac{n-1}{n+1}~~n\ge 2$$
By telescopic product one get,
$$\color{blue}{\frac{b_{n+1}}{b_2}=\prod^{n}_{k=2}\frac{b_{k+1}}{b_k}=\prod^{n}_{k=2}\frac{k-1}{k+1}=\prod^{n}_{k=2}\frac{k-1}{k}\prod^{n}_{k=2}\frac{k}{k+1} =\frac{2}{n(n+1)}}$$
Hence $$b_{n+1}=- \frac{1}{n(n+1)}$$
One can prove see here that $$a_{n+1}=1+ \frac{2}{n(n+1)}$$
hence
$$\color{red}{x_{n+1}=2018^{1+\frac{2}{n(n+1)}}x_1^{-\frac{1}{n(n+1)}}.} \to 2018$$
Also see this
| {
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Searchingthe polar form of $z = 2+i(1+\sqrt{3})$ Find the polar form of:
$$z = 2+i(1+\sqrt{3})$$
Is there any other way to solve that in different way than by finding $|z|$ that is really "ugly" ($|z| = \sqrt{8+2\sqrt{3}}$)?
| you are right $z=\rho \cdot e^{i\cdot \theta}$ with $\rho=|z| = \sqrt{8+2\sqrt{3}}$. But Note that $$ \frac{1 + \sqrt3}{2} = \cos\frac{π}{3}+\sin\frac{π}{3}=\sqrt2\cos(\frac{π}{3}-\frac{π}{4} )=\sqrt2\cos\frac{π}{12}$$
and then ,
$$\theta=\arg (z)=\arctan\left(\frac{1 + \sqrt3}{2}\right)=\color{red}{\arctan\left(\sqrt2\cos\frac{π}{12}\right)}$$
| {
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Equation of Normal I'm struggling to get the same answer as the book on this one. I wonder if someone could please steer me in the right direction?
Q. Find the equation of the normal to $y=x^2 + c$ at the point where $x=\sqrt{c}$
At $x = \sqrt{c}$ then $y = 2c$, so the point given is $(\sqrt{c},2c)$
The gradient of the tangent is $2x$ so the gradient of the normal is $-\frac{1}{2}x$ and the equation of the normal will be $-\frac{1}{2}x + v$
Where $x=\sqrt{c}$
then $y = -\frac{1}{2}\sqrt{c} + v$
We know $y = 2c$ therefore $2c = -\frac{1}{2}\sqrt{c} + v$ and so
$v = 2c + \frac{1}{2}\sqrt{c}$
So the equation of the normal is $y = -\frac{1}{2}x +2c +\frac{1}{2}\sqrt{c}$
We can tidy this up to get $2y = -x + 4c + \sqrt{c}$
Unfortunately the book tells me the answer is $2y\sqrt{c} = -x+\sqrt{c}(4c+1)$
It looks like I lost a $\sqrt{c}$ somewhere? Where did I go wrong?
Thank you
Gary
| Hint: at the point $x_0=\sqrt{c}$ the slope of the tangent is $2x_0=2\sqrt{c}$, the slope of the normal is $k=-\frac{1}{2x_0}=-\frac{1}{2\sqrt{c}}$. Now use the fact that you know the point coordinates to get the normal equation as $y=kx+m$.
| {
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Flip a coin 5 times. What is the probability that heads never occurs twice in a row? Suppose I flip a coin $5$ times in a row. There are $2^5$ possible outcomes, i.e: HHHTH, HTTTT, HTHTH, etc. I want to know the probability that heads never occurs twice in a row.
I drew out $32$ events that can occur, and I found out that the answer was $\cfrac{13}{32}$. But I'm not sure how to do this generally, because say if the coin was flipped $20$ times in a row, I wouldn't write out $2^{20}$ possibilities.
Thanks
| Let head be represented by 0 and tail by 1. Then a sequence of heads and tails can be represented by a binary number.
For 2 tosses ($n = 2$), the possible sequences are:
00
01
10
11
There are $2^2$ total possible sequences and among them the number of sequences where heads never occurs twice in a row is 3. So for $n = 2$, the probability of heads never occurs twice in a row is $P_2=\frac{3}{4}$.
Similarly, for $n = 3$, the sequences are:
000
001
010
011
100
101
110
111
Again there are $2^3$ total possible sequences, and 5 of them with heads never occurs twice in a row. So $P_3=\frac{5}{8}$.
To find the general formula for calculating $P_n$, lets make the following observation:
*
* When adding a new most significant digit to get the binary numbers of $n$ digits, for example, to expend the numbers from $n=2$ to $n=3$ above, we make 2 copies of all of the binary numbers of $n-1$ digits, add $0$s to left of the top copy and $1$s to the bottom copy.
*The top quarter of the new binary numbers all has two $0$s at the beginning.
*The second quarter of the new binary numbers has the same number of sequences, where heads never occurs twice, as that of $n-2$, that is $2^{n-2}P_{n-2}$.
*The bottom half of the new binary numbers has the same number of sequences, where heads never occurs twice, as that of $n-1$, that is $2^{n-1}P_{n-1}$.
Based on these observations we have:
$$
P_n = \frac{2^{n-2}P_{n-2} + 2^{n-1}P_{n-1}}{2^n} = \frac{\frac{1}{2}P_{n-2} + P_{n-1}}{2}
$$
Using this recursive formula, we get:
$$
P_4=\frac{\frac{1}{2}P_2+P_3}{2}=\frac{\frac{1}{2}\frac{3}{4}+\frac{5}{8}}{2}=\frac{1}{2}
$$
$$
P_5=\frac{\frac{1}{2}P_3+P_4}{2}=\frac{\frac{1}{2}\frac{5}{8}+\frac{1}{2}}{2}=\frac{13}{32}
$$
$$
P_6=\frac{\frac{1}{2}P_4+P_5}{2}=\frac{\frac{1}{2}\frac{3}{4}+\frac{5}{8}}{2}=\frac{21}{64}
$$
| {
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limit without expansion I was solving this limit instead of using sum of expansion i did this and got zero but answer is 1/5 by expansion why this is wrong
$$\lim_{n\rightarrow \infty } \frac{1^4 + 2^4 + \ldots + n^4}{n^5}$$
$$\lim_{n\rightarrow \infty } \frac{n^4( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n^5}$$
$$= \lim_{n\rightarrow \infty } \frac{( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n}$$
$$= \frac{(0 + 0 +0 \ldots + 1 )}{n} = 0$$ this is how i got ,
| You were going right till the very last step. You need to apply the limit after computing the sum. The limit converts to Riemann integral:
$$\lim_{n\to \infty}\sum_{k=1}^{n} \frac{k^4}{n^5} = \int_{0}^{1} x^4 dx = \frac{1}{5}$$
Since you are not familiar with integration, another method is to find $\sum_{k=1}^{n} k^4$ explicitly. Only the leading coefficient, ie the coefficient of $n^5$ needs to be found.
$$\sum_{k=1}^nk^4=\frac1{30}n(n+1)(2n+1)(3n^2+3n-1)$$
So we can write $\sum_{k=1}^nk^4 = \frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} -\frac{n}{30}$. Thus if you compute the limit
$$\lim_{n\to \infty}\sum_{k=1}^n\frac{k^4}{n^5} = \lim_{n\to\infty} \frac{\frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} -\frac{n}{30}}{n^5}\\
\lim_{n\to\infty} \frac{\frac{1}{5} + \frac{1}{2n} + \frac{1}{3n^2} -\frac{1}{30n^4}}{1} = \frac{1}{5}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Regular polygon areas in ratio 3:2 Two regular polygons are inscribed in the same circle of radius $r$. First one has $k$ sides and second has $p$ sides. We are given that their areas have a ratio of $1.5$.
Calculate the area of a regular polygon inscribed in the same circle, having number of sides the sum of the other two numbers.
Area of the first polygon:
$$A_k = \frac{1}{2}\cdot k \cdot r^2 \cdot \sin\bigg(\dfrac{2\pi}{k}\bigg)$$
and area of the second:
$$A_p = \frac{1}{2} \cdot p \cdot r^2 \cdot \sin\bigg(\dfrac{2\pi}{p}\bigg)$$
Also $\frac{A_k}{A_p} = 1.5$ (assuming $k>p$ WLOG)
So
$$\frac{A_k}{A_p} = \frac{k}{p} \cdot \frac{\sin\bigg(\dfrac{2\pi}{k}\bigg)}{\sin\bigg(\dfrac{2\pi}{p}\bigg)} = 1.5$$
Now obviously:
$$A_{k+p} = \frac{1}{2} \cdot (k+p) \cdot r^2 \cdot \sin\bigg(\dfrac{2\pi}{k+p}\bigg)$$
But I don't know how to continue, i.e. how to find a relation between the sins.
| There are only limited possibilities for $n$ because we require $n\ge3$ to get any area and also $\frac32\frac n2\sin\frac{2\pi}n<\pi$ because the bigger polygon has to be smaller than the circumcircle. So we enumerate the possibilities:
$$\begin{array}{r|l}n&\text{Area}\\\hline
3&1.299\\
4&2.000\\
5&2.378\\
11&2.974\\
12&3.000\\
13&3.021\\
\end{array}$$
So $n=5$ is too big because $1.5\times2.378>\pi$. $3$ won't work because $1.5\times1.299$ doesn't match anything. Thus we are left with $n=4$ for the smaller polygon, so the area of the bigger polygon is $1.5\times2.000=3.000$, and this matches $n=12$. The final polygon must have $n=4+12=16$ sides with area
$$A=\frac{16}2\sin\frac{2\pi}{16}=8\sin\frac{\pi}8=4\sqrt{2-\sqrt2}$$
So actually @Toby Mak's answer turned out to be correct in that the areas were in fact rational.
EDIT: Given that only one possibility is feasible it is easy to check that for $n=4$
$$A=\frac42\sin\frac{2\pi}4=2\sin\frac{\pi}2=2$$
While for $n=12$
$$A=\frac{12}2\sin\frac{2\pi}{12}=6\sin\frac{\pi}6=3$$
So the identity is exactly satisfied.
| {
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If $a,b,c$ are the roots of $x^3-px+q=0$, then what is the determinant of the given matrix in $a, b, c$?
If $a,b,c$ are the roots of $x^3-px+q=0$, then what is the determinant of
$$
\begin{pmatrix}
a & b & c \\
b & c & a \\
c & a & b \\
\end{pmatrix} \,\,?
$$
(A) $p^2+6q \quad$
(B) $1 \quad$
(C) $p \quad$
(D) $0 \quad$
In this equation given we have product of eigenvalues given as $-q$ and we know product of eigenvalues is determinant then why isn't the determinant is $-q$?
| $$\det\begin{pmatrix}a&b&c\\ b&c&a \\ c&a&b\end{pmatrix}=(a+b+c)\det\begin{pmatrix}1&1&1\\ b&c&a \\ c&a&b\end{pmatrix}=0$$
since $a+b+c=0$.
| {
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