Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Finding the derivative using the definition? Calculate the derivate of the given function directly from the definition of derivative, and express the result using differentials
$$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$
when $f(x)= 1/\sqrt{1+x^2}$
any tips/solutions on how to get started on this one? I am able to do more basic problems, but not with root etc! thanks for tips/advice/solutions!
| $$\eqalign{
\lim_{h \to 0} \frac{f\left(x+h\right)-f\left(x\right)}{h}&=\lim_{h \to 0} \frac{\frac{1}{\sqrt{1+\left(x+h\right)^2}}-\frac{1}{\sqrt{1+x^2}}}{h}\\
&=\lim_{h \to 0} \frac{\sqrt{1+x^2}-\sqrt{1+\left(x+h\right)^2}}{h \sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}}\\
&=\lim_{h \to 0} \frac{\left(\sqrt{1+x^2}-\sqrt{1+\left(x+h\right)^2}\right)\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)}{h \sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)}\\
&=\lim_{h \to 0} \frac{1+x^2-\left(1+\left(x+h\right)^2\right)}{h \sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)}\\
&=\lim_{h \to 0} \frac{-2xh-h^2}{h \sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)}\\
&=\lim_{h \to 0} \frac{-2x-h}{\sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)} \\
&=\frac{-2x}{\left(1+x^2\right) \cdot 2 \cdot \sqrt{1+x^2}} \\
&=-\frac{x}{\left(1+x^2\right)^{{3}/{2}}} .
}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/890397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then $|z|=$? If z is a complex number and $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then find the value of $|z|$ .
I tried to put $ \frac {z^2 + z+ 1} {z^2 -z +1} =k $ then solve for $z$ and tried to find |z|, but it gets messy and I am stuck.
The answer given is |z|=1
| First note:
$$\frac{z^2 + z + 1}{z^2 - z + 1} = 1 + \frac{2z}{z^2 - z + 1}$$
So the given fraction is real if and only if the fraction $\frac{z}{z^2 - z + 1}$ is real. But a fraction is real if and only if its reciprocal is, so we need:
$$\frac{z^2 - z + 1}{z} = z - 1 + z^{-1}$$
To be a real number. So we get:
$$\boxed{\text{The fraction is real if and only if } z + z^{-1} \text{ is real (or } z= 0).}$$
Now if $z = a + bi$, with $z \neq 0$, the imaginary part of $z + z^{-1}$ is $b - \frac{b}{a^2 + b^2} = \frac{b(a^2 + b^2 - 1)}{a^2 + b^2} = \frac{b(|z| - 1)}{|z|}$. Hence we get:
$$\boxed{\text{Either } |z| = 1 \text{ or } z \text{ is real.}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/890723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Prove that $\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}=\frac{(1-\cos\theta)\csc^2\theta}{\csc\theta+1}$ Question:
$$\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}=\frac{(1-\cos\theta)\csc^2\theta}{\csc\theta+1}$$
Prove that L.H.S.=R.H.S.
My Efforts:
L.H.S.$$=\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}\times\frac{1-\cos\theta}{1-\cos\theta}\times\frac{\csc\theta+1}{\csc\theta+1}$$
$$=\frac{\tan^2\theta(1-\cos\theta)(\csc^2\theta-1)}{(1-\cos^2\theta)(\csc\theta+1)}$$
| $$\tan^2\theta(\csc^2\theta-1)=(1-\cos^2\theta)\csc^2\theta,$$
$$\frac{\sin^2\theta}{\cos^2\theta}(\frac1{\sin^2\theta}-1)=(1-\cos^2\theta)\frac1{\sin^2\theta},$$
$$\frac{\sin^2\theta}{\cos^2\theta}\frac{\cos^2\theta}{\sin^2\theta}=\frac{\sin^2\theta}{\sin^2\theta}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/890870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Differential equation: $y + t\frac{dy}{dt} = \frac{y}{y^2 - 1} $ $$y + t\frac{dy}{dt} = \frac{y}{y^2 - 1} \Rightarrow t\frac{dy}{dt} = \frac{-y^3 + 2y}{y^2 - 1} \Rightarrow \frac{1}{t} dt = \frac{y^2 - 1}{-y^3 + 2y}dy \Rightarrow$$
$$\Rightarrow \ln |t| = - \int \frac{y^2 - 1}{y^3 - 2y}dy$$
How to solve the integral in the right part? Had it been $\frac{3y^2 - 2}{y^3 - 2y}$, then the answer would be $\ln |y^3 - 2y|$.
| HINT : We have
$$\frac{y^2-1}{y^3-2y}=\frac 12\left(\frac{1}{y}+\frac{y}{y^2-2}\right)=\frac{1}{2y}+\frac{(y^2-2)'}{4(y^2-2)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/893164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sum of cubes of roots of a quartic equation $x^4 - 5x^2 + 2x -1= 0$
What is the sum of cube of the roots of equation other than using substitution method? Is there any formula to find the sum of square of roots, sum of cube of roots, and sum of fourth power of roots for quartic equation?
| This is precisely what Newton's Sums are for.
Let $S_n$ be the sum of the $n$-th powers of the roots of $a_nx^n+a_{n-1}x^{n-1} + \cdots + a_1x+a_0$.
Then, Newton's Sums tells us that $\displaystyle\sum_{k = 0}^{m-1}\left(a_{n-j}S_{m-j}\right)+ma_{n-m} = 0$ for each positive integer $m$.
We can write out the first few equations explicitly as follows:
$a_nS_1 + a_{n-1} = 0$
$a_nS_2 + a_{n-1}S_1 + 2a_{n-2} = 0$
$a_nS_3 + a_{n-1}S_2 + a_{n-2}S_1 + 3a_{n-3} = 0$
$a_nS_4 + a_{n-1}S_3 + a_{n-2}S_2 + a_{n-3}S_1 + 4a_{n-4} = 0$
and so on. Note that we define $a_j = 0$ for $j < 0$.
Applying those formula to the polynomial $1x^4+0x^3-5x^2+2x+1$ yields the following:
$1 \cdot S_1 + 0 = 0 \leadsto S_1 = 0$ (sum of roots)
$1 \cdot S_2 + 0 \cdot S_1 + 2 \cdot -5 = 0 \leadsto S_2 = 10$ (sum of squares of roots)
$1 \cdot S_3 + 0 \cdot S_2 + - 5 \cdot S_1 + 3 \cdot 2 = 0 \leadsto S_3 = -6$ (sum of cubes of roots)
and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/894826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Determine variables that fit this criterion... There is a unique triplet of positive integers $(a, b, c)$ such that $a ≤ b ≤ c$.
$$
\frac{25}{84} = \frac{1}{a} + \frac{1}{ab} + \frac{1}{abc}
$$
Just having trouble with this Canadian Math Olympiad question. My thought process going into this, is:
Could we solve for $\frac{1}{a}$ in terms of the other variables? Then substitute that value in for each occurrence of $a$, to solve for $a$?
That's all I can really think of right now. It's a question I'm not exactly used to... It's sort of the first of these kinds that I've faced.
Thanks.
| $$
\begin{align}
\frac{25}{84} &= \frac{1}{a} + \frac{1}{ab} + \frac{1}{abc} \\
&= \frac{bc+c+1}{abc} = \frac{(b+1)c+1}{abc}
\end{align}
$$
Now $ 84 = 2\cdot 2 \cdot 3 \cdot 7$ so $b,c$. Let us try $(b+1)c= c = 24 = 2\cdot 2\cdot 2\cdot 3$ and $abc = 84$ (this is just a guess, it can also be a multiple each time). We are looking for $a,b,c$ that are in the factorization of $84$. So $b+1=8$ and $c =3$ would work with $b=7$ since $3$ and $7$ are in the factorization of $84$. So with $abc =84=2\cdot 2 \cdot 3 \cdot 7 = a\cdot 7\cdot 3$ it follows that $a = 4$.
So $\frac{25}{84} = \frac{1}{4} + \frac{1}{28} + \frac{1}{84}$.
Another approach would be trying to find an Eyptian fraction expansion, for example:
$$\frac{25}{84} = \frac{1}{4}+\frac{1}{21} = \frac{1}{4}+\frac{1}{22}+\frac{1}{462} = \frac{1}{4}+\frac{1}{24}+\frac{1}{168} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/895556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Calculate $\int\frac{dx}{x\sqrt{x^2-2}}$. The exercise is:
Calculate:$$\int\frac{dx}{x\sqrt{x^2-2}}$$
My first approach was:
Let $z:=\sqrt{x^2-2}$ then $dx = dz \frac{\sqrt{x^2-2}}{x}$ and $x^2=z^2+2$ $$\int\frac{dx}{x\sqrt{x^2-2}} = \int\frac{1}{x^2}dz = \int\frac{1}{z^2+2}dz = \frac{1}{2}\int\frac{1}{(\frac{z}{\sqrt{2}})^2 + 1}dz = \frac{1}{2} \arctan(\frac{z}{\sqrt{2}})$$
$$ = \frac{1}{2}\arctan(\frac{\sqrt{x^2-2}}{\sqrt{2}})$$
Yet when e.g. Maple derives the expression, it turns out my solution is false. I also cannot find the mistake, and I'd be glad if someone could point it out to me.
I am interested in your solutions, but in another approach I substituted $z:=\frac{\sqrt{2}}{x}$ and multiplied the integrand with $\frac{\sqrt{2}x}{\sqrt{2}x}$ and it worked out just fine.
| If you replace $x=\sqrt{2} \cosh t$, you end with:
$$\begin{eqnarray*} I = \frac{1}{\sqrt{2}}\int \frac{dt}{\cosh t}=\sqrt{2}\arctan\left(\tanh\frac{t}{2}\right)&=&\sqrt{2}\arctan\left(\frac{\sqrt{x^2-2}}{x+\sqrt{2}}\right)\\&=&\frac{1}{\sqrt{2}}\arctan\sqrt{\frac{x^2-2}{2}}.\end{eqnarray*}$$
Your solution is right, except for the $\sqrt{2}$ factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/895602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to simplify the formula for $n$th Fibonacci number when $n=2$? When n is equal to 2 how do I simplify when the $n=2$ is put into the equation below
(by the way I have to prove this formula by induction that when n= any number it will equal that number)
$$F_n=\frac 1{\sqrt 5} \left(\frac{1+\sqrt5}{2}\right)^n − \frac1{\sqrt 5} \left(\frac{1-\sqrt5}{2}\right)^n$$
This is what I did
When $n=2$ is added to the equation
$$\begin{align}F_n&=\frac 1{\sqrt 5} \left(\frac{1+\sqrt5}{2}\right)^2 − \frac1{\sqrt 5} \left(\frac{1-\sqrt5}{2}\right)^2\\
&=\frac{2+\sqrt 5}{2\sqrt 5} - \frac{2-\sqrt 5}{2\sqrt 5}\\
&=\frac{4\sqrt{5}}{2\sqrt 5}=2
\end{align}$$
I don't know if this is right or not so, it would be nice if someone helps me to simply this formula to be (equal 2)
| In the first line, $(1+\sqrt 5)^2=1+2\sqrt 5 + 5=6+2\sqrt 5$ Similarly $(1-\sqrt 5)^2=6-2\sqrt 5$. Also the $2$'s in the denominators get squared. It should be $$\begin{align}F_2&=\frac 1{\sqrt 5} \left(\frac{1+\sqrt5}{2}\right)^2 − \frac1{\sqrt 5} \left(\frac{1-\sqrt5}{2}\right)^2\\
&=\frac{6+2\sqrt 5}{4\sqrt 5} - \frac{6-2\sqrt 5}{4\sqrt 5}\\
&=\frac{4\sqrt{5}}{4\sqrt 5}=1
\end{align}$$ and in the usual definition of the subscripts $F_2=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/895702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int \left(A x^2+B x+c\right) \, dx$ I am asked to find the solution to the initial value problem:
$$y'=\text{Ax}^2+\text{Bx}+c,$$
where $y(1)=1$,
I get:
$$\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d$$
But the answer to this is:
$$y=\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1.$$
Could someone show me what has been done and explain why?
| $$y'=\text{Ax}^2+\text{Bx}+c$$
$$y(x)=\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d \ \ \ (*) $$
Using the initial condition $y(1)=1$ we get:
$$y(1)=1 \Rightarrow \frac{A }{3}+\frac{B }{2}+c +d=1 \Rightarrow d=1-\frac{A}{3}-\frac{B}{2}-c$$
Replacing this at the relation $(*)$ we get:
$$y(x)=\frac{A x^3}{3}+\frac{B x^2}{2}+c x+1-\frac{A}{3}-\frac{B}{2}-c=\frac{A x^3-A}{3}+\frac{B x^2-B}{2} +cx-c+1 \\ \Rightarrow y(x)=\frac{1}{3}A(x^3-1)+\frac{1}{2}B(x^2-1) +c(x-1)+1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Complex Equations The Equation:
$$ z^{4} -2 z^{3} + 12z^{2} -14z + 35 = 0 $$
has a root with a real part 1, solve the equation.
When it says a real part of 1, does this mean that we could use (z-1) and use polynomial division to extract the other rots? Hence:
$$ (z^{4} -2 z^{3} + 12z^{2} -14z + 35 ) / (z-1) $$
But i don't get the right answer, any other approaches I should try?
| If there's a root of that equation of the form $1+ib$ then:
$$(1+ib)^4 - 2(1+ib)^3 -12(1+ib)^2-14(1+ib) + 35 = 0\Longrightarrow \\(1 + 4ib -6b^2 -4ib^3 + b^4) - 2(1 + 3ib - 3b^2 -ib^3) + 12(1 + 2ib - b^2) - 14(1+ib) + 35 =0 \Longrightarrow b^4-12b^2+32 + i(-2b^3+8b) = 0$$
Therefore:
$$\begin{cases}b^4 - 12b^2+32 = 0\\-2b^3+8b = 0 \end{cases}$$
From the second equation we can see that $b\in \lbrace -2,0,2\rbrace$.
But $b=0$ doesn't satisfy the first equation. The other two values do.
Therefore $(1-2i)$ and $(1+2i)$ are roots of the equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/903016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Ellipse like on sphere Find the locus of all points on a sphere such that the sum of geodesic distances from two fixed points $F_1$ and $F_2$ on it is a constant, less than its diameter. ( When radius of sphere goes to infinity, it would look like an ellipse).
Following is image after Equn ($1$) in achille hui note
| WOLOG, consider the case $\begin{cases}
F_1 &= (+\sin\alpha,0,\cos\alpha),\\
F_2 &= (-\sin\alpha,0,\cos\alpha)
\end{cases}$ with $\alpha \in (0,\frac{\pi}{2})$ and the sphere is the unit sphere.
Let $2\theta \ge 2\alpha$ be that constant for the sum of geodesic distances to $F_1$
and $F_2$. For any point $p = (x,y,z)$ on the locus. we can find a $\tau$ such that
$$
\theta + \tau = \text{dist}(p,F_1) = \cos^{-1}(\cos\alpha z + \sin\alpha x)\\
\theta - \tau = \text{dist}(p,F_2) = \cos^{-1}(\cos\alpha z - \sin\alpha x)
$$
This leads to
$$\cos(\theta\pm\tau) = \cos\alpha z \pm \sin\alpha x
\implies
\begin{cases}
\cos\theta\cos\tau &= +\cos\alpha z\\
\sin\theta\sin\tau &= -\sin\alpha x
\end{cases}
$$
and hence
$$\left(\frac{\cos\alpha}{\cos\theta} z\right)^2 + \left(\frac{\sin\alpha}{\sin\theta} x\right)^2 = (\cos\tau)^2 + (\sin\tau)^2 = 1$$
Let $A = \tan\alpha$ and $D = |\tan\theta|$, we can rewrite this as
$$z^2 + \frac{A^2}{D^2} x^2 = \frac{1+A^2}{1+D^2}
\quad\iff\quad \frac{x^2}{D^2} + \frac{y^2}{D^2-A^2} = \frac{1}{1+D^2}\tag{*1}
$$
$\require{enclose}\newcommand{\mybox}[1]{\enclose{roundedbox}{\;#1\;}}$
Geometrically, there are several possibilities
*
*$\mybox{\theta = \alpha}$
$D = A$, $(*1)$ reduces to $y = 0$ and $|x| \le \cos\alpha$.
The locus is a circular arc joining $F_1$, $F2$.
*$\mybox{\alpha < \theta < \frac{\pi}{2}}$
$A < D < \infty$, $(*1)$ reduces to the equation of an ellipse.
The locus lies on the upper-hemisphere. It not only looks like an ellipse, it is an ellipse if you project it to the $xy$-plane.
*$\mybox{\theta = \frac{\pi}{2}}$
$D = \infty$, $(*1)$ reduces to $x^2 + y^2 = 1$. The locus is the equator.
*$\mybox{\frac{\pi}{2} < \theta < \pi - \alpha}$
$A < D < \infty$ again, the locus lies on the lower-hemisphere. Once again, it is an ellipse if you project it to the $xy$-plane.
*$\mybox{\theta = \pi -\alpha}$
$D = A$ and once again, $y = 0$ and $|x| \le \cos\alpha$. The locus is a circular arc joining the antipodal points of $F_1$ and $F_2$.
Update
At the end is a picture showing how the family of loci qualitatively look like.
The plot is generated for $\alpha = 30^\circ$ with $\theta$ start at $30^\circ$
increasing with step $5^\circ$. The loci whose $\theta$ is an integer multiple
of $15^\circ$ is colored in black and the one for $\theta = 90^\circ$ is in white.
Please note that the plot is partially transparent. If you look carefully,
you can see the locus for $\theta = 90^\circ$ behind the back of sphere.
This locus doesn't have any strange feature and it is simply the great circle at equator!
$\hspace1.0in$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/903368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
How to find $\int \frac{x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$ $$I=\int x.\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$$
Try 1:
Put $z= \ln(x+\sqrt{1+x^2})$, $\mathrm dz=1/\sqrt{1+x^2}\mathrm dx$
$$I=\int \underbrace{x}_{\mathbb u}\underbrace{z}_{\mathbb v}\mathrm dz=x\int zdz-\int (z^2/2)\mathrm dz\tag{Wrong}$$
Try 2:
Put
$z= x+\sqrt{1+x^2}$
$$\implies x-z
=\sqrt{1+x^2}\implies x^2+z^2-2xz
=1+x^2\implies x
=\frac{z^2-1}{2z}$$
$$\mathrm dz
=\left(1+\frac{x}{\sqrt{1+x^2}}\right)\mathrm dx
=\frac{z\mathrm dx}{x-z}=\frac{-2z^2\mathrm dx}{1+z^2}$$
$$I
=\int\frac{(z^2-1)\ln z}{2z}.\frac{(1+z^2)\mathrm dz}{-2z^2}$$
$$=\int\frac{(z^4-1)\mathrm dz}{4z^3}
=\frac14\int\left(z-\frac1{z^3}\right)\mathrm dz
=z^2/2+2/z^2+C\tag{Wrong}$$
Try 3:
Put
$z
=\sqrt{1+x^2},\mathrm dx
=x/\sqrt{1+x^2}\mathrm dx$
$$I
=\int \ln(x+z)\mathrm dz
=\int \ln(z+\sqrt{z^2-1})\mathrm dz$$
Don't know how to solve this integral.
[Note that if I take $u=z+\sqrt{z^2-1}$, it is $=\sqrt{1+x^2}+\sqrt{1+x^2-1}=x+\sqrt{1+x^2}$; same as first try.]
What's wrong in try 1 & 2, how to further solve try 3 and the best method to solve this question?
Update: Sorry, I don't know hyperbolic/inverse hyperbolic trigonometry.
| $$
\int \ln(z+\sqrt{z^2-1})dz = \int \mathrm{arcosh}(z) dz\,\,\,z>1.
$$
the equality holds because for real x, $z = \sqrt{x^2+1}>1$.
then using the fact
$$
\int \mathrm{arcosh}(z) dz = z\mathrm{arcosh}(z) +\sqrt{z^2-1} + C
$$
then sub back in your substitution.
$\textbf{brief derivation of the arcosh relation}$
$$
y = \cosh(x) \implies \mathrm{arcosh}(y) = x
$$
now
$$
\cosh(x) = \frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2} = y \tag{*}
$$
I can re-write the expression in Eq.(*) with $\mathrm{e}^{x} = t$ as
$$
\cosh(x) = \frac{t+1/t}{2} = \frac{1}{t}\frac{t^2+1}{2} = y
$$
so solving for $t$ we find
$$
t^2-2yt+1 = 0 \implies t = \frac{2y\pm\sqrt{4y^2-4}}{2} = y\pm\sqrt{y^2-1}
$$
therefore
$$
\mathrm{e}^{x} = t \implies x = \ln\left(y\pm\sqrt{y^2-1}\right) = \mathrm{arcosh}(y)
$$
since we are looking for real valued function then we can ignore the minus sign.
At least in my head anyway.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
To test the convergence of series 2 To test the convergence of series 2
$\displaystyle \frac{a+x}{1!}+\frac{(a+2x)^2}{2!}\frac{(a+3x)^3}{3!}+...\infty$
My Attempt:
$$\begin{align}
\frac{u_n}{u_{n+1}} & = \frac{(a+nx)^n}{n!}\frac{(n+1)!}{(a+(n+1)x)^{n+1}} \\
& = \frac{(a+nx)^n}{1}\frac{n+1}{(a+nx+x)^{n+1}} \\
& = \frac{(x+a/n)^n}{1}\frac{1+1/n}{(x+a/n+x/n)^{n+1}} \\
& = \frac{x^n(1+a/nx)^n}{1}\frac{1+1/n}{x^{n+1}(1+a/nx+1/n)^{n+1}} \\
& = \frac{(1+a/nx)^n}{1}\frac{1+1/n}{(1+a/nx+1/n)^{n+1}}\frac{1}{x} \\
& = ??
\end{align}$$
I see an "e" in the limit $n\to \infty$ here, but can't quite get "there". Any hints?
| Follow Did's advice, you should get something like
$$
a^{\frac{1}{n}}_n = \frac{a+nx}{(n!)^{\frac{1}{n}}} \sim \frac{e(a+nx)}{n} (2 \pi n)^{-\frac{1}{2n}}
$$
The last term converges to 1, so the series converges for $x<e^{-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Infinite Series $\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots$ How do I find the sum of the following infinite series:
$$\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots$$
The series definitely seems to be convergent.
| \begin{array}{l}
s_0 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {\frac{{x^{3k + 3} }}{{3k + 3}}} \Rightarrow s'_0 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {x^{3k + 2} } = \frac{{x^2 }}{{1 - x^3 }} \\
s_1 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {\frac{{x^{3k + 4} }}{{3k + 4}}} \Rightarrow s'_1 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {x^{3k + 3} } = \frac{{x^3 }}{{1 - x^3 }} \\
s_2 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {\frac{{x^{3k + 2} }}{{3k + 2}}} \Rightarrow s_2 ^\prime \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {x^{3k + 1} } = \frac{x}{{1 - x^3 }} \\
s_0 \left( x \right) + s_1 \left( x \right) + s_2 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {\left( {\frac{{x^{3k + 3} }}{{3k + 3}} + \frac{{x^{3k + 4} }}{{3k + 4}} + \frac{{x^{3k + 2} }}{{3k + 2}}} \right)} \\
\frac{x}{{1 - x^3 }} + \frac{{x^3 }}{{1 - x^3 }} - \frac{{2x^2 }}{{1 - x^3 }} = \frac{{x^3 + x - 2x^2 }}{{1 - x^3 }} = \frac{{x\left( {x - 1} \right)}}{{1 + x + x^2 }} \\
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Calculating multi-variable limit. I am struggling to find a way to approach this limit
$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2y+x^2y^3)}{x^2+y^2}$$
I would greatly appriciate if You could explain to me how to solve it or at least show how to start.
| By the inequality $|\sin(\theta)| \leq |\theta|$, which is valid for all real $\theta$, we have
$$\left|\frac{\sin(x^2y + x^2 y^3)}{x^2 + y^2}\right| \leq \left|\frac{x^2y + x^2 y^3}{x^2 + y^2}\right|$$
For $x \neq 0$, the right hand side is equal to
$$\left|\frac{y + y^3}{1 + y^2/x^2}\right| \leq \left|y + y^3\right| $$
where inequality is true because the denominator of the left hand side is $\geq 1$.
On the other hand, if $x=0$ and $y \neq 0$, then
$$\left|\frac{x^2y + x^2 y^3}{x^2 + y^2}\right| = 0$$
Combining these results, we conclude that
$$\left|\frac{\sin(x^2y + x^2 y^3)}{x^2 + y^2}\right| \leq |y + y^3|$$
for all $(x,y) \neq (0,0)$, and the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/910164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Check if two vector equations of parametric surfaces are equivalent Give the vector equation of the plane through these lines:
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}\,\,\,$ and $\,\,\,\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\0\\3\end{pmatrix}+\mu\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}$.
My answer is: $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\-1\\2\end{pmatrix}+\mu\cdot\begin{pmatrix}0\\-3\\1\end{pmatrix}$. The solutions manual suggests the following equation, which is the equation of a straight line (probably a typo): $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}+\mu\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}$
Is my solution the right solution? Could someone provide a general way to check?
| You have two planes that go through the same point (4,1,1). Now all you need is having them parallel. You can check this by forming the cross product of the defining vectors, i.e. (0,-1,2) and (0,-3-1) for the first plane, that will give you a normal to the plane. Do the same for the second plane; if your normal vectors are parallel, so are the planes and hence identical in this case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/912048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to prove this $\frac{\sin{(nx)}}{\sin{x}}\ge\frac{\sqrt{3}}{3}(2n-1)^{\frac{3}{4}}$ let $n<\dfrac{\pi}{2\arccos{\dfrac{c}{2}}},c\in (0,2),c=2\cos{x}$, show that
$$\dfrac{\sin{(nx)}}{\sin{x}}\ge\dfrac{\sqrt{3}}{3}(2n-1)^{\frac{3}{4}}$$
where $0<x<\dfrac{\pi}{2}$
My idea: let $$a_{n}=\dfrac{\sin{(nx)}}{\sin{x}}$$
then for any $k\in[1,n]$, then we have
$$a^2_{k}-a^2_{k-1}=\dfrac{1}{\sin^2{x}}[\sin^2{kx}-\sin^2{(k-1)x}]=\dfrac{\sin{(2k-1)x}\sin{x}}{\sin^2{x}}=\dfrac{\sin{(2k-1)x}}{\sin{x}}$$
since
$$n<\dfrac{\pi}{2\arccos{\dfrac{c}{2}}}=\dfrac{\pi}{2x}\Longrightarrow 0<kx\le nx<\dfrac{\pi}{2}$$
then
$2kx\le 2n<\pi$, so $x\le (2k-1)x<\pi-x$
so $\sin{(2k-1)x}\ge \sin{x}>0$,so
$a^2_{k}-a^2_{k-1}\ge 1$
$$a^2_{n}=\sum_{k=2}^{n}(a^2_{k}-a^2_{k-1})+a^2_{2}\ge n\Longrightarrow a_{n}\ge \sqrt{n}$$
then I can't it,
I'm sorry, I just to eat
,and I'm come back
| This inequality $$\frac{\sin{(nx)}}{\sin{x}}\ge\frac{\sqrt{3}}{3}(2n-1)^{\frac{3}{4}}$$ with $0<x<\dfrac{\pi}{2}$ and $n>1$ makes a serious problem to me.
If we consider $$f_n(x)=\frac{\sin{(nx)}}{\sin{x}}-\frac{\sqrt{3}}{3}(2n-1)^{\frac{3}{4}}$$ we have $$f_n(0)=n-\frac{(2 n-1)^{3/4}}{\sqrt{3}} > 0$$ but it exists $0<x_n<\dfrac{\pi}{2}$ such that $f_n(x_n)=0$ and $x_n$ decreases when $n$ increases.
Built at $x=0$, a second order Taylor expansion gives $$x_n \simeq \frac{\sqrt{6 n-2 \sqrt{3} (2 n-1)^{3/4}}}{\sqrt{n \left(n^2-1\right)}}$$ which for large values of $n$ is $$x_n=\frac{\sqrt{6}}{n}-\sqrt[4]{2}
\left(\frac{1}{n}\right)^{5/4}-\frac{\left(\frac{1}{n}\right)^{3/2}}{2
\sqrt{3}}-\frac{\left(\frac{1}{n}\right)^{7/4}}{6 \sqrt[4]{2}}-\frac{5}{24
\sqrt{6} n^2}+O\left(\left(\frac{1}{n}\right)^{9/4}\right)$$
Solving numerically $f_n(x)=0$ lead to $x_2=0.852587$, $x_3=0.543443$, $x_4=0.412797$, $x_5=0.337444$, $x_6=0.287336$, $x_7=0.251180$, $x_8=0.223690$, $x_9=0.201975$, $x_{10}=0.184339$ and above these values $f_n(x) < 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/912742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Equation $3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$ Solve the equation
$$3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$$
Having a complex root of modulus $1$.
To get the solution, I tried to take a complex root $\sqrt{\frac{1}{2}} + i \sqrt{\frac{1}{2}}$ but couldn't get the solution right. Please help me.
| HINT:
$$3x^4 + 2x^3 + 9x^2 + 4x + 6 = (x^2 + 2)(3x^2 + 2x + 3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/913039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Closed-forms of infinite series with factorial in the denominator How to evaluate the closed-forms of series
\begin{equation}
1)\,\, \sum_{n=0}^\infty\frac{1}{(3n)!}\qquad\left|\qquad2)\,\, \sum_{n=0}^\infty\frac{1}{(3n+1)!}\qquad\right|\qquad3)\,\, \sum_{n=0}^\infty\frac{1}{(3n+2)!}\\
\end{equation}
Of course Wolfram Alpha can give us the closed-forms
\begin{align}
\sum_{n=0}^\infty\frac{1}{(3n)!}&=\frac{e}{3}+\frac{2\cos\left(\frac{\sqrt{3}}{2}\right)}{3\sqrt{e}}\\
\sum_{n=0}^\infty\frac{1}{(3n+1)!}&=\frac{e}{3}+\frac{2\sin\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)}{3\sqrt{e}}\\
\sum_{n=0}^\infty\frac{1}{(3n+2)!}&=\frac{e}{3}-\frac{2\sin\left(\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right)}{3\sqrt{e}}
\end{align}
but how to get those closed-forms by hand? I can only notice that
\begin{equation}
\sum_{n=0}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{(3n)!}+\sum_{n=0}^\infty\frac{1}{(3n+1)!}+\sum_{n=0}^\infty\frac{1}{(3n+2)!}=e
\end{equation}
Could anyone here please help me? Any help would be greatly appreciated. Thank you.
PS: Please don't work backward.
| Related techniques: (I). Here is an approach which enables you to tackle your problems. Let's consider the series
$$ f(x) = \sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!}. $$
Taking the Laplace transform gives
$$ F(s) = \sum_{n=0}^{\infty}\frac{1}{s^{3n+1}} = \frac{s^2}{s^3-1}. $$
To finish the problem you need to find the inverse Laplace of $F(s)$. One technique is partial fraction
$$ F(s) = \frac{1}{3(s-1)} + \frac{1}{3(s+1/2-i\sqrt{3}/2)} + \frac{1}{3(s+1/2+i\sqrt{3}/2)} .$$
Notes:
1) Laplace transform is defined as
$$ F(s) = \int_{0}^{\infty}f(x)e^{-sx}dx. $$
2) Laplace transform of $x^m$ is
$$ \frac{\Gamma(m+1)}{s^{m+1}} $$
3) Laplace transform of $e^{ax}$ is
$$ \frac{1}{s-a}. $$
Or equivalently the inverse Laplace of $\frac{1}{s-a}$ is $e^{ax}$
$$ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/914176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
Addition of fractions repetition and convergence Is this a new mathematical concept?
$$
\frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3} \cdots = \frac{1}{n-1}
$$
If it isn't then what is this called?
I haven't been able to find anything like this anywhere.
| $$\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\dots=\sum_{i=1}^{\infty} \frac{1}{n^i}=\sum_{i=1}^{\infty} \left (\frac{1}{n} \right )^i=\sum_{i=0}^{\infty} \left (\frac{1}{n} \right )^i-1=\frac{1}{1-\frac{1}{n}}-1=\frac{n}{n-1}-1=\frac{n-n+1}{n-1}=\frac{1}{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/915136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
The limit of $(4-\sqrt{16-7\sin(x)})/(8x)$ at zero without using L'Hôpital I stumbled across this silly limit and I am perplexed at how I can arrive to a solution by only relying on the simplest rules of limits.
$$
\lim_{x \to 0}\frac{4-\sqrt{16-7\sin(x)}}{8x}
$$
Any help is appreciated, thanks in advance.
| Multiply the numerator and denominator by $4+\sqrt{16-7\sin x}$ to get
\begin{align}
& \lim_{x \to 0}\dfrac{4-\sqrt{16-7\sin x}}{8x} \\[8pt]
= {} &\lim_{x \to 0}\dfrac{4-\sqrt{16-7\sin x}}{8x} \cdot \dfrac{4+\sqrt{16-7\sin x}}{4+\sqrt{16-7\sin x}} \\[8pt]
= {} & \lim_{x \to 0}\dfrac{4^2 - (16-7\sin x)}{8x(4+\sqrt{16-7\sin x})} \\[8pt]
= {} & \lim_{x \to 0}\dfrac{7\sin x}{8x(4+\sqrt{16-7\sin x})} \\[8pt]
= {} & \lim_{x \to 0}\dfrac{\sin x}{x} \cdot \lim_{x \to 0}\dfrac{7}{8(4+\sqrt{16-7\sin x})}
\end{align}
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/916332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Ad hoc proof of the convergence of $\sum\limits_n\sin\left(\pi\sqrt{n^2+a^2}\right)$ I use the following method to prove that
$$\sum_{n=0}^\infty \sin(\pi\sqrt{n^2+a^2})$$
converges for any $a \in \mathbb{R}$.
First, we can see that
$$\lim_{n\to\infty} \frac{\sqrt{n^2 + a^2}}{n} = 1,$$
which means precisely that $\forall \varepsilon > 0, \exists n_0\ \forall n \ge n_0: \sqrt{n^2+a^2} - n < \varepsilon$. We can therefore find $n_0$ such that $\sqrt{n^2+a^2} - n < \frac\pi2$.
Then we write $$\sum_{n=0}^\infty \sin(\pi\sqrt{n^2+a^2}) = \sum_{n=0}^{n_0} \sin(\pi\sqrt{n^2+a^2})\ + \sum_{n=n_0+1}^\infty \sin(\pi\sqrt{n^2+a^2}),$$
where the first sum contains a finite number of finite terms and evaluates to some $K \in \mathbb{R}$.
The second sum can be rewritten (thanks to the fact that $\sqrt{n^2+a^2} - n < \frac\pi2$) as
$$\sum_{n=n_0+1}^\infty \sin(\pi\sqrt{n^2+a^2}) = \sum_{n=n_0+1}^\infty (-1)^{n+1}|\sin(\pi\sqrt{n^2+a^2})|.$$
We have already shown that $d_n=(\sqrt{n^2+a^2} - n) \to 0$ as $n\to\infty$. $|\sin x|$ has the period $\pi$, and because of that, $|\sin(\pi\sqrt{n^2+a^2})|=|\sin{(n\pi+d_n\pi)}|=|\sin(n\pi + d_n\pi + k\pi)|$ for any $k \in \mathbb{Z}$. Choosing $k=-n$, we get $|\sin(\pi\sqrt{n^2+a^2})| = |\sin{d_n\pi}|$.
As $d_n$ is a monotonic decreasing sequence (we can show easily that $d_{n+1} < d_n\ \forall n$) and $|\sin x|$ is monotonic on $(k\pi,\ k\pi + \frac\pi2)$, we have that $|\sin(\pi\sqrt{n^2+a^2})| \to 0$ is a monotonically decreasing sequence.
The convergence of the sum now follows from the Leibniz's rule.
There is probably a much simpler way. I would like to know whether my solution is correct and if not, why?
| Note that $$\sqrt{n^2+a^2}=n\cdot\sqrt{1+\frac{a^2}{n^2}}=n\cdot\left(1+\frac{a^2}{2n^2}+O\left(\frac1{n^4}\right)\right)=n+u_n,$$
where $$u_n=\frac{a^2}{2n}+O\left(\frac1{n^3}\right).$$ Note also that, when $x\to0$, $$\sin x=x+O(x^3),$$ hence, using this for $x=\pi u_n$, one gets $$\sin\left(\pi\sqrt{n^2+a^2}\right)=\sin(n\pi+\pi u_n)=(-1)^n\cdot\sin(\pi u_n)=\frac{(-1)^n}n \frac{\pi a^2}2+O\left(\frac1{n^3}\right).$$
This shows that the series of interest is the sum of an alternating series and of an absolutely convergent series, hence it converges (but not absolutely).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/917346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How find the least value of the expression: $M = \cot^2 A + \cot^2 B + \cot^2 C + 2(\cot A - \cot B)(\cot B - \cot C)(\cot C - \cot A)$? Consider all triangles $ABC$ where $A < B < C \leq \frac{\pi}{2}$. How find the least value of the expression:
$M = \cot^2 A + \cot^2 B + \cot^2 C + 2(\cot A - \cot B)(\cot B - \cot C)(\cot C - \cot A)$?
| consider $(\cot B - \cot C)(\cot C - \cot A)=-(\cot C-\cot B )(\cot C - \cot A)$
for easy, $x=\cot C,a=\cot A,b=\cot B,ab=-(a+b)x+1 ,f(x)=x^2-(a+b)x+ab=x^2-2(a+b)x+1$ to have max
$x< (a+b) \implies f_{max}=f(0)=1$
$\cot^2 C \ge 0$ so $M$ will get least value when $\cot C=0$
rest is easy.
indeed, without the limitation of $A,B,C$. $M$ will also have least value $1$ but there is one more case is $A=B=C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/918996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integrating $\int\frac{\sqrt{16x^2-9}}x\,\mathrm{d}x$? I am trying to differentiate from my previous question, but I am having trouble in the finishing steps. I have the integral $\int\dfrac{\sqrt{16x^2-9}}x\,\mathrm{d}x$.
$$v=4x\implies\mathrm{d}v=4\,\mathrm{d}x$$
$$\int\frac{\sqrt{v^2-9}}{v}\,\mathrm{d}v\implies a=3,\quad v=3\sec\theta,\quad\mathrm{d}v=3\sec\theta\tan\theta\,\mathrm{d}\theta.$$
$$v^2-9=9\sec^2\theta-9=9\left(\sec^2\theta-1\right)=9\tan^2\theta.$$
$$\int\frac{3\tan\theta\cdot3\sec\theta\tan\theta}{3\sec\theta}\,\mathrm{d}\theta=3\int\tan^2\theta\,\mathrm{d}\theta.$$
$$3\int\left(\sec^2\theta-1\right)\mathrm{d}\theta=3\left(\tan\theta-\theta\right).$$
And this where I feel I am going wrong. So from my understanding since I used $\sec\theta$ for the substitution the triangle I am using to use this integral has a hypotenuse of $v$ or $4x$ since I initially assigned that value to b and the adjacent side of $\theta$ is $a$ or $3$, and for the missing side I got $\sqrt{v^2-9}$ or $\sqrt{16x^2-9}$. So for $\tan\theta$ I got $\dfrac{\sqrt{16x^2-9}}3$ and for $\theta$ I got $\sec^{-1}\left(\dfrac{v}3\right)$ or $\cos\left(\dfrac{4x}3\right)$. So my final answer is $$\sqrt{16x^2-9}-3\cos\left(\frac{4x}3\right)$$
but this is wrong so if someone could tell me where I am going wrong it'd be greatly appreciated. Thanks in advance.
| I would rather like to try to tackle it with rationalization other than substitution. $$
\begin{aligned}
\int \frac{\sqrt{16 x^{2}-9}}{x} d x &=\int \frac{16 x^{2}-9}{x \sqrt{16 x^{2}-9}} d x \\
&=\frac{1}{16} \int \frac{16 x^{2}-9}{x^{2}} d\left(16 x^{2}-9\right) \\
&=\frac{1}{16} \int\left(16-\frac{9}{x^{2}}\right) d\left(\sqrt{16 x^{2}-9}\right) \\
&=\sqrt{16 x^{2}-9}-3 \arctan \left(\frac{\sqrt{16 x^{2}-9}}{3}\right)+C \\
& (\textrm{ OR }\sqrt{16 x^{2}-9}-3 \arccos \left(\frac{3}{4 x}\right)+C)
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim_{x\to\infty}\left[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}\right]$ Evaluate the following the limit:
$$\lim_{x\to\infty}\left[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}\right]$$
I tried expressing the limit in the form $f(x)g(x)\left[\frac{1}{f(x)} - \frac{1}{g(x)}\right]$ but it did not help.
| Write $p(x) = (x-a_{1})(x-a_{2})\cdots (x-a_{n})$. Now
$$
\lim_{x\to\infty}[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}] = \lim_{x \to \infty}[x-\sqrt[n]{p(x)}] \\
= \lim_{x \to \infty}\left((x-\sqrt[n]{p(x)})\frac{x^{n-1}+x^{n-2}\sqrt[n]{p(x)}+\ldots+x(\sqrt[n]{p(x)})^{n-2}+(\sqrt[n]{p(x)})^{n-1}}{x^{n-1}+x^{n-2}\sqrt[n]{p(x)}+\ldots+x(\sqrt[n]{p(x)})^{n-2}+(\sqrt[n]{p(x)})^{n-1}}\right) \\
= \lim_{x \to \infty}\left(\frac{x^{n}-p(x)}{x^{n-1}+x^{n-2}\sqrt[n]{p(x)}+\ldots+x(\sqrt[n]{p(x)})^{n-2}+(\sqrt[n]{p(x)})^{n-1}}\right) \\
= \lim_{x \to \infty}\left(\frac{(a_{1}+a_{2}+\ldots+a_{n})x^{n-1}+q(x)}{x^{n-1}\left(1+\frac{\sqrt[n]{p(x)}}{x}+\ldots+\frac{(\sqrt[n]{p(x)})^{n-2}}{x^{n-2}}+\frac{(\sqrt[n]{p(x)})^{n-1}}{x^{n-1}}\right)}\right) \\
= \lim_{x \to \infty}\left(\frac{a_{1}+a_{2}+\ldots+a_{n}+\frac{q(x)}{x^{n-1}}}{1+\frac{\sqrt[n]{p(x)}}{x}+\ldots+\left(\frac{\sqrt[n]{p(x)}}{x}\right)^{n-2}+\left(\frac{\sqrt[n]{p(x)}}{x}\right)^{n-1}}\right)
$$
where $q(x)$ is a polynomial of degree at most $n-2$.
Now $\frac{\sqrt[n]{p(x)}}{x} \to 1$, and $\frac{q(x)}{x^{n-1}} \to 0$ as $x \to \infty$ so the whole expression tends to
$$
\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find the maximum the value $P_{1}\cdot P_{n}$
Suppose that $n \ge 2$ students attend a test of $m \ge 2$ problems. The scoring rule for each problem is: if $x$ students answer a problem incorrectly, then a correct answer is worth $x$ points and an incorrect answer is worth none. The total score of a student is the sum of scores of all $m$ problems. The total score of all students will be arranged from high to low as $$P_{1}\ge P_{2}\ge P_{3}\ge \cdots\ge P_{n}$$ Find the maximum of the value $P_{1}\cdot P_{n}$.
Let $a_{k}$ students answer the $k$-th problem
then $$p_{1}\le\sum_{k=1}^{m}(n-a_{k}),p_{1}+p_{2}+\cdots+p_{n}=\sum_{k=1}^{m}a_{k}(n-a_{k})$$
so
$$p_{1}\cdot p_{n}\le p_{1}\cdot \dfrac{p_{2}+p_{3}+\cdots+p_{n}}{n-1}$$
I use this idea,at last,I can't solve this problem,(But I fell this problem can use this idea)
PS: Somedays ago, I have solve this problem :
How find this maximum of $P_{1}+P_{n}$
| Let $x_i$ be the score of the $i$-th problem, then
\begin{equation}
\begin{split}
P_1 &\le \sum_{i=1}^{m} x_i \\
P_n &\le \frac{1}{n-1} \Big( \sum_{i=1}^m x_i (n-x_i) - P_1 \Big)
\end{split}
\end{equation}
denote $S = \sum_{i=1}^{m} x_i, T = \sum_{i=1}^m x_i (n-x_i)$, $P_1 P_n \le \frac{1}{n-1} P_1 ( T - P_1 )$.
In case $T \ge 2S$,
\begin{equation}
\begin{split}
P_1 P_n &\le \frac{1}{n-1} S(T-S) \\
& = \frac{1}{n-1} S\Big((n-1)S - \sum x_i^2 \Big) \\
& \le \frac{1}{n-1} S\Big((n-1)S - \frac{S^2}{m} \Big) \\
& = 4m^2(n-1)^2 \cdot \frac{S}{2m(n-1)} \cdot \frac{S}{2m(n-1)} \Big( 1-\frac{S}{m(n-1)} \Big) \\
& \le \frac{4}{27} m^2(n-1)^2.
\end{split}
\end{equation}
The strategy to this bound is: $P_1$ get every problem correct, others try to have equal total score, and every problem has score $\frac{2}{3}(n-1)$. If $n \equiv 1$ (mod $3$), this bound can be achieved, slightly less for other $n$. But to make $T \ge 2S$ possible, we need $n \ge 4$.
In case $T \le 2S$,
\begin{equation}
\begin{split}
P_1 P_n &\le \frac{1}{n-1} \cdot \frac{T^2}{4} \le \frac{1}{n-1} S^2 \le m^2(n-1).
\end{split}
\end{equation}
This is just a rough estimation we can't take equality, but proves for $n \ge 8$, previous strategy is the best.
Actually in case $T \le 2S$, we have $T \le 2m(n-2)$. (for $n \ge 3$)
Rewrite the condition $T \le 2S$ as
\begin{equation}
\sum_{i=2}^{n-1} (i-2)(n-i) \#_i \le (n-1) \#_1
\end{equation}
where $\#_i$ is number of $x_j$ equal to $n-i$. One can check
\begin{equation}
\frac{(i-2)(n-i)}{n-1} \ge \frac{i(n-i)-2(n-2)}{n-3}.
\end{equation}
Hence
\begin{equation}
\sum_{i=3}^{n-1} i(n-i)-2(n-2) \#_i \le (n-3) \#_1, \\
\sum_{i=1}^{n-1} i(n-i) \#_i \le \sum_i 2(n-2) \#_i,
\end{equation}
this is exactly what we want.
This gives a new bound $\frac{m^2(n-2)^2}{n-1}$, can be achieved by taking $x_i = n-2$ for any $i$. This is the previous bound for $n=4$, for $n=5,6,7$, previous bound is larger, for large $m$ previous bound is better, for small $m$, you probably need to compare. For $n=3$, result is $\frac{m^2}{2}$.
For n = 2, result is $\frac{m^2}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/923572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
} |
Find the greatest value of this expression Let $x$, $y$, $z$ be nonnegative real numbers with $x + y + z = 3$. Find the greatest value of the expression
\begin{equation}
P = \sqrt{(x+1)(y^2 + 2)(z^3 + 3)} + \sqrt{(y+1)(z^2 + 2)(x^3 + 3)} + \sqrt{(z+1)(x^2 + 2)(y^3 + 3)}.
\end{equation}
I tried. We have
$$\sqrt{(x+1)(y^2 + 2)(z^3 + 3)}=\dfrac{1}{\sqrt{24}}\sqrt{(x+1)(y^2 + 2)(z^3 + 3)\cdot 24} \leqslant \dfrac{1}{\sqrt{24}}\cdot \dfrac{(x+1)(y^2 + 2)(z^3 + 3)+24}{2}, $$
$$\sqrt{(y+1)(z^2 + 2)(x^3 + 3)}=\dfrac{1}{\sqrt{24}}\sqrt{(y+1)(z^2 + 2)(x^3 + 3)\cdot 24} \leqslant \dfrac{1}{\sqrt{24}}\cdot \dfrac{(y+1)(z^2 + 2)(x^3 + 3)+24}{2}, $$
$$\sqrt{(z+1)(x^2 + 2)(y^3 + 3)}=\dfrac{1}{\sqrt{24}}\sqrt{(z+1)(x^2 + 2)(y^3 + 3)\cdot 24} \leqslant \dfrac{1}{\sqrt{24}}\cdot \dfrac{(z+1)(x^2 + 2)(y^3 + 3)+24}{2}. $$
Therefore
$$P \leqslant \dfrac{1}{\sqrt{24}}\dfrac{ (x+1)(y^2 + 2)(z^3 + 3) + (y+1)(z^2 + 2)(x^3 + 3) +(z+1)(x^2 + 2)(y^3 + 3) + 72 }{2}$$
Expand the above expression, we get
\begin{align*}
P\leqslant \dfrac{1}{\sqrt{24}}(90 + 6(x+y+z) + 2(x^3 + y^3 + z^3) + 2(x^3 y + y^3 z + z^3 x)\\
3(x^2 + y^2 + z^2) + 3(xy^2 + yz^2 + zx^2) + xyz (xy^2 + yz^2 + zx^2).
\end{align*}
I can not solve from here.
|
Hint: Above is plot using Mathematica. Looks like maximum is for $(3,0,0),(0,3,0),(0,0,3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/926415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sqrt{a_n b_n}$ and $\frac{1}{2}(a_n+b_n)$ have the same limit I'm trying to solve the following problem
prove $\sqrt{a_n b_n}$ and $\frac{1}{2}(a_n+b_n)$ have same limit. In this post https://math.stackexchange.com/a/267499, I do not understand the following steps:
$$\frac {1}{2} (\sqrt{b_n} - \sqrt{a_n} ) ( \sqrt{b_n} + \sqrt{a_n} ) = \frac{1}{2}(b_n - a_n) \leq \frac {1}{2^n} (b_1-a_1)$$
Where did the ${2^n}$ come from?
Thank you for your response.
| Look at this chain of inequalities in the post you cited:
$$b_{n+1} - a_{n+1} = \frac {1}{2} (\sqrt{b_{n}} -\sqrt{ a_{n}})^2 \leq \frac {1}{2} (\sqrt{b_n} - \sqrt{a_n} ) ( \sqrt{b_n} + \sqrt{a_n} ) = \frac {1}{2} ( b_n - a_n)$$
So for every $n$, we have
$$b_{n+1} - a_{n+1} \leq \frac{1}{2}(b_n - a_n)$$
Therefore,
$$b_2 - a_2 \leq \frac{1}{2}(b_1 - a_1)$$
$$b_3 - a_3 \leq \frac{1}{2}(b_2 - a_2) \leq \frac{1}{2}\left(\frac{1}{2}(b_1 - a_1)\right) = \frac{1}{2^2}(b_1 - a_1)$$
It's an easy induction argument (or more informally, continuing these steps $n$ times) to conclude that
$$b_{n+1} -a_{n+1} \leq \frac{1}{2^n}(b_1 - a_1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/927192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Orbit and stabaliser of $2\times2 * 2\times1$ matrices I have the group action of matrix multiplication, meaning:
$g((x,y))=\begin{pmatrix}a&0\\0&b\end{pmatrix}$$ \begin{pmatrix}x\\ y\end{pmatrix}$
$G=\left\{\begin{pmatrix}a&0\\0&b\end{pmatrix}:a,b\in\mathbb{R} - \{0\} \right\} g=\begin{pmatrix}a&0\\0&b\end{pmatrix}\in G$ and $X=\{(x,y):x,y\in\mathbb{R}\}$
I want to work out orbits and stabilizer.
I have only done this with permutation groups, so I have no idea how to do this:
Definition of orbit $orb((x,y))=\{g((x,y)):g\in G\}$
So I have $org((x,y))$ equalling the identity matrix?
$stab((x,y))=\{g\in G:g((x,y))=(x,y)\}$
That makes $stab((x,y))$ look like the identity though? Which am I misunderstanding?
| If you write out the group action as $$g * \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax \\ by \end{pmatrix}$$
For the stabilizer, we look for elements $g \in G$ such that $$g * \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$$ This means that we are looking for $a$ and $b$ such that $ax = x$ and $bx = x$. If $x$ is nonzero, then we must have that $a = 1$, and similarly nonzero $y$ implies that our desired $b$ equals $1$. If $x =0$, then we can choose any nonzero $a$ (as $ax = a \cdot 0 = 0$ for each nonzero $a$), and if $y = 0$, we can choose any value of $b$. To summarize this, considering arbitrary nonzero $x$ and $y$, we have
$$\mathrm{stab}_{G}\left( \begin{pmatrix} x \\ y \end{pmatrix} \right) = \left\{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right\}, \: \: \mathrm{stab}_{G}\left( \begin{pmatrix} x \\ 0 \end{pmatrix} \right) = \left\{ \begin{pmatrix} 1 & 0 \\ 0 & b \end{pmatrix} \bigg| b \in \mathbb{R} - \{0\} \right\}$$ $$\mathrm{stab}_{G}\left( \begin{pmatrix} 0 \\ y \end{pmatrix} \right) = \left\{ \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}\bigg| a \in \mathbb{R} - 0 \right\}, \:\mathrm{stab}_{G}\left( \begin{pmatrix} 0 \\ 0 \end{pmatrix} \right) = \left\{ \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \bigg|\: a,b \in \mathbb{R} - 0 \right\}$$
Now, the orbit is the set of elements $\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$ such that there exists $g \in G$ for which $$ g * \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$$ Because $a$ and $b$ can be any nonzero real number, we then have that $$\mathrm{orb}_{G} \left( \begin{pmatrix} x \\ y \end{pmatrix} \right) = \left\{ \begin{pmatrix} ax \\ by \end{pmatrix} \bigg| \: a,b \in \mathbb{R} - \{0\} \right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/928181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\int_0^{\pi}\sin^2x\cos^4x\hspace{1mm}dx$ Find $\int_0^{\pi}\sin^2x\cos^4x\hspace{1mm}dx$
$ $
This appears to be an easy problem, but it is consuming a lot of time, I am wondering if an easy way is possible.
WHAT I DID :
Wrote this as $\int \dfrac{1}{4}(\sin^22x)\cos^2x$
And then I wrote $\cos^2x$ in terms of $\cos(2x)$, I get an Integral which is sum of two known integrals, I did them, I got answer $\dfrac{\pi}{16}$
| $$ 2\sin x \cos x = \sin 2x$$
$$ 2\cos^2 x -1 = \cos 2x \iff \cos 2x = \frac{ \cos 2x+1}{2}$$
$$\sin^2 x \cos^4 x =\sin^2 x \cos^2 x cos^2 x= 1/8 \sin^2 2x ( \cos 2x+1) = 1/8 (\sin^22x\cos2x+\sin^2 2x)$$
$$\int \sin^2 x \cos^4 x dx = 1/8 (\int \sin^22x\cos2xdx+\int \sin^22x dx)$$
denotation:
$$ A = \int \sin^22x\cos2xdx $$
$$ B = \int \sin^22x dx$$
For A, Let$ u = \sin2x$ then $du=2\cos2x dx$ (i.e. $\cos xdx = du/2$)
$$A=\int u^2/2 du = u^3/6 = \sin^3 2x/6$$
For B, note that $ 1-2\sin^2\theta = \cos 2 \theta \iff \sin^2 \theta = (1-cos 2 \theta)/2$
$$B=1/2 (\int dx -\int cos 4x dx) =1/2(x-\frac{\sin4x}{4})$$
In a nutshell:
$$\int \sin^2 x \cos^4 x dx = 1/8 (A+B) = 1/48 \sin^3 2x + 1/16 x - 1/64 \sin 4x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/930421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Taking limits of powers containing 'x'? If we have $lim_{x\rightarrow 0}\sqrt{x^2+sinx-tanx}$ we can write this as
$\sqrt{lim_{x\rightarrow \:0}\left(x^2+sinx-tanx\right)}$
OR If we have $lim_{x\rightarrow \:0}\left(x^3+secx-cosecx\right)^n$. We can write this as
$\left(lim_{x\rightarrow \:\:0}\left(x^3+secx-cosecx\right)\right)^n$
So if we have a limit as follows:
$lim_{x\rightarrow \:\:\:0}\left(\frac{sinx}{x}\right)^{\frac{1}{x^2}}$
We expand the limit as :
$lim_{x\rightarrow 0}\left(\frac{sinx}{x}\right)^{\frac{1}{x^2}}=lim_{x\rightarrow 0}\left(\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}.....}{x}\right)^{\frac{1}{x^2}}=lim_{x\rightarrow 0}\left(1+\left[-\frac{x^2}{3!}+\frac{x^4}{5!}....\right]\right)^{\frac{1}{x^2}}=lim_{x\rightarrow \:0}\left(\left(1+\left[-\frac{x^2}{3!}+\frac{x^4}{5!}....\right]\right)^{\frac{1}{-\frac{x^2}{3!}+\frac{x^4}{5!}....}}\right)^{\frac{-\frac{x^2}{3!}+\frac{x^4}{5!}....}{x^2}}$$=e^{\left(lim_{x\rightarrow 0}\frac{-\frac{x^2}{3!}+\frac{x^4}{5!}....}{x^2}\right)}=e^{\left(lim_{x\rightarrow \:0}-\frac{1}{3!}+\frac{x^2}{5!}....\right)=}e^{-\frac{1}{6}}$
So why can't we take the limit as we did in the initial case as:
$lim_{x\rightarrow \:\:\:0}\left(\frac{sinx}{x}\right)^{\frac{1}{x^2}}$$=lim_{x\rightarrow 0}\left(\frac{sinx}{x}\right).lim_{x\rightarrow 0}\left(\frac{sinx}{x}\right).lim_{x\rightarrow \:0}\left(\frac{sinx}{x}\right)......\left(\frac{1}{x^2}times\right)=\left[lim_{x\rightarrow 0}\left(\frac{sinx}{x}\right)\right]^{lim_{x\rightarrow 0}\left(\frac{1}{x^2}\right)}=\left(1\right)^{lim_{x\rightarrow \:\:0}\frac{1}{x^2}}=1$
which is wrong. I know I am wrong somewhere, please point it out.
| You can't simply take a product $1/x^2$ times when $1/x^2$ may not be an integer. Regardless, your computation leads to $1^{\infty}$, which is an indeterminate form: it can take on any finite value at least $1$, or $\infty,$ which is why you have to compute the limit in a different way.
In both your introductory examples, the exponent $n$ or $1/2$ does not vary with $x$, and the functions $x^n$ and $\sqrt x$ are continuous, so it's valid to interchange them with limits. For a function like $f(x)^{g(x)}$, where $f$ and $g$ are both continuous, we'll have $\lim_{x\to x_0}(f^g)=(\lim_{x\to x_0}f(x))^{\lim_{x\to x_0} g(x)}$ whenever both limits exists. But in your case $g=1/x^2$ does not have a limit at zero, so the manipulation you propose is invalid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A Definite Integral I Given the definite integral
\begin{align}
\int_{0}^{\pi} \frac{1+\cot^{2}(t)}{\cot^{2}(t)} \, \ln\left( \frac{1+2\tan^{2}(t)}{1+\tan^{2}(t)} \right) \, dt = -2\pi
\end{align}
then what is the general value of the indefinite integral
\begin{align}
\int \frac{1+\cot^{2}(t)}{\cot^{2}(t)} \, \ln\left( \frac{1+2\tan^{2}(t)}{1+\tan^{2}(t)} \right) \, dt \,\,\, ?
\end{align}
| $$\begin{align}
\int \frac{1+\cot^{2}(t)}{\cot^{2}(t)} \, \ln\left( \frac{1+2\tan^{2}(t)}{1+\tan^{2}(t)} \right) \, dt
&= \int\sec^2t\ln\left(1+\sin^2t\right) dt\\
&=\ln(1+\sin^2t)\int\sec^2tdt-\int\frac{2\sin t\cos t}{1+\sin^2t}\int\sec^2tdtdt\\
&=\ln(1+\sin^2t)\tan t-\int\frac{2\sin t\cos t\tan t}{1+\sin^2t}dt\\
&=\ln(1+\sin^2t)\tan t-2\int\left(1-\frac{1}{1+\sin^2t}\right)dt\\
&=\ln(1+\sin^2t)\tan t-2t+2\int\frac1{1+\sin^2t}dt
\end{align}$$
Now
$$\begin{align}
\int\frac1{1+\sin^2t}dt
&=\int\frac{\sec^2tdt}{\sec^2t+\tan^2 t}\\
&\stackrel{u=\tan x}\equiv\int\frac{du}{2+u^2}\\
&=\frac1{\sqrt2}\arctan\frac{\tan x}{\sqrt2}+c
\end{align}$$
So
$$
\int \frac{1+\cot^{2}(t)}{\cot^{2}(t)} \, \ln\left( \frac{1+2\tan^{2}(t)}{1+\tan^{2}(t)} \right) \, dt
=\ln(1+\sin^2t)\tan t-2t+\sqrt2\arctan\left(\frac{\tan x}{\sqrt2}\right)+C$$
Also $$\left(
\ln(1+\sin^2t)\tan t-2t+\sqrt2\arctan\left(\frac{\tan x}{\sqrt2}\right)\right)_0^{\pi}
\\=-2(\pi-0)+\sqrt2\left(\arctan(0)-\arctan(0)\right)+\ln(1+0)\tan(\pi)-\ln(1+0)\tan(0)
\\=-2\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/934593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer. How do we prove that $\dfrac{\phi^{400}+1}{\phi^{200}}$ is an integer, where $\phi$ is the golden ratio?
This appeared in an answer to a question I asked previously, but I do not see how to prove this..
| We can prove by induction that
if $x+\dfrac1x$ is an integer, $x^n+\dfrac1{x^n}$ will be an integer
as $$\left(x^n+\frac1{x^n}\right)\left(x+\frac1x\right)=x^{n+1}+\frac1{x^{n+1}}+x^{n-1}+\frac1{x^{n-1}}$$
$$\iff x^{n+1}+\frac1{x^{n+1}}=\left(x^n+\frac1{x^n}\right)\left(x+\frac1x\right)-\left(x^{n-1}+\frac1{x^{n-1}}\right)$$
The base cases being
$n=1\implies x^2+\dfrac1{x^2}=\left(x+\dfrac1x\right)^2-2$ and
$x^3+\dfrac1{x^3}=\left(x+\dfrac1x\right)^3-3\left(x+\dfrac1x\right)$
or $n=2\implies x^3+\dfrac1{x^3}=\left(x^2+\dfrac1{x^2}\right)\left(x+\dfrac1x\right)-\left(x^1+\dfrac1{x^1}\right)$
As Golden Ratio$(\phi)$ satisfies $x^2-x-1=0$
we have $x^2-1=x\implies x-\dfrac1x=1\implies x^2+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2=1^2+2$
Here $n=100$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/936479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 2
} |
How to find these quantities so as to conform to these conditions? Suppose $a \in \mathbb{R}^k$, $b \in \mathbb{R}^k$. Then how to find $c \in \mathbb{R}^k$ and $r > 0$ such that the following holds?
For any $x \in \mathbb{R}^k$, we have $$|x-a| = 2 |x-b|$$ if and only if $$|x-c| = r. $$
| Squaring both sides of the original equation and rewriting both sides as dot products gives
$$(\mathbf{x} - \mathbf{a}) \cdot (\mathbf{x} - \mathbf{a}) = 4 (\mathbf{x} - \mathbf{b}) \cdot (\mathbf{x} - \mathbf{b}).$$ Expanding and rearranging gives
$$3 \mathbf{x} \cdot \mathbf{x} + 2(\mathbf{a} - 4\mathbf{b}) \cdot \mathbf{x} + (4 \mathbf{b} \cdot \mathbf{b} - \mathbf{a} \cdot \mathbf{a}) = 0.$$
Now, clearing the leading coefficient $3$ and completing the square gives:
$$\left(\mathbf{x} - \tfrac{1}{3}(4\mathbf{b} - \mathbf{a})\right) \cdot \left(\mathbf{x} - \tfrac{1}{3}(4\mathbf{b} - \mathbf{a})\right) + \tfrac{1}{3}(4 \mathbf{b} \cdot \mathbf{b} - \mathbf{a} \cdot \mathbf{a}) - \tfrac{1}{9}(4\mathbf{b} - \mathbf{a}) \cdot (4\mathbf{b} - \mathbf{a}) = 0.$$ Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/937787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How find this integral $\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$ let $$D=\{(x,y)|y\ge x^3,y\le 1,x\ge -1\}$$
Find the integral
$$I=\dfrac{1}{2}\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$$
My idea:
$$I=\int_{0}^{1}dx\int_{x^3}^{1}(x^2y+2y^2)dy+\int_{-1}^{0}dx\int_{0}^{-x^3}(xy^2+2x+2y^2)dy$$
so
$$I=\int_{0}^{1}[\dfrac{1}{2}x^2y^2+\dfrac{2}{3}y^3]|_{x^3}^{1}dx+\int_{-1}^{0}[\dfrac{1}{3}xy^3+2xy+\dfrac{2}{3}y^3]|_{0}^{-x^3}dx$$
$$I=\int_{0}^{1}[\dfrac{1}{2}x^2+\dfrac{2}{3}-\dfrac{1}{2}x^8-\dfrac{2}{3}x^9]dx+\int_{-1}^{0}[-\dfrac{1}{3}x^{10}-2x^4-\dfrac{2}{3}x^9]dx$$
so
$$I=\dfrac{5}{6}-\dfrac{1}{18}-\dfrac{2}{30}+\dfrac{1}{33}+\dfrac{1}{10}-\dfrac{2}{30}=\dfrac{67}{90}$$
My question: my reslut is true? can you someone can use computer find it value?
because I use Tom methods to find this reslut is
$$\dfrac{79}{270}$$
which is true? so someone can use maple help me?Thank you
| It seems that you already realize that the for $D$, $D = \{x^3 \leq y \leq 1\} \cap \{-1 \leq x\}$ is the same as $D = \{x^3 \leq y \leq 1\} \cap \{-1 \leq x \leq 1\}$. So, for a function $f(x,y)$, you should have
$$
\iint_D f(x,y) \, dxdy = \int_{-1}^1 \int_{x^3}^1 f(x,y)\,dxdy.
$$
However, you seem to have split up your function $f(x,y)$ over different bounds without reason. You really should be integrating
$$
\int_{-1}^1 \int_{x^3}^1 (x^2 y + xy^2 + 2x + 2y^2)\,dxdy = \int_{-1}^1 \left[
\frac{1}{2}x^2y^2 + \frac{1}{3} xy^3 + 2xy +\frac{2}{3} y^3 \right]_{y=x^3}^{y=1}\,dx= ...
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/937955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Closed Form of Recursion $a_n = \frac{6}{a_{n-1}-1}$ Given that $a_0=2$ and $a_n = \frac{6}{a_{n-1}-1}$, find a closed form for $a_n$.
I tried listing out the first few values of $a_n: 2, 6, 6/5, 30, 6/29$, but no pattern came out.
| The function $f(x) = \dfrac{6}{x-1}$ is a fractional linear transformation.
Composition of these corresponds to matrix multiplication:
$$\text{if}\ T(x) = \dfrac{a_{11} x + a_{12 }}{a_{21} x + a_{22}} \ \text{and} \ S(x) = \dfrac{b_{11} x + b_{12}}{b_{21} x + b_{22}}$$
$$\text{then}\ T(S(x)) = \dfrac{c_{11} x + c_{12}}{c_{21} x + c_{22}}\ \text{where}\
\pmatrix{c_{11} & c_{12}\cr c_{21} & c_{22}\cr} = \pmatrix{a_{11} & a_{12}\cr a_{21} & a_{22}\cr} \pmatrix{b_{11} & b_{12}\cr b_{21} & b_{22}\cr}$$
In your case the matrix is
$$M = \pmatrix{0 & 6\cr 1 & -1\cr}$$
and we'd like to find a closed form for its powers. We diagonalize the matrix as
$ M = E D E^{-1}$ where
$$ E = \pmatrix{-2 & 3\cr 1 & 1\cr}, \ D = \pmatrix{-3 & 0\cr 0 & 2\cr},\
E^{-1} = \dfrac{1}{5}\pmatrix{-1 & 3\cr 1 & 2\cr}$$
so $$M^n = E D^n E^{-1} =\dfrac{1}{5} \left( \begin {array}{cc} 2\, \left( -3 \right) ^{n}+3\cdot{2}^{n}&-6\,
\left( -3 \right) ^{n}+6\cdot{2}^{n}\\ - \left( -3
\right) ^{n}+{2}^{n}&3\, \left( -3 \right) ^{n}+2\cdot{2}^{n}
\end {array} \right)
$$
i.e. $$a_n = \left(\dfrac{ (2 \left( -3 \right) ^{n}+3\cdot{2}^{n}) \cdot 2 -6
\left( -3 \right) ^{n}+6\cdot{2}^{n}}{ \left(- \left( -3
\right) ^{n}+{2}^{n}\right) \cdot 2 + 3 \left( -3 \right) ^{n}+2\cdot{2}^{n}
} \right) = \dfrac{-2(-3)^n + 12 \cdot 2^n}{(-3)^n + 4 \cdot 2^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/939725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Integral of $\sqrt{x^3 + 8}$? I have issues solving the following integral:
$$\int\sqrt{x^3+8}~dx$$
I tried substitution and integration by parts, but with no use.
I'm guessing I have to use some trigonometric substitution.
Can anybody help solve this integral?
| For any real number $x$:
When $|x|\leq2$ ,
$$\begin{array}\int\sqrt{x^3+8}\,dx
&=\int2\sqrt2\sqrt{\dfrac{x^3}{8}+1}\,dx\\
&=\int\sum\limits_{n=0}^\infty\dfrac{2\sqrt2(-1)^n(2n)!x^{3n}}{8^n4^n(n!)^2(1-2n)}\,dx\\
&=\int\sum\limits_{n=0}^\infty\dfrac{2\sqrt2(-1)^n(2n)!x^{3n}}{32^n(n!)^2(1-2n)}\,dx\\
&=\sum\limits_{n=0}^\infty\dfrac{2\sqrt2(-1)^n(2n)!x^{3n+1}}{32^n(n!)^2(1-2n)(3n+1)}+C\end{array}$$
When $|x|\geq2$ ,
$$\begin{array}\int\sqrt{x^3+8}\,dx
&=\int x^\frac{3}{2}\sqrt{1+\dfrac{8}{x^3}}\,dx\\
&=\int x^\frac{3}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!8^n}{4^n(n!)^2(1-2n)x^{3n}}\,dx\\
&=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^nx^{\frac{3}{2}-3n}}{(n!)^2(1-2n)}dx\\
&=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^nx^{\frac{5}{2}-3n}}{(n!)^2(1-2n)\left(\dfrac{5}{2}-3n\right)}+C\\
&=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^{n+1}}{(n!)^2(2n-1)(6n-5)x^{3n-\frac{5}{2}}}+C\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Calculus 2 Integral of$ \frac{1}{\sqrt{x+1} +\sqrt x}$ How would you find $$\int\frac{1}{\sqrt{x+1} + \sqrt x} dx$$
I used $u$-substitution and got this far:
$u = \sqrt{x+1}$ which means $(u^2)-1 = x$
$du = 1/(2\sqrt{x-1}) dx = 1/2u dx$ which means $dx = 2udu$
That means the new integral is $$\int \frac{2u}{u + \sqrt{u^2-1}}du$$
What technique do I use to solve that new integral?
Thanks
| Here is a route.
$$
\begin{align}
\int\frac{1}{\sqrt{x+1} + \sqrt{x}} {\rm d}x &=\int\frac{1}{(\sqrt{x+1} + \sqrt{x})} \times \frac{(\sqrt{x+1} - \sqrt{x})}{(\sqrt{x+1} - \sqrt{x})} {\rm d}x\\\\
&=\int \left(\sqrt{x+1} - \sqrt{x}\right) {\rm d}x \\\\
& = \int(x+1)^{1/2} {\rm d}x-\int x^{1/2} {\rm d}x\\\\
& =\frac{1}{1+1/2}(x+1)^{1+1/2} -\frac{1}{1+1/2}x^{1+1/2} +C\\\\
& =\frac 23 (x+1)^{3/2}-\frac 23 x^{3/2}+C\\\\
\end{align}
$$
where $C$ is any constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Not understanding how to factor a polynomial completely $$P(x)=16x^4-81$$
I know that this factors out as:
$$P(x)=16(x-\frac { 3 }{ 2 } )^4$$
What I don't understand is the four different zeros of the polynomial...I see one zero which is $\frac { 3 }{ 2 }$ but not the three others.
| Try making the substitutions $a=2x$ and $b=3$. Then we have that
$$
16x^4-81=a^4-b^4.
$$
Recalling that $x^2-y^2=(x+y)(x-y)$, we have
$$
a^4-b^4=[a^2-b^2](a^2+b^2)=[(a-b)(a+b)](a^2+b^2)
$$
Plugging $2x$ in for $a$ and $3$ in for $b$ we have
$$
(a-b)(a+b)(a^2+b^2)=(2x-3)(2x+3)((2x)^2+3^2)=(2x-3)(2x+3)(4x^2+9).
$$
If you set each of these three factors equal to zero you will find the four roots you are looking for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
A triangle determinant that is always zero How do we prove, without actually expanding, that
$$\begin{vmatrix}
\sin {2A}& \sin {C}& \sin {B}\\
\sin{C}& \sin{2B}& \sin {A}\\
\sin{B}& \sin{A}& \sin{2C}
\end{vmatrix}=0$$
where $A,B,C$ are angles of a triangle?
I tried adding and subtracting from the rows and columns and I even tried using the sine rule, but to no avail.
| Using the Law of Sines, we can write
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
where $a$, $b$, $c$ are the sides, and $d$ is the circumdiameter, of the triangle. And the Law of Cosines gives us
$$\cos A = \frac{1}{2bc}(-a^2+b^2+c^2) \qquad\text{, etc.}$$
With $\sin 2 x = 2 \sin x \cos x$, we can express the determinant as
$$\left|\begin{array}{ccc}
\frac{a}{d}\frac{-a^2+b^2+c^2}{bc} & \frac{c}{d} & \frac{b}{d} \\[4pt]
\frac{c}{d} & \frac{b}{d}\frac{a^2-b^2+c^2}{ca} & \frac{a}{d} \\[4pt]
\frac{b}{d} & \frac{a}{d} & \frac{c}{d}\frac{a^2+b^2-c^2}{ab}
\end{array}\right|$$
From here, we can "factor-out" $\frac{1}{dbc}$, $\frac{1}{dca}$, $\frac{1}{dab}$ from the first, second, and third rows:
$$\frac{1}{dbc}\frac{1}{dca}\frac{1}{dab}\;\left|\begin{array}{ccc}
a(-a^2+b^2+c^2) & b c^2 & c b^2 \\[4pt]
a c^2 & b (a^2-b^2+c^2) & c a^2 \\[4pt]
ab^2 & b a^2 & c (a^2+b^2-c^2)
\end{array}\right|$$
Then, we factor-out $a$, $b$, $c$ from first, second, and third columns:
$$\frac{a b c}{d^3a^2b^2c^2}\;\left|\begin{array}{ccc}
-a^2+b^2+c^2 & c^2 & b^2 \\[4pt]
c^2 & a^2-b^2+c^2 & a^2 \\[4pt]
b^2 & a^2 & a^2+b^2-c^2
\end{array}\right|$$
Subtracting, say, the first row from the second and third gives
$$\frac{1}{d^3abc}\;\left|\begin{array}{ccc}
-a^2+b^2+c^2 & c^2 & b^2 \\[4pt]
a^2-b^2 & a^2-b^2 & a^2-b^2 \\[4pt]
a^2-c^2 & a^2-c^2 & a^2-c^2
\end{array}\right|$$
which clearly vanishes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/946524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 3
} |
Computing a limit involving Gammaharmonic series It's a well-known fact that
$$\lim_{n\to\infty} (H_n-\log(n))=\gamma.$$
If I use that $\displaystyle \Gamma \left( \displaystyle \frac{1}{ n}\right) \approx n$ when $n$ is large, then I wonder if it's possible to compute the following limit in a closed-form
$$\lim_{n\to \infty}\left(\frac{1}{ \Gamma\left(\displaystyle \frac{1}{1}\right)}+ \frac{1}{ \Gamma\left( \displaystyle \frac{1}{2}\right)}+ \cdots + \frac{ 1}{ \Gamma \left( \displaystyle \frac{1}{ n}\right) }- \log\left( \Gamma\left(\displaystyle\frac{1}{n}\right)\right)\right),$$
where I called $\displaystyle \sum_{k=1}^{\infty}\frac{ 1}{ \Gamma \left( \displaystyle \frac{1}{ k}\right) }$ as Gammaharmonic series.
I can get approximations, but I cannot get the precise limit, and I don't even know if it can be expressed in terms of known constants.
A 500 points bounty moment: I would enjoy pretty much finding a solution (containing a closed-form) for the posed limit, hence the generous bounty. It's unanswered for 3 years and 8 months, and it definitely deserves another chance. Good luck!
| From Wolfram Gamma Function equations (35)-(37) provide
\begin{align}\tag{1}
\frac{1}{\Gamma(x)} = x + \gamma x^{2} + \sum_{k=3}^{\infty} a_{k} x^{k}
\end{align}
where, $a_{1}=1$, $a_{2}=\gamma$,
\begin{align}\tag{2}
a_{n} = n a_{1} a_{n-1} - a_{2} a_{n-2} + \sum_{k=2}^{n} (-1)^{k} \zeta(k) \, a_{n-k}.
\end{align}
Now,
\begin{align}\tag{3}
\sum_{r=1}^{n} \frac{1}{\Gamma\left(\frac{1}{r}\right)} \approx H_{n} + \gamma H_{n,2} + \sum_{k=3}^{\infty} a_{k} H_{n,k},
\end{align}
where $H_{n,r}$ are the generalized Harmonic numbers given by
\begin{align}\tag{4}
H_{n,r} = \sum_{s=1}^{n} \frac{1}{s^{r}}.
\end{align}
Since the limit is for large values of $n$, $n \rightarrow \infty$, then utilize the approximation, Wolfram Harmonic Number Approximations,
\begin{align}\tag{5}
H_{n,r} \approx \frac{(-1)^{r} \psi^{(r-1)}(1)}{(r-1)!} - \frac{1}{(r-1) \, n^{r-1} } \left( 1 + \mathcal{O}\left(\frac{1}{n}\right) \right)
\end{align}
to obtain
\begin{align}\tag{6}
\sum_{r=1}^{n} \frac{1}{\Gamma\left(\frac{1}{r}\right)} \approx H_{n} - \frac{\gamma}{n} + \sum_{k=2}^{\infty} \frac{(-1)^{k} a_{k}}{(k-1)!} \, \psi^{(k-1)}(1) + \mathcal{O}\left(\frac{1}{n^{2}} \right).
\end{align}
Since,
\begin{align}\tag{7}
- \ln \Gamma\left( \frac{1}{n} \right) \approx \frac{\gamma}{n} - \ln(n) + \mathcal{O}\left(\frac{1}{n^{2}}\right)
\end{align}
then
\begin{align}\tag{8}
\sum_{r=1}^{n} \frac{1}{\Gamma\left(\frac{1}{r}\right)} - \ln \Gamma\left( \frac{1}{n} \right) \approx H_{n} - \ln(n) + \sum_{k=2}^{\infty} \frac{(-1)^{k} a_{k}}{(k-1)!} \, \psi^{(k-1)}(1) + \mathcal{O}\left(\frac{1}{n^{2}} \right).
\end{align}
Taking the limit as $n \rightarrow \infty$ and using
\begin{align}
\lim_{n \rightarrow \infty} \left( H_{n} - \ln(n) \right) = \gamma
\end{align}
then
\begin{align}\tag{9}
\lim_{n \rightarrow \infty} \left[ \sum_{r=1}^{n} \frac{1}{\Gamma\left(\frac{1}{r}\right)} - \ln \Gamma\left( \frac{1}{n} \right) \right] = \sum_{k=1}^{\infty} \frac{(-1)^{k} a_{k}}{(k-1)!} \, \psi^{(k-1)}(1).
\end{align}
Since
\begin{align}\tag{10}
\psi^{(m)}(x) = (-1)^{m+1} m! \zeta(m+1, x)
\end{align}
then
\begin{align}\tag{11}
\lim_{n \rightarrow \infty} \left[ \sum_{r=1}^{n} \frac{1}{\Gamma\left(\frac{1}{r}\right)} - \ln \Gamma\left( \frac{1}{n} \right) \right] = \gamma + \sum_{k=2}^{\infty} a_{k} \zeta(k).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/948353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 3
} |
$\int\frac{2x+1}{x^2+2x+5}dx$ by partial fractions $$\int\frac{2x+1}{x^2+2x+5}dx$$
I know I'm supposed to make the bottom a perfect square by making it $(x+1)^2 +4$ but I don't know what to do after that. I've tried to make $x+1= \tan x$ because that's what we did in a class example but I keep getting stuck.
| Using a trigonometric substitution, let $x+1 = 2\tan u$, so that $dx = 2\sec^2 u\,du$ and $2x+1 = 4\tan u-1$. Then we get
\begin{align*}
\int\frac{2x+1}{(x+1)^2+4}\,dx &= \int\frac{4\tan u-1}{4\tan^2 u+4}\cdot 2\sec^2u\,du \\
&= \frac{1}{4}\int\frac{4\tan u-1}{\sec^2u}\cdot 2\sec^2u\,du \\
&= \frac{1}{2}\int (4\tan u-1)\,du \\
&= \frac{1}{2}\left(4\ln\sec u - u\right) + C \\
&= 2\ln\frac{\sqrt{x^2+2x+5}}{2} - \frac{1}{2}\tan^{-1}\frac{x+1}{2} + C \\
&= \ln(x^2+2x+5) - \frac{1}{2}\tan^{-1}\frac{x+1}{2} + C.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/950354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Straight line , gre prolem $170$ score hard What is the distance between the two straight line represented by the equation
$3x+4y=10$ and $6x+8y=10?$ $$A>1$$ $$B>2$$ $$C>\frac43$$ $$D>\frac12$$ $$E>\frac52$$
I try to solve it , firstly find intercept of both line and then find mid point between of them ,and calculate distance between both line , suggest me where I m wrong
| Let me call the lines $l1$ and $l2$, in which:
$l1: y=\frac{-3}{4}x+2.5$
and
$l2: y=\frac{-3}{4}x+1.25$.
With a little notice to the above lines we see that these are the lines mentioned in the question. In another language, we have made the above line equations by altering the shape of the line equations. As we see, both lines have the same slopes, and hence they are parallel to each other and there is no intersection. So we can define distance between them. Let me define $l3$ as:
$l3: y=\frac{4}{3}x$
As we see $l3$ is perpendicular to $l1$ and $l2$ (because the product of their slopes is equal to $-1$). We wanna find the intersection of $l3$ with the both lines. So we have:
$y=\frac{4}{3}x\rightarrow\frac{4}{3}x=\frac{-3}{4}x+2.5\rightarrow x=\frac{30}{25}\rightarrow y=\frac{40}{25}\rightarrow P1=(\frac{30}{25},\frac{40}{25})$
$y=\frac{4}{3}x\rightarrow\frac{4}{3}x=\frac{-3}{4}x+1.25\rightarrow x=\frac{15}{25}\rightarrow y=\frac{20}{25}\rightarrow P2=(\frac{15}{25},\frac{20}{25})$
In which $P1$ and $P2$ are the intersection of $l3$ with $l1$ and $l2$ respectively. The distance between $P1$ and $P2$ is the distance between $l1$ and $l2$, so we have:
$d_{l1,l2}=\sqrt{(\frac{30}{25}-\frac{15}{25})^2+(\frac{40}{25}-\frac{20}{25})^2}=\sqrt{(\frac{15}{25})^2+(\frac{20}{25})^2}=\sqrt{\frac{625}{625}}=1 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/954535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that $ \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x $ In a related question the following integral was evaluated
$$
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
=\int_0^{\pi} \frac{\mathrm{d}x/2}{1 + \cos x \sin x}
=\int_0^{2\pi} \frac{\mathrm{d}x/2}{2 + \sin x} \,\mathrm{d}x
=\int_{-\infty}^\infty \frac{\mathrm{d}x/2}{1+x+x^2}
$$
I noticed something interesting, namely that
$$
\begin{align*}
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
& = \int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \\
& = \int_0^{\pi} \frac{(\cos x)^2}{1 - \cos x \sin x} \,\mathrm{d}x
= \int_0^{\pi} \frac{(\sin x)^2}{1 - \cos x \sin x} \,\mathrm{d}x
\end{align*}
$$
The same trivially holds if the upper limits are changed to $\pi/2$ as well ($x \mapsto \pi/2 -u$).
But I had problems proving the first equality. Does anyone have some quick hints?
| A quick hint is noticing the symmetry. A rigorous proof is that $$\int\limits_a^b {\frac{{f(\cos x)}}{{g(\sin x\cos x)}}dx} = \int\limits_a^b {\frac{{f\left( {\sin \left( {\frac{\pi }{2} - x} \right)} \right)}}{{g\left( {\cos \left( {\frac{\pi }{2} - x} \right)\sin \left( {\frac{\pi }{2} - x} \right)} \right)}}dx} = \int\limits_{\frac{\pi }{2} - b}^{\frac{\pi }{2} - a} {\frac{{f\left( {\sin u} \right)}}{{g\left( {\cos u\sin u} \right)}}du} $$Hope it helps ;)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/955294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
} |
Find all $2\times 2$ matrices that commutes with the matrix $\pmatrix{a & b\\ c & d}$, with $bc\neq 0$ I want to find all $2\times 2$ matrices that commutes with the matrix $\pmatrix{a & b\\ c & d}$, with $bc\neq 0$.
For this, suppose $\pmatrix{e & f\\ g& h}$ such that
$\pmatrix{a & b\\ c & d}\pmatrix{e & f\\ g& h}=\pmatrix{e & f\\ g& h}\pmatrix{a & b\\ c & d}$. Equating both sides, we can get the values of $e,f,g,h$. But this way is complicated. Is there any sophisticated way than the above.
| First we have to find out which matrices commute with a Jordan block $\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Since
$$ \begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}x&y\\z&w\end{pmatrix} = \begin{pmatrix}z&w\\0&0\end{pmatrix}, \begin{pmatrix}x&y\\z&w\end{pmatrix} \begin{pmatrix}0&1\\0&0\end{pmatrix} = \begin{pmatrix}0&x\\0&z\end{pmatrix},$$
we find out that $z = 0$ and $x = w$, so the solution space is two-dimensional.
Second, we do the same for a diagonal matrix with two different eigenvalues: $\begin{pmatrix}0&0\\0&1\end{pmatrix}$:
$$ \begin{pmatrix}0&0\\0&1\end{pmatrix} \begin{pmatrix}x&y\\z&w\end{pmatrix} = \begin{pmatrix}0&0\\z&w\end{pmatrix}, \begin{pmatrix}x&y\\z&w\end{pmatrix} \begin{pmatrix}0&0\\0&1\end{pmatrix} = \begin{pmatrix}0&y\\0&w\end{pmatrix},$$
so the constraints are $y=z=0$, and again the solution space is two-dimensional.
Consider now the Jordan form of the given matrix. It is either a Jordan block of dimension 2, a diagonal matrix with different eigenvalues, or a scalar. In the first two cases, the commutator space has dimension 2, and in the last, it has dimension 4. This explains loup blanc's remark, which can be generalized for arbitrary square matrices.
As loup blanc remarks, since scalars commute with all matrices, the only matrices whose Jordan form is scalar are the scalars themselves, and our matrix can't be scalar since $bc \neq 0$. So the commutator must have dimension 2.
See the comments for why this argument unfortunately doesn't work. Though it almost does.
Let $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, where $bc \neq 0$. In particular, $M$ is not a scalar, so its minimal polynomial has degree two. So $I,M,M^2$ are linearly independent, and span a linear space of dimension three of matrices commuting with $M$. Since the center of the matrix algebra consists only of the scalars, you obtain this way all matrices commuting with $M$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/955466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
The relation of angle between two slant faces of a pyramid and the angles between slant vectors Have any of you seen this theorem of relationship of the angles between two slant faces of a pyramid and the angles between slant vectors, provided that two faces of corresponding to $\phi$ and $\eta$ are perpendicular? I attach the picture
The result of theorem is $\cos(\theta)=\cos(\phi)\cdot cos(\eta)$.
|
Applying Pythagoras theorem to the base, we have $L^2 = a^2 \sin^2 φ + c^2 \sin^2 η$
Applying cosine law to the slant face, we have $L^2 = a^2 + c^2 – 2ac \cos θ$
∴ $a^2 \sin^2 φ + c^2 \sin ^2 η = a^2 + c^2 – 2ac \cos θ$
$2ac \cos θ = a^2 – a^2 \sin^2 φ + c^2 – c^2 \sin^2 η$
$2ac \cos θ = a^2 \cos^2 φ + c^2 \cos^2 η$
$2 ac \cos θ = (a \cos φ) ^2 + (c \cos η) ^2$
$2 ac \cos θ = 2 b^2$
$\cosθ = \frac {b}{a} \frac {b}{c}$
∴ $\cosθ = (\cosφ)( \cos η)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/956983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inequality $\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}<\frac{3}{2}$ I'm trying to prove this by induction: $$\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}<\frac{3}{2}, \quad n\geq1. $$
I got stuck in the inductive step:
$$\frac{3}{2} +\frac{1}{(n+1)^3}\,\ldots$$
Thank you for your help!
| You can use for $n\geq 2$, $$\dfrac{1}{n^3} \leq \dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/959773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Solving $ \frac{1}{ a} = \ \frac{1}{ \ \sqrt{b}} \ +\ \frac{1}{ \ \sqrt{c}} $ with additional conditions How to solve this equation
$$ \frac{1}{a}= \ \frac{1}{ \ \sqrt{b}} \ +\ \frac{1}{ \ \sqrt{c}} $$
where $$ b = \sqrt{ (x-a/2)^2 + y^2 + z^2 )}$$
& $$ c = \sqrt{ (x+a/2)^2 + y^2 + z^2 )}$$
We have to get an equation in x y z and a!
I have tried to rationalize and do squaring but it becomes cumbersome.
Answer given is $y^2+z^2=15/4$
I get this by putting x=0.
| Partial answer: You need a simplification, not a solution.You already have an implicit equation in x, y, z and a ! The surface is symmetric about x = 0 or y-z plane ( change of sign before x makes no change). One needs to study (x,y) intersection only, as it is formed by rotation about x-axis ( due to appearance of $ y^2+z^2$ ).
EDIT: From the graphic, plane x=0 does not cut the surface along any real line of intersection.
It appears like, may be two (higher order?) spheres, centered at $ (\pm a/2,0,0)$ as seen in following Mathematica 3D plot. $a$ is taken as unity.
EDIT1:
Patience with algebraic simplification ( which is the solution of problem of seeing what you are dealing with!) is all that is needed.
$\sqrt{bc}/ ( \sqrt b + \sqrt c) = a $
$ (b c /a^2)^2 + (b+c)^2 - 2 b c (b+c)/ a^2 = b^2 c^2 $
Let $ ( b + c) / ( b c) = u $
$ u^2 + -2 u /a^2 + (1/a^4-1) $
$ u = 1/b + 1/c = 1/a^2 \pm 1 = 1/d $
So due to d there are two surfaces
$ 2 (x^2 + a^2/4 + y^2 + z^2) +2 b c = 1/d^2( (x^2 + a^2/4 + y^2 + z^2)^2 -( 2 a x)^2 $
which shows the two ordinary displaced fourth order spheres.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/960246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Transforming a cartesian equation to a polar one when it has different x and y denominators? $$\frac{x^2}{9}+\frac{y^2}{16}=1$$
Needs to be replaced with an equivalent polar equation. I'm sure the identity I'll have to use will be $$x^2+y^2=r^2$$
though other options are:
$x=rcos\theta$ and $y=r\sin\theta$.
Originally I tried to simply eliminate the denominator. Then I tried to take the square root of both sides to get $\frac{x}{3}+\frac{y}{4}=1$,
to use an angle-sum identity with $r\cos\theta*\frac{1}{3}+r\sin\theta*\frac{1}{4}=1$ to reduce it to $r\cos(\theta+something)$ or $r\sin(\theta+something)$. Neither work.
How should I approach this problem to get it into the form $x^2+y^2=r^2$?
| $$\sqrt{\frac{x^2}{9}+\frac{y^2}{16}=1}$$ does not mean $$\frac{x}{3}+\frac{y}{4}=1$$
as $$\sqrt{\frac{x^2}{9}+\frac{y^2}{16}}\ne\frac x3+\frac y4$$
Set $x=r\cos\theta$ and $y=r\sin\theta$ in $16x^2+9y^2=144$
to get $$r^2(16\cos^2\theta+9\sin^2\theta)=144$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/960453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How find this sum $\frac{1}{1^2}+\frac{2}{2^2}+\frac{2}{3^2}+\frac{3}{4^2}+\frac{2}{5^2}+\frac{4}{6^2}+\cdots+\frac{d(n)}{n^2}+\cdots$ Question:
Find the value
$$\dfrac{1}{1^2}+\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{3}{4^2}+\dfrac{2}{5^2}+\dfrac{4}{6^2}+\cdots+\dfrac{d(n)}{n^2}+\cdots$$
where $d(n)$ is The total number of positive divisors of $n$
I think we can use
$$\zeta{(2)}=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2}+\cdots=\dfrac{\pi^2}{6}$$
I know If the prime factorization of is given by
$$n=p^{a_{1}}_{1}\cdot p^{a_{2}}_{2}\cdots p^{a_{n}}_{n}$$
then the number of positive divisors of is
$$d(n)=(a_{1}+1)(a_{2}+1)\cdots(a_{n}+1)$$
But follow is very ugly,I don't understand @Nate idea(my English is poor),can you post detail? Thank you
| Since $d(n) = \sum_{d|n}1$ you can write
$$\sum_{n=1}^{+\infty} \frac{d(n)}{n^2} = \sum_{n=1}^{+\infty} \sum_{d|n} \frac{1}{n^2}$$
Now, the set $\{ (d,n) \in \mathbb{N}^2 : d|n \}$ can be identified with the set $\{ (k, d)\in \mathbb{N}^2 \}$ by the map $(k,d) \mapsto (d, kd)$ (so $n = kd$). This means that
$$\sum_{n=1}^{+\infty} \sum_{d|n} \frac{1}{n^2}=
\sum_{d=1}^{+\infty} \sum_{k=1}^{+\infty} \frac{1}{(kd)^2} =
\sum_{d=1}^{+\infty} \frac{1}{d^2} \sum_{k=1}^{+\infty} \frac{1}{k^2} = \left(\frac{\pi^2}{6} \right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/961034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Sketch the set of points determined by $\lvert z-i \rvert^2 + \lvert z+i \rvert^2 \le 4$ Sketch the set of points determined by the following conditions $\lvert z-i \rvert^2 + \lvert z+i \rvert^2 \le 4$ So I started by $(z-i)(\bar{z} -\bar{i}) + (z+i)(\bar{z} + \bar{i}) \le 4$
$z\bar{z} - z\bar{i} - i\bar{z} + i\bar{i} + z\bar{z} + z\bar{i} + i\bar{z} + i\bar{i}\le 4$
$=2z\bar{z} + i\bar{i}\le 4$
$=(x+iy)(x-iy) \le 2$
$=(x^2 + y^2)\le2$
$=x^2 + y^2 \le 2$ which gives me a circle with center at the origin with $r= \sqrt2$
But then I thought to do it a different way and then got confused on which way was correct. The second way is as follows.
$\lvert z-i\rvert^2 + \lvert z+i\rvert^2 \le4$
$=\lvert a+(b-1)i\rvert^2 + \lvert a+(b+1)\rvert^2 \le 4$
$=a^2 + (b-1)^2 + a^2 + (b+1)^2 \le 4$
$=a^2 + b^2 -2b+1 + a^2 + b^2 + 2b+1 \le 4$
$=2a^2 +2b^2 \le 4$
$=a^2 + b^2 \le 1$ which gives me a circle with $r=1$
I guess my question is, which one is the right process?
| While the algebraic route is simple, one actually can deduce the result from geometry alone. You have two points, separated by a distance of $c=2$, and you require that their distances $a,b$ to a point $z$ satisfy $a^2+b^2\leq c^2$. For the boundary points, for which this is an equality, we recognize the Pythagorean theorem and conclude that every point is the base point of some right triangle with hypotenuse $c=2$. But from Euclidean geometry we know that the base points of right triangles with a common hypotenuse all lie on a common circle with the hypotenuse as diameter. From such considerations we conclude that the boundary points are the unit circle and the region of interest is the unit disk.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/961111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Tricky sequences and series problem For a positive integer $n$, let $a_{n}=\sum\limits_{i=1}^{n}\frac{1}{2^{i}-1}$. Then are the following true:
$a_{100} > 200$ and
$a_{200} > 100$?
Any help would be thoroughly appreciated. This is a very difficult problem for me. :(
| Have you thought about regrouping some of the terms ?
$\frac{1}{2} \geq \frac{1}{2}$
$\frac{1}{3} + \frac{1}{4} > \frac{1}{2}$ since $\frac{1}{3} > \frac{1}{4}$
$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \frac{1}{2}$ since $\frac{1}{5} > \frac{1}{8}, \frac{1}{6} > \frac{1}{8}...$
...
$\frac{1}{2^{199}+1} + ... +\frac{1}{2^{200}-1} \geq \frac{1}{2} - \frac{1}{2^{200}}$
Finally, we don't forget the $1$ :
$1 = 2*\frac{1}{2}$.
If we sum everything, we get : $1+ \frac{1}{2}+ \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{2^{200}-1} > 201*\frac{1}{2} - \frac{1}{2^{200}} > 100$.
So the answer to your second question is YES.
The other answer is NO : you can prove it by regrouping the terms differently, with inequalities the other way around :
$\frac{1}{2} + \frac{1}{3} < 1$ since $\frac{1}{3} < \frac{1}{2}$
$\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} < 1$ since $\frac{1}{5} < \frac{1}{4}, \frac{1}{6} < \frac{1}{4}...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/963372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$. and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $ If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $.
Progress
We have $\gcd(a,b)=1\implies \exists u,v\in \mathbb Z$ such that $au+bv=1\implies a^2u'+bv'=1, u',v'\in \mathbb Z$.
Let $\gcd(a+b, a-b)=d$. Then $\mid (a+b)x+(a-b)y, \forall x,y\in \mathbb Z$.
How to show $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $.
| If $\gcd(a,b)=1$, then there are $x,y\in\mathbb{Z}$ so that $ax+by=1$.
Then, since
$$
\begin{align}
2
&=\overbrace{[(a+b)+(a-b)]}^{\large2a}\,x+\overbrace{[(a+b)-(a-b)]}^{\large2b}\,y\\[4pt]
&=(a+b)(x+y)+(a-b)(x-y)
\end{align}
$$
we have that
$$
\gcd(a+b,a-b)\mid2
$$
Furthermore
$$
\begin{align}
1
&=(ax+by)^3\\
&=a^2(ax^3+3bx^2y)+b^2(3axy^2+by^3)
\end{align}
$$
implies that $\gcd(a^2,b^2)=1$. Apply the result above to get
$$
\gcd(a^2+b^2,a^2-b^2)\mid2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/964203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Show that $2\cos(x)$ is equal to $2\cos(2x)\sec(x)+\sec(x)\tan(x)\sin(2x)$ This is from the derivative of $\dfrac{\sin(2x)}{\cos x}$
I tried to solve it and arrived with factoring the $\sec(x)$ but I still can't get it to $2\cos(x)$. Could you help me out, please? Thanks
| Another way
$$\begin{align}
2\cos x&=\frac{2\cos^2 x}{\cos x}\\
&=\frac{2(\cos^2 x-\sin^2 x+\sin^2 x)}{\cos x}\\
&=\frac{2(\cos2 x+\sin^2 x)}{\cos x}\\
&=2\left(\frac{\cos2 x}{\cos x}+\frac{\sin x\cdot\sin x}{\cos x}\right)\\
&=2(\cos2 x\sec x+\sin x\tan x)\\
&=2\cos2 x\sec x+2\sin x\cdot\frac{\cos x}{\cos x}\tan x\\
&=2\cos2 x\sec x+\sin 2x\sec x\tan x\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/972819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Limit $\lim\limits_{n \to \infty} n \cdot\ln(\sqrt{n^2+2n+5}-n)$ How should this limit be solved ?
$$\lim_{n \to \infty} n \cdot \ln(\sqrt{n^2+2n+5}-n)$$
I've tried to multiply and at the same time divide $\sqrt{n^2+2n+5}-n$ by $\sqrt{n^2+2n+5}+n$, and then make $n$ as the power of $\frac {2n+5}{\sqrt{n^2+2n+5}+n}$. But I got stuck. I dont think it was the best idea.
| Notice that
$$ \lim_{n \to \infty} n \cdot \ln (\sqrt{n^2+2n+5}-n) = \lim_{n \to \infty} n \cdot \ln (\sqrt{(n+1)^2+4}-n) $$
so we change variable: $a= n+1 \to \infty $ to get
$$ \lim_{a \to \infty} (a-1) \cdot \ln (\sqrt{a^2+4}-a+1) $$
But we know that for $a \to \infty $, for a product $ (a-1)$ behaves as $a$ and that $ \sqrt{a^2+4} $ behaves as $ a + \frac{2}{a} $ and so our limit becomes
$$ \lim_{a \to \infty} a \cdot \ln (a+\frac{2}{a}-a+1) $$ which simplifies to
$$ \lim_{a \to \infty} a \cdot \ln (1+\frac{2}{a}) $$
and since, for small $x$, one has $ \ln(1+x) = x $, the limit becomes
$$ \lim_{a \to \infty} a \cdot \frac{2}{a} = 2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/973637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Solving a system of two cubic equations I'm trying to solve a system of two cubic equations with two variables x and y.
The original problem was to solve the equation $z^3=-4i \overline{z}$. I know how to solve it using polar form.
Now I want to solve it using Cartesian form, say $z=x+yi$.
Doing the algebra and simplifying I got the next system of equations: $$\displaystyle\left\{\begin{matrix}x^3-3xy^2+4y=0\\-y^3+3x^2y+4x=0\end{matrix}\right.$$
It is trivial that $\displaystyle (0,0)$ is a solution, but I couldn't find the other four.
The best I got is $(3x^2-y^2)(x^2-3y^2)=16$, but I don't how to continue.
Please help, thanks!
| $$z^3=-4i\bar{z}\Rightarrow (-i)z^4=4|z|^2\Rightarrow |z|^4=4|z|^2$$
A trivial solution would be $|z|=0\Leftrightarrow z=0$. Assume $|z|\neq 0$ then
$$|z|^2=4 \Rightarrow x^2+y^2=4$$
Expressing the main equation in cartesian coordinates:
$$(x+iy)^3=-4i(x-iy)\Rightarrow x^3+3ix^2y-3xy^2-iy^3=-4ix-4y$$
So one gets after equalizing the real and imaginary parts
$$x^3-3xy^2+4y=0$$
and
$$3x^2y-y^3+4x=0$$
Adding the last two equations together yields
$$x^3-y^3+3xy(x-y)+4(x+y)=0\Rightarrow (x-y)(x^2+xy+y^2)+3xy(x-y)+4(x+y)=0\Rightarrow(x-y)(x^2+4xy+y^2)+4(x+y)=0$$
Using $x^2+y^2=4$ we get
$$(x-y)(1+xy)+(x+y)=0\Rightarrow x+x^2y-y-xy^2+x+y=0\Rightarrow x(2+xy-y^2)=0$$
So $x=0\Rightarrow y=0$ or $2+xy-y^2=0\Rightarrow 2+x\sqrt{4-x^2}-4+x^2=0$
The last result is equivalent
$$(2-x^2)^2=x^2(4-x^2)\Leftrightarrow 4-4x^2+x^4=4x^2-x^4\Leftrightarrow x^4-4x^2+2=0$$
This equation has four roots
$$x=\pm\sqrt{2-\sqrt{2}}$$
and
$$x=\pm\sqrt{2+\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/974838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Logarithmic inequality for a>1 Is $\log_{\sqrt a}(a+1)+\log_{a+1}\sqrt a\ge \sqrt6$ always true for $a>1$?
What is the approach? Do we check the first a's and then form a induction hypothesis?
| rewrite your inequality in the form
$\frac{\ln(a+1)}{\ln(\sqrt{a})}+\frac{\ln(\sqrt{a})}{\ln(a+1)}=$
$\frac{2\ln(a+1)}{\ln(a)}+\frac{\ln(a)}{2\ln(a+1)}\geq 2$
by the theorem that stated
$\frac{x}{y}+\frac{y}{x}\geq 2$ iff $(x-y)^2\geq 0$
for $x,y>0$
if we define $f(a)=\frac{2\ln(a+1)}{\ln(a)}+\frac{\ln(a)}{2\ln(a+1)}$ we get for the first derivative
$f'(a)=-1/2\,{\frac { \left( \ln \left( a \right) -2\,\ln \left( a+1
\right) \right) \left( \ln \left( a \right) +2\,\ln \left( a+1
\right) \right) \left( \ln \left( a \right) a-a\ln \left( a+1
\right) -\ln \left( a+1 \right) \right) }{ \left( a+1 \right)
\left( \ln \left( a \right) \right) ^{2}a \left( \ln \left( a+1
\right) \right) ^{2}}}
$
and $f'a)<0$ for all $a>1$ thus we get by
$\lim_{a \to \infty}f(a)=\frac{5}{2}$ and we get $f(a)\geq \frac{5}{2}$ for $a>1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/979585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Integral $\int\limits_{\sqrt{2}}^{2}\frac1{t^3\sqrt{t^2-1}}$ This is my very first trigonometric substitution integral, so please bear with me here. I would mostly like verification that my methods are sound.
$$\int\limits_{\sqrt{2}}^{2}\dfrac{1}{t^3\sqrt{t^2-1}}\text{ d}t\text{.}$$
Because of the $\sqrt{t^2-1} = \sqrt{t^2-1^2}$ in the integrand, I set $t = 1\sec(\theta) = \sec(\theta)$, and $\theta = \sec^{-1}(t)$. Thus $\text{d}t = \sec(\theta)\tan(\theta)\text{ d}\theta$ and
$$\begin{align}\int\limits_{\sqrt{2}}^{2}\dfrac{1}{t^3\sqrt{t^2-1}} \text{ d}t&= \int\limits_{\pi/4}^{\pi/3}\dfrac{1}{\sec^{3}(\theta)\sqrt{\sec^{2}(\theta)-1}}\sec(\theta)\tan(\theta)\text{ d}\theta \\
&= \int\limits_{\pi/4}^{\pi/3}\dfrac{1}{\sec^{2}(\theta)}\text{ d}\theta \\
&= \int\limits_{\pi/4}^{\pi/3}\cos^{2}(\theta)\text{ d}\theta \\
&= \dfrac{1}{2}\int\limits_{\pi/4}^{\pi/3}[1+\cos(2\theta)]\text{ d}\theta \\
&= \dfrac{1}{2}\left\{\dfrac{\pi}{3}-\dfrac{\pi}{4}+\dfrac{1}{2}\left[\sin\left(\dfrac{2\pi}{3}\right) - \sin\left(\dfrac{\pi}{2}\right)\right]\right\}\\
&= \dfrac{1}{2}\left[\dfrac{\pi}{12}+\dfrac{1}{2}\left(\dfrac{\sqrt{3}}{2}-1\right)\right] \\
&= \dfrac{\pi}{24}+\dfrac{1}{4}\left(\dfrac{\sqrt{3}-2}{2}\right)\text{.}
\end{align}$$
A quick side question: without access to a calculator and assuming I have the values of $\sin(x)$ memorized for $x = 0, \dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{\pi}{3}, \dfrac{\pi}{2}$, what is the easiest way to find $\sin\left(\dfrac{2\pi}{3}\right)$?
| There are no issues with your solution.
You can easily find $\sin \frac{2\pi}{3}$ using $\sin x = \sin (\pi - x)$ for $x = \frac{\pi}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find all positive integers s.t. $10^m-8^n=2m^2$ Find all pairs of positive integers $(m,n)$ such that $10^m-8^n=2m^2$
| Hint:
$${10}^m - 8^n = {2^m}{5^m} - {2^{3n}}$$
so that
$$m^2 = {2^{m-1}}{5^m} - {2^{3n-1}}.$$
In particular:
$${2^{m-1}}{5^m} - {2^{3n-1}} \geq 1,$$
and
$${2^{m-1}}\left({5^m} - {2^{3n-m}}\right)$$
is a square, which means $m \equiv 1 \pmod 2$.
Suppose that $m > 1$. Then the RHS of
$$m^2 = {2^{m-1}}\left({5^m} - {2^{3n-m}}\right)$$
is even while the LHS is odd (since $5^m - {2^{3n-m}}$ is always odd, unless $m = 3n$ [I'll deal with this case later]). A contradiction. Therefore, $m = 1$.
Substituting $m = 1$:
$$10 - {8^n} = 2$$
which gives $n = 1$.
Update: If $m = 3n$, then
$$m^2 = {3^2}{n^2} = 2^{3n - 1}\left(5^{3n} - 1\right).$$
Since $\gcd(3,2) = 1$, then it follows that
$$2^{3n - 1} \mid n^2$$
which implies that $2^{3n - 1} \leq n^2$, contradicting $n^2 < 2^{3n - 1}$.
Therefore, in general, $m = 3n$ cannot occur.
Hence the only solution to the original problem is $(m, n) = (1, 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/982365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field Question:
Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field under the
operations - addition and multiplication where addition is given by:
$(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and
multiplication is given by $(a + b\sqrt{5})(c + d\sqrt{5}) = (ac+5bd)
+ (ad+bc)\sqrt{5}$
My work:
Since $\mathbb Q$ is a field, addition and multiplication defined by $(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and $(a + b\sqrt{5})(c + d\sqrt{5}) = (ac+5bd) + (ad+bc)\sqrt{5}$ will produce elements in $F$ therefore $F$ is closed under multiplication and addition.
Because $F$ is a subset of $\mathbb R$ the operation on $F$ correspond to the usual operations on $\mathbb R$ so the associative, commutative and distributive conditions are inherited from $\mathbb R$
The additive identity is $0+0\sqrt{5}$ because $(a + b\sqrt{5})+(0+0\sqrt{5}) = (a + b\sqrt{5})$
The multiplicative identity is $1+0\sqrt{5}$ because $(a + b\sqrt{5})(1+0\sqrt{5}) = (a + b\sqrt{5})$
The additive inverse is $((-a) + (-b)\sqrt{5})$ because $(a + b\sqrt{5})+((-a) + (-b)\sqrt{5}) = 0$
The multiplicative inverse is where I went wrong, and not quite sure what to do, I think I have to find something such that $(a + b\sqrt{5})\times? = 1 = 1+0\sqrt{5}$ I thought of just putting $$\frac{1+0\sqrt{5}}{a+b\sqrt(5)}$$ but I was told this wasn't right so not sure what else to do.
If anyone can help me with checking the multiplicative inverse condition that would be really appreciated.
| What if there is a condition $a^2+b^2\neq 0?$ How to prove closure under multiplication?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/985020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Simplfy trigonometric functions by only considering integer inputs? I have the below function which only takes integer input,
$$ 2 \sqrt{3} \sin \left(\frac{\pi t}{3}\right)+\sqrt{3} \sin \left(\frac{2 \pi t}{3}\right)-\sqrt{3} \sin \left(\frac{4 \pi t}{3}\right)+6 \cos \left(\frac{\pi t}{3}\right)+\cos \left(\frac{2 \pi t}{3}\right)+\cos \left(\frac{4 \pi t}{3}\right) $$
The resulting output from $t=0$ to $t=10$ is $\{8, 8, -4, -4, -4, -4, 8, 8, -4, -4, -4\}$, and this pattern repeats indefinitely.
Is there a systematic way of reducing this expression using the fact we only consider integer input?
| Since $t$ is an integer,
$$ \sin(\tfrac{4\pi t}{3}) = \sin(2\pi t - \tfrac{2\pi t}{3})
= \sin(2\pi t)\cos(\tfrac{2\pi t}{3}) - \cos(2\pi t)\sin(\tfrac{2\pi t}{3})
= -\sin(\tfrac{2\pi t}{3}) $$
and similarly
$$ \cos(\tfrac{4\pi t}{3}) = \cos(\tfrac{2\pi t}{3}) $$
Thus
\begin{align*}
& 2\sqrt3 \sin(\tfrac{\pi t}{3})
+ \sqrt3\sin(\tfrac{2\pi t}{3})
- \sqrt3\sin(\tfrac{4\pi t}{3})
+ 6\cos(\tfrac{\pi t}{3})
+ \cos(\tfrac{2\pi t}{3})
+ \cos(\tfrac{4\pi t}{3}) \\
&= 2\sqrt3 \sin(\tfrac{\pi t}{3}) + 6\cos(\tfrac{\pi t}{3})
+ 2\sqrt3\sin(\tfrac{2\pi t}{3}) + 2\cos(\tfrac{2\pi t}{3}) \\
&= 4\sqrt3 \left(\tfrac12 \sin(\tfrac{\pi t}{3})
+ \tfrac{\sqrt3}{2}\cos(\tfrac{\pi t}{3})\right)
+ 4 \left(\tfrac{\sqrt3}{2}\sin(\tfrac{2\pi t}{3})
+ \tfrac12\cos(\tfrac{2\pi t}{3}) \right) \\
&= 4\sqrt3 \sin(\tfrac{\pi (t+1)}{3}) + 4 \cos(\tfrac{\pi (2t-1)}{3}) \\
&= 4\sqrt3 \sin(\tfrac{\pi (t+1)}{3}) - 4 \cos(\tfrac{\pi (2t-1)}{3}+\pi) \\
&= 4\sqrt3 \sin(\tfrac{\pi (t+1)}{3}) - 4 \cos(\tfrac{2\pi (t+1)}{3}) \\
&= 4\sqrt3 \sin(\tfrac{\pi (t+1)}{3}) - 4 \cos\big(\pi(t+1)-\tfrac{\pi (t+1)}{3}\big)\\
&= 4\sqrt3 \sin(\tfrac{\pi (t+1)}{3}) - (-1)^{t+1} 4 \cos(\tfrac{\pi (t+1)}{3})\\
&= 8\left(\tfrac{\sqrt3}{2} \sin(\tfrac{\pi (t+1)}{3}) + (-1)^t\tfrac12\cos(\tfrac{\pi (t+1)}{3})\right)\\
&= 8\sin(\tfrac\pi6(2t+2+(-1)^t))
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/987811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result R when tested with various values of $a$ ( where $0<a<1$).
$$
D_1
\, =\,
\int_0^{2\pi}f_1\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R
$$
and
$$D_2\, =\,\int_0^{2\pi}f_2\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R$$
The hypothesis: $D_1$ = $D_2$ has been proved in a separate question Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta = \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta$ .
The remaining hypotheses $D_1$ = $R$ and $D_2$ = $R$ have not been proved. So the question is:-
Prove $D_1$ = $R$ or $D_2$ = $R$.
Only one proof is required because the other can then be obtained from $D_1$ = $D_2$.
For information
WolframAlpha computes expressions for the indefinite integrals $I_1,I_2$ as follows:-
$$I_1
\, =\,
\int\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
$$
$$constant1 + \frac
{a\,\sqrt{a^2-1}\sin\theta\,[-(2a^3+a)\cos^2\theta+3(a^2+1)cos\theta+a(2a^2-5)]}
{2(a^2-1)^{5/2}(a\cos\theta-1)^3}
$$
$$-\frac
{6a\,(a\cos\theta-1)^3\,\tanh^-1
\left(
\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}
\right)
}
{(2(a^2-1)^{5/2}\,(a\cos\theta-1)^3}
$$
and
$$I_2
\, =\,
\int\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
$$
$$constant2 -
\frac
{2a^2\sin\theta-sin\theta}
{2(a^2-1)^2(a\cos\theta-1)}
-\frac
{\sin\theta}
{2(a^2-1)(a\cos\theta-1)^2}
$$
$$
-\frac
{3a\tanh^-1\left(\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}\right)}
{(a^2-1)^{5/2}}
$$
Note that the final terms of each expression ( i.e. the terms involving $\tanh^{-1} $ and $\tan$ ) are equivalent to each other.
Also, note that
$$\int\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta=
\frac{-\sin\theta}{(1-a\cos\theta)^3}
+\int \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta.
$$
Written with StackEdit.
UPDATE 20141028
I have accepted TenaliRaman's answer. I don't yet understand all the steps but his helpful exposition gives me confidence that with time I can understand it because the methods cited (binomials, series) are ones I have learned (at high school).
The answer of M.Strochyk also appears to give a good proof. But the residue method is too advanced for me to understand at present.
UPDATE 20220713
I have now accepted Quanto's answer (because it is simple enough for me to understand). I have also added an answer based on Quanto's but with the intermediate steps written out.
| Lemma 1: If $|x| < 1$, then we know that, $\frac{1}{1 - x} = \sum_{n = 0}^{\infty}x^n$. Using this, we can further show that, $\frac{1}{(1 - x)^3} = \sum_{n = 0}^{\infty}\frac{(n+1)(n+2)}{2}x^n$.
Proof: The first sum is just the GP. The second sum can be obtained by simply counting the different ways to get $x^n$ in $(\sum x^n)(\sum x^n)(\sum x^n)$.
Lemma 2: $$\int_{0}^{2\pi}\cos^{2k}\theta d\theta = {2k \choose k}\frac{2}{4^k}\pi$$
Proof: From Reduction Formula, we can see that,
$$\int_{0}^{2\pi}\cos^{2k}\theta d\theta = \frac{2k - 1}{2k}\int_{0}^{2\pi}\cos^{2k - 2}\theta d\theta$$
Continuing in this fashion, gives us the result.
Lemma 3: $$\int_{0}^{2\pi}\cos^{2k+1}\theta d\theta = 0$$
Proof: This result also follows from Reduction Formula.
Lemma 4: $$\frac{3}{(1 - x)^{5/2}} = \sum_{n = 0}^{\infty} (2n + 3)(2n+1){2n \choose n}\left(\frac{x}{4}\right)^n$$
Proof: We can show that, when $|x| < 1$,
$$\frac{1}{\sqrt{1 - x}} = \sum_{n = 0}^{\infty} {2n \choose n}\left(\frac{x}{4}\right)^n$$
Differentiating once, we obtain,
$$\frac{1}{(1 - x)^{3/2}} = \sum_{n = 1}^{\infty} n{2n \choose n}\left(\frac{x^{n - 1}}{4^n}\right) = \sum_{n = 0}^{\infty} (n + 1){2(n+1) \choose (n+1)}\left(\frac{x^{n}}{4^{n+1}}\right)$$
$$= \frac{1}{2}\sum_{n = 0}^{\infty} (2n + 1){2n \choose n}\left(\frac{x}{4}\right)^n$$
Differentiating again, we get,
$$\frac{3}{(1 - x)^{5/2}} = 2\sum_{n = 1}^{\infty} n(2n + 1){2n \choose n}\left(\frac{x^{n-1}}{4^{n}}\right) = \sum_{n = 0}^{\infty} (2n + 3)(2n+1){2n \choose n}\left(\frac{x}{4}\right)^n$$
Theorem: $$\int_{0}^{2\pi}\frac{\cos\theta}{(1 - a\cos\theta)^3}d\theta = \frac{3a\pi}{(1 - a^2)^{5/2}}$$
Proof: Given, $0 < a < 1$, therefore $|a \cos\theta| < 1$
Hence, from Lemma 1, by replacing $x$ with $a\cos\theta$, we get
$$\frac{1}{(1 - a\cos\theta)^3} = \sum_{n = 0}^{\infty}\frac{(n+1)(n+2)}{2}a^n\cos^n\theta$$
This gives us,
$$\int_{0}^{2\pi}\frac{\cos\theta}{(1 - a\cos\theta)^3}d\theta = \int_{0}^{2\pi} \cos\theta\left(\sum_{n = 0}^{\infty}\frac{(n+1)(n+2)}{2}a^n\cos^n\theta\right)d\theta$$
$$=\sum_{n = 0}^{\infty}\frac{(n+1)(n+2)}{2}a^n\int_{0}^{2\pi}\cos^{n+1}\theta d\theta$$
Using Lemma 2 and 3 and setting $n = 2k + 1$, we get,
$$\sum_{n = 0}^{\infty}\frac{(n+1)(n+2)}{2}a^n\int_{0}^{2\pi}\cos^{n+1}\theta d\theta = \sum_{k = 0}^{\infty}\frac{(2k + 2)(2k + 3)}{2}a^{2k + 1} {2(k+1) \choose k+1}\frac{2}{4^{k+1}}\pi$$
$$=\frac{a\pi}{2}\sum_{k = 0}^{\infty}(k + 1)(2k + 3)a^{2k} {2(k+1) \choose k+1}\frac{1}{4^{k}}$$
$$= \frac{a\pi}{2}\sum_{k = 0}^{\infty}(2k+3)(2k+2)(2k+1)\frac{(2k)!}{k!(k+1)!}\left(\frac{a}{2}\right)^{2k}$$
$$= a\pi\sum_{k = 0}^{\infty}(2k+3)(2k+1)\frac{(2k)!}{k!k!}\left(\frac{a^2}{4}\right)^{k} = a\pi\sum_{k = 0}^{\infty}(2k+3)(2k+1){2k \choose k} \left(\frac{a^2}{4}\right)^{k}$$
Finally, we use Lemma 4 to obtain the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/990813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
} |
Equation: $2\sqrt{1-x}-\sqrt{1+x}+\sqrt{1-x^2}=3-x$ $2\sqrt{1-x}-\sqrt{1+x}+\sqrt{1-x^2}=3-x$
Could someone help me solve this problem?
| Assuming that we are restricted to only real square roots:
Substitute $a = \sqrt{1-x}$ and $b = \sqrt{1+x}$ to get $2a-b+ab = 2+a^2$
Solving for $b$ gives us $b = a-1+\dfrac{1}{a-1}$.
Since $b$ must be non-negative, $a-1$ must be positive. Then, by AM-GM, we get that $b \ge 2$.
However, $b = \sqrt{1+x} \ge 2$ iff $x \ge 3$, which makes $\sqrt{1-x}$ imaginary.
Hence, there are no real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $f(x) = -3x^3 + 2x^2$, find $f(-1)$ and $f(1/2)$. I hope you can help me out
For number 8 and 9 find each value if $f(x) = -3x^3 + 2x^2$
8) $f(-1)$
9) $f(1/2)$
| If you've given some $f(x)$ and asked to find out $f(a)$
then you have to replace all $x$ by $a$ and then simplify ( If you cannot calculate ) resultant expression.
for example
if $$f(x)=-3x^3+2x^2$$
then $$f(-1)=-3(-1)^3+2(-1)^2=5$$
and $$f\left(\frac{1}{2}\right)=-3\left(\frac{1}{2}\right)^3+2\left(\frac{1}{2}\right)^2=\frac{1}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If I roll three dice at the same time, how many ways the sides can sum up to $13$? If I rolled $3$ dice how many combinations are there that result in sum of dots appeared on those dice be $13$?
| Expanding the answer of Henno Brandsma: a generating function is a way to pack a sequence as the coefficients of the power expansion of a function, by example we can pack the Fibonacci sequence as the coefficients on the power expansion of the function $h(x):=\frac1{1-x-x^2}$.
The important point here is that the algebra of generating functions (product, sum, etc.) is a handy way to compose the coefficients packed in them to get a new generating function with coefficients that are of interest to us.
By example the polynomial
$$
p(x):=a_0x^0+a_1x^1+a_2x^2+\ldots+a_n x^n
$$
is the generating function that contains the sequence $a_0,a_1,\ldots,a_n$.
In our case each side of a standard fair dice appears just once in the dice, that is, there is only one side with a given number, from one to six. Therefore the generating function
$$
f(x):= x^1+x^2+x^3+x^4+x^5+x^6
$$
pack the sequence of number of sides of a fair die (note that the power of each monomial represent one of the sides of a dice).
Now: multiplication of generating functions have the effect that the new sequence, after multiplication, is a sum of products of the old ones, where the indices of every product in each sum add up to the exponent of the monomial that will accompany.
Its easy to check that, as we are throwing three dice, then the generating function
$$g(x):=f(x)^3=(x^1+x^2+x^3+x^4+x^5+x^6)^3$$
pack as coefficients the total amounts of different ways to add up to the exponent of each monomial.
Now: the polynomial $f$ can be seen as the partial sum of a geometric series, i.e.
$$
\begin{align*}
f(x)&=x^1+x^2+x^3+x^4+x^5+x^6\\
&=x(x^0+x^1+x^2+x^3+x^4+x^5)\\
&=x\sum_{k=0}^{5}x^k\\
&=x\frac{1-x^6}{1-x}
\end{align*}
$$
Then $$g(x)=x^3\left(\frac{1-x^6}{1-x}\right)^3=x^3\color{red}{(1-x^6)^3}\color{green}{(1-x)^{-3}}$$
The colored expressions (red and green) can be expressed as binomial series[*]. Then
$$\require{cancel} g(x)=x^3\color{red}{\sum_{j=0}^{3}(-1)^j\binom{3}{j}x^{6j}}\color{green}{\sum_{h=0}^{\infty}(-1)^h\binom{-3}{h}x^h}$$
Now: as we know that $\binom{-3}{h}=(-1)^h\binom{3+h-1}{h}=(-1)^h\binom{h+2}{2}$ (to understand this equality you can see here, and remember that $\binom{n}{k}=\binom{n}{n-k}$), then we find that
$$g(x)=x^3\color{red}{\sum_{j=0}^{3}(-1)^j\binom{3}{j}x^{6j}}\color{green}{\sum_{h=0}^{\infty}\cancel{(-1)^h}\cancel{(-1)^h}\binom{h+2}{2}x^h}$$
From here we can build a formula to know the coefficient for any exponent of $x$. First note that any exponent of $x$ will be of the form $S=3+6j+h$, so $h=S-3-6j$, and the coefficient for any sum $S$ will be
$$[x^S]g(x)=1\cdot\sum_{j=0}^{3}\color{red}{(-1)^j\binom{3}{j}}\color{green}{\binom{S-3-6j+2}{2}}\\
=\sum_{j=0}^{3}\color{red}{(-1)^j\binom{3}{j}}\color{green}{\binom{S-1-6j}{2}}$$
where the notation $[x^k]f(x)$ represent the coefficient that the power $x^k$ have in the function $f$.
We can use this last formula to know the amount of ways to obtain a sum $S$ throwing three dice, in our case for $S=13$. Indeed the previous formula can be written in a more precise way: observe that if $S-1-6j<2$ (green binomial) or $j>3$ (red binomial) then the addend will be zero, because if $n<k$ for $n,k\in\Bbb N$ then $\binom{n}{k}=0$. Hence the addends of the sum are not zero when $S-1-6j\ge 2$ and $3\ge j$. And the values of $j$ where the addends are not zero are determined by
$$S-1-6j\geq 2 \implies j\leq\frac{S-3}{6}\le\frac{18-3}6<3\implies j\le 3,\quad S\in\{3,4,\ldots,18\}$$
Then we can re-write $[x^S]g(x)$ as
$$\bbox[5px,border:2px solid gold]{[x^S]g(x)=\sum_{j=0}^{\lfloor\frac{S-3}{6}\rfloor}(-1)^j\binom{3}{j}\binom{S-1-6j}{2}}$$
I hope you understand all information. Anyway surely you must read some more info to understand completely this answer. Just to clarify: the notation $\lfloor x\rfloor$ is the representation of the floor function.
To complete the question, we will evaluate $[x^{13}]g(x)$:
$$
\begin{align*}[x^{13}]g(x)&=\sum_{j=0}^{1}(-1)^j\binom{3}{j}\binom{12-6j}{2}\\
&=\binom{3}{0}\cdot\binom{12}{2}-\binom{3}{1}\binom{6}{2}\\
&=1\cdot \frac{\cancelto{6}{12}\cdot 11}{\cancel{2}}-3\cdot \frac{\cancelto{3}{6}\cdot 5}{\cancel{2}}\\
&=6\cdot 11 - 9\cdot 5\\
&=21
\end{align*}$$
[*] Observe that for $n\in\Bbb N$
$$(x+y)^n=\sum_{k=0}^\infty\binom{n}{k}x^ky^{n-k}=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}$$
then although the second sum is finite it represent a binomial series with infinite addends that are zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/992125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
} |
Solve limit using the concept of equivalent functions How to solve this limit $$\lim_{x\rightarrow 0}{\frac{(2+x)^x-2^x}{x^2}}$$ using the concept of equivalent functions?
For example, if $x\rightarrow 0 $ function $\sin x$ is equivalent to $x$, $\ln(1+x)\sim x$, $a^x-1 \sim x \ln a$, etc.
| we have $$\lim_{x\to 0}{\frac{(2+x)^x-2^x}{x^2}}=\lim_{x\to 0}2^x\lim_{x\to 0}{\frac{(1+\frac{x}{2})^x-1}{x^2}}=\lim_{x\to 0}{\frac{e^{x\ln(1+\frac{x}{2})}-1}{x^2}}$$ therefore we have that$$\lim_{x\to 0}{\frac{(2+x)^x-2^x}{x^2}}\sim\lim_{x\to 0}{\frac{{x\ln(1+\frac{x}{2})}}{x^2}}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/993402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Partial Fractions Decomposition I am failing to understand partial fraction decomposition in cases like the following:
Provide the partial fraction decomposition of the following:
$$\frac{x+2}{(x-4)^3(x^2 + 4x + 16)}$$
I see this and I think of
$$\frac{A}{x-4} + \frac{Bx+C}{(x-4)^2} + \frac{Dx^2 + Ex + F}{(x-4)^3} + \frac{Gx+H}{x^2 + 4x + 16}$$
But I am told that the correct answer is
$$\frac{A}{x-4} + \frac{B}{(x-4)^2} + \frac{C}{(x-4)^3} + \frac{Dx+E}{x^2 + 4x + 16}$$
What exactly is the numerator of each fraction based on?
| Hint. Observe that you have
$$
\frac{Bx+C}{(x-4)^2} =\frac{B(x-4)+4B+C}{(x-4)^2}=\frac{B}{(x-4)}+\frac{4B+C}{(x-4)^2}
$$
and
$$
\begin{align}
\frac{Dx^2 + Ex + F}{(x-4)^3} &=\frac{D(x-4)^2+(E+8D)(x-4)+C+4E+32D}{(x-4)^3}\\\\
&=\frac{D}{(x-4)}+\frac{E+8D}{(x-4)^2}+\frac{C+4E+32D}{(x-4)^3}\\\\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
If $a$ divides $b$, then $a$ divides $3b^3-b^2+5b$.
Prove: Suppose $a$ and $b$ are integers. If $a\mid b$, then $a\mid3b^3-b^2+5b$.
I think I have an idea of how to prove this, but I'm not entirely sure.
I can prove that each individual term in the polynomial $3b^3-b^2+5b$ is a multiple of $a$.
*
*$3b^3 = a(3b^2c) = ad$
*$-b^2 = a(-ac^2) = ae$
*$5b = a(5c) =af$.
Adding together the equalities (1)-(3) we obtain $3b^3-b^2+5b=ad+ae+af =a(d + e + f) $ where $(d + e + f) \in \mathbb{Z}$, demonstrating that $3b^3-b^2+5b$ is a multiple of $a$ and that $a\mid3b^3-b^2+5b$.
This is not my formal proof, I just want to make sure the reasoning is correct.
| if $a|b$ then $b=a.t$ for some integer $t$, so $3b^3-b^2+5b=3(at)^3-(at)^2+5(at)=a(3a^2t^3-at^2+5t)$, and thus clearly $a|3b^3-b^2+5b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$.
Prove that the function $x \mapsto \dfrac 1{1+ x^2}$ is uniformly continuous on $\mathbb{R}$.
Attempt: By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ implies $|f(x) - f(a)| < ε.$
Then suppose $x, a$ are elements of $\Bbb R. $
Now
\begin{align}
|f(x) - f(a)|
&= \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right|
\\&= \left| \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right|
\\&= |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}
\\&≤ |x - a| \frac{|x| + |a|}{(1 + x^2)(1 + a^2)}
\\&= |x - a| \left[\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)}\right]
\end{align}
I don't know how to simplify more. Can someone please help me finish? Thank very much.
| First:$f(x)=\frac{1}{1+x^2}\implies f'(x)=-\frac{2x}{(1+x^2)^2} $
Then observes that; For $|x|\le1$
$$\frac{|x|}{(1+x^2)^2} \le\frac{1}{(1+x^2)^2}\le 1$$
and for $|x|\ge1$
$$|x|\le x^2 \le (1+x^2)^2\implies \frac{|x|}{(1+x^2)^2} \le 1$$
Hence,
$$|f'(x)|=\frac{2|x|}{(1+x^2)^2} \le 2\implies |f(x)-f(y)|\le 2|x-y|$$
This shows that $f$ is Lipschitz which implies, that $f$ is uniformly continuous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/997602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 2
} |
Find $S=\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+...+\frac{2n-1}{2^n}+...$ I'm trying to calculate $S$ where $$S=\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+...+\frac{2n-1}{2^n}+...$$
I know that the answer is $3$, and I also know "the idea" of how to get to the desired outcome, but I can't seem to actually go through with the proof.
The idea for solving this question is: First we will only work on the partial sum $S_n$ and once we find a closed form for it, we will limit $n$ to $\infty$ to find our answer. Notice that for example $\frac{3}{2^2}$ can be rewritten as $\frac{1}{2^2}+\frac{2}{2^2}$, knowing this, we can write $S_n$ a bit differently:
$$S_n=(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^n})+\frac{2}{2^2}+\frac{4}{2^3}+\frac{6}{2^4}+...+\frac{2n-2}{2^n}$$
Notice that the part in the brackets is a finite sum that we can calculate, we know it converges to $1$ when $n$ approaches $\infty$
And now we can repeat the process again for what's not in the brackets and we can repeat this an infinite amount of times.
And if we keep going like that, we will indeed see that the sequence of whats in the bracket is: $1,1,\frac{1}{2},\frac{1}{4},\frac{1}{8}...$ and if we sum them all up it will converge to the desired outcome which is $3$.
And that is exactly what I am having problems showing, that the sequence of whats in the brackets at each step is $\frac{1}{2^k}$.
I hope this was clear enough, it is a bit difficult to explain it, and my english is not perfect so I apologize in advance.
| $$\sum_{n=1}^\infty\frac{2n-1}{2^n}=\sum_{n=1}^\infty\frac n{2^{n-1}}-\sum_{n=1}^\infty\frac1{2^n}=\frac1{\left(\frac12\right)^2}-\frac{\frac12}{1-\frac12}=4-1=3$$
The above follows from
$$|x|<1\implies\frac1{1-x}=\sum_{n=0}^\infty x^n\;,\;\;\left(\frac1{1-x}\right)'=\sum_{n=1}^\infty nx^{n-1}$$
and the splitting of the first sum in the first line above is justified as both series in the right converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/998112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Factoring $x^4 + 4x^2 + 16$ I was putting together some factoring exercises for my students, and came across one that I am unsure of how to factor.
I factored $x^6 - 64$ as a difference of squares, and then tried it as a difference of cubes, but was left with $(x^2 - 4)(x^4 + 4x^2 + 16)$ is there a general method for factoring $x^4 + 4x^2 + 16$?It factors into two irreducible quadratic trinomials, which is where I think the problem is stemming from.
Thanks in advance.
| $$x^4 + 4x^2 + 16$$
Let $y = x^2$, the polynomial is then equal to
$$y^2 + 4y + 16$$
Then, use the quadratic formula:
$$y = \frac{-4 \pm \sqrt{4^2 - 4(1)(16)}}{2(1)}$$
$$y = \frac{-4 \pm \sqrt{-48}}{2}$$
$$y = 2 \pm 2i\sqrt{3}$$
Return to $x$
$$x^2 = 2 \pm 2i\sqrt{3}$$
So, we can now convert $x^4 + 4x^2 + 16$ into factors.
$$x^4 + 4x^2 + 16 = \left(x^2 - {2 + 2i\sqrt{3}}\right)\left(x^2 - {2 - 2i\sqrt{3}}\right)$$
Repeat quadratic formula for each factor.
$$x_{left} = \frac{0 \pm \sqrt{0^2 - 4(1)(2 + 2i\sqrt{3})}}{2}$$
$$x_{left} = \pm \frac{\sqrt{-8 - 8i\sqrt{3}}}{2}$$
$$x_{right} = \frac{0 \pm \sqrt{0^2 - 4(1)(2 - 2i\sqrt{3})}}{2}$$
$$x_{right} = \pm \frac{\sqrt{8i\sqrt{3} - 8}}{2}$$
So, we can now convert $x^4 + 4x^2 + 16$ into factors again.
$$x^4 + 4x^2 + 16 \\ = \left(x - \frac{\sqrt{-8 - 8i\sqrt{3}}}{2}\right)\left(x + \frac{\sqrt{-8 - 8i\sqrt{3}}}{2}\right)\\ \left(x - \frac{\sqrt{8i\sqrt{3} - 8}}{2}\right)\left(x + \frac{\sqrt{8i\sqrt{3} - 8}}{2}\right)$$
None of this is too outlandish for a student to know how to do.
Edit: Corrected the mistake in the quadratic formula.
Edit: Fixed the mistake with naively square-rooting the y-roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If $\gcd(7,abc)=1$ and $a^2+b^2=c^2$, prove that $7$ divides $a^2-b^2$ The only information I have on this problem is that for $a^2+b^2=c^2$ that
$$
a = st, b = \frac{s^2-t^2}{2}, c = \frac{s^2+t^2}{2}
$$
and that $\gcd(7,abc)=1$ gives $7x + abcy = 1$
I have no idea how to proceed, so any help welcomed
| HINT:
$$x\equiv\pm1,\pm2,\pm3\implies x^2\equiv1,4,2\pmod7$$
So, if $a^2\equiv1,b^2$ must be $\equiv1\pmod7$ and so on
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}{4\sqrt{2}}$
Indeed,$\frac{{{a}^{2}}}{1+a}\ge \frac{9{{a}^{2}}+6a+1}{32}$
$\Leftrightarrow 32{{a}^{2}}\ge 9{{a}^{2}}+6a+1+9{{a}^{3}}+6{{a}^{2}}+a$
$\Leftrightarrow 9{{a}^{3}}-17{{a}^{2}}+7a+1\le 0$
$\Leftrightarrow 9{{\left( a-1 \right)}^{2}}\left( a+\frac{1}{9} \right)\le 0$ (!)
It is wrong. Advice on solving this problem.
|
Thus, it remains to prove that $$2\left(\sum_{cyc}(x^2+3xy)\right)^3\geq9\sum_{cyc}xy(x+y)(2z+x+y)^3$$
We have \begin{align*}LHS-RHS&=(x^3+y^3+z^3)\sum x(x-y)(x-z)+(x^2+y^2+z^2-xy-yz-zx)^3\\ &+7(xy+yz+zx)\sum x^2(x-y)(x-z)\\ &+[x^2+y^2+z^2+3(xy+yz+zx)][x^2y^2+y^2z^2+z^2x^2-xyz(x+y+z)]\\ &+ (x+y+z)[4(x^2+y^2+z^2)+9(xy+yz+zx)]\sum z(x-y)^2\\ &+ (x+y+z)(xy+yz+zx)\sum (x+y+7z)(x-y)^2 \ge 0\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 5
} |
solve quadratic equation I'm trying to solve the following equation $2t^2 + t - 3 = 0$
I start by dividing by 2, $t^2 + \frac {t}{2} - \frac {3}{2} = 0$
Then I solve for t $t = - \frac{ \frac {1}{2} }{2} \binom{+}{-} \sqrt{(\frac {1}{2})^2 + \frac {3}{2}}$
$t = - \frac{1}{4} \binom{+}{-} \sqrt{(\frac {1}{4}) + \frac {6}{4}}$
I calculate $t = - \frac{1}{4} \binom{+}{-} \frac {\sqrt7}{4}$
$t_1 = - \frac{1}{4} + \frac {\sqrt7}{4}$
$t_2 = - \frac{1}{4} - \frac {\sqrt7}{4}$
But according to wolframalpha it's suppose to be
$t_1 = 1$
$t_2 = - \frac {3}{2}$
Can't figure out where did I go wrong in my calculation?
| In general, to solve
$$ ax^2+bx+c=0$$
We can use the quadratic formula, which is
$$
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
$$
In your case, we have
$$ 2t^2+t-3=0$$
Which implies that
$$ a=2, b=1, c=-3 $$
Also I'd recommend against your first step of dividing by $2$. This step introduces fractions and the quadratic formula is much simpler when just using integers. So now
$$
t=\frac{-1\pm \sqrt{1^2-4(2)(-3)}}{2(2)}
$$
$$
t=\frac{-1\pm \sqrt{25}}{4}
$$
$$
t=\frac{-1\pm 5}{4}
$$
Therefore
$$
t_1=\frac{-1+ 5}{4}=\frac44=1
$$
$$
t_2=\frac{-1- 5}{4}=-\frac64=-\frac32
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $\int_0^1\frac{\ln2-\ln\left(1+x^2\right)}{1-x}\,dx=\frac{5\pi^2}{48}-\frac{\ln^22}{4}$ How does one prove the following integral
\begin{equation}
\int_0^1\frac{\ln2-\ln\left(1+x^2\right)}{1-x}\,dx=\frac{5\pi^2}{48}-\frac{\ln^22}{4}
\end{equation}
Wolfram Alpha and Mathematica can easily evaluate this integral. This integral came up in the process of finding the solution this question: Evaluating $\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}$. There are some good answers there but avoiding this approach. I have been waiting for a day hoping an answers would be posted using this approach, but nothing shows up. The integral cannot be evaluated separately since each terms doesn't converge. I tried integration by parts but the problem arises when substituting the bounds of integration.
I would appreciate if anyone here could provide an answer where its approach using integral only preferably with elementary ways.
| As M.N.C.E. stated in the comments, integrating by parts shows that $$ \begin{align} \int_{0}^{1} \frac{\log(2) - \log(1+x^{2}) }{1-x} \ dx&= - 2 \int_{0}^{1} \frac{x \log(1-x)}{x^{2}+1} \ dx \\ &= - 2 \ \text{Re} \int_{0}^{1} \frac{\log(1-x)}{x+i} \ dx . \end{align}$$
Letting $ \displaystyle t = \frac{x+i}{1+i} $,
$$ \begin{align} \int \frac{\log(1-x)}{x+i} \ dx &= \int \frac{\log [(1+i)(1-t)]}{t} \ dt \\ &= \log(1+i) \int \frac{dt}{t} + \int\frac{\log(1-t)}{t} \ dt \ \ \left(- \pi < \frac{\pi}{4} + \text{Arg} (1-t) \le \pi \right) \\ &= \log(1+i) \log(t) - \text{Li}_{2} (t) + C \\ &= \log(1+i) \log \left(\frac{x+i}{1+i} \right) - \text{Li}_{2} \left(\frac{x+i}{1+i} \right) + C . \end{align}$$
Therefore,
$$ \begin{align} &\int_{0}^{1} \frac{\log(2) - \log(1+x^{2}) }{1-x} \\ &= - 2 \ \text{Re} \left[- \text{Li}_{2} (1) - \log(1+i) \log \left(\frac{1+i}{2} \right) + \text{Li}_{2} \left(\frac{1+i}{2} \right) \right] \\ &= - 2 \ \text{Re} \left[- \frac{\pi^{2}}{6} - \left(\frac{\log 2}{2} + \frac{i \pi}{4} \right) \left(- \frac{\log 2}{2} + \frac{i \pi}{4} \right) \right] -2 \ \text{Re} \ \text{Li}_{2} \left(\frac{1+i}{2} \right) \\ & = - 2 \left(\frac{\log^{2}(2)}{4} - \frac{5 \pi^{2}}{48} \right) - 2 \left(\frac{5 \pi^{2}}{96} - \frac{\log^{2} (2)}{8} \right) \\ &= \frac{5 \pi^{2}}{48} - \frac{\log^{2}(2)}{4} . \end{align}$$
To show that $$ \text{Re} \ \text{Li}_{2} \left(\frac{1+i}{2} \right) = \frac{5 \pi^{2}}{96} - \frac{\log^{2} (2)}{8}$$ combine the reflection formula for the dilogarithm (5) with the property $\text{Li}_{n}(\bar{z}) = \overline{\text{Li}_{n}(z)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
If $x,y,z.\geq 0$ and $x+y+z = 10$ , Then Max. value of $xyz+xy+yz +zx$, If $x,y,z.\geq 0$ and $x+y+z = 10$ , Then Maximum value of $xyz+xy+yz +zx$, is
$\bf{My\; Try::}$ First we can write the given expression
$xyz+xy+yz+zx = (x+1)(y+1)(z+1)-(x+y+z)-1 = $
$\displaystyle = (x+1)(y+1)(z+1)-\left\{(x+1)+(y+1)+(z+1)\right\}+2$.
So we can write $x+y+z = 10$ as $(x+1)+(y+1)+(z+1) = 13$
Now Let $\displaystyle (x+1)=a\geq 1\;\;,(y+1)=b\geq 1$ and $(z+1)=c\geq 1$
So expression convert into $a+b+c=13\;\;,a,b,c\geq 1$ and Maximize $abc-(a+b+c)+2$
Now How can i solve after that, Help me
Thanks
| Use the fact that $a+b+c=13$ to say you are trying to maximize $abc-14$, which is the same as maximizing $abc$. The AM-GM inequality now tells you that $a=b=c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$ Question:
let $x,y,z>0$ and such $xyz=1$, show that
$$x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$$
My idea: use AM-GM inequality
$$x^3+x^3+1\ge 3x^2$$
$$y^3+y^3+1\ge 3y^2$$
$$z^3+z^3+1\ge 3z^2$$
so
$$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$
But this is not my inequality,so How prove it? I know this condition is very important.but how use this condition? and this inequality is stronger
| Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, our inequality is equivalent to $f(v^2)\geq0$, where $f$ is a linear function.
Hence, $f$ gets a minimal value, when $v^2$ gets an extremal value, which happens when
two numbers from $\{x,y,z\}$ are equal.
Id est, it remains to prove our inequality for $y=x$ and $z=\frac{1}{x^2}$, which gives something obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
} |
Confirming an answer for finding a limit I have to found the limit of the sequence $(x_n)$, which is equal to:
$((2^3-1)/(2^3+1))$ $((3^3-1)/(3^3+1))$ $...$ $((n^3-1)/(n^3+1))$,
and I used the tricks that $(n^3-1)$=$(n-1)(n^2+n+1)$, and that $(n^3+1)$=$(n+1)(n^2-n+1)$, and after cancellation, I found the limit is 2/3. Can anyone confirm this please? Thanks
| Yes you are right!
$$\prod_{k=2}^n \dfrac{k^3 - 1}{k^3 + 1} = \dfrac{2(n^2 +n + 1)}{3n(n+1)} \to \frac{2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving a non-linear congruence How would I go about solving the following:
$$x^2+1\equiv2\pmod8$$
I know I can subtract, and I get:
$$x^2\equiv1\pmod8$$
I'm not sure if I am allowed to square root both sides, or if I should employ a different technique?
| Rewrite the last relation as $x^2-1\equiv0 \pmod 8$ or equivalently as $$(x-1)(x+1)\equiv0 \pmod 8$$ and assume that $x$ is even, i.e. $x=2n$ for $n \in \mathbb N$. Then $$(x-1)(x+1)=(2n-1)(2n+1)=4n^2-1$$ which cannot be a multiple of $8$.
Now assume that $x$ is odd, i.e. $x=2n+1$ for $n \in \mathbb N$. Then $$(x-1)(x+1)=(2n+1-1)(2n+1+1)=2n(2n+2)=4n(n+1)$$ where either $n$ or $n+1$ must be even. Hence $$4n(n+1) | 8$$ which means that $$(x-1)(x+1)\equiv 0 \pmod 8$$ for every odd $x$.
From the last one you have that $$x^2=8k+1$$ which gives $$\begin{array}{rlcl}k=1&&\quad&x^2=9&=3^2\\k=3&=1+2&\quad&x^2=25&=5^2\\k=6&=3+3&\quad&x^2=49&=7^2\\k=10&=6+4&\quad&x^2=81&=9^2\\k=15&=10+5&\quad&x^2=121&=11^2\\...&...&&...&...\end{array}$$ The next $k$ that solve it should be $21, 28, 36, 45, \dots$ with solutions $13^2, 15^2, 17^2, 19^2, \ldots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\sqrt{3}\cos2\theta+\sin2\theta-1=0$ I tried using the identities $\cos2\theta=1-2\sin^2\theta$ and $\sin2\theta=2\sin\theta\cos\theta$. These give
$\sqrt{3}(1-2\sin^2\theta)+2\sin\theta\cos\theta-1=0$
which doesn't seem to lead anywhere. Perhaps I must equate the function to something like $R\sin(2\theta+\alpha)$?
| When you have an equation of the form
$$
a\sin^2\theta+b\sin\theta\cos\theta+c\cos^2\theta+d=0
$$
you can use $1=\cos^2\theta+\sin^2\theta$ and write the equation as
$$
(a+d)\sin^2\theta+b\sin\theta\cos\theta+(c+d)\cos^2\theta=0
$$
If $a+d=0$, this factors; otherwise $\cos\theta=0$ is not a solution and so you can transform it into
$$
(a+d)\tan^2\theta+b\tan\theta+(c+d)=0
$$
that's quadratic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solving a simple first order differential equation I've got the following first order differential equation that is doing something I can't quite figure out. The question is to solve:
$$y' = 2x^2y^2$$
Here is my solution:
$$y' \frac{1}{y^2} = 2x^2$$
$$\int \frac{1}{y^2} y' dx = \int 2x^2 dx$$
$$\int \frac{1}{y^2}dy = \int 2x^2 dx$$
$$\ln|y^2| = \frac{2}{3}x^3 + c$$
$$y^2 = e^{\frac{2}{3}x^3 + c} = e^{\frac{2}{3}x^3}.e^c = ce^{\frac{2}{3}x^3}$$
$$y = \sqrt{ce^{\frac{2}{3}x^3}}$$
However, I know from the textbook that this answer is incorrect. Please help.
| Note that $$\int \frac 1{y^2}\,dy = \int y^{-2} \,dy = -y^{-1} + C$$
So you should obtain $$-\frac 1y = \frac{2}{3}x^3 + c \iff y = -\frac 3{2x^3}\; \text{provided } x\neq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Taylor series expansion for $e^{-x}$ could anyone show me the Taylor series expansion for $e^{-x}$.I was trying to find out how
$e^{-i\theta}$=$\cos\theta-i\sin\theta$.
More specifically could you show me how $e^{-i\theta}$=$\cos\theta-i\sin\theta$ is obtained from Taylor series.
| $$ e^{-x} = \sum\limits_{k=0}^{\infty}\frac{(-x)^k}{k!} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+\dots $$
So plugging in $x=i\theta$ we have that
\begin{align}
e^{-i\theta} &=1-i\theta+\frac{(i\theta)^2}{2!}-\frac{(i\theta)^3}{3!}+\frac{(i\theta)^4}{4!}+\dots\\
&=1-i\theta-\frac{\theta^2}{2!}+i\frac{\theta^3}{3!}+\frac{\theta^4}{4!}+\dots
\end{align}
using the fact that $i^2=-1$, $i^3=-i$ and $i^4=1$, etc.
The taylor expansions of $\sin$ and $\cos$ are
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots\hspace{10px}\text{and}\hspace{10px}\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots $$
So
\begin{align}
e^{-i\theta} &=\left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\dots\right)-i\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\dots\right)\\
&=\cos\theta-i\sin\theta.
\end{align}
As a side-note if you already know that $e^{i\theta}=\cos\theta+i\sin\theta$, then it is easy to show that $e^{-i\theta}$ without using taylor-series using the fact that cosine is even and sine is odd. That is
$$e^{-i\theta} = \cos(-\theta)+i\sin(-\theta)=\cos\theta-i\sin\theta, $$
because $\cos(x)=\cos(-x)$ and $\sin(-x)=-\sin(x).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the positive integer numbers to get $\frac{\pi ^2}{9}$ As we know, there are many formulas of $\pi$ , one of them $$\frac{\pi ^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}......
$$
and this $$\frac{\pi ^2}{8}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}......$$
Now,find the positive integer numbers $(a_{0}, a_{1}, a_{2}....)$ to get $$\frac{\pi^2 }{9}=\frac{1}{a_{0}^2}+\frac{1}{a_{1}^2}+\frac{1}{a_{2}^2}....$$
| This is equal to
$$\frac1{72} \sum_{n=-\infty}^{\infty} \left [\frac1{(n+1/6)^2} + \frac1{(n+5/6)^2}\right ] $$
This may be treated using the following formula from residue theory:
$$\sum_{n=-\infty}^{\infty} f(n) = -\pi \sum_k \operatorname*{Res}_{z=z_k} f(z) \cot{\pi z} $$
where $z_k$ are the poles of $f$. In this case
$$f(z) = \frac1{(z+1/6)^2} + \frac1{(z+5/6)^2} $$
which has double poles at $z=-1/6$ and $z=-5/6$. Then the sum is equal to
$$\pi \left [-\frac{\cot{(\pi/6)}}{(4/6)^2} + \pi \csc^2{(\pi/6)} -\frac{\cot{(5\pi/6)}}{(-4/6)^2} + \pi \csc^2{(5\pi/6)}\right ] = 8 \pi^2$$
The result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
} |
Maximizing the trinomial coefficient Let $\binom{n} {a,b,c} = \frac{n!}{a!\cdot b! \cdot c!}$
In other words, $\binom{n} {a,b,c}$ is the trinomial coefficient.
I am trying to find the triplets $(a,b,c)$ which maximize this trinomial coefficient.
I have determined, by simply plugging in numbers, that:
When $n\bmod 3 = 0$, the trinomial coefficient is maximized when $a = b = c = \frac{n}{3}$
When $n \bmod 3 = 1$, the trinomial coefficient is maximized when $a = b = \frac{(n - 1)}{3} \space \text{and} \space c = \frac{(n + 2)}{3} \impliedby$ Note for this case, any two variables can equal $\frac{(n - 1)}{3}$ while the other equals $\frac{(n + 2)}{3}.$
And when $n \bmod 3 = 2$, the trinomial coefficient is maximized when $a = \frac{(n - 2)}{3}$ and $b = c = \frac{(n + 1)}{3} \impliedby$ Same caveat about the interchanging of variables as $n \bmod 3 = 1$.
While I understand this, I cannot think of a way to rigorously prove this. Could someone help me get started in the right direction?
| Temporarily fix $c$. We want to minimize $x!y!$ given that $x+y=n-c=k$.
Suppose that we have $x\lt y$. Compare $x!y!$ with $(x+1)!(y-1)!$. We have
$$(x+1)!(y-1)!=\frac{x+1}{y}(x!y!).$$
Thus $(x+1)!(y-1)!\lt x!y!$ unless $y=x+1$, in which case we have equality.
So given $a,b,c$ with fixed sum, if two of $a,b,c$ differ by $2$ or more, we can decrease the product. It follows that the minimum product is reached when at least two of $a,b,c$ are equal, and the third differs from them by at most $1$.
Your observations now follow. Suppose for example that $n$ is divisible by $3$. The product $a!b!c!$ is minimized, and therefore $\frac{n!}{a!b!c!}$ is maximized, when at least two of $a,b,c$ are equal, say $a=b$, and the third is $a$, $a+1$, or $a-1$. The sum is divisible by $3$ precisely if they are all equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
what is the sum of square of all elements in $U(n)$? I know that $\sum\limits_{a\in U(n)} a=\frac{n\varphi(n)}{2}$ where $U(n):=\{1\leq r\leq n: (r, n)=1\}$ is a multiplicative group. And I know how to prove this result.
What I was willing to know was this $\sum\limits_{a\in U(n)} a^2$. is it possible to find in closed form?
what I tried is the following:
Let $S=\sum\limits_{a\in U(n)} a^2$. Now $(n, a)=1$ shows that $(n, n-a)=1$ which again under the fact $(a, b)=1, (a, c)=1\Rightarrow (a, bc)=1$, shows that $(n, (n-a)^2)=1$. Hence $\{(n-a_1)^2, \cdots, (n-a_{\varphi(n)})^2\}$ is nothing but a permutation of the original set $\{a_1^2, \cdots, a_{\varphi(n)}^2\}$ in some order. Hence $S=\sum\limits_{a\in U(n)} (n-a)^2$.
In other words we must have:
\begin{align*}
S=&\sum\limits_{a\in U(n)} (n^2-2an+a^2)\\
=&n^2 \sum\limits_{a\in U(n)} 1-2n\sum\limits_{a\in U(n)} a+S\\
=&n^2\varphi(n) -2n\times \frac{n\varphi(n)}{2}+S\\
=&S
\end{align*}
and no result is obtained.
What to do ? Please help me.
Thanks
EDIT:
After the link has been provided below by Robert Israel
the formula reads $ n = p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}$ then $S = n^2\frac{\varphi(n)}{3}+(-1)^r p_1p_2\cdots p_r\frac{\varphi(n)}{6}$ but how to establish this ?
| Lemma 1: For any function $f$ defined on rational values in $[0,1]$, denoting, $\displaystyle F(n) = \sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right)$ and $\displaystyle F^{*}(n) = \sum\limits_{1\le k\le n;(k,n)=1} f\left(\frac{k}{n}\right)$.
We have, $\displaystyle 1*F^{*} = F$
Proof: First we observe that each integer $m \in \{1,2,\cdots,n\}$ has a unique representation in the form $m = k.\frac{n}{d}$, where, $d|n$, $1 \le k\le d$ and $(k,d) = 1$.
Since, $\displaystyle \frac{m}{n} = \frac{k}{d}$, be the reduced fraction i.e., $(k,d) = 1$, each integer $m \le n$ has a representation in the required form, and representation is unique.
Now, $\displaystyle \begin{align} (1*F^{*})(n) = \sum\limits_{d|n}F^{*}(d) = \sum\limits_{d|n}\sum\limits_{1\le k\le n;(k,d)=1} f\left(\frac{k}{d}\right) &= \sum\limits_{d|n}\sum\limits_{1\le k\le n;(k,d)=1} f\left(\frac{k\frac{n}{d}}{n}\right) \\ &= \sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right) = F(n)\end{align}$
Now, take $\displaystyle f(x) = x^2$, we have $\displaystyle F(n) = \frac{\sum\limits_{k=1}^{n} k^2}{n^2} = \sum\limits_{d|n}\sum\limits_{1\le k\le n;(k,d)=1} \left(\frac{k}{d}\right)^2 = \sum\limits_{d|n} \frac{\phi_2(d)}{d^2}$
Where, $\displaystyle \phi_2(n) = \sum\limits_{k \le n;(k,n) = 1} k^2$
That is, $\displaystyle \frac{1}{6}n(n+1)(2n+1) = (\phi_2*f)(n) = g(n)$, (say)
Applying Dirichlet Inversion to the above identity, we have,
$\displaystyle \begin{align} \phi_2(n) = \sum\limits_{d|n} g(d)f^{-1}\left(\frac{n}{d}\right) \end{align}$
Since, $f$ is an absolutely multiplicative function the inverse is given by $f^{-1}(n) = \mu(n)f(n)$
(a proof of this fact can be found in T. Apostol, Introduction to Analytic Number Theory, page 36)
Thus,
$\displaystyle \begin{align} \phi_2(n) &= \sum\limits_{d|n} g(d)\left(\frac{d}{n}\right)^2\mu\left(\frac{n}{d}\right) \\ &= \frac{n^2}{3}\sum\limits_{d|n} d\mu\left(\frac{n}{d}\right) + \frac{n^2}{2}\sum\limits_{d|n} \mu\left(\frac{n}{d}\right) + \frac{n}{6}\sum\limits_{d|n} \left(\frac{n}{d}\right)\mu\left(\frac{n}{d}\right) \tag{1} \\ &= \frac{n^2}{3}\phi(n) + \frac{n}{6}\sum\limits_{d|n}d\mu(d) \end{align}$
Where, in $(1)$, we used $\sum\limits_{d|n} d\mu\left(\frac{n}{d}\right) = \phi(n)$ and $\displaystyle \sum\limits_{d|n} \mu(d) = 0$, for $n > 1$.
Now, $\displaystyle \sum\limits_{d|n}d\mu(d) = \prod\limits_{p^\alpha || n} \left(\sum\limits_{e|p^{\alpha}}e\mu(e)\right) = \prod\limits_{p | n}(1-p)$
Thus, $\displaystyle \phi_2(n) = \frac{n^2}{3}\phi(n) + \frac{n}{6}\prod\limits_{p | n}(1-p)$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluating definite integral I'm having a hard time understanding how to compute this integral.
$$\int_1^4\frac{3x^3-2x^2+4}{x^2}\,\mathrm dx$$
The steps I do is $\dfrac{3x^4}{4} - \dfrac{2x^3}{3} + 4x$ but I don't know how to integrate the $x^2$ in the integral. I know it's suppose to be $\dfrac{x^3}3$.
Is this how the answer is supposed to look like $$\left.\frac{\dfrac{3x^4}{4} - \dfrac{2x^3}{3} + 4x}{\dfrac{x^3}{3}}\right|_1^4?$$
The answer to this equation is $\displaystyle{39\over2}$ and I don't know how they got that answer.
| $$
\int_1^4 \frac{3x^3 - 2x^2 + 4}{x^2} dx = \int_1^4 (\frac{3x^3}{x^2} - \frac{2x^2}{x^2} + \frac{4}{x^2})dx = \int_1^4 (3x - 2 + 4x^-2)dx
$$
$$ = \frac{3}{2}x^2 - 2x - \frac{4}{x}
$$
Now evaluate this function from 1 to 4. First evaluating the function at 4 and then subtracting off the function evaluated at 1.
$$
(\frac{3}{2}(4^2) - 2(4) - \frac{4}{4}) - (\frac{3}{2}(1^2) - 2(1) - \frac{4}{1}) = \frac{39}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Wrong applying of simple Chinese Remainder Theorem problem What am I doing wrong?
So for the following equations
$$
\begin{align}
(*) \left\{
\begin{array}{l}
2x\equiv 3\pmod 5 \\
4x\equiv 2\pmod 6 \\
3x\equiv 2\pmod 7
\end{array} \right.
\end{align}
$$
and $N =\mathrm{lcm}\langle5,6,7\rangle = 210$, giving
$N_1 = \frac{210}{5} = 42, \enspace N_2 = \frac{210}{6} = 35, \enspace N_3 = \frac{210}{7} = 30 $.
$$
\begin{align}
42z_1&\equiv 1\pmod 5\Rightarrow\enspace\enspace\;2z_1\equiv 1\pmod 5\Rightarrow\enspace &&\overline{z_1}=\overline{3}\\
35z_2&\equiv 1\pmod 6\Rightarrow\enspace-1z_2\equiv 1\pmod 6\Rightarrow\enspace &&\overline{z_2}=\overline{-1}\\
30z_3&\equiv 1\pmod 7\Rightarrow\enspace\enspace\;2z_3\equiv 1\pmod 7\Rightarrow\enspace &&\overline{z_1}=\overline{4}
\end{align}
$$
So the solution should be
$$
\begin{align}
\overline{x} &= \overline{3\times42\times3} + \overline{2\times35\times(-1)}+\overline{2\times30\times4}\\
&= \overline{378-70+240}\\
&= \overline{548}\\
&= \overline{128}
\end{align}
$$
Which is clearly wrong, so I'm wondering which additional steps I need to take to get to the correct answer.
Thanks in advance.
| \begin{align}
2x &\equiv 3\pmod 5 \\
4x &\equiv 2\pmod 6 \\
3x &\equiv 2\pmod 7
\end{align}
First, you need to solve each congruence for $x$.
\begin{align}
2x &\equiv 3 \pmod 5 \\
x &\equiv 4 \pmod 5 \\
\hline
4x &\equiv 2 \pmod 6 \\
2x &\equiv 1 \pmod 3 \\
x &\equiv 2 \pmod 3 \\
\hline
3x &\equiv 2 \pmod 7 \\
x &\equiv 3 \pmod 7
\end{align}
Note that the solution set (modulo 6) to $4x \equiv 2 \pmod 6$ is
$\{2,5\}$ and that both solutions are included in $x \equiv 2 \pmod 3.$
summarizing, we get
\begin{align}
x &\equiv 4 \pmod 5 \\
x &\equiv 2 \pmod 3 \\
x &\equiv 3 \pmod 7
\end{align}
One way to compute the solution is this.
\begin{align}
x &\equiv 4 \pmod 5 \\
x &= 5A + 4 \\
\hline
x &\equiv 2 \pmod 3 \\
5A+4 &\equiv 2 \pmod 3 \\
2A &\equiv 1 \pmod 3 \\
A &= 2 + 3B\\
x &= 15B + 14 \\
\hline
x &\equiv 3 \pmod 7 \\
15B + 14 &\equiv 3 \pmod 7 \\
B &\equiv 3 \pmod 7 \\
B &= 7C + 3 \\
x &= 105C + 59 \\
x &\equiv 59 \pmod{105}
\end{align}
Your way of solving is this way.
$N =\mathrm{lcm}\{5,3,7\} = 105$.
$N_1 = 3 \cdot 7z_1 = 21z_1,\enspace
N_2 = 5 \cdot 7z_2 = 35z_2, \enspace
N_3 = 5 \cdot 3z_3 = 15z_3 $.
$21z_1 \equiv 1 \pmod 5 \implies
z_1 \equiv 1 \pmod 5
\implies N_1 = 21$
$35z_2 \equiv 1 \pmod 6 \implies
z_2 \equiv -1 \pmod 6 \implies
N_2 = -35$
$15z_3 \equiv 1 \pmod 7 \implies
z_3 \equiv 1 \pmod 7 \implies
N_3 = 15$
So the solution is
\begin{align}
x &\equiv 4 \cdot 21 - 2\cdot 35 + 3 \cdot 15 \pmod{105}\\
x &\equiv 59 \pmod{105}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3}$ What I attempted thus far:
Multiplying by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} \cdot \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}} = \lim_{x \to 0} \frac{\tan x - \sin x}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
factor out $\sin x$ in the numerator
$$\lim_{x \to 0} \frac{\sin x \cdot (\sec x - 1)}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
simplify using $\lim_{x \to 0} \frac{\sin x}{x} = 1 $
$$\lim_{x \to 0} \frac{\sec x - 1}{x^2 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
From here I don't see any useful direction to go in, if I even went in an useful direction in the first place, I don't know.
I suspect that this could be evaluated using the definition of derivatives, if so, or not, any suggestions?
| If you're willing to use a symbolic differentiation program like Matlab and have forgotten your trig and calculus formulas like I have, you should be able to recover the answer as $$\frac{\frac{d^3(\sqrt{1 + \tan x} - \sqrt{1 + \sin x})}{dx^3}}{\frac{d^3(x^3)}{dx^3}} = \frac{\frac{d^3(\sqrt{1 + \tan x} - \sqrt{1 + \sin x})}{dx^3}}{6}$$ evaluated at $x=0$. This result is derived by applying the L'Hopital rule three times. The denominator will reduce to $3!=6$, and the numerator can be calculated by the computer. Even if you are trying to show your work, it never hurts to have a computer verify your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Limit of square root function at $x \to 6$ I'm trying to find the limit of the following function at $x \to 6$:
$$\frac{x^2-36}{\sqrt{x^2-12x+36}}$$
i've simplified it so that it becomes $\dfrac{(x+6)(x-6)}{\sqrt{(x-6)^2}}$, which simplifies to $x+6$.
the problem is that i shouldn't be getting to $x+6$, because then id be able to plug in $6$, and say that the limit exists for the left hand side and the right hand side of $6$, when clearly i can tell from the graph that the limit does not exist.
What am I doing wrong?
| Hint: $$\sqrt{(x-6)^2}=|x-6|$$ which is in turn equal to $$|x-6|=\begin{cases}-(x-6)=-x+6, & x<6\\ \phantom{+}(x-6)=\phantom{-}x-6,&x>6\end{cases}$$ This implies that $$\lim_{x\to6^-}\dfrac{(x+6)(x-6)}{\sqrt{(x-6)^2}}=\lim_{x\to6^-}\dfrac{(x+6)(x-6)}{-(x-6)}=\lim_{x\to6^-}-(x+6)$$ but $$\lim_{x\to6^+}\dfrac{(x+6)(x-6)}{\sqrt{(x-6)^2}}=\lim_{x\to6^+}\dfrac{(x+6)(x-6)}{(x-6)}=\lim_{x\to6^+}\phantom{+}(x+6)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the $a$ to make the sum of series equal to zero,$\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=0$ Find all values of $a$ which make the sum of series $\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=0$
| Starting from the representation
$$\pi \cot \pi z = \lim_{N\to\infty} \sum_{n=-N}^N \frac{1}{z-n},$$
we obtain
\begin{align}
\pi \tan \pi z &= -\pi \cot \pi\left(z-\tfrac{1}{2}\right)\\
&= \lim_{N\to\infty} \sum_{n=-N}^N \frac{1}{n+\frac{1}{2}-z}\\
&= \lim_{N\to\infty} \left(\frac{1}{N+\frac{1}{2}-z} + \sum_{n=0}^{N-1} \frac{2z}{\left(n+\frac{1}{2}\right)^2-z^2}\right)\\
&= 2z \sum_{n=0}^\infty \frac{1}{\left(n+\frac{1}{2}\right)^2-z^2}
\end{align}
and
$$\frac{\pi \tan \pi z}{6z} = \sum_{n=0}^\infty \frac{1}{3n^2+3n - 3\left(z^2-\frac{1}{4}\right)}.$$
It should not be hard to find the values of $a$ from that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric Limit: $\lim_{x\to 0} \frac{\sqrt {\cos x} - \sqrt[3] {\cos x}}{\sin^2x}$ Compute the following limit:
$$\lim_{x\to 0} \frac{\sqrt {\cos x} - \sqrt[3] {\cos x}}{\sin^2x}$$
How would I go about solving this, I can't used l´Hôpital
| Writing $c$ and $s$ for $\cos x$ and $\sin x$ respectively,
$$\frac{c^{1/2} - c^{1/3}}{s^2} = \frac{c^{1/3}}{s^2} (c^{1/6} - 1) = \frac{c -1}{s^2} \frac{c^{1/3}}{c^{5/6} + c^{4/6} + c^{3/6} + c^{2/6} + c^{1/6} +1} \ \ \ --(*) $$
as $(c^{1/6} - 1)(c^{5/6} + c^{4/6} + c^{3/6} + c^{2/6} + c^{1/6} +1) = c - 1$
Now $$c - 1 = \cos x - 1 = -2\sin^2(x/2) \ \ \hbox{ and} \ \ s^2 = \sin^2 x = 4\sin^2(x/2)\cos^2(x/2)$$ Hence the first fraction of (*)
$$ \frac{c -1}{s^2} = -\frac{1}{2}\cos^2(x/2) \longrightarrow -\frac{1}{2}, \ \ \hbox{ as } x \rightarrow 0$$
The second fraction $\displaystyle \frac{c^{1/3}}{c^{5/6} + c^{4/6} + c^{3/6} + c^{2/6} + c^{1/6} +1} \ \longrightarrow \frac{1}{6}, \ \ \hbox{ as } x \rightarrow 0$
Hence $$\lim_{x\rightarrow 0} \frac{c^{1/2} - c^{1/3}}{s^2} = -\frac{1}{2} . \frac{1}{6} = -\frac{1}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Logarithmic twin integrals $\int_0^1\frac{\ln x\ln(1+x^2)}{1\pm x}dx$ Here is what I have done
\begin{align}
&\int_0^1\frac{\ln(x)\ln(1+x^2)}{1-x}dx\\
=&\int_0^1\frac{(1+x)(1+x^2) \log(x)\log(1-x^4)}{1-x^4} \ dx\\
&-\int_0^1\frac{(1+x)\log(x)\log(1-x^2)}{1-x^2} \ dx
\end{align}
and, then, after letting $x^4\mapsto x$ and $x^2\mapsto x$ respectively, use the beta function. Similarly we proceed with the second twin and we are done.
Please teach me another ways that only use real analysis.
Above also answers the first integral here
| Denote $K_{\pm} = \int_0^1 \frac{\ln t}{1\pm t}dt$ and integrate the twins as follows
\begin{align}
I_\pm =&\int_0^1\frac{\ln x\ln(1+x^2)}{1\pm x}dx \\
=& \int_0^1 \ln(1+x^2)\> d\left(\int_0^x \frac{\ln t}{1\pm t}dt\right)\\
=& \>\ln2 \int_0^1 \frac{\ln t}{1\pm t}dt -\int_0^1 \frac{2x}{1+x^2}\left(\int_0^x \frac{\ln t}{1\pm t}\overset{t=xy}{dt}\right)dx\\
=& \>\ln2 K_{\pm}-2\int_0^1 \int_0^1\frac{x^2\ln(xy)}{(1+x^2)(1\pm xy)}dy\>dx\\
=& \>\ln2 K_{\pm} + 2\int_0^1 \int_0^1\frac1{1+y^2}
\bigg( \frac{(1\mp xy)\ln(xy)}{1+x^2}
-\frac{\overset{t=xy}{\ln(xy)}}{1\pm xy} \bigg)dx\>dy\\
=& \>\ln2 K_{\pm} \pm 2\int_0^1 \frac{x\ln x}{1+x^2} \overset{x^2\to x}{dx}\int_0^1\frac {2y}{1+y^2}dy\\
&+ 4\int_0^1 \frac{\ln x}{1+x^2} dx\int_0^1\frac 1{1+y^2}dy-\int_0^1 \frac{2}{y(1+y^2)}\left(\int_0^y \frac{\ln t}{1\pm t}dt\right)
dy\\
= & \>\ln2 K_{\pm} \pm 2\left(\frac14K_+\right)\ln2
+4\>(-G)\>\frac\pi4
- \int_0^1 d\left(\ln\frac{y^2}{1+y^2}\right)\int_0^y \frac{\ln t}{1\pm t}dt\\
=& \>2\ln2 K_{\pm}\pm \frac12 \ln2 K_+ -\pi G+2\int_0^1 \frac{\ln^2y}{1\pm y}-I_{\pm}\\
\end{align}
Substitute $K_- =-\frac{\pi^2}6$, $K_+ =-\frac{\pi^2}{12}$, $\int_0^1 \frac{\ln^2y}{1-y}dy=2\zeta(3)$, $\int_0^1 \frac{\ln^2y}{1+y}dy=\frac32\zeta(3)$ into above expression to arrive at
\begin{align}
&I_-=\int_0^1\frac{\ln x\ln(1+x^2)}{1-x}dx
= -\frac{3\pi^2}{16}\ln2-\frac\pi2 G+2\zeta(3)\\
&I_+=\int_0^1\frac{\ln x\ln(1+x^2)}{1+x}dx
=-\frac{\pi^2}{16}\ln2-\frac\pi2 G+\frac32\zeta(3)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^{\infty} \log(\sin^2(x))\left(1-x\operatorname{arccot}(x)\right) \ dx$ One of the ways to compute the integral
$$\int_0^{\infty} \log(\sin^2(x))\left(1-x\operatorname{arccot}(x)\right) \ dx=\frac{\pi}{4}\left(\operatorname{Li_3}(e^{-2})+2\operatorname{Li_2}(e^{-2})-2\log(2)-\zeta(3)\right)$$
is to make use of the series of $\log(\sin(x))$, but the result I got after doing that wasn't that friendly.
Is it possible to find a neat way of evaluating the integral?
| First notice that
$$ \begin{align} &\frac{1}{2} \int_{0}^{\infty} \log (4 \sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx \\ &= \frac{\log(4)}{2} \int_{0}^{\infty} \Big( 1- x \, \text{arccot}(x) \Big) \, dx + \frac{1}{2} \int_{0}^{\infty} \log (\sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx \\ &= \frac{\pi \log(2)}{4} + \frac{1}{2} \int_{0}^{\infty} \log (\sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx . \tag{1} \end{align}$$
Now use the fact $$ \text{Re} \log(1-e^{2ix}) = \frac{1}{2} \log(4 \sin^{2} x) $$
and integrate by parts to get
$$ \begin{align} &\frac{1}{2} \int_{0}^{\infty} \log (4\sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx \\ &= \text{Re} \int_{0}^{\infty} \log (1-e^{2ix}) \Big(1- x \, \text{arccot}(x) \Big) \, dx \\ &= \text{Re} \, \Big(1- x \, \text{arccot}(x) \Big) \frac{i \, \text{Li}_{2}(e^{2ix})}{2} \Bigg|^{\infty}_{0}- \text{Re} \, \frac{i}{2} \int_{0}^{\infty} \left(\frac{x}{1+x^{2}} - \text{arccot}(x) \right) \text{Li}_{2} (e^{2ix}) \, dx \\ &= 0 + \frac{1}{2} \int_{0}^{\infty} \Big(\frac{x}{1+x^{2}} - \text{arccot}(x) \Big) \sum_{n=1}^{\infty} \frac{\sin({\color{red}{2}}nx)}{n^{2}} \, dx \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \Big(\frac{x}{1+x^{2}} - \text{arccot}(x) \Big) \sin (2nx) \, dx \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2}} \left(-\frac{\pi}{4n} + \frac{1}{n} \int_{0}^{\infty} \frac{\cos(2nx)}{(1+x^{2})^{2}} \, dx \right) \tag{2} \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2}} \Big(-\frac{\pi}{4n} + \frac{1}{n} \frac{\pi}{4} e^{-2n} (2n+1) \Big) \tag{3} \\ &= - \frac{\pi}{8} \sum_{n=1}^{\infty} \frac{1}{n^{3}} + \frac{\pi}{4} \sum_{n=1}^{\infty} \frac{e^{-2n}}{n^{2}} + \frac{\pi}{8} \sum_{n=1}^{\infty} \frac{e^{-2n}}{n^{3}} \\ &= -\frac{\pi}{8} \zeta(3) + \frac{\pi}{4} \text{Li}_{2}(e^{-2}) + \frac{\pi}{8} \text{Li}_{3} (e^{-2}). \end{align}$$
Therefore,
$$ \int_{0}^{\infty} \log (\sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx = \frac{\pi}{4} \Big(\text{Li}_{3} (e^{-2}) + 2 \text{Li}_{2}(e^{-2}) -2 \log(2) - \zeta(3) \Big).$$
$ $
$(1)$ Simple Integral $\int_0^\infty (1-x\cot^{-1} x)dx=\frac{\pi}{4}$.
$(2)$ Integrate by parts again.
$(3)$ There is probably a question on here about evaluating $\int_{0}^{\infty} \frac{\cos(ax)}{(1+x^{2})^{2}} \, dx$, but I can't find it at the moment. The most direct approach is to use the residue theorem. You could also use the fact that $\int_{0}^{\infty} \frac{\cos (ax)}{b^{2}+x^{2}} \, dx = \frac{\pi}{2b} e^{-ab} \, , \, (a \ge 0,b > 0) $ and differentiate both sides with respect to $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
How find this maximum of this $(1-x)(1-y)(10-8x)(10-8y)$
let $x,y\in (0,1)$, and such
$$(1+x)(1+y)=81(1-x)(1-y)$$
Prove
$$(1-x)(1-y)(10-8x)(10-8y)\le\dfrac{9}{16}$$
I ask $\dfrac{9}{16}$ is best constant?
PS:I don't like Lagrange Multipliers,becasue this is Hight students problem.
My idea:
$$(1-x)(1-y)(10-8x)(10-8y)=\dfrac{1}{64\cdot 81}(8+8x)(8+8y)(10-8x)(10-8y)$$
since
$$(8+8x)(10-8x)\le\dfrac{18^2}{4}=81,(8+8y)(10-8y)\le 81$$
if and only if $$8+8x=10-8x,8+8y=10-8y\Longrightarrow x=y=\dfrac{1}{8}$$
but this not such
$$(1+x)(1+y)=81(1-x)(1-y)$$
so How find this maximum?
| We have $$\frac{81}1=\frac{1+xy+(x+y)}{1+xy-(x+y)}$$
Applying Componendo and dividendo,
$$\frac{1+xy}{x+y}=\frac{41}{40}$$
Let $$\frac{1+xy}{41}=\frac{x+y}{40}=u$$
$$\implies(1-x)(1-y)(10-8x)(10-8y)$$
$$=\{1+xy-(x+y)\}\{36+64(1+xy)-80(x+y)\}$$
$$=(41u-40u)\cdot36(1-16u)=\frac{36}{64}\left[1-\left(8u-1\right)^2\right]\le\frac{36}{64}$$
The equality occurs if $8u-1=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Evaluation of $\int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$ Evaluation of Integral $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$
$\bf{My\; Try::}$ First we will convert $x^4+x^3+x^2+x+1$ into closed form, which is $\displaystyle \left(\frac{x^5-1}{x-1}\right)$
So Integral is $\displaystyle \int_{0}^{\infty}\frac{\sqrt{x-1}}{\sqrt{x^5-1}}dx$
now i did not understand how can i solve it
Help me
Thanks
| We have:
$$I=\int_{0}^{+\infty}\frac{dx}{\sqrt{1+x+x^2+x^3+x^4}}=2\int_{0}^{1}\frac{dx}{\sqrt{1+x+x^2+x^3+x^4}}$$
(just split $[0,+\infty)=[0,1)\cup[1,+\infty)$ and use the substitution $x=1/y$ on the second piece) and since:
$$ \sqrt{1-x}=\sum_{j=0}^{+\infty}\binom{1/2}{j}(-1)^j x^j,\qquad\frac{1}{\sqrt{1-x^5}}=\sum_{j=0}^{+\infty}\binom{-1/2}{j}(-1)^j x^{5j}$$
we can compute the integral by considering the Cauchy product of the last two series and integrating it termwise:
$$ \sqrt{\frac{1-x}{1-x^5}}=\sum_{j=0}^{+\infty}\sum_{k=0}^{\lfloor j/5\rfloor}\binom{-1/2}{k}\binom{1/2}{j-5k}(-1)^j x^j,$$
$$ I = 2\sum_{j=0}^{+\infty}\sum_{k=0}^{\lfloor j/5\rfloor}\binom{-1/2}{k}\binom{1/2}{j-5k}\frac{(-1)^j}{j+1}.$$
As an alternative, since:
$$\int_{0}^{1}x^{5j}(1-x)^{1/2}\,dx = B(5j+1,3/2) = \frac{\sqrt{\pi}\,\Gamma(5j+1)}{2\,\Gamma(5j+5/2)}$$
it follows that:
$$ I = 2\sum_{j=0}^{+\infty}\binom{-1/2}{j}(-1)^j\frac{\sqrt{\pi}\,\Gamma(5j+1)}{2\,\Gamma(5j+5/2)}=\sum_{j=0}^{+\infty}\frac{\Gamma(j+1/2)\Gamma(5j+1)}{\Gamma(j+1)\Gamma(5j+5/2)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Writing $\frac{x^4(1-x)^4}{1+x^2}$ in terms of partial fractions How does one write
$$\frac{x^4(1-x)^4}{1+x^2}$$
in terms of partial fractions?
My Attempt
$$\frac{x^4(1-x)^4}{1+x^2}=\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}$$
$$=Ax^6+Bx^5+Cx^4+Dx^3+Ex^2+Fx+\frac{G}{1+x^2}$$
Multiplying out and comparing coefficient gives:
$$=x^6-4x^5+5x^4-4x^2+0+\frac{0}{1+x^2}$$
This is obviously not correct, how to do this properly?
| You should have tried
$$\frac{x^4(1-x)^4}{1+x^2}=Ax^6+Bx^5+Cx^4+Dx^3+Ex^2+Fx +\mathbf{J}+\frac{\mathbf{G + Hx}}{1+x^2}$$
By the way, multiplying out and comparing coefficients gives:
$$\frac{x^4(1-x)^4}{1+x^2}=x^6-4x^5+5x^4-4x^2+\mathbf{4}+\frac{\mathbf{-4}}{1+x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Binomial identity $\binom{0}{0}\binom{2n}{n}+\binom{2}{1}\binom{2n-2}{n-1}+\binom{4}{2}\binom{2n-4}{n-2}+\cdots+\binom{2n}{n}\binom{0}{0}=4^n.$ Prove the identity
$$\binom{0}{0}\binom{2n}{n}+\binom{2}{1}\binom{2n-2}{n-1}+\binom{4}{2}\binom{2n-4}{n-2}+\cdots+\binom{2n}{n}\binom{0}{0}=4^n.$$
This is reminiscent of the identity $\sum_{i=0}^n\binom{n}{i}^2=\binom{2n}{n}$, which has a nice combinatorial interpretation of choosing $n$ from $2n$ objects. But the identity in question is not easily related to a combinatorial interpretation. Also, for an induction proof, it is not clear how to relate the identity with $n$ to $n+1$.
| By way of enrichment here is another algebraic proof using basic
complex variables. As I pointed out in the comment this identity is
very simple using a convolution, so what follows should be considered
a learning exercise.
We seek to compute
$$\sum_{k=0}^n {2k\choose k} {2n-2k\choose n-k}.$$
Introduce the integral representation
$${2n-2k\choose n-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n-2k}}{z^{n-k+1}} \; dz.$$
We use this to obtain an integral for the sum. Note that when $k>n$
the pole at zero disappears which means that the integral is
zero. Therefore we may extend the sum to infinity, getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n}}{z^{n+1}}
\sum_{k\ge 0} {2k\choose k} \frac{z^k}{(1+z)^{2k}}\; dz.$$
Recall that
$$\sum_{k\ge 0} {2k\choose k} w^k = \frac{1}{\sqrt{1-4w}}$$
so this becomes
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n}}{z^{n+1}} \frac{1}{\sqrt{1-4z/(1+z)^2}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n}}{z^{n+1}} \frac{1+z}{\sqrt{(1+z)^2-4z}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n+1}}{z^{n+1}} \frac{1}{1-z} \; dz.$$
Extracting coefficients we get
$$\sum_{q=0}^n {2n+1\choose q} =
\frac{1}{2} 2^{2n+1} = 2^{2n} = 4^n.$$
Apparently this method is due to Egorychev.
Addendum.
The Lagrange inversion proof goes like this. We seek to compute
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}} \frac{1}{\sqrt{1-4z}} \; dz.$$
Put $1-4z = w^2$ so that
$z= \frac{1}{4} - \frac{1}{4} w^2$ and $dz = -\frac{1}{2} w \; dw$
to get
$$-\frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{4^{k+1}}{(1-w^2)^{k+1}} \frac{1}{w} \frac{1}{2} w \; dw
\\ = -\frac{2^{2k+1}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(1-w)^{k+1}} \frac{1}{(1+w)^{k+1}} \; dw
\\ = -\frac{2^{2k+1}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(1-w)^{k+1}} \frac{1}{(2+w-1)^{k+1}} \; dw
\\ = -\frac{2^k}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(1-w)^{k+1}} \frac{1}{(1+(w-1)/2)^{k+1}} \; dw
\\ = -\frac{2^k (-1)^{k+1}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{k+1}} \frac{1}{(1+(w-1)/2)^{k+1}} \; dw.$$
Extracting coefficients we obtain
$$- 2^k (-1)^{k+1} {k+k\choose k} \frac{(-1)^k}{2^k}
= {2k\choose k}.$$
It is not difficult to see that in the above substitution the image of
a small radius counterclockwise circle around the origin in the $z$
plane is a small radius circle around $w=1$ also traversed
counterclockwise.
A similar calculation may be found at this
MSE link.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Solve for constants: Derivatives using first principles
*
*Question
Find the values of the constants $a$ and $b$ such that $$\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \frac{5}{12}$$
*
*My approach
*
*Using the definition of the derivative, $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
*I view limit as a derivative of a function $f$ at some value, let's call that value $c$, as follows :$$f'(c) = \lim_{x\to 0}\frac{f(c + x) - f(c)}{x} = \frac{\sqrt[3]{ax + b} - 2}{x} = \frac{5}{12}$$
*Now I deduce the following: $$f(c + x) = \sqrt[3]{ax + b}$$ and $$f(c) = 2$$
*Use limits as follows: $$\lim_{x\to 0} f(c + x) = f(c) = 2$$ that is, $$\lim_{x\to 0} f(c + x) = \lim_{x\to 0} \sqrt[3]{ax + b} = \sqrt[3]{b} = 2$$ now solve for $b$, $$\sqrt[3]{b} = 2 \Leftrightarrow b = 8$$
*Since I know that $$f(c + x) = \sqrt[3]{ax + 8}$$, I can solve for $a$, which is $$a = \frac{[f(c+x)]^3 - 8}{x} = \frac{[f(c+x)]^3 - [f(c)]^3}{x}$$
*Let $g(x) = [f(x)]^3$, such that $$g'(x) = 3\cdot [f(x)]^2 \cdot f'(x)$$
so $$g'(c) = 3\cdot [f(c)]^2 \cdot f'(c) = 3 \cdot 4 \cdot \frac{5}{12} = 5$$
*Rephrase $g'(c)$ using first principles such that $$g'(c) = \lim_{x \to 0}\frac{g(c + x)- g(c)}{x}= \lim_{x \to 0}\frac{[f(c + x)]^3 - [f(c)]^3}{x} = \lim_{x \to 0} a = 5$$
*Since $a$ is a constant, $\lim_{x \to 0} a = a$, that is, $$a = 5$$
*My solution: $b = 8, a = 5$.
Please have a look at my approach and give me any hints/suggestions regarding the solution and/or steps taken.
| Using
$$ a^3-b^3=(a-b)(a^2+ab+b^2, $$
one has
\begin{eqnarray}
\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \lim_{x \to 0}\frac{ax+b-8}{x[(\sqrt[3]{ax + b})^2+2\sqrt[3]{ax + b}+4]}.
\end{eqnarray}
Thus if the limit exists, one must have $b=8$, under which it is easy to get
\begin{eqnarray}
\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} &=& \lim_{x \to 0}\frac{ax+b-8}{x[(\sqrt[3]{ax + b})^2+2\sqrt[3]{ax + b}+4]}\\
&=&\frac{a}{[(\sqrt[3]{8})^2+2\sqrt[3]{8}+4]}=\frac{a}{12}.
\end{eqnarray}
Thus $a=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Using the definition of the limit to show I am stuck an how to proceed in my proof. I am suppose to show, using the definition of the limit, that $\lim_{x \rightarrow 2}{(x^3-2x-4)/(x^2-4)=5/2}$.
I did some algebra with the $|f(x)-5/2|$ and got to the point where I have $|f(x)-5/2| \leq (1/2)|x-2||2x+3|$, but I am unsure how to procced in choosing my $\delta$... Any help would be appreciated.
EDIT: Sorry, I should have been more clear; I need to do this using an epsilon and delta argument...
| First $$\begin{align}\left|f(x)-\frac{5}{2}\right|&=\left|\frac{x^2+2x+2}{x+2}-\frac{5}{2}\right|\\&=\left|\frac{x^2+2x+2}{x+2}-\frac{10}{x+2}+\frac{10}{x+2}-\frac{5}{2}\right|\\&\leq\left|\frac{x^2+2x+2}{x+2}-\frac{10}{x+2}\right|+\left|\frac{10}{x+2}-\frac{5}{2}\right|\\&=\frac{|x^2+2x+-8|}{|x+2|}+\frac{|10-5x|}{|2x+4|}\\&=\left(\frac{|x+4|}{|x+2|}+5\frac{1}{|2x+4|}\right)|x-2|\\&\leq\left(\frac{|x-2|+6}{4-|x-2|}+\frac{1}{8-2|x-2|}\right)|x-2|\\&\leq(\frac{6+\delta}{4-\delta}+\frac{1}{8-2\delta})\delta\\&\leq(7/3+1/6)\delta\end{align}$$
Now just put $\delta=\min(1/(7/3+1/6),\epsilon/(7/3+1/6))$.
Above we used things like triangle inequality several times. E.g. $$|x-2+4|\leq |x-2|+4.$$
Or as in $$|x+2|=|4-(-(x-2))|\geq|4|-|x-2|.$$ Or in $$|2x+4|=|2(x-2)+8|\geq 8-2|x-2|.$$
To proceed from where you got it.
You have $$\begin{align}|f(x)-5/2|&\leq(1/2)|x-2||2x+3|\\&=(1/2)|x-2||2(x-2)+7|\\&\leq(1/2)(2|x-2|+7)|x-2|\\&\leq(1/2)(2\delta+7)\delta\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
My attempt regarding finding critical ponts of $(\cos x)(\cos y)(\cos(x+y))$ Given this problem Restrictions on $x$ any are that $x\in[0,\pi]$ , $y\in[0,\pi]$
I have $f_x=-(\cos y)({\sin(2x+y))}--------*$
$f_y=-(\cos x)(\sin x+2y)-----------**$
So from $*$ I get either $\cos y=0$ or $\sin(2x+y)=0$
From $\sin(2x+y)=0$ , I get $2x+y=0,\pi,2\pi,3\pi$ as $2x+y\in[0,3\pi]$
And from second case I get $\cos y=0\implies y=\dfrac{\pi}{2}$
Also from $**$ , I get $x+2y=0,\pi,2\pi,3\pi$ and from $\cos x=0\implies x=\dfrac{\pi}{2}$
Now I have made $16$ cases because of following equations:
$2x+y=0,\pi,2\pi,3\pi$
$x+2y=0,\pi,2\pi,3\pi$
example as like $2x+y=0$ , $x+2y=3\pi$ ....similarly $16$ cases from above equations
and I point $\left(\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
After examining all cases I have got valid critical points only for cases of type
$2x+y=R$
$x+2y=R$
Where $R$ is $0,\pi,2\pi,3\pi$ , taken one by one. So out of $16$ cases I have made above , I got my valid points out of $4$ cases as shown above..
MY QUESTION - Why I am getting points from these cases only ($R$ cases) ...Points from other $12$ cases are either same as that of the ($R$ cases), or they are not in domain
Also am I right in making $16$ cases like this. Is this correct way of solving these equations?
Is there any alternative way??
Thanks for your kind help!!
| You have $f(x,y)=\cos x\cos y\cos(x+y)$, so
$$
f'_x=-\sin x\cos y\cos(x+y)-\cos x\cos y\sin(x+y)=-\cos y\sin(2x+y)
$$
Since interchanging $x$ and $y$ doesn't change the function, we have
$$
f'_y=-\cos x\sin(x+2y)
$$
We have four cases:
$$
\begin{cases}
\cos y=0\\
\cos x=0
\end{cases}
\qquad
\begin{cases}
\cos y=0\\
\sin(x+2y)=0
\end{cases}
\qquad
\begin{cases}
\sin(2x+y)=0\\
\cos x=0
\end{cases}
\qquad
\begin{cases}
\sin(2x+y)=0\\
\sin(x+2y)=0
\end{cases}
$$
The first case gives $x=y=\pi/2$.
The second case gives $y=\pi/2$ and $\sin(x+\pi)=0$, that is, $-\sin x=0$, so critical points at $(0,\pi/2)$ and $(\pi,\pi/2)$.
The third case is symmetric, giving critical points at $(\pi/2,0)$ and $(\pi/2,\pi)$.
Let's tackle the last case. From $\sin(2x+y)=0$ we get $2x+y=m\pi$ for some integer $m$; similarly, $x+2y=n\pi$ for an integer $n$. Now let's solve the linear system
$$
\begin{cases}
2x+y=m\pi\\
x+2y=n\pi
\end{cases}
$$
that gives
$$
\begin{cases}
x=(2m-n)\dfrac{\pi}{3}\\[6px]
y=(2n-m)\dfrac{\pi}{3}
\end{cases}
$$
With the given limitations we must have $0\le 2m-n\le 3$ and $0\le 2n-m\le 3$ and we should determine $m$ and $n$ so that this holds.
If we set $2m-n=h$ and $2n-m=k$, we get
$$
m=\frac{2h+k}{3},\qquad n=\frac{h+2k}{3}
$$
So we cannot have arbitrary choices for $h$ and $k$, since we want integer values of $m$ and $n$. The only values are so
\begin{alignat}{3}
&m=0, n=0&\quad&\text{corresponding to}&\quad& h=0, k=0
\\
&m=1, n=1,&&\text{corresponding to}&& h=1, k=1
\\
&m=1, n=2&&\text{corresponding to}&& h=0, k=3
\\
&m=2, n=1,&&\text{corresponding to}&& h=3, k=0
\\
&m=2, n=2,&&\text{corresponding to}&& h=2, k=2
\\
&m=3, n=3&&\text{corresponding to}&& h=3, k=3
\end{alignat}
and the critical points are
$$
\left(0,0\right)
\qquad
\left(\frac{\pi}{3},\frac{\pi}{3}\right)
\qquad
\left(0,\pi\right)
\qquad
\left(\pi,0\right)
\qquad
\left(\frac{2\pi}{3},\frac{2\pi}{3}\right)
\qquad
\left(\pi,\pi\right)
$$
See the following diagram drawn with Geogebra to see why
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.