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Inverse Fourier transform with a $\delta$-function integrand I'm self-studying math and trying to find the inverse Fourier transform of $\frac{4+w^2}{1+w^2}(4\pi * (\delta(w-2)+\delta(w+2)))$
Based on wolframalpha, the result is $32/5\sqrt{2\pi}\cos(2t)$.
But I can't even find the fourier transform of $\frac{4+w^2}{1+w^2}$ because there doesn't seem to have a Fourier transform pair in my table for $w^2$ in the numerator.
| You don't need to know, just apply the definition
\begin{eqnarray}
f(t) &=& \frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{4 + \omega^2}{1 + \omega^2}[4\pi\delta(\omega - 2) + 4\pi\delta(\omega + 2)] e^{i\omega t}{\rm d}\omega \\
&=& 2\int_{-\infty}^{+\infty} \frac{4 + \omega^2}{1 + \omega^2}4\pi\delta(\omega - 2) e^{i\omega t}{\rm d}\omega + 2\int_{-\infty}^{+\infty} \frac{4 + \omega^2}{1 + \omega^2} \delta(\omega + 2) e^{i\omega t}{\rm d}\omega \\
&=& 2 \frac{4 + 2^2}{1 + 2^2}e^{2it} + 2\frac{4+2^2}{1 + 2^2}e^{-2it} \\
&=&\frac{32}{5}\left(\frac{e^{2it} + e^{-2it}}{2}\right) = \frac{32}{5}\cos(2t)
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3029293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.
Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.
I have tried to solve this question and did pretty well until I reached the end, so I was wondering if I could get help on that part. Here is what I did.
$$a^2+b^2=c^2$$ $$b^2=(c-a)(c+a)$$ Since $a, b, c > 0 \therefore (c+a) \ne 0$ $$\therefore c-a={b^2\over c+a}$$ Similarly we get, $$c-b={a^2\over c+b}$$ $$\therefore {1\over2}(c-a)(c-b)={1\over2}({b^2\over c+a})({a^2\over c+b})$$ $$={(ab)^2\over 2c^2+2ab+2bc+2ca}$$ $$={(ab^2)\over a^2+b^2+c^2+2ab+2bc+2ca}$$ $$={(ab)^2\over (a+b+c)^2}$$ $$=({ab\over a+b+c})^2$$ However, I was unable to prove that ${ab\over a+b+c} \in \mathbb{Z}$ Is there a way to prove it? Thank you
| Alternatively, use Formulas for generating Pythagorean triples
WLOG $a=2pqk, b=(p^2-q^2)k,c=(p^2+q^2)k$
$c-a=k(p-q)^2$
$c-b=2kq^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3029557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Seeking Methods to solve $F\left(\alpha\right) = \int_{0}^{1} x^\alpha \arcsin(x)\:dx$ I'm looking for different methods to solve the following integral.
$$ F\left(\alpha\right) = \int_{0}^{1} x^\alpha \arcsin(x)\:dx$$
For $\alpha > 0$
Here the method I took was to employ integration by parts and then call to special functions, but can this equally be achieved with say a Feynman Trick? or another form integral transform?
My approach in detail:
Employ integration by parts:
\begin{align}
v'(x) &= x^\alpha & u(x) &= \arcsin(x) \\
v(x) &= \frac{x^{\alpha + 1}}{\alpha + 1} & u'(x) &= \frac{1}{\sqrt{1 - x^2}}
\end{align}
Thus,
\begin{align}
F\left(\alpha\right) &= \left[\frac{x^{\alpha + 1}}{\alpha + 1}\cdot\arcsin(x)\right]_0^1 - \int_0^1 \frac{x^{\alpha + 1}}{\alpha + 1} \cdot \frac{1}{\sqrt{1 - x^2}} \:dx \\
&= \frac{\pi}{2\left(\alpha + 1\right)} - \frac{1}{\alpha + 1}\int_0^1 x^{\alpha + 1}\left(1 - x^2\right)^{-\frac{1}{2}} \:dx
\end{align}
Here make the substitution $u = x^2$ to obtain
\begin{align}
F\left(\alpha\right) &= \frac{\pi}{2\left(\alpha + 1\right)} - \frac{1}{\alpha + 1}\int_0^1 \left(\sqrt{u}\right)^{\alpha + 1}\left(1 - u\right)^{-\frac{1}{2}} \frac{\:du}{2\sqrt{u}} \\
&= \frac{\pi}{2\left(\alpha + 1\right)} - \frac{1}{2\left(\alpha + 1\right)}\int_0^1 u^{\frac{\alpha}{2}}\left(1 - u\right) ^{-\frac{1}{2}} \:du \\
&= \frac{1}{2\left(\alpha + 1\right)} \left[ \pi - B\left(\frac{\alpha + 2}{2}, \frac{1}{2} \right) \right]
\end{align}
\begin{align}
F\left(\alpha\right) &=\frac{1}{2\left(\alpha + 1\right)} \left[ \pi - \frac{\Gamma\left(\frac{\alpha + 2}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{\alpha + 2}{2} + \frac{1}{2}\right)} \right] \\
&= \frac{1}{2\left(\alpha + 1\right)} \left[ \pi - \frac{\Gamma\left(\frac{\alpha + 2}{2}\right)\sqrt{\pi}}{\Gamma\left(\frac{\alpha + 3}{2}\right) } \right] \\
&= \frac{\sqrt{\pi}}{2\left(\alpha + 1\right)} \left[ \sqrt{\pi} - \frac{\Gamma\left(\frac{\alpha + 2}{2}\right)}{\Gamma\left(\frac{\alpha + 3}{2}\right) } \right]
\end{align}
Edits:
Correction of original limit observation (now removed)
Correction of not stating region of convergence for $\alpha$.
Correction of 1/sqrt to sqrt in final line.
Thanks to those commentators for pointing out.
| Here is a method that relies on using a double integral.
Noting the integral converges for $\alpha > -2$, recognising
$$\arcsin x = \int_0^x \frac{du}{\sqrt{1 - u^2}},$$
the integral can be rewritten as
$$I = \int_0^1 \int_0^x \frac{x^\alpha}{\sqrt{1 - u^2}} \, du dx.$$
On changing the order of integration, one has
\begin{equation}
\int_0^1 \int_u^1 \frac{x^\alpha}{\sqrt{1 - u^2}} \, dx du. \qquad (*)
\end{equation}
After performing the $x$-integral we are left with
$$I = \frac{1}{\alpha + 1} \int_0^1 \frac{1 - u^{\alpha + 1}}{\sqrt{1 - u^2}} \, du, \quad \alpha \neq -1.$$
Enforcing a substitution of $u \mapsto \sqrt{u}$ results in
$$I = \frac{1}{2(\alpha + 1)} \int_0^1 \left (\frac{1}{\sqrt{u(1 - u)}} - \frac{u^{\alpha/2}}{\sqrt{ 1 - u}} \right ) \, du = I_1 - I_2.$$
The first of the integrals is trivial
$$I_1 = \frac{1}{2(\alpha + 1)} \int_0^1 \frac{du}{\sqrt{\frac{1}{4} - (u - \frac{1}{2})^2}} = \frac{1}{2(\alpha + 1)} \arcsin (2u - 1) \Big{|}^1_0 = \frac{\pi}{2(\alpha + 1)}.$$
For the second of the integrals, it can be evaluated by writing it in terms of the beta function. Here
\begin{align}
I_2 &= \int_0^1 \frac{u^{\alpha/2}}{\sqrt{1 - u}} \, du\\
&= \int_0^1 u^{(\alpha/2 + 1) - 1} (1 - u)^{1/2 - 1} \, du\\
&= \text{B} \left (\frac{\alpha}{2} + 1, \frac{1}{2} \right )\\
&= \sqrt{\pi} \, \frac{\Gamma \left (\frac{\alpha + 2}{2} \right )}{\Gamma \left (\frac{\alpha + 3}{2} \right )}.\\
\end{align}
Thus
$$I = \frac{1}{2(\alpha + 1)} \left [\pi - \sqrt{\pi} \, \frac{\Gamma \left (\frac{\alpha + 2}{2} \right )}{\Gamma \left (\frac{\alpha + 3}{2} \right )} \right ], \qquad \alpha \neq -1.$$
For the case when $\alpha = -1$, the double integral at ($*$) becomes
$$I = \int_0^1 \int_u^1 \frac{1}{x\sqrt{1 - u^2}} \, dx du.$$
After performing the $x$-integral which yields a natural logarithm, one has
$$I = -\int_0^1 \frac{\ln u}{\sqrt{1 - u^2}} \, du.$$
Enforcing a substitution of $u \mapsto \sin u$ leads to
$$I = -\int_0^{\pi/2} \ln (\sin u) \, du. \qquad (**)$$
The integral appearing in ($**$) is quite famous and reasonably (?) well known. Its evaluation can be found either here or here. Thus
$$I = \frac{\pi}{2} \ln 2, \qquad \alpha = -1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Minimizing distance from a point to a parabola Problem: The point on the curve $x^2 + 2y = 0$ that is nearest the point $\left(0, -\frac{1}{2}\right)$ occurs at what value of y?
Using the distance formula, I get my primary equation: $L^2 = (x-0)^2 + \left(y-\left(-\frac{1}{2}\right)\right)^2$
However, when using the secondary equation $x^2 + 2y = 0$ to write $L^2$ as a function of a single variable and then minimizing $L^2$, thus minimizing $L$, I am getting conflicting answers.
Let $D = L^2 = (x-0)^2 + \left(y-\left(-\frac{1}{2}\right)\right)^2$
Approach 1: substituting $y = -\frac{x^2}{2}$ into our primary equation,
$$D = (x-0)^2 + \left(-\frac{x^2}{2}-\left(-\frac{1}{2}\right)\right)^2$$
$$\frac{dD}{dx} = 2x + 2\left(\frac{x^2}{2}-\frac{1}{2}\right)(x)$$
$$\frac{dD}{dx} = x(x^2+1)$$
Therefore, $D$ has a critical point at x = 0. Using the first derivative test, $\frac{dD}{dx} < 0$ for all $x<0$ and $\frac{dD}{dx}>0$ for all $x>0$, therefore a minimum exists at $x = 0$. At $x=0$, $y = 0$.
Approach 2: substituting $x^2 = -2y$ into our primary equation,
$$D = -2y + \left(y-\left(-\frac{1}{2}\right)\right)^2$$
$$\frac{dD}{dy} = -2 + 2\left(y+\frac{1}{2}\right)$$
$$\frac{dD}{dy} = 2y-1$$
Therefore, $D$ has a critical point at $y = \frac{1}{2}$. Using the first derivative test, $\frac{dD}{dy} < 0$ for all $y<\frac{1}{2}$ and $\frac{dD}{dy}>0$ for all $y>\frac{1}{2}$, therefore a minimum exists at $y = \frac{1}{2}$.
Why do the results differ? What faulty assumption(s) have I made? Thanks in advance.
| The correct solution is $(0,0)$. Your first approach is correct.
For the second approach, note that the domain of $D(y)$ is $y \le 0$, so the solution $y= \frac12$ cannot be correct.
Also note that $D$ is always decreasing for $\forall y \le 0$, therefore $D(y)$ attains its minimum at $y=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that point does not belong to a plane Problem
Show that point $\textbf{q}$ does not belong to plane defined by these 3 points:
$$ \textbf{p}_1 = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}, \textbf{p}_2=\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}, \textbf{p}_3 = \begin{bmatrix} 2 \\ 4 \\ 0 \end{bmatrix} $$
Point $\textbf{q}$ is defined as.
$$ \textbf{q}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $$
Attempt to solve
Plane in vector form can be defined as:
$$ \textbf{x} = \textbf{p} + s\textbf{u}+ t\textbf{v}, \{s,t\} \in \mathbb{R} $$
Where vectors $\textbf{u}$ and $\textbf{v} $ define the plane and $\textbf{p}$ is point which belongs to the plane. let $\textbf{p} = \textbf{p}_2$ and vectors $\textbf{u}$ and $\textbf{v}$ can be defined as
$$ \textbf{u} = \textbf{p}_2 - \textbf{p}_1, \textbf{v} = \textbf{p}_2 - \textbf{p}_3 $$
$$ \textbf{u} = \begin{bmatrix} 0 - 2 \\ 1 -1 \\ -1 -1 \end{bmatrix}, \textbf{v} = \begin{bmatrix} 0 - 2 \\ 1 - 4 \\ -1 - 0 \end{bmatrix} $$
$$ \textbf{u} = \begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix}, \textbf{v} = \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} $$
Now we can write the equation as:
$$ \textbf{x}= \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + s \begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix} + t \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} $$
Now i want to show that $$ \nexists \{s,t\} : \textbf{p} + s\text{u} + t \textbf{v} = \begin{bmatrix} 1 \\ 1 \\1 \end{bmatrix} $$
I can form matrix equation:
$$\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} + s \begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix} + t \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $$
$$ \implies s\begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix} + t \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}-\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} $$
$$ \implies s\begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix} + t \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} $$
$$ \begin{bmatrix} -2 & -2 \\ 0 & -3 \\ -2 & -1 \end{bmatrix} \begin{bmatrix} s \\ t \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} $$
Now i can solve the equation with row redcution
$$ \begin{bmatrix} -2 & -2 & 1 \\ 0 & -3 & 0 \\ -2 & -1 & 2 \end{bmatrix} $$
Solution is $$ s = -\frac{15}{19}, t = \frac{1}{19} $$
Problem is solution should be impossible. i Was trying to prove poin $\textbf{q}$ doe $\textbf{not}$ belong to plane. Is my approach correct or did something went wrong with my solution ?
| Let me check your solution:
$$(-2) \left( -\frac{15}{19}\right)+(-2) \left( \frac{1}{19}\right)= \left( \frac{2}{19}\right)(15-1) \ne 1$$
Hence you made a mistake in solving for your $s$ and $t$.
Here is the RREF, we can see that the augmented matrix is not consistent.
octave:1> rref([-2 -2 1; 0 -3 0 ; -2 -1 2])
ans =
1 0 0
0 1 0
0 0 1
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trouble to find the solution to a linear system where a matrix is not invertible Hello community I am new here and I have a question which might be pretty basic.
So I am trying to solve an equation. I have 3 matrices
A = \begin{pmatrix}1&0&0&0&0&0&0&0\\0&1&1&0&0&0&0&0\end{pmatrix}
B = \begin{pmatrix}0&0&0&0&0&0&1&0\\0&0&0&0&0&0&0&1\end{pmatrix}
and
K = \begin{pmatrix}1&0&0&0&0&0&0&0\\0&1&1&0&0&0&0&0\\0&0&0&0&0&0&1&0\end{pmatrix}
H = \begin{pmatrix}A\\B\end{pmatrix}
and Z = \begin{pmatrix}z1\\z2\\z3\end{pmatrix}
And I have linear system
$\begin{pmatrix}H\\K\end{pmatrix} v = \begin{pmatrix}0\\Z\end{pmatrix}$
It says that solution for determined system is
$ v = \begin{bmatrix}\begin{pmatrix}H\\K\end{pmatrix}^T \begin{pmatrix}H\\K\end{pmatrix} \end{bmatrix} ^{-1}\begin{pmatrix}H\\K\end{pmatrix}^T \begin{pmatrix}0\\Z\end{pmatrix} $
Okay when I put this system of matrices in python it shows that matrix inside [] can't be inverse. I am presumping that
H = \begin{pmatrix}A\\B\end{pmatrix}
just means that
H is
\begin{pmatrix}1&0&0&0&0&0&0&0\\0&1&1&0&0&0&0&0\\0&0&0&0&0&0&1&0\\0&0&0&0&0&0&0&1\end{pmatrix}
Two matrices put together. Same I presume for \begin{pmatrix}H\\K\end{pmatrix}
Where am I wrong?
Ps: sorry im new here hope I wrote everything correctly and understandable. Happy holidays everyone.
| For the sake of simpler notations, let us rename $$L:=\begin{pmatrix}H\\K\end{pmatrix} \ \ \text{and} \ \ P:=\begin{pmatrix}0\\Z\end{pmatrix} $$
giving :
$$Lv=P \tag{1}$$
which is a $7 \times 8$ system (thus an underdetermined system : less constraints than the number of unknowns).
Instead of your $$ v = (L^T L)^{-1}L^T P $$
which is a formula valid for overdetermined systems, one must take for an underdetermined system :
$$ v = L^T(L L^T)^{-1} P \tag{2}$$
(see https://see.stanford.edu/materials/lsoeldsee263/08-min-norm.pdf for the difference of formulas between under- and over-determined systems ; in this document, a matrix like $L$ with more columns than lines is called "fat").
This will give you a so-called "mean-squares" solution.
Proof of (1) : Plug (2) into (1), you will get :
$$\underbrace{L L^T(L L^T)^{-1}}_{\text{identity matrix} \ I} P=P...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037758",
"timestamp": "2023-03-29T00:00:00",
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What is the Range of $5|\sin x|+12|\cos x|$ What is the Range of $5|\sin x|+12|\cos x|$ ?
I entered the value in desmos.com and getting the range as $[5,13]$.
Using $\sqrt{5^2+12^2} =13$, i am able to get maximum value but not able to find the minimum.
| If $f(x) = 5|\sin x| + 12 |\cos x|$, then
\begin{align*}
f(x) &= \sqrt{f(x)^2} \\
&= \sqrt{25 \sin^2 x + 144 \cos^2 x + 60 |\sin x \cos x|} \\
&= \sqrt{25 + (144 - 25) \cos^2 x + 60 |\sin x \cos x|} \\
&\ge 5
\end{align*}
with equality obtained when $\cos x = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3040110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many $n$-digit sequences of $0,1 \ \text{or} \ 2$s contain an odd numbers of $0$s? Problem : There are $3^n$ n-digit sequences in which each digit is $0$, $1$ or $2$. How many of these sequences have an odd number of $0$'s ?
Let $o(n)$ = the number of n-digit sequences which have an odd number of $0$'s and
$e(n)$ = the number of n-digit sequences which have an even number of $0$'s.
Obviously $o(n) + e(n) = 3^n$ and by examples I can see that $o(n)=e(n)-1$ but I don't know how to show this.
Moreover if we have $x$ digits to choose from then by examples I see that $o(n) = e(n) - {n}^{x-2}$. And if we consider the number of sequences which have a number congruent to $0$, $1$ respectively $2$ (modulo $3$) of $0$s we see that these numbers are in arithmetic progression.
Can someone prove these assumptions to me or disprove them if they are not right?
| Assume that $n$ is odd, $n = 2s+1$
Note that number of $n$-digit sequences with $k$ number of zeroes is ${N_k}(n) =$ $n \choose k$$2^{n-k}$.
Thus, $$o(n) = \sum_{m=0}^s{N_{2m+1}}(n) = \sum_{m=0}^s{n \choose 2m+1}{2^{n-2m-1}} = \sum_{m=0}^s{n \choose 2m+1}{1^{2m+1} 2^{n-2m-1}} \\ = \frac{(2+1)^n-(2-1)^n}{2} = \frac{3^n-1}{2}$$
If $n$ is even, let $n = 2t$. Then $o(n) = \sum_{m=0}^{t-1}{n \choose 2m+1}{2^{n-2m-1}} = \frac{3^n-1}{2}$, and you are done.
Moreover, $$o(n-1) = \frac{3^{n-1}-1}{2} \implies \fbox{$o(n-1) + 3^{n-1} = o(n)$}$$
Similarly, for $e(n)$, $$e(n)=\begin{cases}{\sum_{m=0}^s{n \choose 2m}{2^{n-2m}}}&n=2s+1 \\ \sum_{m=0}^t{n \choose 2m}{2^{n-2m}}& n = 2t\end{cases}$$
Thus, $$e(n) = \frac{(2+1)^n+(2-1)^n}{2} = \frac{3^n+1}{2}$$
Now, by further computation, one can find out that $\fbox{$e(n)-o(n) = 1$}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3041858",
"timestamp": "2023-03-29T00:00:00",
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Find the value of $1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$
Find the value of this :$$1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$$
Try: We can write the above series as
$${S} = \int^{1}_{0}\bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+\cdots\bigg]dx$$
$$S = \int^{1}_{0}(1-x^6)\bigg[1+x^{8}+x^{16}+\cdots \cdots \bigg]dx$$
So $$S = \int^{1}_{0}\frac{1-x^6}{1-x^{8}}dx = \int^{1}_{0}\frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$
Now i am struck in that integration.
Did not understand how to solve it
could some help me to solve it. Thanks in advance
| Hint: your fraction is $(x^2+1)^{-1}+\frac{x^2}{x^4+1}$, and $\frac{x^2}{x^4+1}=\frac{ax+b}{x^2-\sqrt{2}x+1}+\frac{-ax-b}{x^2+\sqrt{2}x+1}$ for some real numbers $a,b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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} |
Simplify the determinant of a $4 \times 4$ matrix. I have to find the determinant of the following 4x4 matrix:
$\quad A=\begin{bmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{bmatrix}$
So I apply the Gaussian elimination to obtain an upper-triangle matrix:
$$det\begin{bmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{bmatrix}=\begin{vmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{vmatrix}\xrightarrow{3R_3-R_1}\begin{vmatrix}3&0&1&0\\0&2&0&0\\0&0&8&0\\0&0&0&-4\end{vmatrix}$$
Since I know from the solutions that the determinant is -64, I suppose that I need to simplify the third row in the reduced form to $\quad 0 \quad 0 \quad 2 \quad 0 \quad$ and then multiply the elements in the upper-left-to-bottom-right diagonal, which is indeed -64. But this doesn't make much sense since there's also a $-4$ that we can simplify. Can someone explain me the actual rules we need to follow?
| Just expand by the first row:
\begin{align}
\begin{vmatrix}3&0&1&0 \\
0&2&0&0 \\
1&0&3&0 \\
0&0&0&-4
\end{vmatrix}&=
3\,\begin{vmatrix}
2&0&0 \\
0&3&0 \\
0&0&-4
\end{vmatrix}+
1\,\begin{vmatrix}
0&2&0 \\
1&0&0 \\
0&0&-4
\end{vmatrix}
=3(2\cdot3\cdot(-4))+1(-1)\begin{vmatrix}
2&0 \\
0&-4
\end{vmatrix}\\
&=-72+8=-64.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
How to evaluate $\int \frac{x^3}{\sqrt {x^2+1}}dx$ Evaluate $$\int \frac{x^3}{\sqrt {x^2+1}}dx$$
Let $u={x^2}+1$. Then $x=\sqrt {u-1}$ and $dx=\frac{1}{2\sqrt{u-1}}du$.
Therefore, $$\int \frac{(u-1)\sqrt{u-1}}{\sqrt u}\frac{1}{2\sqrt {u-1}}du$$
$$=\frac{1}{2}\int {\sqrt u}-{\frac {1}{\sqrt u}du}$$
$$={\frac{2}{3}}{{(x^2+1})}^{3/2}-{\frac{1}{2}}{\sqrt {x^2+1}}+C$$for some constant $C$
| Just to give another approach, let $x^2+1=u^2$, in which case $xdx=udu$ and we have
$$\int{x^3\over\sqrt{x^2+1}}dx=\int{x^2\cdot xdx\over\sqrt{x^2+1}}=\int{(u^2-1)udu\over u}=\int(u^2-1)du={1\over3}u^3-u+C\\={1\over3}(x^2+1)^{3/2}-(x^2+1)^{1/2}+C$$
The main virtue (if any) of this approach is that it gets rid of the square root symbol for the integration step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
exponential equation has non positive roots
Find real values of $a$ for which the equation
$4^x-(a-3)\cdot 2^x+(a+4)=0$
has non positive roots
Try: Let $2^x=y\in (0,1].$ Then equation convert into
$y^2-(a-3)y+(a+4)=0$
For real roots its discriminant $\geq 0$
So $$(a-3)^2-4(a+4)\geq 0$$
$$a^2-10a-7\geq 0$$
$$a\in \bigg(-\infty,5-4\sqrt{2}\bigg]\cup \bigg[5+4\sqrt{2},\infty\bigg).$$
but answer is different from that , i did not know where i
am missing. could some help me to solve it. thanks
| You want $f(y) = y^2 - (a-3) y + (a+4)$ to have at least one root in $(0,1]$.
Note that $f(0) = a+4$ and $f(1) = 8$, while the minimum value of $f(y)$, occurring at
$y = (a-3)/2$, is $(-a^2 + 10 a + 7)/4$. There are several cases to consider.
*
*If $a < -4$, $f(0) < 0 < f(1)$ so there is a root in $(0,1]$.
*For $a = -4$, $f(y) = y^2 + 7 y$ so the roots $0$ and $-7$ are not in $(0,1]$.
*For $a > -4$, $f(0) > 0$ and $f(1) > 0$ so the only way to have a root in $(0,1]$ would be if the minimum occurs between $0$ and $1$ and the minimum value $\le 0$. The
minimum is at $(a-3)/2$, so this requires $a \in (3,5]$. But for $a$ in this interval,
$-a^2 + 10 a + 7 > 0$.
So the answer is $a < -4$.
Alternative method:
For given $y$, $f(y) = 0$ for $a = A(y) = 4 + y + 8/(y-1)$. Thus you want the set of values $A(y)$ for $0 < y < 1$ (of course $A(1)$ is undefined).
$A'(y) = \dfrac{y^2 - y - 7}{(y-1)^2} < 0$ in this interval, so $A$ is a decreasing function of $y$, with $A(y) \to -\infty$ as $y \to 1-$. Thus the answer is $(-\infty, A(0)) = (-\infty, -4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
what went wrong in solving this singular ODE? I want to solve $$x(2-x) y'' +(1+x)y' -\frac{3}{x} y = 0$$ with singular-regular point $x_0 =0$, the general solution by Frobenius is of the form $y = x^r \sum \limits_{n=0}^{\infty} a_n x^n$. i found that $r = -1, \frac{3}{2}$. substituting the value $r=-1$ and the standard value $a_0 =1$ gave me that $$a_{n+1} = \frac{n^2-4n+3}{2n^2-n-3}a_n$$ which means that $a_1 = a_2 = \cdots =0$
So in the end i have one solution which is $y = \frac{1}{x}(1+0 x^1 +0 x^2+\cdots) = \frac{1}{x}$ but the correct solution is $y = \frac{1}{x}-1$, what went wrong in my solution ?
| You have
$$
a_{n + 1} = a_n\frac{n^2 - 4n + 3}{2n^2 - n - 3}
$$
Using $a_0 = \color{blue}{1}$ you get
$$
a_1 = 1 \frac{3}{(-3)} = \color{red}{-1}
$$
and
$$
a_2 = 0 = a_3 = \cdots
$$
So one of the solutions is
$$
y_1(x) = x^{-1}\left(a_0 + a_1 x + a_2 x^2 + \cdots \right) = x^{-1}\left[\color{blue}{1} + (\color{red}{-1})x\right] = -1 + 1/x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve $T(n) = T(2n/3) + \lg^2 n$ by substitution? My solution through substitution is as follows:
$$T(n) = T(2n/3) + \lg^2 (n)$$
$$T(2n/3) = T(4n/9) + \lg^2 (2n/3)$$
$$T(4n/9) = T(8n/27) + \lg^2 (4n/9)$$
And so on...
But my actual problem is how can I calculate the below step which cause to obtain order of the above expression:
$$\lg^2 \left(n\cdot(2/3)n\cdot(2/3)^2n\cdot(2/3)^3n\cdots\right).$$
Also I know the order is $\theta(\lg^3n)$.
Thanks!
| $$T(n)=T\left(\frac{2n}{3}\right)+\lg^2 n$$
$$T\left(\frac{2n}{3}\right)=T\left(\frac{2\cdot\frac{2n}{3}}{3}\right)+\lg^2\left(\frac{2n}{3}\right)=T\left(\frac{2^2n}{3^2}\right)+\lg^2\left(\frac{2n}{3}\right)$$
$$T\left(\frac{2^2n}{3^2}\right)=T\left(\frac{2^3n}{3^3}\right)+\lg^2\left(\frac{2^2n}{3^2}\right)$$
$$\dots \ T\left(\frac{2^{q-1}\cdot n}{3^{q-1}}\right)=\ T\left(\frac{2^{q}\cdot n}{3^{q}}\right)+\lg^2 \left(\frac{2^{q-1}\cdot n}{3^{q-1}}\right) $$
Now in order to obtain $T(1)$ we consider the limit condition:
$$T\left(\frac{2^q\cdot n}{3^q}\right)=T(1)\Rightarrow \left(\frac23\right)^qn=1\Rightarrow n=\left(\frac{3}{2}\right)^q\Rightarrow \log_{\frac32}n=q$$
$$\Rightarrow T(n)=T(1)+\lg^2 n+\lg^2\left(\frac{2n}{3}\right)+\dots +\lg^2 \left(\frac{2^{q-1}\cdot n}{3^{q-1}}\right) $$
Now we write the above as a sum. Also the time complexity of $T(1)$ is just $\Theta(1)$.
$$T(n)=\Theta(1)+\sum_{k=0}^{q-1}\lg^2\left(\left(\frac{2}{3}\right)^kn\right),\quad q=\log_{\frac32}n$$
Now we have using some basic algebra and properties of the logarithms: $$\left(\lg\left(\left(\frac23\right)^k \cdot n\right)\right)^2=\left(\lg\left(\frac23\right)^k+ \lg n\right)^2=\left(k\lg\left(\frac23\right)+ \lg n\right)^2=$$
$$=k^2 \lg^2\left(\frac23\right)+2k\lg\left(\frac23\right)\lg n+\lg^2n$$
$$\Rightarrow T(n)=\Theta(1)+\lg^2\left(\frac23\right)\sum_{k=0}^{q-1}k^2+2\lg\left(\frac23\right)\lg n\sum_{k=0}^{q-1} k +\lg^2n\sum_{k=0}^{q-1}1$$
$$=\Theta(1)+\lg^2\left(\frac23\right)\frac{(q-1)q(2q-1)}{6}+2\lg\left(\frac23\right)\lg n\frac{(q-1)q}{2}+\lg^2 n (q-1)$$
Now note that $\ \displaystyle{q=\log_{\frac23}{n}=\frac{\lg n}{\lg\frac23}}\,$ and to obtain the time complexity constants don't matter.
$$T(n)= \Theta(1)+c_1 (\lg n-1)\lg n(2\lg n-1)+c_2 \lg n\cdot (\lg n-1)\lg n +c_3 \lg^2n\cdot (\lg n-1) $$
$$T(n)=\Theta(1)+c_1\Theta(\lg^3 n)+c_2\Theta(\lg^3 n)+c_3\Theta(\lg^3n)=\Theta(\lg^ 3 n)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root,then $a+b+c$ can be equal to If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root, then $a+b+c$ can be equal to(where $a,b,c\in R,a\ne0$)
$(1)\;5a$
$(2)\;3b$
$(3)\;2c$
$(4)\;0$
As $x=1$ is the root of the equation $2x^3+x^2+2x-5=0$ and its other two roots are complex in nature. So the common root is $x=1$
Put $x=1$ in $ax^3+(a+b)x^2+(b+c)x+c=0$ we get $2a+2b+2c=0$ so $(4)$ option is correct. But $(1)$ and $(3)$ options are also correct. I don't know how they are correct.
| The equation $2x^3+x^2+2x−5=0$ has 3 solutions, $$1, \quad \frac14 \left(-3-i\sqrt{31} \right), \quad \mbox{and} \quad \frac14 \left(-3+i\sqrt{31} \right).$$ If you put $x=1$ (as you did) you obtained $a+b+c=0$. In the same way if you put $x=\frac14 \left(-3-i\sqrt{31} \right)$ (to check your computation maybe use wolframaplha) you get $2b=3a$ and $2c=5a$, which is equivalent to $$a+b+c=5a.$$ I let you do on your own the last case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Equivalence of two antiderivatives involving trigonometric/hyperbolic functions I am struggling to see how two antiderivatives of the same function—obtained in two different ways—are equivalent (what I mean by equivalent is that they differ from just a constant), if they even are equivalent.
The function in question is
$$
\begin{align}
f \colon \quad \{x\in\mathbb{R} \mid e^{2x} \ge 9\} &\to\mathbb{R}\\
x &\mapsto f(x) = \frac{1}{\sqrt{e^{2x} - 9}}
\end{align}\,.
$$
Wanting to find $I = \displaystyle\int f(x)\mathrm{d}x$, I proceeded as follows.
$$
\int \frac{1}{\sqrt{e^{2x} - 9}}\mathrm{d}x = \frac{1}{3} \int \frac{1}{\sqrt{\left(\dfrac{e^x}{3}\right)^2 - 1}}\mathrm{d}x\,.
$$
Since $\left(\frac{e^x}{3}\right)^2 \in[1,+\infty]$ and the image of $\cosh$ is $[1, +\infty)$,
$$\exists t \in \mathbb{R}\colon\quad \left(\dfrac{e^x}{3}\right)^2 = \cosh^2t\\
\Rightarrow \mathrm{d}x = \tanh t\ \mathrm{d}t\,.$$
Thus,
$$
I = \frac{1}{3} \int \frac{\tanh t}{\sqrt{\cosh^2 t - 1}}\mathrm{d}t = \frac{1}{3}\int \mathrm{sech}\ t\ \mathrm{d}t\,.
$$
Here is where the two different ways diverge. The first thing I tried was using the hyperbolic function's definition.
$$
I = \frac{1}{3}\int \frac{2e^t}{e^{2t} + 1}\ \mathrm{d}t\,;
\qquad\text{let}\quad
\begin{cases}
u = e^t\\
\mathrm{d}u = e^t\mathrm{d}t
\end{cases}\\ \\
\Rightarrow I = \frac{2}{3} \int \frac{1}{u^2 + 1}\ \mathrm{d}u = \frac{2}{3} \arctan u + C = \frac{2}{3} \arctan e^t + C = \\
= \frac{2}{3} \arctan e^{\text{arccosh} \frac{e^x}{3}} + C = \frac{2}{3} \arctan e^{\ln\left(\frac{e^x + \sqrt{e^{2x}-9}}{3}\right)} + C =\\
= \frac{2}{3} \arctan \left(\frac{e^{x} + \sqrt{e^{2x}-9}}{3}\right) + C\,.
$$
The alternative solution is
$$
I = \frac{1}{3}\int \mathrm{sech}\ t\ \mathrm{d}t = \frac{1}{3}\int \frac{\cosh t}{\cosh^2 t}\ \mathrm{d}t = \frac{1}{3}\int \frac{\cosh t}{1 +\sinh^2 t}\ \mathrm{d}t =\\
= \frac{1}{3}\arctan \left(\sinh t\right) + C = \frac{1}{3}\arctan \left(\sinh \left(\mathrm{arccosh} \frac{e^x}{3}\right)\right) + C = \frac{1}{3}\arctan \frac{\sqrt{e^{2x} - 9}}{3} + C\,.
$$
I don't see any obvious way in which $$\frac{2}{3} \arctan \left(\frac{e^x + \sqrt{e^{2x}-9}}{3}\right) + C$$ and $$\frac{1}{3}\arctan \frac{\sqrt{e^{2x} - 9}}{3} + C$$ are equivalent antiderivatives, although they do seem to be.
Update
Following Yuriy S's suggestion, I tried the following. Let
$$
s = \frac{\sqrt{e^{2x} - 9}}{3}\,.
$$
Thus,
$$
\arctan s = 2\arctan \frac{\sqrt{1+s^2}-1}{s} = 2\arctan \frac{\sqrt{1+\frac{e^{2x}-9}{9}}-1}{\frac{\sqrt{e^x - 9}}{3}} = 2\arctan \frac{3e^x-3}{\sqrt{e^x - 9}}\,.
$$
On the other hand, let
$$
r = \frac{e^x + \sqrt{e^{2x} - 9}}{3}\,,
$$
so that
$$
\frac{1}{r} = \frac{3}{e^x + \sqrt{e^{2x} - 9}} \cdot \frac{e^x - \sqrt{e^{2x} - 9}}{e^x - \sqrt{e^{2x} - 9}} = \frac{e^x - \sqrt{e^{2x} - 9}}{3}\,.
$$
I was striving to apply the formula
$$
\arctan x + \arctan \frac{1}{x} = \pm\frac{\pi}{2}
$$
but I got stuck here.
| Whenever you face this problem, you can check the answer by differentiating. If
$$
F(x)=\frac{2}{3} \arctan \left(\frac{e^{x} + \sqrt{e^{2x}-9}}{3}\right) + C
$$
then
\begin{align}
F'(x)
&=\frac{2}{3}\frac{1}{1+\left(\dfrac{e^{x} + \sqrt{e^{2x}-9}}{3}\right)^2}
\frac{1}{3}\left(e^x+\frac{e^{2x}}{\sqrt{e^{2x}-9}}\right) \\[4px]
&=\frac{2}{9+e^{2x}+e^{2x}-9+2e^x\sqrt{e^{2x}-9}}
\frac{e^x(e^x+\sqrt{e^{2x}-9}\,)}{\sqrt{e^{2x}-9}} \\[4px]
&=\frac{1}{\sqrt{e^{2x}-9}}
\end{align}
We can try differentiating the second function
$$
G(x)=\frac{1}{3}\arctan\frac{\sqrt{e^{2x}-9}}{3}+C
$$
and we get
\begin{align}
G'(x)
&=\frac{1}{3}\frac{1}{1+\left(\dfrac{\sqrt{e^{2x}-9}}{3}\right)^2}
\frac{1}{3}\frac{e^{2x}}{\sqrt{e^{2x}-9}} \\[4px]
&=\frac{1}{9+e^{2x}-9}\frac{e^{2x}}{\sqrt{e^{2x}-9}} \\[4px]
&=\frac{1}{\sqrt{e^{2x}-9}}
\end{align}
Good job in both cases.
You can easily compute the constant difference by the limits at $\infty$ (with $C=0$ in both cases):
$$
\lim_{x\to\infty}F(x)=\frac{2}{3}\frac{\pi}{2}=\frac{\pi}{3}
\qquad
\lim_{x\to\infty}G(x)=\frac{1}{3}\frac{\pi}{2}=\frac{\pi}{6}
$$
Alternative solution: substitute $\sqrt{e^{2x}-9}=3t$, so
$$
x=\frac{1}{2}\log(9(t^2+1))
$$
and
$$
dx=\frac{t}{t^2+1}\,dt
$$
so the integral becomes
$$
\int\frac{1}{t^2+1}\,dt=\arctan t+C=\arctan\frac{\sqrt{e^{2x}-9}}{3}+C
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minimizing area of ellipse Find minimum area of an ellipse that can pack three unit circles such that all three touch the ellipse internally:
I took a point H as shown in the diagram and used the fact that the radius of the circle is 1, and that the circle touches the ellipse at point H. I am getting four equations for five unknowns, which means I can derive a relation between a and b of ellipse and use calculus to minimize the area. But those equations are tedious to solve and even after hours, I am not able to solve them.
Is there any easier way to solve this?
Here are the equations which I've got:
| Let me try to help with the discriminant.
Start with $$\frac{x^2}{a^2}+ \frac{y^2}{(1+2\cosθ)^2}=1$$
Put $$y^2=1−(x−2 \sinθ)^2$$
So the equation becomes
$$\frac{x^2}{a^2}+ \frac{1−(x−2 \sinθ)^2}{(1+2\cosθ)^2}=1$$
Compare it with the standard quadratic equation $px^2 + qx + r = 0$. If the discriminant is zero, then the roots are equal and $q^2 = 4pr$
Here, $$p = \frac{1}{a^2} - \frac{1}{(1+2 \cos θ)^2}$$
$$q = \frac{4 \sin θ}{(1+2 \cos θ)^2}$$
$$r = \frac{1- 4 \sin^2 θ}{(1+2 \cos θ)^2}-1$$
Now $q^2 = 4pr \implies$
$$\frac{16 \sin^2 θ}{(1+2 \cos θ)^4} = 4 \left(\frac{1}{a^2} - \frac{1}{(1+2 \cos θ)^2}\right) \left(\frac{1- 4 \sin^2 θ}{(1+2 \cos θ)^2}-1\right) (1)
$$
Next, observe that $$1- 4 \sin^2 θ - (1+2 \cos θ)^2 = -4(1+ \cos θ)$$
Use this in (1) above:
$$\frac{16 \sin^2 θ}{(1+2 \cos θ)^4} = 4 \left(\frac{1}{a^2} - \frac{1}{(1+2 \cos θ)^2}\right) \left(\frac{-4(1+ \cos θ)}{(1+2 \cos θ)^2}\right) (2)
$$
Next cancel $16$ from both LHS and RHS and write $$ \sin^2 \theta = (1 - \cos \theta)(1 + \cos \theta)$$ on the LHS of (2) to obtain
$$\frac{(1 - \cos \theta)(1 + \cos \theta)}{(1+2 \cos θ)^4} = \left(\frac{1}{a^2} - \frac{1}{(1+2 \cos θ)^2}\right) \left(\frac{-(1+ \cos θ)}{(1+2 \cos θ)^2}\right) (3) $$
Next cancel $(1 + \cos \theta)$ from both LHS and RHS of (3) and rearrange to obtain
$$ \frac{1}{a^2} = \frac{\cos \theta}{(1+2 \cos \theta)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Evaluate $\int_0^1 \frac{\operatorname{arctanh}^3(x)}{x}dx$ I'm trying to evaluate the following integral:
$$\int_0^1 \frac{\operatorname{arctanh}^3(x)}{x}dx$$
I was playing around trying to numerically approximate the answer with known constants and found that the integral is almost exactly $\frac{\pi^4}{64}$. The integral seems to break down after about $11$ decimal places.
I have a suspicion that this stems from the integral:
$$\int_0^1 \frac{\operatorname{arctanh}(x)}{x}dx$$
since this is exactly equal to $\frac{\pi^2}{8}$.
Also for the integral:
$$\int_0^1 \frac{\operatorname{arctanh}^5(x)}{x}dx$$
This is suspicously close to $\frac{\pi^6}{128}$ but not exactly.
For some reason the above integrals diverges slightly from the some from of $\frac{\pi^n}{2^m}$ for some $n$ and $m$.
My question is: Why does this happen? And what are the true values of those integrals?
| As MrTaurho pointed out in the comments, we can rewrite $\,\displaystyle{\operatorname{arctanh}x=\frac12 \ln\left(\frac{1+x}{1-x}\right)}$, this gives:
$$I=\int_{0}^1 \frac{\operatorname{arctanh}(x)^3}{x}dx=-\frac18\int_0^1 \frac{\ln^3\left(\frac{1-x}{1+x}\right)}{x}dx$$
And we will substitute $\displaystyle{\frac{1-x}{1+x}=t\Rightarrow dx=-\frac{2}{(1+t)^2}dt}$ in order to get:
$$I=-\frac18\int_0^1 \frac{\ln^3 t}{\frac{1-t}{1+t}}\frac{2}{(1+t)^2}dt=-\frac14 \int_0^1 \frac{\ln^3 t}{1-t^2}dt=-\frac14 \sum_{n=0}^\infty \int_0^1 t^{2n}\ln^3t dt$$
Now we consider the following integral: $$\int_0^1 t^a dt=\frac{1}{a+1}\Rightarrow \int_0^1 t^a \ln^3 tdt=\frac{d^3}{da^3} \left(\frac{1}{a+1}\right)=-\frac{6}{(a+1)^4}$$
$$\Rightarrow I=\frac{6}{4}\sum_{n=0}^\infty \frac{1}{(2n+1)^4}=\frac32\cdot\frac{\pi^4}{96}=\frac{\pi^4}{64}$$
Of course this can be generalized for any power, so let's do that. Consider:
$$I(k)=\int_0^1 \frac{\text{arctanh}^kx}{x}dx=\frac{(-1)^k}{2^k}\int_0^1 \frac{\ln^k\left(\frac{1-x}{1+x}\right)}{x}dx\overset{\large\frac{1-x}{1+x}=t}=\frac{2(-1)^k}{2^k}\int_0^1 \frac{\ln^k t}{1-t^2}dt$$
$$=\frac{(-1)^k}{2^{k-1}}\sum_{n=0}^\infty \int_0^1 x^{2n}\ln^k xdx=\frac{(-1)^k}{2^{k-1}}\sum_{n=0}^\infty \frac{(-1)^k k!}{(2n+1)^{k+1}}=\frac{k!}{2^{k-1}} \sum_{n=0}^\infty \frac{1}{(2n+1)^{k+1}}$$
Where above we used the following result: $$\int_0^1 t^a dt=\frac{1}{a+1}\Rightarrow \int_0^1 t^a \ln^k tdt=\frac{d^k}{da^k} \left(\frac{1}{a+1}\right)=\frac{(-1)^k k!}{(a+1)^{k+1}}$$
And finally it reduces to: $$I(k)=\frac{k!}{2^{k-1}} \left(1-\frac{1}{2^{k+1}}\right)\zeta(k+1)=\boxed{\frac{k!\left(2^{k+1}-1\right)}{4^k}\zeta(k+1)}$$
One can verify the result by comparing to the one announced by Maple: $$I(5)=\int_0^1 \frac{\text{arctanh}^5 x}{x}dx=\frac{5!(2^6-1)}{4^5}\zeta(6)=\frac{945}{128}\cdot\frac{\pi^6}{945}=\frac{\pi^6}{128}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})\cdots(1+\frac{1}{a_n})$, where $a_1=1$, $a_n=n(1+a_{n-1})$
Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})(1+\frac{1}{a_2})\cdots(1+\frac{1}{a_n})$,
where $a_1 = 1$, $a_n = n(1+a_{n-1})$
\begin{align*}
&\lim_{n\to\infty} \left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right)\cdots\left(1+\frac{1}{a_n}\right) \\
&= \lim_{n\to\infty} \frac{1}{a_1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdots \frac{1}{n} \cdot (a_n + 1) \\
&= \lim_{n\to\infty} \frac{a_n + 1}{n!}.
\end{align*}
Then I'm stuck. How to proceed?
| Show by induction that
$a_1=1!/0!$
$a_2=2!/0!+2!/1!$
$a_3=3!/0!+3!/1!+3!/2!$
...
$\color{blue}{a_n=n!/0!+n!/1!+n!/2!+...+n!/(n-1)!}$
Thereby the limit is $1/0!+1/1!+2/2!+...1/n!+...=e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Minimum value of $\sqrt {(x+2)^2+(y+2)^2}+\sqrt {(x+1)^2+(y-1)^2}+\sqrt{ (x-1)^2+(y+1)^2}$ Let $x;y\in R$. Find Minimum value of the function $$\sqrt {(x+2)^2+(y+2)^2}+\sqrt {(x+1)^2+(y-1)^2}+\sqrt{ (x-1)^2+(y+1)^2}$$
My try: By Minkowski inequality:
$LHS=\sqrt{(x+1)^2+(y-1)^2}+\sqrt{(x-1)^2+(y+1)^2}+2\sqrt{\dfrac{1}{4}[(x+2)^2+(y+2)^2]}$
$=\sqrt{(x+1)^2+(y-1)^2}+\sqrt{(x-1)^2+(y+1)^2}+2\sqrt{(\dfrac{x}{2}+1)^2+(\dfrac{y} {2}+1)^2]}$
$\ge \sqrt{(\dfrac{3x}{2}+2)^2+(\dfrac{3y}{2})^2}+\sqrt {(\dfrac{3y}{2}+2)^2+(\dfrac{3x}{2})^2}\ge \sqrt{8}$
And the equality occurs when $x=-y-\frac {4}{3}$
Help me check it. I fear that is not minimum value of the function. THx.
| FERMAT POINT
I get, from the $120^\circ$ construction, that the minimizing point is at $$ \left(\frac{-1}{\sqrt 3},\frac{-1}{\sqrt 3} \right)$$
The main calculation I did was $45^\circ + 60^\circ = 105^\circ$ and $\tan 105^\circ = -2-\sqrt 3.$ The sum of three distances is
$$ \sqrt 6 + \sqrt 8 \approx 5.277916867529368195800661523 $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
How to solve this ode $xy''-(1+x)y'+y=x^2$? Find the general solution of $xy''-(1+x)y'+y=x^2$ knowing that the homogeneous equation has the following solution: $e^{ax}$, where $a$ is a parameter you have to find.
I have found that $a=1$ or $a=1/x$.
| The equation $xy''-(1+x)y'+y=x^2$ can be rewritten as $(xy''-y')+(y-xy')=x^2.$ After division by $x^2$ we have
$$
\frac{xy''-y'}{x^2} + \frac{y-xy'}{x^2} = 1,
$$
which can be rewritten as
$$
\Big( \frac{y'}{x} \Big)' - \Big( \frac{y}{x} \Big)' = 1.
$$
Taking the antiderivative gives
$$
\frac{y'}{x} - \frac{y}{x} = x + C,
$$
which after multiplication with $x$ gives
$$y' - y = x^2+ Cx.$$
Now we multiply with the integrating factor $e^{-x}$:
$$e^{-x} y' - e^{-x} y = x^2 e^{-x} + Cx e^{-x}.$$
The left hand side can be rewritten as a derivative:
$$( e^{-x} y )' = x^2 e^{-x} + Cx e^{-x}.$$
Taking the antiderivative gives
$$e^{-x} y = -(x^2+2x+2) e^{-x} - C(x+1) e^{-x} + D.$$
Thus, all solutions to the original differential equation are given by
$$y = -(x^2+2x+2) - C(x+1) + D e^x,$$
where $C$ and $D$ are two constants.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Evaluating an improper integral $\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$ I tried to solve the integral: $$\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$$
using $ x = \sqrt{\tan(u)}$ and $dx = \frac{ \sec^2(u)}{2\sqrt{\tan(u)}} du,$
but I ended up with an even worse looking integral $$ \int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\tan(u)}}{\sec^2(u)}du.$$
Wolfram gave an answer of $ \dfrac{\pi}{8\sqrt{2}},$ but how would one get to that answer?
| Using the method I employed here: we observe that
\begin{equation}
\int_{0}^{\infty} \frac{x^2}{\left(x^4 + 1\right)^2}\:dx = \frac{1}{4} \cdot 1^{\frac{2 + 1}{4} - 2} \cdot B\left(2 - \frac{2 + 1}{4}, \frac{2 + 1}{4} \right) = \frac{1}{4}B\left(\frac{5}{4}, \frac{3}{4}\right)
\end{equation}
Using the relationship between the Beta and Gamma function:
\begin{align}
\int_{0}^{\infty} \frac{x^2}{\left(x^4 + 1\right)^2}\:dx &= \frac{1}{4}B\left(\frac{5}{4}, \frac{3}{4}\right) = \frac{1}{4}\frac{\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4} + \frac{3}{4}\right)} \\
&= \frac{1}{4}\frac{\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(2\right)} =\frac{\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{3}{4}\right)}{8}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
} |
How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$ Wolfram Alpha shows that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\left(\frac{1}{n}-\frac{1}{n+1}\right)\\
&=H_{1}\left(1-\frac{1}{2}\right)+H_{2}\left(\frac{1}{2}-\frac{1}{3}\right)+H_{3}\left(\frac{1}{3}-\frac{1}{4}\right)\\
&=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{6}+\frac{1}{3}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{9}-\frac{1}{12}
\end{align}$$
I think this method is not quite useful, so I tried another one:
$$H_n=\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
Then,
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\sum_{n=1}^{\infty}\frac{1}{n^2+n}\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
At this point, I do not know how to proceed.
| We have
\begin{align}
\sum_{n=1}^\infty\frac{H_n}{n^2+n}
&=\sum_{n=1}^\infty H_n\left(\frac1n-\frac1{n+1}\right)
\\&=\sum_{n=1}^\infty\sum_{m=1}^n\frac1m\left(\frac1n-\frac1{n+1}\right)
\\&=\sum_{m=1}^\infty\frac1m\sum_{n=m}^\infty\left(\frac1n-\frac1{n+1}\right)
\\&=\sum_{m=1}^\infty\frac1m\cdot\frac1m
\\&=\frac{\pi^2}6.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
} |
Prove the identity for $\tan3\theta$
Prove the identity for $$\tan3\theta= \frac{3\tan\theta - \tan^3 \theta}{1-3\tan^2 \theta}$$
Using de Moivre's theorem I have found that:
$$\cos3\theta = 4\cos^3\theta - 3\cos \theta$$
$$\sin 3\theta = 3\sin \theta-4\sin^3\theta$$
therefore:
$$\tan 3\theta = \frac{\sin 3\theta}{\cos 3 \theta}=\frac{3\sin \theta-4\sin^3\theta}{4\cos^3\theta - 3\cos \theta}$$
To then try and get the whole expression in terms of $\tan\theta$ I multiplied top and bottom of the fraction by $(4\cos^3\theta)^{-1}$. This gave me the following but I'm not now sure how to finish it off
$$\tan3\theta =\frac{\frac{3\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos \theta}{4\cos^3 \theta}}$$
| A slightly different approach using de Moivre's theorem would be: notice that $1 + i \tan \theta = \sec \theta \cdot e^{i \theta}$. Therefore, cubing both sides, $(1 + i \tan \theta)^3 = \sec^3 \theta \cdot e^{3 i \theta}$. Now, if you take the "slope" of this complex number, i.e. the imaginary part divided by the real part, from the right hand side you see the result will be $\tan(3\theta)$. On the other hand, by binomial expansion,
$$(1 + i \tan \theta)^3 = 1 + 3i \tan \theta - 3 \tan^2 \theta - i \tan^3 \theta = (1 - 3 \tan^2 \theta) + i (3 \tan \theta - \tan^3 \theta).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
The closed form representations of Integrals of logarithm functions I wish to find a closed form representations of the following integral
$$\int\limits_{0}^1\frac{\log^p(x)\log^r\left(\frac{1-x}{1+x}\right)}{x}dx=?$$
Here $p\ge 1$ and $r\ge 0$ are nonnegative integers. It can be expressed in terms of a linear combination of well known constants (such as: Riemann zeta values,$\pi$ et. al.)?
| Here is an alternative approach for the case $r = 1, p \in \mathbb{N}$ that has already been given by @Ininterrompue.
Let
$$I_p = \int_0^1 \ln^p x \ln \left (\frac{1 - x}{1 + x} \right ) \frac{dx}{x}.$$
Since
$$\ln \left (\frac{1 - x}{1 + x} \right ) = - 2 \sum_{n = 0}^\infty \frac{x^{2n+1}}{2n + 1}, \qquad |x| < 1,$$
the integral may be written as
$$I_p = -2 \sum_{n = 0}^\infty \frac{1}{2n + 1} \int_0^1 x^{2n} \ln^p x \, dx.$$
Integrating by parts $p$-times gives
\begin{align}
I_p &= - 2 (-1)^p p! \sum_{n = 0}^\infty \frac{1}{(2n + 1)^{p + 2}}\\
&= -\frac{(-1)^p p!}{2^{p + 1}} \sum_{n = 0}^\infty \frac{1}{(n + 1/2)^{p + 2}}\\
&= -\frac{(-1)^p p!}{2^{p + 1}} \cdot \frac{(-1)^{p + 2}}{(p + 1)!} \psi^{(p + 1)} \left (\frac{1}{2} \right )\\
&= -\frac{1}{(p + 1) 2^{p + 2}} \psi^{(p + 1)} \left (\frac{1}{2} \right ). \tag1
\end{align}
Here $\psi^{(m)} (x)$ is the polygamma function with the following series representation of
$$\psi^{(m)} (z) = (-1)^{m + 1} m! \sum_{n = 0}^\infty \frac{1}{(n + z)^{m + 1}},$$
having been used.
Finally, as (see here)
$$\psi^{(m)} \left (\frac{1}{2} \right ) = (-1)^{m + 1} m! (2^{m + 1} - 1) \zeta (m + 1),$$
in terms of the Riemann zeta function, $\zeta (z)$, one can rewrite (1) as
$$I_p = (-1)^{p + 1} p! \left (1 - \frac{1}{2^{p + 2}} \right ) \zeta (p + 2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to find a recurrence relation for counting the number of solutions? Consider the diophantine equation
$$x_1+3x_2+5x_3 = n$$
where $x_i\geq 0$ and $n\geq 1.$
Let $P_n(1,3,5)$ denote the number of solutions to this equation. I want to express $P_n(1,3,5)$ in terms of $P_{1}(1,3,5), P_{2}(1,3,5),\cdots, P_{n-1}(1,3,5).$
Here is what I observed:
If $(x_1,x_2,x_3)$ is a solution to
$$x_1+3x_2+5x_3 = k$$
then $(x_1+1,x_2,x_3)$ is a solution to
$$x_1+3x_2+5x_3 = k+1$$
$(x_1,x_2+1,x_3)$ is a solution to
$$x_1+3x_2+5x_3 = k+3$$
and $(x_1,x_2,x_3+1)$ is a solution to
$$x_1+3x_2+5x_3 = k+5.$$
But I don't know know how to combine this to get the desired relation. Any ideas will be much appreciated.
Edit: Based on the answer given below, we observe that a solution (x_1,x_2,x_3) to
$$x_1+3x_2+5x_3 = n$$
must have $x_1>0$ or $x_2>0$ or $x_3>0.$ If $x_1>0$ then (x_1-1,x_2,x_3) is a solution
$$x_1+3x_2+5x_3 = n-1$$
and there are $P_{n-1}(1,3,5).$ Proceeding in a similar manner for $x_2$ and $x_3$ and applying the inclusion-exclusion principle we get:
$$P_{n}(1,3,5) = P_{n-1}(1,3,5)+P_{n-3}(1,3,5)+P_{n-5}(1,3,5)-P_{n-4}(1,3,5)-P_{n-8}(1,3,5)-P_{n-6}(1,3,5)+P_{n-9}(1,3,5).$$
| One way to solve this is with generating functions. For $x_1$ you get a factor $(1 - z)^{-1}$, $x_2$ gives $(1 - z^2)^{-1}$, $x_3$ adds $(1 - z^3)^{-1}$. Pulling all together:
$\begin{align*}
[z^n] \frac{1}{(1 - z) (1 - z^2) (1 - z^3)}
&= \frac{1}{24} [z^n] \frac{11 + 3 z + 3 z^2}{1 - x^3}
+ \frac{1}{8} [z^n] \frac{1}{1 + z}
+ \frac{17}{72} [z^n] \frac{1}{1 - z}
+ \frac{1}{4} [z^n] \frac{1}{(1 - z)^2}
+ \frac{1}{6} [z^n] \frac{1}{(1 - z)^3}
\end{align*}$
The last expression by partial fraction, but adding back together fractions with denominators $1 + z + z^2$ and $1 - z$ to simplify coefficient extraction.
Now, using the generalized binomial theorem:
$\begin{align*}
(1 - z)^{-m}
&= \sum_{n \ge 0} (-1)^n \binom{-m}{n} z^n \\
&= \sum_{n \ge 0} \binom{n + m - 1}{m - 1} z^n
\end{align*}$
Thus we get:
$\begin{align*}
&\frac{1}{24} (11 [n \equiv 0 \pmod{3}]
+ 3 [n \not\equiv 0 \pmod{3}])
+ \frac{1}{8} (-1)^n
+ \frac{17}{72}
+ \frac{1}{4} \binom{n + 2 - 1}{2 - 1}
+ \frac{1}{6} \binom{n + 3 - 1}{3 - 1} \\
&\quad = \frac{1}{24} (11 [n \equiv 0 \pmod{3}]
+ 3 [n \not\equiv 0 \pmod{3}])
+ \frac{6 n^2 + 36 n + 47 + 9 (-1)^n}{72}
\end{align*}$
Here $[\dotsb]$ is Iverson's convention: 1 if the condition in brackets is true, 0 otherwise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is there a way to evaluate analytically the following infinite double sum? Consider the following double sum
$$
S = \sum_{n=1}^\infty \sum_{m=1}^\infty
\frac{1}{a (2n-1)^2 - b (2m-1)^2} \, ,
$$
where $a$ and $b$ are both positive real numbers given by
\begin{align}
a &= \frac{1}{2} - \frac{\sqrt{2}}{32} \, , \\
b &= \frac{1}{4} - \frac{3\sqrt{2}}{32} \, .
\end{align}
It turns out that one of the two sums can readily be calculated and expressed in terms of the tangente function.
Specifically,
$$
S = \frac{\pi}{4\sqrt{ab}} \sum_{m=1}^\infty
\frac{\tan \left( \frac{\pi}{2} \sqrt{\frac{b}{a}} (2m-1) \right)}{2m-1} \, .
$$
The latter result does not seem to be further simplified. i was wondering whether someone here could be of help and let me know in case there exists a method to evaluate the sum above. Hints and suggestions welcome.
Thank you
PS From numerical evaluation using computer algebra systems, it seems that the series is convergent. This apparently would not be the case if $b<0$.
| Let
$$q=\sqrt{\dfrac{b}{a}},$$
then attack via digamma-function leads to
$$\begin{align}
&S = \dfrac1a\sum\limits_{n=1}^\infty\sum\limits_{m=1}^\infty\dfrac1{(2n-1)^2-q^2(2m-1)^2} \\[4pt]
&= \dfrac1a\sum\limits_{n=1}^\infty\sum\limits_{m=1}^\infty\dfrac1{2(2n-1)}\left(\dfrac1{2n-1-q(2m-1)}+\dfrac1{2n-1+q(2m-1)}\right)\\[4pt]
&= \dfrac1{4aq}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\sum\limits_{m=1}^\infty \left(-\dfrac1{m-\frac{2n-1+q}{2q}}+\dfrac1{m+\frac{2n-1-q}{2q}}\right)\\[4pt]
&= \dfrac1{4aq}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\left(\psi\left(1-\frac{2n-1+q}{2q}\right)-\psi\left(1+\frac{2n-1-q}{2q}\right)\right)\\[4pt]
&= \dfrac1{4aq}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\left(\psi\left(\frac{q-2n+1}{2q}\right)-\psi\left(\frac{q+2n-1}{2q}\right)\right)\\[4pt]
&= \dfrac1{4aq}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\cdot\pi\cot\left(\pi\frac{q-2n+1}{2q}\right)\\[4pt]
&= \dfrac\pi{4\sqrt{ab}}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\tan\left(\frac\pi 2 \sqrt{\frac ab}(2n-1)\right)\\[4pt]
&= \dfrac\pi{4\sqrt{ab}}\sum\limits_{m=1}^\infty\dfrac1{2m-1}\tan\left(\frac\pi 2 \sqrt{\color{red}{\frac ab}}(2m-1)\right).
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3074662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 1
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$\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}}\ge\frac{1}{n}-\frac{1}{n+1}$ Let $m,n\in\mathbb N-\{1,2\}$ such that $m\ne n.$ How to show that
$$\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}}\ge\frac{1}{n}-\frac{1}{n+1}?$$
Please help me. I am clueless.
| I am going to prove this for cases. First if $m < n$:
$$
\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}} = \frac{1}{m}-\frac{1}{n}
$$
For all $n\in \mathbb{N}$
$$
\frac{2}{n+2} > 0
$$
Then
\begin{eqnarray}
(n-1) + \frac{2}{n+2} &>& (n-1) \\
\frac{n^2+n-2 +2}{n+2} &>& n-1 \\
\frac{n^2+n}{n+2} &>& n-1 \\
\end{eqnarray}
For the axioms of Peano (I think), $n-1 \geq m$, then:
\begin{eqnarray}
\frac{n^2+n}{n+2} &\geq & m \\
\frac{1}{m} &\geq & \frac{n+2}{n^2+n} \\
\frac{1}{m} &\geq & \frac{2n+2-n}{n(n+1)} \\
\frac{1}{m} &\geq & \frac{2}{n} - \frac{1}{n+1} \\
\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}} = \frac{1}{m} - \frac{1}{n} &\geq & \frac{1}{n} - \frac{1}{n+1} \\
\end{eqnarray}
And if $m >n$:
$$
\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}} = \frac{1}{n}-\frac{1}{m}
$$
For the axioms of Peano:
\begin{eqnarray}
m &\geq & n+1 \\
\frac{1}{n+1} &\geq & \frac{1}{m} \\
-\frac{1}{m} &\geq & -\frac{1}{n+1} \\
\end{eqnarray}
Then
$$
\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}} = \frac{1}{n}-\frac{1}{m} \geq \frac{1}{n}-\frac{1}{n+1}$$
Thus, $\forall m,n\in\mathbb{N}$ with $m\neq n$
$$\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}}\ge\frac{1}{n}-\frac{1}{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How do I solve for $a$ and $b$ in $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$? Given $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$
I need to solve for $a$ and $b$, so here we go,
$\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right)$
$= \lim\limits_{x \to ∞} x \left(2 +(3+x) \left(\ln (1+\frac{a}{x}) - \ln(1+\frac{b}{x}) \right) \right)$
$=\lim\limits_{x \to ∞} x \left(2 +(3+x)\left( \dfrac{a-b}{x} \right) \right)$
$=\lim\limits_{x \to ∞} \left(2x +(3+x)\left( a-b \right) \right)$
$=\lim\limits_{x \to ∞} \left(2x + 3(a-b) + x(a-b) \right) $
Since the limit exists, $2x$ and $x(a-b)$ must cancel out so $a-b = 2$
But if I do this, I'm left with $3(a-b) = 2$ which gives me a contradiction since $a-b$ is supposed to be $2$.
What do I do now? And where exactly have I gone wrong?
Thank you!
| Set $1/x=h$
$$\lim_{h\to0^+}\dfrac{2h+(3h+1)\{\ln(1+ah)-\ln(1+bh)\}}{h^2}$$
$$=3\left(a\lim_{h\to0^+}\dfrac{\ln(1+ah)-1}{ah}-b\lim_{h\to0^+}\dfrac{\ln(1+bh)-1}{bh}\right)+\lim_{h\to0^+}\dfrac{2h+\ln(1+ah)-\ln(1+bh)}{h^2}$$
$$=3(a-b)+\lim_{h\to0^+}\dfrac{2h+\ln(1+ah)-\ln(1+bh)}{h^2}$$
$$=3(a-b)+a^2\lim_{h\to0^+}\dfrac{\ln(1+ah)-ah}{(ah)^2}-b^2\lim_{h\to0^+}\dfrac{\ln(1+bh)-bh}{(bh)^2}+\lim_{h\to0^+}\dfrac{(2+a-b)h}{h^2}$$
$$=3(a-b)+\dfrac{b^2-a^2}2+\lim_{h\to0^+}\dfrac{(2+a-b)h}{h^2}$$ using Are all limits solvable without L'Hôpital Rule or Series Expansion,
Clearly, we need $2+a-b=0$ and $3(a-b)+\dfrac{b^2-a^2}2=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding remainder when unknown $f(x)$ is divided to $g(x)$ When $f(x)$ is divided by $x - 2$ and $x + 3$, the remainders are 5 and -1, respectively. Find the remainder when $f(x)$ is divided by $x^2 + x - 6$
My method:
Since $x - 2$ and $x + 3$ are linears, dividing by quadratic will leave linear remainders.
Using the remainder theorem:
$$ f(x) = g(x) q(x) + r(x) $$
Where $g(x)$ is a divisor, $q(x)$ is a remainder, and $r(x)$ is a remainder
I let $ax + b$ here be the remainder.
So:
\begin{align}
f(x) &= g(x) (x - 2)(x + 3) + ax + b\\
f(2) &= g(2) (0)(5) + 2a+ b\\
f(-3) &= g(-3) (-5)(0) - 3a + b\\
\\
&5 = 2a + b\\
&-1 = -3a + b\\
\\
&...\\
\\
&a = \frac{6}{5}, b = \frac{13}{5}
\\
\end{align}
Then the remainder is $ax + b = \frac{6}{5}x + \frac{13}{5} $
Is it possible to find the $f(x)$ out of remainders?
| No, there are infinite $f(x)$ which satisfy the given conditions:
$$f(x)=q(x)(x^2 + x - 6)+\frac{6}{5}x + \frac{13}{5}$$
with any polynomial $q(x)$.
| {
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"url": "https://math.stackexchange.com/questions/3080332",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.
I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I wanted.
Now how do I proceed?
| Use that we get from $$a+b=c+d$$ so $$a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$ and $$x^3+y^3=(x+y)(x^2-xy+y^2)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding all roots of $x^6 + 2 x^5 + 2 x^4 + 3 x^2 + 6 x + 6$, knowing that one root is $x=-1+i$.
Find all roots of
$$V(x) = x^6 + 2 x^5 + 2 x^4 + 3 x^2 + 6 x + 6$$
knowing that one root is $x=-1+i$.
Sorry for the picture. I found two roots of the polynomial and also found an equation which can help me to find all of the rest. Can someone help me what do I need to do next?
| To solve $$x^2+2x+2=0$$ use $$x^2+2x+1=-1$$ hence $$(x+1)^2=-1$$ which gives the solutions $\ -1-i\ $ and $\ -1+i\ $
To solve $$x^4+3=0$$ first note that the absolute value of the solutions must be $3^{\frac{1}{4}}$ and then use the $4$ fourth roots of $-1$ , being $$\pm \frac{\sqrt{2}}{2} \pm \frac{\sqrt{2}}{2}i$$ which , multiplied with $3^{\frac{1}{4}}$, give the $4$ other solutions.
| {
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"source": "stackexchange",
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Find $a, b\in\mathbb{R}$ so that $\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2}$ is finite.
Find $a, b\in\mathbb{R}$ so that
$$\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2}$$
is finite.
I tried to use polynomial division, but the computations get tedious really fast. Any suggestions?
| Hint:
Use the following lemma:
Lemma: $\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2}$ is finite if and only if $(x-1)^2$ divides $ax^6+bx^5+1$.
Proof: If $ax^6+bx^5+1 = p(x)\cdot (x-1)^2$ then $\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2} = p(1)$ is finite. For the other direction, if $(x-1)^2$ does not divide $ax^6+bx^5+1$ then one of the following occure
(1) $(x-1)$ does not divide $ax^6+bx^5+1$ in this case $x=1$ is not a root of $ax^6+bx^5+1$ and so the limit is infinite (unformally it equals to "$\frac{a+b+1}{0}$")
(2) $(x-1)$ divides $ax^6+bx^5+1$, let $p(x)$ be such that $ax^6+bx^5+1 = p(x)(x-1)$ then
$$\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2} = \lim_{x\rightarrow 1} \frac{p(x)}{x-1}$$ and $(x-1)$ doesn't divide $p(x)$ so as before the limit is infinite (unformally equals to "$\frac{p(1)}{0}$").
So your goal is to find all the $a,b$ such that $(x-1)^2$ divides $ax^6+bx^5+1$. This means that $x=1$ is a root of this polynomial and then once you divide by $(x-1)$ the integer $1$ is still a root of the polynomial. You will get two equations in the variables $a,b$ which are not hard to solve.
| {
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"source": "stackexchange",
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Proving a strict inequality in the limit I want to prove that
$$ \lim_{k \to \infty} \left( 1 + \frac{1}{2} \right) \left(1 + \frac{1}{4} \right)...\left( 1 + \frac{1}{2^k} \right) < e .$$
Using the $AM-GM$ inequality we arrive at
$$\left( 1 + \frac{1}{2} \right) \left(1 + \frac{1}{4} \right)...\left( 1 + \frac{1}{2^k} \right) < \left(\frac{k + 1 - \frac{1}{2^k} }{k} \right)^k = \left( 1 + \frac{1}{k} - \frac{1}{k2^{k}}\right)^k < \left(1 + \frac{1}{k} \right)^k < e.$$
The first inequality is strict because the terms are different.However, I know that in the limit, strict inequalities can transform into equalities. Since the limit of $\left( 1 + \frac{1}{k} - \frac{1}{k2^{k}}\right)^k$ when $k$ goes to infinity is also $e$, how could I prove a strict inequality?
| $$ \sum_{k\geq 1}\log\left(1+\frac{1}{2^k}\right)<\sum_{k\geq 1}\frac{1}{2^k}=1,$$
then exponentiate both sides. There is also a simple strengthening, namely
$$ \sum_{k\geq 1}\log\left(1+\frac{1}{2^k}\right)<\sum_{k\geq 1}\left(\frac{1}{2^k}-\frac{1-\log 2}{4^k}\right)=\frac{2+\log 2}{3},$$
leading to
$$ \prod_{k\geq 1}\left(1+\frac{1}{2^k}\right) < 2^{1/3}e^{2/3}.$$
$ \prod_{k\geq 1}\left(1+\frac{1}{2^k}\right) <\frac{31}{13}$ is a more challenging inequality.
| {
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Evaluating $\lim_{x\to 1}\frac{^5x-^4x}{(1-x)^5}$, where $^nx$ is the repeated exponent ("tetration") operation
$$\lim_{x\rightarrow1}\frac{x^{x^{x^{x^x}}}-{x^{x^{x^x}}}}{(1-x)^5}$$
My friends told that it appeared on Instagram (some social media network). I tried various methods but failed. (I tried using Taylor series but failed. Tried l'Hopital but failed. Tried substitution but failed. Gave up here.) They also said that some of the comments claim that the answer is $\;-1.$
| Start using composition of Taylor series using mainly $t=e^{\log(t)}$. Then
$$x^x=1+(x-1)+(x-1)^2+\frac{1}{2} (x-1)^3+\frac{1}{3} (x-1)^4+\frac{1}{12}
(x-1)^5+O\left((x-1)^6\right)$$
$$x^{x^x}=1+(x-1)+(x-1)^2+\frac{3}{2} (x-1)^3+\frac{4}{3} (x-1)^4+\frac{3}{2}
(x-1)^5+O\left((x-1)^6\right)$$
$$x^{x^{x^x}}=1+(x-1)+(x-1)^2+\frac{3}{2} (x-1)^3+\frac{7}{3} (x-1)^4+3
(x-1)^5+O\left((x-1)^6\right)$$
$$x^{x^{x^{x^x}}}=1+(x-1)+(x-1)^2+\frac{3}{2} (x-1)^3+\frac{7}{3} (x-1)^4+4
(x-1)^5+O\left((x-1)^6\right)$$ Which shows the limit.
If you continue the expansion, you would get
$$\frac{x^{x^{x^{x^x}}}-x^{x^{x^x}}}{(1-x)^5}=-1-2 (x-1)-\frac{13}{3} (x-1)^2+O\left((x-1)^3\right)$$
Use your computer with $x=1.1$. The "exact" value would be $\approx -1.2525$ while the above expansion gives $-\frac{373}{300}\approx -1.2433$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Cannot find angle for trigonometry problem
A right angle triangle is entrapped within a circle. The triangle entraps within it a circle of its own. The ratio between the big radius and the little radius is $\frac{13}{4}$. What are the angles of the triangle?
Here is the problem as I understand it, given that:
1. $\measuredangle ABC = 90^\circ \rightarrow AC = 2R$
2. The smaller radii are perpendicular to the triangle.
3. The origin of the smaller circle is the intersection between the angle bisectors of triangle ABC.
Let $\measuredangle ACB = 2\alpha \rightarrow \measuredangle ACG = \alpha$
$\measuredangle BAC = 90 - 2\alpha \rightarrow \measuredangle GAC = 45 - \alpha$
From here we can construct the following system:
$$\begin{cases} 2R = \frac{r}{\tan \alpha} + \frac{r}{\tan(45 - \alpha)} \\ \frac{R}{r} = \frac{13}{4} \rightarrow R = \frac{13r}{4} \end{cases}$$
$$\downarrow \\ \frac{26r}{4} = r\left(\frac{1}{\tan \alpha} + \frac{1}{\tan(45 - \alpha)}\right) \quad / \div r \\ 6.5 = \frac{\cos \alpha}{\sin \alpha} + \frac{\cos(45 - \alpha)}{\sin(45 - \alpha)} \quad / {\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \sin\beta\cos\alpha\\ \cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta} \\ 6.5 = \frac{\cos \alpha}{\sin \alpha} + \frac{\cos \alpha \cos 45 + \sin \alpha \sin 45}{\sin 45 \ cos \alpha - \sin \alpha \cos 45} \\ 6.5 = \frac{\cos \alpha}{\sin \alpha } + \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} \cdot \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha} \\ 6.5 = \frac{\cos^2 \alpha - \sin\alpha \cos \alpha + \cos \alpha \sin \alpha + \sin^2 \alpha}{\sin(\cos\alpha - \sin \alpha)} \quad / \sin^2\alpha + \cos ^2 \alpha = 1\\ 6.5 = \frac{1}{\sin\alpha (\cos\alpha - \sin \alpha)}$$
From this point on, I have no identity that I can think of which can be applied here to help me solve for $\alpha$. Amusingly enough, I got an answer which was very close to the correct one, but that was only because I made a mistake in the signs of the identity, and the second fracture turned into 1.
I am aware that they may be an identity which can solve this, but I am restricted to only use the identities available here. I have either made a mistake somewhere, or just don't know how to use the available identities to solve the problem.
| We have
$$2R=\frac{r}{\tan(\alpha/2)}+\frac{r}{\tan(\beta/2)},$$
hence
$$\frac{13}{2}=\frac{1}{\tan(\alpha/2)}+\frac{1}{\tan(\beta/2)}.$$
We know that $\alpha+\beta=\pi/2$, so $\beta/2=\pi/4-\alpha/2$. Let $t=\tan(\alpha/2)$. From the addition theorems for $\sin$ and $\cos$ derive that
$$\tan(\pi/4-\alpha/2)=\frac{1-t}{1+t}.$$ Substitute and solve the quadratic equation for $t$.
| {
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"source": "stackexchange",
"question_score": "2",
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Evaluating $\sum_{n=1}^\infty \frac{\sin(nx)}{n}$ without integrating $\sum_{n=1}^\infty e^{nx}$ I am looking for alternative solutions for finding this sum
$$\sum_{n=1}^\infty \frac{\sin(nx)}{n} $$
My solution proceeds by integrating $$\sum_{n=1}^\infty e^{nx}=\frac{e^{ix}}{1-e^{ix}}$$
With suitable limits and then taking the imaginary part of it.
| Note that\begin{align}\sum_{n=1}^\infty\frac{\sin(nx)}n&=\operatorname{Im}\left(\sum_{n=1}^\infty\frac{e^{inx}}n\right)\\&=\operatorname{Im}\left(\sum_{n=1}^\infty\frac{\left(e^{ix}\right)^n}n\right).\end{align}If $x\in\mathbb{R}\setminus\{0\}$, then $e^{ix}\in S^1\setminus\{1\}$, and therefore$$\sum_{n=1}^\infty\frac{\left(e^{ix}\right)^n}n=\log\left(\frac1{1-e^{ix}}\right),$$(where $\log$ is the principal branch of the logarithm), since\begin{align}\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}x+\cdots&\implies\log(1-x)=-\left(x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+\cdots\right)\\&\implies\log\left(\frac1{1-x}\right)=x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+\cdots\end{align}Therefore\begin{align}\sum_{n=1}^\infty\frac{\sin(nx)}n&=-\arctan\left(\frac{-\sin x}{1-\cos(x)}\right)\\&=\arctan\left(\cot\left(\frac x2\right)\right).\end{align}Of course, if $x=0$, then this equality also holds.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $n$ such that polynomial is divisible Find $n \in N$, such that:
$$(x^2+x+1)^2 | x^n+(x+1)^n+1 = P(x)$$
If I let $Q(x) = x^2 + x + 1$, I have that $x^3 \equiv 1$ mod$Q(x)$, so I can work out the cases for remainder of $n$ when divided by $3$ ..
But what to do next, because I need $Q^2(x)$ dividing $P(x)$ ?
I suppose this could be done with complex zeroes, but is there a faster way ?
| Let's call your polynomial $P_n(x)$, to indicate the dependence on $n$. If $\omega$ is a root of $x^2 + x + 1$, we want
$\omega$ to be a root of $P_n(x)$ of multiplicity $\ge 2$,
which means both $P_n(\omega) = \omega^n + (\omega+1)^n + 1 = 0$ and
$P_n'(\omega) = n \omega^{n-1} + n (\omega+1)^{n-1} = 0$. This last
simplifies to $$\left(1 + \frac{1}{\omega}\right)^{n-1} = -1$$
Now if $\omega^2 + \omega+1 = 0$ we have $\omega + 1 + 1/\omega = 0$, i.e. $1+1/\omega = -\omega$. Since the powers of $\omega$ repeat $1, \omega, \omega^2$, we see that this is true if and only if $n-1 \equiv 3 \mod 6$, i.e. $n \equiv 4 \mod 6$. For such $n$ we also have
$P_n(\omega) = \omega^n + (-\omega^2)^n + 1 = \omega + \omega^2 + 1 = 0$, so both conditions are true.
| {
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"url": "https://math.stackexchange.com/questions/3085998",
"timestamp": "2023-03-29T00:00:00",
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How to find the value this sum converges to?$\sum_{n=2}^{\infty}\frac{n+1}{n(2n-1)(2n+1)} $ How to find the value this sum converges to?$$\sum_{n=2}^{\infty}\frac{n+1}{n(2n-1)(2n+1)} $$
I've tried separating it like
$$\sum_{n=2}^{\infty}\frac{1/2}{(2n+1)}+\frac{3/2}{(2n-1)}-\frac{1}{n}$$
and writing some terms and I get
$$1/6·(1+1/2+...+1/6n+3)+1/2·(1/3+1+1/2+...+1/6n-3)-(1+1/2+...1/n)$$
but I don't know how to end summing it all! Any hint!
FYI I haven't learnt integration and differentation.
| Proceeding on the partial fraction decomposition you already got
$$
\eqalign{
& S = \sum\limits_{2\, \le \,n} {{{n + 1} \over {n\left( {2n - 1} \right)\left( {2n + 1} \right)}}} = \cr
& = \sum\limits_{2\, \le \,n} {{{1/2} \over {\left( {2n + 1} \right)}} + {{3/2} \over {\left( {2n - 1} \right)}} + {1 \over n}} = \cr
& = \sum\limits_{0\, \le \,n} {{{1/2} \over {\left( {2n + 5} \right)}} + {{3/2} \over {\left( {2n + 3} \right)}} - {1 \over {n + 2}}} = \cr
& = \sum\limits_{0\, \le \,n} {{{1/4} \over {\left( {n + 5/2} \right)}} + {{3/4} \over {\left( {n + 3/2} \right)}} - {1 \over {n + 2}}} \cr}
$$
Since
$$
\sum\nolimits_{n = 0}^N {{1 \over {n + a}}} = \sum\nolimits_{n = a}^{N + a} {{1 \over n}} = \psi (N + a) - \psi (a)
$$
then
$$
\eqalign{
& S = \mathop {\lim }\limits_{N\, \to \,\infty } \left( \matrix{
{1 \over 4}\psi (N + 5/2) + {3 \over 4}\psi (N + 3/2) - \psi (N + 2) + \hfill \cr
+ \psi (2) - {1 \over 4}\psi (5/2) - {3 \over 4}\psi (3/2) \hfill \cr} \right) = \cr
& = \psi (2) - {1 \over 4}\psi (5/2) - {3 \over 4}\psi (3/2) = \cr
& = 1 - \gamma - {1 \over 4}\left( {{8 \over 3} - 2\ln 2 - \gamma } \right) - {3 \over 4}\left( {2 - 2\ln 2 - \gamma } \right) = \cr
& = 2\ln 2 - {7 \over 6} \cr}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of series: 9 + 16 + 29 + 54 + 103 + ... The series is neither AP nor GP, so I tried doing some manipulations to get the expression for the general term.
$S = 9 + 16 + 29 + 54 + 103 + ... T_n$
$S = 0 + 9 + 16 + 29 + 54 + ... T_{n-1} + T_n$
Subtracting the two:
$0 = 9 + 7 + 13 + 25 + 49 + ... (T_n - T_{n-1}) - T_n$
This gives:
$T_n = 9 + 7 + 13 + 25 + 49 + ... (T_n - T_{n-1})$
The expression obtained for $T_n$ is neither in AP nor in GP.
I repeated the above process with $T_n$ (which gave me a GP this time) and got this expression finally:
$T_n - T_{n-1} = 7 + 6(2^{n-2} - 1) = 6(2^{n-2}) + 1$
How do I proceed from here to get an expression for the general term as well as the sum of the series?
| From $T_n - T_{n-1} = 6(2^{n-2}) + 1$ you have
$$T_n = T_0 + \sum_{k=1}^n (6(2^{n-2}) + 1) \\
= T_0 + n + 3 (2^n -1)$$
The general approach is:
$$T_n = a+ n + b 2^n$$
so
$$9 = T_1 = a + 1 + 2 b\\
16 = T_2 = a + 2 + 4b
$$
which gives indeed
$b = 3$ and $a = 2$, hence
$$T_n = 2+ n + 3 \cdot 2^n$$
The sum is then
$$
S_n = \sum_{k=1}^n T_k = 2 n + \frac12 n (n+1) + 3 (2^{n+1} -1)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the set of the real numbers $x$ satisfying the inequalities $|x+4|<|2x-1|$ and $|x|+|x+1|<3$
Find the set of the real numbers $x$ satisfying the given inequality and sketch them on the real line:
1) $|x+4|<|2x-1|$
*
*If $x<-4$, we have $x+4<0$ and $2x-1<0$, Then $|x+4|=-x-4$ and $|2x-1|=1-2x$
hence, in this case:
$|x+4|<|2x-1| \iff -x-4<1-2x \iff x<5$.
and we have that, $(-\infty,4)\cap (-\infty,5)=(-\infty,4)$.
So, for any $x \in (-\infty,-4)$, $|x+4|<|2x-1|$.
*
*If $-4\leq x <\frac{1}{2}$, we have that $x+4\geq 0$, and $2x-1<0$, so $|x+4+x+4$ and $|2x-1|=1-2x$
hence, in this case:
$|x+4|<|2x-1| \iff x+4<1-2x \iff 3x<-3 \iff x<-1$.
Thus, for any $x \in [-4,\frac{1}{2}) \cap (-\infty,-1)=[-4,-1)$, $|x+4|<|2x-1|$.
*
*If $x\geq \frac{1}{2}$, we have $x+4>0$, and $2x-1\geq 0$, then $|x+4|=x+4$ and $|2x-1|=2x-1$
Then, in this case
$|x+4|<|2x-1| \iff x+4<2x-1 \iff x>5$.
So, for any $x \in [\frac{1}{2},\infty) \cap (5,\infty)=(5,\infty)$, $|x+4|<|2x-1|$.
Therefore, The solution set of the inequality is $(-\infty,-4) \cup [-4,-1) \cup (5,\infty)=(-\infty,-1) \cup (5,\infty)$.
2) $|x|+|x+1|<3$
*
*If $x<-1$, we have that $x<0$ and $x+1<0$, Then from the Absolute value definition $|x|=-x$ and $|x+1|=-x-1$.
Thus, $|x|+|x+1|=-x-x-1=-2x-1$.
Hence in this case, $|x|+|x+1|<3 \iff -2x-1 < 3 \iff 2x>-4 \iff x>-2$
So, for any $x \in (-2,-1)$, $|x|+|x+1|<3$.
*
*If $-1 \leq x <0$, we have that $|x|=-x$ and $|x+1|=x+1$
we have, $|x|+|x+1|=-x+x+1=1<3$.
Hence, for any $x \in [-1,0)$, $|x|+|x+1|<3$.
*
*If $x \geq 0$, we have that $|x|=x$ and $|x+1|=x+1$
Then in this case, $|x|+|x+1|=x+x+1=2x+1<3 \iff 2x<2 \iff x<1$.
Thus, for any $x \in [0,1)$, $|x|+|x+1|<3$.
Therefore, The solution set of the inequality is $(-2,-1) \cup [-1,0) \cup
[0,1)=(-2,1)$
is it true, please?
| Problem 1.
If you square it you get $$x^2+8x+16<4x^2-4x+1$$
so $$ 3x^2-12x-15>0\implies 3(x-5)(x+1)>0$$
so $x\in(-\infty,-1)\cup (5,\infty)$
And a solution to the second problem you solved is correct.
| {
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Closed form expression for infinite series I was given the following function:
$$ f(x) = x + \frac{2x^3}{1\cdot3} + \frac{2\cdot4x^5}{1\cdot3\cdot5} + \frac{2\cdot4\cdot6x^7}{1\cdot3\cdot5\cdot7}... $$ $$ \forall x \in [0,1) $$
And then I was asked to find the value of $ f(\frac{1}{\sqrt{2}}) $, which obviously requires me to compute the closed form expression of the infinite series.
I tried 'Integration as a limit of sum' but I was unable to modify the expression accordingly. How do I approach the problem?
| I would suggest the following representation
$$\sum_{n=0}^{\infty}\frac{x^{2n+1}(2n)!!}{(2n+1)!!}=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}$$
Plug in $x=\frac{1}{\sqrt{2}}$. The result will be
$$\frac{\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)}{\sqrt{1-\frac{1}{2}}}=\frac{\sqrt{2}\pi}{4}$$
Update:
Your question is similar to problem $6(ii)$ in the eighth William Lowell Putnam Competition. I will present the solution from the book.
Let
$$f(x) = x + \frac{2x^3}{1\cdot3} + \frac{2\cdot4x^5}{1\cdot3\cdot5} + \frac{2\cdot4\cdot6x^7}{1\cdot3\cdot5\cdot7}\cdots $$
Then
$$\begin{align}
f'(x) &= 1+x\left[2x+\frac{2}{3}\cdot 4x^3+\frac{2}{3}\cdot\frac{4}{5}\cdot6x^5 +\cdots \right]\\
\\&= 1+x\frac{d}{dx}\left[x^2+\frac{2}{3}x^4+\frac{2}{3}\cdot\frac{4}{5}x^6+\cdots\right]\\
\\&=1+x\frac{d}{dx}(xf(x))=1+xf(x)+x^2f'(x).
\end{align}$$
Thus $f'(x)$ satisfies the differential equation
$$(1-x^2)f'(x)=1+xf(x)\tag{1}$$
and the initial condition
$$f(0)=0.\tag{2}$$
(We note that the series for $f$ is convergent for $|x| < 1$ so that all formal manipulations are justified.)
Now $(1)$ is a first order linear non-singular differential equation on the interval $(-1,1)$, so it has a unique solution satisfying (2).
The next step is to solve the differential equation,
$$(1-x^2)\frac{df(x)}{dx}=1+xf(x)$$
Rewrite the equation:
$$\frac{df(x)}{dx}-\frac{xf(x)}{1-x^2}=\frac{1}{1-x^2}$$
Multiply both sides by $\sqrt{1-x^2}$
$$\frac{df(x)}{dx}\sqrt{1-x^2}-\frac{xf(x)}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}$$
Substitute $\frac{x}{\sqrt{1-x^2}}=-\frac{d}{dx}\sqrt{1-x^2}$
$$\frac{df(x)}{dx}\sqrt{1-x^2}+\frac{d}{dx}\left(\sqrt{1-x^2}\right)f(x)=\frac{1}{\sqrt{1-x^2}}$$
Apply the reverse product rule $f\frac{dg}{dx}+g\frac{df}{dx}=\frac{d}{dx}(fg)$ to the left-hand side
$$\frac{d}{dx}\left(f(x)\sqrt{1-x^2}\right)= \frac{1}{\sqrt{x^2-1}}$$
Integrate both sides with respect to $x$
$$\int\frac{d}{dx}\left(f(x)\sqrt{1-x^2}\right)dx= \int\frac{1}{\sqrt{1-x^2}}dx$$
Recall fundamental theorem of calculus $\frac{d}{dx}\int f(x) dx =f(x)$ and
$\int\frac{1}{\sqrt{1-x^2}}=\arcsin(x)+C$, then
$$f(x)\sqrt{1-x^2}=\arcsin(x)+C$$
We have the initial condition $f(0)=0$, so
$$0=\arcsin(0)+C$$
$$C=0$$
Therefore,
$$f(x)=\frac{\arcsin(x)}{\sqrt{1-x^2}}$$
| {
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"url": "https://math.stackexchange.com/questions/3091153",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer? For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer?
For example, if x = 3, $y = 4(3^2) + 8(3) + 5$ = 65 which is divisible by 13
if x = 8, y = 325 which is divisible by 13
if x = 16, y = 1157 which is divisible by 13
if x = 21, y = 1937 which is divisible by 13
I am guessing that values of x = 13i + 3 or x = 13i + 8 where i is an integer will result in a value of y that is evenly divisible by 13.
How do you prove that x = 13i + 3 or x = 13i + 8 will result in a value of y that is evenly divisible by 13?
Is there a general proof to find values of x that will result in a value of y that is evenly divisible by an odd integer p?
| Completing the square;
$$4x^2+8x+5=(2x+2)^2+1,$$
so solving the congruence $\;4x^2+8x+5\equiv 0\mod 13$ amounts to solving
$$\bigl(2(x+1)\bigr)^2\equiv -1\mod 13,$$
and ultimately to finding the square roots $y$ of $-1\bmod 13$ (we know this is possible because $13\equiv 1\mod 4$).
Note that $5^2\equiv -1\mod 13$, hence the other square root is $-5\equiv 8$, and since $2^{-1}\equiv 7\mod 13$, the solutions in $x$ are
$$ x\equiv 7\cdot\pm 5 -1\equiv 8,3\mod 13.$$
| {
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Prove $\sqrt{b} - \sqrt{a} < \sqrt{b-a}$
Prove that if $0 < a < b$ then
$$\sqrt{b} - \sqrt{a} < \sqrt{b-a}$$
This is what I have so far:
square both sides to get $a + b -2\sqrt{ab} < b-a$
subtract $b$ from both sides $a-2\sqrt{ab} < -a$
add $a$ to both sides $2a-2\sqrt{ab} < 0$
than add $2 \sqrt{ab}$ to both sides and get $2a < 2\sqrt{ab}$
divide by $2$ and get $a < \sqrt{ab}$
Thus we know $\sqrt{ab}$ is bigger than $a$ so that $\sqrt{a \cdot a} < \sqrt{ab}$ which means $\sqrt{a} < \sqrt{b}$.
Therefore together with the given $a < b$, we have $\sqrt{b} - \sqrt{a} < \sqrt{b - a}$
I'm confused as to where I went wrong and how to fix this.
| It looks like you are assuming your claim: you are showing that your claim implies your assumptions $0 < a < b$, which means you should reverse the argument. Try to do so. Here below is a different approach.
Assuming $0 < a < b$ you have $2\sqrt{a(b-a)}>0$, hence
$$2\sqrt{a(b-a)}+a+(b-a) > b.$$
The left hand side is $(\sqrt{b-a}+\sqrt{a})^2$, so
$$(\sqrt{b-a}+\sqrt{a})^2 > \sqrt{b}^2$$
and since both sides are positive you can take the square root getting
$$\sqrt{b-a}+\sqrt{a} > \sqrt{b},$$
which is your claim.
| {
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"url": "https://math.stackexchange.com/questions/3095630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
maximum and minimum of $x^2+y^4$ for real $x,y$ If $y^2(y^2-6)+x^2-8x+24=0$ then maximum and minimum value of $x^2+y^4$ is
what i try
$y^4-6y^2+9+x^2-8x+16=1$
$(x-4)^2+(y^2-3)^2=1\cdots (1)$
How i find maximum and minimum of $x^2+y^4$ from $(1)$ relation
help me to solve it please
| Use substitution method:
$$(x-4)^2+(y^2-3)^2=1 \Rightarrow y^2=\pm\sqrt{1-(x-4)^2}+3\\
f(x,y)=x^2+y^4 \Rightarrow \\
f(x)=x^2+1-(x-4)^2\pm6\sqrt{1-(x-4)^2}+9\\
f'(x)=8+\frac{\mp6(x-4)}{\sqrt{1-(x-4)^2}}=0 \Rightarrow \\
4\sqrt{1-(x-4)^2}=\pm3(x-4) \Rightarrow \\
16(-x^2+8x-15)=9(x^2-8x+16) \Rightarrow \\
25x^2-200x+384=0 \Rightarrow \\
x_1=\frac{16}{5}; x_2=\frac{24}{5}\\
f''\left(\frac{16}{5}\right)>0 \Rightarrow f\left(\frac{16}{5}\right)=16 \ \text{(min)}\\
f''\left(\frac{24}{5}\right)<0 \Rightarrow f\left(\frac{24}{5}\right)=36 \ \text{(max)}\\$$
| {
"language": "en",
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equating coefficients in algebraic expansion If $\displaystyle \bigg(\frac{1+x}{1-x}\bigg)^n=1+b_{1}x+b_{2}x^2+b_{3}x^3+\cdots +\infty,$ then value of
$(1)\; \displaystyle \frac{3b_{3}-b_{1}}{b_{2}}$
$(2)\; \displaystyle \frac{2b_{4}-b_{2}}{b_{3}}$
$(3)\; \displaystyle \frac{3b_{6}-2b_{4}}{b_{5}}$
$(4)\; \displaystyle \frac{5b_{10}-4b_{8}}{b_{9}}$
Answers given in $n,2n,3n,4n$ formats
What I tried:
$\displaystyle (1+x)^n(1-x)^{-n}=1+b_{1}x+b_{2}x^2+b_{3}x^3+\cdots \cdots \infty$
$\displaystyle \bigg[1+nx+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{6}\cdots \bigg]\bigg[1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+\cdots \bigg]=1+b_{1}x+b_{2}x^2+b_{3}x^3+\cdots $
$\displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$
How do I find other coefficients? Help me, please.
| You made a mistake. It is not $\displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$ , but
$b_{1}=2n$
$b_{2}=2n^2$
$b_{3}=\frac23(n+2n^3)$
To go up to $b_{10}$, the series has to be expended up to $x^{10}$.
By brut force :
$b_{4}=\frac23(2n^2+n^4)$
$b_{5}=\frac{2}{15}(3n+10n^3+2n^5)$
$b_{6}=\frac{2}{45}(23n^2+20n^4+2n^6)$
$b_{7}=\frac{2}{315}(45n+196n^3+70n^5+4n^7)$
$b_{8}=\frac{2}{315}(132n^2+154n^4+28n^6+n^8)$
$b_{9}=\frac{2}{2835}(315n+1636n^3+798n^5+84n^7+2n^8)$
$b_{10}=\frac{2}{14175}(5167n^2+7180n^4+1806n^6+120n^8+2n^{10})$
$\frac{3b_{3}-b_{1}}{b_{2}}=2n$
Then, a supposed recurrence :
$\frac{1b_{2}-0b_{0}}{b_{1}}=n$
$\frac{2b_{4}-1b_{2}}{b_{3}}=n$
$\frac{3b_{6}-2b_{4}}{b_{5}}=n$
$\frac{4b_{8}-3b_{6}}{b_{7}}=n$
$\frac{5b_{10}-4b_{8}}{b_{9}}=n$
$...$
$\frac{k\,b_{2k}-(k-1)b_{2k-2}}{b_{2k-1}}=n$ , not proved.
| {
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Prove that $n^n\left(\frac{n+1}{2}\right)^{2n}\geq (n!)^3$ for natural number $n$
Prove that
$$n^n\left(\frac{n+1}{2}\right)^{2n}\geq (n!)^3$$ for a natural number $n$.
what i try
i have use AM GM inequality
$$\frac{1^3+2^3+3^3+\cdots +n^3}{n}\geq ((n!))^{\frac{1}{3n}}$$
how i prove question inequality help me please
| First of all, by expanding the terms we obtain$$n^n\bigg(\frac{n+1}{2}\bigg)^{2n}{={1\over n^n}\cdot \bigg(\frac{n(n+1)}{2}\bigg)^{2n}\\={1\over n^n}\cdot (1+2^3+\cdots +n^3)^n.}$$By substitution and taking the $n$-th root, we need to prove that $${1+2^3+\cdots+n^3\over n}\ge \sqrt[n]{n!^3}=\sqrt[n]{1^3\cdot 2^3 \cdots n^3},$$which is straight-forward by AM-GM.
| {
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What is the $n$th derivative of $\sin(x)$? I don't understand why for $f(x) = \sin(x)$, $\;f^{(n)}(x) = \sin(n\pi/2 + x)$
It does not make sense to me; can anyone help with the reasoning?
Thank you.
| Part I:
$\sin'(x) = \cos (x)$ and $\cos'x = -\sin x$.
If we iterate these $n$ times then for $f(x) = \sin x$ if
$n = 4k$ for some $k$ then $f^n(x) =\sin x$.
If $n = 4k + 1$ then $f^n(x) = \cos x$
If $n = 4k + 2$ then $f^n(x) = -\sin x$
And if $n = 4k + 3$ then $f^n(x) = -\cos x$.
.....
Now lets walk away and notice.
Part II:
If we have a $\sin x$ and $\cos x$ and we rotate $x$ ninety degrees or by $\frac {\pi}2$ then the result is $\sin \mapsto \cos x$ and $\cos x \mapsto -\sin x$ so $\sin(\frac {\pi}2 + x) = \cos x$.
We rotate again and we get $\sin(\pi + x) = -\sin x$.
And again $\sin(\frac {3\pi}2 + x) = -\cos x$.
And again $\sin (2\pi + x) = \sin x$.
Iterating it:
$\sin (n\frac {\pi}2 + x) = :$
Well if $n = 4k$ for some $k$ then $\sin(n\frac \pi 2 + x) = \sin(4k\frac\pi 2 + x)=\sin(2k\pi + x) = \sin x$.
If $n = 4k + 1$ then $\sin(n\frac \pi 2 + x) = \sin((4k+1)\frac\pi 2 + x)=\sin(2k\pi+\frac \pi 2 + x)= \sin(\frac \pi 2 + x) = \cos x$.
If $n = 4k + 2$ then $\sin(n\frac \pi 2 + x) = \sin((4k+2)\frac\pi 2 + x)=\sin(2k\pi+\pi + x)= \sin( \pi + x) = -\sin x$
And if $n=4k + 3$ then $\sin(n\frac \pi 2 + x) = \sin((4k+3)\frac{3\pi} 2 + x)=\sin(2k\pi+\frac {3\pi 2} + x)= \sin(-\frac \pi 2 + x) = -\cos x$
....
Now we compare Part I to Part II we see
If $n = 4k$ then
$f^n(x) = \sin x= \sin (2k\pi + x) = \sin(n\frac{\pi}2 + x)$
If $n = 4k + 1$
$f^n(x) = \cos x = \sin (2k\pi + \frac \pi 2 + x) = \sin(n\frac{\pi}2 + x)$
If $n = 4k + 2$
$f^n(x) = -\sin x = \sin (2k\pi + \pi + x) = = \sin(n\frac{\pi}2 + x)$
If $n = 4k + 3$
$f^n(x) = -\cos x = \sin (2k\pi + \frac {3\pi}2 + x) = = \sin(n\frac{\pi}2 + x)$
| {
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Find permutations whose third power is known I have to find permutations $a$ such that $a^3=(1 \ 2)(3 \ 4)(5 \ 6)(7 \ 8 \ 9 \ 10)$
and I have to find at least 3 solutions.
So first I must find disjoint cycles.
Those are:
1 [2,2,2][4] and 2 [4,2][4] and 3[6][4]
are there any more cycles?
and what is the order of numbers (1,2,3,...10) in those new cycles?
so is solution (1 2)(3 4)(5 6)(7 10 9 8) sufficient for 1 and (1 3 2 4)(5 6)(7 10 9 8) for 2?
| Yes for $p=(1 \ 2)(3 \ 4)(5 \ 6)(10 \ 9 \ 8 \ 7)$.
No for $p=(1 \ 3 \ 2 \ 4)(5 \ 6)(7 \ 10 \ 9 \ 8)$ ; it isn't a solution because $p^3$ would send $1$ onto $4$ instead of $2$.
Here is a way to find many solutions :
First, as you have well seen it, we must take for the last cycle $\color{red}{the \ reversed \ cycle \ (7 \ 10 \ 9 \ 8)}$ (because for a cycle $c$ on 4 elements, $c^4=id \ \iff \ (c^{-1})^3=c$, and there are no other solutions).
A first global solution is :
$$ (1 \ 2)(3 \ 4)(5 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$
(You had recognized it). Please note that we use the fact that a transposition $t$ is such that $t^3=t$.
A family of 8 other solutions are found by considering an order-$6$ cycle on elements $1 \cdots 6$ :
$$ (\underline{1} \ 3 \ 6 \ \underline{2} \ 4 \ 5)\color{red}{(7 \ 10 \ 9 \ 8)}$$
$$ (\underline{1} \ 4 \ 6 \ \underline{2} \ 3 \ 5)\color{red}{(7 \ 10 \ 9 \ 8)}$$
$$ (\underline{1} \ 5 \ 3 \ \underline{2} \ 6 \ 4)\color{red}{(7 \ 10 \ 9 \ 8)}$$
$$ (\underline{1} \ 6 \ 3 \ \underline{2} \ 5 \ 4)\color{red}{(7 \ 10 \ 9 \ 8)}$$
$$ (\underline{1} \ 3 \ 5 \ \underline{2} \ 4 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$
$$ (\underline{1} \ 4 \ 5 \ \underline{2} \ 3 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$
$$ (\underline{1} \ 5 \ 5 \ \underline{2} \ 6 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$
$$ (\underline{1} \ 6 \ 5 \ \underline{2} \ 5 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$
with the following building recipe :
*
*$1$ and $2$ must be separated by two elements,
*the same for $3$ and $4$,
*the same for $5$ and $6$.
(being understood that "separated" is by reference to a cyclic arrangement).
As the positions of $1$ and $2$ are "frozen", the choice is reduced to the $2\times2\times2 = 8$ solutions given upwards, to which we must add the exceptional first one.
Remarks :
a) about the necessity to gather $1,2 \cdots 6$ into a cycle :
*
*no solution can exist of the form $(1 a b c)(* * )$ for the reason seen upwards : we should have $p=(1 a b 2)(* *)$ but then $p^3=(1 2 a b)(* *)$ which is not what we desire.
*no solution can exist of the form $(a b c d)(1 2)$ for a similar reason.
b) As remarked by @Robert Shore, it is not evident that no other solution exists by grouping for example $7$ with other elements than $8,9,10$, even if our intimate conviction says that there none.
| {
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If $m^2\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}} = n^2$, then find the value of $\frac{m^2 - n^2}{n^2}$ I am a beginner at trigonometry, I want to know the answer to this question.
If $m^2\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}} = n^2$, then find the value of $\frac{m^2 - n^2}{n^2}$.
These are the steps I have tried and got stuck in the middle.
$$m^2\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{7\pi}{15}}\cos{\frac{\pi}{15}} = n^2 $$
$$m^2(\frac{1}{4})(2\cos{\frac{2\pi}{15}}\cos{\frac{\pi}{15}})(2\cos{\frac{4\pi}{15}}\cos{\frac{7\pi}{15}}) = n^2$$
$$m^2(\frac{1}{4})(\cos{\frac{3\pi}{15}} + \cos{\frac{\pi}{15}})(\cos{\frac{11\pi}{15}} + \cos{\frac{3\pi}{15}}) = n^2$$
Fro here onwards I could not continue. These steps may be wrong so please check if my method of solving is correct and help me solve the question. Thanks:)
| Hint:
$\cos\dfrac{14\pi}{15}=\cos\left(\pi-\dfrac\pi{15}\right)=?$
Use $\sin2x=2\sin x\cos x\iff\cos x=\dfrac{\sin2x}{2\sin x}$ repeatedly
Finally here $\sin\dfrac{16\pi}{15}=\sin\left(\pi+\dfrac\pi{15}\right)=?$
| {
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Simultaneous congruences $3x \equiv 2 \pmod{5}$, $3x \equiv 4 \pmod{7}$, $3x \equiv 6 \pmod{11}$ I am stuck in a simultaneous linear congruence problem:
\begin{cases}
3x \equiv 2 \pmod{5} \\[4px]
3x \equiv 4 \pmod{7} \\[4px]
3x \equiv 6 \pmod{11}
\end{cases}
Using the Chinese remainder theorem,
I started with the 'highest' divisor: $11$. Since $(3, 11) =1$ there is a unique solution. $x= 6 \cdot 3 ^{\phi (11) -1} \equiv 6 \cdot 3^4\pmod{11}$ But to be honest, I have no clue how to continue.
Perhaps, cancel out the last equation to: $x \equiv 2 \pmod{11}$?
| It just jumped out at me that if I define $y=x+1$ the first two become
$$3y \equiv 0 \pmod 5\\3y \equiv 0 \pmod 7$$
which clearly calls for $y$ to be a multiple of $35$. Now we have $$3y \equiv 9 \pmod {11}\\y \equiv 3 \pmod {11}$$
We note that $35 \equiv 2 \pmod {11}$ so $7 \cdot 35 = 245 \equiv 7\cdot 2 \equiv 14\equiv 3 \pmod{11}$ and $y=245, x=244$ is a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $a,b$ at $f(x)=\frac{x^2+x-12}{x^2-ax+b}$ An High school question:
Given :$$f(x)=\frac{x^2+x-12}{x^2-ax+b}$$ it's given that $x=3$ is a vertical asymptote find $a$ and $b$.
I tried:
Since $x=3$ is a vertical asymptote then $3^2-3a+b=0$, but now what
| You can try also like that: Since $b= 3a-9$ we have $$f(x) = {(x+4)(x-3)\over x^2-ax+3a-9}= {(x+4)(x-3)\over (x-3)(x+3-a)} ={x+4\over x+3-a}$$
so $3-a=-3\implies a = 6$ and $b= 9$.
| {
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Evaluate the indefinite integral $ I = \int (x^2 + 2x)\cos(x) dx $
Integration by Parts, choose $u$:
$$\begin{align}
u &= \cos(x) \\
dv &= (x^2 + 2x)dx \\
du &= -\sin(x) \\
v &= \frac{1}{3}x^3 + x^2
\end{align}
$$
Substitute into formula:
$$
\begin{align}
\int udu &= uv - \int vdu \\
&= \cos(x)\left(\frac{1}{3}x^3 + x^2\right) - \int\left(\frac{1}{3}x^3 + 2x\right)(-\sin(x)) \\
&= \cos(x)\left(\frac{1}{3}x^3 + x^2\right) + \int\left(\frac{1}{3}x^3 + 2x\right)(\sin(x))
\end{align}
$$
At this point, it doesn't look like I can use the substitution rule on the the right hand integral, so I decide to use the substitution rule again.
Integration by Parts II, choose $u$:
$$\begin{align}
u &= sinx \\
dv &= (\frac{1}{3}x^3 + 2x)dx \\
du &= cosx \\
v &= \frac{1}{12}x^4 + x^2
\end{align}
$$
Substitute into formula:
$$\begin{align}
\int_{}udu &= uv - \int_{}vdu \\
&= (sinx)(\frac{1}{12}x^4 + x^2) - \int_{} (\frac{1}{12}x^4 + x^2)(cosx)dx
\end{align}
$$
Combining the two integration by parts together and I feel like I am no closer to evaluating the integral than whence I started...The integral is still there and I feel another parts by integration won't work.
$$\int(x^2 + 2x)\cos(x) = (\cos(x))\left(\frac{1}{3}x^3 + x^2\right) + (\sin(x))\left(\frac{1}{12}x^4 + x^2\right) - \int\left(\frac{1}{12}x^4 + x^2\right)(\cos(x))dx$$
Did I do the math wrong and make a mistake somewhere? Or am I supposed to approach this differently?
| I would do this way
$$\int(x^2+2x)\cos x\ dx$$
By Integration By Parts: $u=(x^2+2x),v^{\prime}=\cos x$
$$=(x^2+2x)\sin x-\int(2x+2)\sin xdx$$
Again apply Integration By Parts for $\int(2x+2)\sin xdx$ $u=(2x+2), v^{\prime}=\sin x$
and we get
$\int(2x+2)\sin xdx=2\sin x-\cos x(2x+2)$
So, we finally get,
$$=(x^2+2x)\sin x-[2\sin x-\cos x(2x+2)]+C$$
$$=(x^2+2x-2)\sin x+2(x+1)+C$$
$$\int(x^2+2x)\cos x\ dx=(x^2+2x-2)\sin x+2(x+1)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Asymptotic expansion of $\int_0^1 \frac{\operatorname{K}(r x)}{\sqrt{(1-r^2 x^2)(1-x^2)}} \, \mathrm{d} x $ Notation: For $\varphi \in [0,\frac{\pi}{2}]$ and $k \in [0,1)$ the definitions
$$ \operatorname{F}(\varphi,k) = \int \limits_0^\varphi \frac{\mathrm{d} \theta}{\sqrt{1-k^2 \sin^2(\theta)}} = \int \limits_0^{\sin(\varphi)} \frac{\mathrm{d} x}{\sqrt{(1 - k^2 x^2)(1-x^2)}} $$
and $\operatorname{K}(k) = \operatorname{F}(\frac{\pi}{2},k)$ will be used for the elliptic integrals of the first kind.
When answering this question, I came across the function
$$ \psi \colon [0,1) \to (0,\infty) \, , \, \psi(k) = \int \limits_0^1 \frac{\operatorname{K}(k x)}{\sqrt{(1-k^2 x^2)(1-x^2)}} \, \mathrm{d}x = \int \limits_0^{\pi/2} \frac{\operatorname{K}(k \sin(\theta))}{\sqrt{1-k^2 \sin^2 (\theta)}} \, \mathrm{d} \theta \, .$$
While $\psi(k) = \frac{\pi^2}{4} [1+ \frac{3}{8} k^2 + \mathcal{O}(k^4)]$ near $k=0$ is readily found using Maclaurin series, the expansion at $k=1$ is more elusive.
The naive attempt of replacing $\operatorname{K}(kx)$ by $\operatorname{K}(k)$ (since the largest contribution to the integral comes from the region near $x=1$) yields $\psi(k) \simeq \operatorname{K}^2 (k)$, which according to plots is not too far off but also not quite right. Integration by parts reproduces this term, but the remaining integral does not look very nice:
\begin{align}
\psi(k) &= \operatorname{K}^2 (k) - k \int \limits_0^1 \operatorname{K}'(k x) \operatorname{F}(\arcsin(x),k) \, \mathrm{d} x \\
&= \operatorname{K}^2 (k) - \int \limits_0^1 \left[\frac{\operatorname{E}(k x)}{1-k^2 x^2} - \operatorname{K}(kx)\right] \frac{\operatorname{F}(\arcsin(x),k)}{x} \, \mathrm{d} x \, .
\end{align}
The expansion $\operatorname{K}(k) = -\frac{1}{2} \log(\frac{1-k}{8}) + \mathcal{o}(1)$ is useful for the final steps, but I do not know how to extract all the leading terms, so:
How can we find the asymptotic expansion (ideally up to and including the constant terms) of $\psi(k)$ as $k \nearrow 1$ ?
| EllipitcElena's answer and Maxim's correction show that we have
$$ \psi(k) = - \frac{1}{4} \operatorname{K}(k) \log(1-k^2) + 2 \log(2) \operatorname{K}(k) + \chi(k) \, ,$$
where ($\psi_0$ is the digamma function, so $\psi$ was not the best choice)
$$ \chi (k) = \sum \limits_{m=1}^\infty \frac{\left(\frac{1}{2}\right)_m^2}{m!^2} \int \limits_0^1 \frac{(1-k^2 x^2)^{m-\frac{1}{2}}}{\sqrt{1-x^2}} \left[-\frac{1}{2} \log(1-k^2 x^2) + \psi_0(m+1)-\psi_0 \left(m + \frac{1}{2}\right)\right] \, \mathrm{d} x \, . $$
Near $k=1$ we find
$$ \psi (k) = \frac{1}{8} \left[\log^2 \left(\frac{1-k}{32}\right) - 4 \log^2 (2)\right] + \chi(1) + \mathcal{o}(1) \, .$$
$\chi(1)$ can be computed exactly:
\begin{align}
\chi(1) &= \sum \limits_{m=1}^\infty \frac{\left(\frac{1}{2}\right)_m^2}{m!^2} \int \limits_0^1 (1-x^2)^{m-1} \left[-\frac{1}{2} \log(1- x^2) + \psi_0(m+1)-\psi_0 \left(m + \frac{1}{2}\right)\right] \, \mathrm{d} x \\
&= \frac{1}{2} \sum \limits_{m=1}^\infty \frac{\left(\frac{1}{2}\right)_m^2}{m!^2} \left[-\frac{1}{2} \partial_1 \operatorname{B}\left(m,\frac{1}{2}\right) + \left(\psi_0(m+1)-\psi_0 \left(m + \frac{1}{2}\right)\right) \operatorname{B}\left(m,\frac{1}{2}\right) \right] \\
&= \frac{1}{2} \sum \limits_{m=1}^\infty \frac{\left(\frac{1}{2}\right)_m^2}{m!^2} \operatorname{B}\left(m,\frac{1}{2}\right) \left[\frac{1}{2}\left(\psi_0 \left(m + \frac{1}{2}\right) - \psi_0(m)\right) + \psi_0(m+1)-\psi_0 \left(m + \frac{1}{2}\right) \right] \, .
\end{align}
Using $\operatorname{B}(m,\frac{1}{2}) = \frac{(m-1)!}{\left(\frac{1}{2}\right)_m}$, $\left(\frac{1}{2}\right)_m = \frac{(2m)!}{4^m m!}$ and special values of $\psi_0$ this expression can be simplified:
$$ \chi(1) = \sum \limits_{m=1}^\infty \frac{{2m \choose m}}{2m 4^m} \left[\log(2) + H_m - H_{2m-1}\right] = \frac{1}{2} \log^2 (2) \, .$$
The final sum follows from the series
$$ \sum \limits_{m=1}^\infty \frac{{2m \choose m}}{2m 4^m} x^m = \log(2) - \log(1+\sqrt{1-x}) \, , \, x \in [-1,1] \, , $$
and is discussed in this question as well.
Putting everything together we obtain the rather nice result
$$ \boxed{\psi (k) = \frac{1}{8} \log^2 \left(\frac{1-k}{32}\right) + \mathcal{o} (1)} $$
as $k \nearrow 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proving that two determinants are equal without expanding them So I need to prove that
$$
\begin{vmatrix}
\sin^2(\alpha) & \cos(2\alpha) & \cos^2(\alpha) \\
\sin^2(\beta) & \cos(2\beta) & \cos^2(\beta) \\
\sin^2(\gamma) & \cos(2\gamma) & \cos^2(\gamma) \\
\end{vmatrix}
$$
$$
= \begin{vmatrix}
\sin(\alpha) & \cos(\alpha) & \sin(\alpha + \delta) \\
\sin(\beta) & \cos(\beta) & \sin(\beta + \delta) \\
\sin(\gamma) & \cos(\gamma) & \sin(\gamma + \delta) \\
\end{vmatrix}
$$
Now,
$$
\begin{vmatrix}
\sin^2(\alpha) & \cos(2\alpha) & \cos^2(\alpha) \\
\sin^2(\beta) & \cos(2\beta) & \cos^2(\beta) \\
\sin^2(\gamma) & \cos(2\gamma) & \cos^2(\gamma) \\
\end{vmatrix} = \begin{vmatrix}
\sin^2(\alpha) & \cos^2(\alpha) - \sin^2(\alpha) & \cos^2(\alpha) \\
\sin^2(\beta) & \cos^2(\beta) - \sin^2(\beta) & \cos^2(\beta) \\
\sin^2(\gamma) & \cos^2(\gamma) - \sin^2(\gamma) & \cos^2(\gamma) \\
\end{vmatrix}
$$
Adding column $1$ to column $2$ then makes column $2$ and column $3$ equal and hence the first determinant $ = 0$.
I'm stuck in trying to prove that the second determinant is also zero, so I need help in that.
| $$\begin{vmatrix}
\sin(\alpha) & \cos(\alpha) & \sin(\alpha + \delta) \\
\sin(\beta) & \cos(\beta) & \sin(\beta + \delta) \\
\sin(\gamma) & \cos(\gamma) & \sin(\gamma + \delta) \\
\end{vmatrix}=\begin{vmatrix}
\sin(\alpha) & \cos(\alpha) & \sin\alpha\color{red}{\cos\delta}+\color{red}{\sin\delta}\cos\alpha \\
\sin(\beta) & \cos(\beta) & \sin\beta\color{red}{\cos\delta}+\color{red}{\sin\delta}\cos\beta \\
\sin(\gamma) & \cos(\gamma) & \sin\gamma\color{red}{\cos\delta}+\color{red}{\sin\delta}\cos\gamma \\
\end{vmatrix}$$$${}$$
$$=\begin{vmatrix}
\sin(\alpha) & \cos(\alpha) & \sin\alpha \\
\sin(\beta) & \cos(\beta) & \sin\beta\\
\sin(\gamma) & \cos(\gamma) & \sin\gamma \\
\end{vmatrix}\color{red}{\cos\delta}+\begin{vmatrix}
\sin(\alpha) & \cos(\alpha) & \cos\alpha \\
\sin(\beta) & \cos(\beta) & \cos\beta\\
\sin(\gamma) & \cos(\gamma) & \cos\gamma \\
\end{vmatrix}\color{red}{\sin\delta}=0+0=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Asymptotic behavior of the integral $H(\beta)=\frac{2}{\pi}\beta\int_{0}^{\infty}\exp\left(-x^{3/2}\right)\sin(\beta x)xdx$ I found the integral $H(\beta)$ (which is called Holtsmark distribution) in Holtsmark's theory of ion field in plasma. In a book there is its asymptotic representation at small and great $\beta$:
$$
H(\beta)\approx
\left\{\begin{array}{l}
\frac{4\beta^2}{3\pi}\left(1-0,463\beta^2\right) &\beta\ll 1\\
\\
1,496\beta^{-5/2}\left(1+5,107\beta^{-3/2}+14,43\beta^{-3}\right)&\beta\gg 1
\end{array}\right.
$$
I know, how to get the first line: use Taylor series for $\sin(\beta x)$. But what should I do to prove the second line?
I tried to write the integral in the other form:
$$
H(\beta)=\frac{2}{\pi\beta}\int_{0}^{\infty}\exp\left(-\left(\frac{y}{\beta}\right)^{3/2}\right)\sin(y)ydy
$$
and use Taylor series for $e^{-\left(y/\beta\right)^{3/2}}$, but faced with the divergent integral, which appears due to the first term of series:
$$\int_{0}^{\infty}\sin(y)ydy$$
| Well, we are looking (in a more general sense) at the following integral:
$$\mathcal{H}_\text{n}\left(\beta\right):=\frac{2}{\pi}\cdot\beta\cdot\int_0^\infty x\cdot\sin\left(\beta\cdot x\right)\cdot\exp\left(-x^\text{n}\right)\space\text{d}x\tag1$$
Using the 'evaluating integrals over the positive real axis' property of the Laplace transform, we can write:
$$\mathcal{H}_\text{n}\left(\beta\right)=\frac{2}{\pi}\cdot\beta\cdot\int_0^\infty\mathcal{L}_x\left[x\cdot\sin\left(\beta\cdot x\right)\right]_{\left(\sigma\right)}\cdot\mathcal{L}_x^{-1}\left[\exp\left(-x^\text{n}\right)\right]_{\left(\sigma\right)}\space\text{d}\sigma\tag2$$
Using the 'frequency-domain derivative' property of the Laplace transform we can write:
$$\mathcal{L}_x\left[x\cdot\sin\left(\beta\cdot x\right)\right]_{\left(\sigma\right)}=-\frac{\partial}{\partial\sigma}\left\{\mathcal{L}_x\left[\sin\left(\beta\cdot x\right)\right]_{\left(\sigma\right)}\right\}=\frac{2\cdot\beta\cdot\sigma}{\left(\sigma^2+\beta^2\right)^2}\tag3$$
So:
$$\mathcal{H}_\text{n}\left(\beta\right)=\frac{2}{\pi}\cdot\beta\cdot\int_0^\infty\frac{2\cdot\beta\cdot\sigma}{\left(\sigma^2+\beta^2\right)^2}\cdot\mathcal{L}_x^{-1}\left[\exp\left(-x^\text{n}\right)\right]_{\left(\sigma\right)}\space\text{d}\sigma\tag4$$
Finding the other functions, we can write:
$$\mathcal{H}_\text{n}\left(\beta\right)=\frac{2}{\pi}\cdot\beta\cdot\int_0^\infty\frac{2\cdot\beta\cdot\sigma}{\left(\sigma^2+\beta^2\right)^2}\cdot\left\{\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{\sigma^{-1-\text{k}\text{n}}}{\Gamma\left(-\text{k}\cdot\text{n}\right)}\right\}\space\text{d}\sigma=$$
$$\frac{4\cdot\beta^2}{\pi}\cdot\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{1}{\Gamma\left(-\text{k}\cdot\text{n}\right)}\cdot\left\{\int_0^\infty\frac{\sigma^{-\text{k}\text{n}}}{\left(\sigma^2+\beta^2\right)^2}\space\text{d}\sigma\right\}\tag5$$
Using Mathematica I found that:
$$\int_0^\infty\frac{\sigma^{-\text{k}\text{n}}}{\left(\sigma^2+\beta^2\right)^2}\space\text{d}\sigma=\frac{\pi}{4}\cdot\frac{1+\text{kn}}{\beta^{3+\text{kn}}}\cdot\sec\left(\frac{\pi}{2}\cdot\text{kn}\right)\tag6$$
So, in the end we get:
$$\mathcal{H}_\text{n}\left(\beta\right)=\frac{4\cdot\beta^2}{\pi}\cdot\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{1}{\Gamma\left(-\text{k}\cdot\text{n}\right)}\cdot\left\{\frac{\pi}{4}\cdot\frac{1+\text{kn}}{\beta^{3+\text{kn}}}\cdot\sec\left(\frac{\pi}{2}\cdot\text{kn}\right)\right\}=$$
$$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{1}{\Gamma\left(-\text{k}\cdot\text{n}\right)}\cdot\frac{1+\text{kn}}{\beta^{1+\text{kn}}}\cdot\sec\left(\frac{\pi}{2}\cdot\text{kn}\right)\tag7$$
In your example we have $\text{n}=\frac{3}{2}$, so:
$$\mathcal{H}_\frac{3}{2}\left(\beta\right)=\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{1}{\Gamma\left(-\text{k}\cdot\frac{3}{2}\right)}\cdot\frac{1+\text{k}\cdot\frac{3}{2}}{\beta^{1+\frac{3}{2}\text{k}}}\cdot\sec\left(\frac{\pi}{2}\cdot\text{k}\cdot\frac{3}{2}\right)=$$
$$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{1}{\Gamma\left(-\text{k}\cdot\frac{3}{2}\right)}\cdot\frac{1+\text{k}\cdot\frac{3}{2}}{\beta^{1+\frac{3}{2}\text{k}}}\cdot\sec\left(\frac{3\pi}{4}\cdot\text{k}\right)\tag8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Prove that $a^2+b^2+c^2\geqslant\frac{1}{3}$ given that $a\gt0, b\gt0, c\gt0$ and $a+b+c=1$, using existing AM GM inequality Using the AM and GM inequality, given that
$a\gt0, b\gt0, c\gt0$ and $a+b+c=1$ prove that
$$a^2+b^2+c^2\geqslant\frac{1}{3}$$
| In the worst case possible you'd get $$a = b = c = \frac{1}{3} \Longrightarrow a^2 + b^2 + c^2 = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} \geq \frac{1}{3} $$
In the best case possible you'd get $$a = 1, b = c = 0 \Longrightarrow 1^2 + 0^2 + 0^2 = 1 \geq 1/3 $$
Therefore the inequality holds. Didn't use the AM-GM inequality, though.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $x = a \cos t^3 , y = b \sin t^3$ then what is $d^3y/dx^3$? If $ x = a \cos t^3 $, $ y = b \sin t^3 $, then what is $ \frac{d^3y}{dx^3} $?
I tried doing this problem by dividing $ \frac{d^3y}{dt^3} $ by $ \frac{d^3x}{dt^3} $ and got $ \frac{b}{a} $.
However my book says the third derivative doesn't exist. Why is this so?
| Given $x = a \cos t^3,y=b\sin t^3$,
the first derivative:
$$y'_x=\frac{y_t'}{x_t'}=\frac{(b\sin t^3)'}{(a\cos t^3)'}=\frac{b\cos t^3\cdot 3t^2}{-a\sin t^3\cdot 3t^2}=-\frac{b}{a}\cot t^3,$$
which does not exist at $t=\pi k,k\in\mathbb Z$.
The second derivative:
$$y''_{xx}=\frac{(y_x')'_t}{x_t'}=\frac{(-\frac ba\cot t^3)'}{(a\cos t^3)'}=\frac{-\frac ba\cdot \left(-\frac1{\sin ^2 t^3}\right)\cdot 3t^2}{-a\sin t^3\cdot 3t^2}=-\frac{b}{a^2}\csc^3 t^3,$$
The third derivative:
$$y'''_{xxx}=\frac{(y_{xx}'')'_t}{x_t'}=\frac{(-\frac b{a^2}\csc^3 t^3)'}{(a\cos t^3)'}=\\
=\frac{-\frac b{a^2}\cdot 3\csc ^2t^3\cdot (-\cot t^3)\cdot \csc t^3\cdot 3t^2}{-a\sin t^3\cdot 3t^2}=-\frac{3b}{a^3}\csc^4 t^3\cdot \cot t^3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $\cos^220^\circ-\cos20^\circ\sin10^\circ+\sin^210^\circ=\frac34$ The original exercise is to
Prove that
$$4(\cos^320^\circ+\sin^310^\circ)=3(\cos20^\circ+\sin10^\circ)$$
Dividing both sides by $\cos20^\circ+\sin10^\circ$ leads me to the problem in the question title.
I've tried rewriting the left side in terms of $\sin10^\circ$:
$$4\sin^410^\circ+2\sin^310^\circ-3\sin^210^\circ-\sin10^\circ+1\quad(*)$$
but there doesn't seem to be any immediate way to simplify further. I've considered replacing $x=10^\circ$ to see if there was some observation I could make about the more general polynomial $4x^4-2x^3-3x^2-x+1$ but I don't see anything particularly useful about that. Attempting to rewrite in terms of $\cos20^\circ$ seems like it would complicate things by needlessly(?) introducing square roots.
Is there a clever application of identities to arrive at the value of $\dfrac34$? I have considered
$$\cos20^\circ\sin10^\circ=\frac{\sin30^\circ-\sin10^\circ}2=\frac14-\frac12\sin10^\circ$$
which eliminates the cubic term in $(*)$, and I would have to show that
$$4\sin^410^\circ-3\sin^210^\circ+\frac12\sin10^\circ=0$$
$$4\sin^310^\circ-3\sin10^\circ+\frac12=0$$
| A complex numbers proof of the original exercise.
By Euler's formula, we have that
$$\text{Re}\left((\cos 20^\circ+i\sin 20^\circ)^3\right)=\cos(60^\circ)=\frac{1}{2}=\sin(30^\circ)=\text{Im}\left((\cos 10^\circ+i\sin 10^\circ)^3\right)$$
Hence
$$\cos^3 20^\circ-3\cos 20^\circ \sin^2 20^\circ
=3\cos^2 10^\circ\sin 10^\circ -\sin^3 10^\circ$$
or
$$\cos^3 20^\circ-3\cos 20^\circ (1-\cos^2 20^\circ)
=3(1-\sin^2 10^\circ)\sin 10^\circ -\sin^3 10^\circ
$$
which implies
$$4(\cos^320^\circ+\sin^310^\circ)=3(\cos20^\circ+\sin10^\circ).$$
The very same argument leads to the following generalization: if $c^\circ +s^\circ=30^\circ$ then
$$4(\cos^3 c^\circ+\sin^3 s^\circ)=3(\cos c^\circ+\sin s^\circ).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Is my integral in fully reduced form?
I have to integrate this:
$$\int_0^1 \frac{x-4}{x^2-5x+6}\,dx$$
Now
$$\int_0^1 \frac{x-4}{(x-3)(x-2)}\, dx$$
and by using partial fractions we get
$$\frac{x-4}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2}$$
$$x-4 = A(x-2) + B(x-3)$$
$$= Ax - 2A + Bx - 3B$$
$$x-4 = (A+B)x - 2A - 3B$$
so
$$A+B=1$$ or $$2X+2B = 2$$
$-B = -2$ and $B = 2$ and $A = -1$
Then
$$\int_0^1 \frac{x-4}{x^2-5x+6} = \int_0^1 \frac{-1}{x-3}dx + \int_0^1 \frac{2}{x-2} dx$$
usub using $u = x-3$ and $du = dx$
so $$ -1 \ln | x-3 | \rbrack_0^1 + 2 \ln |x-2| \rbrack_0^1$$
$$-1 ( \ln 2 - \ln 3) + 2 (\ln 1 - \ln 2)$$
$$-\ln 2 + \ln 3 + 2\ln 1 - 2\ln 2 = -3 * \ln 2 + \ln 3 + 2\ln 1$$
Is there anyway to simplify this further? Wolfram has the answer at $-\ln(8/3)$ and I'm not sure how to simplify to here? How does my work look?
| Don't forget that when you add two logarithms with the same base, the expressions under the logarithm sign are multiplied: $\log_b{x}+\log_b{y}=\log_b{(xy)}$. When two logarithms with the same base are subtracted, you divide: $\log_b{x}-\log_b{y}=\log_b{\frac{x}{y}}$. And also don't forget that the power on the expression under the logarithm sign can always be brought in or out of the logaritm: $\log_b{x^a}=a\log_b{x}$.
$$
-( \ln 2 - \ln 3) + 2 (\ln 1 - \ln 2)=-\ln\frac{2}{3}+2\ln\frac{1}{2}=
-\ln\frac{2}{3}+\ln\left(\frac{1}{2}\right)^2=\\
\ln\frac{1}{4}-\ln\frac{2}{3}=
\ln\left(\frac{1}{4}\div\frac{2}{3}\right)=
\ln\frac{3}{8}=
\ln\left(\frac{8}{3}\right)^{-1}=
-\ln\frac{8}{3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3118767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to evaluate $\int_{0}^{1}{\frac{x\ln(1+x)\ln(1+x^2)}{1+x^2}}dx$
How to evaluate
$$\int_{0}^{1}{\frac{x\ln(1+x)\ln(1+x^2)}{1+x^2}}dx$$
My attempt
$$\begin{align}\int_{0}^{1}{\frac{x\ln(1+x)\ln(1+x^2)}{1+x^2}}dx&=\frac{1}{4}\int_{0}^{1}{\ln(1+x)}d(\ln^2(1+x^2))\\ &=\frac{1}{4}(\ln(1+x)\ln^2(1+x^2)|_0^1-\int_{0}^{1}{\ln^2(1+x^2)}d\ln(1+x))\\ &=\frac{1}{4}(\ln^32-\int_{0}^{1}\frac{\ln^2(1+x^2)}{1+x}dx)\\
\end{align}$$
$$\begin {align}
&\int_{0}^{1} \frac{\ln ^{2}\left(1+x^{2}\right)}{1+x} d x\\=&\int_{0}^{1} \frac{\ln ^{2}\left(1-y^{2}\right)}{1+i y} i d y-\int_{0}^{\frac{\pi}{2}} \frac{\ln ^{2}\left(1+e^{i 2 \theta}\right)}{1+e^{i \theta}} i e^{i \theta} d\theta\\
=&\int_{0}^{1} \frac{y \ln ^{2}\left(1-y^{2}\right)}{1+y^{2}} dy+\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \tan \left(\frac{\theta}{2}\right) \ln ^{2}(2 \cos \theta)d\theta+\int_{0}^{\frac{\pi}{2}} \theta \ln (2 \cos \theta)d\theta\\&-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \theta^{2} \tan \left(\frac{\theta}{2}\right)d\theta
\end {align}$$
But how to evaluate
$$\int_{0}^{1}\frac{\ln^2(1+x^2)}{1+x}dx$$
The answer from Mathematica is $$\frac{1}{96}(24\pi\mathbf{G}+\pi^2\ln2+8\ln^32-60\zeta(3))$$
$\mathbf{G}$ is Catalan's constant.
My answer
| You already used contour integration to split the integral into four parts:
$$\begin {align}
&\int_{0}^{1} \frac{\ln ^{2}\left(1+x^{2}\right)}{1+x}dx\\
=&\int_{0}^{1} \frac{x \ln ^{2}\left(1-x^{2}\right)}{1+x^{2}} dx+\frac{1}{2} \int_{0}^{\pi/2} \tan \left(\frac{x}{2}\right) \ln ^{2}(2 \cos x)dx+\int_{0}^{\pi/2} x \ln (2 \cos x)dx\\&-\frac{1}{2} \int_{0}^{\pi/2} x^{2} \tan \left(\frac{x}{2}\right)dx
\end {align}$$
Denote those four integrals by $I_1,\cdots, I_4$.
$$I_1 = \frac{1}{2}\int_0^1 \frac{\ln^2(1-x)}{1+x}dx = \frac{1}{2}\int_0^1 \frac{\ln^2(x)}{2-x}dx = \text{Li}_3(\frac{1}{2}) = \frac{\ln^2 3}{6}-\frac{\pi^2}{12}\ln 2 +\frac{7}{8}\zeta(3)$$
For $I_2$, tangent-half substitution gives
$$I_2 = \int_0^1\frac{2t\ln^2(2\frac{1-t^2}{1+t^2})}{1+t^2}dt = \int_0^1 \frac{\ln^2(2\frac{1-t}{1+t})}{1+t}dt = \int_0^1 \frac{\ln^2 (2t)}{1+t}dt = \frac{3\zeta(3)}{2}+\ln^3 2 -\frac{\pi^2}{6}\ln 2$$
where the penultimate step involves $t\mapsto (1-t)/(1+t)$.
For $I_3$, integration by part gives
$$I_3 = \int_0^{\pi/2} (\frac{\pi}{2}-x)\ln(2\sin x) dx = -\frac{\pi^2}{8}\ln 2+\frac{1}{2}\int_0^{\pi/2} x^2 \cot x dx $$
Use the identity
$$\sum_{n=1}^N \sin(nx) = \frac{1}{2}\cot\frac{x}{2}-\frac{\cos(N+1/2)x}{2\sin(x/2)}$$
letting $N\to \infty$, and the remainder $\to 0$ by Riemann-Lebesuge lemma, so
$$\int_0^{\pi/2} x^2\cot x dx = 2\sum_{n=1}^\infty \int_0^{\pi/2} x^2\sin(2nx) dx = \frac{\pi^2}{4}\ln 2 -\frac{7\zeta(3)}{8}$$ Note that each summand can be evaluated explicitly. Therefore $I_3 = -7\zeta(3)/16$.
For $I_4$, use the same identity, gives
$$I_4 = 2\sum_{n=1}^\infty \int_0^{\pi/2} x^2\sin[n(\pi-x)]dx = 2\sum_{n=1}^\infty (-1)^{n+1} \int_0^{\pi/2} x^2\sin (nx)dx$$ the series definition of Catalan constant immediately gives
$$I_4 = 2\pi G - \frac{21\zeta(3)}{8} - \frac{\pi^2}{4}\ln 2$$
Combining all these should give you the result.
| {
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"url": "https://math.stackexchange.com/questions/3124493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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Find shortest distance from point on ellipse to focus of ellipse.
Find the point $(x,y)$ on the ellipse $b^2x^2 + a^2y^2 = a^2b^2$ such that the distance to the focus $(c,0)$ is a minimum.
My book I got this problem out of gave a suggestion saying to express the distance as a function of $x$ and work the problem and then express the distance as a function of $y$ and work the problem.
So, I tried to work it out:
$$x^2 = \frac{a^2b^2 - a^2y^2}{b^2}$$
$$y^2 = \frac{a^2b^2 - b^2x^2}{a^2}$$
Now, I write the distance function:
$$d = \sqrt{(x-c)^2 + y^2}$$
Then, I express the distance function as a function of x:
$$d = \sqrt{(x-c)^2 + \frac{a^2b^2 - b^2x^2}{a^2}}$$
Then, I took the derivative w.r.t $x$:
$$d' = 2x - 2c - \frac{2b^2}{a^2}$$
Now, I did as the book told me, and expressed the distance function as a function of y:
$$d' = -\frac{2a^2}{b^2} + \frac{4a^2c}{bx} + 2y$$
I tried to set both derivatives to zero and attempted to solve, but I couldn't really figure out what to do (my algebra is poor).
Any help would be appreciated.
Thanks.
Edit: I need a calculus solution (I'm working out of a calculus book).
| The hint.
We need to find a a values of $d>0$, for which the system
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ and
$$(x-c)^2+y^2=d^2$$ has solutions.
We obtain:
$$\frac{x^2}{a^2}+\frac{d^2-(x-c)^2}{b^2}=1$$ or
$$\left(\frac{1}{a^2}-\frac{1}{b^2}\right)x^2+\frac{2cx}{b^2}+\frac{d^2-c^2}{b^2}-1=0$$ or
$$-\frac{c^2x^2}{a^2b^2}+\frac{2cx}{b^2}+\frac{d^2-a^2}{b^2}=0$$ or
$$c^2x^2-2ca^2x+a^2(a^2-d^2)=0$$ or
$$(cx-a^2)^2=a^2d^2,$$ which gives
$$x=\frac{a^2+ad}{c},$$ which is impossible or
$$x=\frac{a^2-ad}{c}.$$
Now, we need that the equation $$\frac{\left(\frac{a^2-ad}{c}\right)^2}{a^2}+\frac{y^2}{b^2}=1$$ has real roots for which we need
$$1-\frac{\left(\frac{a^2-ad}{c}\right)^2}{a^2}\geq0$$ or
$$c^2-(a-d)^2\geq0,$$ which gives
$$a-c\leq d\leq a+c.$$
Id est, $d\geq a-c$.
The equality occurs for the point $(a,0)$ on the ellipse, which says that $a-c$ it's the answer.
Another way.
Let $F_1(c,0)$, $F_2(-c,0)$, $A(a,0)$ and let $M(x,y)$ be a point on the ellipse.
Thus, $$MF_1+MF_2=2a$$
and by the triangle inequality $$MF_1+F_1F_2\geq MF_2$$ or
$$2MF_1+F_1F_2\geq MF_1+MF_2$$ or
$$2MF_1+2c\geq2a,$$ which gives
$$MF_1\geq a-c.$$
The equality occurs, when $M\equiv A$, which says that $a-c$ is a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
integrate sin(x)cos(x) using trig identity. Book tells me the answer is:
$$ \int \sin(x)\cos(x) dx = \frac{1}{2} \sin^{2}(x) + C $$
however, I get the result:
$$
\sin(A)\cos(B) = \frac{1}{2} \sin(A-B)+\frac{1}{2}\sin(A+B)
$$
$$
\begin{split}
\int \sin(x)\cos(x) dx
&= \int \left(\frac{1}{2}\sin(x-x) + \frac{1}{2}\sin(x+x)\right) dx \\
&= \int \left(\frac{1}{2}\sin(0) + \frac{1}{2}\sin(2x)\right) dx \\
&= \int \left(\frac{1}{2}\sin(2x)\right) dx \\
&= -\frac{1}{2} \frac{1}{2}\cos(2x) +C\\
&= -\frac{1}{4} \cos(2x) +C
\end{split}
$$
How did the book arrive at the answer $\frac{1}{2}\sin^2(x)$?
| To answer the actual question (“How did the book”, etc.): probably just by inspection (chain rule backwards), or else via the substitution $u=\sin x$, $du = \cos x \, dx$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
} |
Solve for $x$ : $\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$? I want to solve the following equation for $x$ :
$$\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$$
My approach:
Let the given eq.:
$$\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9 \tag {i}$$
On rearranging, we get:
$$\sqrt{x-6} \, + \, \sqrt{x+6} = 9 \, - \, \sqrt{x-1} $$
On Squaring both sides, we get:
$$(x-6) \, + \, (x+6) + 2 \,\,. \sqrt{(x^2-36)} = 81 + (x-1)\, - 18.\, \sqrt{x-1}$$
$$\implies 2x + 2 \,\,. \sqrt{(x^2-36)}= 80 + x - 18.\, \sqrt{x-1}$$
$$\implies x + 2 \,\,. \sqrt{(x^2-36)}= 80 - 18.\, \sqrt{x-1} \tag{ii}$$
Again we are getting equation in radical form.
But, in Wolfram app, I am getting its answer as $x=10$, see it in WolframAlpha.
So, how to solve this equation? Please help...
| Hint:
Clearly,the LHS is increasing function of $x$
so,we cannot have multiple roots
For real solution, $x\ge6$
Also, $3\sqrt{x+6}>9>3\sqrt{x-6}$
$\implies x+6>9>x-6\iff3< x<15\implies6\le x<15$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Is this sequence following some rule: $(1+3)^2 + (1+2)^2=(2+3)^2$ , $(1+3+5)^2 + (3+4+5)^2=(4+5+6)^2$ , ...? I found its interesting but don't know whether there is any rule that is governing its next term.
The sequence is given as follows:
$$(1+3)^2 + (1+2)^2=(2+3)^2$$
$$(1+3+5)^2 + (3+4+5)^2=(4+5+6)^2$$
$$(1+3+5+7)^2 + (6+7+8+9)^2=(7+8+9+10)^2$$
$$(1+3+5+7+9)^2 + (10+11+12+13+14)^2=(11+12+13+14+15)^2$$
$$\dots$$
$$\text{And so on...}$$
I know how it's coming,
i.e, Pythagorean triplets
For eg.:$(1+3)^2 + (1+2)^2=(2+3)^2$ = ($3,4,5$) Pythagorean triplet
But my question is "Does it following any rule for its next term" ?
| A simple induction gives $\left(\frac{n^3+n}{2}\right)^2$ for the right-hand side. On the left-hand side we have $(n^2)^2 + \left(\frac{n^3+n}{2}- n\right)^2$. Since
$$(n^2)^2 + \left(\frac{n^3+n}{2}- n\right)^2 = n^4 + \left(\frac{n^3+n}{2}\right)^2 - 2\frac{n^4+n^2}{2} + n^2 = \left(\frac{n^3+n}{2}\right)^2,$$
they are the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Find $\max\{y-x\}$
If $x+y+z=3, $ and $x^2+y^2+z^2=9$ , find $\max\{y-x\}$.
I tried to do this geometrically, $x+y+z=3$ is a plane in $\Bbb{R}^3$ and $x^2+y^2+z^2=9$ is a ball with radius 3 and center of origin . So the candidate points for $y-x$ are on the intersection of the plane and the ball. But now I am confused how to choose to make $y-x$ maximized.
| Note that $y^2=(3-x-z)^2=9-x^2-z^2\implies z^2+(x-3)z+(x^2-3x)=0$ so $$2z=3-x\pm\sqrt3\sqrt{3+x-x^2}\implies y=3-x-z=\frac{3-x}2\mp\frac{\sqrt3}2\sqrt{3+2x-x^2}$$ giving $$\max\{y-x\}=\max\left\{\frac32-\frac32x+\frac{\sqrt3}2\sqrt{3+2x-x^2}\right\}$$ and differentiating gives $$-\frac32+\frac{\sqrt3}4\cdot\frac{2-2x}{\sqrt{3+2x-x^2}}=0\implies (1-x)^2=3(3+2x-x^2)$$ so $x^2-2x-2=0\implies x=1\pm\sqrt3$, and $\max\{y-x\}=2\sqrt3$ when $x=1-\sqrt3$.
| {
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"url": "https://math.stackexchange.com/questions/3132235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Leibniz integral rule for solving integrals I am giving as homework to solve these integral
1) $\int \limits_{0}^{1} \frac{\arctan(y x)}{x \sqrt{1-x^2}}dx$
2)$\int \limits_{0}^{\frac{\pi}{2}} \frac{x}{\tan x}dx $ with the hint of $0\leq y \leq 1$ and $\int \limits_{0}^{\frac{\pi}{2}} \frac{\arctan( y \tan x)}{\tan x}dx $
If any one could solve any one of them because i am new to this and don't have a clue on how to proceed.
Thanks.
| $$I_1(y)=\int_0^1 \frac{\arctan(yx)}{x\sqrt{1-x^2}}dx\Rightarrow I_1(0)=0$$
$$I_2(y)=\int_0^{\pi/2}\frac{\arctan(y\tan x)}{\tan x}dx\Rightarrow I_2(0)=0$$
First for $I_1$, take $\frac{d}{dy}$ on both sides (and use the Leibniz integral rule)
$$I_1'(y)=\int_0^1\frac1{x\sqrt{1-x^2}}\frac{\partial}{\partial y}\arctan(yx)dx$$
$$I_1'(y)=\int_0^1\frac1{x\sqrt{1-x^2}}\frac{x}{1+y^2x^2}dx$$
$$I_1'(y)=\int_0^1\frac{dx}{(1+y^2x^2)\sqrt{1-x^2}}$$
Then set $x=\sin(u)$:
$$I_1'(y)=\int_0^{\pi/2}\frac{du}{1+y^2\sin^2u}$$
Then set $u=t/2$:
$$I_1'(y)=\int_0^\pi\frac{dt}{2+y^2-y^2\cos t}$$
Then let $x=\tan(t/2)$:
$$I_1'(y)=2\int_0^\infty \frac{1}{2+y^2+y^2\frac{x^2-1}{x^2+1}}\frac{dx}{x^2+1}$$
$$I_1'(y)=\int_0^\infty \frac{dx}{(1+y^2)x^2+1}$$
And since it is easily shown that
$$\int_0^\infty \frac{dx}{ax^2+1}=\int_0^\infty \frac{dx}{x^2+a}=\frac\pi{2\sqrt{a}}$$
We have that
$$I_1'(y)=\frac\pi{2\sqrt{1+y^2}}$$
And since $I_1(0)=0$ we have that
$$I_1(y)=\frac\pi2\int_0^y \frac{da}{\sqrt{1+a^2}}=\frac\pi2\sinh^{-1}(y)$$
Like with $I_1$, we compute $I_2$ by taking $d/dy$ on both sides which gives
$$I_2'(y)=\int_0^{\pi/2} \frac1{\tan x}\frac{\partial}{\partial y}\arctan(y\tan x)dx$$
$$I_2'(y)=\int_0^{\pi/2} \frac1{\tan x}\frac{\tan x}{1+y^2\tan^2x}dx$$
$$I_2'(y)=\int_0^{\pi/2}\frac{dx}{1+y^2\tan^2x}$$
We can re write this as
$$I_2'(y)=\int_0^{\pi/2}\frac{\sec(x)^2dx}{(1+\tan(x)^2)(1+y^2\tan(x)^2)}$$
We then use $u=\tan(x)$ to get
$$I_2'(y)=\int_0^\infty \frac{du}{(1+u^2)(1+y^2u^2)}$$
Then we can use partial fractions and a trig sub to get
$$I_2'(y)=\frac\pi{2y+2}$$
And since $I_2(0)=0$, we have that
$$I_2(y)=\frac\pi2\int_0^y \frac{dt}{t+1}=\frac\pi2\ln(y+1)$$
So we plug in $y=1$ to get that
$$\int_0^{\pi/2}\frac{x}{\tan x}=\frac\pi2\ln2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3136159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrals involving multi-valued functions Show that
$$\int_{0}^{\infty}{\frac{\cos{x}}{x^\alpha}dx}=\mathrm{\Gamma}\left(1-a\right)\sin{\left(\frac{\mathrm{\pi\alpha}}{2}\right)}\ ,\ \ \ \ 0<\alpha<1$$
what would be the contour of this integration if I want to solve this integration using residue theorem
| Assuming you wish to show
$$\int_0^\infty \frac{\cos x}{x^\alpha} \, dx = \Gamma (1 - \alpha) \sin \left (\frac{\pi \alpha}{2} \right ), \qquad 0 < \alpha < 1,$$
what follows is a real method that can be used.
Let
$$I = \int_0^\infty \frac{\cos x}{x^\alpha} \, dx.$$
By employing the following useful property for the Laplace transform:
$$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (t) \cdot \mathcal{L}^{-1} \{g(x)\} (t) \, dt.$$
Noting that
$$\mathcal{L} \{\cos x\}(t) = \frac{t}{1 + t^2},$$
and
$$\mathcal{L}^{-1} \left \{\frac{1}{x^\alpha} \right \} (t)= \frac{1}{\Gamma (\alpha)} \mathcal{L}^{-1} \left \{\frac{\Gamma (\alpha)}{x^{(\alpha - 1) + 1}} \right \} (t) = \frac{t^{\alpha - 1}}{\Gamma (\alpha)},$$
then
\begin{align}
I &= \int_0^\infty \cos x \cdot \frac{1}{x^\alpha} \, dx\\
&= \int_0^\infty \mathcal{L} \{\cos x\} (t) \cdot \mathcal{L}^{-1} \left \{\frac{1}{x^\alpha} \right \} (t) \, dt\\
&= \frac{1}{\Gamma (\alpha)} \int_0^\infty \frac{t^\alpha}{1 + t^2} \, dt.
\end{align}
Setting $u = t^2$, one has
\begin{align}
I &= \frac{1}{2 \Gamma (\alpha)} \int_0^\infty \frac{u^{\frac{\alpha}{2} - \frac{1}{2}}}{1 + u} \, du\\
&= \frac{1}{2 \Gamma (\alpha)} \operatorname{B} \left (\frac{1}{2} - \frac{\alpha}{2}, \frac{1}{2} + \frac{\alpha}{2} \right ) \tag1\\
&= \frac{1}{2 \Gamma (\alpha)} \Gamma \left (\frac{1}{2} - \frac{\alpha}{2} \right ) \Gamma \left (\frac{1}{2} + \frac{\alpha}{2} \right ) \tag2\\
&= \frac{1}{2 \Gamma (\alpha)} \Gamma \left [1 - \left (\frac{1}{2} + \frac{\alpha}{2} \right ) \right ] \Gamma \left (\frac{1}{2} + \frac{\alpha}{2} \right )\\
&= \frac{1}{2 \Gamma (\alpha)} \frac{\pi}{\sin \left (\frac{\pi}{2} - \frac{\pi \alpha}{2} \right )} \tag3\\
&= \frac{1}{2 \Gamma (\alpha)} \frac{\pi}{\cos \left (\frac{\pi \alpha}{2} \right )}\\
&= \frac{\Gamma (1 - \alpha) \sin (\pi \alpha)}{2 \pi} \cdot \frac{\pi}{\cos \left (\frac{\pi \alpha}{2} \right )} \tag4\\
&=\frac{\Gamma (1 - \alpha) \sin \left (\frac{\pi \alpha}{2} \right ) \cos \left (\frac{\pi \alpha}{2} \right )}{\cos \left (\frac{\pi \alpha}{2} \right )}\tag5\\
&= \Gamma (1 - \alpha) \sin \left (\frac{\pi \alpha}{2} \right ),
\end{align}
as required to show.
Explanation
(1) Using $\operatorname{B} (x,y) = \displaystyle{\int_0^\infty \frac{t^{x - 1}}{(1 + t)^{x + y}} \, dt}$.
(2) Using $\operatorname{B}(x,y) = \dfrac{\Gamma (x) \Gamma (y)}{\Gamma (x + y)}$.
(3) Using the reflexion formula for the gamma function: $\Gamma (1 - z) \Gamma (z) = \dfrac{\pi}{\sin (\pi z)}$.
(4) Again using the reflexion formula for the gamma function.
(5) Using the double angle formula for sine.
| {
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"timestamp": "2023-03-29T00:00:00",
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solve $\int_{-\sqrt{3}}^{\sqrt{3}} 4 \sqrt{3-y^2}dy$ $\int_{-\sqrt{3}}^{\sqrt{3}} 4 \sqrt{3-y^2}dy$
trig sub
$y = \sqrt{3}\sin(u)$
$dy = \sqrt{3}\cos(u)du$
\begin{align}\int_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sqrt{3})}& 4 \sqrt{3-3\sin^2(u)}\sqrt{3}\cos(u)\,du = 12 \int_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sqrt{3})} \cos^2(u)\,du\\
&= 6\int_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sqrt{3})} (1+\cos(2u))du\\
&=\left[6u + 3\sin(2u)\right]_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sqrt{3})}
\end{align}
If I plug in the limits to you I get some insanely low number while the answer should be $6 \pi$
what do I do ?
| Say $x(y) = \sqrt{3-y^2}$. This is a formula for a semi-circle of radius $\sqrt{3}$, valid from $y=-\sqrt{3}$ to $y = \sqrt{3}$.
The area of this semi-circle is $\pi r^2 / 2 = 3\pi/2$, which is $\int_{-\sqrt{3}}^{\sqrt{3}}x(y)\,dy$.
Your integral has an extra factor of four in it, so you need to multiply the result by 4 to get $6\pi$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\int\frac{\sin x+\tan x}{\cos x+\csc x}dx$ Finding $\displaystyle \int\frac{\sin x+\tan x}{\cos x+\csc x}dx$
what i try
$\displaystyle \Lambda =\int\frac{\sin^2 x(1+\cos x)}{\cos x(\sin x\cos x+1)}dx$
$\displaystyle \Lambda=\int\frac{\sin^4 x}{\cos x(1-\cos x)(\sin x\cos x+1)}dx$
How do i solve it help me please
| Hint:
$$\text{Define }I=\underbrace{\int\dfrac{\tan x}{\cos x+\csc x}\mathrm dx}_{I_1}+\underbrace{\int\dfrac{\sin x}{\cos x+\csc x}\mathrm dx}_{I_2}$$
For $I_1$, substitute $u=\tan(x/2)$ (Weierstrass substitution) and perform Partial Fraction Decomposition twice.
For $I_2$, rewrite everything in terms of $\tan x$ and $\sec x$, let $v=\tan x$ and perform Partial Fraction Decomposition.
The antiderivative is not very pleasing, have a look: $$I=\ln\left(\left|\sec\left(x\right)\right|\right)-\dfrac{\ln\left(\tan^2\left(x\right)+\tan\left(x\right)+1\right)}{2}+\dfrac{1}{\sqrt{3}}\left[\arctan\left(\frac{2\tan\left(x\right)+1}{\sqrt{3}}\right)-\ln\left(\left(\tan\left(\frac{x}{2}\right)-1\right)\left(\tan\left(\frac{x}{2}\right)+\sqrt{3}\right)+2\right)\\ +\ln\left(\left(\tan\left(\frac{x}{2}\right)-1\right)\left(\tan\left(\frac{x}{2}\right)-\sqrt{3}\right)+2\right)\right]+\ln\left(\left|\tan\left(\dfrac{x}{2}\right)+1\right|\right)-\ln\left(\left|\tan\left(\dfrac{x}{2}\right)-1\right|\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140892",
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How to find the roots of $\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$? Find all $x$ in the interval $(0,\pi/2)$ such that $\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$.
The options are (i)$\pi/9,2\pi/7$, (ii)$\pi/36,11\pi/12$ (iii)$\pi/12,11\pi/36$ (iv) All
I have been able to find one value of $x$, $\pi/12$. How do I find the other root(s)?
My attempt:
$\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$
or, $\frac{\sin\pi/3-\sin\pi/6}{\sin x}+\frac{\cos\pi/6+\cos\pi/3}{\cos x}=2\sqrt{2}$
or, $\frac{\sin(\pi/4)cos(\pi/12)}{\sin x}+\frac{\cos(\pi/4)cos(\pi/12)}{\cos x}=\sqrt{2}$
or, $\sin(x+\pi/12)=\sin2x$
or, $x=\pi/12$
| Use $$\sin15^{\circ}=\frac{\sqrt3-1}{2\sqrt2}$$ and
$$\cos15^{\circ}=\frac{\sqrt3+1}{2\sqrt2}.$$
We obtain:
$$\sin(15^{\circ}+x)=\sin2x.$$
Thus, $$15^{\circ}+x=2x+360^{\circ}k,$$ where $k$ is an integer number, or
$$15^{\circ}+x=180^{\circ}-2x+360^{\circ}k.$$
Can you end it now?
| {
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"url": "https://math.stackexchange.com/questions/3142283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Equilateral triangle with vertices on 3 concentric circles Is my idea correct? 3 concentric circles of radius 1, 2 and 3 are given. An equilateral triangle is formed having its vertices lie on the side of the three concentric circles. What is the length of the equilateral triangle?
My idea is to set a point at the middle of the triangle, then use the distance of it to the vertices given that the three concentric circles are set as $$x^2 + y^2 = 1$$$$x^2 + y^2 = 4$$ and $$x^2 + y^2 = 9$$ I will manipulate the formula afterwards,,,
| Prompted by an additional point in a duplicate question asking whether the construction is always possible for arbitrary concentric circles, the following is about the general case of radii $\,A,B,C\,$ (where OP's question here is about $\,A=1,B=2,C=3\,$).
Let $\,a,b,c\,$ be points in the complex plane on the respective circles, and assume WLOG that $\,a=A \in \mathbb R\,$, since the problem is invariant to rotations. Let $\,\omega\,$ be a complex cube root of unity, then $\,\omega^2+\omega+1=0\,$, and the condition for the triangle $\,a,b,c,\,$ to be equilateral is:
$$
c-a + \omega(b-a)=0 \;\;\iff\;\; c = - \omega b + (1+\omega)a \tag{1} \;\;\iff\;\; c = - \omega b - \omega^2a
$$
Multiplying by the conjugate, and using that $\,\omega\overline\omega=|\omega|^2=1\,$, $\,\overline\omega=\omega^2\,$:
$$
\begin{align}
|c|^2 &= |b|^2+|a|^2 + a \left(\omega\overline\omega^2b+\overline\omega\omega^2\overline b\right) = |b|^2+|a|^2 + 2a \,\text{Re} \left(\omega^2 b\right)
\end{align}
$$
It follows that:
$$
\text{Re} \left(\omega^2 b\right) \,=\,\frac{|c|^2-|a|^2-|b|^2}{2a}=\frac{C^2-A^2-B^2}{2A} \tag{2}
$$
Since $\,\left|\omega^2 b\right|=|b|\,$, the condition for existence of a solution is:
$$
\left|\text{Re} \left(\omega^2 b\right)\right| \le |b| \quad\iff\quad \left(C^2-A^2-B^2\right)^2 \le 4 A^2 B^2 \tag{3}
$$
In that case:
$$
\text{Im} \left(\omega^2 b\right) \,=\,\pm \sqrt{|b|^2 - \left(\text{Re} \left(\omega^2 b\right)\right)^2}=\pm\sqrt{B^2-\left(\frac{C^2-A^2-B^2}{2A}\right)^2} \tag{4}
$$
When a solution exists according to $\,(3)\,$, equations $(2),(4)$ give $\,b\,$ (up to a reflection about the real axis), then $\,(1)\,$ gives $\,c\,$.
Quick verification for OP's case $\,a=A=1, B=2, C=3\,$, which satisfies $\,(3)\,$ so solutions exist.
*
*From $\,(2)\,$: $\;\text{Re}\left(\omega^2b\right)=\frac{9-1-4}{2}=2\,$.
*From $\,(4)\,$: $\;\text{Im}\left(\omega^2b\right)=\pm\sqrt{4-2^2}=0\,$.
*Then $\,\omega^2 b=\text{Re}(\omega^2b)+i\,\text{Im}(\omega^2b)=2\,$, so $\,b = 2\omega\,$.
*From $\,(1)\,$: $\;c=-\omega\cdot 2 \omega - \omega^2 \cdot 1 =-3\omega^2\,$.
Therefore the solution is $\,\big\{\,1, \,2 \omega, \,-3\omega^2 \,\big\}\,$ up to rotations and reflections, and it can be easily verified that the side length of the equilateral triangle is $\,|b-1|=\sqrt{7}\,$.
[ EDIT ] $\;$ Condition $\,(3)\,$ can be written as $\,\left(2AB\right)^2 - \left(A^2+B^2-C^2\right)^2 \ge 0\,$ and, after factoring the differences of squares twice, is equivalent to:
$$
(A+B+C)(A+B-C)(A-B+C)(-A+B+C) \ge 0 \tag{3'}
$$
This matches $\,(5)\,$ in Jean Marie's answer and, when $\,A \le B \le C\,$, is equivalent to $\,C \le A + B\,$ which matches the answer to another related question linked in Jean Marie's comment.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to approximate $49^{4}81^{5}$? I am doing this exercise for the GMAT test
Which of the following option is closest to $49^{4}81^{5}$?
A. $8^{18}$
B. $8^{19}$
C. $8^{20}$
D. $8^{21}$
E. $8^{22}$
My attempt:
$49^{4}81^{5} = (7^2)^4(9^2)^5 = 7^{8}9^{10} = (8-1)^{8}(8+1)^{10} \approx 8^{8}8^{10} = 8^{18}$
But I am not sure how $(8-1)^{8}(8+1)^{10}$ is closer to $8^{18}$ than to $8^{19}$.
Please shed me some lights. Thank you for your help!
| Let's take it exactly for a few more steps:
$$
(8-1)^{8}(8+1)^{10} =(8-1)^8(8+1)^8(8+1)^2\\
=(8^2-1)^8\cdot9^2<8^{16}\cdot 9^2
$$
And $9^2$ is much closer to $8^2$ than to $8^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A pizza parlor has 5 meat and 5 veggie toppings and 3 different sizes. How many pizzas are there with at least 1 meat and at least 1 veggie topping? I managed to get
$$3(2^{10} -1) - 3(2^{5}-1) - 3(2^{5}-1) $$
or
$$3(2^{10} -1) - 3(2^{5} + 2^{5}-1) $$
However, I am a little confused why the answer isn't
$$3(2^{10} -1) - 3(2^{5}) - 3(2^{5}) $$
Why do we have to have $-1$ in all cases? Wouldn't it be enough to have $-1$ only once so we get rid of the empty case once?
| Let's use the Inclusion-Exclusion Principle to solve the problem.
There are $3$ ways to choose the size of the pizza and $2^{10}$ ways to choose which subset of toppings will be placed on the pizza. Hence, if there were no restrictions, we could select a pizza in $3 \cdot 2^{10}$ ways.
From these, we must subtract those selections in which no meat toppings or no veggie toppings are selected.
No meat toppings: There are again $3$ ways to choose the size of the pizza. If there are no meat toppings, only the five veggie toppings can be selected. There are $2^5$ ways to choose a subset of the veggie toppings. Thus, there are $3 \cdot 2^5$ ways to choose a pizza with no meat toppings.
No veggie toppings: By symmetry, there are $3 \cdot 2^5$ ways to choose a pizza with no veggie toppings.
If we subtract the number of pizzas with no meat toppings and the number of pizzas with no veggie toppings from the total, we will have subtracted those pizzas with no meat and no veggie toppings twice. We only want to subtract them once, so we must add them back.
No meat and no veggie toppings: There are only three such pizzas, one of each size.
Total: There are $$3 \cdot 2^{10} - 3 \cdot 2^5 - 3 \cdot 2^5 + 3 = 3(2^{10} - 2^5 - 2^5 + 1)$$
pizzas that have at least one meat topping and at least one veggie topping. Notice that
$$3(2^{10} - 2^5 - 2^5 + 1) = 3(2^{10} - 1) - 3(2^5 - 1) - 3(2^5 - 1)$$
Why should this be the case?
We want pizzas with at least one meat topping and at least one veggie topping. The term $3(2^{10} - 1)$ counts all pizzas with at least one topping. One of the terms $3(2^5 - 1)$ counts all pizzas with at least one topping that contain no meat toppings, and the other one counts all pizzas with at least one topping that contain no veggie toppings. The $2^5 - 1$ terms are necessary since you started by counting all pizzas with at least one topping. If you instead start with all possible pizzas, you arrive at the answer obtained above with the Inclusion-Exclusion Principle.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for $a,p,q \in \Bbb R$ the solutions of: $\frac{1}{x-p} + \frac{1}{x-q} = \frac {1}{a^2}$ are real numbers. Prove that for $a,p,q \in \Bbb R$ the solutions of: $$\frac{1}{x-p} + \frac{1}{x-q} = \frac {1}{a^2}$$
are real numbers.
I tried manipulating the expression, getting rid of the denominators, but i can't factor $x$.
Any hints?
| Observe that your equation is equivalent to
$$ a^2(x-q)+a^2(x-p)=(x-p)(x-q)$$ $$\iff a^2x-a^2q+a^2x-a^2p-x^2+(p+q)x-pq=0$$ $$\iff-x^2+(2a^2+p+q)x-(a^2q+a^2p+pq)=0$$
Which is a quadratic equation with discriminant $$\color{red}{D=(2a^2+p+q)^2-4(a^2p+a^2q+pq)=4a^4+p^2+q^2-2pq>0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $f(a)=f(b)=0$ and $|f''(x)|\le M$ prove $|\int_a^bf(x)\mathrm{d}x| \le \frac{M}{12}(b-a)^3$ If $f(a)=f(b)=0$ and $|f''(x)|\le M$. Prove
$$|\int_a^bf(x)\mathrm{d}x| \le \frac{M}{12}(b-a)^3$$
I have thought about that since $f(a) = f(b) = 0 $ there is $\xi$ such that $f'(\xi) = 0$. Then when $x \le \xi$, $|f'(x)| \le (\xi - x)M$ and when $x \ge \xi$, $|f'(x)| \le (x-\xi)M$. After that $|f(x)| \le \frac{\xi^2 - (\xi -x)^2}{2}$ when $x \le \xi$ and $|f(x)| \le \frac{(b-\xi)^2 - (x - \xi)^2}{2}$ when $x \ge \xi$. Therefore
$$|\int_a^b f(x)\mathrm{d}x| \le \int_a^\xi |f(x)| + \int_\xi^b |f(x)| = \frac{\xi^3}{3} + \frac{(b-\xi)^3}{3}$$
If $x = \frac{a+b}{2}$, we have
$$\frac{x^3}{3}+\frac{(b-x)^3}{3} = \frac{(b-a)^3}{12}$$
But in this situation $\frac{x^3}{3}+\frac{(b-x)^3}{3}$ is the minimal value. So I can't go on.
| For simplicity let $a = -1, b = 1$. Let $\varepsilon > 0$. Consider the two functions $g_+(x) = \frac{M + \varepsilon}{2}(1-x^2)$ and $g_-(x) = -g_+(x)$. Then
$$
g_\pm(-1) = g_\pm(1) = 0, \; g_+''(x) = -M - \varepsilon, \; g_-''(x) = +M + \varepsilon
$$
and
$$
\int_{-1}^1 g_+(x) = 2\frac{M+ \varepsilon}{3} = - \int_{-1}^1 g_-(x) dx \, .
$$
The claim now is that $g_-(x) \le f(x) \le g_+(x)$ for all $x$.
Indeed, suppose $h_-(x) = f(x) - g_-(x)$ is negative somewhere in $(-1,1)$. Then there is an absolute minimum $c \in (-1,1)$ at which $h_-''(c) \ge 0$ (second derivative test). But
$$
h_-''(c) = f''(x) - g_-''(c) = f''(c) - M - \varepsilon \le |f''(c)| - M - \varepsilon \le - \varepsilon < 0
$$
and therefore this is impossible.
Consequently $g_-(x) \le f(x)$ for all $x$. Similarly $f(x) \le g_+(x)$ for all $x$. It follows that
$$
-2\frac{M + \varepsilon}{3} \le \int_{-1}^1 f(x) dx \le 2\frac{M + \varepsilon}{3} \, .
$$
Since $\varepsilon > 0$ was arbitrary, the desired estimate follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the closed form of infinite sum $x+x^4+x^7+x^{10}+...$? I know that $1+x+x^2+x^3+...=\frac{1}{1-x}$, $1+x^2+x^4+...=\frac{1}{1-x^2}$. It seems to follow some rules of geometric series. But I'm confused how to get the title from here. Thanks
| It's always nice, if we want to see if the sum is geometric, to do factoring to ensure that the first term of the sum is $1$, so factor out an $x$:
$$x + x^4 + x^7 + x^{10} + ... = x(1 + x^3 + x^6 + x^9 + ...)$$
The parenthetical expression is a geometric series in which the ratio is $x^3$. Then as a result, by the formula for an infinite geometric sum,
$$1 + x^3 + x^6 + x^9 + ... = \frac{1}{1-x^3}$$
Thus, bringing in the factored-out $x$,
$$x + x^4 + x^7 + x^{10} + ... = \frac{x}{1-x^3}$$
Note that the formula for a geometric sum only holds when the ratio is less than $1$ in magnitude. Thus we require $|x^3| < 1$. (This can be simplified to $|x| < 1$ as $|x^n|=|x|^n$ and then you just take the cube root.)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $0$ an eigen value of $A$
Consider the following matrix:
$A=\begin{bmatrix} 9&1&1&1&1&1&2&2\\1& 9&1&1&1&1&2&2\\1&1&9&1&1&1&2&2\\1&1&1&9&1&1&2&2\\1&1&1&1&9&1&2&2\\1&1&1&1&1&9&2&2\\2&2&2&2&2&2&13&1\\2&2&2&2&2&2&1&13\end{bmatrix}$.
I got the eigen values of the matrix
\begin{bmatrix} 9&1&1&1&1&1\\1& 9&1&1&1&1\\1&1&9&1&1&1\\1&1&1&9&1&1\\1&1&1&1&9&1\\1&1&1&1&1&9\end{bmatrix}
and the eigen values of the matrix
\begin{bmatrix} 13&1\\1&13\end{bmatrix}.
I have the following two questions:
*
*Is $0$ an eigen value of $A$
*Find the eigen values of $A$.
Note that $0$ is not an eigen value of both the submatrices listed above.
Can we conclude that $0$ is not an eigen value of $A$ from above?If yes,How?
I am unable to solve Part(2) of the question.Please help.
| Subtracting $8$ and $12$ from the main diagonal, it is easily seen that $8$ is an eigenvalue of multiplicity $5$ and $12$ is an eigenvalue of multiplicity $1$. Hence we are still missing two eigenvalues. The trace tells you that the sum of the remaining two eigenvalues has to be $80-52=28$. If one is $0$, then the other has to be $28$. But $28$ cannot be an eigenvalue as the diagonal of $A-28I$ is dominating. Therefore $0$ is not an eigenvalue of $A$. This answers 1. To get the remaining two eigenvalues, you'd have to compute the characteristic polynomial and divide out $(\lambda-8)^5(\lambda-12)$. This gets you (using computer algebra) $\lambda^2-28\lambda+148$, which can be solved for $\lambda$.
| {
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Convergence of $\sqrt[k]{z+\sqrt[k]{z+\sqrt[k]{z+\cdots}}}$, where $z=(1+x)^k-(1+x)$ If one writes
$$1+x=\sqrt{(1+x)^2}=\sqrt{1+2x+x^2}=\sqrt{x+x^2+(1+x)}$$
then one has a recursive definition of the function $1+x$ which can be used to write $1+x$ as the infinite nested radical:
$$1+x=\sqrt{x+x^2+\sqrt{x+x^2+\sqrt{x+x^2+\sqrt{\cdot\cdot\cdot}}}}$$
But this definition relies on the fact that
$$
1+x=\sqrt{(1+x)^2}
$$
which is only true for $x \ge-1$. In general one could state that
$$
1+x=\sqrt[n]{(1+x)^n}=\sqrt[n]{(1+x)^n-(1+x)+\sqrt[n]{(1+x)^n-(1+x)+\sqrt[n]{\cdots}}}
$$
But the RHS does not converge to $1+x$ for most values of $x\in\mathbb{C}$. So, my question is, what is the actual closed form of the following function? For what values of $x\in\mathbb{C}$ does the following function converge?
$$
\sqrt[k]{(1+x)^k-(1+x)+\sqrt[k]{(1+x)^k-(1+x)+\sqrt[k]{\cdots}}}
$$
If unclear the above nested radical can be defined by $\lim_{n\to\infty}a_n$ where
$$
a_1=\sqrt[k]{(1+x)^k-(1+x)},\quad
a_n=\sqrt[k]{(1+x)^k-(1+x)+a_{n-1}}.$$
| The expression
$$
y=\sqrt[k]{(1+x)^k-(1+x)+\sqrt[k]{(1+x)^k-(1+x)+\cdots}}
$$
where $x\ge 0$, represents the limit of the recursive sequence
$$
a_1=\sqrt[k]{(1+x)^k-(1+x)}, \quad a_{n+1}=\sqrt[k]{(1+x)^k-(1+x)+a_n},
\quad n\in\mathbb N.
$$
if such a limit exists.
Clearly, the sequence $\{a_n\}$ is increasing. (The fact $a_n\le a_{n+1}$ can be shown inductively.)
Next, we show inductively that $\{a_n\}$ is upper bounded by $1+x$.
Clearly,
$$
a_1=\sqrt[k]{(1+x)^k-(1+x)}\le\sqrt[k]{(1+x)^k}=1+x.
$$
Assume that $a_n\le 1+x$. Then
$$
a_{n+1}=\sqrt[k]{(1+x)^k-(1+x)+a_n}\le
\sqrt[k]{(1+x)^k-(1+x)+(1+x)}=1+x.
$$
Therefore, $\{a_n\}$ is increasing and upper bounded and thus it converges. Let $y=\lim a_n$.
If $x=0$, then observe that $a_n=0$, for all $n$, and hence $y=0$.
If $x>0$, then clearly $y>0$, and hence
$$
y \leftarrow a_{n+1}=\sqrt[k]{(1+x)^k-(1+x)+a_n}\to
\sqrt[k]{(1+x)^k-(1+x)+y}
$$
and thus
$$
y^k-y=(1+x)^k-(1+x)
$$
Now the function $g(z)=z^k-z$ is strictly increasing, and hence one-to-one when $g'(z)=kz^{k-1}-1>0$ equivalently when $z>k^{-1/(k-1)}$. So if we show that $y>k^{-1/(k-1)}$, then we will shown that $y=1+x$.
We have
$$
a_1=\sqrt[n]{(1+x)^k-(1+x)}=\sqrt[n]{(1+kx+\cdots)-(1+x)} \\ \ge
\sqrt[n]{(k-1)x}\ge x^{1/k}
$$
then
$$
a_2=\sqrt[n]{(1+x)^k-(1+x)+a_1}\ge a^{1/k}_1\ge x^{1/{k^2}}
$$
and in general
$$
a_n\ge x^{1/{k^n}}\to 1,
$$
and hence
$$
y=\lim a_n\ge 1>k^{-1/(k-1)},
$$
which implies that $a_n\to 1+x$.
Note. If $x\in [-1,0]$, then $a_n\to 0$.
| {
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induction proof: number of compositions of a positive number $n$ is $2^{n-1}$ So my problem was to find the formula for the number of composition for a positive number $n$. Then prove its validity. I found the formula which was $2^{n-1}$. Having trouble with the induction.
$$Claim:2^0+2^1+2^2+...+(2^{n-2}+1)=2^{n-1}$$
When $n=5$ $$ 1+2+4+(2^{5-2}+1)=2^{5-1}$$
$$ 1+2+4+(2^{3}+1)=2^{4}$$
$$ 1+2+4+(8+1)=16$$
$$ 16=16$$
Assume that $n=k$
$$2^0+2^1+2^2+...+(2^{k-2}+1)=2^{k-1}$$
Now show for $k+1$
$$2^0+2^1+2^2+...+(2^{k-1}+1)=2^{k}$$
Don't know what to do after. Help is appreciated.
$$2^0+2^1+2^2+...+(2^k)+(2^{k-1}+1)=2^{k}$$
| You're almost there. You just have to use your hypothesis in your final case for $k+1$.
You found for $n=k$ that
$$2^0+2^1+2^2+...+2^{k-2}+1=2^{k-1}.$$
You want to prove for $n=k+1$ that
$$2^0+2^1+2^2+...2^{k-2}+2^{k-1}+1=2^{k},$$
Fill in the first equation into the second
\begin{align*}
2^0+2^1+2^2+...+ 2^{k-2}+&(2^0+2^1+2^2+...+2^{k-2}+1)+1\\
=2&(2^0+2^1+2^2+...+2^{k-2}+1)
\end{align*}
Now using the induction hypothesis again
\begin{align*}
&=2(2^{k-1})\\&=2^{k}.
\end{align*}
By mathematical induction we have now proven this for all $n\geq 5$.
The theorem does however hold for any $n\geq 2$, so I suggest making your basecase be $n=2$.
| {
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How to determine the smallest value of $N=n^4+6n^3+11n^2+6n$ if 13 and 19 both divide N? I tried to solve for an integer solution by making N equal to multiples of 247 but this is not leading me anywhere. I then tried using the tests for divisibility which did not seem to lead me anywhere either.
| $N=n^4 + 6n^3 + 11n^2 + 6n = n(n+1)(n+2)(n+3)$ so $n, n+1, n+2$ and $n+3$ all divide $N$.
$19$ and $13$ are prime so we need $19|K$ and $13|M$ where $K$ and $M$ are each one of $n,n+1,n+2,$ or $n+3$
So basically we need $|13a + 19b| \le 3$ and the smallest value that is so.
Do a hobbled Euclid's Algorithm.
$19 - 13 = 6$ that's too big.
But $13 = 2*6 + 1$ so $13 = 2(19 - 13) + 1$ so $3*13-2*19= 1$. So for $38,39 \in \{n,n+1,n+2, n+3\}$ will be a solution. Obviously the smallest product will but $n+3 = 39$ and $n+2 =38$ and $n = 36$.
Is there any smaller?
Well, the next smaller multiple of $13$ is $26$ and $26 - 1*19 = 7> 3$ and $2*19-26=12 > 3$ so none of $26= n,n+1,n+2,n+3$ will involve $26$. ($19$ divides none of $23... 29$.)
And the next smaller multiple of $13$ is $13$ and $19-13 =6 > 3$ so none of $13=n, n+1,n+2,n+3$ will have solution.
So $n = 36$ and $N= 36*37*38*39$ is the smallest (positive) such value.
If $N \le 0$ is allowed then $N=0$ is a smaller solution. As is $-38,-39\in \{n,n+1,n+2,n+3\}$. For $n<-3$, $N > 0$ anyway, the value of $N$ when $n\le -39$ is the same as $N$ when $n \ge 36$.
SO $N= 0$ is the smallest integer value.
| {
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How to prove the inequality $a/(b+c)+b/(a+c)+c/(a+b) \ge 3/2$ Suppose $a>0, b>0, c>0$.
Prove that:
$$a+b+c \ge \frac{3}{2}\cdot [(a+b)(a+c)(b+c)]^{\frac{1}{3}}$$
Hence or otherwise prove:
$$\color{blue}{\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}}$$
| Hint:
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}\iff \frac{a+b+c}{b+c}+\frac{b+a+c}{a+c}+\frac{c+a+b}{a+b}\ge \frac{3}{2}+3$$
$$\iff \big(a+b+c\big)\cdot \bigg(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\bigg)\ge \frac{9}{2}\iff \color{blue}{\frac{2\cdot (a+b+c)}{3}\ge \frac{3}{\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}}}$$ Which is trivial by the AM-HM inequality. Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solve the equation $13x + 2(3x + 2)\sqrt{x + 3} + 42 = 0$.
Solve the equation $13x + 2(3x + 2)\sqrt{x + 3} + 42 = 0$.
Let $y = \sqrt{x + 3} \implies 3 = y^2 - x$.
$$\large \begin{align}
&13x + 2(3x + 2)\sqrt{x + 3} +42\\
= &14(x + 3) + (6x + 4)y - x\\
= &14y^2 + [6(x + 3) - 14]y - x\\
= &14y(y - 1) - (y^2 - x - 9)y^3 - x\\
= &14y(y - 1) + x(y^3 - y) - y^3(y^2 - 1) + 8y^3\\
= &14y(y - 1) + (xy^2 + xy + x)(y - 1) - (y^4 + y^3)(y - 1) + 8y^3\\
= &(-y^4 + y^3 + xy^2 + xy + x + 14y)(y - 1) + 8y^3\\
\end{align}$$
And I'm stuck here.
| Part of the issue seems to be mixing up $x$ and $y$. Usually once you identify a substitution it's easiest to bite the bullet and change the variable over completely. Here, $y=\sqrt{x+3}$ implies $x=y^2-3$ and then
$13(y^2-3)+2(3y^2-7)y+42=0$
$6y^3+13y^2-14y+3=0$
Trying out rational root candidates with the Rational Root Theorem we identify $y=1/2$ as one of these roots. Thereby
$(2y-1)(3y^2+8y-3)=0$
and solving the quadratic factor gives the additional roots $y=-3, y=1/3$. The former is thrown out as the original substitution, by definition, requires a nonnegative square root. Each of the other roots $y=1/2, y=1/3$ will give a valid solution for $x=y^2-3$ to the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find (a,b,c) if $ax^2-bx+c=0$ have roots lying in $(0,1)$ where $a,b,c\in\mathbb{Z_+}$
Let the equation $ax^2-bx+c=0$ have distinct real roots both lying in the open interval $(0,1)$ where $a,b,c$ are given to be positive integers. Then the value of the ordered triplet $(a,b,c)$ can be
$$
a)\quad (5,3,1)\;,\quad b)\quad(4,3,2)\;,\quad c)\quad(5,5,1)\;,\quad d)\quad(6,4,1)
$$
The solution given in my reference is $(5,5,1)$.
My Attempt
$$
a,b,c\in\mathbb{Z_+}\\
\alpha,\beta=\frac{b\pm\sqrt{b^2-4ac}}{2a}\in(0,1)\\
b+\sqrt{b^2-4ac}<2a\implies\sqrt{b^2-4ac}<2a-b\\
b^2-4ac<4a^2+b^2-4ab\implies-ac<a^2-ab\\
a=0\text{ (or) }-c<a-b\implies a+c>b\\
\Big[a=0\text{ (or) } a+c>b\Big]\; \&\;b^2\geq4ac
$$
Does that constitute the complete condition for the given problem or am I missing something in my attempt ?
| You have that $b^2-4ac\gt0$, since the roots are real. Only $(5,5,1)$ meets this condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3169073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that the equation $x^5+x^4=1$ has a unique solution. Show that the equation $x^5 + x^4 = 1 $ has a unique solution.
My Attempt:
Let $f(x)=x^5 + x^4 -1 $
Since $f(0)=-1$ and $f(1)=1$, by the continuity of the function $f(x)$ has at least one solution in $(0,1)$.
| Let $x\geq0$ and $f(x)=x^5+x^4.$
Thus, since $f$ increases, continuous, $f(0)<1$ and $f(1)>1$, we obtain that our equation has an unique root.
Let $x<0$.
Thus, $-x>0$.
We'll prove that the equation $-x^5+x^4=1$ has no roots for $x>0$.
Indeed, by AM-GM we obtain:
$$x^5+1-x^4=4\cdot\frac{x^5}{4}+1-x^4\geq5\sqrt[5]{\left(\frac{x^5}{4}\right)^4\cdot1}-x^4=\left(\frac{5}{\sqrt[5]{256}}-1\right)x^4>0$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Trigonometry AS I've been trying this question for a while but still don't understand it. Do you use trigonometry to form a equation for the sides and then add them up as you know the perimeter is 40?
| Construct a perpendicular from $B$ to $AC$ such that it touches $AC$ at point $D$.
Using the fact that Perimeter is $40cm$
$$a+b+14=40 \\ \Rightarrow a=26-b\\ $$
Using Pythagoras Theorem on Triangle $ABD$, $$\ \ \ \ \ (26-b)^2=BD^2+AD^2 \\\Rightarrow (26-b)^2 -AD^2=BD^2 \\ \\ \\ \\ \\ \\$$
Using Pythagoras Theorem on Triangle $CBD$,
$$\begin {align}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b^2&=(26-b)^2-AD^2+DC^2\\
&=(26-b)^2-(14-DC)^2+DC^2\\
&=676+b^2-52b-196-DC^2+28DC+DC^2\\
\\ \Rightarrow 0&=480 -52b+28DC\\
\Rightarrow DC&=\frac{52b-480}{28}=\frac {13b}{7}-\frac {120}{7}\\
\Rightarrow \frac {DC}{b} &=\cos \theta= \frac {13}{7}-\frac {120}{7b}
\end {align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I solve this absolute value equation? This is the equation:
$|\sqrt{x-1} - 2| + |\sqrt{x-1} - 3| = 1$
Any help would be appreciated. Thanks!
| Let $x$ be a solution of the equation. Notice $x \geq 1$, since $\sqrt{x-1}$ has its domain as $x \geq 1$.
If $x \geq 10$, then each of the absolute value is just the term inside (i.e.$|\sqrt{x-1}-2| = \sqrt{x-1}-2$ and similarly $|\sqrt{x-1}-3| =\sqrt{x-1}-3$) so that the given equation in this case becomes $2\sqrt{x-1}-5=1$. Solve for $x$ you get $x=10$. Thus we have shown that IF there is a solution that is greater than or equal to $10$ then $x$ must be $10$; we have not yet proven that $10$ is a solution. However, we can check that $x=10$ is a solution by plugging it back in.
If $10 > x \geq 5$ then $|\sqrt{x-1}-2| = \sqrt{x-1}-2$ and $|\sqrt{x-1}-3| =-\sqrt{x-1}+3$ so that the given equation becomes (after simplification) $1=1$; this does not give us a new information, but we can check that any number $x$ such that $10 > x \geq 5$ satisfies the original equation.
If $5 > x \geq 1$ then $|\sqrt{x-1}-2| = -\sqrt{x-1}+2$ and $|\sqrt{x-1}-3| = -\sqrt{x-1}+3$, so that the given equation becomes (after simplification) $\sqrt{x-1}=2$, implying $x=5$, contradicting our assumption that $5 > x$. Thus, there cannot be any solution in this case.
Hence, the solution are $5 \leq x \leq 10$.
In general, any time you see an absolute value in a given equation, it's a good idea to divide into cases according to whether each expression in an absolute value is negative or non-negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Roots of a Complex Number expression If $z_1$, $z_2$, $z_3$, $z_4$ are roots of the equation $z^4+z^3+z^2+z+1=0$,
then what is the least value of $\lfloor mod(z_1 + z_2)\rfloor + 1$?
($\lfloor.\rfloor$ denotes Greatest Integer Function)
| The roots are
$$w_k = e^{\frac{2\pi k}{5}i} \quad(k = 1,2,3,4)$$
$$\begin{align*}
\min\left\lfloor\left|w_m+w_n\right|\right\rfloor+1
&= \min\left\lfloor\left|e^{\frac{2\pi m}{5}i}+e^{\frac{2\pi n}{5}i}\right|\right\rfloor+1\\
&= \min \left\lfloor\left|e^{\frac{2\pi m}{5}i}\right|\left|1+e^{\frac{2\pi (n-m)}{5}i}\right|\right\rfloor+1\\
&=\min\left\lfloor\left|1+e^{\frac{2\pi (n-m)}{5}i}\right|\right\rfloor+1\\
&=\min\left\lfloor\sqrt{\left[1+\cos{\frac{2\pi(n-m)}{5}}\right]^2 + \left[\sin{\frac{2\pi(n-m)}{5}}\right]^2}\right\rfloor+1\\
&=\min\left\lfloor\sqrt{2+2\cos{\frac{2\pi(n-m)}{5}}}\right\rfloor+1\\
&=\min\left\lfloor\sqrt{4\cos^2{\frac{2\pi(n-m)}{10}}}\right\rfloor+1\\
&=\min\left\lfloor2\left|\cos{\frac{2\pi(n-m)}{10}}\right|\right\rfloor+1\\
\end{align*}$$
(The identity $\left|1+e^{i\theta}\right| \equiv 2\left|\cos\frac \theta2\right|$ also has a geometric interpretation)
The least value is when the $\cos$ is closest to zero, when $(n-m) = 2$ (or 3 or -2 or -3):
$$\begin{align*}
\min\left\lfloor\left|w_m+w_n\right|\right\rfloor+1
&=\left\lfloor2\left|\cos{\frac{2\pi\cdot2}{10}}\right|\right\rfloor+1\\
&=\left\lfloor2\left|\cos{\frac{2\pi}{5}}\right|\right\rfloor+1\\
\end{align*}$$
Without knowing the actual value of $\cos \frac{2\pi}{5}$, note that
$$\frac{2\pi}{6}<\frac{2\pi}{5}<\frac{2\pi}{4}\\
\frac12 > \cos \frac{2\pi}{5} > 0\\
1 > 2 \left|\cos \frac{2\pi}{5}\right| > 0\\
\left\lfloor2 \left|\cos \frac{2\pi}{5}\right|\right\rfloor = 0$$
So the least value is
$$\begin{align*}
\min\left\lfloor\left|w_m+w_n\right|\right\rfloor+1
&= 0+1\\
&= 1
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3178747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to get an equation system from a Simplex table Let's assume I already have a simplex table (with an optimal solution):
$$\left(\begin{array}{ccccc|c}
& x_1 & x_2 & S_1 & S_2 & \\
S_1 & 0 & 2 & 0 & 1 & 2 \\
S_2 & 1 & 0 & \frac{1}{5} & 0 & 4 \\
\hline
P & 5 & 0 & \frac{1}{2} & 1 & 1
\end{array}\right)$$
note: the above table is not a real problem.
How do I get the original problem model from this?
| If you write a better and correct example, you can find it easier. Just make sure that:
1) The matrix in front of your basic variables should be the identity matrix.
2) The objective function row in the optimal tableau will have 0s for basic variables.
I'll change your example to the following and show you the steps to retrieve your original problem from the optimal tableau.
The original problem (which we want to get at the end):
Maximize Z = 3x + 2y
subject to:
2x + y ≤ 18
2x + 3y ≤ 42
3x + y ≤ 24
x ≥ 0 , y ≥ 0
Optimal Tableau:
\begin{array}{cccccc|c}
& x & y & S_1 & S_2 & S_3 & \\
y & 0 & 1 & \frac{-1}{2} & \frac{1}{2} & 0 & 12 \\
S_3 & 0 & 0 & \frac{-7}{4} & \frac{1}{4} & 1 & 3\\
x & 1 & 0 & \frac{3}{4} & \frac{-1}{4} & 0 & 3 \\
\hline
Z & 0 & 0 & \frac{5}{4} & \frac{1}{4} & 0 & 33
\end{array}
Your original tableau will have a form like this:
\begin{array}{rr|l}
* & * & 0 \\ \hline
N & I & b
\end{array}
and your optimal tableau will be:
\begin{array}{rr|l}
* & * & * \\ \hline
B^{-1} N & B^{-1} & B^{-1} b
\end{array}
So, under our slack variables (here $S_1, S_2, S_3$), we can find $ B^{-1} $ matrix:
\begin{array}{ccc}
\frac{-1}{2} & \frac{1}{2} & 0 & \\
\frac{-7}{4} & \frac{1}{4} & 1 & \\
\frac{3}{4} & \frac{-1}{4} & 0 & \\
\end{array}
Now, the inverse of your $ B^{-1} $ matrix is your $B$ matrix. So, $B$ is:
\begin{array}{ccc}
1 & 0 & 2 & \\
3 & 0 & 2 & \\
1 & 1 & 3 & \\
\end{array}
Multiply $B$ by $[12, 3, 3]$ ($B^{-1} b$ in the optimal tableau) and you get the original $b$:
$[18, 42, 24]$
and multiply it by $B^{-1} N$ and you get your original $N$:
\begin{array}{cc}
2 & 1 & \\
2 & 3 & \\
3 & 1 & \\
\end{array}
You can also refer to this answer for a more in-depth explanation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the last three digits of $p$ if the equations $x^6 + px^3 + q = 0$ and $x^2 + 5x - 10^{2013} = 0$ have common roots.
Find the last three digits of $p$ if the equations $x^6 + px^3 + q = 0$ and
$x^2 + 5x - 10^{2013} = 0$ have common roots.
Let $a,b $ be the solutions of second equation, then by Vieta we have $a+b =-5$ and $ab=-10^{2013}$
Since $ a^6+a^3p+q=0$ and $b^6+b^3p+q=0$ we have $ a^6-b^6+p(a^3-b^3)=0$ so $ -p = a^3+b^3 = (a+b)(a^2-ab+b^2)$
So $ p = 5((a+b)^2-3ab) = 5(25+3\cdot 10^{2013})$ and thus last three digits are $125$.
| Your proof is entirely correct, though you might want to clarify two things:
*
*First, that if the two polynomials have a common root, then they have two common roots.
*Second, that $a^3-b^3\neq0$ when dividing by $a^3-b^3$ to deduce that $-p=a^3+b^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Use generating functions to solve the recurrence relation Use generating functions to solve $a_n = 3a_{n-1} - 2a_{n-2} + 2^n + (n+1)3^n$.
What I have so far, not sure if I forgot to do something or am missing out on something obvious:
Define $$G(x) = \sum_{n=0}^{\infty} a_nx^n$$
Then, $G(x) = a_0 + a_1x + a_2x^2 + \sum_{n=3}^{\infty} a_nx^n$
= $a_0 + a_1x + a_2x^2 + \sum_{n=3}^{\infty} (3a_{n-1} - 2a_{n-2} + 2^n + (n+1)3^n)x^n$
= $a_0 + a_1x + a_2x^2 + \sum_{n=3}^{\infty} 3a_{n-1}x^n - \sum_{n=3}^{\infty} 2a_{n-2}x^n + \sum_{n=3}^{\infty} 2^nx^n + \sum_{n=3}^{\infty} (n+1)3^nx^n$
= $a_0 + a_1x + a_2x^2 + 3\sum_{n=3}^{\infty} a_{n-1}x^n - 2\sum_{n=3}^{\infty} a_{n-2}x^n + \sum_{n=3}^{\infty} (2x)^n + \sum_{n=3}^{\infty} (n+1)(3x)^n$
=$a_0 + a_1x + a_2x^2 + 3x\sum_{n=3}^{\infty} a_{n-1}x^{n-1} - 2x^2\sum_{n=3}^{\infty} a_{n-2}x^{n-2} + \sum_{n=3}^{\infty} (2x)^n +
\sum_{n=3}^{\infty} (n+1)(3x)^n$
= $a_0 + a_1x + a_2x^2 + 3x\sum_{n=2}^{\infty} a_nx^n - 2x^2\sum_{n=1}^{\infty} a_nx^n + \sum_{n=3}^{\infty} (2x)^n +
\sum_{n=3}^{\infty} (n+1)(3x)^n$
| Given that $G(x) = \sum_{n=0}^{\infty} a_nx^n$, you can write $a_n$ as $[x^n]G(x)$. Then the recurrence becomes $$[x^n]G(x) = 3[x^{n-1}]G(x) - 2[x^{n-2}]G(x) + 2^n + (n+1)3^n$$
If we introduce functions $P(x) = \sum_{n=0}^\infty 2^n x^n$ and $Q(x) = \sum_{n=0}^\infty (n+1)3^n x^n$ then, by the linearity of coefficient extraction, $$[x^n]G(x) = 3[x^n]xG(x) - 2[x^n]x^2G(x) + [x^n]P(x) + [x^n]Q(x)$$ and if this holds for every $n$ then $$G(x) = 3xG(x) - 2x^2G(x) + P(x) + Q(x)$$ which rearranges to $$G(x) = \frac{P(x) + Q(x)}{2x^2 - 3x + 1}$$
Finding closed forms for $P$ and $Q$ is left as an exercise...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $f^{(22)}(0)$ for $f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$
Find $f^{(22)}(0)$ for $f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$
I know that I should use Taylor's theorem and create power series. However I don't have idea how I can find $a_{n}$ such that $f(x)=\sum_{n=1}^{+\infty} a_{n}x^{n}$. I know only that $f(x)=\frac{x}{(x-1)(x^{2}+2)}$ but don't know how I can continue it because $f(x)=\frac{A}{(x^2+2)}+\frac{B}{(x-1)}$ and I get that $A=1, B=0, A=2B$ so it is conflict.Have you some tips how I can write this function and finish this task? Updated: Thanks to @gt6989b I know that $f(x)=\frac{\frac{-1}{3}x+\frac{2}{3}}{x^{2}+2}+\frac{1}{3} \cdot \frac{1}{1-x}=\frac{\frac{-1}{3}x+\frac{2}{3}}{x^{2}+2}+\frac{1}{3} \cdot (1+x+x^{2}+...)$ but I still have a problem with $\frac{\frac{-1}{3}x+\frac{2}{3}}{x^{2}+2}$
| $$
\begin{split}
f(x) &= \frac{x}{x^3-x^2+2x-2}
= \frac{x}{x^2(x-1)+2(x-1)}
= \frac{x}{(x^2+2)(x-1)}
\end{split}
$$
and use partial fractions. Then use
$$
\frac{1}{1-u} = 1 + u + u^2 + \ldots
$$
and
$$
\frac{1}{1+u} = \frac{1}{1-(-u)} = 1 -u + u^2-u^3 \pm \ldots
$$
UPDATE
From partial fractions,
$$
\frac{Ax+B}{x^2+2} + \frac{C}{x-1}
= \frac{(Ax+B)(x-1) + C(x^2+2)}{(x^2+2)(x-1)}
$$
expanding the numerator and equating to the desired expression you get
$$
x = (Ax+B)(x-1) + C(x^2+2)
= x^2(A+C) + x(-A+B) -B+2C
$$
which results in the system of 3 equations and 3 unknowns
$$
\begin{cases}
A & & + C & = 0 \\
-A& +B & & = 1 \\
& -B & +2C & = 0
\end{cases}
$$
Adding all three together yields $C = 1/3$ which implies $A = -1/3$ and $B = 2/3$.
Can you now finish this problem?
UPDATE 2
Another hint:
$$
\frac{Ax+B}{x^2+2}
= \left[\frac{Ax}{2} + \frac{B}{2}\right]
\frac{1}{1 + (x/\sqrt{2})^2}
$$
and now the fraction expands as $(1+u)^{-1}$ and then expand the bracket by multiplying by the resulting Taylor series, getting 2 different series, i.e.
$$
(ax+b) \sum_{k=0}^\infty a_k x^k
= ax \sum_{k=0}^\infty a_k x^k + b \sum_{k=0}^\infty a_k x^k
= \sum_{k=0}^\infty (a a_k) x^{k+1} + \sum_{k=0}^\infty (b a_k) x^k
$$
and now change the index on the left summation and combine to get $\sum_{j=0}^\infty c_j x^j$ and you need $c_{22}$...
| {
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"url": "https://math.stackexchange.com/questions/3184465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\tan(\sin(x))=x-\frac{x^{3}}{6}+o(x^{3})$
Prove that $\tan(\sin(x))=x-\frac{x^{3}}{6}+o(x^{3})$
I know that when I calculate derivative then I get true answer. However I want to know why the way that I will present soon does not work. My try:$$\tan(x)=x+r_{1}(x), \quad r_{1}(x)=o(x)$$ $$\tan(\sin(x))=\sin(x)+r_{1}(\sin x),\quad r_{1}(\sin x)=o(\sin x)$$ $$\tan(\sin(x))=x-\frac{x^{3}}{6}+r_{2}(x)+r_{1}(x-\frac{x^{3}}{6}+r_{2}(x)),\quad r_{2}(x)=o(x^{3})$$ In this moment I should find: $$o(?)=r_{1}(x-\frac{x^{3}}{6}+r_{2}(x))$$I suspect that $o(?)=o(x^{3})$ so I write: $$\frac{r_{1}(x-\frac{x^{3}}{6}+r_{2}(x))}{(x-\frac{x^{3}}{6}+r_{2}(x))}\cdot \frac{(x-\frac{x^{3}}{6}+r_{2}(x))}{x^{3}}$$ But in this moment I have $0 \cdot (+\infty)$. Moreover I see that $$r_{1}(x-\frac{x^{3}}{6}+r_{2}(x))=o(x)$$ because $$\frac{r_{1}(x-\frac{x^{3}}{6}+r_{2}(x))}{(x-\frac{x^{3}}{6}+r_{2}(x))}\cdot \frac{(x-\frac{x^{3}}{6}+r_{2}(x))}{x} \rightarrow 0$$Can someone tell me where I'm making a mistake, that I'm getting $ o (x) $ instead of $ o (x ^ {3}) $?
| The problem statement contains a sign error.
In any case you want to compute $\tan \sin x$ modulo $o(x^3)$, so you should compute $\sin x$ and $\tan x$ modulo $o(x^3)$, not $o(x)$. Indeed,
$$\sin x = x - \frac{1}{6} x^3 + o(x^3), \qquad \tan y = y + \frac{1}{3} y^3 + o(y^3) ,$$
so substituting gives $$\color{#df0000}{\boxed{\tan \sin x = \left(x - \frac{1}{6} x^3\right) + \frac{1}{3} \left(x - \frac{1}{6} x^3\right)^3 + o(x^3) = x + \frac{1}{6} x^3 + o(x^3)}} .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How does this infinite series $1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\cdots$ simplify to an integral $\int_0^1\frac{dx}{1+x^3}$? How does the infinite series below simplify to that integral?
$$1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\cdots=\int_0^1\frac{dx}{1+x^3}$$
I thought of simplifying the series to the sum to infinity of $\frac{1}{6n-5} - \frac{1}{6n-2}$, but this did not help.
| for $x$ real, $n\geq 0$ integer
\begin{align}\frac{1}{1+x^3}&=\frac{1-(-x^3)^{n+1}}{1-(-x^3)}+\frac{(-x^3)^{n+1}}{1-(-x^3)}\\
&=\frac{1-(-x^3)^{n+1}}{1-(-x^3)}+\frac{(-x^3)^{n+1}}{1+x^3}\\
\end{align}
For $x\neq 1$, $n\geq 0$ integer, \begin{align}\sum_{k=0}^n x^k=\frac{1-x^{n+1}}{1-x}\end{align}
Therefore,
\begin{align}\int_0^1 \frac{1}{1+x^3}\,dx&=\int_0^1 \left(\sum_{k=0}^n (-x^3)^k\right)\,dx+\int_0^1 \frac{(-x^3)^{n+1}}{1+x^3}\,dx\\
&=\sum_{k=0}^n \left(\int_0^1 (-x^3)^k\,dx\right)+\int_0^1 \frac{(-x^3)^{n+1}}{1+x^3}\,dx\\
&=\sum_{k=0}^n \frac{(-1)^k}{3k+1}+\int_0^1 \frac{(-x^3)^{n+1}}{1+x^3}\,dx\\
\end{align}
For $x\in[0;1],n\geq 0$, integer,
\begin{align}\frac{x^{3(n+1)}}{1+x^3}\leq x^{3(n+1)}\end{align}
and,
\begin{align}\int_0^1 x^{3(n+1)}\,dx=\frac{1}{3n+4}\end{align}
Therefore,
\begin{align}\left|\int_0^1 \frac{(-x^3)^{n+1}}{1+x^3}\,dx\right|\leq \frac{1}{3n+4}\end{align}
\begin{align}\left|\int_0^1 \frac{1}{1+x^3}\,dx-\sum_{k=0}^n \frac{(-1)^k}{3k+1}\right|\leq \frac{1}{3n+4}\end{align}
Therefore,
\begin{align}\boxed{\int_0^1 \frac{1}{1+x^3}\,dx=\sum_{k=0}^\infty \frac{(-1)^k}{3k+1}}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Definite Integral of $\int_0^1\frac{dx}{\sqrt {x(1-x)}}$ We have to calculate value of the following integral :
$$\int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \qquad \qquad (2)$$
What i've done for (2) :
\begin{align}
& = \int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \\
& = \int_0^1\cfrac{dx}{\sqrt {x-x^2}} \\
& = \int_0^1\cfrac{dx}{\sqrt {(x^2-x+\frac 14)-\frac 14 }} \\
& = \int_0^1\cfrac{dx}{\sqrt {(x-\frac 12)^2-(\frac 12)^2 }} \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {(\frac 12\sec \theta)^2-(\frac 12)^2 }} I\ used\ trigonometric\ substitution \ u=a\sec \theta, by \ it's \ form \ u^2-a^2 \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {(\frac 14\sec^2 \theta)-\frac 14 }} \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {\frac 14(\sec ^2\theta-1)}} \ using \\tan^2\theta=\sec^2\theta-1 \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {\frac 12(\sqrt{\tan^2\theta) }}} \\
& = \int_0^1\sec\theta d\theta = \sec\theta \tan \theta |_0^1 \\
\end{align}
But i got problems calculating $\theta$ value, using trigonometric substitution, any help?
| As $0\le x\le1$
WLOG $x=\sin^2t;0\le t\le\dfrac\pi2,dx=?$
So $\sqrt x=+\sin t,\sqrt{1-x}=?$
Alternatively $4x(1-x)=1-(2x-1)^2$
Set $2x-1=\sin y$ or $\cos y$
Observe that $2x-1=-\cos2t$ in the first method
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How many columns of this matrix are linearly independent? \begin{matrix}
1 & 2 & 0 & 1 \\
2 & 4 & 1 & 4 \\
3 & 6 & 3 & 9 \\
\end{matrix}
I have tried to transpose it and then reduce it by row echelon form and i get zeros on the last two rows. But i can't grasp if i should be doing that or doing it another way.
| If you look at your matrix
$$ A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 2 & 4 & 1 &4 \\ 3 & 6 & 3 & 9 \end{bmatrix} $$
The rank can be no more than $3$. Do the following
Subtract $2 R_{1} $ from $R_{2}$
$$ R_{2} - 2 R_{1} = \begin{bmatrix} 2 & 4 & 1 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 4 & 0 & 2\end{bmatrix} = \begin{bmatrix} 0 & 0& 1 & 2\end{bmatrix} $$
$$ A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 &2 \\ 3 & 6 & 3 & 9 \end{bmatrix} $$
Now we need to zero the third row. So subtract $3 R_{1} $ from $R_{3}$
$$ R_{3} - 3R_{1} = \begin{bmatrix} 3 & 6 & 3 & 9 \end{bmatrix} - \begin{bmatrix} 3 & 6 & 0 & 3\end{bmatrix} = \begin{bmatrix} 0 & 0& 3 & 6\end{bmatrix} $$
$$ A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 &2 \\ 0 & 0 & 3 & 6 \end{bmatrix} $$
Now we need to zero that $3$ in row $3$ . So we subtract $3R_{2}$ from $R_{3}$
$$ R_{3} - 3R_{2} = \begin{bmatrix} 0 & 0 & 3 & 6 \end{bmatrix} - \begin{bmatrix} 0 & 0 & 3 & 6\end{bmatrix} = \begin{bmatrix} 0 & 0& 0 & 0\end{bmatrix} $$
You're left with this matrix
$$ A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 &2 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$
There are only $2$ non-zero rows and they aren't multiples of eachother. The rank is $2$
If you want to look at column-wise then $2C_{1} = C_{2}$ and $C_{4} = C_{1}+2C_{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Combinatorics problem on counting. How many positive integers n are there such that all of the following take place:
1) n has 1000 digits.
2) all of the digits are odd.
3) the absolute value of the difference of any two consecutive (neighboring) digits is equal to 2.
Please help. I don’t even know how to start.
| Define $n_i=2i-1$ (so a bijection between 1,2,3,4,5 with 1,3,5,7,9).
Consider the 5x5 matrix $A=(a_{i,j})$ with $a_{i,j}=1$ if $n_i$ and $n_j$ differ by 2 and $a_{i,j}=0$ otherwise. Then, the number of positive integers with "m" digits satisfying your properties is the sum of entries of $A^{m-1}$. So you want to find the sum of entries of $A^{999}$. I don't know if this is easy to compute without computers.
Edit:
We have $$A=\left(\begin{array}{ccccc}
0&1&0&0&0\\
1&0&1&0&0\\
0&1&0&1&0\\
0&0&1&0&1\\
0&0&0&1&0\\
\end{array}
\right)$$
So, thanks to @Mike's comment, it shouldn't be difficult to find the entries of $A^{999}$ we have that $A=PDP^{-1}$ with
$$D=\left(\begin{array}{ccccc}
-1&0&0&0&0\\
0&0&0&0&0\\
0&0&1&0&0\\
0&0&0&-\sqrt{3}&0\\
0&0&0&0&\sqrt{3}\\
\end{array}
\right)$$
$$P=\left(\begin{array}{ccccc}
-1&1&-1&1&1\\
1&0&-1&-\sqrt{3}&\sqrt{3}\\
0&-1&0&2&2\\
-1&0&1&-\sqrt{3}&\sqrt{3}\\
1&1&1&1&1\\
\end{array}
\right)$$
So, we can compute $A^{999}=PD^{999}P^{-1}$ whose entries will be a linear combination of $(-1)^{999}, (1)^{999}, (-\sqrt{3})^{999},(\sqrt{3})^{999}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How does one prove such an equation? The problem occurred to me while I was trying to solve a problem in planimetry using analytic geometry.
for $b$ between $-\frac{1}2$ and $1$ :
$\sqrt{2+\sqrt{3-3b^2}+b} = \sqrt{2-2b}+ \sqrt{2-\sqrt{3-3b^2}+b}$
| First, we rearrange the equation into the equivalent equation
$$
\sqrt{2+b+\sqrt{3-3b^2}}-\sqrt{2+b-\sqrt{3-3b^2}} \overset?= \sqrt{2-2b}
$$
which we want to prove. Both sides are positive (keep in mind we have not yet proven the equality), so we can square both sides, and equivalence still holds:
$$
4+2b-2\sqrt{(2+b)^2-(3-3b^2)}
= \left(\sqrt{2+b+\sqrt{3-3b^2}}-\sqrt{2+b-\sqrt{3-3b^2}}\right)^2
\overset?= 2-2b
$$
Ie, $x^2=y^2$ implies $x=y$ if we know $x,y\ge 0$, otherwise the squares could be equal, but the two sides have opposite signs.
Now, simplify the left-hand side,
$$
2-2b = 4+2b-2\sqrt{1+4b+4b^2}= 4+2b-2\sqrt{(2+b)^2-(3-3b^2)} \overset?= 2-2b,
$$
and we see that there actually is equality.
Note that I have used throughout the assumption that the expression inside square roots are non-negative numbers. That's where limitations on $b$ come from.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Complex number cannot arive at $\frac{9}{2}-\frac{9}{2}i$ with problem $\frac{4+i}{i}+\frac{3-4i}{1-i}$ I am asked to evaluate: $\frac{4+i}{i}+\frac{3-4i}{1-i}$
The provided solution is: $\frac{9}{2}-\frac{9}{2}i$
I arrived at a divide by zero error which must be incorrect. My working:
$\frac{4+i}{i}$, complex conjugate is $-i$ so:
$\frac{-i(4+i)}{-i*i}$
= $\frac{-4i+i^2}{i^2}$
= $\frac{-4i--1}{-1}$
= $-4i+1$
Then the next part:
$\frac{3-4i}{1-i}$ complex conjugate is $1+i$ so:
$\frac{(1+i)(3-4i)}{(1+i)(1-i)}$
= $\frac{3-4i+3i-4i^2}{1-i^2}$
= $\frac{7-i}{0}$ # 1 + -1 = 0
How can I arrive at $\frac{9}{2}-\frac{9}{2}i$?
| Here is your error $1-i^2=1-(-1)=2\ne 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How can I find the P matrix in $P^{-1}AP $= D? I have the following exercise:
Let a matrix A = $\begin{pmatrix}
4 & 0 & 1\\
2 & 2 & 1\\
0 & 4 & 1
\end{pmatrix}$
1) Determine its eigenvalues and their multiplicity.
2) Give a basis of the eigen spaces associated with each of the
distinct eigenvalues of A.
3) find P and D such that $P^{-1}AP$ = D is diagonal.
I've done 1) and 2) and found the basis
B ={ $\begin{pmatrix}
-1 \\
-1 \\
4
\end{pmatrix}$ , $\begin{pmatrix}
-2 \\
1 \\
4
\end{pmatrix}$ , $\begin{pmatrix}
-1 \\
1 \\
1
\end{pmatrix}$ }
Now, in my notes they say that the matrix P is constituted of the vectors of the basis found.
So, P = $\begin{pmatrix}
-1 & -2 & -1\\
-1 & 1 & 1\\
4 & 4 & 1
\end{pmatrix}$
Also, because we have a 3x3 matrix and we've found three distinct eigen values, the matrix D exists and as such we put the eigen values on the diagonal line of the matrix. ($E_(\lambda_1) = 0$, $E_(\lambda_2) = 2$, $E_(\lambda_3) = 5$ )
D = $\begin{pmatrix}
0 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 5
\end{pmatrix}$
Finally to verify that this is all true I compute $AP = PD$. However I didn't find the same result on the left side and the right side of the equation. Where did I go wrong?
$AP \ne PD$
$\begin{pmatrix}
0 & -4 &-3\\
0 & 2 & 1 \\
0 & 8 & 5 \\
\end{pmatrix} \ne \begin{pmatrix}
0 & -4 &-5 \\
0 & 2 & 5 \\
0 & 8 & 5 \\
\end{pmatrix} $
| You got the last eigenvector wrong.
If you take $A$ and multiply with $(-1,1,1)$ you will get $(-3,1,5)$.
I believe you need to take $(1,1,1)$ instead. Then everything will work.
Good Luck!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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