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Simplify $2 \sqrt[3]{50x^2 z^5} × 3 \sqrt[3]{15 y^3 z}$ Image of problem
As you can see the answer is $30 y z^2 \sqrt[3]{6x^2}$ . I understand mostly everything in the problem, but one thing that I am having confusion on is where the "30" came from in the answer. I tried many ways to find out how the "30" got there.
| $$2\,\sqrt[3]{50\,x^2 z^5} \cdot 3\,\sqrt[3]{15\,\color{red}{y^3} z}=$$
$$=2 \cdot 3 \cdot \color{red}{y} \cdot \sqrt[3]{50\,x^2 z^5} \cdot \sqrt[3]{15\,z}=$$
$$=6\,y \cdot \sqrt[3]{(50\,x^2 z^5)(15 z)}=$$
$$=6\,y \cdot \sqrt[3]{750\,x^2 \color{magenta}{z^6}}=$$
$$=6\,y\color{magenta}{z^2} \cdot \sqrt[3]{2 \cdot 3 \cdot \color{blue}{5^3} \cdot x^2}=$$
$$=6 \cdot \color{blue}{5} \cdot yz^2 \cdot \sqrt[3]{2 \cdot 3 \cdot x^2}=$$
$$=30\,yz^2 \sqrt[3]{6\,x^2}.$$
As you see, $30$ is due to $6\cdot\sqrt[3]{5^3}$, emphatized in blue.
| {
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"url": "https://math.stackexchange.com/questions/1536032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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System of equations based question in matrix form If $\begin{bmatrix}1^2&2^2&3^2\\2^2&3^2&4^2\\3^2&4^2&5^2\\4^2&5^2&6^2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\4\\10\\\lambda\end{bmatrix}$,then $\lambda=$
$(A)17\hspace{1cm}(B)18\hspace{1cm}(C)19\hspace{1cm}(D)20$
I tried solving this question.
$1^2x+2^2y+3^2z=1$
$2^2x+3^2y+4^2z=4$
$3^2x+4^2y+5^2z=10$
I solved this system to get $x=\frac{25}{8},y=\frac{-5}{2},z=\frac{7}{8}$
$4^2x+5^2y+6^2z=\lambda$
I substituted $x,y,z$ in the above equation to get $\lambda=19$.
But this method is long and tedious.My book gives one method but that i cannot understand.Please help me understand this.Book's method is not self explanatory.
Book's solution:
$(n+3)^2=n^2-3(n+1)^2+3(n+2)^2$
$\lambda=1-3\times4+3\times 10=19$
OR if someone knows any better and elegant way to solve this question,please tell me,like using eigen values etc.Thanks.
| Each column of the matrix is of the form $[n^2,(n+1)^2,(n+2)^2,(n+3)^2]^T$. The book’s expansion of $(n+3)^2$ shows you that the third row is a linear combination of the first three, which means that $\lambda$ must be the same linear combination of the first three elements of the vector.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sqrt{2}x-\sqrt{x^{2}+1} \geq \frac{\sqrt{2}}{2}\ln{(x)}$ How can I prove that
$$
x\sqrt{2}-\sqrt{x^{2}+1} \geq \frac{\sqrt{2}}{2}\ln{(x)}
$$
It's a derivation-based process if I remember correctly, however I was unable to prove it correctly.
| Let $f(x)=x\sqrt{2}-\sqrt{x^{2}+1} -\dfrac{\sqrt{2}}{2}\ln{(x)}$. Then, $$f'(x)=\sqrt{2}-\frac{x}{\sqrt{x^2+1}}-\frac{1}{x\sqrt{2}}=\frac{2x\sqrt{x^2+1}-x^2\sqrt{2}-\sqrt{x^2+1}}{x\sqrt{2(x^2+1)}}.$$
For $x>0$, the denominator is positive, so we just need to check the sign of the numerator: $$2x\sqrt{x^2+1}-x^2\sqrt{2}-\sqrt{x^2+1}\geq2\sqrt{2}-\sqrt{2}-\sqrt{2}=0 \;\;\text{ for } \; x\geq1.$$
Thus, $f(x)$ is an increasing function on $[1,\infty)$.
Since $f(1)=0$, it follows that $f(x) \geq 0$ for all $x \in [1,\infty)$. Hence, $$x\sqrt{2}-\sqrt{x^{2}+1} -\dfrac{\sqrt{2}}{2}\ln{(x)} \geq0 \implies x\sqrt{2}-\sqrt{x^{2}+1} \geq \frac{\sqrt{2}}{2}\ln{(x)}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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integer solutions of $z=\frac{x^2}{4}+\frac{x}{y}$ Let $x,y,z$ be positive integers and $2012\leq z\leq 2555$.
How to solve the equation $z=\dfrac{x^2}{4}+\dfrac{x}{y}$ ?
By wolfram alpha, I get $x=88,y=1,z=2024$.
Can it be solved by manual?
Thank you.
| $$\frac{x^2}4<z<\frac{x^2}4+x+1\\
2012<\left(\frac x2+1\right)^2,\left(\frac x2\right)^2<2555\\
2\sqrt{2012}-2<x<2\sqrt{2555}$$
If $x$ is even, then $y$ can be any factor of $x$.
If $x$ is odd, then $x/y$ must be $n+\frac34$, so $x/(y/4)$ can be any factor of $x$ which is $3\pmod4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the limit of a fraction: $\lim_{x \to 3} \frac{x^3-27}{x^2-9}$
Find $$\lim_{x \to 3} \frac{x^3-27}{x^2-9}$$
What I did is:
\begin{align}
\lim_{x \to 3} \frac{x^3-27}{x^2-9} &= \lim_{x \to 3} \frac{(x-3)^3+9x-27x}{(x-3)(x+3)} = \lim_{x \to 3} \frac{(x-3)^3+9(x-3)}{(x-3)(x+3)} \\
&= \lim_{x \to 3} \frac{(x-3)^3}{(x-3)(x+3)} + \lim_{x \to 3} \frac{9(x-3)}{(x-3)(x+3)} \\
&= \lim_{x \to 3} \frac{(x-3)^2}{(x+3)} + \lim_{x \to 3} \frac{9}{(x+3)} =0 + \frac{9}{6}
\end{align}
Wolfram factor the numerator to $(x-3)(x^2+3x+9)$ is there a quick way to find this?
|
Wolfram factor the numerator to $(x-3)(x^2+3x+9)$ is there a quick way to find this?
Generally, you have
$$
a^{n+1}-b^{n+1}=(a-b)(a^n+a^{n-1}b+a^{n-2}b^2+\cdots+a^2b^{n-2}+ab^{n-1}+b^n)
$$ (to see this expand the right hand side, then terms telescope).
Then apply it to $a=x$, $b=3$, $n=2$ giving
$$
x^3-27=(x-3)(x^2+3x+9).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the sum of $\frac {3}{4} + \frac {3}{28} + \frac {3}{70} + \frac {3}{130} .... + \frac {3}{9700}$ What will be the sum of :
$\frac {3}{4} + \frac {3}{28} + \frac {3}{70} + \frac {3}{130} .... + \frac {3}{9700}$ ?
| $\frac {3}{4} + \frac {3}{28} + \frac {3}{70} + \frac {3}{130} .... + \frac {3}{9700}=\frac {4-1}{4} + \frac {7-4}{28} + \frac {10-7}{70} + \frac {13-10}{130} .... + \frac {100-97}{9700}=\left(\frac {4}{4} + \frac {7}{28} + \frac {10}{70} + \frac {13}{130} .... + \frac {100}{9700}\right)-\left( \frac {1}{4} + \frac {4}{28} + \frac {7}{70} + \frac {10}{130} .... + \frac {97}{9700} \right)=1-\frac{1}{100}=\frac{99}{100}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Taylor's Remainder $x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) Prove that
$\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) <x-\frac{x^2}{2}+\frac{x^3}{3}$
My attempt: I can prove this by taking one side at a time and assuming $\displaystyle f(x) = \log (1+x)-x+\frac{x^2}{2}-\frac{x^3}{3}$
and then proving that it is a decreasing function and then the same for the other side.
But I am looking for a better solution using Lagrange's Mean Value Theorem or using Taylor's remainder.
| This is shown in
section 14 of
Heinrich Dorrie's
"100 Great Problems of
Elementary Mathematics",
which I highly recommend.
The proof,
for general $n$,
goes like this:
Let
$f(t) = \dfrac1{1+t}$.
Then,
as can be shown by induction,
$$f(t)
=\sum_{k=0}^{n-1} (-1)^{k}t^k+(-1)^nt^nf(t)
$$
Integrating from
$0$ to $x$,
\begin{align}\\
\ln(1+x)
&=\int_0^x \frac{dt}{1+t}\\
&=\int_0^x \left(\sum_{k=0}^{n-1} (-1)^{k}t^k+(-1)^nt^nf(t)\right)dt\\
&=\sum_{k=0}^{n-1} \int_0^x(-1)^{k}t^kdt+\int_0^x(-1)^nt^nf(t)dt\\
&=\sum_{k=0}^{n-1} (-1)^{k}\frac{x^{k+1}}{k+1}+(-1)^n\int_0^x\frac{t^ndt}{1+t}\\
&=\sum_{k=0}^{n-1} (-1)^{k}\frac{x^{k+1}}{k+1}+(-1)^nr_n(t)\\
\end{align}
where
$$r_n(t)
=\int_0^x\frac{t^ndt}{1+t}
$$
Since
$1 \le 1+t
\le 1+x
$ for
$0 \le t \le x$,
$$t^n
\ge \frac{t^n}{1+t}
\ge \frac{t^n}{1+x}
$$
so that,
since
$$\int_0^x t^n dt
=\frac{x^{n+1}}{n+1}
$$
$$\frac{x^{n+1}}{n+1}
\ge r_n(x)
\ge \frac{x^{n+1}}{(n+1)(1+x)}
$$
Therefore
$$r_n(x)
=\frac{x^{n+1}}{(n+1)(1+v)}
$$
where
$0 < v < x
$.
Therefore
$$\ln(1+x)
=\sum_{k=0}^{n-1} (-1)^{k}\frac{x^{k+1}}{k+1}
+(-1)^n\frac{x^{n+1}}{(n+1)(1+v)}
$$
where
$0 <v < x$.
| {
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"url": "https://math.stackexchange.com/questions/1546172",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Showing that a polynomial is a unit in a quotientring In this exercise I have a polynomial ring over a finite field $F_2$:$(F_2[X],+,*)$ There is then given an ideal : I $=<X^3+X+1>$.
I am then trying to show that $X^2+X+1 + I$ is a unit in $F_2[X]/I$ by using the extended Euclidean algorithm on $X^3+X+1$ and $X^2+X+1$.
I tried using this algorithm and then got that $(X^3+X+1)*r(X)+(X^2+X+1)*s(X)=gcd(X^3+X+1,X^2+X+1) => (X^3+X+1)*(X+1)+(X^2+X+1)*X^2=1$
I can't really figure out if this is right, and if it is what I am suppose to do from here to show that the given expression is an unit.
| What you have is correct: Since $+1 = -1$, you can write it as
$$
(x^3 + x + 1)(x+1) - (x^2 + x + 1)x^2 = x^4 + x^3 + x^2 + 2x + 1 - x^4 - x^3 - x^2 = 1
$$
since $2x=0$
When you apply the quotient map $\pi : F_2[x] \to F_2[x]/(x^3+x+1)$ to this expression, you get
$$
\pi(x^2+x+1)\pi(x^2) = \pi(1)
$$
Hence, $\pi(x^2+x+1)$ is invertible in the quotient.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On the decreasing sequences I do not know how to prove that these sequences are decreasing:
$$
a_n=\frac{1}{\sqrt{n}(n+1)}-\frac{1}{\sqrt{n+1}(n+2)},
$$
$$
b_n=\frac{1}{n(\sqrt{n}+1)}-\frac{1}{(n+1)(\sqrt{n+1}+1)}.
$$
Thank you for all kind help and comments.
My attemption. I considered the following function
$$
f(x)=\frac{1}{\sqrt{x}(x+1)}-\frac{1}{\sqrt{x+1}(x+2)},
$$
and calculated its derivative. But its derivative is rather complicated
$$
\nabla f(x)=\frac{3x+1}{2\sqrt{x}x(x+1)^2}-\frac{3x+4}{2\sqrt{x+1}(x+1)(x+2)^2}
$$
| Both sequences can be written as $a_n = f(n) - f(n+1)$ where $f$
is a (stricly) convex function.
Then
$$
a_{n+1} - a_n = \bigl(f(n+1) - f(n+2)\bigr) - \bigl((f(n) - f(n+1)\bigr) = 2f(n+1)- f(n) - f(n+2) < 0
$$
because
$$
f(n+1) < \frac12 \bigl(f(n) + f(n+2) \bigr)
$$
holds for a strictly convex function.
For your first sequence, $f(x)=\frac{1}{\sqrt{x}(x+1)}$
is convex because
the second derivative
$$
f''(x) = \frac{5x^2 + 10 x + 3}{4 x^{5/2}(x+1)^3}
$$
is positive for $x > 0$.
For the second sequence, $f(x) = \frac{1}{x (\sqrt{x} + 1)}$
and
$$
f''(x) = \frac{15 x + 21 \sqrt x +8}{4 (x^{3/2} + x)^3}
$$
is positive for $x > 0$.
(All derivatives calculated with Wolram Alpha :)
| {
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"timestamp": "2023-03-29T00:00:00",
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integate $\int\sqrt{x^2+3x+3} dx$ Is there any universal method how to solve integrals like this?
$$\int\sqrt{x^2+3x+3} dx$$
Or this?
$$\int\sqrt{-x^2+3x+3} dx$$
I tried use first Euler subs, but it was not good idea.
$$\sqrt{ax^2+bx+c} = t\pm\sqrt{ax}$$
| For any integral of the forms, you have mentioned,
transform the integrand into the sum or difference of two perfect square expressions and the resulting expression will look like $\sqrt{(x\pm a)^2+(b)^2}$ ,$\sqrt{(x\pm a)^2-(b)^2}$ or $\sqrt{(b)^2-(x\pm a)^2}$.
And then use the following formulae which can be deduced with the help of integration by parts:
*
*$$\sqrt{(x\pm a)^2+(b)^2}dx=\frac{(x\pm a)\sqrt{(x\pm a)^2+(b)^2}}{2}+\frac{b^2}{2}\ln |(x\pm a)+\sqrt{(x\pm a)^2+(b)^2}|$$
*$$\sqrt{(x\pm a)^2-(b)^2}dx=\frac{(x\pm a)\sqrt{(x\pm a)^2-(b)^2}}{2}-\frac{b^2}{2}\ln |(x\pm a)+\sqrt{(x\pm a)^2-(b)^2}|$$
*$$\sqrt{(b)^2-(x\pm a)^2}dx=\frac{(x\pm a)\sqrt{(b)^2-(x\pm a)^2}}{2}+\frac{b^2}{2}\sin \frac{x\pm a}{b}$$
Hope you can apply this to your problem and solve it now.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $\sinh^{-1} (\tan x)=\log \tan (\frac{\pi}{4}+\frac{x}{2})$ Like the question says
How to prove $$\sinh^{-1} (\tan x)=\log \tan (\frac{\pi}{4}+\frac{x}{2})$$
I have tried using many identity but in vain
For reference
$$\tanh ^{-1} x=\frac{1}{2} \log \frac{1+x}{1-x}$$
and $$\sinh^{-1} x=\log (x+\sqrt{x^2+1})$$
| Notice, $$\sinh^{-1}(\tan x)=\log(\tan x+\sqrt{\tan^2 x+1})$$
$$=\log(\tan x+\sec x)$$$$=\log\left(\tan x+\frac{1}{\cos x}\right)$$
$$=\log\left(\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}+\frac{1+\tan^2\frac{x}{2}}{1-\tan^2\frac{x}{2}}\right)$$
$$=\log\left(\frac{\left(1+\tan\frac{x}{2}\right)^2}{\left(1-\tan\frac{x}{2}\right)\left(1+\tan\frac{x}{2}\right)}\right)$$
$$=\log\left(\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right)$$
$$=\log\left(\frac{\tan\frac{\pi}{4}+\tan\frac{x}{2}}{1-\tan\frac{\pi}{4}\tan\frac{x}{2}}\right)$$
$$=\color{red}{\log\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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About the integral $\int_{-\pi}^{\pi} \frac{(1+2\cos x)^n}{5+4\cos x} \cos nx\space dx$ This equality, I read in a Russian book of problems, attracts me but I cannot verify it yet. If someone helps me, thank you.
$$\int_{-\pi}^{\pi} \frac{(1+2\cos x)^n}{5+4\cos x} \cos nx\space dx=\frac{2\pi}{3}\left ( \frac 34\right)^n$$
It seems to be advisable to use complex numbers and Cauchy integration (I don’t remember very well this topic now).
| Suppose we seek to evaluate
$$\int_{-\pi}^{\pi} \frac{(1 + 2\cos x)^n}{5 + 4\cos(x)}
\cos(nx) dx.$$
Introduce $z=\exp(ix)$ as suggested in the comments so that $dz=iz \;
dx$ to get
$$\int_{|z|=1}
\frac{(1 + z + 1/z)^n}{5 + 2z + 2/z}
\frac{z^n + 1/z^n}{2} \frac{dz}{iz}
\\ = \frac{1}{2i} \int_{|z|=1}
\frac{(1 + z + 1/z)^n}{2z^2 + 5z + 2}
(z^n + 1/z^n) \; dz
\\ = \frac{1}{2i} \int_{|z|=1}
\frac{(z^2 + z + 1)^n}{2z^2 + 5z + 2}
(1 + 1/z^{2n}) \; dz.$$
There are two poles at $z=-1/2$ and $z=-2$ of the fractional term of
which only the first one is inside the contour for a residue of
$$\left.\frac{(z^2+z+1)^n}{4z+5} (1 + 1/z^{2n})\right|_{z=-1/2}
= \frac{(3/4)^n}{3} (1 + 2^{2n}).$$
The remaining contribution is from the pole at zero:
$$\frac{1}{2i} \int_{|z|=1} \frac{1}{z^{2n}}
\frac{(z^2 + z + 1)^n}{2z^2 + 5z + 2}
\; dz.$$
We evaluate this using the fact that the residues at the poles sum to
zero so we compute the residues at $z=-1/2$ and $z=-2.$
We obtain
$$\left.\frac{(z^2+z+1)^n}{2n z^{2n-1} (2z^2+5z+2) + z^{2n} (4z+5)}
\right|_{z=-1/2}
= \frac{(3/4)^n}{3/2^{2n}} = 3^{n-1}$$
and
$$\left.\frac{(z^2+z+1)^n}{2n z^{2n-1} (2z^2+5z+2) + z^{2n} (4z+5)}
\right|_{z=-2}
= \frac{3^n}{2^{2n} (-3)}.$$
Adding the three contributions we obtain
$$2\pi i \times \frac{1}{2i}
\left(\frac{3^{n-1}}{4^n} + 3^{n-1}
- 3^{n-1} + \frac{3^{n-1}}{2^{2n}}\right)
= \pi \times 2 \frac{3^{n-1}}{4^n}
= \pi \times \frac{2}{3} \left(\frac{3}{4}\right)^n.$$
Remark. To be perfectly rigorous here we also need to show that
the residue at infinity is zero. We get
$$-\mathrm{Res}_{z=0} \frac{1}{z^2}
\frac{(1/z^2 + 1/z + 1)^n}{2/z^2 + 5/z + 2}
z^{2n}
= -\mathrm{Res}_{z=0}
\frac{(1/z^2 + 1/z + 1)^n}{2 + 5z + 2z^2}
z^{2n}
\\ = -\mathrm{Res}_{z=0}
\frac{(1 + z + z^2)^n}{2 + 5z + 2z^2} = 0$$
as claimed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the differential equation for obtaining $x$ as a relation of $t$: $\frac{d^2x}{dt^2}=\alpha\sqrt{x}$ Question: Solve the differential equation for obtaining $x$ as a relation of $t$:
$$\frac{d^2x}{dt^2}=\alpha\sqrt{x}$$
My attempt: $$\frac{d^2x}{dt^2}=\alpha\sqrt{x}$$
$$\Rightarrow 2\frac{dx}{dt}\frac{d^2x}{dt^2}=\alpha\sqrt{x}\cdot 2\frac{dx}{dt}$$
$$\Rightarrow \frac{d}{dt}\left[\left(\frac{dx}{dt}\right)^2\right]=\frac{4}{3}\alpha\cdot \frac{d}{dt}\left(x^\frac{3}{2}\right)$$
$$\Rightarrow \left(\frac{dx}{dt}\right)^2=\frac{4}{3}\alpha x^\frac{3}{2}+c_1$$
Now we have $$\frac{dx}{dt}=\sqrt{\frac{4}{3}\alpha x^\frac{3}{2}+c_1}=\sqrt{k x^\frac{3}{2}+c_1}$$
$$\Rightarrow \frac{dx}{\sqrt{k x^\frac{3}{2}+c_1}}=dt$$
Can anyone suggest how to proceed? Any substitutions?
| HINT:
$$x''(t)=\alpha\sqrt{x(t)}\Longleftrightarrow$$
$$\frac{\text{d}^2x(t)}{\text{d}t^2}=\alpha\sqrt{x(t)}\Longleftrightarrow$$
$$\frac{\text{d}^2x(t)}{\text{d}t^2}\cdot\frac{\text{d}x(t)}{\text{d}t}=\alpha\sqrt{x(t)}\cdot\frac{\text{d}x(t)}{\text{d}t}\Longleftrightarrow$$
$$\int\space\frac{\text{d}^2x(t)}{\text{d}t^2}\cdot\frac{\text{d}x(t)}{\text{d}t}\space\text{d}t=\int\space\alpha\sqrt{x(t)}\cdot\frac{\text{d}x(t)}{\text{d}t}\space\text{d}t\Longleftrightarrow$$
$$\frac{1}{2}\left(\frac{\text{d}x(t)}{\text{d}t}\right)^2=\frac{2\alpha x(t)^{\frac{3}{2}}}{3}+\text{C}\Longleftrightarrow$$
$$\left(\frac{\text{d}x(t)}{\text{d}t}\right)^2=\frac{4\alpha x(t)^{\frac{3}{2}}}{3}+2\text{C}\Longleftrightarrow$$
$$\frac{\text{d}x(t)}{\text{d}t}=\pm\sqrt{\frac{4\alpha x(t)^{\frac{3}{2}}}{3}+2\text{C}}\Longleftrightarrow$$
$$\frac{\frac{\text{d}x(t)}{\text{d}t}}{\sqrt{\frac{4\alpha x(t)^{\frac{3}{2}}}{3}+2\text{C}}}=\pm 1\Longleftrightarrow$$
$$\int\space\frac{\frac{\text{d}x(t)}{\text{d}t}}{\sqrt{\frac{4\alpha x(t)^{\frac{3}{2}}}{3}+2\text{C}}}\space\text{d}t=\pm\int\space 1\space\text{d}t\Longleftrightarrow$$
$$\frac{_2\text{F}_1\left(\frac{1}{2},\frac{2}{3};\frac{5}{3};-\frac{2\alpha x(t)^{\frac{3}{2}}}{3\text{C}}\right)\cdot x(t)\sqrt{\frac{2\alpha x(t)^{\frac{3}{2}}}{\text{C}}+3}}{\sqrt{6\text{C}+4\alpha x(t)^{\frac{3}{2}}}}=\pm t+\text{F}$$
With $\text{C}$ and $\text{F}$ is an arbitrary constants
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553616",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Is there an algorithm to split a number into the sum of powers of 2? Am I able to split, lets say 76, into the sum of powers of two, through an algorithm and without cycling through possible combinations?
For the example above, the answer would be '2^6+2^3+2^2' or just simply the exponents, so '6,3,2'
Thanks in advance.
| Make successive divisions by $2$ and note the remainders, until the quotient is $0$:
$$\begin{array}{r|cc}
76&0\\38&0\\19&1\\9&1&\uparrow\\4&0\\2&0\\1&1
\end{array}$$
The binary digits of $76$ are $\;\color{red}{1001100}_2$. In other words
$$76=2^6+2^3+2^2.$$
This is because, if you write the Euclidean division equalities for each of these divisions, you have (Horner scheme)
\begin{align*}
76&=2\cdot 38 =2(2\cdot 19))=2(2(2\cdot 9+1))=2(2(2(2\cdot 4+1)+1))
\\&=2(2(2(2(2\cdot 2)+1)+1))=2(2(2(2(2(2\cdot 1))+1)+1))
\end{align*}
| {
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} |
integer solution of $\frac{x^3+y^3+z^3-xy(x+y)-yz(y+z)-xz(x+z)-2xyz}{(x+y+z)(x+y-z)(x-y+z)(x-y-z)}=\frac{1}{2016}$ Let $x,y,z$ be positive integers such that $\frac{x^3+y^3+z^3-xy(x+y)-yz(y+z)-xz(x+z)-2xyz}{(x+y+z)(x+y-z)(x-y+z)(x-y-z)}=\frac{1}{2016}$.
How to find all solutions ?
I have no any idea.
Thanks in advance.
| Hint:
Since$$\frac{x^{3}+y^{3}+z^{3}-xy(x+y)-yz(y+z)-zx(z+x)-2xyz}{(x+y+z)(x+y-z)(x-y+z)(x-y-z)}=\frac{1}{2}\left(\frac{1}{-x+y+z}+\frac{1}{x-y+z}+\frac{1}{x+y-z}+\frac{1}{x+y+z}\right),$$
we are trying to solve $$\frac{1}{-x+y+z}+\frac{1}{x-y+z}+\frac{1}{x+y-z}+\frac{1}{x+y+z}=\frac{1}{1008}.$$
This is equivalent to solving the equation $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c}=\frac{1}{1008}.$$
| {
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"source": "stackexchange",
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Find the minimal polynomial of $(x-1)^2(x-2)^3(x-3)$? Given ch. polynomial is $(x-1)^2(x-2)^3(x-3)$ and for the eigenvalue $1$, we have one eigenvector, for eigenvalue $2$ we have $2$ eigenvector and for $3$ we have $1$ eigenvector. Now determine the minimal polynomial?
I know that minimal polynomial divides the ch. polynomial, so I list the probable candidates as $$(x-1)(x-2)(x-3)$$ $$(x-1)(x-2)^2(x-3)$$ $$(x-1)^2(x-2)^2(x-3)$$ and $$(x-1)^2(x-2)(x-3).$$ But how to determine which one of these serves the purpose? There is no additional information given. Help me out
| In fact there exist operators having your prescribed characteristic polynomial for each of the possible minimal polynomials provided:
Indeed,
*
*for $(x−1)(x−2)(x−3)$ we have the diagonal matrix diag$(1,1,2,2,2,3)$
*for $(x−1)^2(x−2)^2(x−3)$;
$$\left(\begin{smallmatrix}1 & 1& 0& 0&0 &0\\ 0& 1& 0& 0&0 &0 \\ 0& 0& 2& 1 & 0 & 0 \\0 & 0 & 0& 2 & 0 & 0 \\ 0 & 0 & 0 & 0& 2 & 0 \\ 0 & 0 & 0 & 0 &0 &3 \end{smallmatrix}\right)$$
which is unique up to permutation of the Jordan blocks.
*for $(x−1)(x−2)^2(x−3)$;
$$\left(\begin{smallmatrix}1 & 0& 0& 0&0 &0\\ 0& 1& 0& 0&0 &0 \\ 0& 0& 2& 1 & 0 & 0 \\0 & 0 & 0& 2 & 0 & 0 \\ 0 & 0 & 0 & 0& 2 & 0 \\ 0 & 0 & 0 & 0 &0 &3 \end{smallmatrix}\right)$$
again unique up to permutation of the Jordan blocks.
and finally
*
*for $(x−1)^2(x−2)^3(x−3)$;
$$\left(\begin{smallmatrix}1 & 1& 0& 0&0 &0\\ 0& 1& 0& 0&0 &0 \\ 0& 0& 2& 1 & 0 & 0 \\0 & 0 & 0& 2 & 1 & 0 \\ 0 & 0 & 0 & 0& 2 & 0 \\ 0 & 0 & 0 & 0 &0 &3 \end{smallmatrix}\right)$$
up to permutation of the Jordan blocks.
Additionally, the combinatorial theory behind these ideas is very interesting. What you have specified in your problem (which is merely a characteristic polynomial) is
*
*the dimension of the vector space (a.k.a. the size of the matrix in question); this is the degree of the characteristic polynomial.
*the algebraic multiplicity of each of the eigenvalues occurring in its upper triangular form (i.e. exactly which values will appear on the diagonal of the Jordan form)
*The different possible minimal polynomials hold the geometric multiplicity of each of the eigenvalues which specify the size of the maximum Jordan block for each of the eigenvalues in question.
| {
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"source": "stackexchange",
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} |
Chain rule with fraction In the case of $$f(x)=\ln\big(x+\sqrt{1+x^2}\big)$$
in the derivative we multiply $$f'(x)=\frac{1}{x+\sqrt{1+x^2}}\bigg(1+\frac{2x}{2\sqrt{1+x^2}}\bigg)$$ when the expression multiply the numerator?
| Following up on the tip of exponentiation of both side.
$$
\mathrm{e}^{f(x)} = \mathrm{e}^{\ln\left(x+\sqrt{1+x^2}\right)} = x+\sqrt{1+x^2}
$$
taking the derivative of both sides
$$
\mathrm{e}^{f(x)} f'(x) = 1 + \frac{1}{2}\frac{2x}{\sqrt{1+x^2}} = 1+\frac{x}{\sqrt{1+x^2}} = \frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2}}
$$
bow the l.h.s we have
$$
\mathrm{e}^{f(x)} f'(x) = \left(x+\sqrt{1+x^2}\right)\cdot f'(x)
$$
equating the last term in the previous two equations we get
$$
\left(x+\sqrt{1+x^2}\right)\cdot f'(x) = \left(x+\sqrt{1+x^2}\right)\cdot \frac{1}{\sqrt{1+x^2}}
$$
thus we get
$$
f'(x) = \frac{1}{\sqrt{1+x^2}}
$$
the trick used here reduces the pain of doing derivations of the form
$$
f(x) = g(u(x))
$$
where the inverse of $g$ yields forms that are easier to handle in the differentiation such as $\ln (x) $ and $\mathrm{e}^x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Quotient Rings example I have a field = $F_2$ and a polynomial $x^2 + x + 1$ over that field.
I understand that since the polynomial is degree 2, it has no root in $F_2$, so it is irreducible. But why is $F_2[X]/(x^2+x+1)$ a field with 4 elements? (0, 1, x, x+1)
Also when I do the multiplication table of the elements, why is xx = (x+1) and why is x(x+1) = 1?
| $\newcommand{\FF}{\mathbb{F}}$
Concretely, the field has an explicit presentation as equivalence classes of polynomials in $F := \FF_2[x]$ modulo the relation: $f(x)\equiv g(x)$ if $f(x)-g(x)\equiv 0\mod x^2+x+1$.
It's clear that $0,1,x,x+1$ are inequivalent elements of $F$. On the other hand, any other element is a polynomial of degree at least 2, and thus it starts $x^n + \cdots$. But modulo $x^2+x+1$, you know that $x^2 \equiv -x-1 \equiv x+1$. This is because in $\FF_2$, $-1 = 1$, and thus
$$x^2 - (x+1) = x^2 - x - 1 = x^2 + (-1)x + (-1) = x^2 + x + 1\equiv 0 \mod x^2+x+1$$
This shows that $x^2 - (x+1)\equiv 0\mod x^2+x+1$, and so $x^2\equiv x+1\mod x^2+x+1$.
Thus, any polynomial that starts $x^n + \cdots$ is equivalent to $x^{n-2}(x+1) + \cdots$ mod $x^2+x+1$, which has degree $n-1$. By induction this shows that any polynomial of degree at least 2 is equivalent to one with degree less than 2. This shows that $0,1,x,x+1$ are a complete set of representatives for elements of $F$.
As an example, consider $f = x^3+x^2+x+1$. Using $x^2 \equiv x+1$, you get
$$f \equiv x(x+1) + (x+1) + x + 1 = x^2 + x + x + 1 + x + 1 = x^2 + x \equiv (x+1) + x\equiv 1\mod x^2+x+1$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Adding vector to span a space
Let $A=\left\{\left(\begin{array}{c} 18 \\ 6 \\ -4 \\ 12 \end{array}\right),\left(\begin{array}{c} 6 \\ 2 \\ 2 \\ -6 \end{array}\right)\right\}$
find the vectos to be added so A will span $\mathbb{R}^4$?
So what I did is :
\begin{pmatrix}
18 & 6 & -4 & 12 \\
6 & 2 & 2 & -6 \\
\end{pmatrix}
the row reduced echelon form is:
\begin{pmatrix}
1 & \frac{1}{3} & 0 & 0 \\
0 & 0 & 1 & -10 \\
\end{pmatrix}
What should I do next? I need to write the vectors that should be added with 1 and 0 only?
| Observe that the "core" elements that make those two vectors linearly independent is the tiny little matrix formed by their second and third elements:
$$
M(2,3) = \begin{matrix}
6 & 2 \\
-4 & 2 \\
\end{matrix}
$$
which has non-zero determinant.
If we can add two other vectors such that the determinant of the four column vectors is reduced to the calculation of the determinant of $M(2,3)$, which is $\neq 0$, then we have found a span of $\mathbb{R}^4$.
Fortunately that's easy:
$$
\begin{vmatrix}
18 & 6 & 0 & 1\\
6 & 2 & 0 & 0\\
-4 & 2 & 0 & 0\\
12 & -6 & 1 & 0\\
\end{vmatrix} =
-\begin{vmatrix}
6 & 2 & 0\\
-4 & 2 & 0\\
12 & -6 & 1\\
\end{vmatrix} =
-\begin{vmatrix}
6 & 2\\
-4 & 2\\
\end{vmatrix}\neq 0
$$
So the vectors that you can add are the last two columns of the big determinant.
| {
"language": "en",
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} |
$n^{th}$ determinant Find determinant $D_n$ of matrix $$
\begin{bmatrix}
1 & 1 & \cdots & 1 & -n \\
1 & 1 & \cdots & -n & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
1 & -n & \cdots & 1 & 1 \\
-n & 1 & \cdots & 1 & 1
\end{bmatrix}
$$
After multiplying first row by $-1$ and adding to $n-1$ rows:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & 0 & \cdots & -n-1 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & -n-1 & \cdots & 0 & n+1 \\
-n & 1 & \cdots & 1 & 1
\end{vmatrix}
$$
Second and $n-1$ row are changing place:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
-n & 1 & \cdots & 1 & 1
\end{vmatrix}
$$
Multiplying first row by $n$ and adding it to $n$-th row:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
0 & n+1 & \cdots & n+1 & 1-n^2
\end{vmatrix}
$$
Adding second to $n$-th row:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
0 & 0 & \cdots & n+1 & (1+n)(2-n)
\end{vmatrix}
$$
Adding $n-1$ to $n$-th row:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
0 & 0 & \cdots & 0 & (1+n)(3-n)
\end{vmatrix}
$$
Product on main diagonal gives $$D_n=(n+1)(3-n)(-n-1)^{n-2}$$
This is not correct. What is wrong with this upper triangular transformation?
| Another solution, perhaps with an unusual method.
Let $T_m(x)$ the determinant of the $m\times m$ matrice, with entries $x$ on the antidiagonal, and $1$ otherwise. Clearly, your determinant $D_n$ is equal to $T_n(-n)$. The determinant $T_m(x)$ is a polynomial in $x$. The derivative of $T_m$ is obtained by differentiating the $k$-th column, computing the new determinant, and adding all these determinants. The computation of these determinant is easy: they are all equal to $(-1)^{m-1}T_{m-1}(x)$. Hence we get $T_m^{\prime}(x)=(-1)^{m-1}m T_{m-1}(x)$. In addition it is clear that $T_m(1)=0$. Put $S_m(x)=T_m(1+x)$. We get $S_m^{\prime}(x)=(-1)^{m-1}m S_{m-1}(x)$ and $S_m(0)=0$. We have $S_1(x)=1+x$, $S_2(x)=-2x-x^2$. An easy induction show that $S_m(x)=(-1)^{m(m-1)/2}(m x^{m-1}+x^m)$. Now to finish
$D_n=T_n(-n)=S_n(-n-1)$, and we are done.
| {
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"source": "stackexchange",
"question_score": "1",
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Recurrence relation $T(n+1)=T(n)+⌊\sqrt{n+1}⌋$? Consider the following recurrence relation
$T(1)=1$
$T(n+1)=T(n)+⌊\sqrt{n+1}⌋$ for all $n≥1$
The value of $T(m^2)$ for $m≥1$ is
*
*$(m/6) (21m – 39) + 4$
*$(m/6) (4m^2 – 3m + 5)$
*$(m/2) (m^{2.5} – 11m + 20) – 5$
*$(m/6) (5m^3 – 34m^2 + 137m – 104) + (5/6)$
My attempt :
I've used counter example to choose correct option.
$m = 3$
$T(9) = T(4) + 2*5 + 1 = 5 + 10 + 1 = 16$
Both options $(1)$ & $(2)$ produces $16$.
$m = 4$
$T(16) = T(9) + 3*7 + 1 = 16 + 21 + 1 = 38$
Only option $(2)$ produces $38$, $(1)$ produces $34$ which doesn't match.
Can you explain in formal way, please?
| Note that $T(1)=\left\lfloor\sqrt1\right\rfloor$; from this and the recurrence it’s straightforward to see that
$$T(n)=\sum_{k=1}^n\left\lfloor\sqrt k\right\rfloor$$
for all $n\ge 1$. Thus
$$T(m^2)=\sum_{k=1}^{m^2}\left\lfloor\sqrt k\right\rfloor\;.\tag{1}$$
Now for $k=1,\ldots,m^2$ the possible values of $\left\lfloor\sqrt k\right\rfloor$ are the integers $1,\ldots,m$, and we can calculate $(1)$ quite easily if we can determine how often each of these integers appears. We have $\left\lfloor\sqrt k\right\rfloor=n$ if and only if $n^2\le k<(n+1)^2=n^2+2n+1$, so there are $2n+1$ integers $k$ such that $\left\lfloor\sqrt k\right\rfloor=n$. This holds for $k=1,\ldots,m-1$, and we have one value of $k$, $k=m^2$, for which $\left\lfloor\sqrt k\right\rfloor=m$. Thus, the sum in $(1)$ is equal to
$$\begin{align*}
m+\sum_{n=1}^{m-1}n(2n+1)&=m+2\sum_{n=1}^{m-1}n^2+\sum_{n=1}^{m-1}n\\
&=m+2\cdot\frac{(m-1)m(2m-1)}6+\frac{(m-1)m}2\\
&=m+\frac{m(m-1)\big(2(2m-1)+3\big)}6\\
&=m+\frac{m(m-1)(4m+1)}6\\
&=\frac{m}6\big(6+(m-1)(4m+1)\big)\\
&=\frac{m}6(4m^2-3m+5)\;.
\end{align*}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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What to do with this Prove that
$$
\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x(x+1)}dx=\frac{1}{2} (\ln 2)^2-\ln \pi \ln2
$$
I separated them
$$
\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x}dx-\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x+1}dx
$$
For the former integral i tried to use differentiation under integration but got stuck and i have no idea about the latter one. Plz help!
| \begin{align}
&\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x}dx-\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x+1}dx=\\
&\int^{1}_{0} \frac{\ln\frac{\sin \pi x}{2\sin(\pi x/2)}}{x}dx-\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x+1}dx=\\
&\lim_{\varepsilon\to 0}\left(\int^{1}_{\varepsilon} \frac{\ln\frac{\sin\pi x}{2}}{x}dx-\int^{1}_{\varepsilon} \frac{\ln \left(\sin \left(\frac{\pi x}{2} \right)\right)}{x}dx-\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x+1}dx\right)=\\
&\lim_{\varepsilon\to 0}\left(\int^{1}_{\varepsilon} \frac{\ln\frac{\sin\pi x}{2}}{x}dx-\int^{1}_{\varepsilon} \frac{\ln \left(\sin \left(\frac{\pi x}{2} \right)\right)}{x}dx-\int^{2}_{1} \frac{\ln \left(\sin \left(\frac{\pi x}{2} \right)\right)}{x}dx\right)=\\
&\lim_{\varepsilon\to 0}\left(\int^{1}_{\varepsilon} \frac{\ln\frac{\sin\pi x}{2}}{x}dx-\int^{2}_{\varepsilon} \frac{\ln \left(\sin \left(\frac{\pi x}{2} \right)\right)}{x}dx\right)=\\
&\lim_{\varepsilon\to 0}\left(\int^{1}_{\varepsilon} \frac{\ln\frac{\sin\pi x}{2}}{x}dx-\int^{1}_{\varepsilon/2} \frac{\ln \left(\sin \pi x\right)}{x}dx\right)=\\
&\lim_{\varepsilon\to 0}\left(-\ln 2\int^{1}_{\varepsilon}\frac{dx}{x}-\int^{\varepsilon}_{\varepsilon/2} \frac{\ln \left(\sin \pi x\right)}{x}dx \right)=\\
&\lim_{\varepsilon\to 0}\left(-\ln 2\int^{1}_{\varepsilon}\frac{dx}{x}-\int^{\varepsilon}_{\varepsilon/2} \frac{\ln \left(\pi x\right)}{x}dx \right)=\\
&\lim_{\varepsilon\to 0}\left(\ln 2\ln\varepsilon-\ln\pi\ln 2-\frac{1}{2}\left[(\ln\varepsilon)^2-(\ln\varepsilon/2)^2\right]\right)=\frac{1}{2} (\ln 2)^2-\ln \pi \ln2
\end{align}
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Finding the distribution law Throwing a cube twice,
let $X$ be the sum of the two throws
Find the distribution law
My attempt
$$
\begin{array}{c|lcr}
&2&3&4&5&6&7&8&9&10&11&12 \\
\hline
\text{P}_\text{X}(x) & 1/12 & 1/12 & 1/12&1/12&1/12&1/12&1/12&1/12&1/12&1/12&1/12 \\
\end{array}
$$
But shouldn't the sum be $1$? Currently the sum is $11/12$
| \begin{array}{c|lcr}
&2&3&4&5&6&7&8&9&10&11&12 \\
\hline
\text{P}_\text{X}(x) & 1/36 & 2/36 & 3/36&4/36&5/36&6/36&5/36&4/36&3/36&2/36&1/36 \\
\end{array}
Now the sum is $1$.
| {
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"source": "stackexchange",
"question_score": "3",
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Prove the identity $\tanh\left(\frac{x}{2}\right)=\frac{\cosh(x)-1}{\sinh(x)}$
Prove that $$\tanh\left(\frac{x}{2}\right)=\frac{\cosh(x)-1}{\sinh(x)}$$
I have started with:
$\tanh\left(\frac{x}{2}\right)=\sqrt{\tanh^2\left(\frac{x}{2}\right)}=\sqrt{1-\cosh^{-2}\left(\frac{x}{2}\right)}=\sqrt{\cosh^2{\frac{x}{2}}-\sinh^2{\frac{x}{2}}-\cosh^{-2}\left(\frac{x}{2}\right)}$
I am stuck here.
| $\frac{cosh x-1}{sinh x}=\frac{1+2sinh^2\frac{x}{2}-1}{2sinh\frac{x}{2}cosh\frac{x}{2}}$, (by using double angle whereby,$cosh x=1+2sinh^2\frac{x}{2}$)
=$\frac{sinh\frac{x}{2}}{cosh\frac{x}{2}}$
=$tanh\frac{x}{2}$. Hence shown.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Factoring $x^3-3x-1$ in terms of $\alpha$ unknown
I never got a satisfactory answer here: Factoring $x^3-3x-1\in \Bbb Q[x]$ in terms of a unknown root
But all context is contained below
I want to factor $x^3-3x-1\in \Bbb Q[x]$ in terms of an unknown root $\alpha$, i.e. over $\Bbb Q(\alpha)$.
So since $\alpha$ is a root of $x^3-3x-1$ we can divide it out, obtaining:
$(x-\alpha)(x^2+\alpha x+(\alpha^2 -3))$ and obtaining $\alpha^3-3\alpha-1=0$.
Now my next thought would be to apply the quadratic formula to the quadratic factor, but this doesn't yield anything useful, atleast not seemingly, even with me manipulating what I believe is the key, that $\alpha^3-3\alpha-1=0$
We can do other manipulations:
$$x^2+\alpha x+(\alpha^2-3))=x^2+\alpha x + \frac1\alpha=0\implies \alpha x^2 +\alpha^2 x +1 = 0$$
$$\implies \frac{-\alpha^2\pm \sqrt{\alpha^4-4\alpha}}{2\alpha}=-\frac{\alpha}{2}\pm \frac{\sqrt{3\alpha^2-3\alpha}}{2\alpha}$$
Now I am lost again. I am meant to get the other two roots as:
$$\alpha^2-\alpha-2,\quad 2-\alpha^2$$
| OK, here's one way to extend your approach to reach the answer needed.
Suppose $a$ is a root. Then $a^3 = 3a+1$. This implies we can write higher powers of $a$ as quadratics of $a$:
$$a^4 = 3a^2+a, \quad a^5 = a^2+9a+3, \quad a^6 = 9a^2+6a+1, \quad \dots$$
Further $x^3-3x-1 = (x-a)(x^2+ax+a^2-3) \implies$ the other roots are $r_{\pm} = \dfrac{-a\pm \sqrt{12-3a^2}}2$, as you noted.
One way from here is to write out a series expansion for the square root term and reduce all the higher order terms to quadratics to simplify. However often there is an easier way, with a bit of manipulation:
$$12-3a^2 = 16-4(1+3a)+4(3a^2+a)-15a^2+8a \\ =16+8a-15a^2-4(a^3)+4(a^4) \\= (2a^2-a-4)^2$$
So $r_{\pm} = \dfrac{-a\pm (2a^2-a-4)}2 \implies r_+ = a^2-a-2, \quad r_-= 2-a^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $|z|<1$ then $\prod_{k=0}^\infty (1+z^{2^k})$ = $(1-z)^{-1}$ Prove that if $|z|<1$ then $\prod_{k=0}^\infty (1+z^{2^k})$ converges and is equal to $(1-z)^{-1}$.
My attempt:
Note that
\begin{equation}
\begin{aligned}
(1-z)\prod_{k=0}^N (1+z^{2^k}) &= (1-z)(1+z)(1+z^2)(1+z^{2^2})\cdots(1+z^{2^N})\\
&= (1-z^2)(1+z^2)(1+z^{2^2})(1+z^{2^3})\cdots(1+z^{2^N})\\
&= (1-z^{2^2})(1+z^{2^2})(1+z^{2^3})(1+z^{2^4})\cdots(1+z^{2^N})\\
&= (1-z^{2^{N+1}})
\end{aligned}
\end{equation}
For |z|<1
\begin{equation}
\lim_{N\to\infty} (1-z)\prod_{k=0}^N (1+z^{2^k})= \lim_{N\to\infty}(1-z^{2^{N+1}}) =1.
\end{equation}
Is this enough to solve the problem?
| You can see by induction on $N$ that
$$
\prod_{k=0}^{N}\left(1+z^{2^{k}}\right)=\sum_{k=0}^{2^{N+1}-1}z^{k}
$$
and passing to the limit, you can deduce that
$$
\prod_{n=0}^{+\infty}\left(1+z^{2^{n}}\right)=
\sum_{n=0}^{+\infty}z^{n}=\frac{1}{1-z},\;\;\;\forall |z|<1
$$
(this last one is the geometric series, which you should know!)
| {
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Solve $ \frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}} = 27\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$. Is my solution correct? Find the roots of the following equation, if any:
$$
\frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}
=
27\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}.
$$
My approach:
The following constraints should hold jointly for x:
*
*$1-x^2\geq0\iff x\in[-1,1]$
*$1+x\geq0 \iff x\geq-1$
*$1-x\geq0 \iff x\leq1$
*$\sqrt{1+x}\ne\sqrt{1-x}\iff x\ne0$
Consequently, $x\in[-1,0)\cup(0,1]$ and for such x's I solve and I reach at the following equation
$$
\left(\frac{x}{1+\sqrt{1-x^2}}\right)^3=27\implies
\frac{x}{1+\sqrt{1-x^2}}=3,
$$
which gives
$$
x = 3 + 3\sqrt{1-x^2}\implies
3\sqrt{1-x^2} = x-3\implies
9-9x^2=x^2-6x+9\implies
10x^2-6x=0\implies
x(10x-6)=0,
$$
and thus $x=0$ or $x=\frac{3}{5}$. The first one is rejected, but $x=\frac{3}{5}\in[-1,0)\cup(0,1]$.
However, if we plug $x=\frac{3}{5}$ into the original equation, we find out that this is impossible. So we should reject $x=\frac{3}{5}$. The original equation is impossible.
Is my approach correct? Thank you very much in advance.
| Let $2y=\arccos x\implies0\le2y\le\pi$
$\implies x=\cos2y$ and $\sqrt{1-x^2}=|\sin2y|=+\sin2y$ as $0\le2y\le\pi$
So, we have $$\dfrac{1-\sin2y}{1+\sin2y}=27\cdot\dfrac{\cos y+\sin y}{\cos y-\sin y}$$
$$\left(\dfrac{1-\tan y}{1+\tan y}\right)^2=27\cdot\dfrac{1+\tan y}{1-\tan y}$$
$$\iff\left(\dfrac{1-\tan y}{1+\tan y}\right)^3=27$$
As $\tan y$ is real, $$\dfrac{1-\tan y}{1+\tan y}=3\iff\tan y=-\dfrac12$$
But $\tan y\ge0$ as $0\le y\le\dfrac\pi2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Integrate $\oint\frac{z}{\cos z-1}dz$ with residue theorem $$\oint\limits_{|z-3|=4}^{}\frac{z}{\cos z-1}dz$$
My attempt:
$$\cos z=1$$
$$z=2\pi k$$
The set includes only $z=0$ and $z=2\pi$. What next?
| $$\DeclareMathOperator*{\res}{Res}
\oint_C\frac{z}{\cos z-1}\,dz=2\pi i\res_{z=0}\frac{z}{\cos z-1}+2\pi i\res_{z=2\pi}\frac{z}{\cos z-1}$$
Use the fact that around $z=0$ we have $$\cos(z)\simeq 1-\tfrac12z^2$$
Then $$\frac{z}{\cos z-1}\simeq \frac{z}{1-\tfrac12z^2-1}=-\frac{2}{z}$$
$$\res_{z=0}\frac{z}{\cos z-1}=\res_{z=0}\frac{-2}{z}=-2$$
And around $z=2\pi$:
$$\cos(z)\simeq 1-\tfrac12(z-2\pi)^2$$
$$\res_{z=2\pi}\frac{z}{\cos z-1}=\res_{z=2\pi}\frac{z}{1-\tfrac12(z-2\pi)^2-1}=-\res_{z=2\pi}\frac{2z-4\pi+4\pi}{(z-2\pi)^2}$$
$$=-2-\res_{z=2\pi}\frac{4\pi}{(z-2\pi)^2}=-2$$
Hence $$\oint_C\frac{z}{\cos z-1}\,dz=-8\pi i$$
Why does Taylor approximation work?
$$\cos(z)=1-\tfrac12z^2+\tfrac{1}{24}z^4+...$$
$$\frac{z}{\cos(z)-1}=\frac{z}{-\tfrac12z^2+\tfrac{1}{24}z^4+...}$$
$$=\frac{-2}{z-\tfrac{1}{12}z^3+...}=\frac{-2}{z(1-\tfrac{1}{12}z^2+...)}=g(z)$$
We can break this into
$$g(z)=\frac{A}{z}+\frac{f(z)}{1-\tfrac{1}{12}z^2+...}$$
$$-2\equiv A(1-\tfrac{1}{12}z^2+...)+z\cdot f(z)$$
Especially for $z=0$:
$$-2=A$$
So $$\res_{z=0} g(z)=\res_{z=0} \frac{-2}{z}+\res_{z=0}\frac{f(z)}{1-\tfrac{1}{12}z^2+...}=-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is series convergent? $$ \sum_{n=1}^{\infty } (-1)^{\frac{n(n+1)}{2}} \cdot \frac{1}{n}$$
Is it convergent? Of course it's not absolutely convergent, but I'm not sure about conditional convergence. Leibnitz seems not to work here.
| \begin{align}
\dfrac {n(n+1)}2&=\\
&n=4m+1&\rightarrow (4m+1)(2m+1)= odd\\
&n=4m+2&\rightarrow (2m+1)(4m+3)= odd\\
&n=4m+3&\rightarrow (4m+3)(2m+2)= even\\
&n=4m+0&\rightarrow 2m(4m+1)= even\\
\end{align}
\begin{align}
\therefore -\sum_{n=1}^{\infty } \frac{(-1)^{\frac{n(n+1)}{2}}}{n}&=\sum_{m=0}^{\infty} \left(\frac1{4m+1}-\frac1{4m+3}+\frac1{4m+2}-\frac1{4m+4}\right)\\
&=\sum_{m=0}^{\infty} \left(\frac{2}{(4m+1)(4m+3)}+\frac{2}{(4m+2)(4m+4)}\right)\\
&<\frac23+\frac14+\sum_{m=1}^{\infty} \left(\frac{2}{(4m)(4m)}+\frac{2}{(4m)(4m)}\right)
\\
&=\frac{11}{12}+\frac14\sum_{m=1}^{\infty}\frac1{m^2}=\frac{11}{12}+\frac14\cdot\frac{\pi^2}6\\
\end{align}
Dr. MV also mentioned that, this series can be expressed in a closed form as $\frac{\pi}4+\frac12\cdot\log 2$.
Then it came to me that we could do this way too (although dividing it into two series may be arguable):
$$\left(\frac11-\frac13+\frac15-\frac17+\frac19\cdots\right)+\left(\frac12-\frac14+\frac16-\frac18+\frac1{10}\cdots\right)$$ $$=\frac{\pi}4+\frac12\left(\frac11-\frac12+\frac13-\frac14+\frac15\cdots\right)=\frac{\pi}4+\frac12\cdot\log 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1570067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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A simple expansion Where I am doing wrong? if any one can tell!!!
I have done this equation as
$$y ={x\over 1+(x-x^4+x^7)^3} = x [1+(x-x^4+x^7)^3]^{-1}$$
$$y= x[ 1-(x-x^4+x^7)^3+...]$$
$$y= x[ 1-(x^3+x^{12}+x^{21}-3x^6+3x^{9}+3x^9+3x^{15}+3x^{15}-3x^{18})+...] $$
$$y= x[ 1-x^3-x^{12}-x^{21}+3x^6-3x^9-3x^9-3x^{15}-3x^{15}+3x^{18}+...]$$
$$y= x-x^4+3x^7+...$$
But in book it is written as,
$$y= x-x^4+4x^7+...$$
| This may be a typo in the book. If you expand the reciprocal one term further there is a term $x^7.$ Perhaps the author mistakenly wrote your $3x^6$ as $3x^7$ and then grouped it with the extra $x^7.$
Do you want to share the book's reference?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $y' + y^2 - 2y\sin x + \sin^2x = \cos x$ How to solve the following equation?
$$y' + y^2 - 2y\sin x + \sin^2x = \cos x$$
It is necessary to determine the type and total solution.
Help me please.
| $$y' + y^2 - 2y\sin x + \sin^2x = \cos x$$
$$(y' - \cos x) + (y-\sin x)^2 = 0$$
Say $z=y- \sin x$
Then $\frac{dz}{dx}= y'-\cos x$
So we have $$\frac{dz}{dx}+z^2=0$$
$$\int z^{-2} dz = -\int dx$$
$$\frac{1}{z}=x+c$$
$$\frac{1}{y- \sin x}=x+c$$
$$y = \sin x + \frac{1}{x+c}$$
This is the required solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Proof using formal definition: Infinite limit I was wondering how get the proof of this limit:
$$\lim\limits_{x\to -\infty}\dfrac{{x^2} - x + 1}{x + 4} = -\infty$$
The problem is that I don't know what to do for find the appropriated values to make valid the implication of the formal definition (epsilon-delta).
I would appreciate if somebody can help me.
| Note that
$$\frac{x^2-x+1}{x+4}=\frac{(x+4)(x-5)+21}{x+4}.$$
You need to show that given $M<0$, there exists $N<0$ such that $\frac{x^2-x+1}{x+4}<M$ for all $x<N$. Let $N=\min\{M,-4\}.$ Then if $x<N$,
\begin{align*}
\frac{x^2-x+1}{x+4}&=\frac{(x+4)(x-5)+21}{x+4}\\
&=(x-5) + \frac{21}{x+4}\\
&<(N-5) + \frac{21}{x+4}\\
&<M+\frac{21}{x+4}\\
&<M.
\end{align*}
We can drop the $\frac{21}{x+4}$ because $x<N \leq -4$, so $x+4$ is negative, so $\frac{21}{x+4}<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Complex Numbers (Locus) Find the Cartesian equation of $z$ described by $$\arg\left(\frac{z-2}{z+5}\right)=\frac{\pi}{4}$$
So what I have done is let $z = x+iy$
$$\frac{z-2}{z+5} $$
$$ \frac{x+iy-2}{x+iy+5} $$
$$ \frac{x-2+iy}{x+5+iy} \frac{x+5-iy}{x+5-iy} $$(rationalizing here)
$$ \frac{(x-2)(x+5)+y^2+iy(x+5)-iy(x-2)}{(x+5)^2+y^2}$$
As $\arg\left(\frac{z-2}{z+5}\right)=\frac{\pi}{4}$
$$\arctan(\frac{Im(z)}{Re(z)}) = \frac{\pi}{4}$$
$$\frac{Im(z)}{Re(z)} = \tan(\frac{\pi}{4}) = 1$$
Therefore $$\frac{Im(z)}{Re(z)} = 1$$
Now comparing $im(z)$ and $re(z)$
$$\frac{y(x+5) - y(x-2)}{(x-2)(x+5)+y^2} = 1$$
$$7y = x^2 + 5x - 2x - 10 + y^2$$
$$x^2 + 3x - 10 + y^2 - 7y = 0$$
Now completing the square
$$(x+1.5)^2 + (y - 3.5)^2 = 14.5$$
Therefore locus is a circle with a centre of $(-1.5$ and $3.5i)$ with radius of $\sqrt{14.5}$
Is this correct?
| Yes ! It is very easy if you can interpret geometrically . See the red lines in the picture .
| {
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Confusion in evaluating the limit $\lim_{x\to-\infty}\sqrt{x^2+ax}-\sqrt{x^2+bx}$ I was solving a question related to functions and i come across a limit which i cannot understand.The question is
If $a$ and $b$ are positive real numbers such that $a-b=2,$ then find the smallest value of the constant $L$ for which $\sqrt{x^2+ax}-\sqrt{x^2+bx}<L$ for all $x>0$
First i found the domain of definition of the function in question $\sqrt{x^2+ax}-\sqrt{x^2+bx}$.The domain is $x\geq0 \cup x\leq -a$.
Then i found the horizontal asymptotes as $x\to \infty$.
$\lim_{x\to\infty}\sqrt{x^2+ax}-\sqrt{x^2+bx}=\lim_{x\to\infty}\frac{(a-b)x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}=\lim_{x\to\infty}\frac{2x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}$
$=\lim_{x\to\infty}\frac{2}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}=\frac{2}{2}=1$
Then i found the horizontal asymptotes as $x\to -\infty$.
$\lim_{x\to-\infty}\sqrt{x^2+ax}-\sqrt{x^2+bx}=\lim_{x\to-\infty}\frac{(a-b)x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}=\lim_{x\to-\infty}\frac{2x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}$
$=\lim_{x\to-\infty}\frac{2}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}=\frac{2}{2}=1$
But when i drew the graph using graphing calculator,the horizontal asymptote as $x\to-\infty$ was $-1$I do not understand what mistake i made in calculating the limit as $x\to -\infty$
Then i guessed i should have made the substitution $x=-t$ and
$\lim_{x\to-\infty}\sqrt{x^2+ax}-\sqrt{x^2+bx}=\lim_{t\to\infty}\sqrt{t^2-at}-\sqrt{t^2-bt}=\lim_{t\to \infty}\frac{(b-a)t}{\sqrt{t^2-at}+\sqrt{t^2-bt}}=\lim_{t\to\infty}\frac{-2t}{\sqrt{t^2-at}+\sqrt{t^2-bt}}$
$=\lim_{t\to\infty}\frac{-2}{\sqrt{1-\frac{a}{t}}+\sqrt{1-\frac{b}{t}}}=\frac{-2}{2}=-1$
I want to ask why the answer came wrong in the first method and correct in the second method.
Is it always necessary to put $x=-t$ and change the limit to plus
infinity while calculating the limit as $x\to-\infty$
Please help me.Thanks.
| The short answer to your question is that $x=\sqrt{x^2}$ only for $x > 0$. For $x<0$, we have $x=-\sqrt{x^2}$.
In the first method, the fact that
$$\dfrac{2x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}} = \dfrac2{\sqrt{1+a/x}+\sqrt{1+b/x}}$$
is true only for $x > 0$. For $x<0$, we have $x = -\sqrt{x^2}$, and hence we have
$$\dfrac{2x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}} = -\dfrac2{\sqrt{1+a/x}+\sqrt{1+b/x}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Convergence of $\sum_{n=1}^\infty \frac{(-1)^n3^n}{2n+1}z^{2n+1}$ for $|z|= \frac{\sqrt{3}}{3}$ I have shown that the series $$\sum_{n=1}^\infty \frac{(-1)^n3^n}{2n+1}z^{2n+1}$$
converges for $|z|<\frac{\sqrt{3}}{3}$. Can anyone help me to study the convergence for $|z|= \frac{\sqrt{3}}{3}$?
| HINT:
Notice:
$$\sum_{n=1}^\infty \frac{(-1)^n3^n}{2n+1}z^{2n+1}=\frac{\sqrt{3}\arctan(z\sqrt{3})-3z}{3}\space\space,\text{when}\space\left|z\right|<\frac{1}{\sqrt{3}}$$
So if $z=\frac{1}{\sqrt{3}}$:
$$\sum_{n=1}^\infty \frac{(-1)^n3^n}{2n+1}\left(\frac{1}{\sqrt{3}}\right)^{2n+1}=\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{3}(1+2n)}=\lim_{m\to\infty}\sum_{n=1}^m\frac{(-1)^n}{\sqrt{3}(1+2n)}=$$
$$\lim_{m\to\infty}\frac{2(-1)^m\Phi\left(-1,1,m+\frac{3}{2}\right)+\pi-4}{4\sqrt{3}}=$$
$$\lim_{m\to\infty}\left(\frac{(-1)^m\Phi\left(-1,1,m+\frac{3}{2}\right)}{2\sqrt{3}}+\frac{\pi}{4\sqrt{3}}-\frac{1}{\sqrt{3}}\right)=$$
$$\frac{\pi}{4\sqrt{3}}-\frac{1}{\sqrt{3}}+\lim_{m\to\infty}\frac{(-1)^m\Phi\left(-1,1,m+\frac{3}{2}\right)}{2\sqrt{3}}=$$
$$\frac{\pi}{4\sqrt{3}}-\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}\lim_{m\to\infty}(-1)^m\Phi\left(-1,1,m+\frac{3}{2}\right)=$$
$$\frac{\pi}{4\sqrt{3}}-\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}\cdot0=$$
$$\frac{\pi}{4\sqrt{3}}-\frac{1}{\sqrt{3}}+0=$$
$$\frac{\pi}{4\sqrt{3}}-\frac{1}{\sqrt{3}}=$$
$$\frac{\pi-4}{4\sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1577225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}$ using the function $f(x)=\frac{1+x}{(1-x)^3}$ Find $\sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}$ using the function $f(x)=\frac{1+x}{(1-x)^3}$
Power series representation of $f(x)$ is $\sum\limits_{n=1}^{\infty}n^2x^{n-1}$.
Question: Why is $\sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}=\frac{1}{2}f\left(\frac{1}{2}\right)$? Doesn't it has to be $f\left(\frac{1}{2}\right)$?
| You need the extra factor of $2$:
$$f(1/2) = \sum_{n=1}^{\infty}\frac{n^2}{2^{n-1}},$$
but
$$\frac{1}{2}f(1/2) = \sum_{n=1}^{\infty}\frac{n^2}{2 \cdot 2^{n-1}} = \sum_{n=1}^{\infty}\frac{n^2}{2^{n}}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Computing $7$th root of $2$ modulo $33$
*
*$\varphi(33) = 20$
*$ed = 1 \pmod {20}$, and $d$ is $7$, so $e$ is $3$. $(3 \times 7 \mod 20 \equiv 1)$
So $x^e \pmod {33}$ is the seventh root. But how do you compute $x$?
Later: My error was thinking x needed to be computed; it was in fact given as 2.
if ed = 1 mod Phi(N), then raising to the dth power is the inverse of raising to the eth power.
$(x^e)^d$ = x mod N, and $(x^d)^e$ = x mod N.
$x^d$ is the e-th root of x modulo N.
We know d is 7 and x is 2. The value of e may be computed:
$ed$ = 1 mod $\varphi(N)$, so $e * 7$ = 1 mod 20 = 1
$e * 7$ mod 20 = 1 so $e$ is 3.
And indeed, $2^3$ mod 20 is 1. So the answer is $2^3$ or 8.
| More generally, if $\gcd(k,\phi(n))=1$, then $$x^k\equiv a\pmod{n}\iff x\equiv a^{k^{-1}\bmod \phi(n)}\pmod{n}$$
In your case, let $(k,a,n)=(7,2,33)$. Then $7^{-1}\equiv 3\pmod{\phi(33)}$, so raise both sides of $x^7\equiv 2\pmod{33}$ to the power of $3$ (as the other answers have noticed) to get $x\equiv 2^3\equiv 8\pmod{33}$.
Another method to solve $x^k\equiv a\pmod{n}$ without necessarily knowing $\gcd(k,\phi(n))=1$ can be using primitive roots.
You're solving $x^7\equiv 2\pmod {33}$, i.e. the system $\begin{cases}x^7\equiv 2\pmod{3} \\x^7\equiv 2\pmod{11}\end{cases}$
Clearly $x^7\equiv 2\pmod{3}\iff x\equiv 2\pmod{3}$ (there are only the cases $x\equiv \{0,1,2\}\pmod{3}$ to check).
To solve $x^7\equiv 2\pmod{11}$, instead of checking $x\equiv \{0,1,2,\ldots,10\}\pmod{11}$ you can note that $2$ is a primitive root mod $11$, because $\gcd(2,11)=1$ and $\phi(11)=2\cdot 5$ and $2^5\not\equiv 1\pmod{11}$ and $2^2\not\equiv 1\pmod{11}$.
Let $x\equiv 2^k\pmod{11}$. Then you're solving $2^{7k-1}\equiv 1\pmod{11}$, i.e. $7k\equiv 1\equiv 21\pmod{10}$, i.e. $k\equiv 3\pmod{10}$, so $x\equiv 2^k\equiv 8\pmod{11}$.
$$\begin{cases}x^7\equiv 2\pmod{3} \\x^7\equiv 2\pmod{11}\end{cases}\iff \begin{cases}x\equiv 2\pmod{3}\\x\equiv 8\pmod{11}\end{cases}\iff x\equiv 8\pmod{33},$$
where in the last step you can apply the Chinese Remainder Theorem.
| {
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"url": "https://math.stackexchange.com/questions/1578854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the equation: $ \sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos3x$ How do I solve this equation:
$$ \sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos3x$$
| First $$\sin(x)+\sin(y)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) \\ \cos(x)+\cos(y)=2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$ Therefore $$\sin(x)+\sin(3x)=2\sin(2x)\cos(x) \ , \ \cos(x)+\cos(3x)=2\cos(2x)\cos(x)$$Hence, we can write $$2\sin(2x)\cos(x)-3\sin(2x)=2\cos(2x)\cos(x)-3\cos(2x)$$Factoring out we get $$\sin(2x)\left[2\cos(x)-3\right]=\cos(2x)\left[2\cos(x)-3\right]$$Eventually $\sin(2x)=\cos(2x) \text{ Or } 2\cos(x)-3=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving a functional equation $f(x)+f\left(\frac{1}{1-x}\right)=x$ I was given the following homework:
list all functions $f:\mathbb{R}\setminus\{0,1\}\rightarrow\mathbb{R}$ such that $f(x)+f\left(\frac{1}{1-x}\right)=x$. And obviously have I no idea what should I do here. A step-by-step explanation welcome.
| The system matrix is
$\left( {\begin{array}{*{20}{c}}
1&1&0&& x\\
0&1&1&& {\frac{1}{{1 - x}}}\\
1&0&1&& {\frac{{x - 1}}{x}}
\end{array}} \right)$
The solution of the system is
$\left\{ {\frac{{1 - x + {x^3}}}{{2\left( { - 1 + x} \right)x}},\frac{{ - 1 + x - 2{x^2} + {x^3}}}{{2\left( { - 1 + x} \right)x}},\frac{1}{2}\left( {1 + \frac{1}{{1 - x}} - \frac{1}{x} - x} \right)} \right\}$
so $f(x)={\frac{{1 - x + {x^3}}}{{2\left( { - 1 + x} \right)x}}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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$\lim_{x\to 0}\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}$ without series expansion or L Hospital rule
If $$f(x)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^5},$$ $x\neq 0$, is continuous at $x=0$, then find $A,B$ and $f(0)$. Do not use series expansion or L Hospital's rule.
As $f(x)$ is continuous at $x=0$,therefore its limit at $x=0$ should equal to its value.
Note that this question is to be solved without series expansion or L Hospital's rule,
I tried to find the limit $\lim_{x\to 0}\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}$
$\lim_{x\to 0}\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}=\lim_{x\to 0}\frac{3\sin x-4\sin^3x+2A\sin x\cos x+B\sin x}{x^5}=\lim_{x\to 0}\frac{3-4\sin^2x+2A\cos x+B}{x^4}\times\frac{\sin x}{x}$
$=\lim_{x\to 0}\frac{3-4\sin^2x+2A\cos x+B}{x^4}$
As the denominator is zero,so numerator has to be zero,in order the limit to be finite.
So, $3+2A+B=0. (1)$
I tried but I could not get the second equation between $A$ and $B$. I am stuck here. How do I continue?
| Using trigonometric identities we have
\begin{align}
3-4\sin^2 x+2A\cos x+B&=4(1-\sin^2 x)+2A\cos\left(2\cdot\frac{x}{2}\right)+B-1\\
&=4\cos^2\left(2\cdot\frac{x}{2}\right)+2A\cos\left(2\cdot\frac{x}{2}\right)+B-1\\
&=4\left(1-2\sin^2\frac{x}{2}\right)^2+2A\left(1-2\sin^2 \frac{x}{2}\right)+B-1\\
&=16\sin^4\frac{x}{2}-16\sin^2\frac{x}{2}-4A\sin^2\frac{x}{2}+2A+B+3\\
&=16\sin^4\frac{x}{2}-4(A+4)\sin^2\frac{x}{2}+2A+B+3\\
\end{align}
In order to make the limit finite we must have
$$A+4=0\quad\text{and}\quad 2A+B+3=0\qquad\iff\qquad \color{blue}{A=-4}\quad\text{and}\quad \color{blue}{B=5}$$
By taking those values we get
\begin{align}
\lim_{x\to 0}\frac{\sin 3x\color{blue}{-4}\sin 2x+\color{blue}{5}\sin x}{x^5}&=\left(\lim_{x\to 0}\frac{16\sin^4\frac{x}{2}}{x^4}\right)\left(\lim_{x\to 0}\frac{\sin x}{x}\right)\\
&=\left(\lim_{x\to 0}\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^4\left(1\right)\\
&=\color{blue}{1}
\end{align}
Since $f$ is continuous at $0$ it follows $f(0)=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
Finding the Jordan Basis and the Canonical form corresponding to the Jordan basis \begin{pmatrix} 4 & 0 &0 \\ 2 &1 &3 \\ 5& 0 &4 \end{pmatrix}
I know that the Characteristic polynomial is : $$(t-4)^2(t-1)$$
I started with eigenvalues $λ=1$
and got in the null space:
\begin{pmatrix} 0\\ 1 \\ 0 \end{pmatrix}
$λ=4$
and got :
\begin{pmatrix} 0\\ -1 \\ 1 \end{pmatrix},
in $A- 4I$
and got
\begin{pmatrix} -1\\ 1 \\ 0 \end{pmatrix},
\begin{pmatrix} 1\\ 0 \\ 1 \end{pmatrix},
in $(A- 4I)^2$
Which vector(s) do I choose correctly for the Jordan Basis?
I chose $[0,1,0],[1,0,1]$ and got $b_3$ as $A-4I[b_2] = [0,5,5]$
What procedures should I use for choosing the correct Jordan basis vectors?
Secondly am I right?
Lastly, How do I get the Canonical form "corresponding to the basis"?
is it \begin{pmatrix} 1 & 0 &0 \\ 0 &4 &1 \\ 0& 0 &4 \end{pmatrix}
or \begin{pmatrix} 4 & 1 &0 \\ 0 &4 &0 \\ 0& 0 &1 \end{pmatrix}
How do I set it up correctly?
| You have the right idea, but you made a slight computational error: the $4$-eigenspace of $A$ is spanned by $(0,1,1)$, not $(0,-1,1)$. You correctly picked a vector $(1,0,1) \in \ker (A-4I)^2$ that doesn't belong to $\ker(A-4I)$, which let you construct the Jordan chain $((1,0,1),(A-4I)(1,0,1)) = ((1,0,1),(0,5,5))$. (This now works because $(0,5,5)$ is an eigenvector for the eigenvalue $4$.)
Then you just place the Jordan chains as columns of the change of basis matrix (reading a chain backwards if you want $1$s on the superdiagonal):
$$P = \begin{pmatrix} 0 & 0&1\\ 1&5&0 \\ 0&5&1 \end{pmatrix}.$$
And then we get
$$ P^{-1}AP = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & 1 \\ 0 & 0 & 4 \end{pmatrix}.$$
If instead you chose
$$P = \begin{pmatrix} 0&1&0\\ 5&0&1 \\ 5&1&0 \end{pmatrix},$$
you'd get
$$ P^{-1}AP = \begin{pmatrix} 4 & 1 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1581737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\sum_{n=1}^{\infty} \frac{4(-1)^n}{1-4n^2}$ I recently found a series representation for 1 from the calculation of a Fourier series:
$$1 = \frac{2}{\pi} + \sum_{n=1}^{\infty} \frac{4(-1)^n}{\pi(1-4n^2)}$$
From this, I can easily find that
$$\sum_{n=1}^{\infty} \frac{4(-1)^n}{1-4n^2} = \pi - 2,$$
but what other methods are there to evaluate the sum? I don't even know where to start.
| $$\frac{4}{1-4 n^2} = \frac{2}{1-2 n} + \frac{2}{1+2 n}$$
Thus,
$$\begin{align}\sum_{n=1}^{\infty} \frac{4 (-1)^n}{1-4 n^2} &= 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2 n-1}+2 \sum_{n=1}^{\infty} \frac{(-1)^n}{2 n+1} \\ &= 2 \frac{\pi}{4} + 2 \left (\frac{\pi}{4}-1 \right ) \\ &= \pi-2 \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the maximum value of $ab+ac+ad+bc+bd+3cd$
If $a,b,c,d>0$ satisfy the condition ${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }=1$, find the maximum value of $ab+ac+ad+bc+bd+3cd$.
I'm not progress in this inequality problem. Please help.
Thank you.
| Let $ab+ac+ad+bc+bd+3cd=k$.
Hence, $k>0$ and $ab+ac+ad+bc+bd+3cd=k(a^2+b^2+c^2+d^2)$ or
$ka^2-(b+c+d)a+k(b^2+c^2+d^2)-bc-bd-3cd=0$.
Hence, $(b+c+d)^2-4k(k(b^2+c^2+d^2)-bc-bd-3cd)\geq0$ or
$(4k^2-1)b^2-2(2k+1)(c+d)b+(4k^2-1)(c^2+d^2)-2(6k+1)cd\leq0$.
If $0<k\leq\frac{1}{2}$ so the last inequality is obviously true.
Let $k>\frac{1}{2}$. Hence, $(2k+1)^2(c+d)^2-(4k^2-1)\left((4k^2-1)(c^2+d^2)-2(6k+1)cd\right)\geq0$ or
$(2k+1)(c+d)^2-(2k-1)\left((4k^2-1)(c^2+d^2)-2(6k+1)cd\right)\geq0$ or
$(2k^2-k-1)c^2-(6k-1)cd+(2k^2-k-1)d^2\leq0$.
If $\frac{1}{2}<k\leq1$ so the last inequality is obviously true.
Let $k>1$. Hence, $(6k-1)^2-4(2k^2-k-1)^2\geq0$, which gives $k\leq1+\frac{\sqrt5}{2}$.
Easy to see that for $k=1+\frac{\sqrt5}{2}$ the equality indeed occurs.
Id est, the answer is $1+\frac{\sqrt5}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Prove that $\cosh^{-1}(1+x)=\sqrt{2x}(1-\frac{1}{12}x+\frac{3}{160}x^2-\frac{5}{896}x^3+....)$
How can we prove the series expansion of
$$\cosh^{-1}(1+x)=\sqrt{2x}\left(1-\frac{1}{12}x+\frac{3}{160}x^2-\frac{5}{896}x^3+...\right)$$
I know the formula for $\cosh^{-1}(x)=\ln(x+\sqrt{x^2-1})$ so, $$\cosh^{-1}(1+x)=\ln(1+x+\sqrt{x^2+2x}).$$ I tried to apply Maclaurin series but i could not find the $f(0),f'(0),f''(0),f'''(0)$
Is there any other method available to prove this series expansion like Laurent series etc.
Please help me.Thanks.
| The trick is to recognize the $1/\sqrt x$ singularity in the derivative of the function of interest and transform the series into a series in $\sqrt x$.
To proceed, we substitute $z=\sqrt{x}$. Then, we have
$$\begin{align}
\cosh^{-1}(1+x)&=\cosh^{-1}(1+z^2)\\\\
&=\log\left((z^2+1)+z\sqrt{z^2+2}\right)\tag 1
\end{align}$$
Now, expanding $f(z)=\log\left((z^2+1)+z\sqrt{z^2+2}\right)$ in a series around $z=0$ reveals
$$\begin{align}
f'(z)&=\frac{1}{(z^2+1)+z\sqrt{z^2+2}}\left(2z+\sqrt{z^2+2}+\frac{z^2}{\sqrt{z^2+2}}\right)\\\\
&=\frac{2}{\sqrt{z^2+2}}
\end{align}$$
Then, continuing to differentiate, we find that
$$\begin{align}
f^{(2)}(z)&=-2z(z^2+2)^{3/2}\\\\
f^{(3)}(z)&=4(z^2-1)(z^2+2)^{-5/2}\\\\
f^{(4)}(z)&=-12z(z^2-3)(z^2+2)^{-7/2}\\\\
f^{(5)}(z)&=24(2z^4-12z^3+3)(z^2+2)^{-9/2}
\end{align}$$
Evaluating these at $z=0$ reveals
$$\begin{align}
f(z)&=f(0)+f'(0)z+\frac12f^{(2)}(0)z^2+\frac16f^{(3)}(0)z^3+\frac1{24}f^{(4)}(0)z^4+\frac1{120}f^{(5)}(0)z^5+O(z^7)\\\\
&=\sqrt 2\,z-\frac{\sqrt{2}}{12}z^3+\frac{3\sqrt 2}{160}z^5+O(z^7)\\\\
&=\sqrt{2x}\left(1-\frac1{12}x+\frac{3}{160}x^2+O(x^3)\right)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Describe the nonzero integer solutions to the equation $a^3 + b^3 + c^3 + d^3 + e^3 + f^3 + g^3 =0$ Can someone describe all the integer solutions to the above equation such that $abcdefg\neq 0$ ?
| Regarding equation, (a^3 + b^3 + c^3 + d^3 + e^3 + f^3 = g^3 )
The above equation has parametrization given below:
(a,b,c,d,e,f)^3=
[(6k^2+12k-18),(k^2-2k+49),(5k^2+38k+5),(8k^2+16k-24),
(7k^2-14k-41),(10k^2+20k-30)]^3 = [(13k^2+22k+13)]^3= (g)^3
For k=2 we have:
(30,49,101,40,-41,50)^3=(109)^3
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving logarithm problem, Found the value This is my first question in this place, I don't know how to solve my problem.
I have this equation, I need to find the central value based on this equation:
$k = 0.2$
$$k = 2^{(\frac{1}{24} + \frac{1}{24\cdot2^x})}-2^{(\frac{1}{24} - \frac{1}{24\cdot2^x})}$$
I need to figure out the x value, the known value is k.
$$x=?$$
if I define y like: $y = \frac{1}{24\cdot2^{x}}= 2^{-x}/24$, and l like: $l = k/2^{\frac{1}{24}}$
$$k = 2^{\frac{1}{24}}\cdot2^y-2^{\frac{1}{24}}\cdot2^{-y}$$
$$l = 2^y-2^{-y}$$
$$\log_{2}(l)=\text{ ?}$$
$$y=\text{ ?}$$
Please help me.
| $$k=2^{\frac{1}{24}+\frac{1}{24\cdot2^x}}-2^{\frac{1}{24}-\frac{1}{24\cdot2^x}}\Longleftrightarrow$$
$$k=2^{\frac{2^{-x}+1}{24}}-2^{\frac{1-2^{-x}}{24}}\Longleftrightarrow$$
$$k=2\cdot2^{\frac{1}{24}}\sinh\left(\frac{2^{-3-x}\ln(2)}{3}\right)\Longleftrightarrow$$
$$\frac{k}{2\cdot2^{\frac{1}{24}}}=\sinh\left(\frac{2^{-3-x}\ln(2)}{3}\right)\Longleftrightarrow$$
$$\text{arcsinh}\left(\frac{k}{2\cdot2^{\frac{1}{24}}}\right)=\frac{2^{-3-x}\ln(2)}{3}\Longleftrightarrow$$
$$3\text{arcsinh}\left(\frac{k}{2\cdot2^{\frac{1}{24}}}\right)=2^{-3-x}\ln(2)\Longleftrightarrow$$
$$\frac{3\text{arcsinh}\left(\frac{k}{2\cdot2^{\frac{1}{24}}}\right)}{\ln(2)}=2^{-3-x}\Longleftrightarrow$$
$$\frac{\ln\left(\frac{3\text{arcsinh}\left(\frac{k}{2\cdot2^{\frac{1}{24}}}\right)}{\ln(2)}\right)}{\ln\left(2\right)}=-3-x\Longleftrightarrow$$
$$\frac{\ln\left(\frac{3\text{arcsinh}\left(\frac{k}{2\cdot2^{\frac{1}{24}}}\right)}{\ln(2)}\right)}{\ln\left(2\right)}+3=-x\Longleftrightarrow$$
$$x=-\left[\frac{\ln\left(\frac{3\text{arcsinh}\left(\frac{k}{2\cdot2^{\frac{1}{24}}}\right)}{\ln(2)}\right)}{\ln\left(2\right)}+3\right]\Longleftrightarrow$$
$$x=-\frac{\ln\left(\frac{3\text{arcsinh}\left(\frac{k}{2\sqrt[24]{2}}\right)}{\ln(2)}\right)}{\ln(2)}-3\Longleftrightarrow$$
$$x=-\frac{\ln\left(\frac{24\text{arcsinh}\left(\frac{k}{2\sqrt[24]{2}}\right)}{\ln(2)}\right)}{\ln(2)}$$
EDIT:
In general:
$$2^{\frac{a+1}{x}}-2^{\frac{1-a}{x}}=2^{\frac{1-a}{x}}\left(4^{\frac{a}{x}}-1\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to prove $\left(1+\frac{1}{\sin a}\right)\left(1+\frac{1}{\cos a}\right)\ge 3+2\sqrt{2}$? Prove the inequality:
$$\left(1+\dfrac{1}{\sin a}\right)\left(1+\dfrac{1}{\cos a}\right)\ge 3+2\sqrt{2}; \text{ for } a\in\left]0,\frac{\pi}{2}\right[$$
| I think it should be $(0,\frac{\pi}{2})$, so that both $\sin a$ and $\cos a$ are positive and well defined.
$$\bigg (1+\frac{1}{\sin a} \bigg)\bigg (1+\frac{1}{\cos a} \bigg) = 1+ \frac{1}{\sin a} + \frac{1}{\cos a} + \frac{1}{\sin a\cos a}$$
$$\frac{1}{\sin a} + \frac{1}{\cos a} \geq \frac{4}{\sin a + \cos a} \geq \frac{4}{\sqrt{2(\sin^2 a + \cos^2 a)}} = 2\sqrt 2$$
$$\frac{1}{\sin a\cos a} \geq \frac{2}{\sin^2 a + \cos^2 a} = 2$$
So,
$$\bigg (1+\frac{1}{\sin a} \bigg)\bigg (1+\frac{1}{\cos a} \bigg) = 1+ \frac{1}{\sin a} + \frac{1}{\cos a} + \frac{1}{\sin a\cos a} \geq 3+ 2\sqrt 2$$
Equality occurs when $a = \frac{\pi}{4}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\left (\frac{1}{a}+1 \right)\left (\frac{1}{b}+1 \right)\left (\frac{1}{c}+1 \right) \geq 64.$
Let $a,b,$ and $c$ be positive numbers with $a+b+c = 1$. Prove that $$\left (\dfrac{1}{a}+1 \right)\left (\dfrac{1}{b}+1 \right)\left (\dfrac{1}{c}+1 \right) \geq 64.$$
Attempt
Expanding the LHS we obtain $\left (\dfrac{1+a}{a} \right)\left (\dfrac{1+b}{b} \right)\left (\dfrac{1+c}{c} \right)$. We are given that $a+b+c = 1$, so substituting that in we get $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right)$. Then do I say $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right) \geq 64$ and see if I can get a true statement from this?
| Probably you mean $a,b,c\geq 0$. You want to find the miminmum of
$$\ln (1+\frac{1}{a})+\ln(1+\frac{1}{b})+\ln(1+\frac{1}{1-a-b})$$
for $a,b\geq 0$. The partial derivatives are zero iff
$$
-\frac{1}{a(a+1)}+\frac{1}{(1-a-b)(2-a-b)}=0
$$
and a similar equation in which $a$ and $b$ will be replaced. By subtracting, you get $a(a+1)=b(b+1)$ which implies (given that all is positive) $a=b$. But that's the minimum, $a=b=c=1/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How find the sum of the last two digits of $(x^{2})^{2013} + \frac{1}{(x^{2})^{2013}}$ for $x + \frac{1}{x} = 3$? Let x be a real number so that $x + \frac{1}{x} = 3$. How find the sum of the last two digits of $(x^{2})^{2013} + \frac{1}{(x^{2})^{2013}}$?
| Square your equation to get:
$$x^2+\frac{1}{x^2}=3^2 - 2$$
And again:
$$x^4+\frac{1}{x^4}=(3^2 - 2)^2-2$$
So:
$$x^{2^n}+\frac{1}{x^{2^n}}=(\ldots((3^2 - 2)^2-2)^2 -2)\ldots)^2 -2$$
Define:
$$a_{n+1} = a_n^2 - 2$$
Now let $a_n = 100 b_n + y_n$. So that:$$a_{n+1} = 100^2 +200 b_n y_n+ y_n^2 - 2$$
$$y_{n+1} = y_n^2 - 2\mod 100$$
Start with $y_1=7$. $y_2 = 47$ and $y_3 = 107 \equiv 7$.
Edit:
This works to get us a answer if we had $2048$ instead of $2013$.
Working on an alternate solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if c ϵ C and $x = \frac{(c+\sqrt{c^2+4})}{2}$ so $x-\frac{1}{x} $ ϵ C. Prove that if c ϵ C and $x = \frac{(c+\sqrt{c^2+4})}{2}$ so $x-\frac{1}{x} $ ϵ C. I have no idea how do this. Please help me.
| We have $\dfrac1x = \dfrac2{c+\sqrt{c^2+4}} = \dfrac2{c+\sqrt{c^2+4}}\cdot \dfrac{c-\sqrt{c^2+4}}{c-\sqrt{c^2+4}} = \dfrac{\sqrt{c^2+4}-c}2$
Hence, we have
$$x-\dfrac1x = \dfrac{c+\sqrt{c^2+4}}2 - \dfrac{\sqrt{c^2+4}-c}2 = c$$
| {
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Prove that the values of $x$ for which $x = \frac{x^2+1}{198}$ lie between $\frac{1}{198}$ and $199.99 \overline{49}$.
Without using a calculator, prove that the values of $x$ for which $x = \dfrac{x^2+1}{198}$ lie between $\dfrac{1}{198}$ and $199.99 \overline{49}$. Then use this result to prove that $\sqrt{2} < 1.41 \overline{421356}$. Finally, is it true that that $\sqrt{2} < 1.41421356$?
For the first part we have that the roots are $99-70\sqrt{2}$ and $99+70\sqrt{2}$ by the quadratic formula. How do I show these roots lie in this range? Also is using the quadratic formula here a good idea?
| we have $$\frac{1}{198}<x<199.99\overline{49}=\frac{39599}{198}$$
with $x=\frac{x^2+1}{198}$ we get $1<x^2+1$ which is true.on the other hand we have $$x^2+1<39599$$ this is true since $x^2<198^2<39598$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{a}{(1+b)(1+c)}+\frac{b}{(1+a)(1+c)}+\frac{c}{(1+a)(1+b)} \geq \frac{3}{4}.$
Let $a,b,c$ be positive numbers that satisfy $abc = 1$, prove that $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} \geq \dfrac{3}{4}.$$
Attempt
I tried doing $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} = \dfrac{1}{bc(1+b)(1+c)}+\dfrac{1}{ac(1+a)(1+c)}+\dfrac{c}{bc(1+a)(1+b)}$$ then using the fact that $a^2+b^2+c^2 \geq ab+bc+ca$ but that only seemed to get me $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} \geq \dfrac{1}{bc+bc^2+c+1}+\dfrac{1}{ac+a^2c+c+1}+\dfrac{1}{ab+ab^2+b+1}.$$
| Hint: Multiply both sides by $(1+a)(1+b)(1+c)$, and expand the aforementioned expression as follows: $1+abc+a+b+c+ab+ac+bc$.
You'll notice that you can substitute the value $1$ for $abc$; I think you should be able to tackle it from here by application of RM-AM-GM-HM inequality (Only AM-GM necessary though).
| {
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Find the range of $m$ for which atleast one of the following equations $ax^2+bx+cm=0,bx^2+cx+am=0,cx^2+ax+bm=0$ have real roots. Let $a,b,c$ and $m\in R^+$.Find the range of $m$ for which atleast one of the following equations $ax^2+bx+cm=0,bx^2+cx+am=0,cx^2+ax+bm=0$ have real roots.
Either one or two or all of the three equations $ax^2+bx+cm=0,bx^2+cx+am=0,cx^2+ax+bm=0$ have the real roots.
But i do not know how to find the range of $m.$Any help will be appreciated.Thanks.
| Claim: $$m \leq \dfrac{1}{4}$$
For if all of them do not have real roots then : $a^2-4bcm < 0, b^2-4acm < 0, c^2-4abm < 0 \Rightarrow a^2+b^2+c^2 < 4m(ab+bc+ca) \Rightarrow m > \dfrac{a^2+b^2+c^2}{4ab+4bc+4ca} \geq \dfrac{ab+bc+ca}{4ab+4bc+4ca} = \dfrac{1}{4}$. Thus if $m \leq \dfrac{1}{4}$, then at least one equation has real roots.
| {
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Solve $ \int{\frac{7x^2 + 1}{(x+1)(x-1)(x+3)}}\,dx $ I don’t know how to solve this integral:
$$\int{\frac{7x^2 + 1}{(x+1)(x-1)(x+3)}}\,dx$$
I know this is a rational integral but I don’t know how to write it in a different way
| As every one suggests, do the partial fraction decomposition. However, since
*
*the roots of the denominator are all simple
*the numerator has lower degree than denominator.
you can read off the decomposition directly.
$$\require{cancel}
\newcommand{\xxx}[2]{\color{red}{\cancelto{#2}{\color{gray}{#1}}}}
\frac{7x^2+1}{(x+1)(x-1)(x+3)}\\
= \frac{\xxx{7(-1)^2+1}{8}}{(x+1)\xxx{(-1-1)(-1+3)}{-4}}
+ \frac{\xxx{7(1)^2+1}{8}}{(x-1)\xxx{(1+1)(1+3)}{8}}
+ \frac{\xxx{7(-3)^2+1}{64}}{\xxx{(-3+1)(-3-1)}{8}(x+3)}\\
= \frac{-2}{x+1}+\frac{1}{x-1}+\frac{8}{x+3}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Centroid of a Triangle and Cosine Law In $\triangle ABC$, $M$ and $N$ are midpoints of $BC$ and $CA$ respectively such that $AM=14$ and $BN=8$. If $\angle C= 60^{\circ}$, find the length of $AB$.
For simplicity sake, let $x=AB$, $y=BM=MC$ and $z=AN=NC$. Then by Cosine Law, we have
$$8^2=z^2+4y^2-4yz\cos 60^{\circ},$$
$$14^2=4z^2+y^2-4yz\cos 60^{\circ},$$ and
$$x^2=4z^2+4y^2-8yz\cos 60^{\circ}.$$
Subtracting the 2nd equation from 1st gives $z^2-y^2=(14^2-8^2)/3=44,$ and subtracting the 1st and the 2nd from the 3rd gives $x^2+y^2+z^2=14^2+8^2=260.$
But then I couldn't get further. I observed that the intersection of $BN$ and $AM$ is the centroid of the triangle but I don't know if this information would help, please help, thanks!
| $$z^2-y^2=44 \tag{1}$$ Adding first two equations we get
$$5(y^2+z^2)-260=4yz \tag{2}$$ ans third equation can be written as
$$4(y^2+z^2)-x^2=4yz \tag{3} $$ Now subtracting $(3)$ from $(2)$ we get
$$y^2+z^2=260-x^2 \tag{4}$$
From $(1)$ and $(4)$ we get
$$2z^2=304-x^2 \tag{5}$$ and
$$2y^2=216-x^2 \tag{6}$$
Now substitute $y^2+z^2$ from $(4)$ in $(2)$ we get
$$1040-5x^2=4yz$$ squaring above equation both sides we get
$$(1040-5x^2)^2=4(2y^2)(2z^2)$$ $\implies$
$$(1040-5x^2)^2=4(304-x^2)(216-x^2)$$ from which you will get quadratic in $x^2$
| {
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How prove $\sum\limits_{cyc} \frac{ab}{c} \geq 3.$ Let $a,b,c>0 ,\sum\limits_{cyc} a^{3}=3$. How prove $\sum \frac{ab}{c} \geq 3.$
My try:
$ a^3+b^3+c^3=3$
$a^2b^2+a^2c^2+b^2c^2 \ge 3abc$
so
$ (a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc=3$
$(ab+ac+bc)^2-2(a+b+c)abc \ge 3abc$
| The inequality is false: take $a=\frac{1}{\sqrt[3] 6}$,$b=\frac{1}{\sqrt[3] 6}$,$c=\frac{\sqrt[3] 8}{\sqrt[3] 3}$, you obtain
$$ \sim 2.9918... \ge 3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$-\sin x -\cos x = -\sqrt{2}\sin (x+{\pi\over4})$
$-\sin x -\cos x = -\sqrt{2}\sin (x+{\pi\over4})$
How does the cosine disappear and how did sin x turn into $\sin (x+{\pi\over4})$?
| A very useful formula:
$$a\sin x+b\cos x=\sqrt{a^2+b^2}\cos(x-\arctan (a/b))$$
$$\sqrt{a^2+b^2}\cos(x)\cos(\arctan a/b)+\sin (x)\sin(\arctan a/b)\space\space (1)$$
$$=\frac{b\sqrt{a^2+b^2}\cos(x)}{\sqrt {a^2+b^2}}+ \frac{a\sqrt{a^2+b^2}\sin x}{\sqrt{a^2+b^2}}\space\space (2) \space\space (3)$$
$$=b\cos x+a\sin x$$
$\cos(A-B)=\cos A\cos B+\sin A\sin B $ $(1)$ [A and B are arbitrary]
$\cos(\arctan a/b)=\frac{b}{\sqrt{a^2+b^2}} \space (2);\sin(\arctan a/b)=\frac{a}{\sqrt{a^2+b^2}}\space (3)$
| {
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solve$ \int \frac{(1+e^{-5x})^{\frac{1}{3}}}{e^{5x}}dx$
$$\int \frac{(1+e^{-5x})^{\frac{1}{3}}}{e^{5x}}dx$$
$u=1+e^{-5x}$
$du=-5e^{-5x}dx$
so the expression is $\int \frac{u^{\frac{1}{3}}}{-\frac{1}{5}}du=-5u^{\frac{1}{3}}du$?
| $$\begin{align}\int \frac{(1+e^{-5x})^{\frac{1}{3}}}{e^{5x}}dx &= \int (1+e^{-5x})^{\frac{1}{3}}e^{-5x}dx \\ &= \int (1+e^{-5x})^{\frac{1}{3}}\left(\frac{-5}{-5}e^{-5x}\right)dx \\ &=-\frac{1}{5}\int \color{red}{(1+e^{-5x})^{\frac{1}{3}}}\color{blue}{(-5e^{-5x})dx} \\ &\stackrel{sub}= -\frac 15\int \color{red}{u^{1/3}}\color{blue}{du}\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the integral $\int \frac{1}{\sqrt[3]{(x+1)^2(x-2)^2}}\mathrm dx$ What substitution is useful for this integral?
$$\int \frac{1}{\sqrt[3]{(x+1)^2(x-2)^2}}\mathrm dx$$
Substitutions $u=x^{\frac{2}{3}},u=(x+1)^{\frac{2}{3}},u=(x-2)^{\frac{2}{3}}$ are not working.
Can't find useful trigonometric substitution.
| I found an interesting substitution by a bit of trial and error.
$$\begin{align}
x&=\frac{1+3\sin(u)}{2}\\
\therefore dx&=\frac{3\cos(u)}{2}du\\
x+1&=\frac{3+3\sin(u)}{2}&&=\frac{3(1+\sin(u))}{2}\\
x-2&=\frac{3\sin(u)-3}{2}&&=\frac{3(\sin(u)-1)}{2}\\
\therefore (x+1)(x-2)&=\frac{9(\sin^2(u)-1)}{4}&&=-\frac{9\cos^2(u)}{4}
\end{align}$$
Using these leads to this interesting final integral:$$\left(\frac{2}{3}\right)^{\frac{1}{3}}\int\frac{du}{\cos^{\frac{1}{3}}(u)}$$
| {
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Prove that $\frac1{a^3(b+c)}+\frac1{b^3(a+c)}+\frac1{c^3(a+b)} \geq \frac32$
Let $a,b,c$ be positive real numbers with $abc = 1$. Prove that $$\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)} \geq \dfrac{3}{2}.$$
Attempt
We can rewrite it as $\dfrac{bc}{a^2(b+c)}+\dfrac{ac}{b^2(a+c)}+\dfrac{ab}{c^2(a+b)}$. Then, using AM-GM we have $$\dfrac{bc}{2a^2\sqrt{bc}}+\dfrac{ac}{2b^2\sqrt{ac}} \dfrac{ab}{2c^2\sqrt{ab}} \geq \dfrac{bc}{a^2(b+c)}+\dfrac{ac}{b^2(a+c)}+\dfrac{ab}{c^2(a+b)}.$$
I get stuck here.
| Note that $LHS=\dfrac{b^2c^2}{ab+ac}+\dfrac{a^2c^2}{ab+bc}+\dfrac{a^2b^2}{bc+ac}\ge\dfrac{(bc+ac+ab)^2}{2(ab+bc+ac)}=\dfrac{ab+bc+ac}{2}\ge\dfrac{3}{2}$, where equality holds iff $\dfrac{b^2c^2}{ab+ac}=\dfrac{a^2c^2}{ab+bc}=\dfrac{a^2b^2}{bc+ac}$ and $ab=bc=ac$. As $abc=1$, then equality holds iff $a=b=c=1$.
First inequality is due to Minimum Principle of Arthur Engel. Second one is MA-MG.
| {
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Prove that $\dfrac{1}{2}\sqrt{4\sin^2(36^{\circ})-1} = \cos(72^{\circ})$
Prove that $\dfrac{1}{2}\sqrt{4\sin^2(36^{\circ})-1} = \cos(72^{\circ})$
This question seemed pretty simple so I first started out by turning the left-hand side into terms of $\cos(x)$. We have $\dfrac{1}{2}\sqrt{4(1-\cos^2(36^{\circ}))-1} = \dfrac{1}{2}\sqrt{3-4\cos^2(36^{\circ})}$. Then I would use $\cos(72^{\circ}) = 2\cos^2(36^{\circ})-1$. Thus, $\dfrac{1}{2}\sqrt{3-4\cos^2(36^{\circ})} = \dfrac{1}{2}\sqrt{3-2(\cos(72^{\circ})+1)} = \dfrac{1}{2}\sqrt{1-2\cos(72^{\circ})}$. I get stuck here.
| Note that we have the system of equations $$\begin{cases}\cos 72^\circ&=2\cos^2 36^\circ-1,\\\cos 36^\circ&=1-2\cos^2 72^\circ.\end{cases}$$ Adding the two equations together yields $$\cos 72^\circ+\cos 36^\circ=2(\cos^236^\circ-\cos^272^\circ)\implies\cos 36^\circ-\cos 72^\circ=\frac12.$$ Now subtracting the first equation from the second gives $$\cos 36^\circ-\cos 72^\circ=2-2(\cos^236^\circ+\cos^272^\circ)\implies\cos^236^\circ+\cos^272^\circ=\frac34.$$ To finish, note that this equality rearranges to $$1-\cos^2 36^\circ-\cos^272^\circ=\sin^236^\circ-\cos^272^\circ=\frac14.$$ Hence $4\sin^236^\circ-1=4\cos^272^\circ$, and taking the square root of both sides and dividing by $2$ gives the desired result.
| {
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How to prove this inequality given the equation Given $ a>0 $ and $ a^5-a^3+a=3$, how can I prove the inequality: $ a^6 \ge 5 $ ?
I have tried factorizing the equation, solving for $ x, x^2, x^3 $ and then adding equations made together. Last but not least I tried, to solve for $x$ and then raise it to the power of $ x^6 $, but none of these seem to work
| The discriminant of $z^2-z+1$ is negative, hence $a(a^4-a^2+1)=3$ implies $a>0$.
Since $z^2-z+1$ attains its minimum at $z=\frac{1}{2}$, $f(z)=z^4-z^2+1$ attains its minimum, $\frac{3}{4}$, at $z=\frac{1}{\sqrt{2}}$. Moreover, $0<f(z)\leq 1$ if $z\in[0,1]$, hence $z\cdot f(z)=3$ implies $z>1$.
$g(z)=z\cdot f(z)$ is an increasing function over $[1,+\infty)$ as the product of two positive increasing functions. Since $g(5^{1/6})<3$ by direct computation, in order to have $g(z)=3$
$$ z > 5^{1/6} $$
must hold. It may be useful to notice that:
$$ g(z) = \frac{z(z^6+1)}{z^2+1}, $$
hence if $z=5^{1/6}$ we have $g(z)=\frac{6z}{z^2+1}<3$ by the AM-GM inequality.
| {
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What's the value of $\int_0^1\frac{1}{2y} \ln(y) \ln^2(1-y) \, dy$? I came across this integral while doing a different problem:
$$ \int_0^1\frac{1}{2y} \ln (y)\ln^2(1-y) \, dy$$
I think we can evaluate this integral by differentiating the common integral representation of the beta function, but it seems to get a bit messy.
| I offer up yet another approach, this one relying on the Maclaurin series expansion for $\ln^2 (1 - x)$. It is similar to that used by @Frank W.
As was shown here
$$\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n}, \qquad |x| < 1.$$
Here $H_n$ denotes the Harmonic number. So for the integral we may write
\begin{align}
\int_0^1 \frac{1}{2x} \ln x \ln^2 (1 - x) \, dx &= \sum_{n = 2}^\infty \frac{H_{n - 1}}{n} \int_0^1 x^{n - 1} \ln x \, dx = -\sum_{n = 2}^\infty \frac{H_{n - 1}}{n^3},
\end{align}
after integrating by parts. From properties for the Harmonic number, since
$$H_n = H_{n - 1} + \frac{1}{n},$$
the infinite sum can be rewritten as
\begin{align}
\int_0^1 \frac{1}{2x} \ln x \ln^2 (1 - x) \, dx &= \sum_{n = 2}^\infty \frac{1}{n^4} - \sum_{n = 2}^\infty \frac{H_n}{n^3} = \sum_{n = 1}^\infty \frac{1}{n^4} - \sum_{n = 1}^\infty \frac{H_n}{n^3}.
\end{align}
Values for each of these sums are well known. For the first
$$\sum_{n = 1}^\infty \frac{1}{n^4} = \zeta (4) = \frac{\pi^4}{90}.$$
For the second sum
$$\sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{\pi^4}{72}.$$
(for various proofs of this result, see here) Thus
$$\int_0^1 \frac{1}{2x} \ln x \ln^2 (1 - x) \, dx = \frac{\pi^4}{90} - \frac{\pi^4}{72} = -\frac{\pi^4}{360}.$$
| {
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Cosine rule problem Question:
In triangle ABC, BC = a, AC= b, AB = c, and BD is perpendicular to AC.
If $\angle ABC = 60^{\circ}$, prove that $c = \dfrac{1}{2}a \pm \sqrt{b^2-\dfrac{3}{4}a^2}$
My approach: $b^2 = a^2 + c^2 - 2ac\cos B$
$b^2 = a^2 + c^2 - 2ac \cos{60}^\circ$
$b^2 = a^2 + c^2 - ac$
$b^2 = (a-c)^2+ac$
After this I don't know how to isolate the $c$ term.
| If you are not familiar with the standard formula for a quadratic equation then you can approach it like this:
$$\begin{align}
b^2 &= a^2 + c^2 - ac\\
\therefore c^2-ac&=b^2-a^2\\
\therefore \left(c-\frac{a}{2}\right)^2-\frac{a^2}{4}&=b^2-a^2\\
\therefore \left(c-\frac{a}{2}\right)^2&=\frac{a^2}{4}+b^2-a^2=b^2-\frac{3a^2}{4}\\
\therefore c-\frac{a}{2}&=\pm\sqrt{b^2-\frac{3a^2}{4}}\\
\therefore c&=\frac{a}{2}\pm\sqrt{b^2-\frac{3a^2}{4}}
\end{align}$$
This technique is called completing the square.
| {
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Find the value of the infinite product $\sqrt\frac12\cdot\sqrt{\frac12+\sqrt\frac12}\cdot\sqrt{\frac12+\sqrt{\frac12+\sqrt\frac12}}\cdots$ Find $\sqrt{\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}}....\infty$
Let $x=\sqrt{\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}}....\infty$
$\log x=\frac{1}{2}\log(\frac{1}{2})+\frac{1}{2}\log(\frac{1}{2}+\sqrt\frac{1}{2})+\frac{1}{2}\log(\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}})+....$
I do not know how to solve it further.
| Let $$x = \prod_{n = 1}^{\infty} y_n \qquad \text{and} \qquad y_n = \sqrt {\frac {1} {2} + \sqrt {\frac {1} {2} + \cdots + \sqrt {\frac {1} {2}}}},$$ and $$y = \lim_{n \to \infty} y_n.$$ We have $$y_{n + 1} = \sqrt {\frac {1} {2} + y_n},$$ which, passing to limit, gives $$y = \sqrt {\frac {1} {2} + y}.$$ We have $y = \frac {\sqrt {3}} {2} + \frac {1} {2}$, which is $> 1$, and that's why $x$ diverges to $+ \infty$.
| {
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integrate $\int \frac{x^4}{(1-x)^3}dx$
$$\int \frac{x^4}{(1-x)^3}$$
Because the degree of the numerator is bigger than the denominator I need to do a long division, should I open $(1-x)^3$ or is there a shorter way?
I got to $-\int x-\int 3-\int \frac{6x^2-10x-3}{(x-1)^3}$
So $6x^2-10x-3=A(x-1)^2+B(x-1)+C$
So $A=6$ And $B=2$ AND $C=-9$ which is wrong
| First, substitute $u = 1 -x$ into the integral:
$$u = 1 - x$$
$$x = 1 - u$$
$$du = -dx$$
$$\int \frac{x^4}{(1-x)^3}dx$$
$$\int \frac{(1-u)^4}{u^3}(-du)$$
$$-\int \frac{(1-u)^4}{u^3}du$$
$$-\int \frac{(u-1)^4}{u^3}du$$
$$-\int \frac{u^4 - 4u^3 + 6u^2 - 4u + 1}{u^3}du$$
$$-\int \left(u - 4 + \frac{6}{u} - \frac{4}{u^2} + \frac{1}{u^3}\right)du$$
Can you take it from here?
| {
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Locus of point of intersection of tangents at $A$ and $B$ From a Point $P$ on $C_1 \equiv x^2+y^2=9$ two tangents are drawn to $C_2 \equiv x^2+y^2=1$ which meets $C_1$ at $A$ and $B$. Find the Locus of point of intersection of tangents at $A$ and $B$ on $C_1$.
My Try: Let $P(3\cos\alpha, 3\sin\alpha)$, $A(3\cos\beta, 3\sin\beta)$ and $B(3cos\gamma, 3sin\gamma)$. Now $PA$ is chord of $C_1$ and tangent to $C_2$ say at point $M$. Since center of two circles is same, let it be $O$.
Now $OM$ is perpendicular to $PA$ at $M$ and since $PA$ is chord of $C_1$, $M$ is Mid Point of $PA$. So coordinates of $M$ is $M(\frac{3\cos\alpha+3\cos\beta}{2},\frac{3\sin\alpha+\sin\beta}{2})$
Now since $M$ lies on $C_2$ we have
$$\left(\frac{3\cos\alpha+3\cos\beta}{2}\right)^2+\left(\frac{3\sin\alpha+3\sin\beta}{2}\right)^2=1$$ Simplifying we get
$$\cos(\alpha-\beta)=\frac{-7}{9} $$ By similar analysis
$$\cos(\alpha-\gamma)=\frac{-7}{9} $$ which means
$$(\alpha-\beta)=2\pi+(\alpha-\gamma) \implies \gamma-\beta=2\pi \tag{1}$$
Now Equations of tangent to $C_1$ at $A$ and $B$ are
$$x\cos\beta+y\sin\beta=3$$ and $$x\cos\gamma+y\sin\gamma=3$$ Their Point of Intersection is
$$(h,k) =\left(\frac{3\cos\left(\frac{\gamma+\beta}{2}\right)}{\cos\left(\frac{\gamma-\beta}{2}\right)},\frac{-3\sin\left(\frac{\gamma+\beta}{2}\right)}{\cos\left(\frac{\gamma-\beta}{2}\right)}\right)$$
Now $$h^2+k^2=\frac{18}{1+cos(\gamma-\beta)}$$ Now using $(1)$ i will get the locus as $$h^2+k^2=9$$ which is the same circle $C_1$. Can i know where i went wrong and any better approaches?
|
We know that the required locus is a circle centered at O(0, 0). We only need to determine it radius.
For a particular case, that circle will go through Q(?, 0) when point P is at (3, 0).
The value of $\beta$ can be found from the brown triangle.
In the green right angled triangle, one angle is $2 \beta$.
The required radius = OQ can then be found.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with sum of coefficients please! Problem 1: "Imagine that the polynomial $(1 + x - y)^3$ is converted to the standard form. What is the sum of its coefficients?"
Problem 2 (continued): "What is the sum of the coefficients of the terms not containing $y$?"
Problem 3 (continued): "What is the sum of the coefficients of the terms containing $x$?"
I'm able to solve problem 1 & 2, but I've been stuck on 3 for hours..I'd really appreciate it if someone could explain how to solve it :D
| One should expand $(1+x-y)^3$ as follows $$(1+x-y)^3$$$$=(1+x-y)(1+x-y)^2$$
$$=(1+x-y)(1+x^2+y^2+2x-2xy-2y)$$
$$=1+x^2+y^2+2x-2xy-2y+x(1+x^2+y^2+2x-2xy-2y)-y(1+x^2+y^2+2x-2xy-2y)$$
$$=1+x^2+y^2+2x-2xy-2y+x+x^3+xy^2+2x^2-2x^2y-2xy-y-x^2y-y^3-2xy+2xy^2+2y^2$$
$$=\color{red}{x^3-y^3-3x^2y+3xy^2+3x^2+3y^2-6xy+3x-3y+1}$$
I hope you can take it from here
| {
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"question_score": "2",
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Solving second order differential equation - Important for thesis Thank you so much in advance for any help that you might give me. I'm stuck and can't solve the following equation
$y(x)+C_1y'(x) x+C_2 y''(x)x²=0$
Where $C_1$ and $C_2$ are constants.
More importantly,
$-r y(x)+\theta (\mu - x)y'(x)+\frac{\sigma ²}{2}y''(x)x²=0$
Any help?
| Looking at the first equation:
$0
=y(x)+cy'(x) x+d y''(x)x^2
$.
Let
$y(x)
=\sum_{n=0}^{\infty} a_nx^n
$.
$y'(x)
=\sum_{n=1}^{\infty} na_nx^{n-1}
$
so
$cxy'(x)
=\sum_{n=1}^{\infty} cna_nx^{n}
$.
$y''(x)
=\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}
$
so
$dx^2y''(x)
=\sum_{n=2}^{\infty} dn(n-1)a_nx^{n}
$.
Therefore
$\begin{array}\\
0
&=y(x)+cy'(x) x+d y''(x)x^2\\
&=a_0+a_1x+ca_1x +\sum_{n=2}^{\infty}a_nx^n(1+cn+dn(n-1))\\
\end{array}
$
Therefore, no solution of this type.
If $y = ax^b$,
$cxy' = abcx^{b}$
and
$dx^2y'' = ab(b-1)dx^{b}$,
so
$0
=x^b(a+abc+ab(b-1)d)
=ax^b(1+bc+b(b-1)d)
$
so
$\begin{array}\\
0
&=1+bc+b(b-1)d\\
&=1+bc-bd+b^2d\\
&=1+b(c-d)+b^2d\\
\end{array}
$
Therefore,
if $d \ne 0$,
$\begin{array}\\
b
&=\frac{d-c\pm\sqrt{(c-d)^2-4d}}{2d}\\
&=\frac{d-c\pm\sqrt{c^2-6cd+d^2}}{2d}\\
&=\frac{1-r\pm\sqrt{r^2-6r+1}}{2}
\qquad\text{where }r = \frac{c}{d}\\
\end{array}
$
If $d = 0$,
$b = -\frac1{c}
$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin^{6} x+\cos^{6} x)$ Solve $3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin^{6} x+\cos^{6} x)$
$a.)\ 1 \\
\color{green}{b.)\ 13} \\
c.)\ 15 \\
d.)\ 16 $
$ 3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin^{6} x+\cos^{6} x)\\
=3(1-\sin (2x))^{2}+6(1+\sin 2x)+4(\sin^{6} x+\cos^{6} x)\\
=3(1-2\sin (2x)+\sin^{2} (2x))+6(1+\sin 2x)+4(\sin^{6} x+\cos^{6} x)\\
=9+3\sin^{2} 2x+4(\sin^{6} x+\cos^{6} x)\\
=9+12\sin^{2}x\times \cos^{2}x+4(\sin^{6} x+\cos^{6} x)\\$
I am stucked.
I look for a short and simple way.
I have studied maths upto $12$th grade.
| Note that
$$\begin{align}&\sin^6x+\cos^6x\\&=(\sin^2x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)\\&=1\cdot ((\sin^2x+\cos^2x)^2-3\sin^2x\cos^2x)\\&=1-3\sin^2x\cos^2x\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding roots of a Complex Polynomial
Question
Given that x = 3 is a solution to $$ x^3 - (7+3i)x^2 + (16+15i)x - 6(2+3i) = 0 $$ find the other two solutions.
What I have attempted;
If $x = 3$ is a root then $(x-3)$ is a factor
so $$ (x-3)(x^2 + Ax + B) = 0 $$
$$ x^3 + (A-3)x^2 + (B-3A)x - 3B = 0 $$
Equation Coefficients
$$ x^3 - (7+3i)x^2 + (16+15i)x - 6(2+3i) = 0 $$
$$ x^2 :−3+A=−7−3i $$
$$ ⇒ A=−4−3i $$
$$\text{Constant: } -3B = -6(2+3i) $$
$$ ⇒ B=4+6i $$
$$ (x-3)(x^2 -(4+3i)x + 4+6i) = 0 $$
$$ x = {-b\pm\sqrt{b^2-4ac} \over 2a} $$
$$ x = {4+3i\pm\sqrt{(4+3i)^2-4(4+6i)} \over 2} $$
$$ x = {4+3i\pm\sqrt{-9} \over 2} $$
$$ x = {4+3i±3i\over 2} $$
$$ x_1 = 3 $$
$$ x_2 = 2+3i $$
$$ x_3 = 2 $$
Hopefully this is correct , and if so I was wondering if there is an alternative method in solving this.
I was thinking because we are given x = 3 is a root and (x-3) is a factor we can do polynomial division with complex numbers? Is it possible to do polynomial division with complex numbers? If so can someone show how to get the same answer by using that method..
| Call the cubic $p(x).$ $\bullet$ Method 1: We have $p(3)=0.$ We have $$p(x)=p(x)-p(3)=(x-3)p'(3)+(x-3)^2p''(3)/2+(x-3)^3p'''(3)/6=$$ $$=[x-3]\cdot [p'(3)+(x-3)p''(3)/2+(x-3)^2p'''(3)/6].$$ For computation, let $x=y+3$ and find the solutions, in $y$, of $ p'(3)+y p''(3)/2+y^2 p'''(3)/6=0.$ $\bullet$ Method 2. For any cubic $p$ with leading term $A x^3$ we have $p(x)=(x-3) A x^2 +q(x)$ where $q$ is of degree $2$ or less, with leading term $B x^2$ ($B$ may be $0$). We have $q(x)=(x-3)B x+r(x)$ where $r(x)=Cx +D$ with constant $C,D.$ We have $r(x)= (x-3)C +E$ with constant $E.$ If $p(3)=0$ then $E=0,$ and $$p(x)=(x-3)A x^2 +q(x)=(x-3)A x^2+(x-3)B x+r(x)=$$ $$=(x-3)A x^2+(x-3)B x +(x-3)C=(x-3)(A x^2+B x +C).$$ In practice this method is fast and easy.
| {
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Determine (or evaluate) the sum of the series $\sum_{j=0}^\infty (-1)^j \frac{\frac{3}{2}}{\frac{3}{2}+j} \frac{x^{2j}}{(2j)!}, \ \ \ x\in\mathbb R$. Determine (or evaluate) the sum of the series
$$\sum_{j=0}^\infty (-1)^j \frac{\frac{3}{2}}{\frac{3}{2}+j} \frac{x^{2j}}{(2j)!}, \ \ \ x\in\mathbb R$$
and
$$\sum_{j=0}^\infty (-1)^j \frac{\frac{5}{2}}{\frac{5}{2}+j} \frac{x^{2j}}{(2j)!}, \ \ \ x\in\mathbb R$$
The best that I have managed to do is to notice that
$$\left|\frac{\frac{1}{2}+k}{\frac{1}{2}+k+j}\right|\leq 1, \ \ \ \text{for all}\ j,k\in\mathbb N.$$
Then the sum is estimated from above by an hyperbolic cosine.
Thanks in advance.
| For the first problem (assuming $x\neq0$),
$$\sum\limits_{j=0}^{\infty}(-1)^j\frac{\frac{3}{2}}{\frac{3}{2}+j}\frac{x^{2j}}{(2j)!}$$
Define $f(x)=\sum\limits_{j=0}^{\infty}(-1)^j\frac{1}{2j+3}\frac{x^{2j+3}}{(2j)!}$. Then, the given sum is equal to $3f(x)/x^3$ for all $x\neq0$.
$$f'(x)=\sum\limits_{j=0}^{\infty}(-1)^j\frac{x^{2j+2}}{(2j)!}=x^2\cos x$$
Integrating by parts, we have $f(x)=x^2\sin x + 2x\cos x - 2\sin x + c$ for some constant $c$.
So, we have
$$\sum\limits_{j=0}^{\infty}(-1)^j\frac{\frac{3}{2}}{\frac{3}{2}+j}\frac{x^{2j}}{(2j)!}=\frac{3(x^2\sin x + 2x\cos x - 2\sin x + c)}{x^3}$$
As $x$ goes to $0$, LHS goes to $1$. In order for the limit to exist on the RHS, we must have $c=0$. We can then evaluate the limit of the RHS by L'Hospital's rule to verify that it equals $1$. Hence, for $x\neq 0$, sum of the series is
$$\frac{3(x^2\sin x + 2x\cos x - 2\sin x)}{x^3}$$
Proceeding in exactly the same manner, the sum for the second problem evaluates to (for $x\neq 0$):
$$\frac{5(x^4\sin x +4x^3\cos x-12x^2\sin x-24x\cos x+24\sin x)}{x^5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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mysterious sum of two sequences Let
$$S_1 = \sum_{n=1}^\infty \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \cdots$$
$$S_2 = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \cdots$$
So
\begin{align}
S_1 - S_2 = {} & 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \cdots \\
& {} - \left[ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \cdots \right] \\
= {} & 0 + 1 + 0 + \frac{1}{2} + 0 + \frac{1}{3} + 0 + \frac{1}{4} + \cdots \\
= {} & S_1
\end{align}
That makes $S_2$ zero, right?
The reason I am asking is because I was under the impression that $S_2 = \ln 2$.
What did I miss?
Is it that $S_1$ is divergent and $S_2$ is convergent and it's not allowed?
|
That makes $S_2$ zero, right ?
Not if $|S_1|=\infty.~$ ;-$)~$ Basically, infinity is for addition what $0$ is for multiplication. So, for instance, just like $S_1\cdot S_2=S_1$ does not imply $S_2=1$ when $S_1=0,~($since one can't exactly divide by $0),$ so also $S_1+S_2=S_1$ does not imply $S_2=0$ when $S_1\to\infty.~$ A simple method of evaluating $S_2$ can be found here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\sin(3x)=\cos(2x)$
Question: Solve $\sin(3x)=\cos(2x)$ for $0≤x≤2\pi$.
My knowledge on the subject; I know the general identities, compound angle formulas and double angle formulas so I can only apply those.
With that in mind
\begin{align}
\cos(2x)=&~ \sin(3x)\\
\cos(2x)=&~ \sin(2x+x) \\
\cos(2x)=&~ \sin(2x)\cos(x) + \cos(2x)\sin(x)\\
\cos(2x)=&~ 2\sin(x)\cos(x)\cos(x) + \big(1-2\sin^2(x)\big)\sin(x)\\
\cos(2x)=&~ 2\sin(x)\cos^2(x) + \sin(x) - 2\sin^2(x)\\
\cos(2x)=&~ 2\sin(x)\big(1-\sin^2(x)\big)+\sin(x)-2\sin^2(x)\\
\cos(2x)=&~ 2\sin(x) - 2\sin^3(x) + \sin(x)- 2 \sin^2(x)\\
\end{align}
edit
\begin{gather}
2\sin(x) - 2\sin^3(x) + \sin(x)- 2 \sin^2(x) = 1-2\sin^2(x) \\
2\sin^3(x) - 3\sin(x) + 1 = 0
\end{gather}
This is a cubic right?
So $u = \sin(x)$,
\begin{gather} 2u^3 - 3u + 1 = 0 \\
(2u^2 + 2u - 1)(u-1) = 0
\end{gather}
Am I on the right track?
This is where I am stuck what should I do now?
| You have made some errors in your calculations (or some typos here).
$$\sin(3x)=\cos(2x)$$
$$ \sin(2x+x) = \cos(2x)$$
$$\sin(2x)\cos(x) + \cos(2x)sin(x) = \cos(2x) $$
$$ 2\sin(x)\cos(x)\cos(x) + (1-2\sin^2(x))\sin(x)) = \cos(2x) $$
$$ 2\sin(x)\cos^2(x) + \sin(x) - 2\sin^{\bf{3}}(x) = \cos(2x) $$
$$ 2\sin(x)(1-\sin^2(x))+\sin(x)-2\sin^{\bf{3}}(x)=\cos(2x) $$
$$ 2\sin(x) - 2\sin^3(x) + \sin(x)- 2 \sin^{\bf{3}}(x) = \cos(2x) $$
$$3\sin(x)-4\sin^3(x)=\cos(2x)$$
Then recall that $\cos(2x)=1-2\sin^2(x)$ to give:
$$3\sin(x)-4\sin^3(x)=1-2\sin^2(x)$$
This is a cubic in $\sin(x)$. For simplicity write $y=\sin(x)$ to get:
$$-4y^3+2y^2+3y-1=0$$
$$-(y-1)(4y^2+2y-1)=0$$
So $\sin(x)=1$ or $\sin(x)=\frac{-1\pm\sqrt{5}}{4}$
So $x=\frac{\pi}{2}$ or $x=\frac{\pi}{10}$, $\frac{9\pi}{10}$, $\frac{13\pi}{10}$, $\frac{17\pi}{10}$
| {
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"url": "https://math.stackexchange.com/questions/1617890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Given the lengths of two sides of a triangle, is it possible to obtain (sharp) bounds for the average length of the sides of the triangle? The title says it all.
Given the lengths of two sides of a triangle, is it possible to obtain (sharp) bounds for the average length of the sides of the triangle?
Let the lengths of two sides of the triangle be $a$ and $b$. Then the length $c$ of the third side is bounded, by the Triangle Inequality, as follows:
$$|a - b| < c < a + b.$$
Another way to state it is
$$\max(a, b, c) < a + b + c - \max(a, b, c)$$
implying
$$2\cdot\max(a, b, c) < a + b + c.$$
Consequently, a lower bound for the average length of the sides of the triangle is given by
$$\frac{2}{3}\cdot\max(a, b, c) < \frac{a + b + c}{3}.$$
My questions are as follows:
*
*Is it possible to obtain a sharper lower bound for $(a + b + c)/3$?
*Is it possible to obtain a sharp upper bound for $(a + b + c)/3$?
Thanks!
| According to the law of cosines given $a,b,\gamma$ to be respectively two sides ($a \ge b$) and the angle in between you get that
$$c^2 = a^2+b^2-2ab\cos \gamma$$
so the average of the three sides is
$$\frac{a+b+c}{3} = \frac{a+b+\sqrt{a^2+b^2-2ab\cos \gamma}}{3}$$
Now note that
*
*the minimum of the average is when $\cos \gamma = 1 \Rightarrow$
$$\frac{a+b+c}{3} = \frac{a+b+\sqrt{a^2-2ab+b^2}}{3} = \frac{a+b+\sqrt{(a-b)^2}}{3} = \frac{2a}{3}$$
*the maximum is when $\cos \gamma = -1 \Rightarrow$
$$\frac{a+b+c}{3} = \frac{a+b+\sqrt{a^2+b^2+2ab}}{3} = \frac{a+b+\sqrt{(a+b)^2}}{3} = \frac{2(a+b)}{3}$$
so
$$\frac{2a}{3} \le \frac{a+b+c}{3} \le \frac{2(a+b)}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\lim _{x\to 3}\left(3x-8\right)^{\frac{x}{\sin\left(x-3\right)}}$ without L'Hôspital I have to calculate this limit $$\lim _{x\to 3}\left(3x-8\right)^{\frac{x}{\sin (x-3)}}$$
without L'Hôpital's rule.
These are my steps
$$\left(3x-8\right)^{^{\frac{x}{\sin\left(x-3\right)}}}\:=\:e^{\ln\left(\left(3x-8\right)^{^{\frac{x}{\sin\left(x-3\right)}}}\right)}=e\:^{\frac{x\ln\left(3x-8\right)}{\sin\left(x-3\right)}}$$
now i will caculate only $\frac{x\left(3x-8\right)}{\sin\left(x-3\right)}$
$$ t = x-3 \Rightarrow x = t+3$$
$\lim _{x\to 3}\left(\frac{x\ln\left(3x-8\right)}{\sin\left(x-3\right)}\right)\:=\lim \:_{t\to \:0}\:\frac{\left(t+3\right)\ln\left(3(t+3)-8\right)}{\sin\left(t\right)}$
And here im stack.
Help someone ?
Thanks.
| Using the Taylor expansion note that $$
\begin{align}
\lim _{x\to 3}\left(\frac{x\ln\left(3x-8\right)}{\sin\left(x-3\right)}\right)\ &= \lim_{t\to 0}\frac{\left(t+3\right)\ln\left(3(t+3)-8\right)}{\sin\left(t\right)}\\
&= \lim_{t\to 0}\frac{(t+3)(3t-\frac{3t^2}{2}+\frac{3t^3}{3!}+o({t^4)}}{t-\frac{t^3}{3!}+o(t^4)}
=9\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619856",
"timestamp": "2023-03-29T00:00:00",
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Factor $(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$ to get $(b+c-a) (a+b-c) (a-b+c) (a+b+c)$
Is there an easy way to factor $(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$ to get $(b+c-a) (a+b-c) (a-b+c) (a+b+c)$?
I have tried various methods such as completing the square, but I wasn't able to figure it out as it became very messy.
| Expand and notice that it's "trying to be" $-(\pm a^2 \pm b^2 \pm c^2)^2$
for some choice of $\pm$ signs. None of the sign choices quite works, but they're each off by a square, e.g.
$$
(a^2+b^2+c^2)^2-2(a^4+b^4+c^4) = 4a^2b^2 - (a^2+b^2-c^2)^2.
$$
So you can factor it as the difference of two squares, and then each factor is itself a difference of two squares and thus splits further into linear forms, giving the desired factorization.
Note that the factored form is within a constant factor $16$ of Heron's formula for the square of a triangle of side lengths $a,b,c$.
| {
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"timestamp": "2023-03-29T00:00:00",
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solve in positive integers $x$ and $y$ Solve the equation for positive integers $x,y$
$$y^3+3y^2+3y=x^3+5x^2-19x+20$$
i have written the equation as
$$(y+1)^3=x^3+5x^2-19x+21$$ which is
$$(y+1)^3-x^3=5x^2-19x+21$$ and by using $a^3-b^3$ we get
$$(y+1-x)((y+1)^2+(y+1)x+x^2)=5x^2-19x+21$$
now since the discriminant of $5x^2-19x+21$ is negative, the quadratic is always positive, so $y+1 \gt x$
Now if RHS is a prime number then the only possibility is
$y+1-x=1$ i.e., $y=x$ and substituting $x=y$ in original equation we get $2x^2-22x+20=0$ $\implies$ $x^2-11x+10=0$
Hence $x=1,y=1$ and $x=10,y=10$ are Possible solutions.
But if RHS is not a prime number how can i proceed?
| You did a great job that LHS is a perfect cube. Now your task is to figure out when is $x^3+5x^2-19x+21$ also a perfect cube. Notice that for $x>10$, you have $(x+1)^3 < x^3+5x^2-19x+21 < (x+2)^3$, therefore it can't be a perfect cube. So you have to check the cases when $x\in\{1,2,\ldots,10\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I prove that $\frac{x}{x^2 - 4}$is invertible?
Prove that $$f(x)=\frac{x}{x^2-4}$$ is invertible in $(-2,2)$.
I´m trying to proof this using the definition of injective function:
$$\frac{x}{x^2-4}=\frac{y}{y^2-4} \rightarrow x=y$$ but I couldn´t prove this definition.
| For $x$ and $y$ in our interval, we have
$$\frac{x}{x^2-4}=\frac{y}{y^2-4}\quad\text{if and only if}\quad x(y^2-4)=y(x^2-4).$$
This can be rewritten as $xy^2-yx^2+4(y-x)=0$, and then as
$$(xy+4)(y-x)=0.$$
In our interval, we cannot have $xy+4=0$, so $y=x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the value of $xyz$? Given an expression as :
$$xyz+xyz+xyz=zzz$$ where $x,y,z$ are integers and $xyz$ represents a number for example $236$ (not to be confused with $x\times y\times z$), what is the number $xyz$?
| From the last digit we have $z+z+z=z$ or $z+z+z=10+z$ or $z+z+z=20+z$, so $z=0$ or $z=5$.
If $z=0$ then we have $y+y+y=0$ or $10+0$ for the second digit, so $z=0,y=0$ and similarly $x=0$.
If $z=5$ then there's a carry to the 2nd digit and $y+y+y+1$ = $5$ (no solution) or $y+y+y+1=14$ (no solution) or $y+y+y+1=25$, i.e. $y=8$, with a carry of $2$ to the first digit. Then $x+x+x+2=5$, so $x=1$.
So $xyz$ = $000$ or $185$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $G$ is a group show that if $(a \cdot b)^2 = a^2 \cdot b^2$ then $G$ must be abelian.
If $G$ is a group show that if $(a \cdot b)^2 = a^2 \cdot b^2$ then $G$ must be abelian.
$\begin{aligned}(a \cdot b)^2 = a^2 \cdot b^2 & \iff (a\cdot b)\cdot(a \cdot b) = (a \cdot a)\cdot (b \cdot b) \\& \iff a \cdot (b \cdot (a \cdot b)) =a(a \cdot (b\cdot b)) \\& \iff (a^{-1} \cdot a) \cdot (b \cdot (a \cdot b)) =(a^{-1}\cdot a)(a \cdot (b\cdot b))\\& \iff (b \cdot (a \cdot b)) =(a \cdot (b\cdot b)) \\& \iff (b \cdot a) \cdot b =(a \cdot b)\cdot b\\& \iff (b \cdot a) \cdot (b \cdot b^{-1}) =(a \cdot b)\cdot (b \cdot b^{-1}) \\& \iff b \cdot a = a \cdot b\end{aligned}$
Thus $G$ must be abelian. Is this right? Is there less clunky way to write it if it's?
| More simply, $(ab)^2=abab=a^2b^2$
Cancel $a$ from the left and $b$ from the right to get $ba=ab$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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In a $\triangle ABC,$ Evaluation of minimum value of $\cot^2 A+\cot^2 B+\cot^2 C$
In a $\triangle ABC,$ Evaluation of minimum value of $\cot^2 A+\cot^2 B+\cot^2 C$,
Given $A+B+C = \pi$
$\bf{My\; Try::}$ Using $\bf{A.M\geq G.M}$
$$\frac{\cot^2 A+\cot^2 B}{2}\geq \cot A\cdot \cot B\Rightarrow \cot^2 A+\cot^2B \geq 2\cot A\cdot \cot B$$
Similarly $$\cot^2 B+\cot^2 C\geq 2\cot B\cdot \cot C$$
and $$\cot^2 C+\cot^2 A\geq 2\cot C\cdot \cot A$$
So $$\cot^2 A+\cot^2 B+\cot^2 C\geq \cot A\cdot \cot B+\cot B\cdot \cot C+\cot C\cdot \cot A=1$$
Using $$A+B+C = \pi\Rightarrow A+B=\pi-C$$
So $$\cot\left(A+B\right) = \cot\left(\pi-C\right)\Rightarrow \frac{\cot B\cdot \cot A-1}{\cot B+\cot A} = -\cot C$$
So we get $$\cot A\cdot \cot B+\cot B\cdot \cot C+\cot C\cdot \cot A=1$$
My question is Instead of using $\bf{A.M\geq G.M}$ Inequality, Can we use Jensen inequality directly
$$\frac{\cot^2A+\cot^2 B+\cot^2 C}{3}\geq \cot^2\left(\frac{A+B+C}{3}\right)$$
So we can Write it as $$\cot^2A+\cot^2 B+\cot^2 C \geq 1$$
If no, then what wrong with it, Thanks
| You can use Jensen's inequality directly, if you first prove that the function
$$f(x):=\cot^2 x\qquad(0<x<\pi)$$
is convex. To this end compute $$f''(x)={2(1+2\cos^2 x)\over\sin^4 x}$$
and verify that $f''(x)$ is $>0$ on the interval $\ ]0,\pi[\ $.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Limit of: $ -x+\sqrt{x^2+x} $ for $ x\to\infty $ Why the solution of this limit: $$\lim_{x\to\infty}-x+\sqrt{x^2+x} $$ is $\frac{1}{2}$?
I tried to do De L'hopital to $$\frac{-x^2+x\sqrt{x^2+x}}{x}$$ but whitout success.
| $$\begin{align}
-x+\sqrt{x^2+x}&=-x+\sqrt{x^2+x}\left(\frac{x+\sqrt{x^2+x}}{x+\sqrt{x^2+x}}\right)\\\\
&=\frac{x}{x+\sqrt{x^2+x}}\\\\
&=\frac{1}{1+\sqrt{1+1/x}}\to 1/2\,\,\text{as}\,\,x\to \infty
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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An Expression of $\ln(2)$ I saw online that the following infinite series has a value of $\ln(2)$:
$\sum_{n=0}^\infty \left(\dfrac{1}{n+1}-\dfrac{1}{n+2}
+\dfrac{1}{n+3}-\cdot\cdot\cdot\right)^2$
I started off by defining
$b_n=\sum_{k=0}^\infty \dfrac{(-1)^k}{n+k+1}$
I see that $b_0=ln(2)$ but that didn't help me in finding the sum of the
original question.
| Rewrite as
$$ \begin{align}S &=\sum_{n=0}^{\infty} \left [\int_0^1 dt \left (t^n - t^{n+1} + t^{n+2} - \cdots \right ) \right ]^2 \\ &=\sum_{n=0}^{\infty} \left [\int_0^1 dt \frac{t^n}{1+t} \right ]^2 \\ &= \sum_{n=0}^{\infty} \int_0^1 du \frac{u^n}{1+u} \, \int_0^1 dv \frac{v^n}{1+v} \\&= \int_0^1 \frac{du}{1+u} \, \int_0^1 \frac{dv}{1+v} \sum_{n=0}^{\infty} (u v)^n \\ &= \int_0^1 \frac{du}{1+u} \, \int_0^1 \frac{dv}{1+v} \frac1{1-u v} \\ &= \int_0^1 \frac{du}{(1+u)^2} [\log{2} - \log{(1-u)}] \\ &= \frac12 \log{2} - \int_0^1 du \frac{\log{u}}{(2-u)^2}\\ &= \frac12 \log{2} + \frac12 \log{2}\end{align}$$
Thus, $S=\log{2}$ as conjectured.
Note that
$$\begin{align}\int_0^1 \frac{dv}{1+v} \frac1{1-u v} &= \frac1{1+u}\int_0^1 dv \left (\frac{u}{1-u v} + \frac1{1+v} \right )\\ &= -\frac{\log{(1-u)}}{1+u}+\frac{\log{2}}{1+u} \end{align}$$
and
$$\begin{align}\int_0^1 du \frac{\log{u}}{(2-u)^2} &= \frac14 \sum_{k=0}^{\infty}(k+1) \frac1{2^k} \int_0^1 du \, u^k \log{k} \\ &= -\frac14 \sum_{k=0}^{\infty}(k+1) \frac1{2^k}\frac1{(k+1)^2} \\ &= -\frac12 \sum_{k=1}^{\infty} \frac{(1/2)^k}{k} \\ &= \frac12 \log{(1/2)} \end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the value of $\lim _{x\to \infty }\left(x^2\ln \left(\frac{\frac{1}{x}}{\sin \left(\frac{1}{x}\right)}\right)\right)=$? $$\lim _{x\to \infty }\left(x^2\ln \left(\frac{\frac{1}{x}}{\sin \left(\frac{1}{x}\right)}\right)\right)=$$
I think it can be solved using L'Hospitals rule, but is there a simpler method?
| A solution using Taylor expansions:
We will use the fact that, when $u\to 0$,
$$
\begin{align}
\sin u &= u -\frac{u^3}{6} + o(u^3) \\
\ln(1+u) &= u + o(u)
\end{align}
$$
Since $x\to\infty$, we have $\frac{1}{x}\to 0$, and we can apply the above with "$u=\frac{1}{x}$" (for the $\sin$) and "$u=\frac{-1}{6x^2} + o\left(\frac{1}{x^2}\right)$" (in the middle, for the $\ln$):
$$
\begin{align}
\ln \frac{\frac{1}{x}}{\sin \frac{1}{x}}
&= - \ln \frac{\sin \frac{1}{x}}{\frac{1}{x}}
= - \ln \frac{\frac{1}{x} - \frac{1}{6x^3} + o\left(\frac{1}{x^3}\right)}{\frac{1}{x}} \\
&= - \ln\left( 1- \frac{1}{6x^2} + o\left(\frac{1}{x^2}\right) \right) \\
&= \frac{1}{6x^2} + o\left(\frac{1}{x^2}\right) \\
\end{align}
$$
so that
\begin{align}
x^2\ln \frac{\frac{1}{x}}{\sin \frac{1}{x}}
&= \frac{1}{6} + o\left(1\right) \xrightarrow[x\to\infty]{} \frac{1}{6} \\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Elementary symmetric polynomial task with three variables Can anyone help me to wite this as sum or product of elementary symmetric polynomial.
$$\frac xy+\frac yx +\frac xz + \frac zx +\frac yz + \frac zy =7$$
I tried to set under one fraction, but I didn't go much far.
| In the LHS of
$$x^2 y+x^2 z+x y^2+x z^2+y^2 z+y z^2= 7xyz$$
you recognize all the binary terms from the development of $(x+y+z)^3$. To complete the cube,
$$(x+y+z)^3=\color{blue}{x^3+y^3+z^3 + 3}(x^2 y+x^2 z+x y^2+x z^2+y^2 z+y z^2)\color{blue}{+6xyz}\\= x^3+y^3+z^3 +27xyz.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a MacLaurin expansion of a function I am being asked to Find the MacLaurin expansion of the following function:
$f(x) = \frac{2x-8}{x^2-8x+12}$
I was not given a point about which to expand so I assume to use x=0. I know I can begin taking derivatives but this seems inefficient, but I do not see any way to reduce it to relate it to a more common series. Is there something I am missing, or is taking derivatives the only option?
| Taking partial fractions helps put it in a neater form. Here's my solution below:
\begin{align*}
\frac{2x-8}{x^2-8x+12}&=2\left(\frac{x-4}{(x-6)(x-2)}\right).
\end{align*}
Now we assume we can break this apart using partial fractions:
\begin{align*}
LHS&=2\left(\frac{a}{x-6}+\frac{b}{x-2}\right)\\
\end{align*}
Rewriting this as a single fraction gives
\begin{align*}
LHS&=2\left(\frac{a(x-2)+b(x-6)}{(x-6)(x-2)}\right)\\
&=2\left(\frac{(a+b)x-2a-6b)}{(x-6)(x-2)}\right)\\
\end{align*}
Equating coefficients, we see $a+b=1$, and $-2a-6b=-4$. Solving this system of questions give $a=b=\frac{1}{2}$. Substituting this back in for $a$ and $b$ gives
\begin{align*}
LHS&=2\left(\frac{1}{2(x-6)}+\frac{1}{2(x-2)}\right)\\
&=\frac{1}{x-6}+\frac{1}{x-2}
\end{align*}
Next we take the $n^\text{th}$ derivative of a function of the form $f=\frac{1}{x-a}$. We're going to use the mathematical principle of induction here to find a formula for all natural numbers. If you aren't familiar with this concept, we show something holds when $n=1$, then show that if it holds for some arbitrary value $k$ then $k+1$ also holds.
For $n=1$, $f'=-\frac{1}{(x-a)^2}$. Assume that $f^{(k)}=\frac{(-1)^kk!}{(x-a)^{k+1}}$. Consider $f^{(k+1)}$. This is the same as
\begin{align*}
f^{(k+1)}&=\frac{d}{dx}(f^{(k)})
\end{align*}
From our induction hypothesis,
\begin{align*}
f^{(k+1)}&=\frac{d}{dx}\left(\frac{(-1)^kk!}{(x-a)^{k+1}}\right)\\
&=(-1)^kk!\frac{d}{dx}\left(\frac{1}{(x-a)^{k+1}}\right)\\
&=(-1)^kk!\left(\frac{-(k+1)(x-a)^{k}}{(x-a)^{2(k+1)}}\right)\\
&=\frac{(-1)^{k+1}(k+1)!}{(x-a)^{2(k+1)-k}}\\
&=\frac{(-1)^{k+1}(k+1)!}{(x-a)^{k+2}}
\end{align*}
Thus $k+1$ holds and $f^{(n)}=\frac{(-1)^nn!}{(x-a)^{n+1}}$ is true for all positive integers. Therefore, we can write the Maclaurian expansions of these two functions as
\begin{align*}
\frac{1}{x-6}+\frac{1}{x-2}&=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n+\sum_{n=0}^\infty\frac{g^{(n)}(0)}{n!}x^n\\
&=\sum_{n=0}^\infty\frac{f^{(n)}(0)+g^{(n)}(0)}{n!}x^n\\
&=\sum_{n=0}^\infty\frac{\frac{(-1)^nn!}{(-6)^{n+1}}+\frac{(-1)^nn!}{(-2)^{n+1}}}{n!}x^n\\
&=\sum_{n=0}^\infty\left(\frac{(-1)^n}{(-6)^{n+1}}+\frac{(-1)^n}{(-2)^{n+1}}\right)x^n\\
&=\sum_{n=0}^\infty\left(\frac{1}{6^{n+1}}+\frac{1}{2^{n+1}}\right)x^n
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to find pythagorean triples and n-tuples A pythagorean triple is any set of three positive integers $(a,b,c)$ where $a^2 + b^2 = c^2$
I'm wondering, is there a formula to find all pythagorean triples, and can it be generalized to $n$-tuples? (i.e $a_1^2+a_2^2+a_3^2+a_4^2+....+a_n^2=x^2$)
| we can create an n-tuple of any length by finding an $A$ to match the $C$ of any previous triangle.
For $A=m^2-n^2$, we let $n=\sqrt{m^2-A}$ where $\lceil\sqrt{A+1}\space\rceil\le m\le \bigl\lceil\frac{A}{2}\bigr\rceil$. If any $m$ yields a positive integer $n$, we have $(m,n)$ for a Pythagorean triple.
The simplest triple is $3,4,5$ and we let $A=\sqrt{m^2-5}$ where;
$$m_{min}=\lceil\sqrt{5+1}\space\rceil=3\quad M_{max}=\lceil\frac{5}{2}\space\rceil=3$$
This one was easy: $n=\sqrt{3^2-5}=2\quad f(3,2)=(5,12,13)$
Let's try $(21,20,29)$. We have $A=29\implies 6\le m \le 15$ and we find $only$ $f(15,14)=(29,420,421)$
If we continue the process begun with $(3,4,5)$ we can find $(3,4,5)\rightarrow(5,12,13)\rightarrow(13,84,85)\rightarrow(85,132,157)\rightarrow(157,12324,12325)\text{ and so on.}$
So far, we have $3^2+4^2+12^2+84^2+132^2+12324^2=12325^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric Substitution (integration) $$\int\frac{dx}{(x^{2}-36)^{3/2}}$$
My attempt:
the factor in the denominator implies
$$x^{2}-36=x^{2}-6^{2}$$
substituting $x=6\sec\theta$, noting that $dx=6\tan\theta \sec\theta$
$$x^{2}-6^{2}=6^{2}\sec^{2}\theta-6^{2}=6^{2}\tan^{2}\theta$$
$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\int\frac{6\tan\theta \sec\theta}{36\tan^{2}\theta}=\frac{1}{6}\int\frac{\sec\theta}{\tan\theta}$$
using trig identities:
$$\frac{1}{6}\int\frac{\sec\theta}{\tan\theta}=\frac{1}{6}\int \sin^{-1}\theta$$
now using integration by parts:
$$\frac{1}{6}\int \sin^{-1}\theta$$
$$u=\sin^{-1}\theta, du=\frac{1}{\sqrt{1-\theta^{2}}}, dv=1, v=\theta$$
using $uv-\int{vdu}$
$$\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\int{\frac{\theta}{\sqrt{1-\theta^{2}}}}d\theta\bigg)$$
now using simple substitution:$$z=1-\theta^{2}, dz=-2\theta d\theta, -\frac{1}{2}du=\theta d\theta$$
it is apparent that
$$\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\bigg(-\frac{1}{2}\int{\frac{dz}{\sqrt{z}}}\bigg)\bigg)$$
$$=\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\bigg(-\frac{1}{2}(2\sqrt{z})\bigg)\bigg)=\frac{1}{6}\bigg(\theta \sin^{-1}\theta+\sqrt{1+\theta^{2}}\bigg)$$
$$=\frac{1}{6}\theta \sin^{-1}\theta+\frac{1}{6}\sqrt{1+\theta^{2}}+C$$
I have the following questions:
1.This integral seems tricky and drawn out to me, is there another method that reduces the steps/ methods of integration? I had to use trig substitution, integration by parts, and substitution in order to solve the integral, what can I do to find easier ways to complete integrals of this type?
2.Is this solution even correct? wolfram alpha says the solution to this integral is $-\frac{x}{36\sqrt{x^{2}-36}}+C$ how can i determine equivalence?
| Without substitution:
$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\frac1{36}\int\frac{(x^2-(x^2-36))\,dx}{(x^{2}-36)^{3/2}}=\frac1{36}\int\frac{x^2\,dx}{(x^{2}-36)^{3/2}}-\frac1{36}\int\frac{dx}{(x^{2}-36)^{1/2}}.$$
Then by parts on the first term,
$$\int\frac{x\cdot x\,dx}{(x^{2}-36)^{3/2}}=-\frac x{(x^{2}-36)^{1/2}}+\int\frac{dx}{(x^{2}-36)^{1/2}}.$$
After scaling the variable with a factor $6$, you recognize a known derivative,
$$\int\frac{dx}{\sqrt{x^2-1}}=\text{arcosh}(x).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the following limit $\lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{\frac{1}{a^2}e^{-\frac{(x-a)^2}{2}}+e^{-\frac{x^2}{2}}} dx$ Does the following limit exists
\begin{align}
\lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{\frac{1}{a^2}e^{-\frac{(x-a)^2}{2}}+e^{-\frac{x^2}{2}}} dx
\end{align}
Observe that if we do
\begin{align}
\lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{\frac{1}{a^2}e^{-\frac{(x-a)^2}{2}}+e^{-\frac{x^2}{2}}} dx \le \lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{e^{-\frac{x^2}{2}}} dx = \lim_{a \to \infty} e^{a^2}=\infty
\end{align}
or another upper bound is
\begin{align}
\lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{\frac{1}{a^2}e^{-\frac{(x-a)^2}{2}}+e^{-\frac{x^2}{2}}} dx \le \lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{\frac{1}{a^2}e^{-\frac{(x-a)^2}{2}}} dx = \lim_{a \to \infty} a^2 \sqrt{2 \pi}=\infty
\end{align}
For lower bounds we get
\begin{align}
\lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{\frac{1}{a^2}e^{-\frac{(x-a)^2}{2}}+e^{-\frac{x^2}{2}}} dx \ge \lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{\frac{1}{a^2}+1} dx = \lim_{a \to \infty} \int e^{-(x-a)^2}dx= \lim_{a \to \infty}\frac{\sqrt{2 \pi}}{\frac{1}{a^2}+1} =\sqrt{2 \pi}
\end{align}
Yet on the other had we get
\begin{align}
\int \lim_{a \to \infty} \frac{e^{-(x-a)^2}}{\frac{1}{a^2}+e^{-\frac{x^2}{2}}} dx =0
\end{align}
This together with the lower bound suggests that switching limit and integration is not allowed. So, I don't think that DCT or MCT apply?
Can someone explain how to approach this problem?
| Call your integral $I_a.$ Let $x=y+a$ in the integral, then reduce the interval of integration to $[0,1].$ Then we have
$$\tag 1 I_a > \int_0^1 \frac{e^{-y^2}}{(e^{-y^2/2})/a^2 + e^{-(y+a)^2/2}}\, dy > \int_0^1 \frac{e^{-y^2}}{(e^{-y^2/2})/a^2 + e^{-a^2/2}}\, dy$$ $$ = a^2\int_0^1 \frac{e^{-y^2}}{e^{-y^2/2} + a^2e^{-a^2/2}}\, dy $$
Now $a^2e^{-a^2/2} \to 0$ as $a\to \infty.$ So a very simple application of the DCT shows the last integral $\to \int_0^1 e^{-y^2/2}\, dy.$ Therefore $I_a$ is at least a positive constant times $a^2$ for large $a,$ hence $I_a \to \infty.$
| {
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I want to solve $\int \frac{2}{x^2(x^2+1)^2}dx$ I want to solve this primitive $$I=\int \frac{2}{x^2(x^2+1)^2}dx.$$
I substitute $u=x^2$ then,
$$I=\int \frac{2}{x^2(x^2+1)^2}dx=\int \frac{du}{u^{3/2}(u+1)^2}=\cdots$$
How do I use partial fraction decomposition? Is there another idea to compute this primitive?
| $\displaystyle\frac{1}{x^2(x^2+1)^2}=\frac{x^2+1}{x^2(x^2+1)^2}-\frac{x^2}{x^2(x^2+1)^2}=\frac{1}{x^2(x^2+1)}-\frac{1}{x^2+1)^2}$
$\hspace{.9 in}\displaystyle=\left[\frac{x^2+1}{x^2(x^2+1)}-\frac{x^2}{x^2(x^2+1)}\right]-\left[\frac{x^2+1}{(x^2+1)^2}-\frac{x^2}{(x^2+1)^2}\right]$
$\hspace{.9 in}\displaystyle=\frac{1}{x^2}-\frac{2}{x^2+1}+\frac{x^2}{(x^2+1)^2}$,
so $\displaystyle\int\frac{2}{x^2(x^2+1)^2}dx=-\frac{2}{x}-4\tan^{-1}x+\int\frac{2x^2}{(x^2+1)^2}dx$.
Now let $\displaystyle u=x,\; dv=\frac{2x}{(x^2+1)^2}dx,\;\; du=dx,\; v=-\frac{1}{x^2+1}$ to get
$\hspace{.3 in}\displaystyle-\frac{2}{x}-4\tan^{-1}x-\frac{x}{x^2+1}+\tan^{-1}x+C=\color{blue}{-\frac{2}{x}-3\tan^{-1}x-\frac{x}{x^2+1}+C}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Tough definite integral: $\int_0^\frac{\pi}{2}x\ln^2(\sin x)~dx$ Any ideas on evaluating the definite integral
$$\int_0^\frac{\pi}{2}x\ln^2(\sin x)\ dx$$
The best numerical approximation I could get is $0.2796245358$.
Is there even a closed form solution?
| Solution by real methods:
From here we have
$$\frac23\arcsin^4x=\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}(2x)^{2n}}{n^2{2n\choose n}}=\sum_{n=1}^\infty\frac{H_{n}^{(2)}(2x)^{2n}}{n^2{2n\choose n}}-\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^4{2n\choose n}}$$
Set $x=1$ we get
$$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=\sum_{n=1}^\infty\frac{4^nH_{n}^{(2)}}{n^2{2n\choose n}}-\frac{15}{4}\zeta(4)\tag1$$
In this question we showed
$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}=-\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}+12\ln^2(2)\zeta(2)\tag2$$
Adding $(1)$ and $(2)$ yields
$$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=12\ln^2(2)\zeta(2)-\frac{15}{4}\zeta(4)-\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}$$
By using the Fourier series of $\tan x\ln(\sin x)$, we showed in this solution:
$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}=-8\text{Li}_4\left(\frac12\right)+\zeta(4)+8\ln^2(2)\zeta(2)-\frac{1}{3}\ln^4(2)$$
substitute this result we get
$$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=8\text{Li}_4\left(\frac12\right)-\frac{19}{4}\zeta(4)+4\ln^2(2)\zeta(2)+\frac{1}{3}\ln^4(2)\tag3$$
Now we use the well-known series expansion of $\arcsin^2 x$:
$$\arcsin^2(x)=\frac12\sum_{n=1}^\infty\frac{4^n x^{2n}}{n^2{2n\choose n}}$$
Multiply both sides by $-\frac{\ln x}{x}$ then $\int_0^1$ and use that $-\int_0^1 x^{2n-1}\ln xdx=\frac{1}{4n^2}$ we get
$$\frac18\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=-\int_0^1\frac{\ln x\arcsin^2(x)}{x}dx$$
$$\overset{IBP}{=}\int_0^1\frac{\ln^2x\arcsin(x)}{\sqrt{1-x^2}}dx\overset{x=\sin\theta}{=}\int_0^{\pi/2}x\ln^2(\sin x)dx\tag4$$
From $(3)$ and $(4)$ we obtain
$$\int_0^{\pi/2} x\ln^2(\sin x)dx=\frac{1}{2}\ln^2(2)\zeta(2)-\frac{19}{32}\zeta(4)+\frac{1}{24}\ln^4(2)+\operatorname{Li}_4\left(\frac{1}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
} |
Prove that the altitudes of an acute triangle intersect inside the triangle.
Prove that the altitudes of an acute triangle intersect inside the triangle.
I can pretty easily see that this is true by a pythagorean theorem argument. Given any two sides, the smaller length to the side will be closer on the base so thus the orthocenter must lie inside the triangle. What would be a more mathematical way of proving this?
| Here is a proof
by analytic geometry.
Assume that the vertices are
$A=(0, 0), B=(1, 0)$,
and
$C=(a, 1)$.
It turns out that
the condition for
the intersection of the altitudes
to be inside the triangle
is $0 < a < 1$.
Looking at the triangle,
it is intuitively obvious
that this makes the triangle acute,
and it is nice that
this falls out of the proof.
The vertical altitude
through $C$ is
$x = a$.
The line $BC$
has slope
$\frac{1-0}{a-1}
=\frac{1}{a-1}
$.
Therefore the altitude through $A$
has slope
$1-a$,
so its equation is
$y = (1-a)x$.
Therefore, its intersection with
the altitude through $C$
has
$x=a, y=a(1-a)$,
so the intersection of the altitudes
is it
$V=(a, a(1-a))$.
A point inside the triangle is
$D
=(A+B+C)/3
=(a+1, 1)/3
=((a+1)/3, 1/3)
$.
Therefore $V$ is inside the triangle
if and only if it is
on the same side of each side
as $D$.
The equation of side
$AB$ is
$y=0$.
Evaluating this at $D$
is
$1/3$
and at $V$ is
$a(1-a)$.
Since $1/3 > 0$,
we must have
$a(1-a) > 0$
or
$0 < a < 1$.
The equation of side
$AC$ is
$y - x/a=0$.
Evaluating this at $D$
is
$\frac13-\frac{a+1}{3a}
=\frac{a-(a+1)}{3a}
=\frac{-1}{3a}
$
and at $V$ is
$a(1-a)-a/a
=a-a^2-1
$.
Since $\frac{-1}{3a}$
has the opposite sign of $a$,
and we now know that
$0 < a < 1$,
we must have
$a-a^2-1 < 0$
or
$a(1-a) < 1$.
Since we know that
$0 < a < 1$,
this is true.
The equation of side
$BC$ is
$\frac{1}{a-1}
=\frac{y}{x-1}$
or
$x-1-y(a-1) = 0$.
Evaluating this at $D$
is
$0
=\frac{a+1}{3}-1-\frac13(a-1)
=\frac{a+1-3-(a-1)}{3}
=\frac{-1}{3}
$
and at $V$ is
$x-1-y(a-1)
=a-1-a(a-1)(a-1)
=(a-1)(1-a(a-1))
$.
Since
$\frac{-1}{3} < 0$,
we must have
$(a-1)(1-a(a-1))
< 0$.
Since $0 < a < 1$,
$a-1 < 0$
and
$a(a-1) < 0$
so
$1-a(a-1) > 1 > 0$,
so we do have
$(a-1)(1-a(a-1))
< 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1641167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
The closed form of $\int^\infty_{B}e^{-(x+\frac{A}{x})}\,dx$, where $A>0$, $B>0$. What tools, ways would you propose for getting the closed form of this integral?
$$\int^\infty_{B}e^{-\left(x+\frac{A}{x}\right)}\,dx,$$ where $A>0$, $B>0$.
When $B=0$, from Table of Integrals, Series, and Products, we can use this to calculate $$\int^\infty_0 \exp \left(-\frac{\beta}{4x}-\gamma x\right)\,dx= \sqrt{\frac{\beta}{\gamma}} K_1(\sqrt{\beta\gamma})$$ with $\operatorname{Re} \beta\geq 0$, $\operatorname{Re} \gamma > 0 $.
But when $B>0$, how to calculate this intergral?
Thank you everyone.
| Let $u=x+A/x$ so that $x^2-u x+A=0$, or
$$x = \frac{u}{2} \pm \frac12 \sqrt{u^2-4 A} $$
$$dx = \frac12 \left (1 \pm \frac{u}{\sqrt{u^2-4 A}} \right ) du$$
Note that the integration limits provided by the mapping $x \mapsto u$ depend on whether the point $x=B$ is less than or greater than the minimum of $u$ at $x=\sqrt{A}$. Let's assume the former, i.e., $B \lt \sqrt{A}$; then the integral over $u$ is
$$\begin{align} I(A,B) &= \frac12 \int_{B+A/B}^{2 \sqrt{A}} du \, \left (1 - \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u} + \frac12 \int_{2 \sqrt{A}}^{\infty} du \, \left (1 + \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u}\\ &= \frac12 e^{-\left (B+\frac{A}{B} \right )} + 2 \sqrt{A} K_1 \left (2 \sqrt{A} \right ) - \sqrt{A}\int_{\alpha}^{\infty} dt \, \cosh{t} \; e^{-2 \sqrt{A} \cosh{t}} \end{align} $$
where
$$\alpha = \operatorname{arccosh}{\left (\frac{B+\frac{A}{B}}{2 \sqrt{A}} \right )}$$
If however $B \ge \sqrt{A}$,
$$\begin{align} I(A,B) &= \frac12 \int_{B+A/B}^{\infty} du \, \left (1 + \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u} \\ &= \frac12 \int_{B+A/B}^{2 \sqrt{A}} du \, \left (1 + \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u} + \frac12 \int_{2 \sqrt{A}}^{\infty} du \, \left (1 + \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u} \\ &= \frac12 e^{-\left (B+\frac{A}{B} \right )} + \sqrt{A} \int_{\alpha}^{\infty} dt \, \cosh{t} \; e^{-2 \sqrt{A} \cosh{t}}\end{align} $$
For either case, let's consider
$$\begin{align} \int_{\alpha}^{\infty} dt \, \cosh{t} \; e^{-2 \sqrt{A} \cosh{t}} &= \int_{\alpha}^{\infty} dt \, \coth{t} \sinh{t} \; e^{-2 \sqrt{A} \cosh{t}} \\ &= \frac1{2 \sqrt{A}} \coth{\alpha}\; e^{-2 \sqrt{A} \cosh{\alpha}} - \frac1{2 \sqrt{A}}\int_{\alpha}^{\infty} dt \, \frac1{\sinh^2{t}} e^{-2 \sqrt{A} \cosh{t}}\\ &= \frac1{2 \sqrt{A}} \coth{\alpha}\; e^{-2 \sqrt{A} \cosh{\alpha}} - \frac1{4 A} \frac1{\sinh^3{\alpha}} \; e^{-2 \sqrt{A} \cosh{\alpha}} + \frac{3}{4 A} \int_{\alpha}^{\infty} dt \, \frac{\cosh{t}}{\sinh^4{t}} e^{-2 \sqrt{A} \cosh{t}} \end{align} $$
We could go on and develop more terms in the series by further integration by parts, but let's take stock. Recall that
$$\cosh{\alpha} = \frac{B+\frac{A}{B}}{2 \sqrt{A}} $$
$$\sinh{\alpha} = \frac{B+\frac{A}{B}}{2 \sqrt{A}} \left [1-\frac{4 A}{\left (B+\frac{A}{B} \right )^2} \right ]^{1/2} $$
Then
$$\frac1{2 \sqrt{A}} \coth{\alpha}\; e^{-2 \sqrt{A} \cosh{\alpha}} = \frac1{2 \sqrt{A}} \left [1-\frac{4 A}{\left (B+\frac{A}{B} \right )^2} \right ]^{-1/2} e^{-\left ( B+\frac{A}{B} \right )} $$
$$\frac1{4 A} \frac1{\sinh^3{\alpha}} \; e^{-2 \sqrt{A} \cosh{\alpha}} = \frac{2 \sqrt{A}}{\left ( B+\frac{A}{B} \right )^3} \left [1-\frac{4 A}{\left (B+\frac{A}{B} \right )^2} \right ]^{-3/2} e^{-\left ( B+\frac{A}{B} \right )} $$
Note that additional terms will produce smaller contributions, so we can stop here for the moment and appreciate two uses of this series:
1. Perturbation for $0 \lt B \lt \sqrt{A}$ from the exact Bessel function result at $B=0$:
Using the assumption that $B \gt 0$ is a small perturbation, we expand out to $O(B^3)$, requiring the first two terms of the above expansion. Any further terms requires more integration by parts. Note that, in this region, it is the $1/B$ term that dominates. We find that
$$I(A,B) = 2 \sqrt{A} K_1 \left (2 \sqrt{A} \right ) - \left [ \frac1{2 A} B^2 + \frac1{2 A^2} B^3 +O(B^4)\right ] e^{-B-\frac{A}{B}}$$
Note that the perturbation term for small $B$ is extremely small. This has been verified numerically, although I caution that the extreme smallness of the perturbation makes it difficult to verify the exact expansion terms in $B$.
2. Global approximation of $I(A,B)$ for all $B$
We may accomplish this by considering expansions for both small $B$ and large $B$. Note that we use either one of the two cases delineated above according to whether $B$ is less than or greater than $\sqrt{A}$. I will not push this further, and I anticipate some difficulties because of the exponential term in the expansion, but I believe a two-point Pade approximant may do the trick.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1641929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Can we say that $\det(A+B) = \det(A) + \det(B) +\operatorname{tr}(A) \operatorname{tr}(B) - \operatorname{tr}(AB)$? Let $A,B \in M_n$. Is this formula true?
$$\det(A+B) = \det(A) + \det(B) + \operatorname{tr}(A) \operatorname{tr}(B) - \operatorname{tr}(AB).$$
| Here's an attempt to make the $2\times 2$ case seem just slightly horrible, while practicing to master the align* environment.
Let $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ and $B = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}$.
Then the left-hand side expands to eight terms:
\begin{align*}
\det (A+B) ={}& (a_{11}+b_{11})(a_{22}+b_{22}) - (a_{12}+b_{12})(a_{21}+b_{21}) \\
={}& \hphantom{(}a_{11}a_{22} - a_{12}a_{21} \\
&+ b_{11}b_{22} - b_{12}b_{21} \\
&+ a_{11}b_{22} - a_{21}b_{12} \\
&+ a_{22}b_{11} - a_{12}b_{21}
\end{align*}
The first two lines of the latter expression are respectively $\det A = a_{11}a_{22} - a_{12}a_{21}$ and $\det B=b_{11}b_{22} - b_{12}b_{21}$.
That leaves only the expansion and cancellation for the trace expressions:
\begin{align*}
(\operatorname{tr} A)(\operatorname{tr} B) - \operatorname{tr}(AB)
={}& (a_{11}+a_{22})(b_{11}+b_{22}) \\
&- (a_{11}b_{11} + a_{12}+b_{21} + a_{21}+b_{12} + a_{22}+b_{22}) \\
={}& a_{11}b_{22} - a_{21}b_{12} \\
&+ a_{22}b_{11} - a_{12}b_{21}
\end{align*}
Thus in the $2\times 2$ cases the identity is valid:
$$ \det A+B = \det A + \det B + (\operatorname{tr} A)(\operatorname{tr} B) - \operatorname{tr}(AB) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 2
} |
Proof that $\frac{2}{3} < \log(2) < \frac{7}{10}$ Positive integrals
$$\int_{0}^{1}\frac{2x(1-x)^2}{1+x^2}dx=\pi-3$$
and
$$\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi
$$
(https://math.stackexchange.com/a/1618454/134791)
prove that $$3<\pi<\frac{22}{7}$$
Is there a similar argument for the following $\log (2)$ inequality?
$$\frac{2}{3}<\log(2)<\frac{7}{10}$$
| There are positive integrals that relate $\log(2)$ to its first four convergents: $0,1,\frac{2}{3},\frac{7}{10}$.
$$
\begin{align}
\int_0^1\frac{2x}{1+x^2}dx
&=
\log\left(2\right) \\
\int_0^1\frac{(1-x)^2}{1+x^2}dx
&=
1-\log\left(2\right) \\
\int_0^1\frac{x^2(1-x)^2}{1+x^2}dx
&=
\log\left(2\right)-\frac{2}{3} \\
\int_0^1\frac{x^4(1-x)^2}{1+x^2}dx
&=\frac{7}{10}-\log\left(2\right) \\
\end{align}
$$
Therefore,
$$-\int_0^1\frac{x^2(1-x)^2}{1+x^2}dx<0<\int_0^1\frac{x^4(1-x)^2}{1+x^2}dx$$
$$\frac{2}{3}-\log(2)<0<\frac{7}{10}-\log\left(2\right)$$
$$\frac{2}{3}<\log(2)<\frac{7}{10}$$
A similar set is available with denominators $(1+x)$:
$$\begin{align}
\int_0^1 \frac{1}{1+x}dx
&=
\log(2) \\
\int_0^1 \frac{x}{1+x}dx
&=
1-\log(2)\\
\frac{1}{2}\int_0^1 \frac{x^2(1-x)}{1+x} dx
&=
\log(2)-\frac{2}{3} \\
\frac{1}{2}\int_0^1 \frac{x^5(1-x)}{1+x} dx
&=
\frac{7}{10}-\log(2)
\end{align}$$
and series versions are given by
$$\begin{align}
\log(2)-\frac{2}{3}
&=
\sum_{k=1}^\infty \frac{1}{(2k+1)(2k+2)(2k+3)} \\
\frac{7}{10}-\log(2)
&=
\sum_{k=2}^\infty \frac{1}{(2k+2)(2k+3)(2k+4)} \\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that $f(n) =\frac{(n+1)^n}{n^{n+1}}$ is Monotonic
Prove that for each $n \in N$,
$$f(n) =\frac{(n+1)^n}{n^{n+1}}$$
is monotonic.
First, I can tell that the function is decreasing. If I take $\frac{1}{n}$, the function looks like $\frac{1}{n}(1+\frac{1}{n})^n$. Can this help?
| $$f(n)=\frac{1}{n}\bigg(1+\frac{1}{n}\bigg)^n\longrightarrow f(n+1)=\frac{1}{n+1}\bigg(1+\frac{1}{n+1}\bigg)^{n+1}$$
$$\frac{f(n+1)}{f(n)}=\frac{n}{n+1}\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^{n}}=\frac{n}{n+1}\bigg(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)=\bigg(\frac{n(n+2)}{(n+1)^2}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)\frac{n}{n+1}$$
$$ =\bigg(\frac{n(n+2)}{(n+1)^2}\bigg)^n\frac{n(n+2)}{(n+1)^2}<1\cdot 1=1$$
Because $\frac{n(n+2)}{(n+1)^2}=\frac{n^2+2n}{n^2+2n+1} < 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Showing Trigonometric Identity Prove that:
$$\cos^2\theta\sin^4\theta=\frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta)$$
Attempt:
\begin{align*}
L.H.S & = \cos^2\theta\sin^4\theta\\
& = \cos^2\theta\sin^2\theta\sin^2\theta\\
& = \frac{1+\cos2\theta}{2}.\frac{1-\cos2\theta}{2}.\frac{1-\cos2\theta}{2}\\
& = \frac{1}{8} (1-\cos^22\theta)(1-\cos2\theta)
\end{align*}
Now, what should I do?
| With $z=e^{i\theta}$,
$$\left(\frac{z+z^{-1}}2\right)^2\left(\frac{z-z^{-1}}{2i}\right)^4=\frac1{32}\left(\frac{z^6+z^{-6}-z^2-z^{-2}+4-2z^4-2z^{-4}}2\right).$$
For convenience, multiply by $64z^6$, set $t:=z^2$ and rewrite
$$(t^2-1)^2(t-1)^2=t^6+1-t^4-t^2+4t^3-2t^5-2t.$$
By direct expansion of the LHS you get the equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Proof of limit of $x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)$ correct? I want to show that $$\lim_{x\to\infty} x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)= 1 $$
But I'm not sure if it's correct:
Because $\arcsin:[-1;1]\to \mathbb{R}$ is continuous and differentiable on $(-1,1)$, the mean value theorem says that there is a $c$ in $\left(\frac{1}{x+1} , \frac{1}{x}\right)$ for $f(t)=\arcsin(t)$ such that
$$x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)= f'(c_x) {\frac{x}{1+x}} $$
then
$f'(c_x)=\frac{1}{(\sqrt{1-c_x})}(\frac{x}{x+1}))\leqslant \frac{1}{(\sqrt{1-c_x})}\times\frac{x}{x} = \frac{1}{(\sqrt{1-c_x})} $
with Sandwichlemma for $x\to\infty, c_x\to0$
so the limit of $(x^2(\arcsin{\frac{1}{x}} -\arcsin{\frac{x}{1+x}} )= 1$
| Your proof is not complete: to use the Squeeze Theorem, you need both an upper and a lower bound converging to the same value (i.e., here $1$). But you only provided an upper bound.
Now, you have $c_x \in \left(\frac{1}{x+1}, \frac{1}{x}\right)$, so
$$f^\prime(c_x) = \frac{1}{\sqrt{1-c_x^2}} \geq \frac{1}{\sqrt{1-\frac{1}{(x+1)^2}}}
$$
and therefore
$$f^\prime(c_x)\cdot \frac{x}{x+1} \geq \frac{1}{\sqrt{1-\frac{1}{(x+1)^2}}}\cdot \frac{x}{x+1} \xrightarrow[x\to\infty]{}1.
$$
(In your question, you appear to have missed the square in the derivative of $\arcsin$.) Also, note another mistake in what you wrote (for the upper bound): $c_x\xrightarrow[x\to\infty]{}0$, not to $\infty$, since $c_x \in \left(\frac{1}{x+1}, \frac{1}{x}\right)$.
For reference, using Taylor approximations: namely, we shall use the facts that, when $u\to 0$
$$\begin{align}
\arcsin u &= u + o(u^2) \\
\frac{1}{1+u} &= 1-u + o(u)
\end{align}$$
(Note that, as you'll see, we need the $o(u^2)$ in the first expansion, $o(u)$ is not enough.)${}^{(\dagger)}$
Taking $\frac{1}{x}\xrightarrow[x\to\infty]{} 0$ as "our $u$":
$$\begin{align}
\arcsin \frac{1}{x} &= \frac{1}{x} + o\!\left(\frac{1}{x^2}\right) \\
\arcsin \frac{1}{x+1} &= \arcsin \left(\frac{1}{x}\left(\frac{1}{1+\frac{1}{x}}\right)\right)
= \arcsin \left(\frac{1}{x}\left(1-\frac{1}{x} + o\!\left(\frac{1}{x}\right) \right)\right)\\
&= \arcsin \left(\frac{1}{x}-\frac{1}{x^2} + o\!\left(\frac{1}{x^2}\right)\right) \\&= \frac{1}{x}-\frac{1}{x^2} + o\!\left(\frac{1}{x^2}\right)
\end{align}$$
so that
$$\begin{align}
\arcsin \frac{1}{x} - \arcsin \frac{1}{x+1}
&= \frac{1}{x} + o\!\left(\frac{1}{x^2}\right) - \left( \frac{1}{x}-\frac{1}{x^2} + o\!\left(\frac{1}{x^2}\right) \right) = \frac{1}{x^2} + o\!\left(\frac{1}{x^2}\right)
\end{align}$$
and to conclude
$$\begin{align}
x^2\left( \arcsin \frac{1}{x} - \arcsin \frac{1}{x+1} \right)
&= x^2\left( \frac{1}{x^2} + o\!\left(\frac{1}{x^2}\right) \right) = 1 + o(1)
\xrightarrow[x\to\infty]{} 1.
\end{align}$$
$(\dagger)$ If we only used $\arcsin u = u+o(u)$, then we would get $\arcsin \frac{1}{x} = \frac{1}{x} + o\!\left(\frac{1}{x}\right)$ alright, but only $\arcsin \frac{1}{x+1} = \arcsin \left(\frac{1}{x}-\frac{1}{x^2} + o\!\left(\frac{1}{x^2}\right)\right) = \frac{1}{x}-\frac{1}{x^2} + o\!\left(\frac{1}{x^2}\right) + o\!\left(\frac{1}{x}\right) =\frac{1}{x} + o\!\left(\frac{1}{x}\right)$ as well. This is no longer sufficient to conclude, as now all we can say is
$$\begin{align}
x^2\left( \arcsin \frac{1}{x} - \arcsin \frac{1}{x+1} \right)
&= x^2\left( o\!\left(\frac{1}{x}\right) \right) = o(x)
\end{align}$$
which could converge (or diverge) to basically anything.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Convergence and value of infinite product $\prod^{\infty}_{n=1} n \sin \left( \frac{1}{n} \right)$? Since the limit $\frac{\sin(x)}{x}=1$ for $x \rightarrow 0$, I wondered about the infinite product:
$$\prod^{\infty}_{n=1} n \sin \left( \frac{1}{n} \right)=\sin(1) \cdot 2 \sin\left( \frac{1}{2} \right) \cdot 3 \sin\left( \frac{1}{3} \right) \dots$$
By numerical experiment in Mathematica it seems to converge, even if very slowly (I mean to non-zero value):
$$P(14997)= 0.755371783$$
$$P(14998)= 0.755371782$$
$$P(14999)= 0.755371782$$
$$P(15000)= 0.755371781$$
I can prove the convergence by integral test for the series:
$$\sum^{\infty}_{n=1} \ln\left( n \sin \left( \frac{1}{n} \right) \right)$$
$$\int^{\infty}_{1} \ln\left( x \sin \left( \frac{1}{x} \right) \right) dx=\int^{1}_{0} \frac{1}{y^2} \ln \left( \frac{\sin (y)}{y} \right) dy=-0.168593$$
I think the integral test can work with negative function as long as it's monotone, otherwise I can just put the minus sign before the infinite sum.
By the way, this is a related question about the convergence of the sum above.
But I'm more interested in the infinite product itself.
I'm not sure if the value of this infinite product can be found and how to go about it. Is it zero or not? Any thoughts would be appreciated
| Thanks for all the comments. The closed form is unlikely, but some very good estimates can be provided, using Taylor series for $\sin$:
$$\prod_{n=1}^{\infty} n \sin \frac{1}{n}<\prod_{n=1}^{\infty} 1=1$$
$$\prod_{n=1}^{\infty} n \sin \frac{1}{n}>\prod_{n=1}^{\infty} \left(1-\frac{1}{6n^2} \right)=\frac{\sqrt{6}}{\pi} \sin \frac{\pi}{\sqrt{6}}=0.747529$$
$$\prod_{n=1}^{\infty} n \sin \frac{1}{n}<\prod_{n=1}^{\infty} \left(1-\frac{1}{6n^2}+\frac{1}{120n^4} \right)< \exp \left(-\frac{\pi^2}{36}+\frac{\pi^4}{10800} \right)=0.767101$$
Actually, for the last product Mathematica gives the closed form:
$$\prod_{n=1}^{\infty} \left(1-\frac{1}{6n^2}+\frac{1}{120n^4} \right)=\frac{5\sqrt{5}}{\pi^2} \sin \left(\frac{\pi \sqrt{1-\frac{i}{\sqrt{5}}}}{2 \sqrt{3}}\right) \sin
\left(\frac{\pi \sqrt{1+\frac{i}{\sqrt{5}}}}{2 \sqrt{3}}\right)=$$
See this answer.
$$=\frac{\sqrt{30}}{\pi^2} \left(\cosh \left( \pi \sqrt{\frac{1}{\sqrt{30}}-\frac{1}{6}} \right)-\cos \left( \pi \sqrt{\frac{1}{\sqrt{30}}+\frac{1}{6}} \right) \right)=0.755542$$
We get the estimation:
$$\prod_{n=1}^{\infty} n \sin \frac{1}{n}<0.755542$$
Which gives at least three (maybe four) correct digits for the numerical value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
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