Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the limit $L=\lim_{n\to \infty} \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\cdots+\sqrt[n]{\frac{1}{n}}}}$ Find the limit following:
$$L=\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots+\sqrt[\Large n]{\frac{1}{n}}}}$$
P.S
I tried to find the value of $\:L$, but I found myself stuck into the abyss of incertitude.
Thus, any help to get me out of this rift is more than welcome!
| Actually, there is another way which gives better upper boundary. I'm posting it as a separate answer because of the size.
First we notice that:
$$\lim_{n \to \infty} \left( \frac{1}{n} \right)^{\frac{1}{n}}=\lim_{n \to \infty} \left(1- \left(1- \frac{1}{n} \right) \right)^{\frac{1}{n}}=1$$
This is not a proof, but the fact is well known. Now let's consider the following:
$$1< \left(\frac{1}{n-1}+1 \right)^{\frac{1}{n-1}}<1+\frac{1}{(n-1)^2}$$
$$1< \left(\frac{1}{n-2}+1+\frac{1}{(n-1)^2} \right)^{\frac{1}{n-2}}<1+\frac{1}{(n-2)^2}+\frac{1}{(n-2)(n-1)^2}$$
On the next step we get:
$$\dots 1+\frac{1}{(n-3)^2}+\frac{1}{(n-3)(n-2)^2}+\frac{1}{(n-3)(n-2)(n-1)^2}$$
In the end we obtain the following inequality:
$$\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots}}<1+\sum^{\infty}_{k=2} \frac{1}{k ~k!}=Ei(1)-\gamma=1.3179$$
We can increase precision by moving the truncated series under the radical (and we should not forget to get rid of $2$ in every denominator):
$$1+2\sum^{\infty}_{k=3} \frac{1}{k ~k!}=1+2(Ei(1)-\gamma-1-1/4)=1.135804$$
$$L < \sqrt{\frac{1}{2}+1.135804}=1.27899$$
$$1+2\cdot 3 \sum^{\infty}_{k=4} \frac{1}{k ~k!}=1+6(Ei(1)-\gamma-1-1/4-1/18)=1.074080$$
$$L < \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1.074080}}=1.27305$$
Now for the lower boundary the better estimation would be:
$$L > \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1}}=1.26517$$
This is not ideal, but much more accurate than just truncating the sequence.
| {
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"timestamp": "2023-03-29T00:00:00",
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Math Induction Proof: $(1+\frac1n)^n < n$ So I have to prove:
For each natural number greater than or equal to 3,
$$(1+\frac1n)^n<n$$
My work:
Basis step: $n=3$
$$\left(1+\frac13\right)^3<3$$
$$\left(\frac43\right)^3<3$$
$$\left(\frac{64}{27}\right)<3$$
which is true.
Now the inductive step, assume $P(k)=\left(1+\frac1k\right)^k<k$ to be true and prove $P(k+1)=\left(1+\frac1{k+1}\right)^{k+1}<k+1$.
This is where I am stuck because usually you add or multiply by $k+1$ or some similar term.
| Two approaches
Bernoulli-like Inductive Step
$$
\begin{align}
\frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^n}
&=\overbrace{\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}}^{\lt1}\frac{n+1}n\\
&\lt\frac{n+1}n
\end{align}
$$
This shows that if $\left(1+\frac1n\right)^n\lt n$, then $\left(1+\frac1{n+1}\right)^{n+1}\lt n+1$
Non-Inductive Bernoulli Approach
Bernoulli's Inequality says that for $n\ge2$,
$$
\begin{align}
\left(1-\frac1{n+1}\right)^n
&=\left(\left(1-\frac1{n+1}\right)^{n/2}\right)^2\\
&\ge\left(1-\frac{n/2}{n+1}\right)^2\\[3pt]
&=\left(\frac{n/2+1}{n+1}\right)^2
\end{align}
$$
which is the reciprocal of
$$
\begin{align}
\left(1+\frac1n\right)^n
&\le\left(\frac{n+1}{n/2+1}\right)^2\\[6pt]
&\lt2^2
\end{align}
$$
Thus, $\left(1+\frac1n\right)^n\lt n$ if $n\ge4$.
We just need to verify the inequality for $n=3$:
$$
\left(1+\frac13\right)^3=\frac{64}{27}\lt3
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing $x^4-24x^2+4$ is irreducible over $\mathbb{Q}$
I want to show that $x^4-24x^2+4$ is irreducible over $\mathbb{Q}$.
I have tried Eisenstein's criterion using $p=2$ and substituting $x+1$ and $x-1$ but it won't work. Then what I tried was using the rational root test. I got as my possible rational roots $\pm 1, \pm 2, \pm4$. But each one does not work. From here I can consider $x^4-24x^2+4$ over $\mathbb{Z}$. So suppose that $x^4-24x^2+4$ is reducible over $\mathbb{Z}$. Then $$x^4-24x^2+4= (x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+ac+d)x^2+(bc+ad)x+bd.$$ By matching up the coefficients we get $$a+c=0$$ $$b+ac+d=-24$$ $$bd=4$$ $$bc+ad=0$$
Then we see $a+c=0\implies a=-c$. Then $$bc+ad=0\implies bc-cd=0 \implies bc=cd\implies b=d.$$ Since $bd=4$ then $b^2=4\implies b=\pm 2$ and $d=\pm 2$. We get two cases: $b=d=2$ and $b=d=-2$. In the first case $b=d=2$ we get $b+ac+d=-24\implies 4-c^2=-24\implies c^2=28$ (I substituted $a=-c$ into $b+ac+d=-24$), a contradiction since $c$ is an integer. Similarly in the case $b=d=-2$ we have $ b+ac+d=-24\implies -4-c^2=-24\implies c^2=20$. Again a contradiction. Would this be correct?
| $x^4-24x^2+4=t^2-24t+4=0 \Rightarrow t=12\pm 2\sqrt{35}\Rightarrow x^2-(12\pm 2\sqrt{35})=0$
I was a bit excited (Not careless ) and concluded that given polynomial is irreducible.
But It is not proper to conlcude at this stage but rather find all other roots and then check for possible combinations.
$x^2=12+2\sqrt{35}\Rightarrow x= \pm(\sqrt{7}+\sqrt{5})$ and $x^2=12-2\sqrt{35}\Rightarrow x= \pm(\sqrt{7}-\sqrt{5})$
Now we have all roots of $x^4-24x^2+4$ as $\pm(\sqrt{7}\pm\sqrt{5})$
It would be now your duty to check if :
$(x-a)(x-b)\in\mathbb{Q}[x]$ where $a,b \in \{\pm(\sqrt{7}\pm\sqrt{5})\}$ and $a\neq b$.
So, now i will leave this to you conclude this polynomial is irreducible/not in $\mathbb{Q}[x]$
Danger : This worked out only because you have only $x^4$ and $x^2$ and no $x^3$.
This is done under experts supervision(It is not me). please do not repeat this at home. :D
| {
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Find $x$ such that $\sqrt{x+\sqrt{x+7}}\in \mathbb{N}$ Find $x$ such that $$\sqrt{x+\sqrt{x+7}}\in \mathbb{N}$$
I tried many ways: $$\sqrt{x+\sqrt{x+7}}=n$$
$$\sqrt{x+\sqrt{x+7}}^2=n^2$$
$$x+\sqrt{x+7}=n^2$$
then solve for $x$ but didn't do with success.
I think this is the most difficult problem in my lifetime
Also $x$ must be made
of $2$ digit.
Thanks to everybody for helping me understand this problem and its solution!
| To see what's going on in this problem, it helps to generalize it slightly. (This should also help put martini's wonderful answer in context.) Let's look for integers $x$ that make an expression of the form
$$\sqrt{x+\sqrt{x+A}}\in\mathbb{N}$$
for an arbitrary $A\in\mathbb{N}$.
As other answers have shown, setting the expression in question equal to $k$ leads to the quadratic equation
$$x^2-(2k^2+1)x+(k^4-A)=0$$
for which the discriminant is
$$\Delta = (2k^2+1)^2-4(k^4-A)=4k^2+4A+1$$
For $x$ to be an integer, the discrimant must be a square, say $\Delta = n^2$. This implies
$$(n+2k)(n-2k)=4A+1$$
Now any factorization of the integer $4A+1$, say $4A+1=ab$, gives integer values for $n$ and $k$ (with non-negative $k$ if we take $a\ge b$): Setting $(n+2k)=a$ and $(n-2k)=b$ implies
$$n={a+b\over2}\quad\text{and}\quad k={a-b\over4}$$
(The denominators here, especially the $4$, may appear problematic, but note that $4A+1$ is congruent to $1$ mod $4$, hence its factors must be odd numbers that are both congruent to either $1$ or $3$ mod $4$.) Each factorization gives two values for $x$:
$$x={2k^2+1-n\over2}\quad\text{and}\quad x={2k^2+1+n\over2}$$
However, only the first of these is a solution of $\sqrt{x+\sqrt{x+A}}=k$; the other is a solution of $\sqrt{x-\sqrt{x+A}}=k$.
In particular, we always have the factorization $a=4A+1$, $b=1$, and this gives $n=2A+1$, $k=A$, leading to $x=A^2-A$. This is the solution in martini's wonderful answer.
For $A=7$, the number $4A+1=29$ is prime, so this is the only factorization. But for $A=11$, for example, we wind up with three values of $x$ that make $\sqrt{x+\sqrt{x+11}}$ an integer: $x=110$, $x=5$, and $x=-2$, corresponding to the factorizations $45\cdot1$, $15\cdot3$, and $9\cdot5$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $x^4+y^4$ and $x^3+y^3$ if $x+y=2$ and $x^2+y^2=8$ Find $x^4+y^4$ if $x+y=2$ and $x^2+y^2=8$
So i started the problem by nothing that $x^2+y^2=(x+y)^2 - 2xy$ but that doesn't help!
I also seen that $x+y=2^1$ and $x^2+y^2=2^3$ so maybe $x^3+y^3=2^5$ and $x^4+y^4=2^7$ but i think this is just coincidence
So how can i solve this problem?
PLEASE i need some help and thanks for all!!
| Solve the given equations simultaneously to obtain $x=1-\sqrt 3, y=1+\sqrt 3$ or vice versa. Then just compute $x^4+y^4$ directly.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
Series $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ I want to check, whether $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ converges or diverges.
I tried to use the Ratio test:
$|\frac{a_{n+1}}{a_n}|= \frac{n+2}{2^{n+1}} \frac{2^n}{n+1} = \frac{1}{2} \frac{n+1+1}{n+1} = \frac{1}{2} (1+ \frac{1}{n+1})$
$\lim\limits_{n \rightarrow \infty}{{(\frac{1}{2}(1+ \frac{1}{n+1})) = \frac{1}{2}}} \leq 1$
So $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ converges.
Could somebody please check my solution?
| You can also break it down into a sum of two series $1/2^n $ and $n/2^n$ , each of which converges; first is a geometric series (with $r<1$), the second converges by , e.g., the zeta test; $n/2^n < 1/n^2$ for all $n>N$ ; $N$ a finite value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/595981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Trigonometric equation: $2(\sin^6 x+\cos^6 x)-3(\sin^4 x+\cos^4 x)+1=0$ I'm new here, but I need your help so much to solve an equation:
$$2(\sin^6 x+\cos^6 x)-3(\sin^4 x+\cos^4 x)+1=0$$
I tried a lot like making $2[(\sin^2 x)^3 + (\cos^2 x)^3$
| Hint: $x^6 + y^6 = (x^2 + y^2)(x^4 - x^2y^2 + y^4)$. Then look for squares.
Also, it is not equal to zero, but to $-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$ Show that $$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$$
$$\sqrt{2}=\mathbf{2}^{1/2}$$
$$\sqrt{2\sqrt{2}}=\mathbf{2}^{1/2+1/2^2}$$
$$\sqrt{2\sqrt{2\sqrt{2}}}=\mathbf{2}^{1/2+1/2^2+1/2^3}$$
Show the limit of $$\mathbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+\dotsb+\frac{1}{2^n}=1$$ when $n\to\infty$
$$\textbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}$$
$$\Rightarrow \frac{1}{2}\textbf{S}_{n}=\frac{1}{2}(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n})$$
$$\Rightarrow \frac{1}{2}\textbf{S}_{n}=(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n+1}})$$
$$\Rightarrow (1)-(2)=\textbf{S}_{n}-\frac{1}{2}\textbf{S}_{n}=\frac{1}{2}-\frac{1}{2^{n+1}}$$
$$\Rightarrow \textbf{S}_{n}(1-\frac{1}{2})=\frac{1}{2}-\frac{1}{2^{n+1}}$$
$$\Rightarrow \frac{1}{2^{n+1}}\rightarrow\textbf{0}\quad\textit{when n}\rightarrow\infty$$
$$\Rightarrow \textbf{S}_{n}\rightarrow\textbf{1}\quad\textit{when n}\rightarrow\infty$$
$$\Rightarrow \lim_{n \to \infty}\textbf{2}^{\textbf{S}_{n}}=2\quad\textit{when n}\rightarrow\infty$$
| Your proof is correct!
Here is a different proof.
First note that the sequence, let's call it $(a_n)_{n\in\mathbb N}$, can be defined recursively as $a_1=\sqrt{2}$ and $a_{n+1}=\sqrt{2a_n}$. Then you can show that it is increasing and bounded above by 2, therefore it converges.
Its limit, $\ell$, must satisfy $\ell=\sqrt{2\ell}$, and therefore $\ell=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many Positive integer solutions does the equation $x + y + z + w = 15$ have? How many Positive integer solutions does the equation $x + y + z + w = 15$ have?
Attempt:
Let $x = m + 1, y = n + 1, z = o + 1, w = p + 1 $
Then, $ m + 1 + n + 1 + o + 1 + p + 1 = 15$
$ m + n + o + p = 11 $
| Hint:
Find the coefficient of $x^{15}$ of the following function:
$$f(x)=(x+x^2+x^3+\cdots+x^{15})^4$$
Why this gives you the number of integer solution of your equation?
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the determinant of $3\times 3$ matrix with $\sin x$ and powers of $\cos x$ How to calculate the determinant of this matrix
$A=\begin{bmatrix}
\sin x & \cos^2x & 1 \\
\sin x & \cos x & 0 \\
\sin x & 1 & 1
\end{bmatrix}$
$$\left[A\right]=\begin{vmatrix}
\sin x & \cos^2x & 1 \\
\sin x & \cos x & 0 \\
\sin x & 1 & 1
\end{vmatrix}=\\=\sin x\begin{vmatrix}
\cos x& 0\\
1 & 1
\end{vmatrix}-\cos^2x\begin{vmatrix}
\sin x & 0 \\
\sin x & 1
\end{vmatrix}+\begin{vmatrix}
\sin x & \cos x\\
\sin x & 1\\\end{vmatrix}=\\=\sin x\cos x-\cos^2x\sin x+\sin x-\sin x\cos x =\\=\sin x\left(\cos x-\cos^2x+1-\cos x\right)=\sin x \left(1-\cos ^2x\right)=\\=\sin x\cdot \sin^2x=\sin^3x$$
The path is something like this? I'm using the wrong rule?
| $$\left[A\right]=\begin{vmatrix} \sin x & \cos^2x & 1\\ \sin x & \cos x & 0 \\ \sin x & 1 & 1 \end{vmatrix}\begin{vmatrix} \sin x & \cos^2x \\ \sin x & \cos x \\ \sin x & 1 \end{vmatrix}=\sin x\cos x+\sin x-\sin x\cos x -\sin x\cos^2x=\\=\sin x\left(1-\cos^2 x\right)=\\=\sin x\cdot \sin^2x=\sin^3 x$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How do I solve this limit: $\lim _{x \to 0} \left(\frac{ \sin x}{x}\right)^{1/x}$? $$\lim _{x \rightarrow 0} \left(\frac{
\sin x}{x}\right)^{1/x}$$
I have spent an hour on the above limit and have no work to show. I tried using L'Hopital's Rule, but just kept going around in circles. Any help would be appreciated. Thank you.
| Hint : Take its logarithm, use the fact that $\ln a^b=b\ln a=\dfrac{\ln a}{1/b}$, and apply l'Hopital $3$ or $4$ times.
$$\ln L=\lim_{x\to0}\frac1x\cdot\ln\frac{\sin x}x=\lim_{x\to0}\frac{\ln\sin x-\ln x}x=\lim_{x\to0}\frac{\dfrac{\cos x}{\sin x}-\dfrac1x}1=\lim_{x\to0}\frac{x\cdot\cos x-\sin x}{x\cdot\sin x}=$$
$$=\lim_{x\to0}\frac{(1\cdot\cos x-x\cdot\sin x)-\cos x}{1\cdot\sin x+x\cdot\cos x}=-\lim_{x\to0}\frac{x\cdot\sin x}{\sin x+x\cdot\cos x}=$$
$$=-\lim_{x\to0}\frac{1\cdot\sin x+x\cdot\cos x}{\cos x+(1\cdot\cos x-x\cdot\sin x)}=-\lim_{x\to0}\frac{\sin x+x\cdot\cos x}{2\cdot\cos x-x\cdot\sin x}=-\frac{0+0\cdot0}{2\cdot1-0\cdot0}=-\frac02$$
$$=0\iff L=e^0=1.$$
| {
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If $abc=1$ and $a,b,c$ are positive real numbers, prove that ${1 \over a+b+1} + {1 \over b+c+1} + {1 \over c+a+1} \le 1$.
If $abc=1$ and $a,b,c$ are positive real numbers, prove that $${1 \over a+b+1} + {1 \over b+c+1} + {1 \over c+a+1} \le 1\,.$$
The whole problem is in the title. If you wanna hear what I've tried, well, I've tried multiplying both sides by 3 and then using the homogenic mean. $${3 \over a+b+1} \le \sqrt[3]{{1\over ab}} = \sqrt[3]{c}$$ By adding the inequalities I get $$ {3 \over a+b+1} + {3 \over b+c+1} + {3 \over c+a+1} \le \sqrt[3]a + \sqrt[3]b + \sqrt[3]c$$ And then if I proof that that is less or equal to 3, then I've solved the problem. But the thing is, it's not less or equal to 3 (obviously, because you can think of a situation like $a=354$, $b={1\over 354}$ and $c=1$. Then the sum is a lot bigger than 3).
So everything that I try doesn't work. I'd like to get some ideas. Thanks.
| let $$a=x^3,b=y^3,c=z^3\Longrightarrow xyz=1$$
since
$$y^3+z^3\ge y^2z+yz^2$$
so
$$\dfrac{1}{1+b+c}=\dfrac{xyz}{xyz+y^3+z^3}\le\dfrac{xyz}{xyz+y^2z+yz^2}=\dfrac{x}{x+y+z}$$
so
$$\sum_{cyc}\dfrac{1}{1+b+c}\le\sum_{cyc}\dfrac{x}{x+y+z}=1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The roots of the equation $z^n=(1+z)^n$... The complex roots of the equation
$$z^{n}=(1+z)^{n}$$
$A.$ are vertices of a regular polygon
$B.$ lie on a circle
$C.$ are collinear
$D.$ none of these
Don't know where to start.. Please help.!
NOTE: $z$ is a complex number.
| Clearly, $z\ne0,$
So dividing either sides by $z^n,$ we have $$\left(1+\frac1z\right)^n=1$$
Using de Moivre's formula & Euler's formula, $$1+\frac1z=e^{\frac{2m\pi i}n}=\cos\frac{2m\pi}n+i\sin\frac{2m\pi }n$$ where $m$ is any integer
$$\implies \frac1z=\cos\frac{2m\pi}n+i\sin\frac{2m\pi }n-1=i2\sin\frac{m\pi}n\cos\frac{m\pi}n-2\sin^2\frac{m\pi}n$$
$$=2i\sin \frac{m\pi}n\left(\cos\frac{m\pi}n+i\sin \frac{m\pi}n\right)$$ (using $\cos2A=1-2\sin^2A$)
$$z=\frac1{2i\sin \frac{m\pi}n\left(\cos\frac{m\pi}n+i\sin \frac{m\pi}n\right)}=\frac{{\cos\frac{m\pi}n-i\sin \frac{m\pi}n}}{2i\sin \frac{m\pi}n}=-\frac12-i\frac{\cot\frac{m\pi}n}2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find equation of a tangent on $y= \sin2x$ Find equation of a tangent on $y= \sin2x$ in intersections with $y=\frac{1}{2}$
What I calculated:
Intersections: $$\sin2x= \frac{1}{2}$$
...
$$0=tg^2x-4tgx-1$$
$$tgx_{1}=2+\sqrt{3}$$
$$x_{1}=75+k\pi$$
$$tgx_{2}=2-\sqrt{3}$$
$$x_{2}=15+ k\pi$$
$$T_{1}(75+k\pi,\frac{1}{2})$$ and $$T_{2}(15+k\pi,\frac{1}{2})$$
Derivative of $y= sin2x$ is: $y'= 2cos2x$
$$ y'(75)= -\sqrt {3} =k_{t_{1}} $$
$$y'(15)= \sqrt {3} =k_{t_{2}}$$
formula for calculating tangent that I used: $y-y_{1}=k_{t}(x-x_{1})$
1.)$T_{1}(75+k\pi,\frac{1}{2})$
$k_{t_{1}}=-\sqrt {3}$
$y_{1}-\frac{1}{2}=-\sqrt {3}(x-75-k\pi)$
$y_{1}=-\sqrt {3}x+75\sqrt {3}+\sqrt {3}k\pi+\frac{1}{2}$
2.)$T_{2}(15+k\pi,\frac{1}{2})$
$k_{t_{2}}=\sqrt {3}$
$y_{2}-\frac{1}{2}=\sqrt {3}(x-15 -k\pi)$
$y_{2}=\sqrt {3}x-15\sqrt {3}-\sqrt {3}k\pi+\frac{1}{2}$
Both solutions are wrong; what I should actually get:
$y_{1}=\sqrt{3}x-\frac{\sqrt{3}pi-6+12kpi\sqrt{3}}{12} $ and $y_{2}=-\sqrt{3}x-\frac{5\sqrt{3}pi+6+12kpi\sqrt{3}}{12};k=Z $
How could I possibly get that ?!
| \begin{align*}
\ \sin{2x} &= \frac{1}{2} \\
\ \arcsin{\big(\sin{2x}}\big) &= \arcsin{\frac{1}{2}} \\
\ 2x &= \frac{\pi}{6} \\
\ x &= \frac{\pi}{12} \qquad\text{to be more precise you can say:}\\
\ x &= \frac{\pi}{12} + \pi k\quad ,\quad k\in\mathbb Z\\
\end{align*}
So we know $x=\dfrac{\pi}{12}$ when they intersect.
$y=a(x-x_0) + y_0$ is the formula for the tangent, where $a=f'(x_0)$.
So, $x_0 = \dfrac{\pi}{12} \land y_0=\dfrac{1}{2}$.
$f(x)=\sin{2x} \implies f'(x)=2\cos{2x}$
$f'(x_0) = f'(\dfrac{\pi}{12}) = 2\cos{\big(\dfrac{\pi}{6}\big)} = \sqrt{3} $
Thus, the tangent is $$y=\underline{\underline{\sqrt{3}\big(x-\frac{\pi}{12}\big) + \dfrac{1}{2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/610616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How can I show that $\left|\sum_{n=1}^\infty\frac{x}{n^2+x^2}\right|\leq\frac{\pi}{2}$ for any $x\in{\bf R}$?
Show that
$$\left|\sum_{n=1}^\infty\frac{x}{n^2+x^2}\right|\leq\frac{\pi}{2}$$
It is true for $x=0$. But I don't see how it is true for any $x\in{\bf R}$. The identity $\frac{\pi}{2}=\int_0^\infty\frac{1}{1+x^2}dx$ may help, I think. But I don't know how to go on.
| We know that $$\left|\sum_{n=1}^\infty\frac{x}{n^2+x^2}\right| < \int_0^\infty{\frac{x}{n^2+x^2}dn}$$ for any $x\in{\Bbb R}$
Evaluating the integral we get $$\begin{align}
\int_0^\infty{\frac{x}{n^2+x^2}dn} &= tan^{-1}\bigg(\frac{n}{x}\bigg)\bigg|_0^\infty \\
&= \lim_{t\to\infty} tan^{-1}\bigg(\frac{t}{x}\bigg) \\
&= \frac{\pi}{2}
\end{align}$$
Therefore $$\left|\sum_{n=1}^\infty\frac{x}{n^2+x^2}\right| < \frac{\pi}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/611186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Prove Trig Identity
For any three angles $\alpha,\beta,\gamma$, show that $$\sin(\alpha-\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}2\sin\frac{\alpha-\gamma}2\cos\frac{\beta-\gamma}2$$
This is what I've tried:
$$2\sin\overbrace{\left(\frac{A-C}2\right)}^x\cos\overbrace{\left(\frac{A-B}2\right)}^y2\cos\left(\frac{B-C}2\right)\\
\sin\left(\frac{A-C}2+\frac{A-B}2\right)+\sin\left(\frac{A-C}2-\frac{A-B}2\right)\\
\left(\frac{B-C}2\right)\left[\sin\left(\frac{2A-B-C}2\right)+\sin\left(\frac{B-C}2\right)\right]\\
\cos\left(\frac{B-C}2\right)\sin\left(\frac{2A-B-C}2\right)+\cos\left(\frac{B-C}2\right)\sin\left(\frac{B-C}2\right)$$
I know it's basically turning sum into product and I previously tried to derive the LHS the same way the normal sum to product identity works but I couldn't figure out how. I tried to use the sum-product identity on RHS but couldn't quite make it work.
| $$
\begin{align}
4\cos(\phi)\cos(\theta)\sin(\phi+\theta)
&=2\big[\cos(\phi-\theta)+\cos(\phi+\theta)\big]\sin(\phi+\theta)\tag{1}\\
&=\sin(2\phi)+\sin(2\theta)+\sin(2\phi+2\theta)\tag{2}
\end{align}
$$
Justification:
$(1)$: $2\cos(\phi)\cos(\theta)=\cos(\phi-\theta)+\cos(\phi+\theta)$
$(2)$: $\begin{array}{}2\cos(\phi-\theta)\sin(\phi+\theta)=\sin(2\phi)+\sin(2\theta)\\2\cos(\phi+\theta)\sin(\phi+\theta)=\sin(2\phi+2\theta)\end{array}$
Now set $\phi=\frac{\alpha-\beta}{2}$ and $\theta=\frac{\beta-\gamma}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/611324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Integral $-\int_{0}^{a}{at^{3}\,{\rm d}t\over\,\sqrt{\,\left(a^{2} -t^{2}\right)\left(b^{2}t^{2} +a^{4}-a^{2}t^{2}\right)\,}\,}$ This problem's taking me a lot longer than I like (probably doing things the hard way...). I have 9 different terms to integrate, some of which are messier than others. This one, though, is messier than most.
$$-{\Large\int}_{\!0}^{a}
\!\raise 0.8ex {at^{3}\,{\rm d}t
\over
\,\sqrt{\,\left(a^{2} -t^{2}\right)\left(b^{2}t^{2} +a^{4}-a^{2}t^{2}\right)\,}\,}
$$
I tried a whole series of substitutions on this. First $t=a\sin\left(u\right)$, then $v=\cos\left(u\right)$ and split the resulting integral into two, only to find (after substitution number three) that the first half diverged (haven't tried the second half). In any case, I'm assuming there has to be a better way to attempt this. Any suggestions appreciated.
| Using the substitution $u=a^2-t^2$ as suggested by André Nicolas, we get:
$\dfrac{\mathrm du}{\mathrm dt}=-2t$. So,
$\begin{align}&\int-\frac{at^3}{\sqrt{(a^2-t^2)(b^2t^2+a^4-a^2t^2)}}\mathrm dt\\
&=\frac{a}2\int\frac{t^2(-2t)}{\sqrt{(a^2-t^2)((b^2-a^2)t^2+a^4)}}\mathrm dt\\
&=\frac{a}2\int\frac{(a^2-u)}{\sqrt{u((b^2-a^2)(a^2-u)+a^4)}}\mathrm du\\
&=\frac{a}2\int\frac{(a^2-u)}{\sqrt{u(m-nu)}}\mathrm du
\end{align}$
where $m=a^2b^2$ and $n=(b^2-a^2)$
$\begin{align}
&=\frac{a^3}2\int\frac{1}{\sqrt{u(m-nu)}}\mathrm du - \frac{a}2\int\frac{\sqrt{u}}{\sqrt{m-nu}}\mathrm du\\
&=\frac{a^3}2I_1 - \frac{a}2I_2.
\end{align}$
where I have substituted $I_1$ and $I_2$ for the first and second integral respectively.
For $I_1$, substitute $v=\sqrt{u}$, so that $\dfrac{\mathrm dv}{\mathrm du}=\dfrac1{2\sqrt{u}}$. Then,
$\displaystyle I_1=2\int \frac1{\sqrt{m-nv^2}}\mathrm dv
=\frac2{\sqrt{m}}\int \frac1{\sqrt{1-{\left(\sqrt{\frac{n}{m}}v\right)}^2}}\mathrm dv
=\dfrac{2}{\sqrt{m}}\sqrt{\dfrac{m}{n}}\arcsin \sqrt{\dfrac{n}{m}}v+C$.
Now for $I_2$,
$\begin{align}
I_2=&\int\frac{\sqrt{u}}{\sqrt{m-nu}}\mathrm du
=\frac1{\sqrt{m}}\sqrt{\frac{m}{n}}\int\frac{\sqrt{\frac{n}{m}u}}{\sqrt{1-\frac{n}{m}u}}\mathrm du\\
&=\frac1{\sqrt{m}}\sqrt{\frac{m}{n}}\frac{m}{n}\int\frac{\sqrt{x}}{\sqrt{1-x}}\mathrm dx
\qquad\qquad{\left(x=\dfrac{n}{m}u\right)}
\end{align}$
To solve $\displaystyle\int\frac{\sqrt{x}}{\sqrt{1-x}}\mathrm dx$, let $\theta=\arcsin \sqrt{x} (\implies x=\sin^2 \theta)$. We get, $\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac1{\sqrt{1-x}}\cdot\dfrac1{2\sqrt{x}}$.
So,
$\begin{align}&\int \sqrt{\frac{x}{1-x}} \mathrm dx
=\int 2x \dfrac1{\sqrt{1-x}}\cdot\dfrac1{2\sqrt{x}}\mathrm dx\\
&=\int 2\sin^2\theta \;\mathrm d\theta
=\theta-\dfrac{\sin2\theta}{2}+C\\
&=\arcsin \sqrt{x}-\dfrac12\sin(2\arcsin \sqrt{x})+C=\arcsin \sqrt{x}-\sin(\arcsin \sqrt{x})\cos (\arcsin \sqrt{x})+C\\
&=\arcsin \sqrt{x}-\sin(\arcsin \sqrt{x})\sqrt{1-\sin^2 (\arcsin \sqrt{x})}+C\\
&=\arcsin \sqrt{x}-\sqrt{x}\sqrt{1-x}+C.
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/614425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Function : Find the range of $f(x) = \sin^{-1}x +\tan^{-1}x +\cos^{-1}x$ Problem :
Find the range of $f(x) = \sin^{-1}x +\tan^{-1}x +\cos^{-1}x$
Solution : Since, $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$
Since range of $\tan^{-1}x$ is $ (\frac{-\pi}{2}, \frac{\pi}{2})$
$\therefore, \frac{-\pi}{2} \leq \tan^{-1}x \leq \frac{\pi}{2}$
= $ \frac{-\pi}{2} + \frac{\pi}{2} \leq \tan^{-1}x + \frac{\pi}{2} \leq \frac{\pi}{2} + \frac{\pi}{2}$.
= $0 \leq \tan^{-1}x+ \frac{\pi}{2} \leq \pi $
Is it correct.. please suggest thanks....
| This is not correct because you need to restrict $\tan^{-1}$ to the same domain as the other two functions: $[-1,1]$. The extreme vales that $\tan^{-1}(x)$ takes on $[-1,1]$ are $\pm\frac{\pi}{4}$. So the range of your $f$ is $\left[\frac{\pi}{2}-\frac{\pi}{4},\frac{\pi}{2}+\frac{\pi}{4}\right]=\left[\frac{\pi}{4},\frac{3\pi}{4}\right]$ (and its domain is $[-1,1]$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Cubic equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots. How can I find their value in terms of $a,b,c$? If the equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots, then equal root must
be equal to $\displaystyle \frac{bc-ad}{2(ac-b)^2}.$
My Try:: Let $x=\alpha,\alpha,\beta$ be the roots of given equation. Then using Vieta's formula
$$ \alpha+\alpha+\beta = -\frac{3b}{a}\Rightarrow 2\alpha +\beta = -\frac{3b}{a}$$
$$ \alpha \cdot \alpha +\alpha \cdot \beta +\alpha \cdot \beta = \frac{3c}{a}\Rightarrow \alpha^2+2\alpha \cdot \beta = \frac{3c}{a}$$
$$\alpha \cdot \alpha \cdot \beta = -\frac{d}{a}\Rightarrow \alpha^2 \cdot \beta = -\frac{d}{a}.$$
Now I did not understand how can I find the value of $\alpha$ in terms of $a,b$ and $c$.
Help is required.
Thanks
| Since $\beta=-2\alpha-\frac{3b}{a}$, we have
$${\alpha}^2+2\alpha \left(-2\alpha-\frac{3b}{a}\right)=\frac{3c}{a}\Rightarrow a{\alpha}^2+6b\alpha+c=0\Rightarrow 2a{\alpha}^3+12b{\alpha}^2+2c\alpha=0,$$
$${\alpha}^2\left(-2\alpha-\frac{3b}{a}\right)=-\frac da\Rightarrow 2a{\alpha}^3+3b{\alpha}^2-d=0.$$
Hence, we have
$$9{\alpha}^2+2c\alpha+d=0\Rightarrow 9a{\alpha}^2+2ca\alpha+ad=0.$$
Hence, we've already have
$$9a{\alpha}^2+54b\alpha+9c=0,$$
we have
$$(2ca-54b)\alpha=9c-ad$$
Hence, we have
$$\alpha=\frac{9c-ad}{2(ca-27b)}.$$
I don't know how to reach your value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Solutions for $\frac{3}{x+1}\le\frac{2}{2x+5}$ Im in search of the solutions for:
$$\frac{3}{x+1}\le\frac{2}{2x+5}$$
So first i tried to combine the two sites:
$$\frac{6x + 15 - 2x + 2}{2x^2 +7x + 5}\le{0}$$
$$\frac{4x + 17}{2x^2 +7x +5}\le{0}$$
My problem is that now i have two solutions for the denominator and i dont know how to continue:
$2x^2+7x+5 = -1 \text{ and } -2.5$
The solution should be: $(-2.5;-1) \cup (-\infty;-3.25)$
| Write $$\frac{3}{x+1}-\frac{2}{2x+5}=\frac{4x+13}{\left(x+1\right)\left(2x+5\right)}\leq0$$ There are
'special' values $-1,-2.5$ and $-3.25$, found if one of the factors in numerator or denominator is asked to equalize $0$.
Ask yourself the question:
What is the sign of this fraction if I fill in a value $x>-1$ or
$-2.5<x<-1$ or $-3.25<x<-2.5$ or $x\leq-3.25$? The answers to these questions lead you to the final answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
} |
$\omega^2+\omega+1$divides a polynomial The question is
Show that $f(n)=n^5+n^4+1$ is not prime for $n>4$.
The solution is given as
Let $\omega$ be the third root of unity. Then $\omega^2+\omega+1=0$. Since $\omega^5+\omega^4+1=\omega^2+\omega+1$, we see that $\omega^2+\omega+1$ is a *factor of the polynomial. So *$n^2+n+1|n^5+n^4+1$.
Which polynomial are we referring to in the bold typeface above? And how is $n^2+n+1|n^5+n^4+1$ true due to $\omega^5+\omega^4+1=\omega^2+\omega+1$?
| This is rather poor phrasing (in the highlighted solution). What it is saying is that that $\omega,$ which is a root of the irreducible polynomial $x^2+x+1$ is also a root of $x^5+x^4+1,$ and so the gcd of the two polynomials is not $1,$ and since the first polynomial is irreducible, it must divide the second. Now, that said, you can verify this without any fanciness by long division.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Help with Evaluating triple integral Evaluate $\int \int \int_B xyz^2 dV$, where B is a cuboid bounded by the regions $ 0 \le x \le 1 $, $ -1 \le y \le 2 $, $ 0 \le z \le 3 $.
I keep getting $ \frac{27}{4}$ as my answer but apparently it's incorrect...Any help would be appreciated.
| This is equivalent to
\begin{align*}
\int_0^1 x dx \int_{-1}^2 y dy \int_0^3 z^2 dz &= \left(\frac{1}{2} x^2 \Big|_0^1\right) \left(\frac 1 2 y^2\Big|_{-1}^2\right) \left(\frac 1 3 z^3 \Big|_0^3 \right) \\
&= \frac{1}{2} \cdot \left(\frac{4}{2} - \frac{1}{2}\right)\cdot \frac{3^3}{3} \\
&= \frac{1}{2} \cdot \frac{3}{2} \cdot 9 \\
&= \frac{27}{4}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How does mod multiplication work? For example, $10^{10} \equiv 4\pmod{6}$
If I used $\pmod{2}$ and $\pmod{3}$, how does the multiplication process work?
Since $10^{10} \equiv 0 \pmod{2}$ and $10^{10}\equiv 1\pmod{3}$,
$$
10^{10}\equiv (0,1) \pmod{(2,3)}
$$
how do we get the value $4$ at the end? do we list out the possible values of $0\pmod{2}$ and $1\pmod{3}$?
$$
1\pmod{3} = 1, 4, 7, 10
$$
so on.
Since only $4, 10$ and so on satisfy $\pmod{2}$, only values that satisfy both criteria can be used.
In general, can we do this for $\pmod{n}$, $n$ being any integer?
| This is called the "Chinese Remainder Theorem". You are given two congruent condition and you have to find the answer. I will show an example using slightly bigger numbers.
Suppose we are given
$$ n \equiv 3 \mod 7, ~~\hbox{and}~~ n\equiv 5\mod 11$$
From the first we can write
$$ n = 7 k + 3$$
Substituting in the second, we get
$$ 7 k + 3\equiv5 \mod 11$$
or
$$ 7 k \equiv2 \mod 11 \Rightarrow k \equiv (7^{-1}) 2 \mod 11$$
where the inverse is modulo $11$. One way to find $7^{-1} \mod 11$ is to use Euclid's algorithm. For small numbers it is easy to look at 7,14,21 etc.
Actually since $21=-1 \mod 11$ we can stop and deduce that $$7 \cdot 3\equiv -1 \mod 11$$ or
$$7 \, (-3) \equiv 1 \mod 11$$ If you prefer positive numbers, you can use $-3 \equiv 11-3 \mod 11$.
Hence
$$ 7^{-1} \equiv 8 \mod 11$$
Hence
$$ k = 8 \cdot 2 = 16\equiv 5 \mod 11$$
Hence $k = 11 m + 5$ and finally
$$ n =7 \, (11 m + 5) + 3 = 77 m + 38$$
You can easily extend this to any number of congruents.
Added in response to OP's request
We want $$n \equiv 0 \mod 2, ~~~ n \equiv 1 \mod 3$$
From second condition (OP already did it in the other order, so I want to provide a second method)
$$ n = 3 k + 1$$
substituting in the first condition
$$ 3 k + 1 = 0 \mod 2 \Rightarrow k + 1 = 0 \mod 2 ~~~\hbox{since $3 \equiv 1 \mod 2$}$$
So $k = -1 \mod 2$ or if you prefer positive numbers
$$ k = -1 + 2 \mod 2 \Rightarrow k = 2 m + 1$$
Substituting in the formula for $n$ we have
$$ n = 3 (2 m + 1) + 1 = 6m +4$$
So the possible values of $n$ are 4, 10, 16, 22, -2, -8, -14, -20 etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Find the exact value of $\tan\left ( \sin^{-1} \left ( \sqrt 2/2 \right )\right )$
Find the exact value of $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )$ without using a calculator.
I started by finding $\sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )=\dfrac{\pi}{4}$
So, $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )=\tan\left( \dfrac{\pi}{4}\right)$.
The answer is $1$. Can you show how to solve $\tan\left( \dfrac{\pi}{4}\right)$ to get $1$? Thank you.
| Hint: $$\tan\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/623703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How find this ODE:$y''-y'+y=x^2e^x\cos{x}$ Find this follow ODe solution
$$y''-y'+y=x^2e^x\cos{x}$$
I konw solve this follow three case
$$y''-y'+y=x^2\cos{x}$$
$$y''-y'+y=e^x\cos{x}$$
$$y''-y'+y=x^2e^x$$
But for $f(x)=x^2e^x\cos{x}$, I can't.
Thank you very much!
| One more way to solve it is using the Laplace transformation $\mathcal{L}.$
Let $\mathcal{L}(y)=Y(p).$ Then
$$
\mathcal{L}(y')=p Y(p)-y(0),\\
\mathcal{L}(y'')=p^2 Y(p)-y'(0)-p y(0),\\ \text{and, using the standard rules for L.t.}\\
\mathcal{L}({x}^{2}{{\rm e}^{x}}\cos \left( x \right))=\frac{1}{\left( p-1-i \right) ^{3}}+ \frac{1}{\left( p-1+i \right) ^{3}}
$$
We have now
$$
\left( {p}^{2}-p+1 \right) Y \left( p \right) -py \left( 0 \right) +y
\left( 0 \right) - y' \left( 0 \right) =
\frac{1}{\left( p-1-i \right) ^{3}}+ \frac{1}{\left( p-1+i \right) ^{3}},
$$
or
$$
Y(p)={\frac {1}{ \left( p-1-i \right) ^{3} \left( {p}^{2}-p+1 \right) }}+{\frac {1}{ \left( p-1+i \right) ^{3} \left( {p}^{2}
-p+1 \right) }}-{\frac {-py \left( 0 \right) +y \left( 0 \right) -
y'' \left( 0 \right) }{{p}^{2}-p+1}}.
$$
To find $y$ we compute the inverse Laplace transform for $Y(p).$ After some calculation we get
$$
y(x)=\mathcal{L}^{-1}(Y(p))=\left( 2\,\cos \left( x \right) \left( -3+x \right) +\sin \left( x \right) \left( 6-4\,x+{x}^{2} \right) \right) {{\rm e}^{x}}+1/3\,
\left( 3\,\cos \frac{\sqrt {3}x}{2} \left( 6+y \left( 0
\right) \right) +\sin \frac{\sqrt {3}x}{2}
\left( -10+2\, y'' \left( 0 \right) -y \left( 0
\right) \right) \right) {{\rm e}^{\frac 12\,x}}.
$$
After simplification taking into account that $y(0), y'(0)$ are constant we get
$$
y(x)={{e}^{\frac 12\,x}}(\sin \frac{\sqrt {3}x}{2} { C_1}+\cos \frac{\sqrt {3}x}{2} {C_2})+ \left( 2\,\cos \left( x \right) \left( x-3
\right) +\sin \left( x \right) \left( 6-4\,x+{x}^{2} \right)
\right) {{ e}^{x}},
$$
for some constant $C_1, C_2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Roots of unity and a system of equations by Ramanujan Is it immediately apparent that the solution to the system of equations,
$$\begin{aligned}
x_1^2 &= x_2+2\\
x_2^2 &= x_3+2\\
x_3^2 &= x_4+2\\
&\vdots\\
x_n^2 &= x_1+2\\
\end{aligned}\tag{1}$$
can be given by the roots of unity? Specifically,
$$x =\frac{y_k^2+1}{y_k}\tag{2a}$$
where the $y_k$ are,
$$\begin{aligned}
y_k &= \exp\Big(\frac{2\pi i k}{2^n-1}\Big),\; k = 0\dots 2^{n-1}-1\\
y_k &= \exp\Big(\frac{2\pi i k}{2^n+1}\Big),\; k = 1\dots 2^{n-1}\\
\end{aligned}\tag{2b}$$
Example. Let $n=4$. Then (1) is equivalent to,
$$x = (((x^2-2)^2-2)^2-2)^2-2\tag{3}$$
Expanded out, (3) is a $2^4=16$-deg polynomial and its 16 roots are given by (2) where,
$$y_k = \exp\Big(\frac{2\pi i k}{15}\Big),\; k = 0\dots 7$$
$$y_k = \exp\Big(\frac{2\pi i k}{17}\Big),\; k = 1\dots 8$$
Ramanujan considered the system (1) for $n=3,4$ in the general case and also as nested radicals. For $x = (((x^2-a)^2-a)^2-a)^2-a$, see this related post. (Interestingly, $n=5$ in the general case is no longer completely solvable in radicals.)
Question:
I observed (2) empirically. How do we prove from first principles that this is indeed the solution?
| Let $f(x)=x^2-2$. Then $$f(2\cos u)=4\cos^2u-2=2(2\cos^2u-1)=2\cos2u$$ so $$f^n(2\cos u)=2\cos2^nu$$ Then $f^n(x)=x$ becomes $$\cos u=\cos2^nu$$ for which solutions are given by $$(2^n-1)u=2k\pi,k=0,\pm1,\pm2,\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/626550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\binom{2n}{n}\frac{1}{n+1} = \prod \limits_{i = 2}^n \frac{2i-1}{i+1} $? How to prove this closed form involving Catalan numbers? $$\binom{2n}{n}\frac{1}{n+1} = \prod \limits_{i = 2}^n \frac{2 \times (2i-1)}{i+1} $$
I have seen this being used here. Not sure how to derive it. Any ideas?
| Note that $\binom{2n}{n}=\frac{(2n)!}{n!n!}$. But
$$(2n)!=\left(1\cdot 3\cdot 5\cdots \cdot(2n-1)\right)\left(2\cdot 4\cdot 6\cdots\cdot(2n)\right)$$
(we separated the odd numbers and the even numbers).
The product of the even numbers $2\cdot 4\cdot 6\cdots(2n)$ is $2^n n!$.
From this it follows that
$$\frac{1}{n+1}\binom{2n}{n}=\frac{2^n}{(n+1)!}\left(1\cdot 3\cdot5 \cdots \cdot(2n-1)\right).$$
That is equal to the expression of the OP. For it can be written as
$$\prod_{i=1}^n \frac{2\cdot(2i-1)}{i+1}.$$
Starting at $i=2$, as the revised OP does, eliminates one factor of $2$ from both the numerator and the denominator, so makes no difference.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/627026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
How would you explain to a 9th grader the negative exponent rule? Let us assume that the students haven't been exposed to these two rules: $a^{x+y} = a^{x}a^{y}$ and $\frac{a^x}{a^y} = a^{x-y}$. They have just been introduced to the generalization: $a^{-x} = \frac{1}{a^x}$ from the pattern method: $2^2 = 4, 2^1 = 2, 2^0 = 1, 2^{-1} = \frac{1}{2}$ etc. However, some students confuse $2^{-3}$ to be $(-2)(-2)(-2)$ since they are familiar with $2^{3} = 2 \cdot 2 \cdot 2$. This is a low-income urban school and most kids in this algebra class struggle with math dealing with exponents, fractions and decimals. What would be the best approach to reach all 32 students?
| Multiplication of exponents:
$4^2 \times 4^3 = 4^5$
The powers are added: $2 + 3 = 5$
Division of exponents:
$\dfrac{4^2}{4^3}\ = 4^{-1}$
The powers are subtracted: $2 - 3 = -1$
Now tell them to try and find the value of $4^{-1}$
Since $4^2 = 16$, and $4^3 = 64$
$\dfrac{16}{64}\ = 1/4 = 4^{-1}$
This is the way I would have wanted to have been taught it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/629740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "61",
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"answer_id": 12
} |
How to prove the inequality: $\frac{(1+x)^2}{2x^2+(1-x)^2}+\frac{(1+y)^2}{2y^2+(1-y)^2}+\frac{(1+z)^2}{2z^2+(1-z)^2}\leq 8$ Prove that:
$$\frac{(1+x)^2}{2x^2+(1-x)^2}+\frac{(1+y)^2}{2y^2+(1-y)^2}+\frac{(1+z)^2}{2z^2+(1-z)^2}\leq 8$$
subject to the constraints:
$$x,y,z >0$$
and
$$x+y+z=1.$$
| if $a+b+c=1,a,b,c>0$
$$\Longleftrightarrow \sum_{cyc}\dfrac{a^2+2a+1}{3a^2-2a+1}\le 8$$
since
$$\dfrac{a^2+2a+1}{3a^2-2a+1}\le 4a+\dfrac{4}{3}$$
this is true becasuse
$$\Longleftrightarrow 3(4a+1)(a-\dfrac{1}{3})^2\ge 0$$
so
$$\sum_{cyc}\dfrac{(a+1)^2}{2a^2+(1-a)^2}\le 4(a+b+c)+4=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/629811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the integral: $ \int \frac{2x}{\sqrt {4x-1}}\, \mathrm{d}x\;?$ My problem, is how to find the integral of $$ \int \frac{2x}{\sqrt {4x-1}}\, \mathrm{d}x$$
Can I do this?
$$ \int \frac{\frac{1}{2}\cdot4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x$$
$$ \frac{1}{2}\int \frac{4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x$$
$$ \frac{1}{2}\int \frac{4x-1}{\sqrt {4x-1}}+\frac{1}{\sqrt {4x-1}}\, \mathrm{d}x$$
$$ \frac{1}{2}\int \sqrt {4x-1}+\frac{1}{\sqrt {4x-1}}\, \mathrm{d}x$$
then
$$ 4x-1= u/'$$
$$ 4dx= du$$
$$ dx= \frac{1}{4}du$$
insert that in the integral
$$ \frac{1}{2}\int \sqrt {u}+\frac{1}{\sqrt {u}}\cdot\frac{1}{4}\, \mathrm{d}u$$
$$ \frac{1}{8}\int \sqrt {u}+\frac{1}{\sqrt {u}}\, \mathrm{d}u$$
and I get
$$\frac{1}{12}\cdot u\cdot \sqrt{u}+ \frac{1}{4}\cdot \sqrt{u}+C$$
$$\frac{1}{12}\cdot(4x-1)\cdot \sqrt{4x-1}+ \frac{1}{4}\cdot \sqrt{4x-1}+C$$
and that's the correct solution. But can I just rewrite $2x$ as $\frac{1}{2}\cdot4x-1+1$ and then put $\frac{1}{2}$ before the integral? Thank you in advance!
| You've got the right idea, but there are two places you erred:
$$ \int \frac{\frac{1}{2}\cdot4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x = \int \frac {2x - 1 + 1}{\sqrt{4x - 1}}\,dx \neq \frac{1}{2}\int \frac{4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x$$
You need to have: $$\int \dfrac{2x}{\sqrt{4x-1}}\,dx = \int \frac{\frac{1}2(4x-1) +\frac 12}{\sqrt {4x-1}}\, \mathrm{d}x = \frac 12\int \frac{(4x - 1) + 1}{\sqrt{4x-1}}$$
Here you also erred: $$ \frac{1}{2}\int\sqrt {u}+\frac{1}{\sqrt {u}}\cdot\frac{1}{4}\, \mathrm{d}u = \int \left(\frac 12 \sqrt u + \frac 1{8\sqrt u}\right)\,du \neq \frac{1}{8}\int \sqrt {u}+\frac{1}{\sqrt {u}}\, \mathrm{d}u $$
If you got the correct answer, it's because the second error reversed the first error.
If you go back and fix the first error, and work from there with the same idea (just be careful to note the scope of a a factor, and to distribute the factor, when needed!) you should obtain the correct result, and do so correctly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/630684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to calculate limit without L'Hopital How can I calculate limit without using L'Hopital rule?
$\displaystyle\lim_{x\to 0 }\frac{e^{\arctan(x)}-e^{\arcsin(x)}}{1-\cos^3(x)}$.
| Let $\arctan x = t$ so that $x = \tan t$ so that $\sin t = x/\sqrt{1 + x^{2}}$ or $\arctan x = t = \arcsin (x/\sqrt{1 + x^{2}})$.
Now we have
$\displaystyle\begin{aligned}\arctan x - \arcsin x &= \arcsin (x/\sqrt{1 + x^{2}}) - \arcsin x\\
&= \arcsin\left(x\sqrt{\frac{1 - x^{2}}{1 + x^{2}}} - \frac{x}{\sqrt{1 + x^{2}}}\right)\\
&= \arcsin\left\{\frac{x}{\sqrt{1 + x^{2}}}\left(\sqrt{1 - x^{2}} - 1\right)\right\}\\
&= \arcsin\left\{\frac{-x^{3}}{\sqrt{1 + x^{2}}\left(\sqrt{1 - x^{2}} + 1\right)}\right\}\\
&= \arcsin y = f(x) \text{ (say)}\end{aligned}$
Next we have
$\displaystyle\begin{aligned}\lim_{x \to 0}\frac{e^{\arctan x} - e^{\arcsin x}}{1 - \cos^{3}x} &= \lim_{x \to 0}e^{\arcsin x}\cdot\frac{e^{f(x)} - 1}{1 - \cos^{3} x}\\
&= \lim_{x \to 0}1\cdot\frac{e^{f(x)} - 1}{f(x)}\cdot\frac{f(x)}{1 - \cos^{3}x}\\
&= \lim_{x \to 0}1\cdot 1\cdot\frac{f(x)}{(1 - \cos x)(1 + \cos x + \cos^{2}x)}\\
&= \frac{1}{3}\lim_{x \to 0}\frac{f(x)}{1 - \cos x}\\
&= \frac{1}{3}\lim_{x \to 0}\frac{\arcsin y}{y}\cdot \frac{y}{1 - \cos x}\\
&= \frac{1}{3}\lim_{x \to 0}\frac{y}{2\sin^{2}(x/2)}\\
&= \frac{1}{6}\lim_{x \to 0}\frac{y}{(x/2)^{2}}\cdot \frac{(x/2)^{2}}{\sin^{2}(x/2)}\\
&= \frac{1}{6}\lim_{x \to 0}\frac{y}{(x/2)^{2}}\cdot 1\\
&= \frac{2}{3}\lim_{x \to 0}\frac{y}{x^{2}}\\
&= \frac{2}{3}\lim_{x \to 0}\frac{-x}{\sqrt{1 + x^{2}}\left(\sqrt{1 - x^{2}} + 1\right)}\\
&= \frac{2}{3}\cdot 0 = 0\end{aligned}$
No Taylor or L'Hospital is required. We just need algebraic and trigonometric manipulation combined with the use of standard limits.
| {
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"url": "https://math.stackexchange.com/questions/632993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit of two sequences
*
*$ \lim_{n\to \infty} \sqrt[n]{3^n+4^n} $ . I think the limit is $4$. I did :
$ \sqrt[n]{3^n+4^n} = 4 \sqrt[n]{(\frac{3}{4}) ^n+1}$ .Am I right?
*$ \lim_{n\to \infty} \frac{1}{1\cdot 4 } + \frac{1}{4\cdot 7} +...+\frac{1}{(3n-2)(3n+1)} $. I know that for each $k$ , this sequence is the sum of $k$ terms, the smallest one is $ \frac{1}{(3n-2)(3n+1)} $, and the largest is $ \frac{1}{1. 4 }$ . But when substituting and trying to use the squeeze thm, I get that the limit should be between $0$ and $\infty$, which gives me nothing.
Thanks in advance.
| Hint:
For the second one use this fact that :
$$\frac{1}{(3n+1)(3n-2)}=\frac{-1}{3(3n+1)}+\frac{1}{3(3n-2)}$$ So the sum can be reduced to this one: $$S_n=\frac{1}3\left(\frac{1}{1}+\frac{1}{(3n+1)(3n-2)}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the minimum value of $x^2+y^2$, where $x,y$ are non-negative integers and $x+y$ is a given positive odd integer Let $k$ be a fixed positive odd integer. Find the minimum value of $x^2+y^2$, where $x,y$ are non-negative integers and $x+y=k$
My approaches:
*
*Since $k$ is odd, $x$ and $y$ have different parity. I consider, $k=2m+1$, so that $x+y=2m+1$.
I also consider $x$ to be even and $y$ to be odd.
So, $2p+2q+1=2m+1$.
Also, $4p^2+4q^2+4q+1+8pq+4p=4m^2+4m+1$ which apparently doesn't lead me anywhere.
*$x+y=k$. Then
$x^2+y^2+2xy=k^2$.
Now I am stuck!
Please help!
| Assume wlog that $x < y$ (they cannot be equal since $k$ is odd). Let's compare $x^2 + y^2$ to $(x + 1)^2 + (y-1)^2$.
$$
(x + 1)^2 + (y-1)^2 - (x^2 + y^2) = 2x -2y+ 2
$$
Since $x < y$, the result is non-positive, and as long as $x \neq y-1$, it is strictly negative, mening the net value has decreased. So as long as $x \neq y - 1$, you can do strictly better by increasing $x$ and decreasing $y$. Therefore $x = y - 1$ (or $x = y + 1$) is the optimal value for minimizing the sum of the squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/633675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Are there real numbers such that ... $x^2 + xy =3$ and $x - y^2 = 2$?
I graphed it and there are no intersections so obviously there is no real number solutions, but is there a "mathier" (read algebraic) way to prove this?
| From the second equation, $x = 2 + y^2$. Substituting this into the first equation gives
$$3 = x^2 + xy = (2 + y^2)^2 + (2 + y^2) y = (2 + y^2)(2 + y^2 + y) $$
Now $2 + y^2 \geq 2$ and $$y^2 + y + 2 = \left(y + \frac 1 2\right)^2 +\frac7 4 \geq \frac 74$$
So the right hand side of the above equation is at least
$$2 \cdot \frac 7 4 = 3.5 \gt 3$$ So there's no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/636456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to prove that a $3 \times 3$ Magic Square must have $5$ in its middle cell? A Magic Square of order $n$ is an arrangement of $n^2$ numbers, usually distinct integers, in a square, such that the $n$ numbers in all rows, all columns, and both diagonals sum to the same constant.
How to prove that a normal $3\times 3$ magic square must have $5$ in its middle cell?
I have tried taking $a,b,c,d,e,f,g,h,i$ and solving equations to calculate $e$ but there are so many equations that I could not manage to solve them.
| Let
\begin{array}{|c|c|c|}
\hline
a & b & c\\
\hline
d & e & f \\
\hline
g & h & i \\
\hline
\end{array}
be a magic square of order three with sum row, column, and diagonal sums equal to S. then
\begin{align}
a+e+i &= S\\
b+e+h &= S\\
c+e+g &= S\\
d+e+f &= S \\
\hline
(a+b+c)+(d+e+f)+(g+h+i) + 3e &= 4S\\
3S + 3e = 4S\\
e = \dfrac 13 S
\end{align}
So, if the top row is $a,b,c$, then $e = \dfrac{a+b+c}{3}$.
We end up with
\begin{array}{|c|c|c|}
\hline
a & b & c\\
\hline
\dfrac{-2a+b+4c}{3} & \dfrac{a+b+c}{3} & \dfrac{4a+b-2c}{3} \\
\hline
\dfrac{2a+2b-c}{3} & \dfrac{2a-b+2c}{3} & \dfrac{-a+2b+2c}{3} \\
\hline
\end{array}
This is OK, but it would be nice to have a representation that doesn't use fractions. This works.
\begin{array}{|c|c|c|}
\hline
a & b & -a-b+3e\\
\hline
-2a-b+4e & e & 2a+b-e \\
\hline
a+b-e & -b+2e & -a+2e \\
\hline
\end{array}
There are other representations.
For programmers
Every order-3 magic square can be rotated and reflected so that the smallest element is in the center of the left column and the next-smallest element in the bottom of the right column. Having made this transformation, and ordered the elements from 1 to 9, there are two possible outcomes for the 'arrays of orders'.
$$\text{Type 1: } \;
\begin{bmatrix}
8 & 3 & 4\\
1 & 5 & 9 \\
6 & 7 & 2 \\
\end{bmatrix}
\qquad
\text{Type 0: } \;
\begin{bmatrix}
8 & 4 & 3\\
1 & 5 & 9 \\
7 & 6 & 2 \\
\end{bmatrix}$$
Note that the first array of orders is in fact also a magic square while the second isn't. An example of a magic square of type $0$ is
$$\begin{bmatrix}
8 & 4 & 3\\
0 & 5 & 10 \\
7 & 6 & 2 \\
\end{bmatrix}$$
Using
$\quad \begin{bmatrix}
a & b & c\\
d & e & f \\
g & h & i \\
\end{bmatrix} \quad$
to reference the elements of a magic square, it would be tempting, noticing the locations of the elements of orders $1,2$ and $3$ to hope that elements $\{d,i,b\}$ or elements $\{d,i,c\}$ could be used to generate the other six elements of the magic square. Unfortunately, elements $\{d,i,b\}$ do not since, for all magic squares $b+d = 2i$. But, choosing elements $\{d,i,c\}$, we get
\begin{array}{|c|c|c|}
\hline
2c-2d+i & -d+2i & c\\
\hline
d & c-d+i & 2c-3d+2i \\
\hline
c-2d+2i & 2c-d & i \\
\hline
\end{array}
So, to generate positive integer magic squares with distinct elements, choose integer $0 < d < i < c$, with $2i-d > 0$ and use the above formula to fill in the rest.
| {
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Integration method for $\int_0^\infty\frac{x}{(e^x-1)(x^2+(2\pi)^2)^2}dx=\frac{1}{96} - \frac{3}{32\pi^2}.$ The following definite integral is obtained directly from Hermite's integral representation of the Hurwitz zeta function. But is it possible to obtain the same result via the residue calculus or another technique?
$$\int_{0}^{\infty}
{x
\over
\left({\rm e}^{x} - 1\right)\left[x^{2} + \left(2\pi\right)^{2}\right]^{2}}
\,{\rm d}x
={1 \over 96} - {3 \over 32\pi^{2}}.
$$
| This integral may be done via the residue theorem, by considering the integral
$$\oint_C dz \frac{z \log{z}}{(e^z-1)(z^2+4 \pi^2)^2}$$
where $C$ is a keyhole contour about the positive real axis, having an outer circle of radius $R$ and inner circle of radius $\epsilon$, each centered about the origin. One may show that the integrals about each of the circular arcs vanish as $R\to\infty$ and $\epsilon \to 0$. In this case, then, by the residue theorem, we have
$$\int_0^{\infty} dx \frac{x}{(e^x-1) (x^2+4 \pi^2)^2} = -\sum_{k \ne 0, k \in \mathbb{Z}} \operatorname*{Res}_{z=i 2 \pi k} \frac{z \log{z}}{(e^z-1)(z^2+4 \pi^2)^2}$$
When $k \ne 1$, the poles are simple. Noting that
$$\lim_{z\to i 2 \pi k} \frac{z-i 2 \pi k}{e^z-1}=1$$
we have that, when $k \gt 1$
$$-\operatorname*{Res}_{z=i 2 \pi k} \frac{z \log{z}}{(e^z-1)(z^2+4 \pi^2)^2} = \frac{k}{16 \pi^2 (k^2-1)^2}- i \frac{k \log{(2 \pi k)}}{8 \pi^3 (k^2-1)^2}$$
When $k \lt 0$, however, we must be careful with the argument of $-i$ as defined by the keyhole contour we are using; in this case, $\arg{(-i)} = 3 \pi/2$ and not $-\pi/2$, and therefore, when $k \lt -1$
$$-\operatorname*{Res}_{z=-i 2 \pi k} \frac{z \log{z}}{(e^z-1)(z^2+4 \pi^2)^2} = -\frac{3 k}{16 \pi^2 (k^2-1)^2}+ i \frac{k \log{(2 \pi k)}}{8 \pi^3 (k^2-1)^2}$$
Thus the sum of the residues when $k \ne \pm 1$ is
$$-\frac1{8 \pi^2} \sum_{k=2}^{\infty} \frac{k}{(k^2-1)^2} $$
The sum may be evaluated by partial fractions:
$$\sum_{k=2}^{\infty} \frac{k}{(k^2-1)^2} = \sum_{k=1}^{\infty} \frac{k+1}{((k+1)^2-1)^2} = \sum_{k=1}^{\infty}\frac14 \left [\frac1{k^2}-\frac1{(k+2)^2} \right ] = \frac{5}{16}$$
Putting this together gives the sum of the residues for $|k|\gt 1$ as $-5/(128 \pi^2)$.
For $|k|=1$, however, we have a triple pole. The calculation in this case is far uglier and I will spare you the details:
$$\begin{align}-\operatorname*{Res}_{z=i 2 \pi} \frac{z \log{z}}{(e^z-1)(z^2+4 \pi^2)^2} &= -\frac12 \left [\frac{d^2}{dz^2}\frac{z (z-i 2 \pi) \log{z}}{(e^z-1)(z+i 2 \pi)^2} \right ]_{z=i 2 \pi}\\ &= -\frac1{192}-\frac{9}{256 \pi^2} -i \frac{(3+4 \pi^2) \log{(2 \pi)}}{384 \pi^3} \end{align}$$
Similarly, and keeping in mind the argument of $-i$ as above, we have
$$\begin{align}-\operatorname*{Res}_{z=-i 2 \pi} \frac{z \log{z}}{(e^z-1)(z^2+4 \pi^2)^2} &= -\frac12 \left [\frac{d^2}{dz^2}\frac{z (z+i 2 \pi) \log{z}}{(e^z-1)(z-i 2 \pi)^2} \right ]_{z=-i 2 \pi}\\ &= \frac{3}{192}-\frac{5}{256 \pi^2} +i \frac{(3+4 \pi^2) \log{(2 \pi)}}{384 \pi^3} \end{align}$$
We may put this all together, and we finally have for the integral
$$\int_0^{\infty} dx \frac{x}{(e^x-1) (x^2+4 \pi^2)^2} = \frac1{96}-\frac{7}{128 \pi^2} - \frac{5}{128 \pi^2} = \frac1{96}-\frac{3}{32 \pi^2}$$
as was to be shown.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
My work:
$(\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b})=\frac{1}{a+b}$
By squaring both sides, we get,
$\frac{\sin^8 x}{a^2}+\frac{\cos^8 x}{b^2}+2\frac{\sin^4 x \cos^4 x}{ab}=\frac{1}{(a+b)^2}$
$\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
So, now, we have to prove that,
$-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=0$
I cannot do this. Please help!
| $\cos^4x=(\cos^2x)^2=(1-\sin^2x)^2=1+ \sin^4x-2\sin^2x$
Now,
$\dfrac{\sin^4x}{a} +\dfrac{cos^4x}{b}=\dfrac{1}{a+b}$
$\Rightarrow \dfrac{\sin^4x}{a} + \dfrac{(1+\sin^4x-2\sin^2x)}{b} = \dfrac{1}{a+b}$
$\dfrac{[b\cdot\sin^4x + a(\sin^4x-2\sin^2x+1)]}{ab} = \dfrac{1}{a+b}$
$\Rightarrow \dfrac{[(a+b)\sin^4x-2a \sin^2x+a]}{ab} = \dfrac{1}{a+b}$
$\Rightarrow (a+b)^2 \sin^4x - 2a(a+b)\sin^2x + a(a+b) =ab$
$\Rightarrow (a+b)^2 \sin^4x - 2a(a+b)\sin^2x + a^2=0$
$\Rightarrow [(a+b)\sin^2x-a]^2 = 0$
$\Rightarrow (a+b)\sin^2x - a = 0$
$\Rightarrow \sin^2x=\dfrac{a}{(a+b)}$
Taking third power on both side
$\sin^6x=\dfrac{a^3}{(a+b)^3}$
Dividing by both sides by $a^2$
$\dfrac{\sin^6x}{a^2}=\dfrac{a}{(a+b)^3}$
Now,
$\cos^2x=1 - \sin^2x=1-\dfrac{a}{(a+b)}=\dfrac{b}{(a+b)}$
$\Rightarrow \cos^2x=\dfrac{b}{(a+b)}$
Taking third power of both side
$\cos^6x=\dfrac{b^3}{(a+b)^3}$
Dividing both sides by $b^2$,
$\dfrac{\cos^6x}{b^2} =\dfrac{b}{(a+b)^3}$
$\Rightarrow \dfrac{\sin^6x}{a^2} + \dfrac{\cos^6x}{b^2}=\dfrac{a}{(a+b)^3} + \dfrac{b}{(a+b)^3} =\dfrac{(a+b)}{(a+b)^3}=\dfrac{1}{(a+b)^2}$
Hence proved.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Show that $\forall p\in \mathbb{R}_3[X]$ we have $p(a) + p(b) + 4p(\frac{a + b}{2}) = \frac{6}{b - a}\int_{a}^{b} p(t)dt$ Let $a, b$ and $c$ three reals distinct.
*
*Let $P(X) = (X-a)(X-b)(X-c)$
determine a necessery and suffecient condition for $$\int_{a}^{b} p(t)dt = 0$$
*Show that $\forall p\in \mathbb{R}_3[X]$ we have $$p(a) + p(b) + 4p(\frac{a + b}{2}) = \frac{6}{b - a}\int_{a}^{b} p(t)dt$$
My work :
I consedered the following map $\Phi$ from $\mathbb{R}_3[X]$ to $\mathbb{R}$ such that $\Phi (P) = \int_{a}^{b} p(t)dt$ and the necessery and suffecient condition is $P \in Ker(\Phi)$ is that correct?
but for 2. Itried to find the matrix of $\Phi$ in the canonical base of $\mathbb{R}_3[X]$ but I find myself nowhere. please help me?
| *
*Using 2. for this particular $p(X)=(X-a)(X-b)(X-c)$, as $p(a)=p(b)=0$, a necessary and sufficient condition for the integral to vanish would be
$$4p\left(\frac{a+b}2\right)=0$$
that is, $\displaystyle\frac{a+b}2$ would be also a root of $p$, but that means exactly $c=\displaystyle\frac{a+b}2$.
*
The matrix of $\Phi$ has only one row as it maps to one dimension, and the $i$th element of the row is $p(X^i)$ (for $i=0..3$), so it is
$$[\Phi]\ =\ \pmatrix{b-a&\frac{b^2-a^2}2& \frac{b^3-a^3}3 &\frac{b^4-a^4}4 } $$
So, for a $p(X)=u_0+u_1X+u_2X^2+u_3X^3$, we have
$$ \Phi(p)=[\Phi]\cdot\pmatrix{u_0\\u_1\\u_2\\u_3}\ = \quad\quad (1)\\
=\ (b-a)\,\left(
u_0+\frac{b+a}2 u_1+\frac{b^2+ab+a^2}3u_2+\frac{b^3+ab^2+a^2b+a^3}4u_3\right)$$
And, $p(a)+p(b)=2u_0+u_1(a+b)+u_2(a^2+b^2)+u_3(a^3+b^3)$ and
$$4p\left(\frac{a+b}2\right)=4u_0+2u_1(a+b)+u_2(a+b)^2+u_3\frac{(a+b)^3}2\,. $$
Adding these, we get $6u_0+3u_1(a+b)+u_2(2a^2+2ab+2b^2)+\displaystyle\frac{u_3}2\left(
3a^3+3a^2b+3ab^2+3b^3\right)$.
Divide it by $6$ to arrive at (1).
| {
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Hints on evaluating this $\frac{3}{2\pi}\int_0^{2\pi}\frac{e^{-ikx}}{5 - 4\cos(x)} \mathrm dx$? I have the following integral I'm trying to solve:
$$\frac{3}{2\pi}\int_0^{2\pi}\frac{e^{-ikx}}{5 - 4\cos(x)} \mathrm dx, \quad k \in \mathbb{Z}.$$
I've tried writing the exponential in terms of sines and cosines and then using the usual rational trig function substitutions, i.e. rewriting the integrand as
$$\frac{\cos(kx)}{5 - 4\cos(x)} - \frac{i\sin(kx)}{5 - 4\cos(x)}$$
and then using the substitution $t = \tan{x/2}$, but this doesn't result in anything useful that I could come up because of the different fequencies in the trig functions. I also tried writing the cosine in terms of complex exponentials, but didn't end up with anything useful either.
If the only solution involves residue calculus then please let me know, as I'm not familiar with that subject so I haven't tried anything like that. Thanks!
UPDATE: As was suggested I wrote the $\frac{1}{5 - 4\cos(x)}$ term as a geometric series $\frac{1}{5}\sum\limits_{n = 0}^\infty (\frac{4}{5}\cos(x))^n$, so now I am trying to evaluate
$$\int_0^{2\pi} \cos^n(x) e^{-ikx} dx,$$
but am stuck here again. I am trying to do this without using the residue theorem, but if that's the only way then so be it!
| For $k\ge 1$ put $z=e^{ix}$ to obtain
$$\frac{3}{2\pi} \int_{|z|=1} \frac{1/z^k}{5-2z-2/z} \frac{1}{iz} dz
= 3\times \frac{1}{2\pi i} \int_{|z|=1} \frac{1/z^k}{5z-2z^2-2} dz
\\= 3\times \frac{1}{2\pi i} \int_{|z|=1} \frac{1/z^k}{(1-2z)(z-2)} dz
= -\frac{3}{2}\times \frac{1}{2\pi i} \int_{|z|=1} \frac{1/z^k}{(z-1/2)(z-2)} dz.$$
There are two poles inside the unit circle, one at $z=1/2$
with residue $2^k\times 1/(-3/2).$
For the residue at the other pole which is at $z=0$ observe that
$$\frac{1}{(z-1/2)(z-2)} = \frac{4}{3}\frac{1}{1-2z} - \frac{1}{3}\frac{1}{1-z/2}.$$
This gives the residue $$\frac{4}{3} 2^{k-1} -\frac{1}{3} \frac{1}{2^{k-1}}
= \frac{2}{3} 2^k -\frac{2}{3} \frac{1}{2^k}.$$
The first term cancels with the residue at $z=1/2$ and the second term is the only contribution for a final answer of
$$-\frac{3}{2} \times \left( -\frac{2}{3} \frac{1}{2^k} \right)= \frac{1}{2^k}.$$
| {
"language": "en",
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Finding amplitude of oscillation So I know how to find the amplitude of oscillation. It's just the coefficient of the trig function. In general it would be $$ x = A\sin\left(\omega t +\phi_0\right) $$ and $A$ would be the amplitude. But my problem is asking for the amplitude of $$ x=-\cos(t)+3\sin\left(t-\frac{\pi}{6}\right) $$I don't exactly know what to do if there are 2 trig functions. The amplitude is $3.606$ but I don't know how to get that.
| Expanding the explanations given in the comments by Artem and André Nicolas to the question.
We can always find constants $A,B$ and $C$ such that the equation
\begin{equation*}
x(t)=A\cos \omega t+B\sin \omega t
\end{equation*}
is identical to the equation
\begin{equation*}
x(t)=C\sin \left( \omega t+\phi _{0}\right) =\left( C\sin \phi _{0}\right)
\cos \omega t+\left( C\cos \phi _{0}\right) \sin \omega t.
\end{equation*}
To expand $C\sin \left( \omega t+\phi _{0}\right) $ we applied the
trigonometric identity
\begin{equation*}
\sin (a+b)=\sin a\cos b+\cos a\sin b.
\end{equation*}
Equating the coefficients of $\cos \omega t$ and $\sin \omega t$ gives
\begin{equation*}
A=C\sin \phi _{0},\qquad B=C\cos \phi _{0},
\end{equation*}
while squaring and adding these last equations gives
\begin{equation*}
C=\sqrt{A^{2}+B^{2}}\geq 0.
\end{equation*}
Dividing one by the other yields
\begin{equation*}
\tan \phi =\frac{A}{B}.
\end{equation*}
In this case we have that
\begin{equation*}
x(t)=C\sin \left( \omega t+\phi _{0}\right) =-\cos t+3\sin (t-\frac{\pi }{6}
)=-\frac{5}{2}\cos t+\frac{3\sqrt{3}}{2}\sin t.
\end{equation*}
We changed the notation of the amplitude of the wave $x(t)$ to $C$
instead of $A\ $as in the question. We see that $\omega =1$. The expansion
of $\sin (t-\frac{\pi }{6})$ follows from the identity
\begin{equation*}
\sin (a-b)=\sin a\cos b-\cos a\sin b
\end{equation*}
and the trigonometric values
\begin{equation*}
\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2},\quad \sin \frac{\pi }{6}=\frac{1}{2}.
\end{equation*}
From the numeric values
\begin{equation*}
A=-\frac{5}{2},\qquad B=\frac{3\sqrt{3}}{2},
\end{equation*}
we find that the amplitude is
\begin{equation*}
C=\sqrt{A^{2}+B^{2}}=\sqrt{13}\approx 3. 6056,
\end{equation*}
the same value of yours. If we wanted to compute the phase angle $\phi _{0}\ $we would get
\begin{eqnarray*}
\tan \phi _{0} &=&\frac{A}{B}=\frac{-5\sqrt{3}}{9}, \\
\phi _{0} &=&\arctan \frac{-5\sqrt{3}}{9}\approx -0.766\,16\text{ }\mathrm{
rad}\approx -43.898{{}^\circ}.
\end{eqnarray*}
Graph of
\begin{equation*}
x(t)=\sqrt{13}\sin \left( t-0.766\,16\right)
\end{equation*}
| {
"language": "en",
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"answer_id": 0
} |
Determine if $\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$ converges or diverges. Another series I found I'm struggling with.
Determine if the following series converges or diverges.$$\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$$
Ratio test and n-th root test are both inconclusive, Leibniz - criterion cannot be applied since the sequence given is not in the form of $(-1)^na_n$. I am sure the problem can be solved with the limit comparison test, though, the $n^{(...)}$ look pretty inviting after all. Let $a_n:=\frac{(-1)^nn^2+n}{n^3+1}$ then $|a_n|= \frac{n^2+n}{n^3+1}$. A try showing that the series diverges using the divergence of $\sum\frac{1}{n}$. For $n≥1$ it is clear that
$$ \frac{n^3+n^2}{n^3+1} ≥ 1 $$ dividing by $n$ yields:
$$ \frac{n^2+n}{n^3+1} = |a_n| ≥ \frac{1}{n}$$
Thus the series diverges. (?)
EDIT: Hints you gave me yield:
$$\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1} = \sum_{n=1}^{\infty}(-1)^n\frac{n^2}{n^3+1} +
\sum_{n=1}^{\infty}\frac{n}{n^3+1}$$ With the first series converging by Leibniz-theorem and the second by limit comparison test with $\frac{1}{n^2}$.
| Here is a different approach.
The sum of each pair of terms is
$$
\begin{align}
a_{2n-1}+a_{2n}
&=\frac{-(2n-1)^2+(2n-1)}{(2n-1)^3+1}+\frac{(2n)^2+2n}{(2n)^3+1}\\
&=\frac{4n^3-6n^2+5n-1}{n(4n^2-2n+1)(4n^2-6n+3)}\\
&=\frac1{4n^2}\frac{1-\frac3{2n}+\frac5{4n^2}-\frac1{4n^3}}{\left(1-\frac1{2n}+\frac1{4n^2}\right)\left(1-\frac3{2n}+\frac3{4n^2}\right)}\\
&=\frac1{4n^2}+O\left(\frac1{n^3}\right)
\end{align}
$$
Thus, the sum of $a_{2n-1}+a_{2n}$ converges by comparison to $\frac1{n^2}$.
Since the terms go to $0$, convergence of the partial sums is the same as convergence of every other partial sum.
Let $s_n=\sum\limits_{k=1}^n a_k$ and $p_n=\sum\limits_{k=1}^n(a_{2k-1}+a_{2k})$. Then $s_{2n}=p_n$ and $s_{2n+1}=p_n+a_{2n+1}$
Since $\lim\limits_{n\to\infty}a_n=0$, if $p_n$ converges, then $s_n$ converges.
Since $p_n$ is a subsequence of $s_n$, if $s_n$ converges, then $p_n$ converges.
That is, if $\lim\limits_{n\to\infty}a_n=0$, then
$$
\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty(a_{2n-1}+a_{2n})
$$
| {
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"answer_count": 4,
"answer_id": 2
} |
Double integral -- tricky?
If $f(x,y) = x^2+y^2$ and $D=\{(x,y)\in\mathbb{R}^2:x^2+y^2\geq1, x^2+y^2-2x\leq0 \text{ and } y\geq0\}$, find $\displaystyle\int\displaystyle\int_D f$.
$D$ looks like the intersection between two circumferences, as I draw below:
Polar coordinates seems a obvious choice, but using them I'd have unwanted bits, this is, if I split the region in two sections (by $x=\frac{1}{2}$) after computing the red and blue region I would have twice $R1$ and $R2$, a problem that could be solved computing each one of them and then substract them of the final answer -this is, after adding the integral of the red and blue sections-.
I believe the answer would be given by
$$\displaystyle\int \displaystyle\int_D f = \displaystyle\int\displaystyle\int _{\text{RED}} f + \displaystyle\int\displaystyle\int _{\text{BLUE}} f - \displaystyle\int\displaystyle\int _{\text{R1}} f - \displaystyle\int\displaystyle\int _{\text{R2}} f$$
$R1$ and $R2$ could be described as follows:
$$R1 = \{(x,y):0\leq y\leq \sqrt{3}x,\;0\leq x\leq 1/2\}$$
$$R2 = \{(x,y):0\leq y\leq \sqrt{3}-\sqrt{3}x,\;1/2 \leq x\leq 1\}$$
Which gives
$$\displaystyle\int\displaystyle\int _{\text{R1}} f =\displaystyle\int_0^{1/2}\displaystyle\int_0^{\sqrt{3}x} (x^2+y^2)\; dy dx = 2\sqrt{3}\displaystyle\int_0^{1/2}x^3 = \displaystyle\frac{1}{52\sqrt{3}}.$$
$$\displaystyle\int\displaystyle\int_{R2} f = \displaystyle\int_{1/2}^1\displaystyle\int_0^{\sqrt{3}-\sqrt{3}x}(x^2+y^2)\; dydx \\ = \displaystyle\int_{1/2}^1(x^2(\sqrt{3}-\sqrt{3}x)dx + \frac{1}{3}\int_{\frac{1}{2}}^1 (\sqrt{3}-\sqrt{3}x)^3 dx \\ = \left[\sqrt{3}\displaystyle\int_{1/2}^1 x^2 -\sqrt{3}\displaystyle\int_{1/2}^1 x^3\right] + \frac{1}{3}\int_{\frac{1}{2}}^1 (\sqrt{3}-\sqrt{3}x)^3 dx \\ = \sqrt{3}\left[\left(\displaystyle\frac{1}{3}-\displaystyle\frac{1}{24}\right)+\left(\displaystyle\frac{1}{4}-\displaystyle\frac{1}{64}\right) \right] + \displaystyle\frac{1}{3}\left(\displaystyle\frac{1}{2}\right) \\= \frac{\sqrt{3}}{16}+\frac{1}{6} = \frac{1}{48}(3\sqrt{3}+8).$$
Now changing to polar coordinates $x=r\cos \theta$, $y= r\sin \theta$ to compute the red and blue:
$$\displaystyle\int\displaystyle\int _{\text{RED}} f = \displaystyle\int_0^{\pi/3}\displaystyle\int_0^1 r^3\;drd\theta = \frac{1}{4}\displaystyle\int_0^{\pi/3}d\theta = \frac{1}{12}\pi$$
Setting auxiliary axis $u = x-1, v = y$ I have $\displaystyle\frac{\partial (x,y)}{\partial (u,v)}=1$ and after with $u=r\cos \theta$, $v=r\sin \theta$
$$\displaystyle\int\displaystyle\int _{\text{BLUE}} f = \displaystyle\int_{2\pi/3}^{\pi}\displaystyle\int_0^1 r^3\;drd\theta = \frac{1}{12}\pi.$$
Finally
$$\displaystyle\int \displaystyle\int_D f = \frac{1}{6}\pi -\displaystyle\frac{1}{52\sqrt{3}} -\frac{1}{48}(3\sqrt{3}+8) .$$
Now I'd like to know if the solution is right, but fundamentally, can this integral be computed without splitting the $D$?
P.S: Legitimate question, but couldn't resist the pun.
| There is no need to split this integral into pieces. The inequality $x^2 + y^2 \geq 1$ is the same as $r \geq 1$, and the inequality $x^2 + y^2 - 2x \leq 0$ is the same as $r^2 - 2 r \cos \theta \leq 0$ or just $r \leq 2\cos \theta$. The circles $r = 1$ and $r = 2 \cos \theta$ intersect at $\theta = \frac{\pi}{3}$, so the integral is:
$$\int_0^{\pi/3} \int_1^{\cos \theta} r^2 \cdot r\,dr\,d\theta$$
| {
"language": "en",
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Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$
For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.
My attempt: Let
$$\begin{align*}
f_n(x)
&= \frac{\ln\left(1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\left(1-2a\cos x+a^2\right)}{\frac{1}{n}}\\
&=\frac{\ln\left(\displaystyle\frac{1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2}{1-2a\cos x+a^2}\right)}{\frac{1}{n}}\\
&=\frac{\ln\left(1+\dfrac{1}{n}\left(\displaystyle\frac{2a-2\cos x+\frac{1}{n}}{1-2a\cos x+a^2}\right)\right)}{\frac{1}{n}}.
\end{align*}$$
Now it is easy to see that $f_n(x) \to \frac{2a-2\cos x}{1-2a\cos x+a^2}$ as $n \to \infty$. $|f_n(x)|\le \frac{2a+2}{(1-a)^2}$ RHS is integrable so $\lim_{n\to\infty}\int_0^\pi f_n(x)dx = \int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2} dx=I'(a)$. But
$$\int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2}=\int_0^\pi\left(1-\frac{(1-a)^2}{1-2a\cos x+a^2}\right)dx.$$ Consider
$$\int_0^\pi\frac{dx}{1-2a\cos x+a^2}=\int_0^\infty\frac{\frac{dy}{1+t^2}}{1-2a\frac{1-t^2}{1+t^2}+a^2}=\int_0^\infty\frac{dt}{1+t^2-2a(1-t^2)+a^2(1+t^2)}=\int_0^\infty\frac{dt}{(1-a)^2+\left((1+a)t\right)^2}\stackrel{(*)}{=}\frac{1}{(1-a)^2}\int_0^\infty\frac{dt}{1+\left(\frac{1+a}{1-a}t\right)^2}=\frac{1}{(1-a)(1+a)}\int_0^\infty\frac{du}{1+u^2}=\frac{1}{(1-a)(1+a)}\frac{\pi}{2}.$$
So $$I'(a)=\frac{\pi}{2}\left(2-\frac{1-a}{1+a}\right)\Rightarrow I(a)=\frac{\pi}{2}\left(3a-2\ln\left(a+1\right)\right).$$
It looks too easy, is there any crucial lack?
$(*)$ — we have to check $a=1$ here by hand and actually consider $[0,1), (1,\infty)$ but result on these two intervals may differ only by constant - it may be important but in my opinion not crucial for this proof.
| If
$$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)\ \mathrm{d}x,$$ then
\begin{align}
I'(a)&=\int_{0}^{\pi}\frac{2a-2\cos x}{1-2a\cos x+a^2} \ \mathrm{d}x,\\
I'(a) & = \frac{1}{a}\int_{0}^{\pi}\frac{2a^2-2a\cos x}{1-2a\cos x+a^2} \ \mathrm{d}x \\
& = \frac{1}{a}\int_{0}^{\pi}\frac{1-1+a^2+a^2-2a\cos x}{1-2a\cos x+a^2} \ \mathrm{d}x \\
& = {\pi \over a} + \frac{1}{a}\int_{0}^{\pi}\frac{a^2-1}{1-2a\cos x+a^2} \ \mathrm{d}x
\end{align}
and making the Weierstrass substitution, $$\cos x = \frac{1 - t^2}{1 + t^2}, $$
$$\mathrm{d}x = \frac{2 \,\mathrm{d}t}{1 + t^2}.$$
$$I'(a)={\pi \over a} + \frac{2}{a}\int_{0}^{\infty}\frac{a^2-1}{(1+a^2)(1+t^2)-2a(1-t^2)}\ \mathrm{d}t $$ $$I'(a)={\pi \over a} + \frac{2}{a}\int_{0}^{\infty}\frac{a^2-1}{(1-a)^2 + (1+a)^2t^2} \ \mathrm{d}t$$
$$I'(a)={\pi \over a} + \frac{\pi}{a} \operatorname{sgn} (a^2-1),$$
so for $a > 1$,
$$I'(a)={2\pi \over a},$$
$$I(a)={2 \pi \log a},$$ given that $\displaystyle \lim_{a \rightarrow 1^+} I(a)=0$. You have to be careful to show this last statement I believe, but you can see the result here - this integral is easy to evaluate:
\begin{align}
\int_0^{\pi} \log(2-2\cos x) \ \mathrm{d}x &= \int_0^{\pi} \log(4 \sin^2 x) \ \mathrm{d}x \\
& = \pi \log 4 + 2\int_0^{\pi} \log(\sin x) \ \mathrm{d}x \\
& = 2\pi \log 2 + 4\int_0^{\pi/2} \log(\sin x) \ \mathrm{d}x \\
\end{align}
That final integral can be found here, which gives us the final result.
| {
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"answer_id": 5
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Equality of triangle inequality in complex numbers $z$ and $w$ be nonzero complex numbers. How do I show that $|z+w|=|z|+|w|$ if and only if $z=sw$ for some real positive number $s$.
I approached this by letting $z=a+ib$, and $w=c+id$, and kinda play around with it. I also tried to square both sides when proving forward direction, but I could not get it to work. Can anyone get me some ideas, maybe?
Thank you!
| First try to prove it in one direction.
Let $z = a + bi$ and $w = sa + sbi$. Then we have:
$$\mid z + w \mid = \mid a + bi + sa + sbi \mid = \mid |a(1+s) + (1+s)bi\mid = \sqrt{(1+s)^2a^2 + (1+s)^2b^2} = (1+s)\sqrt{a^2 + b^2} = \sqrt{a^2 + b^2} + \sqrt{s^2a^2 + s^2b^2} = \mid z \mid + \mid w \mid$$
Now for the other direction. Let $z=a + bi$ and $w = c + di$. Then we have:
$$\mid z + w \mid = \mid a + bi + c + di \mid = \mid (a+c) + (b+d)i \mid = \sqrt{(a+c)^2 + (b+d)^2}$$
$$\mid z \mid + \mid w \mid = \mid a + bi \mid + \mid c + di \mid = \sqrt{a^2 + b^2} + \sqrt{c^2 + d^2}$$
Now since $\mid z + w \mid = \mid z \mid + \mid w \mid$ we have:
$$\sqrt{a^2 + c^2 + 2ac + b^2 + d^2 + 2bd} = \sqrt{a^2 + b^2} + \sqrt{c^2 + d^2}$$
Square both sides and we have:
$$a^2 + c^2 + 2ac + b^2 + d^2 + 2bd = a^2 + b^2 + c^2 + d^2 + 2\sqrt{(a^2 + b^2)(c^2 + d^2)}$$
$$ac + bd = \sqrt{(a^2 + b^2)(c^2 + d^2)}$$
Now square again and multiply out:
$$a^2c^2 + b^2d^2 + 2abcd = a^2c^2 + b^2c^2 + a^2d^2 + b^2d^2$$
$$2abcd = b^2c^2 + a^2d^2$$
$$b^2c^2 + a^2d^2 - 2abcd = 0$$
$$(cb - ad)^2 = 0 \implies cb = ad$$
So let $c = sa$, then we have:
$$scb = sad \implies sb = d$$
Hence $z = a + bi$ and $w = c + di = as + sbi$. Q.E.D.
| {
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"question_score": "7",
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Partial Fraction Solution? This is the original fraction : $$\dfrac{8s^2}{(2s-3)^3}$$
From this I get :
$$4As^2 - 12As + 9A + 2Bs - 3B + C = 8s^2$$
On the solution it says $4A = 8, A=2$, okay I understand this, $4As^2 = 8s^2$, and the rest is cancelled in the equation since there is only one $s^2$, but how $B=12$ and $C = 18$, I don't understand, can somebody explain please?
| As alternative to expanding polynomial expression and equating coefficients as you have done, you can exploit the fact that two polynomials are equal iff they have the same values for all possible values of the variable. So starting with $8s^2 = A(2s-3)^2 + B(2s-3) + C$, we can choose values for $s$ and equate both sides. This can be a much faster method for partial fractions with lots of factors in the denominator. For your problem, I'd pick $s=\frac{3}{2}, 1, 2$ and evaluate them in that order. The reason for those numbers is that $\frac{3}{2}$ is the solution to the linear factors present so it makes for easy evaluation. The other two are chosen because they yield easy to work with results from $2s-3$.
$s=\frac{3}{2} \Longrightarrow 8(\frac{3}{2})^2 = 18 = A(0)^2 + B(0)^2 +C \Longleftrightarrow C = 18$
$s = 1 \Longrightarrow 8(1)^2 = A(-1)^2 + B(-1) + C \Longleftrightarrow 8 = A - B + 18 \Longleftrightarrow A - B = -10$
$s = 2 \Longrightarrow 8(2)^2 = A(1)^2 + B(1) + C \Longleftrightarrow 32 = A + B + 18 \Longleftrightarrow A + B = 14$
A very quick solve the system of equation generated from $s=1$ and $s=2$ yields that $A = 2$ and $B=12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/653362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can I calculate the maximum and minimum of this function?
This is incorrect and I have a feeling I am not doing this correctly.
| I understand your unease. The origin is a saddle point, as the Hessian can say:
$$H(0,0) = \begin{vmatrix} f_{xx} (0,0) & f_{xy} (0,0) \\ f_{xy} (0,0) & f_{yy} (0,0) \end{vmatrix} = \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} = -1.$$
As for nonzero points, I have continued from your system the following way: we can write
$$\lambda = \frac{x}{2y} \text{ and } \lambda = \frac{y}{18x},$$
therefore
$$\frac{x}{y} = \frac{y}{9x} \text{ and } 9x^2 =y^2, \text{ hence } 3x = \pm \, y.$$
Using this on the restriction $9x^2 +y^2 = 5$ yields
$$9x^2 + 9x^2 = 18x^2 = 5 \text{ and } x = \pm \frac{\sqrt{5}}{3 \sqrt{2}}.$$
In total we have four points:
$$
\begin{align}
(x_1, y_1) & = \frac{\sqrt{5}}{\sqrt{2}} \left( \frac{1}{3}, 1 \right), \\
(x_2, y_2) & = \frac{\sqrt{5}}{\sqrt{2}} \left( \frac{1}{3}, -1 \right), \\
(x_3, y_3) & = \frac{\sqrt{5}}{\sqrt{2}} \left( -\frac{1}{3}, -1 \right), \\
(x_4, y_4) & = \frac{\sqrt{5}}{\sqrt{2}} \left( -\frac{1}{3}, 1 \right).
\end{align}
$$
For them find
$$f(x_1,y_1) = f(x_3,y_3) = \frac{5}{6} \text{ and } f(x_2,y_2) = f(x_4,y_4) = - \frac{5}{6},$$
as respectively maximum and minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/655549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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is there a nicer way to $\int e^{2x} \sin x\, dx$? I have to solve this:
$\int e^{2x} \sin x\, dx$
I managed to do it like this:
let $\space u_1 = \sin x \space$ and let $\space \dfrac{dv}{dx}_1 = e^{2x}$
$\therefore \dfrac{du}{dx}_1 = \cos x \space $ and $\space v_1 = \frac{1}{2}e^{2x}$
If I substitute these values into the general equation:
$\int u\dfrac{dv}{dx}dx = uv - \int v \dfrac{du}{dx}dx$
I get:
$\int e^{2x} \sin x dx = \frac{1}{2}e^{2x}\sin x - \frac{1}{2}\int e^{2x}\cos x\, dx$
Now I once again do integration by parts and say:
let $u_2 = \cos x$ and let $\dfrac{dv}{dx}_2 = e^{2x}$
$\therefore \dfrac{du}{dx}_2 = -\sin x$ and $v_2 = \frac{1}{2}e^{2x}$
If I once again substitute these values into the general equation I get:
$\int e^{2x}\sin x dx =\frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x - \frac{1}{4}\int e^{2x}\sin x dx$
$\therefore \int e^{2x}\sin x dx = \frac{4}{5}(\frac{1}{2}e^{2x}\sin x -
\frac{1}{4}e^{2x}\cos x) + C$
$\therefore \int e^{2x}\sin x\, dx = \frac{2}{5}e^{2x}\sin x - \frac{1}{5}e^{2x}\cos x + C$
I was just wondering whether there was a nicer and more efficient way to solve this?
Thank you :)
| we have
$$
\int e^{2x}\sin x \mathrm d x = (A\cos x + B\sin x)e^{2x}
$$
Equation is the same at both ends of the derivative $x$
$$
e^{2x}\sin x = 2e^{2x}(A\cos x +B \sin x) + (B\cos x -A\sin x)e^{2x}
$$
Finishing
$$
e^{2x}\sin x =（2B-A)\sin x\,e^{2x} +(2A+B)\cos x\,e^{2x}
$$
Compare coefficient，we have
$$
2A+B = 0\\
2B-A = 1
$$
Solutions have
$$
A =-\frac{1}{5} \\ B=\frac{2}{5}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/657389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Proof sought for a sum involving binomials that simplifies to 1/2 A proof of:
$$\begin{align*}(1/2)^{2m+1} \sum_{k=0}^{m} \binom{m}{k} \sum_{j=0}^{k} \binom{m+1}{j} = \frac{1}{2} \end{align*} $$
Conjecture based on the following Maple code:
Q := (1/2)^(2*m+1) * sum( binomial(m, k) * sum(binomial(m+1, j), j = 0 .. k), k = 0 .. m):
simplify([seq(Q, m = 1 .. 20, 1)]);
[1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2]
| Suppose we seek to verify that
$$\sum_{k=0}^m {m\choose k}
\sum_{j=0}^k {m+1\choose j} = 2^{2m}.$$
We use the integral
$${m+1\choose j} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+1}}{z^{j+1}} \; dz.$$
This yields for the inner sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+1}}{z}
\sum_{j=0}^k \frac{1}{z^j}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+1}}{z}
\frac{1/z^{k+1}-1}{1/z-1}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{m+1}
\frac{1/z^{k+1}-1}{1-z}\; dz.$$
Now the second term in the difference does not have a pole at zero so
we are left with
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+1}}{z^{k+1}} \frac{1}{1-z} \; dz.$$
This yields for the remaining sum the integral
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+1}}{z} \frac{1}{1-z}
\sum_{k=0}^m {m\choose k} \frac{1}{z^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+1}}{z} \frac{1}{1-z}
\left(1+\frac{1}{z}\right)^m \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2m+1}}{z^{m+1}} \frac{1}{1-z} \; dz.$$
Extracting the residue we find
$$\sum_{q=0}^m {2m+1\choose q} =
\frac{1}{2} \times 2^{2m+1} = 2^{2m},$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/658502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
How to maximize $\left({a+b \choose a} 2^{-a-b}\right)$? How can you maximize $\left({a+b \choose a} 2^{-a-b}\right)$ assuming, $a,b \geq 0$ and $0< (a+b) \leq n$, where all the variables are non-negative integers? Is the maximum when $a=b=n/2$, assuming $n$ is even?
| The expression can be viewed as the probaility of having exactly $a$ heads in $a+b$ fair coin tosses, so intuitively, this should be maximized if $a=b$. Moreover, the more coin tosses we make, the less likely it will become to obtain exactly (rounded) half of them heads.
Formally, with $f(a,b)={a+b\choose a}2^{-a-b}$ we have
$$ \frac{f(a+1,b-1)}{f(a,b)}=\frac{a+b\choose a+1}{a+b\choose a}=\frac{a!b!}{(a+1)!(b-1)!}=\frac{b}{a+1}$$
so that the maximum is certainly not assumed at a pair $(a,b)$ with $b>a+1$. By symmetry, i.e. $f(a,b)=f(b,a)$, we can also exclude $a>b+1$. Thus we only have to maximize among $f(a,a)$ with $0<a\le\frac n2$ and $f(a,a+1)$ with $0\le a\le\frac{n-1}2$.
From the additive recrusion for binmoials,
$$ \begin{align}{2a+2\choose a+1}&={2a+1\choose a}+{2a+1\choose a+1}\\&={2a\choose a-1}+2{2a\choose a}+{2a\choose a+1}\\&=\left(\frac{a-1}{a}+2+ \frac{a-1}{a}\right){2a\choose a}\\&<4{2a\choose a}\end{align}$$
so that we see that the maximuum among all $f(a,a)$ is $f(1,1)=\frac12$.
From ${2a+2\choose a+1}=2{2a+1\choose a}$ we see that $f(a,a+1)=f(a+1,a+1)$, so the maximum of these values is $f(0,1)=f(1,1)=\frac12$.
To summarize: We have $f(0,1)=f(1,0)=f(1,1)=\frac12$ and $f(a,b)<\frac 12$ for all other choices of $a,b$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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radius of a sphere containing a circle Consider the circle $C$ which is the intersection of the sphere $x^2+y^2+z^2-x-y-z=0$ and the plane $x+y+z=1$. The radius of the sphere with centre at the origin, containing the circle $C$ is
a. $1$
b. $2$
c. $3$
d. $4$
| When we complete the squares in the equation for the sphere, we find
$$ (x^2 - x + \frac{1}{4}) + (y^2 - y + \frac{1}{4}) + (z^2 - z + \frac{1}{4}) \ = \frac{3}{4} $$
$$ \Rightarrow \ (x - \frac{1}{2})^2 \ + \ (y - \frac{1}{2})^2 \ + \ (z - \frac{1}{2})^2 \ = \frac{3}{4} \ \ . $$
An immediate way to find the circle would be to first consider the line from the origin through the center of the sphere, $ \ x = y = z = t \ . $ The center of the circle is also on this line: using the equation of the plane, we have $ \ t + t + t \ = \ 1 \ \Rightarrow \ t \ = \ \frac{1}{3} \ . $ The center of the circle is thus at $ \ (\frac{1}{3} , \frac{1}{3} , \frac{1}{3}) \ . $
Now, the distance from the center of the sphere to the center of this circle is $ \ s \ = \ \sqrt{(\frac{1}{2} - \frac{1}{3})^2 + (\frac{1}{2} - \frac{1}{3})^2 + (\frac{1}{2} - \frac{1}{3})^2} \ = \ \sqrt{\frac{1}{12}} \ . \ $ The radius of the circle is one leg of a right triangle with a hypotenuse equal to the radius of the sphere, $ \ R \ = \ \frac{\sqrt{3}}{2} \ $ and the other leg being $ \ s \ . \ $ So the radius of the circle is $ \ \sqrt{(\sqrt{\frac{3}{4}})^2 - (\sqrt{\frac{1}{12}})^2} \ = \ \sqrt{\frac{2}{3}} \ . \ $
There is also a right triangle formed by the radius of the circle, the segment from the origin to the center of the circle, and the radius of the origin-centered sphere, which is the hypotenuse. The center of the circle is at a distance $ \ \sqrt{( \frac{1}{3})^2 + (\frac{1}{3})^2 + (\frac{1}{3})^2} \ = \ \sqrt{\frac{1}{3}} \ $ from the origin. Thus the radius of our sphere is $ \ \sqrt{(\sqrt{\frac{1}{3}})^2 + (\sqrt{\frac{2}{3}})^2} \ = \ 1 \ . \ \ $ [choice (a)]
(RicardoCruz got his illustration posted first; here's another view.)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Pre-Calculus: Fractions with Exponents If $x = \frac{2}{3}$ and $y = \frac{1}{9}$ find the value of $\dfrac{x^3y^2}{xy^5}$
I've tried working it out multiple ways, but my answer is wayyyy off. I'm not sure how to deal with this problem.
Regards,
| We could simply, but to avoid this we can also just do the math. Since $x=\frac{2}{3}$ and $y=\frac{1}{9}$, we have
$$
x^3=\left(\frac{2}{3}\right)^3=\frac{8}{27}\text{ and }y^2=\left(\frac{1}{9}\right)^2=\frac{1}{81} \text{ and }y^5=\left(\frac{1}{9}\right)^5=\frac{1}{59049}
$$
then we have
$$
\frac{x^3y^2}{xy^5}=\frac{\frac{8}{27}\frac{1}{81}}{\frac{2}{3}\frac{1}{59049}}=\frac{\frac{8}{2187}}{\frac{2}{177147}}=\frac{8}{2}\cdot \frac{177147}{2187}=4\cdot 81=324
$$
While this works, simplfying works better first.
$$
\frac{x^3y^2}{xy^5}=x^3y^2(xy^5)^{-1}=x^3y^2x^{-1}y^{-5}=x^{3-1}y^{2-5}=x^2y^{-3}=\frac{x^2}{y^3}
$$
But since we have the values of $x$ and $y$, we have
$$
\frac{(2/3)^2}{(1/9)^3}=\frac{\frac{4}{9}}{\frac{1}{729}}=\frac{4}{9}\cdot \frac{729}{1}=4\cdot 81=324
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a non-diagonizable matrix that commutes with $\begin{bmatrix}0 & 0& -1\\1 & 1 &1\\0 & 0 & 1\end{bmatrix}$ I'm really lost on this one, with no clue on where to begin. Any help would be appreciated.
| Not a general approach, but one solution is easy to see. You have for example
$$\begin{pmatrix} 0&0&0 \\ 0&0&1\\0&0&0\end{pmatrix}\cdot
\begin{pmatrix}0&0&-1\\1&1&1\\0&0&1\end{pmatrix} = \begin{pmatrix} 0&0&0 \\ 0&0&1\\0&0&0\end{pmatrix}=\begin{pmatrix}0&0&-1\\1&1&1\\0&0&1\end{pmatrix}\cdot\begin{pmatrix} 0&0&0 \\ 0&0&1\\0&0&0\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/666413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$\lim\limits_{(x,y)\rightarrow(0,0)} \frac{7 \sin(2 x) x^2 y}{2 x^3 + 2 x y^2}$ Does the following limit exist?
$$\lim_{(x,y)\rightarrow(0,0)} \frac{7 \sin(2 x) x^2 y}{2 x^3 + 2 x y^2}$$
I've shown that it exist but i'm unsure about how to find the value of the limit.
| Note that
$$\frac{7 \sin(2 x) x^2 y}{2 x^3 + 2 x y^2}=7\cdot\frac{\sin(2x)}{2x}\cdot\frac{x^2y}{x^2+y^2}.$$
When $(x,y)\to (0,0)$, we have $x\to 0$ and $r=\sqrt{x^2+y^2}\to 0$, therefore, we have
$$\left|\frac{\sin(2x)}{2x}\right|\to 1\mbox{ as }x\to 0$$
and
$$\left|\frac{x^2y}{x^2+y^2}\right|=\left|\frac{r^3\cos^2\theta\sin\theta}{r^2}\right|\leq r\to 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/667306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how can I show that the sequence $x_{n}=\frac{n}{n^{2}+1}+\frac{n}{n^2+2}+...+\frac{n}{n^2+n},n=1,2,...$ converges? How can I check if the sequence $x_{n}=\frac{n}{n^{2}+1}+\frac{n}{n^2+2}+...+\frac{n}{n^2+n},n=1,2,...$ converges?Is there a theorem that I could use?
| Notice the following:
Every two integers $n,k$ such that $1\leq k \leq n$ satisfy
$$\frac {1}{n+1} = \frac{n}{n^2+n} \leq \frac{n}{n^2+k} \leq \frac{n}{n^2}=\frac{1}{n}$$
So we get
$$ \frac{1}{1+\frac{1}{n}} = n \cdot \frac {1}{n+1} \leq x_n \leq n \cdot \frac{1}{n} =1$$
Thus by the Sandwich Theorem we get $x_n\rightarrow 1$.
| {
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"url": "https://math.stackexchange.com/questions/669532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to resolve this integration $\int\frac{dx}{1+x^2+\sin^2x}$? I have tried Trigonometric Substitution, but I can´t get an already known function to be easy for integrate:
$$\int\frac{dx}{1+x^2+\sin^2x}$$
I entered this on Wolfram and it gave me the same function. I'm not asking for the exact solution, just a good way to solve it.
Regards.
| Hint:
$\because0\leq\dfrac{\sin^2x}{x^2+1}<1$ $\forall x\in\mathbb{R}$
$\therefore\int\dfrac{dx}{1+x^2+\sin^2x}$
$=\int\dfrac{dx}{(x^2+1)\left(1+\dfrac{\sin^2x}{x^2+1}\right)}$
$=\int\dfrac{1}{x^2+1}\sum\limits_{m=0}^\infty\dfrac{(-1)^m\sin^{2m}x}{(x^2+1)^m}dx$
$=\int\dfrac{1}{x^2+1}dx+\int\sum\limits_{m=1}^\infty\dfrac{(-1)^m\sin^{2m}x}{(x^2+1)^{m+1}}dx$
$=\int\dfrac{1}{x^2+1}dx+\int\sum\limits_{m=1}^\infty\dfrac{(-1)^m(2m)!}{4^m(m!)^2(x^2+1)^{m+1}}dx+\int\sum\limits_{m=1}^\infty\sum\limits_{n=1}^m\dfrac{(-1)^{m+n}(2m)!\cos2nx}{2^{2m-1}(m-n)!(m+n)!(x^2+1)^{m+1}}dx$
$=\int\dfrac{1}{x^2+1}dx+\int\sum\limits_{m=1}^\infty\dfrac{(-1)^m(2m)!}{4^m(m!)^2(x^2+1)^{m+1}}dx+\int\sum\limits_{m=1}^\infty\sum\limits_{n=1}^m\sum\limits_{p=0}^\infty\dfrac{(-1)^{m+n+p}(2m)!4^pn^{2p}x^{2p}}{2^{2m-1}(m-n)!(m+n)!(2p)!(x^2+1)^{m+1}}dx$
| {
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"url": "https://math.stackexchange.com/questions/672802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Given positive real numbers $a, b, c$ with $aI am trying to prove the following:
$$\frac{a}{1+a} < \frac{b}{1+b} + \frac{c}{1+c}$$
given that $a, b, c > 0$ and $a < b+c$. I tried various rearrangements but can't seem to get anywhere with it.
| By C-S
$$\frac{b}{1+b}+\frac{c}{c+1}=\frac{b^2}{b^2+b}+\frac{c^2}{c^2+c}\geq\frac{(b+c)^2}{b^2+c^2+b+c}\geq\frac{(b+c)^2}{(b+c)^2+b+c}=$$
$$=\frac{b+c}{b+c+1}=1-\frac{1}{b+c+1}>1-\frac{1}{a+1}=\frac{a}{a+1}.$$
Done!
| {
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"question_score": "1",
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A coordinate geometry question. On pg. 46 of "Coordinate geometry" by S.L. Loney, the following question has been posed:
Show that the equations to the straight lines passing through the point $(3,-2)$ and inclined at $60^\circ$ to the line $\sqrt{3}x+y=1$ are $y+2=0$ and $y-\sqrt{3}x+2+3\sqrt{3}=0$.
I am getting different equations for those two lines.
My method:
$L_1: \frac{y+2}{x-3}=\frac{-\sqrt{3}+\tan(60^\circ)}{1-(-\sqrt{3})\tan(60^\circ)}$
$L_2: \frac{y+2}{x-3}=\frac{-\sqrt{3}-\tan(60^\circ)}{1+(-\sqrt{3})\tan(60^\circ)}$
On simplifying, $L_1$ and $L_2$ are coming out different as compared to the given equations.
Any help would be much appreciated.
| Let $\theta$ be the angle between two lines whose slopes are $m_1,m_2$. Then we have:
$$\tan \theta = |\frac{m_1-m_2}{1+m_1m_2}|$$
As we know $m_1$, we can just plug in the value and get $m_2$:
$$\sqrt{3} = |\frac{-\sqrt{3}-m_2}{1-\sqrt{3}{m_2}}|$$
$$\sqrt{3}-3m_2 = -\sqrt{3}-m_2 \implies m_2 = \sqrt{3}$$ or
$$-\sqrt{3}+3m_2 = -\sqrt{3}-m_2 \implies m_2 = 0$$
$L_2$: $y+2 = 0(x-3) \implies y+2=0$
$L_1$: $y+2 = \sqrt{3}(x-3) \implies y-\sqrt{3}x+2+3\sqrt{3} = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/673437",
"timestamp": "2023-03-29T00:00:00",
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How to evaluate the integral: $I=\int_0^3\frac{x\sqrt{x+1}dx}{x^2+x+1}$ Evaluating this integral: $I=\int_0^3\frac{x\sqrt{x+1}dx}{x^2+x+1}$
I've tried:
Set : $\sqrt{x+1}=t\mapsto dx=2t\,dt \Longrightarrow I=2\int_1^2\frac{t^4-t^2}{t^4-t^2+1}dt=2\int_1^2(1-\frac{1}{t^4-t^2+1})dt$
And then, I set $(t^2-\frac{1}{2})=\frac{\sqrt{3}}{2}\tan v$ ... But I can't continue to solve this integral...
| HINT:
As $\displaystyle t^4-t^2+1=(t^2+1)^2-3t^2=(t^2+\sqrt3t+1)(t^2-\sqrt3t+1)$
$$\frac1{t^4-t^2+1}=\frac1{2\sqrt3t}\left(\frac1{t^2-\sqrt3t+1}-\frac1{t^2+\sqrt3t+1}\right)$$
Using Trigonometric substitution ,
$$t^2-\sqrt3t+1=\frac{4t^2-4\sqrt3t+4}4=\frac{(2t-\sqrt3)^2+1}4$$
So set, $2t-\sqrt3=\tan\phi$ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/674050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Proving $\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)= -\frac{\log^33}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6}$? Ramanujan gave the following identities for the Dilogarithm function:
$$
\begin{align*}
\operatorname{Li}_2\left(\frac{1}{3}\right)-\frac{1}{6}\operatorname{Li}_2\left(\frac{1}{9}\right) &=\frac{{\pi}^2}{18}-\frac{\log^23}{6} \\
\operatorname{Li}_2\left(-\frac{1}{3}\right)-\frac{1}{3}\operatorname{Li}_2\left(\frac{1}{9}\right) &=-\frac{{\pi}^2}{18}+\frac{1}{6}\log^23
\end{align*}
$$
Now, I was wondering if there are similar identities for the trilogarithm? I found numerically that
$$\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)\stackrel?= -\frac{\log^3 3}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6} \tag{1}$$
*
*I was not able to find equation $(1)$ anywhere in literature. Is it a new result?
*How can we prove $(1)$? I believe that it must be true since it agrees to a lot of decimal places.
| By "anywhere in literature" did you include L. Lewin, Polylogarithms and Assoicated Functions ?
In the (fortcoming) solution for Monthly problem 11654, you will see that very equation. For the proof, our solver (Richard Stong) used these identities:
$$
\mathrm{Li}_3\left(\frac{1-z}{1+z}\right) - \mathrm{Li}_3\left(\frac{z-1}{z+1}\right) =
2\;\mathrm{Li}_3(1-z) + 2\;\mathrm{Li}_3\left(\frac{1}{z+1}\right) -
\frac{1}{2}\;\mathrm{Li}_3(1-z^2) - \frac{7}{4}\zeta(3)-
\frac{1}{3}\big(\log(1+z)\big)^3 + \frac{\pi^2}{6}\log(1+z)
$$
at $z=2$ and
$$
\mathrm{Li}_3(z) = \mathrm{Li}_3\left(\frac{1}{z}\right) -
\frac{1}{6}\big(\log(-z)\big)^3 - \frac{\pi^2}{6}\log(-z)
$$
at $z=-3$. Stong cites the Lewin book.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/676124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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} |
Prove that for every integer greater than 1. $\dbinom{n}{1}-2\dbinom{n}{2}+3\dbinom{n}{3}+.....+(-1)^{n-1}n\dbinom{n}{n}=0$ How do you show the following.
Prove that for every integer greater than 1.
$\dbinom{n}{1}-2\dbinom{n}{2}+3\dbinom{n}{3}+.....+(-1)^{n-1}n\dbinom{n}{n}=0$
My idea is that is that the whole sequence seem to alternate but it end as a negative.
So if the whole sequence adds up it will probably be negative one since it starts positive and ends negative.
And the the greatest integer function of (-1) is zero.
But I am not sure how to prove the following.
I did the binomial theorem for $(1-x)^n$ and got
$\dbinom{n}{0}1^{n}(-x)^{(0)}+\dbinom{n}{1}1^{n-1}(-x)^{(1)}+\dbinom{n}{2}1^{n-2}(-x)^{(2)}+\dbinom{n}{3}1^{n-3}(-x)^{(3)}+.....+\dbinom{n}{k}1^{n-k}(-x)^{(k)}$
but I find myself a bit stuck as the odds and even do not match.
then I got
$\dbinom{n}{0}1^{n}+\dbinom{n}{1}1^{n-1}(-x)+\dbinom{n}{2}1^{n-2}(x^2)+\dbinom{n}{3}1^{n-1}(-x^3)+\dbinom{n}{k} 1^{n-k}(x^k)$
ok I differentiated and I got $\dbinom{n}{1}-2\dbinom{n}{2}x+3\dbinom{n}{3}x^2-4x^3\dbinom{n}{4}+.....+\dbinom{n}{k} k x^{k-1}$
| Hint: Expand $(1-x)^n$ using the Binomial Theorem and differentiate term by term.
Alternately, note that $k\binom{n}{k}=n\binom{n-1}{k-1}$.
Details: For the first hint, let $n\gt 1$, and let $f(x)=(1-x)^n$. Then $f'(x)=n(1-x)^{n-1}$ and therefore $f'(1)=0$.
We now compute $f'(x)$ another way, by first expanding $f(x)$ using the Binomial Theorem. We have
$$f(x)=1-\binom{n}{1}x+\binom{n}{2}x^2 -\binom{n}{3}x^3 +\cdots +(-1)^n \binom{n}{n}x^n.$$
Differentiating, we get
$$f'(x)=-(1)\binom{n}{1}+(2)\binom{n}{2}x^1 -(3)\binom{n}{3}x^2+\cdots +(n)(-1)^n
\binom{n}{n}x^{n-1}.$$
Put $x=1$. We get
$$f'(1)=-(1)\binom{n}{1}+(2)\binom{n}{2} -(3)\binom{n}{3}+\cdots +(n)(-1)^n
\binom{n}{n}.$$
We already saw that the above expression is $0$ if $n\gt 1$. But the above expression is the negative of the expression you were asked to evaluate.
The other hint gives another approach to the same problem. Using it, we find that
$$\sum_1^n (-1)^{k-1}k \binom{n}{k}=n\sum_{1}^n(-1)^{k-1}\binom{n-1}{k-1}.$$
The expression on the right is equal to
$$n\sum_{j=0}^{n-1}(-1)^j \binom{n-1}{j}.$$
Using the Binomial Theorem, we find this is $n(1-1)^{n-1}$, which is $0$. There are also more "combinatorial" ways of seeing that the sm of the binomial coefficients with alternating signs is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/677807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Let $\theta=\frac{2\pi}{5}$. Show $2\cos(2\theta)+2\cos(\theta)+1=0$. I have been at this for a while. Any ideas?
| If $\displaystyle5\theta=2\pi, 3\theta=2\pi-2\theta$
$\displaystyle\implies\cos3\theta=\cos(2\pi-2\theta)=\cos2\theta$
Using Triple and Double angle formulas,
$$4c^3-3c=2c^2-1\iff 4c^3-2c^2-3c+1=0\ \ \ \ (1)$$
where $ c=\cos\theta$
Again, $\displaystyle\cos3\theta=\cos2\theta$
$\displaystyle\implies3\theta=2n\pi\pm2\theta$ where $n$ is any integer
Taking the '+' sign, $\displaystyle\implies3\theta=2n\pi+2\theta\iff\theta=2n\pi\implies\cos\theta=\cos2n\pi=1$
Taking the '-' sign, $\displaystyle\implies3\theta=2n\pi-2\theta\iff\theta=\frac{2n\pi}5$ where $n\equiv0,1,2,3,4\pmod5$
Again as $\displaystyle\cos(2\pi-y)=\cos y,\cos\frac{2\pi}5=\cos\left(2\pi-\frac{2\pi}5\right)=\cos\frac{8\pi}5$ and $\displaystyle\cos\frac{4\pi}5=\cos\frac{6\pi}5$
Clearly, $c-1$ is a factor of $(1)$ and $\displaystyle\cos\frac{2\pi}5,\cos\frac{4\pi}5\ne1$
So,
$\displaystyle\cos\frac{2\pi}5=\cos\frac{8\pi}5$ and $\displaystyle\cos\frac{4\pi}5=\cos\frac{6\pi}5$ are the roots of $$\frac{4c^3-2c^2-3c+1}{c-1}=0\iff 4c^2+2c-1=0$$
Again use Double formula $\displaystyle\cos2\theta=2\cos^2\theta-1\implies c^2=\cos^2\theta=\frac{1+\cos2\theta}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/678956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 7
} |
Is there an easy way to prove that $3a^2+2=b^2$ does not have any rational solutions? I am a physicist who needs to prove (for his research) that there are no $a,b\in\mathbb{Q}$ such that $3a^2+2=b^2$. Is there an easy way to do this?
| If there is a rational solution $(a,b)$, it can be expressed as $a=\frac{A}{z}$, $b=\frac{B}{z}$ where $A$ and $B$ are integers, and $z$ is a non-zero integer. Clearing denominators, we get
$$3A^2+2z^2=B^2.$$
If the above equation has solutions with $B\ne 0$, there is a triple of integers, not all $0$, such that $B$ is positive, $3A^2+2z^2=B^2$, and $B$ is the smallest positive integer for which the equation $3x^2+2y^2=B^2$ has a solution.
Note that $B$ must be divisible by $3$. For if it is not, we have $2z^2\equiv 1\pmod{3}$, which is impossible.
Let $B=3B_1$. Since $3$ divides $3A^2$, it follows that $3$ divides $2z^2$, and therefore $3$ divides $z$. So $z=3y$ for some integer $y$.
Then $9$ divides $3A^2$, so $3$ divides $A$, say $A=3x$. Substituting and dividing through by $9$, we get
$$3x^2+2y^2=B_1^2.$$
This contradicts the choice of $B$ as the smallest positive integer for which the equation $3x^2+2y^2=B^2$ has a solution.
Remark: The same argument can be rewritten in various ways. One can make it look like one of the standard proofs of the irrationality of $\sqrt{2}$, by assuming that $a$ can be expressed as $\frac{A}{z}$, $b$ as $\frac{B}{z}$, where $z\gt 0$ is as small as possible.
Or else we can rewrite the proof as a Fermat Infinite Descent argument. We showed that if there is a solution $A,z,B$ with $B\ne 0$, then there is a solution $A_1,z_1,B_1$ with $B_1\lt B$. Continue. We obtain an infinite descending chain $B\gt B_1\gt B_2\gt \cdots$, which is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/679442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
2014 AMC 12 B problem 25 What is the sum of all positive real solutions $x$ to the following equation? $$2\cos(2x)\left( \cos(2x) - \cos{\left(\frac{2014\pi^2}{x^2}\right)} \right) = \cos(4x) - 1 $$
| Using $\cos2A=2\cos^2A-1,$
$$2\cos^22x-2\cos2x\cos\left(\frac{2014\pi^2}{x^2}\right)=2\cos^22x-1-1$$
$$\implies\cos2x\cos\left(\frac{2014\pi^2}{x^2}\right)=1$$
As for real $y,-1\le\cos y\le1;$ We need
$$\cos2x=\cos\left(\frac{2014\pi^2}{x^2}\right)=\pm1$$
Taking the '+' sign,
$\displaystyle\cos2x=1\implies 2x=2n\pi\iff x=n\pi$ where $n$ is an integer
and $\displaystyle\cos\left(\frac{2014\pi^2}{x^2}\right)=1\implies \frac{2014\pi^2}{x^2}=2m\pi\iff x^2=\frac{1007\pi}m$ where $m$ is an integer
$\displaystyle\implies \frac{1007\pi}m=(n\pi)^2\iff\pi=\frac{1000}{mn^2}$ which is rational unlike $\pi$
Can you manage the '-' sign from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/683585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Minimum value of the function $\sqrt{(1+1/m)(1+1/n)}$ If $m, n$ are positive real variables whose sum is a constant $k$, then what is the minimum value of $$\sqrt{\bigg(1 + \frac{1}{m}\bigg)\bigg(1 + \frac{1}{n}\bigg)}$$
| $\sqrt{1+\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{mn}}=\sqrt{1+\dfrac{m+n}{mn}+\dfrac{1}{mn}}=\sqrt{1+\dfrac{k+1}{mn}}\ge\sqrt{1+\dfrac{k+1}{\bigg(\dfrac{m+n}{2}\bigg)^2}}$
$1+\dfrac{4(k+1)}{k^2}=\dfrac{k^2+4k+4}{k^2}=\bigg(\dfrac{k+2}{k}\bigg)^2$
So the least value of your desired expression is $\dfrac{k+2}{k}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/684637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Checking multivariable function's differentiability at $(0,0)$ I have a function: $$f(x,y) = \begin{cases} (x^2+y^2)\sin (x^2+y^2)^{-1} \ \ \ &\text{, if } x^2+y^2 \neq 0 \\ 0 &\text{, if } x^2+y^2=0 \end{cases}$$
I calculated the partial derivatives:
$$f_x(x,y)=2x\left( \sin (x^2+y^2)^{-1}-\frac{(x^2+y^2)\cos (x^2+y^2)^{-1}}{(x^2+y^2)^2} \right)$$
$$f_y(x,y)=2y\left( \sin (x^2+y^2)^{-1}-\frac{(x^2+y^2)\cos (x^2+y^2)^{-1}}{(x^2+y^2)^2} \right)$$
I'm stuck at checking the function's differentiability at point $(0,0)$, I use following definition :
$f$ is differentiable at point $(a,b)$ and $M,N\in \mathbb{R}$ $\leftrightarrow$
$\lim_{h_1,h_2\to 0} \frac{f(a+h_1,b+h_2)-f(a,b)-Mh_2-Nh_2}{\sqrt{h_1^2+h_2^2}}=0$
Here's what I've done so far:
$$\lim_{h_1,h_2\to 0} \frac{(h_1^2+h_2^2)\sin(h_1^2+h_2^2)^{-1}-Mh^1-Nh_2}{\sqrt{h_1^2+h_2^2}}=$$
$$= \lim_{h_1,h_2\to 0} \frac{\sqrt{h_1^2+h_2^2}\sqrt{h_1^2+h_2^2}\sin(h_1^2+h_2^2)^{-1}-Mh^1-Nh_2}{\sqrt{h_1^2+h_2^2}}=$$
$$=\lim_{h_1,h_2\to 0} \frac{\sqrt{h_1^2+h_2^2}\left(\sqrt{h_1^2+h_2^2}\sin(h_1^2+h_2^2)^{-1}-\frac{Mh_1}{\sqrt{h_1^2+h_2^2}}-\frac{Nh_2}{\sqrt{h_1^2+h_2^2}}\right)}{\sqrt{h_1^2+h_2^2}}=$$
$$=\lim_{h_1,h_2\to 0} \sqrt{h_1^2+h_2^2}\sin(h_1^2+h_2^2)^{-1}-\frac{Mh_1-Nh_2}{\sqrt{h_1^2+h_2^2}}=$$
| Hint: You can use Taylor series as
$$ \sin(h_1^2+h_2^2) = (h_1^2+h_2^2)-\frac{(h_1^2+h_2^2)^3}{3!}+\dots\,. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/693217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to show whether a $3\times 4$ matrix has no solution, a unique solution or infinitely many solutions? The system is :
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
-2 & 5 & -4 & -1 & | & b \\
1 & -10 & 22 & 8 & | & c
\end{matrix}
$$
After Gaussian elimination, I found that
$$
\begin{array}{cccc|cc}
1 & -4 & 6 & a & & 0 \\
0 & 1 & -\tfrac{8}{3} & - \left( 2a- \tfrac{1}{3} \right) & & - \tfrac{1}{3}b \\
0 & 0 & 0 & 10-5a & & c-2b
\end{array}
$$
Is it correct and I can continue to determine whether there is no solution, a unique solution or infinitely many solutions?
Here are the operations:
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
-2 & 5 & -4 & -1 & | & b \\
1 & -10 & 22 & 8 & | & c
\end{matrix}
$$
$R_2+2R_1\rightarrow R_2$
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
0 & -3 & 8 & 2a-1 & | & b \\
1 & -10 & 22 & 8 & | & c
\end{matrix}
$$
$R_3-R_1\rightarrow R_3$
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
0 & -3 & 8 & 2a-1 & | & b \\
0 & -6 & 16 & 8-a & | & c
\end{matrix}
$$
$R_3-2R_2\rightarrow R_3$
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
0 & -3 & 8 & 2a-1 & | & b \\
0 & 0 & 0 & 10-5a & | & c-2b
\end{matrix}
$$
$-\frac 13(R_2)\rightarrow R_2$
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
0 & 1 & -\frac 83 & -\tfrac{2a-1}{3} & | & -\frac 13b \\
0 & 0 & 0 & 10-5a & | & c-2b
\end{matrix}
$$
| Constraints on $a$
In general, we have existence of a solution when the data vector is in the column space of the target matrix $\mathbf{A}$. That is, if the data vector can be written as a linear combination of the fundamental columns of $\mathbf{A}$.
Additionally, if the nullspace $\mathcal{N} \left( \mathbf{A}^{*} \right)$ is trivial, then the solution is also unique.
Look for the fundamental columns. Start with the reduced row echelon form:
$$
\begin{align}
\mathbf{A} &\mapsto \mathbf{E}_{\mathbf{A}} \\
%
\left[
\begin{array}{rrrr}
1 & -4 & 6 & a \\
-2 & 5 & -4 & -1 \\
1 & -10 & 22 & 8 \\
\end{array}
\right]
%
&\mapsto
%
\left[
\begin{array}{rrrr}
\color{blue}{1} & 0 & -\frac{14}{3} & 0 \\
0 & \color{blue}{1} & -\frac{8}{3} & 0 \\
0 & 0 & 0 & \color{blue}{1} \\
\end{array}
\right]
%
\end{align}
$$
The fundamental columns are marked by blue pivots. Because the 3rd column is linearly dependent, we can ignore it and study a simpler, equivalent form:
$$
\hat{\mathbf{A}} =
%
\left[
\begin{array}{rrrr}
1 & -4 & a \\
-2 & 5 & -1 \\
1 & -10 & 8 \\
\end{array}
\right]
$$
The two matrices have the same range space:
$$
\mathcal{R} \left( \mathbf{A} \right) =
\mathcal{R} \left( \hat{ \mathbf{A} } \right)
$$
The determinant will determine when there is a nullspace.
$$
\det \hat{\mathbf{A}} = 15 \left( a - 2 \right)
$$
When $a=2$ there are no longer three linearly independent columns. If a solution exists, it will not be unique.
Constraints on $b$, $c$
If $b=c=0$, we are out of the range space and probing the null space. When
$a\ne2$ the nullspace is
$$
\mathcal{N} \left( \mathbf{A}^{*} \right) = \text{span }
%
\left\{ \,
\left[ \begin{array}{c}
14 \\ 8 \\ 3 \\ 0
\end{array} \right]
\, \right\}
$$
When
$a=2$ the dimension of the nullspace increases
$$
\mathcal{N} \left( \mathbf{A}^{*} \right) = \text{span }
%
\left\{ \,
\left[ \begin{array}{c}
14 \\ 8 \\ 3 \\ 0
\end{array} \right]
,\,
\left[ \begin{array}{c}
2 \\ 1 \\ 0 \\ 1
\end{array} \right]
\, \right\}
$$
When $a\ne 2$ and either $b\ne0$ or $c\ne0$ we are guaranteed a unique solution because $\mathcal{R} \left( \mathbf{A} \right) = \mathbb{C}^{3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Infinite series for partial sums of square roots. Can you prove these infinite series for partial sums of square roots?
$$\sqrt{1}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$$
$$\sqrt{1}+\sqrt{2}=\sum\limits_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}-\frac{2}{\sqrt{n+2}}\right)$$
$$\sqrt{1}+\sqrt{2}+\sqrt{3}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}-\frac{3}{\sqrt{n+3}}\right)$$
$$\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}=\sum _{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\frac{1}{\sqrt{n+3}}-\frac{4}{\sqrt{n+4}}\right)$$
$$\cdots$$
And is there some easy cancellation that I have missed on the right hand side?
Mathematica:
Clear[s, i, n, j]
s = 1/2;
i = 1;
j = 0;
Sum[1/(n + 0)^s - 1/(n + 1)^s, {n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s - 2/(n + 2)^s + 1/(n + 0)^s, {n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s + 1/(n + 2)^s - 3/(n + 3)^s + 1/(n + 0)^s,
{n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s + 1/(n + 2)^s + 1/(n + 3)^s - 4/(n + 4)^s +
1/(n + 0)^s, {n, 1, Infinity}]
N[%, 20]
N[Accumulate[Sqrt[Range[4]]], 20]
| You have a very elegant demonstration given by Sami Ben Romdhane.
If, instead of summing to $\infty$, you sum to $m$, you will get the following formulas for the different sums
$$\sqrt{1}-\frac{1}{\sqrt{m+1}}$$
$$\sqrt{1}+\sqrt{2}+H_m^{\left(\frac{1}{2}\right)}+H_{m+1}^{\left(\frac{1}{2}\right)}-2
H_{m+2}^{\left(\frac{1}{2}\right)}$$
$$\sqrt{1}+\sqrt{2}+\sqrt{3}+H_m^{\left(\frac{1}{2}\right)}+H_{m+1}^{\left(\frac{1}{2}\right)}+H_{m+2}^{\left(\frac{1}{2}\right)}-3 H_{m+3}^{\left(\frac{1}{2}\right)}$$
$$\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+H_m^{\left(\frac{1}{2}\right)}+H_{m+1}^{\left(\frac{1}{2}\right)}+H_{m+2}^{\left(\frac{1}{2}\right)}+H_{m+3}^{\left(\frac{1}{2}\right)}-4
H_{m+4}^{\left(\frac{1}{2}\right)}$$ Now, push $m$ to $\infty$ and use the properties of the harmonic numbers.
| {
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"url": "https://math.stackexchange.com/questions/695073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Do not understand formula... How does;
$x^{n} -y^{n}=(x-y)(x^{n-1} + x^{n-2}y+...+x y^{n-2}+ y^{n-1} )$
work on $x^{2} - y^{2}$
When I attempt to apply the formula on $x^{2} - y^{2}$
I get the following
$x^{2} - y^{2} =(x-y)( x^{1} + x^{0}y+...+x y^{0} + y^{1} )$
$x^{2} - y^{2} =(x-y)(2x+2y)$
which is obviously false. What is the correct way to use the formula?
| You have too many terms, $x^{n-n}y^{n-1}$ is the last, which in this case is your second term $x^{n-2}y$. The third term would in your case be $x^{n-3}y^{n-2}$ which we don't want since $n=2$ and the exponent of $x$ would become negative.
To more easily see this, write the factorization in this form:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \dots + x^{n-(n-1)}y^{n-2} + x^{n-n}y^{n-1})$$
In this case it becomes
$$x^n - y^n = (x-y)(x^1y^0 + x^0y^1) = (x-y)(x + y)$$
which of course is the correct factorization.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/695164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
How to find a matrix $C$ such that $C^{-1}AC$ is in Jordan block form. $A:=\begin{bmatrix}
6 & -1\\
4 & 2
\end{bmatrix}$
Now, just to show I've done some working, at least to find $A$'s eigenvalues and deduced that it's not diagonalisable:
Any hints/advice?
Thanks
| For a $2 \times 2$ (say complex) matrix $A$ which is not diagonalizable, there is a method to determine the conjugating matrix $C$ such that $C^{-1}AC$ is in Jordan form. The only way that $A$ can fail to be diagonalizable is if the characteristic polynomial of $A$ has a repeated root $s,$ but $A \neq sI.$ Then the only possibility for the Jordan form is $\left( \begin{array}{clcr} s & 1\\0&s \end{array} \right)$. In order to have trace $2s$ and determinant $s^{2},$ the matrix $A$ must either have the form $\left( \begin{array}{clcr} s+a & ta\\\frac{-a}{t} & s-a \end{array} \right)$ for some $a,t \neq 0,$ or have one of the forms $\left( \begin{array}{clcr} s & t\\0&s \end{array} \right)$ or $\left( \begin{array}{clcr} s & 0\\t&s \end{array} \right)$ for some non-zero $t.$ We have to solve the equation $AC = C \left( \begin{array}{clcr} s & 1\\0&s \end{array} \right)$ with ${\rm det} C \neq 0.$ This reduces to solving one of the equations $\left( \begin{array}{clcr} a & ta\\\frac{-a}{t} & -a \end{array} \right)C = C\left( \begin{array}{clcr} 0 & 1\\0&0 \end{array} \right)$ or $\left( \begin{array}{clcr} 0 & t\\0 & 0 \end{array} \right)C = C\left( \begin{array}{clcr} 0 & 1\\0&0 \end{array} \right)$ or $\left( \begin{array}{clcr} 0 & 0\\t & 0 \end{array} \right)C = C\left( \begin{array}{clcr} 0 & 1\\0&0 \end{array} \right).$ This can always be done.
In your case, you have $s = 4, a = 2$ and $t = \frac{-1}{2}.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
How to find multiplicative orders of all elements in field $\Bbb F$ (say $\Bbb F_{13}$)? I am working on some finite fields and I was referring to some online class material. Is there any way to find the multiplicative orders of all elements in a field $\Bbb F$?
| Just compute the powers:
\begin{align}
2^2&\equiv_{13}4 &2^3&\equiv_{13}8 &
2^4=4^2&\equiv_{13}16\equiv_{13}3&
2^5&\equiv_{13}6&2^6&\equiv_{13}12\equiv_{13}-1\\
3^2&\equiv_{13}9\equiv_{13}-4 &3^3&\equiv_{13}-12\equiv_{13}1\\
5^2&\equiv_{13}-1\\
6^2&\equiv_{13}-3&6^3&\equiv_{13}-5&6^4&\equiv_{13}-4&6^5&\equiv_{13}2&6^6&\equiv_{13}-1
\end{align}
and $7\equiv_{13}-6$ etc. It follows that
*
*$2,6,7,11$ have order $12$,
*$3,9$ have order $3$,
*$4,10$ have order $6$,
*$5,8$ have order $4$,
*$12$ has order $2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integration of $1/(1+\sin x)$ I solved it using $t=\tan(\frac{x}{2})$ substitution and got $-2/(1+\tan(x/2))+C$, but in my math book solution is $\tan(x/2-\pi/4)+C$. Are those the same expressions and if they are, how do I transform from one to another, or are one(or both) solutions incorrect ?
| $\bf{My\; Solution::}$ Let $\displaystyle I = \int\frac{1}{1+\sin x}dx = \int\frac{1}{1+\cos \left(\frac{\pi}{2}-x\right)}dx = \int\frac{1}{2\cos^2 \left(\frac{\pi}{4}-\frac{x}{2}\right)}dx$
So $\displaystyle I = \frac{1}{2}\int \sec^2 \left(\frac{\pi}{4}-\frac{x}{2}\right)dx = -\frac{1}{2}\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)+\mathbf{C} = -\frac{1}{2}\cdot \left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right)+\mathbb{C}$
So $\displaystyle I = \int\frac{1}{1+\sin x} = \frac{1}{2}\left(\frac{\tan \frac{x}{2}-1}{1+\tan \frac{x}{2}}\right)+\mathbb{C} = \left(\frac{-2}{1+\tan \frac{x}{2}}\right)+1+\mathbb{C}$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 0
} |
Integral $\int_0^a \ln \left( \frac{b-\sqrt{a^2-x^2}}{b+\sqrt{a^2-x^2}} \right)dx$ Hi I am trying to calculate,
$$
\int_0^a \ln \left( \frac{b-\sqrt{a^2-x^2}}{b+\sqrt{a^2-x^2}} \right)dx
$$
where $a,b$ are positive real constants. I Know $\ln(xy)=\ln x +\ln y$, but I do not know how to evaluate this integral then.
I need to find a closed form for the indefinite integral
$$
\int \ln \big(b\pm \sqrt{a^2-x^2}\big) dx
$$
so this is really the problem I am facing. The closed form exists and is in terms of elementary functions. Thanks!
| This is a possible way (I assume $a\geq 0$, $b\geq 0$):
1) substitute $x= a \sin t $.
2) integrate the resulting integral by parts to get rid of $\ln$
3) you should end up with $$-\int_0^{\pi/2}\frac{2 a^2 b \sin^2 t\,dt }{b^2 -a^2 \cos^2 t} .$$
4) it is possible to integrate the last integral by elementary means or using the residue theorem. You should obtain $$\pi \left(\sqrt{b^2 -a^2} -b\right).$$
Edit:
Here are the steps in a bit more detail ($0\leq a \leq b$):
1) after subsitution $x= a \sin t$ with $dx = a \cos t\, dt$ we obtain
$$ a \int_{0}^{\pi/2} \!dt\,\cos t\, \ln \left( \frac{b -a \cos t}{b+a \cos t} \right). $$
2) Integration by parts (where $\cos t$ is integrated and the rest differentiated) leads directly to
$$ a \sin t \, \ln \left( \frac{b -a \cos t}{b+a \cos t} \right) \Bigg|_{0}^{\pi/2} - a \int_0^{\pi/2}\!dt\, \sin t \frac{b+ a \cos t}{b -a \cos t} \frac{2 a b}{(b+ a \cos t)^2} \sin t\\
= - 2 a^2 b \int_0^{\pi/2}\!dt\, \frac{\sin^2 t}{b^2 -a ^2 \cos^2 t}.$$
3) Now use the substitution $s= \tan t$ with $ds = (1+s^2) dt =\cos^{-2} t\, dt$ as proposed by sos440. This leads to
$$ -2 a^2 b \int_0^{\infty}\!\frac{ds}{1+s^2}\, \frac{\sin^2 t}{b^2 -a^2\cos^2 t}
= -2 a^2 b \int_0^{\infty}\!ds\, \frac{s^2}{(1+s^2)[b^2 (1+s^2)-a^2]}.$$
4) Employing the partial fraction expansion on the last integrand, you obtain
$$ -2 b \int_0^\infty\frac{ds}{1+s^2} + \frac{2(b^2-a^2)}{b} \int_0^\infty \frac{ds}{s^2 +1 - (a/b)^2} .$$
Thus, we need to evaluate the integral
$$\int_0^\infty \frac{ds}{s^2+\alpha^2}
= \alpha^{-1} \arctan(s/\alpha) \big|_{0}^{\infty} = \frac{\pi}{2\alpha},$$
and we obtain the final result (with $\alpha=1$ and $\alpha= \sqrt{1-(a/b)^2}$)
$$ -\pi b + \frac{(b^2-a^2)\pi}{b \sqrt{1-(a/b)^2}} .$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Values of the limit $\lim\limits_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$ Hi I have a question regarding finding the values of limit for the following question.
Let $a, b \in \mathbb R$. Find the limit
$$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$
| it is easily shown that as $z$ approaches zero
$$
\log(1+z) = z + O(z^2)
$$
so we may write
$$
\log \sqrt{(x+a)(x+b)} =\log x + \log \sqrt{(1+\frac{a}x)(1+\frac{b}x)} \\= \log x + \frac{a+b}{2x} + O(x^{-2})
$$
taking exponentials
$$
\sqrt{(x+a)(x+b)} = x\bigg(1+O(x^{-2})\bigg) \exp\frac{a+b}{2x} \\
= (x+O(x^{-1})\bigg(1+ \frac{a+b}{2x} + O(x^{-2}\bigg) \\
= x + \frac{a+b}2 +O(x^{-1})
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
"Identity" of volume of a solid of revolution (Limited Weierstrass function) The question is: Is this correct?
This is pretty much the first thing I've tried to come up with by myself. I wanted to see what would happen if I tried to calculate volume of a solid of revolution of the Weierstrass-like function. I'm pretty bad at calculus so I'm pretty certain there are lot's errors but I wanted to share it anyway and maybe get some feedback.
Please, let the bashing begin!
Let $w(x) := \displaystyle\sum_{k= 0}^{\infty} a^k\cos(b^k\pi x) $ where $0 < a < 1$ and $ab > 1+ \frac{3\pi}{2}, b$ is an odd integer greater than $1$.
Let us limit $w(x)$ to $n$ so that,
$w(x)_n := \displaystyle\sum_{k= 0}^{n} a^k\cos(b^k\pi x) $. Consider the volume of the solid formed by rotating the area of $w(x)_n$ on the x-axis in the interval $[\alpha,\beta]$ given by,
$V_{w(x)_n} =\pi\displaystyle\int_{\alpha}^{\beta} (w(x)_n)^2dx = \pi\displaystyle\int_{\alpha}^{\beta} (\displaystyle\sum_{k= 0}^{n} a^k\cos(b^k\pi x) )^2dx =$
$\pi\displaystyle\int_{\alpha}^{\beta} (\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x))^2dx =$
$\pi\displaystyle\int_{\alpha}^{\beta} (\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x))(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x))dx =$
$\pi\displaystyle\int_{\alpha}^{\beta} [ \cos(\pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) +$
$\cos(\pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) + $
$a\cos(b \pi x) (\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) + $
$a^2\cos(b^2 \pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) + $$\ldots +$
$a^n\cos(b^n \pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) ] dx =$
$\pi\displaystyle\int_{\alpha}^{\beta}[ (\cos^2(\pi x) + a\cos(\pi x)\cos(b \pi x) + a^2\cos(\pi x)\cos(b^2 \pi x) + \ldots + a^n\cos(\pi x)\cos(b^n \pi x)) +$
$( a^2\cos^2(b \pi x)+a\cos(b \pi x) \cos(\pi x) + a^3\cos(b \pi x)\cos(b^2 \pi x) + \ldots + a^{n+1}\cos(b \pi x)\cos(b^n \pi x)) +$
$(a^4\cos^2(b^2 \pi x) + a^2\cos(b^2 \pi x) \cos(\pi x) + a^3\cos(b^2 \pi x) \cos(b \pi x) + \ldots + a^{n+2}\cos(b^2 \pi x) \cos(b^n \pi x)) + \ldots + $
$( a^{2n}\cos^2(b^n \pi x)) + a^n\cos(b^n \pi x)) \cos(\pi x) + a^{n+1}\cos(b^n \pi x)) \cos(b \pi x) + a^{n+2}\cos(b^n \pi x)) \cos(b^2 \pi x) + \ldots + a^{2n-1}\cos(b^n \pi x)) \cos(b^{n-1} \pi x)) ]dx = $
$ \pi\displaystyle\int_{\alpha}^{\beta} ([ \displaystyle\sum_{k= 1}^{n+1}a^{2(k-1)}\cos^2(b^{k-1}\pi x) ] + [\bigcup _{i,j=0,i\neq j}^{n}a^{i+j}\cos(b^i\pi x)\cos(b^j\pi x)])dx =$
$ \pi[\displaystyle\int_{\alpha}^{\beta} \displaystyle\sum_{k= 1}^{n+1}a^{2(k-1)}\cos^2(b^{k-1}\pi x)dx + \displaystyle\int_{\alpha}^{\beta} \bigcup _{i,j=0,i\neq j}^{n}a^{i+j}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$
[ A miracle happens here When can a sum and integral be interchanged? If this is correct, how can I motivate this? ]
$ \pi[ \displaystyle\sum_{k= 1}^{n+1}\displaystyle\int_{\alpha}^{\beta} a^{2(k-1)}\cos^2(b^{k-1}\pi x)dx + \displaystyle\int_{\alpha}^{\beta} \bigcup _{i,j=0,i\neq j}^{n}a^{i+j}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)}\displaystyle\int_{\alpha}^{\beta} \cos^2(b^{k-1}\pi x)dx + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \displaystyle\int_{\alpha}^{\beta}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$
Note that
$\int \cos^2(\gamma_1 x)dx = \frac{2\gamma_1 x + \sin(2\gamma_1x )}{4\gamma_1} + c$
and
$\int \cos(\gamma_1 x)cos(\gamma_2 x)dx = \frac{\gamma_1 \sin(\gamma_1 x)\cos(\gamma_2 x) - \gamma_2 \cos(\gamma_1 x)\sin(\gamma_2 x) }{\gamma_1^2 -\gamma_2^2} + c, \gamma_1, \gamma_2 \in \mathbb{R} $
Thus,
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)}\displaystyle\int_{\alpha}^{\beta} \cos^2(b^{k-1}\pi x)dx + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \displaystyle\int_{\alpha}^{\beta}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \Big| \frac{2\pi b^{k-1}x + \sin(2\pi b^{k-1}x)}{4\pi b^{k-1}} \Big|_\alpha^\beta + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{(-b^j -b^i)\sin(\pi x b^i - \pi x b^j) + (b^j -b^i)\sin(\pi x b^j + \pi x b^i)} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] $
Now restrict $\alpha, \beta \in \mathbb{R}: \alpha, \beta \in \mathbb{Z}$
Now, note that,
$ \sin(2\pi b^{k-1}x) = \sin(\pi y) = 0, \forall y \in \mathbb{Z}$
$ \sin(\pi x b^i - \pi x b^j) = \sin(\pi (x b^i - x b^j)) = \sin(\pi y) = 0 \forall y \in \mathbb{Z}$ and
$\sin(\pi x b^j + \pi x b^i) = \sin(\pi (x b^j + x b^i)) = \sin(\pi y) = o \forall y \in \mathbb{Z}$
Which implies that,
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \Big| \frac{2\pi b^{k-1}x + \sin(2\pi b^{k-1}x)}{4\pi b^{k-1}} \Big|_\alpha^\beta +
\bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{(-b^j -b^i)\sin(\pi x b^i - \pi x b^j) + (b^j -b^i)\sin(\pi x b^j + \pi x b^i)} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] =$
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \Big| \frac{2\pi b^{k-1}x + 0}{4\pi b^{k-1}} \Big|_\alpha^\beta +
\bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{(-b^j -b^i)0 + (b^j -b^i)0} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] =$
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta +
\bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{0} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] =$
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta +
\bigcup _{i,j=0,i\neq j}^{n}a^{i+j}0 ] = \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta + 0] =$
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta] = \pi \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta $
Thus
$V_{w(x)_n} = \pi\displaystyle\int_{\alpha}^{\beta} (w(x)_n)^2dx = \pi\displaystyle\int_{\alpha}^{\beta} (\displaystyle\sum_{k= 0}^{n} a^k\cos(b^k\pi x) )^2dx = \pi \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta, \alpha,\beta \in \mathbb{Z} $
Can this be "extended" to $w(x)$?
|
please do not up- or downvote this answer. It's just impossible to do this by pure commenting. I copy pasted your calculus and added some remarks.
$V_{w(x)_n} =\pi\displaystyle\int_{\alpha}^{\beta} (w(x)_n)^2dx = \pi\displaystyle\int_{\alpha}^{\beta} (\displaystyle\sum_{k= 0}^{n} a^k\cos(b^k\pi x) )^2dx =$
$\pi\displaystyle\int_{\alpha}^{\beta} (\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x))^2dx =$
$\pi\displaystyle\int_{\alpha}^{\beta} (\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x))(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x))dx =$
here you wrote the first line doulbe, must be a typo:
$\pi\displaystyle\int_{\alpha}^{\beta} [ \cos(\pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) +$
$\cos(\pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) + $
$a\cos(b \pi x) (\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) + $
$a^2\cos(b^2 \pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) + $$\ldots +$
$a^n\cos(b^n \pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) ] dx =$
then you sorted some special summands to the beginning which is fine due to the fact that you have a finite sum
$\pi\displaystyle\int_{\alpha}^{\beta}[ (\cos^2(\pi x) + a\cos(\pi x)\cos(b \pi x) + a^2\cos(\pi x)\cos(b^2 \pi x) + \ldots + a^n\cos(\pi x)\cos(b^n \pi x)) +$
$( a^2\cos^2(b \pi x)+a\cos(b \pi x) \cos(\pi x) + a^3\cos(b \pi x)\cos(b^2 \pi x) + \ldots + a^{n+1}\cos(b \pi x)\cos(b^n \pi x)) +$
$(a^4\cos^2(b^2 \pi x) + a^2\cos(b^2 \pi x) \cos(\pi x) + a^3\cos(b^2 \pi x) \cos(b \pi x) + \ldots + a^{n+2}\cos(b^2 \pi x) \cos(b^n \pi x)) + \ldots + $
$( a^{2n}\cos^2(b^n \pi x)) + a^n\cos(b^n \pi x)) \cos(\pi x) + a^{n+1}\cos(b^n \pi x)) \cos(b \pi x) + a^{n+2}\cos(b^n \pi x)) \cos(b^2 \pi x) + \ldots + a^{2n-1}\cos(b^n \pi x)) \cos(b^{n-1} \pi x)) ]dx = $
from here on I assume by $\cup$ you mean $\Sigma$
$ \pi\displaystyle\int_{\alpha}^{\beta} ([ \displaystyle\sum_{k= 1}^{n+1}a^{2(k-1)}\cos^2(b^{k-1}\pi x) ] + [\bigcup _{i,j=0,i\neq j}^{n}a^{i+j}\cos(b^i\pi x)\cos(b^j\pi x)])dx =$
$ \pi[\displaystyle\int_{\alpha}^{\beta} \displaystyle\sum_{k= 1}^{n+1}a^{2(k-1)}\cos^2(b^{k-1}\pi x)dx + \displaystyle\int_{\alpha}^{\beta} \bigcup _{i,j=0,i\neq j}^{n}a^{i+j}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$
[ A miracle happens here When can a sum and integral be interchanged? If this is correct, how can I motivate this? ]
finite sum and integral are always commuting - since you are not considering any limits this is fine.
$ \pi[ \displaystyle\sum_{k= 1}^{n+1}\displaystyle\int_{\alpha}^{\beta} a^{2(k-1)}\cos^2(b^{k-1}\pi x)dx + \displaystyle\int_{\alpha}^{\beta} \bigcup _{i,j=0,i\neq j}^{n}a^{i+j}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)}\displaystyle\int_{\alpha}^{\beta} \cos^2(b^{k-1}\pi x)dx + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \displaystyle\int_{\alpha}^{\beta}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$
the following claim I just believe you. (maybe I or someone else will check it later)
Note that
$\int \cos^2(\gamma_1 x)dx = \frac{2\gamma_1 x + \sin(2\gamma_1x )}{4\gamma_1} + c$
and
$\int \cos(\gamma_1 x)cos(\gamma_2 x)dx = \frac{\gamma_1 \sin(\gamma_1 x)\cos(\gamma_2 x) - \gamma_2 \cos(\gamma_1 x)\sin(\gamma_2 x) }{\gamma_1^2 -\gamma_2^2} + c, \gamma_1, \gamma_2 \in \mathbb{R} $
ok up to this point
Thus,
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)}\displaystyle\int_{\alpha}^{\beta} \cos^2(b^{k-1}\pi x)dx + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \displaystyle\int_{\alpha}^{\beta}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \Big| \frac{2\pi b^{k-1}x + \sin(2\pi b^{k-1}x)}{4\pi b^{k-1}} \Big|_\alpha^\beta + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{(-b^j -b^i)\sin(\pi x b^i - \pi x b^j) + (b^j -b^i)\sin(\pi x b^j + \pi x b^i)} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] $
Now restrict $\alpha, \beta \in \mathbb{R}: \alpha, \beta \in \mathbb{Z}$
you are restricting to the case $x\in\mathbb{R}:\forall i\in\mathbb{Z}:x\cdot b^{i}\in\mathbb{Z}$. since $b$ is odd (and hence not zero) this is means $x\in\mathbb{Z}$ (take $i=0$).
Now, note that,
$ \sin(2\pi b^{k-1}x) = \sin(\pi y) = 0, \forall y \in \mathbb{Z}$
$ \sin(\pi x b^i - \pi x b^j) = \sin(\pi (x b^i - x b^j)) = \sin(\pi y) = 0 \forall y \in \mathbb{Z}$ and
$\sin(\pi x b^j + \pi x b^i) = \sin(\pi (x b^j + x b^i)) = \sin(\pi y) = o \forall y \in \mathbb{Z}$
Which implies that,
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \Big| \frac{2\pi b^{k-1}x + \sin(2\pi b^{k-1}x)}{4\pi b^{k-1}} \Big|_\alpha^\beta +
\bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{(-b^j -b^i)\sin(\pi x b^i - \pi x b^j) + (b^j -b^i)\sin(\pi x b^j + \pi x b^i)} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] =$
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \Big| \frac{2\pi b^{k-1}x + 0}{4\pi b^{k-1}} \Big|_\alpha^\beta +
\bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{(-b^j -b^i)0 + (b^j -b^i)0} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] =$
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta +
\bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{0} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] =$
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta +
\bigcup _{i,j=0,i\neq j}^{n}a^{i+j}0 ] = \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta + 0] =$
$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta] = \pi \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta $
Thus
$V_{w(x)_n} = \pi\displaystyle\int_{\alpha}^{\beta} (w(x)_n)^2dx = \pi\displaystyle\int_{\alpha}^{\beta} (\displaystyle\sum_{k= 0}^{n} a^k\cos(b^k\pi x) )^2dx = \pi \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta, \alpha,\beta \in \mathbb{Z} $
Can this be ""extended" to $w(x)$?
since you made a very strong restriction on $x$ generalization will be hard. I have a suggenstion but that might also be a dead end: You might try to write $x=\frac{p}{q}$ and then try to apply integral substitution to get rid of $q$ and then be able to apply your formula. after this (if this really should work) you could try to find a density argument since $\mathbb{Q}$ is dense in $\mathbb{R}$ (not even talking about the limit $n\rightarrow\infty$; therefore you could use that the sum on the right hand side is geometric and then you would probably have to apply dominated convergence on the integral at the left hand side of this expression)
| {
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Induction proof of $n^{(n+1) }> n(n+1)^{(n-1)}$ The question statement from my homework booklet goes:
Prove by mathematical induction that $n^{n+1} > n(n+1)^{n-1}$
is true for all integers $n \geq 2$.
I've managed to come up with this for the induction step (the base case is trivial), but I am not sure what to do from here:
Assume true for n=k. For n=k+1,
\begin{align*}
&k^{k+1} > k(k+1)^{k+1} \\
&(k+1)k^{k+1} > k(k+1)^k \\
&(k+1)^{k+2} < (k+1)^3 k^k < (k+1)^3 (k+2)^k \\
&(k+1)^{k-1} < (k+2)^k
\end{align*}
I would greatly appreciate any help with how to solve this. Thanks in advance.
| For $k$ we have
$\large k^{k+1}>k(k+1)^{k-1}$
which we hold to be true.
For $k+1$ we have
$\large(k+1)^{k+2}=(k+1)^{k+1}(k+1)=k^{k+1}(1+\frac{1}{k})^{k+1}(k+1)
\\\large>k(1+k)^k(1+\frac{1}{k})^{k+1}$
as $k^{k+1}>k(k+1)^{k-1}$
Now
$\large k(k+1)^k(1+\frac{1}{k})^{k+1}=k(k+1)^k(1+\frac{1}{k})^{k+1}(\frac{k+2}{k+2})^k
\\\large =k(k+1)^k(\frac{k+1}{k})^{k+1}(\frac{k+2}{k+2})^k=(\frac{(k+1)^2}{k(k+2)})^k(k+1)(k+2)^k\\\large>(k+1)(k+2)^k$
where we have used the fact that
$(k+1)^2=k(k+2)+1>k(k+2)$
so that
$\large (\frac{(k+1)^2}{k(k+2)})^k>\frac{(k+1)^2}{k(k+2)}>1$
Thus we have
$\large (k+1)^{k+2}>k(1+k)^k(1+\frac{1}{k})^{k+1}>(k+1)(k+2)^k$
Culminating in
$\large (k+1)^{k+2}>(k+1)(k+2)^k$
| {
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Need help solving this equation with complex numbers $$(z^3 + 1)^3 = 1$$
where $z$ is an element of the complex number system.
Can someone show me the most efficient way of finding all the solutions for $z$ here and also if possible please demonstrate how both sides of the equation can be converted into polar coordinates and then the equation be solved?
Thanks a lot!!
| As $\displaystyle(z^3+1)^3=1=\cos0+i\sin 0,$
using this, $\displaystyle z^3+1=\cos\dfrac{(2k\pi+0)}3+i\sin\dfrac{(2k\pi+0)}3$ where $k=0,1,2$
$\displaystyle\implies z^3=\cos\dfrac{(2k\pi+0)}3+i\sin\dfrac{(2k\pi+0)}3-1$
Using Double-Angle Formulas,
$\displaystyle z^3=2i\sin\frac{k\pi}3\cos\frac{k\pi}3-2\sin^2\frac{k\pi}3$
$\displaystyle=2i\sin\frac{k\pi}3\left(\cos\frac{k\pi}3+i\sin\frac{k\pi}3\right)$
$\displaystyle=2\sin\frac{k\pi}3\left(\cos\frac\pi2+i\sin\frac\pi2\right)\left(\cos\frac{k\pi}3+i\sin\frac{k\pi}3\right)$ as $\displaystyle\cos\frac\pi2+i\sin\frac\pi2=i$
$\displaystyle\implies z^3=2\sin\frac{k\pi}3\left[\cos\left(\frac\pi2+\frac{k\pi}3\right)+i\sin\left(\frac\pi2+\frac{k\pi}3\right)\right]$
Again apply this
| {
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Prove or disprove: $99^{100}+100^{101}+101^{99}+1$ is a prime number
Prove or disprove: $$99^{100}+100^{101}+101^{99}+1$$ is a prime number.
My idea: let $100^{101}=x^{x+1}$,then
$$99^{100}+100^{101}+101^{99}+1=(x-1)^{x}+x^{x+1}+(x+1)^{x-1}+1$$ is prime number？
I have solved following problem before:
Show that $$5^{100}+5^{75}+5^{50}+5^{25}+1$$ is not prime number.
Let $x=5^{25}$, then
$$x^4+x^3+x^2+x+1=(x^2+3x+1)^2-5x(x+1)^2$$
Note that
$$5x(x+1)^2=5^{26}(x+1)^2=[5^{13}(x+1)]^2$$
But for my problem it looks I can't use this approach.
| Similar Mathematica code is FactorInteger[99^100 + 100^101 + 101^99 + 1, 2] which returns $825277$ and $\frac{99^{100} + 100^{101} + 101^{99} + 1}{825277}$.
| {
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If $x,y,z \in \mathbb{R^+}$ such that $x+y+z=3$. Prove the inequality $\sqrt x+\sqrt y+\sqrt z\ge xy+yz+zx$ If $x,y,z \in \mathbb{R^+}$ such that $x+y+z=3$. Prove the inequality $\sqrt x+\sqrt y+\sqrt z\ge xy+yz+zx$.
My work: We have
$$3(x+y+z)=x^2+y^2+z^2+2(xy+yz+zx) \implies (xy+yz+zx)=\dfrac12(3x-x^2+3y-y^2+3z-z^2)$$
So, we have to prove, $\sqrt x+\sqrt y+\sqrt z-\dfrac12(3x-x^2+3y-y^2+3z-z^2)\ge 0$
Now, I cannot proceed further. Please help.
| Hint: Now, one way to proceed: show $x^2-3x+2\sqrt x>0$. Let $x=w^2$ and factor.
| {
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How to find continued fraction of pi I have always been amazed by the continued fractions for $\pi$. For example some continued fractions for pi are:
$\pi=[3:7,15,1,292,.....]$ and many others given here.
Similarly some nice continued fractions for $e$ and it's derivatives are given here.I have tried to prove that the are indeed the continued fractions but did not get very far.If any one can can help it would be great(especially if there is some easy way to get continued fraction of derivatives of e from the original continued fraction of e).Here derivatives of e I mean $\sqrt e,\frac{e-1}{e+1} $ etc.
| A neat method to construct a continued fraction for $\pi$ is to use the addition formula for $\arctan$:
$$\arctan(x) + \arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$$
which can also be written
$$\arctan\left(\frac{1}{x}\right) + \arctan\left(\frac{1}{y}\right) = -\arctan\left(\frac{x+y}{1-xy}\right) = \arctan\left(\frac{1}{x - \frac{1+x^2}{x+y}}\right)$$
Applying this formula one more time gives
$$\arctan\left(\frac{1}{x}\right) + \arctan\left(\frac{1}{y}\right) + \arctan\left(\frac{1}{z}\right) = \arctan\frac{1}{z-\frac{1+z^2}{z + x - \frac{1+x^2}{x+y}}}$$
and by induction we arrive at
$$\arctan\left(\frac{1}{x_1}\right)-\arctan\left(\frac{1}{x_2}\right)+\arctan\left(\frac{1}{x_3}\right) - \ldots = \arctan\cfrac{1}{x_1+\cfrac{1+x_1^2}{x_2-x_1 + \cfrac{1+x_2^2}{x_3 - x_2 + \ddots}}}$$
If we now let $x_k = \frac{2k-1}{\epsilon}$ form an arithmetic series then
$$\sum_{k=1}^\infty (-1)^{k-1}\arctan\left(\frac{\epsilon}{2k-1}\right) = \arctan\cfrac{\epsilon}{1+\cfrac{1+\epsilon^2}{2 + \cfrac{9+\epsilon^2}{2 + \cfrac{25+\epsilon^2}{2 + \ddots}}}}$$
Dividing by $\epsilon$ and taking $\epsilon\to 0$ gives us
$$1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots = \cfrac{1}{1+\cfrac{1}{2 + \cfrac{9}{2 + \cfrac{25}{2 + \ddots}}}} = \frac{\pi}{4}$$
since the series on the left is just Leibniz formula for $\frac{\pi}{4}$. This can be used to generate other simple continued fractions with known sum by taking $x_k = \frac{ak+b}{\epsilon}$ (as long as we are able to evaluate $\sum(-1)^{k-1}\frac{1}{ak+b}$). The value of this particular continued fraction was first found by Brouncker and I came across it, and the method used to generate it, in an article by Viggo Brun in an old mathematics journal from the 1950's (it was not specified what method Brouncker used to derive it, other than that he derived it after learning about Wallis product formula).
| {
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Partial Fractions - $\frac{x^3}{x^2 + 12x +36}$ Ok, so I know that since the numerator has a higher power that long division is needed. So after doing that, the main fraction is $\frac{-6x-36}{x^2 + 12x + 36}$. I think that's right. But my problem is that after you factor the denominator they're both equal to $(x+6)$. So how would you use partial fraction decomposition?
| After the observation that the denominator is equal to $(x+6)^2$ (which was mentioned in other answer), this problem can be solved relatively easy using the substitution $y=x+6$.
$$
\begin{align*}
\frac{x^3}{(x+6)^2}
&=\frac{(y-6)^3}{y^2}=\\
&=\frac{y^3-3\cdot 6 y^2+3\cdot 6^2 y - 6^3}{y^2}=\\
&=\frac{y^3-18y^2+108y-216}{y^2}=\\
&=y-18+\frac{108}y-\frac{216}{y^2}=\\
&=x-12+\frac{108}{x+6}-\frac{216}{(x+6)^2}
\end{align*}$$
You can check the result using WolframAlpha.
| {
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Resolving a Limit I'm studying for college exams and I don't know how to solve this type of limit:
$$\begin{align}
\lim_{x\to -3} \sqrt{\frac{x^2-9}{2x^2+7x+3}}
\end{align}$$
Any help?
Update: I know that the solution is:
$$\begin{align}
\frac{1}{5} \sqrt{30}
\end{align}$$
| We have by the L'Hôpital theorem:
$$\begin{align}
\lim_{x\to -3} \frac{x^2-9}{2x^2+7x+3}=\lim_{x\to -3}\frac{2x}{4x+7}=\frac{6}{5}\end{align}$$
Notice that
$$\lim_{x\to -3}{\sqrt{\frac{x^2-9}{2x^2+7x+3}}}=\sqrt{\lim_{x\to -3}\frac{x^2-9}{2x^2+7x+3}}\quad\text{Why?}$$
| {
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Find complicated Taylor Series According to some software, the power series of the expression,
$$\frac{1}{2} \sqrt{-1+\sqrt{1+8 x}}$$
around $x=0$ is
$$\sqrt{x}-x^{3/2}+\mathcal{O}(x^{5/2}).$$
When I try to do it I find that I can't calculate Taylor because there are divisions by zero. Also I do not understand how Taylor could give non integer powers.
Does anybody know how this expression is calculated?
|
$$f(x) =\frac{1}{2} \sqrt{-1+\sqrt{1+8 x}}$$
The argument of the square root is 0, hence $f$ has no derivative in 0. That is why you get non integer powers.
First, get the equivalent:
$$\sqrt{1+8 x}-1 \sim 4x\\
f(x) = \frac 12\sqrt{-1+\sqrt{1+8 x}}\sim \frac 12\sqrt{4x} = \sqrt{x}
$$
and then:
$$
f(x) -\sqrt{x}
= \frac 12\sqrt{-1+\sqrt{1+8 x}} - \frac 12\sqrt{4x}
\\= \frac 12 \frac{-1+\sqrt{1+8 x} +4x}
{\sqrt{-1+\sqrt{1+8 x}} - \sqrt{4x}}
\sim
\frac 12 \frac{-\frac 18 (8x)^2}{4\sqrt{x}}=-x\sqrt{x}
$$
| {
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Help with surface integral question Find the surface area of the plane
$$x+\frac{1}{\sqrt{2}}y+\frac 14 z=1$$
limited by the coordinate system planes
My findings :
I suppose we should express the scalar $z=f(x,y) \rightarrow z=4-4x-2\sqrt{2}y$ and then $\mathbf{r}(s,t)$ is
$$\mathbf{r}(s,t)=s\mathbf{i}+t\mathbf{j}+(4-4s-2\sqrt{2}t)\mathbf{k}$$
and normal is
$$\mathbf{n}(s,t)=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -4 \\ 0 & 1 & -2\sqrt{2} \end{vmatrix}=4\mathbf{i}+2\sqrt{2}\mathbf{j}+\mathbf{k}$$
Good so far? I guess I should find $(u,s)\in \text{"limits"}$ and use the right integral. Any hints/help?
$$\int \int_S gdS = \int \int_D g(\mathbf{r}(s,t)) \sqrt{(\frac{\partial{f}}{\partial{s}})^2+(\frac {\partial{f}}{\partial{t}})^2+1}dsdt$$
| The area of a surface is $$\iint\limits_{\Sigma} dS,$$ where $\Sigma$ is the surface and $dS$ is the surface area element. It is computed by finding the length of the normal vector to the surface. In your case we have the graph of a function, therefore
$$dS = \sqrt{ 1 + \left( \frac{\partial f}{\partial s} \right)^2 + \left( \frac{\partial f}{\partial t} \right)^2 } \, dA,$$
where $dA$ is the area element in a projected region, assuming parameters $s,t$ as you did. Writing
$$z = 4 - 4x - 2 \sqrt{2} y = 4 - 4s - 2 \sqrt{2}t = f(s,t)$$
I am taking $x=s$ and $y=t$ (arbitrarily). The projected region will be when $z=0$, yielding $4s + 2\sqrt{2} t = 4$ or $2s + \sqrt{2} t = 2$. Since $s,t \geq 0$ we have $s \in [0,1]$ and $t \in [0, \sqrt{2}]$.
The length of the normal vector is
$$\sqrt{ 1 + \left( \frac{\partial f}{\partial s} \right)^2 + \left( \frac{\partial f}{\partial t} \right)^2 } = \sqrt{ 1 + (-4)^2 + (-2 \sqrt{2})^2 } = \sqrt{1 + 16 + 8} = 5.$$
Our surface area will be
$$\iint\limits_{\Sigma} dS = \iint\limits_{R} 5 \,dA = 5 \iint\limits_{R} \, dA,$$
where $R$ is the region projected in the $xy$ plane. Setting up the integral gives
$$5 \iint\limits_{R} \, dA = 5 \int_0^{\sqrt{2}} \hspace{-5pt} \int_0^1 dx \, dy.$$
You can compute the integrals or realize this is a right triangle with sides 1 and $\sqrt{2}$ and therefore the area is $1 \cdot \frac{\sqrt{2}}{2}$ and the surface area is
$$\iint\limits_{\Sigma} dS = \frac{5 \sqrt{2}}{2}.$$
| {
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How to prove the formula of altitude from this following triangle? Given: Right triangle $\triangle ABC$ with $A$ as right angle.
If $t_A$ is altitude that drawn from point $A$ to $\overline{BC}$, called $\overline{AD}$.
Prove that $t_A = \sqrt{2}\cdot\dfrac{bc}{b+c}$
Thanks
| By the picture that you drew, we have $\frac{AD}{c}=\frac{b}{\sqrt{b^2+c^2}}$, which gives $AD=\frac{bc}{\sqrt{b^2+c^2}}$.
This is hardly ever equal to $\sqrt{2}\frac{bc}{b+c}$. For if we had equality, we would have $\frac{\sqrt{2}}{b+c}=\frac{1}{\sqrt{b^2+c^2}}$, or equivalently $2(b^2+c^2)=(b+c)^2$. This simplifies to $b^2-2bc+c^2=0$, that is, $b=c$.
So the proposed formula only gives the right answer if our triangle is right-angled and isosceles.
| {
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Integers that can be expressed as $a^3+b^3+c^3-3abc$ $$S=\{a^3+b^3+c^3-3abc|a,b,c\in\Bbb Z\}$$
Can we decide $S$? that is, we want to find all integers of the form $a^3+b^3+c^3-3abc$.
obviously,
*
*if $m,n\in S$, then $mn\in S$, so we only need to consider primes;
*if $n\in S$, then $-n\in S$.
Let $f(a,b,c)=a^3+b^3+c^3-3abc$. $f(0,0,0)=0$, $f(1,0,0)=1$, $f(1,1,0)=2$, we get that $0,1,2\in S$
p.s. I find a solution for $a,b,c \geq0$: $n=2^rp_1^{r_1}\dotsb p_s^{r_s}$, $p_1< p_2<...$ are odd primes, then a suficient and necessary condition for $n$ can be expressed as $a^3+b^3+c^3-3abc$($a,b,c \geq0$) is $p_1>3$ or $p_1=3$ with $r_1\geq2$
| Note that $$(a\pm1)^3+a^3+a^3-3(a\pm1)a^2=3a^3\pm3a^2+3a\pm1-3a^3\mp3a^2=3a\pm1$$
and
$$(a+1)^3+a^3+(a-1)^3-3(a+1)a(a-1)=3a^3+6a-3a^3+3a=9a$$
Suppose $n=a^3+b^3+c^3-3abc$ where $n\in\mathbb Z$.
So we can get all $n$ such that $3\nmid n$ or $9\mid n$.
Note that $(3p+k)^3\equiv k^3\equiv k\ \ (\text{mod 9})$ for $k\in\{-1,0,1\}$.
We can write $a=3p+k$, $b=3q+l$ and $c=3r+m$, where $k,l,m\in\{-1,0,1\}$. Then $$a^3+b^3+c^3-3abc\equiv k+l+m-3(3p+k)(3q+l)(3r+m)\equiv k+l+m-3klm\ \ (\text{mod }9)$$
Additionally
$$k+l+m-3klm\equiv k+l+m\ \ (\text{mod 3})$$
Suppose $3\mid n$. Then $3\mid k+l+m$.
If $k=0$, $l=0$ or $m=0$ the other two must be opposite, so $k+l+m-3klm=0$ and $9\mid n$. Else we can only have $k,l,m=-1$ or $k,l,m=1$. But in both of these cases we get $9\mid n$ too.
We can conclude the only possible values are such $n\in\mathbb Z$ that $3\nmid n$ or $9\mid n$.
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Minimum area of a triangle In triangle inscribed circle with radius $r = 1$ and one of it sides $a=3$. Find the minimum area of triangle? Ans = 5.4
My reasonings:
$BC = a$, $AC = b$, $AB = c$
$AD=AF=x$
$FC=CE=y$
$BD=BE=z$
$a=z+y$, $b=x+y$, $c=x+z$
The radius of the incircle is $$r =\frac{A_{ABC}}{s}$$ where $s = \frac{a+b+c}{2} = x + y+ z$. By condition $z+y=3$ so $s=x+3$
By Heron's formula, the area of the triangle is $A=\sqrt{s(s-a)(s-b)(s-c)}$
In other side $A=sr = x+3$.
What is next? I think that I should get a fucntion for which I will can find a minimum, but I don't know how.
| Trigonometric approach:
In your notation,
$AD=AF=x$, $FC=CE=y$, $BD=BE=z$,
denote in addition
$\alpha=\angle OAF=\angle OAD$,
$~~~\beta=\angle OCF=\angle OCE$,
$~~~\gamma=\angle OBE=\angle OBD$.
So, if $r=1$, then
$x=\dfrac{1}{\tan\alpha}$,
$~~~y=\dfrac{1}{\tan\beta}$,
$~~~z=\dfrac{1}{\tan\gamma}$.
Then, as you said,
$$
A=r(x+y+z)=r(x+3)=x+3.
$$
$$
x = \dfrac{1}{\tan(90^\circ - \beta-\gamma)} = \tan(\beta+\gamma) = \dfrac{\tan\beta+\tan\gamma}{1-\tan\beta\tan\gamma}.
$$
Dividing numerator and denominator by $(\tan\beta\tan\gamma)$, we get:
$$
x=\frac{z+y}{yz-1}=\frac{3}{yz-1}.
$$
$$
A=x+3=\frac{3}{yz-1}+3.
$$
Further thoughts - as in answer of ElThor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/725056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Solve $\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx$ In solving the definite integral $$\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx$$I tried to do $$\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx=\int{\frac{1}{\sqrt{3x^2-4}}}\;dx-\int{\frac{1}{x\sqrt{3x^2-4}}}\;dx\Longrightarrow$$ If $x=\frac{2}{\sqrt{3}}\sec(u)\Longrightarrow dx=\frac2{\sqrt3}\tan(u)\sec(u)\;du$, then
$$\int{\frac{1}{\sqrt{3x^2-4}}}\;dx-\int{\frac{1}{x\sqrt{3x^2-4}}}\;dx=$$ $$\frac{2}{\sqrt3}\int\frac{\tan(u)\sec(u)}{2\sqrt{\left(\frac{\sqrt3}{2}\sec(u)\right)^2-1}}\;du-\frac1{2}\int\frac{\frac{\sqrt3}{2}}{\frac{\sqrt3}{2}x\sqrt{\left(\frac{\sqrt3}{2}x\right)^2-1}}\;dx=$$ $$\frac{1}{\sqrt3}\int\sec{u}\;du-\frac12\sec^{-1}\left(\frac{\sqrt3}{2}x\right)$$
In my feedback says that the result is $$\frac{1}{\sqrt3}\log\left|\sqrt3x+\sqrt{3x^2-4}\right|-\frac12\sec^{-1}{\left(\frac{\sqrt3}{2}x\right)}+C$$
It also says that I have to use this method of trigonometric substitution.
The path is this?
I know $\int\sec(t)dt=\log|\sec t+\tan t|$
| If $\tan t=\frac{\sqrt{3x^2-2}}{2}$ and $\cos t=\frac{2}{\sqrt{3}x} $, then $x=\frac{2}{\sqrt{3}}\sec t \Rightarrow dx=\frac{2}{\sqrt{3}}\sec t \tan t dt$ and so
$$\int \frac{x-1}{x\sqrt{3x^2-4}}dx=\int \frac{(\frac{2}{\sqrt{3}}\sec t-1)}{\frac{2}{\sqrt{3}}\sec t\cdot 2\tan t} \cdot \frac{2}{\sqrt{3}}\sec t\tan t dt=\frac{1}{\sqrt{3}}\int\sec t dt- \frac{1}{2}\int dt=$$
$$=\frac{1}{\sqrt{3}}\ln |\sec t+\tan t|-\frac{1}{2}t +c= $$
$$\frac{1}{\sqrt{3}}\ln |\frac{\sqrt{3}x}{2}+\frac{\sqrt{3x^2-4}}{2}|-\frac{1}{2}\arccos(\frac{2}{\sqrt{3}x})+c. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/725168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Help finishing proof via induction for a summation So I have to prove the following equation using induction for n >= 2:
$$
\sum\limits_{i=1}^n 4/5^i < 1
$$
However the question asks me to prove something stronger such as this:
$$
\sum\limits_{i=1}^n 4/5^i <= 1 - \frac{1}{5^n}
$$
first to imply the first equation is true.
So far I have the following:
Base Case:
Let n = 2
$$
\sum\limits_{i=1}^2 4/5^i = \frac{4}{5} + \frac{4}{25} = \frac {24}{25}
$$
then I also applied it to
$$ 1 - \frac{1}{5^n} \rightarrow 1 - \frac{1}{5^2} = \frac{24}{25}$$
Therefore I can make the following assumptions yes?
Inductive Hypothesis
for all 2 <= n <= k it is
$$
\sum\limits_{i=1}^n 4/5^i = 4\frac{\frac{1}{5^n} - 1}{\frac{1}{5} - 1} = 1 - \frac{1}{5^n} < 1
$$
Inductive Step
Hopefully I'm ok up to here, I'll show what I have so far for this step.
$$
\sum\limits_{i=1}^{k+1} 4/5^i = \frac{\frac{1}{5^{k+1}} - 1}{\frac{1}{5} - 1} = 4\frac{(\frac{1}{5^k}-1) * \frac{1}{5} - \frac{4}{5}}{\frac{1}{5} -1} $$
$$
= \frac{1}{5} * 4\frac{(\frac{1}{5^k}) - 1}{\frac{1}{5} -1} - 4\frac{\frac{4}{5}}{\frac{1}{5} - 1}
$$
so here I have:
$$
4\frac{(\frac{1}{5^k}) - 1}{\frac{1}{5} -1}
$$
which I know is:
$$
= \sum\limits_{i=1}^k 4/5^i
$$
which is my inductive hypothesis, I am unsure of how to finish my proof from here... any help correcting or finishing the proof is very much appreciated
| $$\sum_{i=1}^{3}1/5^i=(1/5+1/5^2+1/5^3)*(1-1/5)/(1-1/5)=(1-1/5^3)/4 $$
By induction it is easy to see that
$$\sum_{i=1}^{n}1/5^i = (1-1/5^n)/4 $$
And so
$$\sum_{i=1}^{n}4/5^i = (1-1/5^n) $$
Which is strictly less than one.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the largest circle that fits in $\sin(x)?$ Imagine dropping a circle into the trough of $\sin(x)$. Would it reach the bottom or get wedged between two points on the curve? Depends on the size of the circle.
So, what is the radius of the largest circle that will reach the bottom of the curve $y=\sin(x)$?
This problem was inspired by this similar one I found in a calculus textbook: Find the radius of the largest circle that will reach the bottom of the curve $y=x^2$ without getting stuck. I was intrigued by the answer of $r=1/2$.
I have tried attacking this problem from several angles but all have failed. Perhaps an exact numerical solution is not obtainable, I don't know. All help will be appreciated.
Thanks!
| Nice question.
If we are dropping the circle so that its center is right above $\frac{3\pi}{2}$ then the coordinates of the center of the dropped circle will be $(x_0,y_0) = (3\pi/2,r-1)$ where $r$ is the radius of the circle. Now, the distance from this point to any point $(x,y)$ on the curve of $\sin$ is
$$
\sqrt{(x_0 - x)^2 + (y_0 - y)^2}
= \sqrt{ (3\pi/2-x)^2 + (r-1)^2 - 2(r-1)\sin x + \sin^2 x},
$$
and this can be no less than $r$. The resulting inequality is
$$
2r(1 + \sin x)
\leq (3\pi/2-x)^2 + 1 + 2\sin x + \sin^2 x
$$
which leads to
$ r \leq ((3\pi/2-x)^2 + (1+\sin x)^2)/2(1+\sin x)$. Then the answer should be
\begin{align}
r
&= \inf \{ ((3\pi/2-x)^2 + (1+\sin x)^2)/2(1+\sin x) \mid x \in (0,\pi) \} \\
&= \frac{1}{2}\inf \{ (3\pi/2-x)^2/(1+\sin x) + (1+\sin x) \mid x \in (0,\pi) \}.
\end{align}
Let the function in the infimum be $f$. Differentiating yields
$$
\frac{-2(3\pi/2 - x)(1+\sin x) - \cos x(3\pi/2-x)^2}{(1+\sin x)^2} + \cos x
$$
and if you solve for where this is $0$ you get
$$
\cos x((1 + \sin x)^2 - (3\pi/2-x)^2)
= 2(3\pi/2-x)(1+\sin x).
$$
One solution is at $x = 3\pi/2$, using L'hopital's rule,
$$
\lim_{x \to 3\pi/2} f(x)
= \frac{1}{2} \lim_{x \to 3\pi/2} \frac{-2(3\pi/2-x)}{\cos x}
= \frac{1}{2} \lim_{x \to 3\pi/2} \frac{2}{-\sin x}
= 1.
$$
This is $r$ (it just remains to show that the critical point is the minimum for $f$, and the first derivative test verifies this).
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "17",
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show that $a^3+4a=b^2$ then exist $M\in\mathbb{N}$ s.t $a=2M^2$ I would appreciate if somebody could help me with the following problem:
Q: show that
If $a,b$ satisfy equation
$$a^3+4a=b^2(a,b\in\mathbb{N})$$ then there exist $M\in\mathbb{N}$ s.t $a=2M^2$
| Let $d = \gcd(a,b)$. So $a = xd$, and $b = yd$. Thus: $x^3d^3 + 4xd = y^2d^2$. So $d^2x^3 + 4x = dy^2$. This gives: $x$ divides $dy^2$. But $\gcd(x,y) = 1$ so $\gcd(x,y^2) = 1$. Thus $x$ divides $d$. So $d = kx$. So: $k^2x^5 + 4x = kxy^2$. Thus: $k^2x^4 + 4 = ky^2$. Hence: $4 = k(y^2 - kx^4)$. This means $k$ divides $4$. So $k = 1, 2, 4$.
Case 1: $k = 1$. So: $4 = y^2 - x^4 = (y - x^2)(y + x^2)$. So $y - x^2 = 1, y + x^2 = 4$. So $2y = 5$. This can't happen since $y$ has to be an integer.
Case 2: $k = 2$. So: $2 = y^2 - 2x^4$. So $y^2 = 2(x^4 + 1)$. So $x^4 + 1 = 2m^2$. Observe that this equation has initial solution: $x = 1 = m$ and it will have infinitely many solutions.
Case 3: $k = 4$. So: $1 = y^2 - 4x^4 = (y - 2x^2)(y + 2x^2)$. This can't happen. So we have: $k = 2$ is the only acceptable solution and this means: $a = 2x^2 = 2M^2$ with $M = x$. Done.
| {
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"timestamp": "2023-03-29T00:00:00",
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A question on multinomial theorem using binomial theorem $(3x^2+2x+c)^{12}=\sum A_r x^r$
and
$\frac{A_{19}}{A_5}=\frac 1 {2^7}$
Find $c$. I really have no idea what to do with this. This was on a test. I have studied only binomial theorem. So, please use it only(combinatorics approach is always welcome).
| By comparing coefficients
$$
A_{19} = \frac{12!}{9! \cdot 1! \cdot 2!} \cdot 3^9 \cdot 2 \cdot c^2 + \frac{12!}{8! \cdot 3! \cdot 1!} 3^8 \cdot 2^3 \cdot c + \frac{12!}{7! \cdot 5! \cdot 0!} 3^7 \cdot 2^5 = 3^7\left(\frac{12!}{9! \cdot 1! \cdot 2!}18c^2 + \frac{12!}{8! \cdot 3! \cdot 1!}24c + \frac{12!}{7! \cdot 5! \cdot 0!} 32\right) \;,
$$
and
$$
A_5 = \frac{12!}{7! \cdot 5! \cdot 0!} 2^5 \cdot c^7 + \frac{12!}{8! \cdot 3! \cdot 1!} 3 \cdot 2^3 \cdot c^8 + \frac{12!}{9! \cdot 1! \cdot 2!} 3^2 \cdot 2 \cdot c^9 = c^7\left(\frac{12!}{9! \cdot 1! \cdot 2!}18c^2 + \frac{12!}{8! \cdot 3! \cdot 1!}24c + \frac{12!}{7! \cdot 5! \cdot 0!} 32\right) \;.
$$
Thus, by dividing, we get
$$
\frac{A_5}{A_{19}} = \left(\frac{c}{3}\right)^7 \;,
$$
which implies $c = 6$ by the given condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/731335",
"timestamp": "2023-03-29T00:00:00",
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Show that $\int_{0}^{1}\{P(x)\}^{2}\,dx = (n + 1)^{2}\left(\int_{0}^{1}P(x)\,dx\right)^{2}$
Suppose $P(x)$ is a polynomial of degree $n \geq 1$ such that $\int_{0}^{1}x^{k}P(x)\,dx = 0$ for $k = 1, 2, \ldots, n$. Show that $$\int_{0}^{1}\{P(x)\}^{2}\,dx = (n + 1)^{2}\left(\int_{0}^{1}P(x)\,dx\right)^{2}$$
If we assume that $P(x) = a_{0}x^{n} + \cdots + a_{n - 1}x + a_{n}$ then we can easily see that $$\int_{0}^{1}\{P(x)\}^{2}\,dx = a_{n}\int_{0}^{1}P(x)\,dx$$ and therefore to solve the given problem we need to show that $$\int_{0}^{1}P(x)\,dx = \frac{a_{n}}{(n + 1)^{2}}$$ Direct integration of the polynomial gives the expression $$\frac{a_{0}}{n + 1} + \frac{a_{1}}{n} + \cdots + \frac{a_{n - 1}}{2} + a_{n}$$ and simplifying this to $a_{n}/(n + 1)^{2}$ does not seem possible. I think there is some nice "integration by parts" trick which will give away the solution, but I am not able to think of it.
| Since, $\displaystyle \int_0^1 x^kP(x)\,dx=\frac{a_0}{n+k+1}+\frac{a_1}{n+k}+\ldots+\frac{a_n}{k+1}=0$, for each $k=1,2,\ldots,n$
Then $f(x)=\dfrac{a_0}{n+x+1}+\dfrac{a_1}{n+x}+\ldots+\dfrac{a_n}{x+1}=\dfrac{Q(x)}{(n+x+1)\ldots(x+1)}$ (where, Q is a polynomial of degree atmost n), has $n$ zeros $x=1,2,\ldots,n$.
Thus, $Q(x)=c(x-1)(x-2)\ldots(x-n)$, for some constant $c$.
Also, $(x+1)f(x)=\dfrac{a_0(x+1)}{n+x+1}+\dfrac{a_1(x+1)}{n+x}+\ldots+a_n=\dfrac{Q(x)}{(n+x+1)\ldots(x+2)}$
Setting, $x=-1$ in the above expression $a_n = \dfrac{Q(-1)}{n!}=\dfrac{c(-1)^n(n+1)!}{n!}=c(-1)^n(n+1)$
and, setting $x=0$ we have $\displaystyle \int_0^1 P(x)\,dx = \dfrac{a_0}{n+1}+\ldots+a_n=\dfrac{Q(0)}{(n+1)!}=\dfrac{c(-1)^n}{n+1}$.
Thus $a_n=\displaystyle (n+1)^2\int_0^1 P(x)\,dx$, implying $\displaystyle \int_{0}^{1}\{P(x)\}^{2}\,dx = (n + 1)^{2}\left(\int_{0}^{1}P(x)\,dx\right)^{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to solve $x^4-8x^3+24x^2-32x+16=0$ How can we solve this equation?
$x^4-8x^3+24x^2-32x+16=0.$
| You could factorise it, in the manner of $(x-2)^4=0$. I saw those factors immediately.
One process is to note that $16$ has divisors, and one can try various combinations of this such that the sum gives eight.
Possibilities include $2, 2, 2, 2$ and $4, 4, 1, -1$. However, one can not produce the second set to give +16, so trying $(x-2)(x-2)(x-2)(x-2)$ is more likely than $(x-4)(x-4)(x-1)(x+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/732334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The derivative of $2^{x+1}$ in the point $-3$ They ask me to compute the derivative of $2^{x+1}$ in the point $-3$. Since the function is continous in that point, all I have to do is to compute the lateral derivatives. The left one first:
$$\lim_{x \to -3^-} \frac{2^{x+1} - \frac{1}{4}}{x+3}$$
I've tried to solve this limit using this formula:
$$\frac{a^{u(x)} - 1}{u(x)} = \ln{a}$$
But the result seems to be incorrect. How should I solve that limit?
| $$\lim_{x \to -3^-} \frac{2^{x+1} - \frac{1}{4}}{x+3} = \lim_{x \to -3^-} \frac{2^{x+1}\frac{2^2}{2^2} - \frac{1}{4}}{x+3} = \lim_{x \to -3^-} \frac{\frac{2^{x+3}}{4} - \frac{1}{4}}{x+3} =$$
$$= \frac{1}{4}\lim_{x \to -3^-} \frac{2^{x+3}- 1}{x+3} = \frac{1}{4}\ln{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/738128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to integrate $\int_0^{\theta}\frac{x}{(1-x^2)(1-x)}dx$?
How to integrate $\displaystyle\int_0^{\theta}\frac{x}{(1-x^2)(1-x)}dx$ ?
If $\theta\in(0,\frac{1}{2})$
$$
\begin{align}
& \int_0^{\theta}\frac{x}{(1-x^2)(1-x)}\,dx \\[8pt]
= & \int_0^{\theta}\frac{x}{(1-x)(1+x)(1-x)}\,dx \\[8pt]
= & \int_0^{\theta}\frac{1}{(1-x)^2}\frac{x}{1+x}dx \\[8pt]
= & \int_0^{\theta}\frac{1}{(1-x)^2}\Big(1-\frac{1}{1+x}\Big) \, dx \\[8pt]
= & \int_0^{\theta}\frac{1}{(1-x)^2}-\displaystyle\int_0^{\theta}\frac{1}{(1-x^2)(1-x)} \, dx
\end{align}
$$
LHS is OK, but RHS ?
| How about partial fraction decomposition?
$$\begin{align} \displaystyle\int_0^{\theta}\frac{x}{(1-x^2)(1-x)}dx & = \int_0^\theta \frac x{(1-x)(1+x)(1 - x)} \,dx \\ \\ & = \int_0^\theta \frac x{(1+x)(1-x)^2}dx \\ \\ & = \int_0^\theta \left[ \frac A{1+x} + \frac B{1-x} + \frac C{(1-x)^2}\right]dx\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/738424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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coefficient of $x^2$, in $(1+x+x^2)^{10}$
How to find coefficient of $x^2$, in $(1+x+x^2)^{10}$, without actually expanding it?
I think the fact $\dfrac{1-x^3}{1-x}=1+x+x^2$ may help.
But can't use it!
| Method $\#1:$
Following your process, $$(1+x+x^2)^{10}=\left(\frac{1-x^3}{1-x}\right)^{10} =(1-x^3)^{10}(1-x)^{-10}$$
$$=\left[1-\binom{10}1x^3+\binom{10}2x^6-\cdots\right]\left[1-10(-x)+\frac{-10(-10-1)}2(-x)^2+\cdots\right]$$
Clearly the coefficient of $x^2$ is $\displaystyle \frac{-10(-10-1)}2=55$
Method $\#2:$
For a clearer version of user$125053$'s answer
$$(1+x+x^2)^{10}=\left[(1+x)+x^2\right]^{10}$$
$$=\binom{10}0(1+x)^{10}+\binom{10}1(1+x)^9(x^2)^1+\binom{10}1(1+x)^8(x^2)^2\cdots$$
So, the required coefficient of $x^2$ is
the coefficient of $x^2$ in $\displaystyle\binom{10}0(1+x)^{10}+$ the coefficient of $x^0$ in $\displaystyle\binom{10}1(1+x)^9$
$\displaystyle=\binom{10}0\cdot\binom{10}2+\binom{10}1\cdot\binom90 =55 $ again
Method $\#3:$
$$(1+x+x^2)^{10}=\left[1+x(1+x)\right]^{10}$$
$$=1+\binom{10}1x(1+x)+\binom{10}2\{x(1+x)\}^2+\binom{10}3\{x(1+x)\}^3+\cdots$$
$$=1+\binom{10}1(x+x^2)+\binom{10}2\{x^2(1+x)^2\}+\binom{10}3\{x(1+x)\}^3+\cdots$$
So, the required coefficient of $x^2$ is $\displaystyle\binom{10}1+\binom{10}2=55$
Method $\#4:$
Using Multinomial Theorem,
the general term of the expansion of $\displaystyle(1+x+x^2)^{10}$
is $\displaystyle\frac{10!}{a!b!c!}(1)^a(x)^b(x^2)^c=\frac{10!}{a!b!c!}x^{b+2c}$ where $a,b,c\ge0$ and $a+b+c=10$
Now we need $b+2c=2\iff2c=2-b\le2$ as $b\ge0$
$\displaystyle\implies b\le1\implies b=0,1$
If $b=0,\cdots$
If $b=1,\cdots$
| {
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if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$ This is supposed to be an application of AM-GM inequality.
if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$
First of all,
$a^2+b^2+c^2\ge 3$
by a direct application of AM-GM.Also,we have
$a^2+b^2+c^2\ge ab+bc+ca$
Next,we consider the expression
$(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc\le a+b+c+a^2+b^2+c^2+1$
but that hardly helps.I know that
$3(a^2+b^2+c^2)\ge (a+b+c)^2$
From the first derived inequality,we know that the left hand side of the above inequality is greater than or equal to 9.But I can't see how that can be used here.I know that we can get $a+b+c\ge 3$ using AM-GM but that does not take me a step closer to finding the solution(from what I can understand).Also,
$a^3+b^3+c^3\ge a^2b+b^2c+c^2a$
$(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]\ge a^2b+b^2c+c^2a$
A hint will be appreciated at this point.
| Notice that for all real $a,b,c$,
$$(a - 1)^2 + (b-1)^2 + (c - 1)^2 \ge 0.$$
$$a^2 + b^2 + c^2 - 2a - 2b - 2c + 3 \ge 0.$$
$$a^2 + b^2 + c^2 \ge -3 + (a + b + c) + (a + b + c).$$
But by AM-GM, $a + b + c \ge 3\sqrt[3]{abc} = 3$. So,
$$a^2 + b^2 + c^2 \ge -3 + 3 + (a + b + c).$$
$$a^2 + b^2 + c^2 \ge a + b + c.$$
Equality is attained when $a = b = c = 1$.
(I really wish to put everything after the first line as a spoiler so that this answer becomes a hint, but I don't know how D:)
| {
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"answer_id": 4
} |
Find domains of convergence of the series I tried to solve this question but the final solution which I obtain is not the as same as in the text book
Find domains of convergence of the series
$$\sum_{n=1}^ \infty \frac{1.3.5...(2n-1)}{n!}(\frac{1-z}{z})^n$$
The book answer is $|z-\frac{4}{5}|\lt \frac{2}{3}$
But my answer is $$|\frac{1-z}{z}|\lt \frac{1}{2}$$
Is this true ??
How can I obtain the solution as in the book .
| @DonAntonio Your solution is correct :
$$\left|\frac{1-z}z\right|<\frac12\iff|2-2z|<|z|$$
Now put $\;z=x+iy\;,\;\;x,y\in\Bbb R\implies 2-2z=(2-2x)-2yi\;$ , so::
$$|2-2z|^2<|z|^2\iff(2-2x)^2+4y^2<x^2+y^2\iff$$
$$4-8x+4x^2+4y^2<x^2+y^2\iff3x^2-8x+3y^2+4<0\iff$$
$$3\left(x-\frac43\right)^2-\frac{16}3+3y^2+4<0\iff 3\left(x-\frac43\right)^2+3y^2<\frac43\iff$$
$$\left(x-\frac43\right)^2+y^2<\left(\frac23\right)^2\ldots$$
Thus
$$|z-\frac43|\lt \frac23 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/740918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Use the integration formula $\frac{1}{a}\arctan\frac{x}{a}$ to solve $\frac{1}{2} \int_{-1}^1 \mathrm{ \frac{dx}{1+\sqrt{2}x+x^2} }\, $ As question states, I am trying to figure out how to use the integration formula to solve the integral. My issue is that the integral isn't of the form $\frac{dx}{a^2+x^2}$
| If I amy interfer in this discussion, you have been shown by various answers and comments that through the appropriate change of variable dictated by completing the square
$$\frac{1}{2} \int_{} \frac{dx}{1+\sqrt{2}x+x^2} =\frac{\tan ^{-1}\left(x\sqrt{2} +1\right)}{\sqrt{2}}$$ So, putting the bounds $$\frac{1}{2} \int_{-1}^1 \frac{dx}{1+\sqrt{2}x+x^2} =\frac{1}{\sqrt{2}}[\tan^{-1}(1+\sqrt{2})-\tan^{-1}(1-\sqrt{2})]$$ Now, use the fact that $$\tan^{-1}(a)-\tan^{-1}(b)=\tan ^{-1}\left(\frac{a-b}{1+a \times b}\right)$$ Put $a=1+\sqrt{2}$, $b=1-\sqrt{2}$ and you will find that $\frac{a-b}{1+a\times b}$ is infinity and for such an argument, the $\tan ^{-1}$ is equal to $\frac{\pi}{2}$. You then end with $$\frac{1}{2} \int_{-1}^1 \frac{dx}{1+\sqrt{2}x+x^2} =\frac{\pi }{2 \sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/741497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
solve the equation in Z Solve the equation over $\textbf{Z}$ :
2$x^2$ - 2$xy$ - 5$x$ - $y$ + 19 = 0
I tried to obtain some $(A+B)^2$ terms, but I didn't make it.
Thanks for your time!
| $2x^2 - 2xy - 5x - y + 19 = 0 \to y = \dfrac{2x^2 - 5x + 19}{2x + 1} = x - 3 + \dfrac{22}{2x + 1}$. So $2x + 1 | 22$, and $2x + 1 = 1, -1 , 11, -11$. we can then find $y$ for each value of $x$ listed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/743126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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