Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Limit, Riemann Sum, Integration, Natural logarithm For any natural number $m$, $\lim_{n\rightarrow \infty }\left ( \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots +\frac{1}{mn} \right )=\ln (m)$.
I tried to prove the statement in the following way.
Proof:
$$\lim_{n\rightarrow \infty }\left ( \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots +\frac{1}{mn} \right )=\lim_{n\rightarrow \infty }\sum_{r=1}^{(m-1)n}\frac{1}{n+r}$$
Dividing the numerator and the denominator of $\frac{1}{n+r}$ by $n$, we get $\frac{1/n}{1+r/n}$.
Therefore,
$$\lim_{n\rightarrow \infty }\sum_{r=1}^{(m-1)n}\frac{1}{n+r}=\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{r=1}^{(m-1)n}\frac{1}{1+r/n}$$
this is a Riemann sum, so replacing $\frac{1}{n}$ with $dx$, $\frac{r}{n}$ with $x$, and integrating between the limits $x=0$ and $x=m-1$
we get $
$$\int_{0}^{m-1}\frac{dx}{1+x}=\ln(1+x)|_{0}^{m-1}=\ln(m)-\ln(1)=\ln(m)\blacksquare$$
Is this a valid way?
| A simpler and more direct choice is to write
\begin{align}\lim_{n \to \infty} \left( \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{mn} \right) &= \lim_{n \to \infty} \sum_{r=n+1}^{mn} \frac{1}{r} \\&= \lim_{n \to \infty} \frac{1}{n} \sum_{r=n+1}^{mn} \frac{1}{r/n} \\&= \int_{x=1}^m \frac{1}{x} \, dx \\&= \log m,\end{align} but your solution is valid.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding determinant of $3\times3$ matrix $$A = \left(\begin{matrix}
\lambda - 1 & -1 & -1 \\
1 & \lambda - 3 & 1 \\
-3 & 1 & \lambda + 1 \\
\end{matrix}\right)$$
$$\det A = \begin{vmatrix}A\end{vmatrix} =
(\lambda - 1) \begin{vmatrix}
\lambda - 3 & 1 \\
1 & \lambda + 1 \\
\end{vmatrix} + 1\begin{vmatrix}
1 & 1 \\
-3 & \lambda + 1 \\
\end{vmatrix} -
1\begin{vmatrix}
1 & \lambda - 3 \\
-3 & 1 \\
\end{vmatrix}\\
= ((\lambda - 1)(\lambda - 3)(\lambda + 1)) - (1 + (\lambda + 1) + 3) + (1 +3\lambda - 9) \\
= (\lambda - 1)(\lambda - 3)(\lambda + 1) + (\lambda + 1) + 2(\lambda - 3)$$
The solution says that it is $(\lambda - 2)(\lambda + 2)(\lambda - 3)$. I feel like I am so close, but I don't get what I am supposed to do to get to the solution.
| $$=(\lambda - 1)\Big((\lambda - 3)(\lambda + 1)-1\Big) + ((\lambda + 1) + 3) -(1 +3\lambda - 9) $$
$$=(\lambda - 1)\Big(\lambda ^2 - 2\lambda -4\Big) + (\lambda + 4) -(3\lambda - 8) $$
$$=(\lambda - 1)\Big(\lambda ^2 - 2\lambda -4\Big) - 2\lambda +12 $$
$$=\lambda ^3 - 3\lambda^2 -2\lambda +4 - 2\lambda +12 $$
$$=\lambda ^3 - 3\lambda^2 -4\lambda +16$$
| {
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Let $aThe prove I tried is the following. I really wish someone can check if I made some logical mistake, especially the last part I found myself diffident proving $a$ can only be $1$ or $5$.
Because $b\neq c$ and $b<c$, otherwise switch the value of $b,c$. Becuase $a\mid (3b+2c)$ and $a \mid (3c+2b), \exists q_1,q_2 \in N$ that $$aq_1=(3b+2c), aq_2=(3c+2b), a(q_2-q_1)=c-b.$$
Therefore we have $a\mid (c-b)$ and $a \mid 3(c-b)+(3b+2c)$ which is $a \mid 5c$. And we have $a \mid 2(c-b)+(3c+2b)$, $a\mid 5c$. Hence $a \mid 5b$ and $a \mid 5c$. Because $b,c$ are distinct odd primes. This is possible only when $a\mid b$ and $a \mid c$ or $a \mid 5$. The only number that divides two primes is $1$. And the two numbers that divides $5$ are $1$ and $5$. Hence the two possible values for $a$ are $1$ and $5$.
| You proof is fine, but you are over elaborating at certain points. Do you need $q_1,q_2$, for example? You do not seem to have used them.
Indeed, the simplest proof is that if $a|3b+2c$ and $a | 3c+2b$, then $a | 3(3c+2b) - 2(3b+2c) = 5c$, and $a | 3(3b+2c) - 2(3c + 2b) = 5b$.
Therefore, $a | 5c$ and $a | 5b$, which means that $a$ divides the greatest common divisor of $5c$ and $5b$. But, $(5c,5b) = 5(c,b) = 5 \times 1 = 5$ (as $c,b$ are prime), so $a | 5$.
You have done the same thing, but with a little more elaboration.
Also, it is useful to show that both cases are attained : for example, $b = 2,c=7$ gives $3b+2c = 20$, and $3c+2b = 25$.
However, note that if $5 | 3c+2b$ and $5 | 3b+2c$, then $5 | b-c$. Given that $b,c$ are primes, this forces one of $b,c$ to be even. Thus, $b = 2$ and $c = 7$ is forced.
Conclusion : if $b \neq 2, c \neq 7$, then in fact $a = 1$ is forced given $b < c$.
Try this more general question : given $ b < c$ natural numbers not necessarily prime, and integers $d,e$ , if $a | db+ce$ and $a | eb+cd$, what can you say about $a$?
| {
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Finding the first non-zero terms of a power series I have the function:
$f(x) = \frac{30}{(x^2 + 1)(x^2-9)}$
I need to find the first four non-zero terms of the power series centered at zero. I have not had much experience with power series so I am not sure how to start/complete this problem.
| $f(x)=\frac{-3}{x^2+1}+\frac{3}{x^2-9}$
$\frac{-3}{x^2+1}=-3(1-x^2+x^4-x^6+....)$
$\frac{3}{x^2-9}=\frac{-1/3}{1-x^2/9}=(-1/3)(1+x^2/9+x^4/81+x^6/729+.....)$
Combine to get $f(x)=-4/3+(80/27)x^2-........$
I'll let you finish.
| {
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Evaluate $\int_{-\infty}^\infty \frac{1}{\sqrt{z^2 + 1}}\frac{1}{z - \alpha} dz$.
Evaluate $$\int_{-\infty}^\infty \frac{1}{\sqrt{z^2 + 1}}\frac{1}{z - \alpha} dz\,.$$
What is an elegant way to evaluate this integral for Im $\alpha >0$? I imagine using residue theorem will lead to an elegant solution, such as in these related questions [1,2,3]. However I've been unable to adapt them to this line integral.
One requirement is that $\frac{1}{\sqrt{z^2 + 1}}$ be analytic in a strip around the real line $(-\infty,\infty)$. In my eyes this implies that the branch cuts can not cross the real line. For example the principal branches (parallel to the real line) or the branches $[\mathrm{i},\mathrm{i}\infty)$ and $(-\mathrm{i} \infty, -\mathrm{i}]$.
| This is probably not elegant, but you can probably find a place to use the residue theorem. Let $I$ denote the integral
$$\int_{-\infty}^{\infty}\frac{1}{\sqrt{x^2+1}\ (x-\alpha)}dx.$$
Then, $I$ equals
$$\int_0^\infty\frac{1}{\sqrt{x^2+1}}\left(\frac{1}{x-\alpha}-\frac{1}{x+\alpha}\right)dx=2\alpha\int_0^\infty\frac{1}{\sqrt{x^2+1}\ (x^2-\alpha^2)}dx$$
Take $x$ to be $\sinh(t)$. Then
$$I=2\alpha\int_0^\infty\frac{1}{\sinh^2(t)-\alpha^2}dt.$$
Since $\sinh(t)=\frac{e^t-e^{-t}}{2}$, by setting $s=e^t$, we have
$$I=2\alpha\int_1^\infty\frac{1}{\frac{1}{4}\left(s-\frac1s\right)^2-\alpha^2}\frac{ds}{s}=2\alpha\int_0^1\frac{1}{\frac{1}{4}\left(s-\frac1s\right)^2-\alpha^2}\frac{ds}{s}.$$
That is,
$$I=4\alpha\int_0^\infty\frac{s}{s^4-(4\alpha^2+2)s^2+1}ds.$$
Using partial fractions,
$$I=\int_0^\infty\left(\frac{1}{s^2-2\alpha s-1}-\frac{1}{s^2+2\alpha s-1}\right)ds.$$
(This is probably the place you can use the residue theorem but I am not too competent with that. Maybe you need to use a logarithm factor, and something like a keyhole contour.)
Since $$s^2-2\alpha s-1=(s-\alpha-\sqrt{\alpha^2+1})(s-\alpha+\sqrt{\alpha^2+1})$$ and $$s^2+2\alpha s-1=(s+\alpha-\sqrt{\alpha^2+1})(s+\alpha+\sqrt{\alpha^2+1})$$ (using the principal branch of $\sqrt{\phantom{a}}$), we get
\begin{align}I&=\frac{1}{2\sqrt{\alpha^2+1}}\int_0^\infty\left(\frac{1}{s-\alpha-\sqrt{\alpha^2+1}}-\frac{1}{s-\alpha+\sqrt{\alpha^2+1}}\right)ds\\
&\phantom{aaa}-\frac{1}{2\sqrt{\alpha^2+1}}\int_0^\infty\left(\frac{1}{s+\alpha-\sqrt{\alpha^2+1}}-\frac{1}{s+\alpha+\sqrt{\alpha^2+1}}\right)ds
\\&=-\frac{1}{2\sqrt{\alpha^2+1}}\ln\left(\frac{\alpha+\sqrt{\alpha^2+1}}{\alpha-\sqrt{\alpha^2+1}}\right)+\frac{1}{2\sqrt{\alpha^2+1}}\ln\left(\frac{\alpha-\sqrt{\alpha^2+1}}{\alpha+\sqrt{\alpha^2+1}}\right)\\&=\frac{1}{\sqrt{\alpha^2+1}}\ln\left(\frac{\alpha-\sqrt{\alpha^2+1}}{\alpha+\sqrt{\alpha^2+1}}\right)=-\frac{2\ln(\alpha+\sqrt{\alpha^2+1})}{\sqrt{\alpha^2+1}}=-\frac{2\operatorname{arccosh}(-i\alpha)}{\sqrt{\alpha^2+1}}.\end{align}
The particular case $\alpha=i$ yields $I=2i$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does trying to compute $\lim_{x\to-\infty} {2x-1\over \sqrt{3x^2+x+1}}$ result in the negative of the answer given? My textbook asks me to evaluate the limit $$\lim_{x\to-\infty} {2x-1\over \sqrt{3x^2+x+1}}$$ which evaluates to $-2\over\sqrt{3}$. The method in the book is to factor out $x^2$ from the root in the denominator:
$$\begin{align}
\lim_{x\to-\infty} {2x-1\over \sqrt{3x^2+x+1}} & = \lim_{x\to-\infty} {2x-1\over \sqrt{x^2\left(3+\frac{1}{x}+\frac{1}{x^2}\right)}} \\
& = \lim_{x\to-\infty} {2x-1\over -x\sqrt{3+\frac{1}{x}+\frac{1}{x^2}}} \\
& = \lim_{x\to-\infty} {-2+\frac{1}{x}\over \sqrt{3+\frac{1}{x}+\frac{1}{x^2}}} \\
& = {-2\over\sqrt{3}}
\end{align}$$
the second step is justified because $x\to-\infty$ implies $x\lt0$, so $\sqrt{x^2}=-x$.
For my attempt I ended up with the negative of the correct answer:
$$\begin{align}
\lim_{x\to-\infty} {2x-1\over \sqrt{3x^2+x+1}} & = \lim_{x\to-\infty} \left({2x-1\over \sqrt{3x^2+x+1}}\cdot\frac{\frac{1}{x}}{\frac{1}{x}}\right) \\
& = \lim_{x\to-\infty} {2-\frac{1}{x}\over \sqrt{\frac{1}{x^2}\left(3x^2+x+1\right)}} \\
& = \lim_{x\to-\infty} {2-\frac{1}{x}\over \sqrt{3+\frac{1}{x}+\frac{1}{x^2}}} \\
& = {2\over\sqrt{3}}
\end{align}$$
Where have I gone wrong? I suspect the mistake lies in my second step, but I'm unable to identify what went wrong exactly.
| Your mistake is in writing
$$\frac 1 x = \sqrt{\frac{1}{x^2}}.$$
Since $x < 0$, the correct version includes a negative sign.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Eigenvalues and Eigenvectors of Sum of Symmetric Matrix Question:
Let A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}
Find all eigenvalues and eigenvectors of the martrix:
$$\sum_{n=1}^{100} A^n = A^{100} +A^{99} +...+A^2+A$$
I know that the eigenvectors of A are \begin{bmatrix} 1 \\ 1 \end{bmatrix} and \begin{bmatrix} 1 \\ -1 \end{bmatrix}
But I do not see any sort of correlation with the sum term and A's eigenvectors.
| Hint: If $$A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}$$then we have $$A^2 = \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}=\begin{bmatrix} 2 & 2 \\ 2 & 2 \\ \end{bmatrix}\\A^3=\begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}\begin{bmatrix} 2&2 \\ 2&2 \\ \end{bmatrix}=\begin{bmatrix} 4&4 \\ 4&4 \\ \end{bmatrix}\\A^4=\begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}\begin{bmatrix} 4&4 \\ 4&4 \\ \end{bmatrix}=\begin{bmatrix}8&8 \\ 8&8 \\ \end{bmatrix}\\.\\.\\.\\.$$and you can prove by induction that $$A^k=\begin{bmatrix} 2^{k-1}&2^{k-1} \\ 2^{k-1}&2^{k-1}\\ \end{bmatrix}$$can you finish now?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I find the expectation and variance of $Z=\max\{X,Y\}$ where $X$ and $Y$ are defined through joint probability distribution? Random variables $X$ and $Y$ and have the joint distribution below, and $Z=\max\{X,Y\}$
$$
\begin{array}{c|lcr}
\text{X\Y} & \text{1} & \text{2} & \text{3} \\
\hline
1 & 0.12 & 0.08 & 0.20 \\
2 & 0.18 & 0.12 & 0.30 \\
\end{array}
$$
Find $E[Z]$ and $V[Z]$
I am unable to understand that if $Z=\max\{X,Y\}$ then how will we take the pairs? Or just $Y=3$ ? Because it is the only maximum. Please explain I am just stuck here.
| If $Z = \max(X, Y)$, then the following is true:
$Z = 1$ iff $X = 1$ and $Y = 1$.
$Z = 2$ iff either $X = 1$ and $Y =2$ or $X = 2$ and $Y = 1$, or $X = 2$ and $Y = 2$.
$Z = 3$ iff $X = 3$.
So $P(Z = 1) = 0.12$
$P(Z = 2) = 0.08 + 0.12 + 0.18 = 0.38$
$P(Z = 3) = 0.2 + 0.3 = 0.5$.
Thus $E(Z) = 0.12 + 2*0.38 + 3*0.5 = 0.12 + 0.76 + 1.5 = 2.38$
$E(Z^2) = 0.12 + 4*0.38 + 9*0.5 = 0.12 + 1.52+ 4.5 = 6.04$
That means $V(Z) = E(Z^2) - (E(Z))^2 = 6.04 - 5.6644 = 0.3756$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the $L^{p}$ norm $\|f\|_{L^{p}} := \big( \int^{b}_{a} |f(x)|^p\big)^{1/p}$ is not induced by a scalar product for $p \neq 2$.
On $X = C^0\big([a,b]\big)$, for any $p \in \mathbb{R}$, $p>1$, we define the $L^p$ norm by,
$$\|f\|_{L^{p}}:=\big(\int^{b}_{a}|f(x)|^{p}dx \big)^{1/p}.$$
Show that for $p\neq 2$, this norm is not induced by a scalar product.
My method of trying to prove this was to prove a contradiction to the parallelogram rule,
$$ \|f+g\|^{2}_{p} + \|f-g\|^{2}_{p} = 2\|f\|^{2}_{p} + 2\|g\|^{2}_{p}, \tag{$1$}$$
where $f,g \in C^{0}([a,b])$.
So I defined the following functions;
$$f(x):=\frac{a+b}{2}-x$$
$$g(x) := \begin{cases}\frac{a+b}{2}-x, \ \ for \ \ a \leq x \le \frac{a+b}{2}. \\
x-\frac{a+b}{2}, \ \ for \ \ \frac{a+b}{2} < x \le b \end{cases}$$
which gives
$$f(x)+g(x) = \begin{cases} a+b-2x, \ \ & for \ \ a\le x \le \frac{a+b}{2}. \\ 0, & for \ \ \frac{a+b}{2} < x \le b\end{cases}$$
$$f(x)-g(x) = \begin{cases} 0, & for \ \ a \le x \le \frac{a+b}{2}. \\
2x - (a+b), \ \ & for \ \ \frac{a+b}{2} < x \le b \end{cases}$$
Then I proceeded to calculate each term of the parallelogram rule,
$$\|f+g\|^{2}_{p} = \bigg( \int^{\frac{a+b}{2}}_{a}|a+b-2x|^{p}\bigg)^{2/p} = \frac{(b-a)^{\frac{2(p+1)}{p}}}{(2(p+1))^{2/p}} $$
$$ \|f-g\|^{2}_{p} = \bigg( \int_{\frac{a+b}{2}}^{b}|2x- (a+b)|^{p}\bigg)^{2/p} = \frac{(b-a)^{\frac{2(p+1)}{p}}}{(2(p+1))^{2/p}}$$
$$2\|f\|^{2}_{p} = 2 \bigg( \int^{b}_{a}| \frac{a+b}{2}-x|^{p} dx \bigg)^{2/p} = 2 \cdot \frac{2^{2/p}(\frac{b-a}{2})^{\frac{2(p+1)}{p}}}{(p+1)^{2/p}} $$
$$\begin{align}2 \|g\|^{2}_{p} & = 2 \bigg(\int^{\frac{a+b}{2}}_{a} |\frac{a+b}{2} - x|^{p} dx \ + \ \int^{b}_{\frac{a+b}{2}}|x- \frac{a+b}{2}|^{p} dx\bigg)^{2/p} \\ & =2 \cdot \frac{2^{2/p}(\frac{b-a}{2})^{\frac{2(p+1)}{p}}}{(p+1)^{2/p}} \end{align}$$
Plugging into $(1)$ we then get
$$2 \cdot \frac{(b-a)^{\frac{2(p+1)}{p}}}{(2(p+1))^{2/p}} = 4 \cdot \frac{2^{2/p}(\frac{b-a}{2})^{\frac{2(p+1)}{p}}}{(p+1)^{2/p}}$$
which simplifies quite nicely to
$$2^{p} = 4.$$
So the equality only holds for $p = 2$.
Is what i've done correct? is there another way of proving the question which is better?
| Your idea looks fine. However, here's a simpler approach: Let $f = \chi_A$ and $g = \chi_B$ be the indicator functions of two disjoint sets*. Then
$$\|f + g\|_p^2 + \|f - g\|_p^2 = 2 (|A| + |B|)^{2/p}$$
by a direct calculation. On the other hand,
$$2 \|f\|_p^2 + 2\|g\|_p^2 = 2 (|A|^{2/p} + |B|^{2/p}).$$
This would imply that for all real numbers $a, b \ge 0$ we have
$$(a + b)^{2/p} = a^{2/p} + b^{2/p}.$$
For any $a, b$ which are both positive, this implies $p = 2$ as a consequence of Jensen's inequality. For a specific example, $a = b = 1$ implies $2^{2/p} = 2$, so $p = 2$.
*The point of this answer is that many integral inequalities can be studied purely from the point of view of testing against sets. Although $\chi_A$ and $\chi_B$ aren't continuous, they can be approximated arbitrarily well in $L^p$ by smooth functions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate:$\frac{1}{2^{101}}\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{51}{k}\binom{50}{r}$ Evaluate:$$\frac{1}{2^{101}}\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{51}{k}\binom{50}{r}$$
My Attempt:
I did try writing the series $(1+x)^{50}$and $(1+x)^{50}$ separately,then multiplied but could not determine the power of $x$ whose coefficient should be found.Can Vandermonde's Identity be put to use here.
| Using the identity $$
\binom{51}{k}=\binom{50}{k}+\binom{50}{k-1},
$$ we get
$$
\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{51}{k}\binom{50}{r}=\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{50}{k}\binom{50}{r}+\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{50}{k-1}\binom{50}{r}.$$ Observe that the former is equal to
$$
\sum_{k=0}^{50} \sum_{r=0}^{k-1}\binom{50}{k}\binom{50}{r}= \sum_{ 0\leq r<k\leq 50}\binom{50}{k}\binom{50}{r} = \sum_{ 0\leq k<r\leq 50}\binom{50}{k}\binom{50}{r}=\sum_{k=0}^{50} \sum_{r=k+1}^{50}\binom{50}{k}\binom{50}{r}.
$$ Also note that the latter is equal to
$$\begin{eqnarray}
\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{50}{k-1}\binom{50}{r}=\sum_{k=0}^{50}\sum_{r=0}^{k}\binom{50}{k}\binom{50}{r}.
\end{eqnarray}$$ Gathering them together yields
$$
\sum_{k=0}^{50} \sum_{r=k+1}^{50}\binom{50}{k}\binom{50}{r}+\sum_{k=0}^{50}\sum_{r=0}^{k}\binom{50}{k}\binom{50}{r}=\sum_{k=0}^{50} \sum_{r=0}^{50}\binom{50}{k}\binom{50}{r} = 2^{50}\cdot2^{50} = 2^{100}.
$$ Hence the answer is $\frac{1}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute the limit $\lim_{n\to\infty} I_n(a)$ where $ I_n(a) :=\int_0^a \frac{x^n}{x^n+1}\,\mathrm{d}x, n\in N$. For $a>0$ we define
$$\space I_n(a)=\int_0^a\frac{x^n}{x^n+1}\,\mathrm{d}x , n\in N.$$
*
*Prove that $0\le I_n(1) \le \frac{1}{n+1}$
*Compute $\lim_{n\to\infty} I_n(a)$
My attempt:
*
*I regard $I_n(1)=\int_0^1\frac{x^n}{x^n+1}$. If $x\in (0,1)$ then $x^n\in(0,1)$ and $x^n+1\in(1,2)$.
$$x^n>0 \Rightarrow x^n+1>1 \Rightarrow 1>\frac{1}{1+x^n }\Rightarrow x^n>\frac{x^n}{x^n+1}\Rightarrow \int_0^1\frac{x^n }{x^n+1}dx<\int_o^1 x^n \mathrm{d}x\\ \Rightarrow \int_0^1\frac{x^n }{x^n+1}dx<\frac{1}{n+1} \\ 0\le\frac{x^n}{x^n+1} \\ \text{In concusion } 0\le I_n(1) \le \frac{1}{n+1}.$$
*first case $a\in(0,1) \Rightarrow \lim_{n\to\infty} I_n(a) =0$. $I_n(a)\le\frac{1}{n+1})\text{case 2 . }a\in(1,\infty) \Rightarrow$ ???????
I don't believe the limit is $\infty$ because $\frac{x^n }{x^n+1}\le 1$.
I would appreciate some hints.
| Note that we have
$$\begin{align}
\int_0^a \frac{x^n}{1+x^n}\,dx&=\int_0^1 \frac{x^n}{1+x^n}\,dx+\int_1^a \frac{x^n}{1+x^n}\,dx\\\\
&=\int_0^1 \frac{x^n}{1+x^n}\,dx+(a-1)-\int_1^a \frac{1}{1+x^n}\,dx
\end{align}$$
For $x\in [0,1]$, $0\le \frac{x^n}{1+x^n}\le x^n$ and for $x\in[1,a]$, $\frac{1}{1+x^n}\le \frac1{x^n}$. Therefore,
$$\left|\int_0^1 \frac{x^n}{1+x^n}\,dx\right|\le \frac1{n+1}$$
and
$$\left|\int_1^a \frac{1}{1+x^n}\,dx\right|\le \frac{1-a^{1-n}}{n-1}$$
Can you finish now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving $\int\limits_{-\infty}^\infty \frac{1}{x^8+1}dx$ through Glasser's Master Theorem Trying to find a way to solve
$$\int_{-\infty}^\infty \frac{1}{x^8+1}dx$$
through Glasser's Master Theorem, more specifically the Cauchy–Schlömilch substitution. Preferably, I'm looking for the closed form solution, and I am already aware of how to attain this through contour integration.
Solution: $$\frac{\pi}{4\sin(\frac{\pi}{8})}$$
Link to general closed form solution: solutions to $\int_{-\infty}^\infty \frac{1}{x^n+1}dx$ for even $n$
| I believe the most direct way for tackling such integrals is to exploit Euler's Beta function and the reflection formula for the $\Gamma$ function: assuming $m>1$,
$$ \int_{0}^{+\infty}\frac{dx}{1+x^m}\stackrel{\frac{1}{1+x^m}\to u}{=}\frac{1}{m}\int_{0}^{1}u^{-\frac{m-1}{m}}(1-u)^{-\frac{1}{m}}\,du =\frac{\Gamma\left(\tfrac{1}{m}\right)\Gamma\left(\tfrac{m-1}{m}\right)}{m\,\Gamma(1)}=\frac{\pi/m}{\sin(\pi/m)}.$$
Another way is to exploit the reflection formula for the digamma function, in the form
$$ \sum_{n\geq 0}\left[\frac{1}{an+b}+\frac{1}{an+(a-b)}\right]=\frac{\pi}{a}\,\cot\left(\frac{\pi b}{a}\right)$$
via
$$\begin{eqnarray*} \int_{0}^{+\infty}\frac{dx}{1+x^m}&=&\int_{0}^{1}\frac{1+x^{m-2}}{1+x^m}\,dx=\int_{0}^{1}\frac{1+x^{m-2}-x^m-x^{2m-2}}{1-x^{2m}}\,dx\\
&=& \sum_{n\geq 0}\left[\color{blue}{\frac{1}{2mn+1}}+\color{red}{\frac{1}{2mn+m-1}}-\color{red}{\frac{1}{2mn+m+1}}-\color{blue}{\frac{1}{2mn+2m-1}}\right]\\&=&\frac{\pi}{2m}\left[\cot\left(\frac{\pi}{2m}\right)+\tan\left(\frac{\pi}{2m}\right)\right].\end{eqnarray*}$$
Through Glasser's master theorem, in the $m=8$ case we may state
$$\begin{eqnarray*} \int_{-\infty}^{+\infty}\frac{dx}{1+x^8}&\stackrel{x\mapsto z\cdot 2^{1/4}}{=}&\int_{-\infty}^{+\infty}\frac{dz}{(1-\sqrt{2}z^2+z^4)(1+\sqrt{2}z^2+z^4)} \\&=&2\int_{0}^{+\infty}\frac{1}{2\sqrt{2}z^2}\left[\frac{1}{1-\sqrt{2}z^2+z^4}-\frac{1}{1+\sqrt{2}z^2+z^4}\right]\,dz
\\&\stackrel{z\mapsto 1/z}{=}&2\int_{0}^{+\infty}\frac{z^4}{2\sqrt{2}}\left[\frac{1}{1-\sqrt{2}z^2+z^4}-\frac{1}{1+\sqrt{2}z^2+z^4}\right]\,dz
\\&=&\frac{1}{\sqrt{2}}\int_{0}^{+\infty}\frac{1}{z^4}\left[\frac{1}{\frac{1}{z^2}-\sqrt{2}+z^2}-\frac{1}{\frac{1}{z^2}+\sqrt{2}+z^2}\right]\,dz
\\&=&\frac{1}{\sqrt{2}}\int_{0}^{+\infty}z^2\left[\frac{1}{\frac{1}{z^2}-\sqrt{2}+z^2}-\frac{1}{\frac{1}{z^2}+\sqrt{2}+z^2}\right]\,dz\end{eqnarray*}$$
then invoke integration by parts and averaging in order to convert the original integral into
$$\begin{eqnarray*}\int_{0}^{+\infty}r\left(x^2+\frac{1}{x^2}\right)\,dx &=& \frac{1}{2}\int_{-\infty}^{+\infty}r\left(\left(x-\frac{1}{x}\right)^2+2\right)\,dx\\&\stackrel{\text{GMT}}{=}&\frac{1}{2}\int_{-\infty}^{+\infty}r(x^2+2)\,dx=\int_{0}^{+\infty}r(x^2+2)\,dx\end{eqnarray*}$$
with $r$ being a rational function. On the other hand, this approach looks pretty forced/artificial, especially if compared to the previous ones.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
If $xy$ divides $x^2 + y^2$ show that $x=\pm y$ Problem statement : Let $x,y$ be integers, show that if $xy$ divides $x^2 + y^2$ then $x=\pm y.$
What I have tried:
I can reduce this to the case where $\gcd(x,y)=1$, since if $x$ and $y$ have a common factor, $d$ say, then $d^2$ divides through both $xy$ and $x^2 + y^2$
This then allows me to introduce another equation $1=ax+by$ for some $a, b.$
But I then get stuck ...
| Suppose that $\gcd(x,y)=1$ and $(xy)\mid(x^2+y^2)$.
Then $y^2\equiv0\pmod x$. If $1=ax+by$ then $by\equiv1\pmod x$ and so
$1\equiv(by)^2=b^2y^2\equiv0\pmod x$. So $x=\pm1$. Likewise, $y=\pm1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Solve the differential equation $\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$ Solve the differential equation $$\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$$
My try:
we can write the equation as:
$$\frac{dy}{dx}=\frac{1}{y^3}\frac{\left(1+2y^4\right)}{1+\frac{4x}{y^4}}$$
Multiplying both sides with $\frac{1}{y^5}$ we get:
$$\frac{1}{y^5}\frac{dy}{dx}=\frac{1}{y^8}\frac{y^4(2+\frac{1}{y^4})}{1+\frac{4x}{y^4}}=\frac{1}{y^4}\frac{(2+\frac{1}{y^4})}{1+\frac{4x}{y^4}}$$
Now letting $$\frac{1}{y^4}=t$$ we get
$$\frac{-1}{4}\frac{dt}{dx}=\frac{t^2+2t}{4tx+1}$$
Any way further to convert in to variable separable?
| $$\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$$
$$(y+2y^5)\frac{dx}{dy}-4x=y^4$$
Considering the function $x(y)$ this is a first order linear ODE. Solving such kind of ODE is classic. The result is :
$$x(y)=\frac{y^4}{2y^4+1}\ln(c\:y)$$
$y(x)$ is the inverse function. It cannot be expressed with a finite number of elementary functions. A closed form requires a special function W(X) the Lambert W function.
$$y(x)=\left(\frac{4x}{\text{W}(C\:xe^{-8x})}\right)^{1/4}$$
$C$ is an arbitrary constant, to be determined according to some boundary condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
calculating eigen values from an equation
I'm trying to use this equation(in yellow) to calculate the eigen values of B = \begin{pmatrix}
1&1&1\\
1&1&1\\
1&1&1\\
\end{pmatrix}
but I'm getting $$-λ^3+3λ^2-3λ$$
and the proper answer using the other method is
$$det( \begin{pmatrix}
1-λ&1&1\\
1&1-λ&1\\
1&1&1-λ\\
\end{pmatrix})$$
$$=-λ^3+3λ^2$$
Anyone see where I could have gone wrong or if the yellow equation only works in certain situations?
| $\DeclareMathOperator\tr{tr}$The full proper formula for $n=3$ is:
$$\det(M-\lambda I_3)=(-1)^3\lambda^3 + (-1)^2\tr(M)\lambda^2 + (-1)\cdot \frac 12\big[(\tr M)^2-\tr(M^2)\big]\lambda + \det(M)$$
In this case:
$$\tr\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}^2
=\tr \begin{pmatrix}3&3&3\\3&3&3\\3&3&3\end{pmatrix}
=9$$
So the result is indeed $-\lambda^3+3\lambda^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases}$$
Solve in the set of real numbers.
The furthest I have got is summing the equations, and I got
$$x^3+(y-3)x^2+(y^2+2)x+y^3-3y^2-4y=0.$$
But I have no idea how to solve this problem. How can I solve it?
| Hint: Write your equations in the form $$\frac{3x-y}{3-x}=\frac{x+3y}{y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Stuck finding the sum of two series This is the first time I ever make a post on Stack Exchange (and the last time I stop lurking it!), so apologies in advance if I caused any problems.
I am completely stuck at finding the sum of the two following series:
$$\sum_{n=1}^∞(2n+1)x^n$$
$$\sum_{n=0}^∞\frac{x^n}{(n+1)2^n}$$
I understand that I should be using this series to find out their value: $$\sum_{n=0}^∞x^n=\frac{1}{1-x}$$
I already used it for a couple of simpler series, but I cannot get my head around the two ones above
| *
*Set $f(x) = \sum_{n=1}^\infty (2n+1)x^n$. Then
$$
f(t^2)
= \sum_{n=1}^\infty (2n+1)t^{2n}
= \sum_{n=1}^\infty \frac{d}{dt}t^{2n+1}
= \frac{d}{dt} \sum_{n=1}^\infty t^{2n+1}
= \frac{d}{dt} \left( t \sum_{n=1}^\infty (t^2)^n \right)
= \frac{d}{dt} \left( \frac{t^3}{1-t^2} \right) \\
= \frac{3t^2(1-t^2)-t^3(-2t)}{(1-t^2)^2}
= \frac{3t^2-t^4}{(1-t^2)^2}.
$$
Thus,
$$
f(x) = \frac{3x-x^2}{(1-x)^2}.
$$
*Set $g(x) = \sum_{n=0}^\infty \frac{x^n}{(n+1)2^n}.$ Then,
$$
\frac{d}{dx} \left( x \, g(x) \right)
= \frac{d}{dx} \left( 2 \sum_{n=0}^\infty \frac{(x/2)^{n+1}}{(n+1)} \right)
= 2 \sum_{n=0}^\infty \frac{1}{2} (x/2)^n
= \frac{1}{1-x/2}
= \frac{2}{2-x}.
$$
Therefore, for some constant $C$,
$$x \, g(x) = C - 2 \ln(2-x).$$
The constant can be determined by $0 = 0 \, g(0) = C - 2 \ln 2,$ i.e. $C = 2 \ln 2.$
Thus,
$$g(x) = \frac{2 \ln 2}{x} - 2 \frac{\ln(2-x)}{x} = 2 \frac{\ln 2 - \ln(2-x)}{x} = 2 \frac{\ln(2/(2-x))}{x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Coordinates of a vector FFT I was reading a paper about DFT, at the end he got the relations $Y=\frac{1}{\sqrt N}W_n \cdot y$ and $y=\frac{1}{\sqrt N}W_n \cdot Y$, where $y$ is a vector and $Y$ is the coordinates of that vector in the basis. $W_n$ stands for the transformation matrix.
Later, the author shows how to reduce the following into smaller matrices but I don't really understand what is he doing nor how is he doing it.
\begin{align*}
W_4 \begin{bmatrix} y_0\\y_1\\y_2\\y_3 \end{bmatrix} &= \begin{bmatrix}
\begin{bmatrix} 1&1\\1&\omega^2 \end{bmatrix}
\begin{bmatrix} y_0\\y_2 \end{bmatrix}
+\begin{bmatrix} 1&1\\\omega&\omega^3 \end{bmatrix}
\begin{bmatrix} y_1\\y_3 \end{bmatrix}
\\
\begin{bmatrix} 1&\omega^4\\1&\omega^6 \end{bmatrix}
\begin{bmatrix} y_0\\y_2 \end{bmatrix}
+\begin{bmatrix} \omega^2&\omega^6\\\omega^3&\omega^9 \end{bmatrix}
\begin{bmatrix} y_1\\y_3 \end{bmatrix}
\end{bmatrix}
\\ &=
\begin{bmatrix}
\begin{bmatrix} 1&1\\1&\omega^2 \end{bmatrix}
\begin{bmatrix} y_0\\y_2 \end{bmatrix}
+\begin{bmatrix} 1\\&\omega \end{bmatrix}
\begin{bmatrix} 1&1\\1&\omega^2 \end{bmatrix}
\begin{bmatrix} y_1\\y_3 \end{bmatrix}
\\
\begin{bmatrix} 1&1\\1&\omega^2 \end{bmatrix}
\begin{bmatrix} y_0\\y_2 \end{bmatrix}
-\begin{bmatrix} 1\\&\omega \end{bmatrix}
\begin{bmatrix} 1&1\\1&\omega^2 \end{bmatrix}
\begin{bmatrix} y_1\\y_3 \end{bmatrix}
\end{bmatrix}
\end{align*}
| In the first step, the author is just exploiting some associativity of the matrix-vector product. For instance, the first two rows of $W_4y$ are:
\begin{align*}
\begin{bmatrix} 1&1&1&1\\ 1&\omega&\omega^2&\omega^3 \end{bmatrix} \begin{bmatrix} y_0\\y_1\\y_2\\y_3 \end{bmatrix}
&=
\begin{bmatrix} y_0+y_1+y_2+y_3\\ y_0+\omega y_1+\omega^2y_2+\omega^3y_3 \end{bmatrix}
\\ &=
\begin{bmatrix} (y_0+y_2)+(y_1+y_3)\\ (y_0+\omega^2y_2)+(\omega y_1+\omega^3y_3) \end{bmatrix}
\\ &=
\begin{bmatrix} y_0+y_2\\ y_0+\omega^2y_2 \end{bmatrix}
+
\begin{bmatrix} y_1+y_3\\ \omega y_1+\omega^3y_3 \end{bmatrix}
\\ &=
\begin{bmatrix} 1&1\\1&\omega^2 \end{bmatrix} \begin{bmatrix} y_0\\y_2 \end{bmatrix} +\begin{bmatrix} 1&1\\\omega&\omega^3 \end{bmatrix} \begin{bmatrix} y_1\\y_3 \end{bmatrix}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\sum\limits_{j = 0}^{m - 1} \left(c_j + 1\right)\ln\left(c_j + 1\right)$ where $c_j = \cos\left(\frac{\pi}{2m}\left(1 + 2j\right) \right)$ As part of solving:
\begin{equation}
I_m = \int_0^1 \ln\left(1 + x^{2m}\right)\:dx.
\end{equation}
where $m \in \mathbb{N}$. I found an unresolved component that I'm unsure how to start:
\begin{equation}
G_m = \sum_{j = 0}^{m - 1} \left(c_j + 1\right)\ln\left(c_j + 1\right),
\end{equation}
where $c_j = \cos\left(\frac{\pi}{2m}\left(1 + 2j\right) \right)$
I'm just looking for a starting point. Any tips would be greatly appreciated.
By the way, I was able to show (and this was part of the solution too) :
\begin{equation}
\sum_{j = 0}^{m - 1} c_j = 0
\end{equation}
Edit: For those that may be interested. In collaboration with clathratus, we found that for $n > 1$
\begin{equation}
\int_{0}^{1} \frac{1}{t^n + 1}\:dt = \frac{1}{n}\left[\frac{\pi}{\sin\left(\frac{\pi}{n} \right)}- B\left(1 - \frac{1}{n}, \frac{1}{n}, \frac{1}{2}\right)\right]
\end{equation}
Or for any positive upper bound $x$:
\begin{align}
I_n(x) &= \int_{0}^{x} \frac{1}{t^n + 1}\:dt = \frac{1}{n}\left[\Gamma\left(1 - \frac{1}{n} \right)\Gamma\left(\frac{1}{n} \right)- B\left(1 - \frac{1}{n}, \frac{1}{n}, \frac{1}{x^n + 1}\right)\right]
\end{align}
Here though, I was curious to investigate when $n$ was an even integer. This is my work:
Here we will consider $r = 2m$ where $m \in \mathbb{N}$. In doing so, we observe that the roots of the function are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:
\begin{align}
x^{2m} + 1 = 0 \rightarrow x^{2m} = e^{\pi i}
\end{align}
By De Moivre's formula, we observe that:
\begin{align}
x = \exp\left({\frac{\pi + 2\pi j}{2m} i} \right) \mbox{ for } j = 0\dots 2m - 1,
\end{align}
which we can express as the set
\begin{align}
S &= \Bigg\{ \exp\left({\frac{\pi + 2\pi \cdot 0}{2m} i} \right) , \:\exp\left({\frac{\pi + 2\pi \cdot 1}{2m} i} \right),\dots,\:\exp\left({\frac{\pi + 2\pi \cdot (2m - 2)}{2m} i} \right)\\
&\qquad\:\exp\left({\frac{\pi + 2\pi \cdot (2m - 1)}{2m} i} \right)\Bigg\},
\end{align}
which can be expressed as the set of $2$-tuples
\begin{align}
S &= \left\{ \left( \exp\left({\frac{\pi + 2\pi j}{2m} i} \right) , \:\exp\left({\frac{\pi + 2\pi(2m - 1 - j )}{2m} i} \right)\right)\: \bigg|\: j = 0 \dots m - 1\right\}\\
& = \left\{ (z_j, c\left(z_j\right)\:|\: j = 0 \dots m - 1 \right\}
\end{align}
From here, we can factor $x^{2m} + 1$ into the form
\begin{align}
x^{2m} + 1 &= \prod_{r \in S} \left(x + r_j\right)\left(x + c(r_j)\right) \\
&= \prod_{i = 0}^{m - 1} \left(x^2 + \left(r_j + c(r_j)\right)x + r_j c(r_j)\right) \\
&= \prod_{i = 0}^{m - 1} \left(x^2 + 2\Re\left(r_j\right)x + \left|r_j \right|^2\right)
\end{align}
For our case here $\left|r_j \right|^2 = 1$ and $\Re\left(r_j\right) = \cos\left(\frac{\pi + 2\pi j}{2m} \right)= \cos\left(\frac{\pi}{2m}\left(1 + 2j\right)\right) = c_j$
\begin{align}
\int_0^1 \log\left( x^{2m} + 1\right)\:dx &= \int_0^1 \log\left(\prod_{r \in S} \left(x^2 + 2c_jx+ \left|r_j \right|^2\right)\right)\\
&= \sum_{j = 0}^{m - 1} \int_0^1 \log\left(x^2 + 2c_jx + 1 \right)\\
&= \sum_{j = 0}^{m - 1} \left[2\sqrt{1 - c_j^2}\arctan\left(\frac{x + c_j}{\sqrt{1 - c_j^2}}\right) + \left(x + c_j\right)\log\left(x^2 + 2c_jx + 1\right) - 2x \right]_0^1 \\
&= \sum_{j = 0}^{m - 1} \left[ 2\sqrt{1 - c_j^2}\arctan\left(\sqrt{\frac{1 - c_j}{1 + c_j}} \right) + \log(2)c_j + \left(\log(2) - 2\right) + \left(c_j + 1\right)\log\left(c_j + 1\right) \right] \\
&= 2\sum_{j = 0}^{m - 1}\sqrt{1 - c_j^2}\arctan\left(\sqrt{\frac{1 - c_j}{1 + c_j}} \right) + \log(2)\sum_{j = 0}^{m - 1} c_j + m\left(\log(2) - 2\right)\\
&\qquad+ \sum_{j = 0}^{m - 1}\left(c_j + 1\right)\log\left(c_j + 1\right)
\end{align}
Thus,
\begin{align}
\int_0^1 \log\left( x^{2m} + 1\right)\:dx &=\sum_{j = 0}^{m - 1}c_j\sin\left(\frac{\pi}{2m}\left(1 + 2j\right)\right) + \log(2)\sum_{j = 0}^{m - 1} c_j + m\left(\log(2) - 2\right)\\
&\qquad+ \sum_{j = 0}^{m - 1}\left(c_j + 1\right)\log\left(c_j + 1\right)
\end{align}
| I did it!
I actually have no idea whether or not this works, but this is how I did it.
$n\in\Bbb N$
Define the sequence $\{r_k^{(n)}\}_{k=1}^{k=n}$ such that
$$x^n+1=\prod_{k=1}^{n}\big(x-r^{(n)}_{k}\big)$$
We then know that $$r_k^{(n)}=\exp\bigg[\frac{i\pi}{n}(2k-1)\bigg]$$
Then we define
$$S_n=\{r_k^{(n)}:k\in[1,n]\cap\Bbb N\}$$
So we have that
$$\frac1{x^n+1}=\prod_{r\in S_n}\frac1{x-r}=\prod_{k=1}^n\frac1{x-r_k^{(n)}}$$
Then we assume that we can write
$$\prod_{r\in S_n}\frac1{x-r}=\sum_{r\in S_n}\frac{b(r)}{x-r}$$
Multiplying both sides by $\prod_{a\in S_n}(x-a)$,
$$1=\sum_{r\in S_n}b(r)\prod_{a\in S_n\\ a\neq r}(x-a)$$
So for any $\omega\in S_n$,
$$1=b(\omega)\prod_{a\in S_n\\ a\neq \omega}(\omega-a)$$
$$b(\omega)=\prod_{a\in S_n\\ a\neq \omega}\frac1{\omega-a}$$
$$b(r_k^{(n)})=\prod_{p=1\\ p\neq k}^n\frac1{r_k^{(n)}-r_p^{(n)}}$$
So we know that
$$I_n=\int_0^1\frac{\mathrm{d}x}{1+x^n}=\sum_{k=1}^{n}b(r_k^{(n)})\int_0^1\frac{\mathrm{d}x}{x-r_k^{(n)}}$$
$$I_n=\sum_{k=1}^{n}b(r_k^{(n)})\log\bigg|\frac{r_k^{(n)}-1}{r_k^{(n)}}\bigg|$$
$$I_n=\sum_{k=1}^{n}\log\bigg|\frac{r_k^{(n)}-1}{r_k^{(n)}}\bigg|\prod_{p=1\\ p\neq k}^n\frac1{r_k^{(n)}-r_p^{(n)}}$$
So we have
$$\int_0^1\log(1+x^n)\mathrm{d}x=\log2-n+n\sum_{k=1}^{n}\log\bigg|\frac{r_k^{(n)}-1}{r_k^{(n)}}\bigg|\prod_{p=1\\ p\neq k}^n\frac1{r_k^{(n)}-r_p^{(n)}}$$
along with a plethora of other identities...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Limit by polar coordinates: $\frac{2x(1-\cos(x-y))}{x^2+y^2}$ I need to prove that:
$$\lim_{(x,y)\to (0,0)}\frac{2x(1-\cos(x-y))}{x^2+y^2} = 0$$
I have tried using polar coordinates, but I'm not comfortable with my approach. What I did was saying:
$$|f(\rho\cos\theta,\rho\sin\theta)| = \frac{2\cos\theta(1-\cos(\rho(\cos\theta-\sin\theta))}{\rho} \leq \frac{2(1-\cos(\sqrt{2}\rho))}{\rho}$$
and this last expression converges to $0$ when $\rho$ goes to $0$ uniformly in $\theta$. Is this correct? How can I do it better? I'm also looking for a way to solve this limit without using polar coordinates ( the easier, the better) . Any ideas?
| Solution without the use of polar coordinates, as asked by OP at the end of their post.
Rewrite $$\begin{aligned}1-\cos(x-y)&=1-\cos x \cos y - \sin x \sin y \\ &=1-\cos x + \cos x (1-\cos y)- \sin x \sin y, \end{aligned}$$
then $$\begin{aligned}\frac{2x(1-\cos(x-y))}{x^2+y^2}&= 2x\cdot \frac{1-\cos x}{x^2} \cdot\frac{x^2}{x^2+y^2}\\ &+ 2x\cos x \cdot \frac{1-\cos y}{y^2}\cdot\frac{y^2}{x^2+y^2}\\&- 2y\cdot \frac{\sin x}{x}\cdot\frac{\sin y}{y}\cdot\frac{x^2}{x^2+y^2}\end{aligned}$$
From where is
$$\lim_{(x,y)\to (0,0)}\frac{2x(1-\cos(x-y))}{x^2+y^2} = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Tangents to parabola $y^{2}=4ax$ meet hyperbola $x^2/a^2-y^2/b^2=1$ at $A$ and $B$. Find the locus of intersections of the tangents at $A$ and $B$.
If tangents to the parabola $y^{2} = 4ax$ intersect the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ at $A$ and $B$, then find the locus of point of intersection of tangents at $A$ and $B$.
I know that tangent to parabola is $y = mx + a/m$ ($m$ being the slope), but I am not able to figure out how to take out point of intersections.
| $$y^2=4ax \tag{1}$$
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \tag{2}$$
Let $P(X,Y)$ be the required locus.
*
*For hyperbola $(2)$, $(X,Y)$ is the pole of the polar $AB$ (i.e. chord $AB$ for the hyperbola).
Equation of $AB$ is
$$\frac{X x}{a^2}-\frac{Y y}{b^2}=1 \tag{3}$$
*Equation of tangent of $(1)$ at $C(x_1,y_1)$
$$y_1 y=2a(x+x_1)$$
Rearranging, we have
$$-\frac{x}{x_1}+\frac{y_1 y}{2a x_1}=1 \tag{4}$$
*Identifying $(3)$ and $(4)$, we get
$$(X,Y)=\left( -\frac{a^2}{x_1}, -\frac{b^2 y_1}{2a x_1} \right)$$
$$(x_1,y_1)=\left( -\frac{a^2}{X}, \frac{2a^3 Y}{b^2 X} \right)$$
But $$y_1^2=4a x_1$$
$$\left( \frac{2a^3 Y}{b^2 X} \right)^2=4a\left( -\frac{a^2}{X} \right)$$
The locus of $P$ is
$$\fbox{$a^3 Y^2+b^4 X=0$}$$
Useful fact:
Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x_1,y_1)$ is given by
$$ax_1 x+h(y_1 x+x_1 y)+by_1 y+g(x+x_1)+f(y+y_1)+c=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3055972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove Nice Determinant Equations I often come across these kinds of problems in A-level exam papers:
Prove that
$$
\begin{vmatrix}
(a+b)^2 & 1 & 1 \\
a^2 & (1+b)^2 & a^2 \\
b^2 & b^2 & (1+a)^2
\end{vmatrix}
= 2ab(1+a+b)^3
$$
or
Prove that
$$
\begin{vmatrix}
1 & 1 & 1 \\
x & y & z \\
yz & xz & xy
\end{vmatrix}
= (x-y)(y-z)(z-x)
$$
Questions which involve nice determinant results, but which are quite a pain to prove by expanding.
Are there any tricks that one can use to prove such results? I'm familiar with Vandermonde matrices for example, but I haven't come across anything that might help me with these, especially the first one.
| The idea is to use elementary row-column operations to have a simpler determinant. Since this is $3\times 3$, our goal is to get two successive zeros in a single row/column.
You have
$$A=\begin{bmatrix}(a+b)^2 & 1 & 1 \\a^2 & (1+b)^2 & a^2 \\b^2 & b^2 & (1+a)^2 \end{bmatrix}$$
Then
\begin{align}
\det A&=\det\begin{bmatrix}(a+b)^2-1 & 0 & 1 \\0 & (1+b)^2-a^2 & a^2 \\b^2-(1+a)^2 & b^2-(1+a)^2 & (1+a)^2 \end{bmatrix}\qquad\quad[C_j'=C_j-C_3\,,j=1,2]
\\\\&=\det\begin{bmatrix}(a+b+1)(a+b-1) & 0 & 1 \\0 & (a+b+1)(1+b-a) & a^2 \\(a+b+1)(b-1-a) & (a+b+1)(b-1-a) & (1+a)^2 \end{bmatrix}
\\\\&=(a+b+1)^2\det\begin{bmatrix}a+b-1 & 0 & 1 \\0 & 1+b-a & a^2 \\ b-1-a & b-1-a & (1+a)^2 \end{bmatrix}
\\\\&=(a+b+1)^2\det\begin{bmatrix}a+b-1 & 0 & 1 \\0 & 1+b-a & a^2 \\ -2a & -2 & 2a \end{bmatrix}\qquad\qquad[R_3'=R_3-(R_1+R_2)]
\\\\&=\frac{2(a+b+1)^2}{a}\det\begin{bmatrix}a+b-1 & 0 & 1 \\0 & a+ab-a^2 & a^2 \\ -a & -a & a \end{bmatrix}\qquad\qquad\quad[C_2'=aC_2]
\\\\&=2(a+b+1)^2\det\begin{bmatrix}a+b-1 & 0 & 1 \\0 & a+ab-a^2 & a^2 \\ -1 & -1 & 1 \end{bmatrix}
\\\\&=2(a+b+1)^2\det\begin{bmatrix}a+b & 1 & 1 \\a^2 & a+ab & a^2 \\ 0 & 0 & 1 \end{bmatrix}\qquad\qquad\qquad[C_j'=C_j+C_3\,,j=1,2]
\end{align}
Now expand with respect to the third row.
| {
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"url": "https://math.stackexchange.com/questions/3056130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A Series For the Golden Ratio
Question: Can we show that $$\phi=\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2} $$; where $\phi={1+\sqrt{5} \above 1.5pt 2}$ is the golden ratio ?
Some background and motivation:
Wikipedia only provides one series for the golden ratio - see also the link in the comment by @Zacky. I became curious if I could construct another series for the Golden Ratio based on a slight modification to a known series representation of the $\sqrt{2}.$ At first I considered $$\sqrt{2}=\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n+1)!!}{(2n)!!}$$; which series can be accelerated via an Euler transform to yield $$\sqrt{2}=\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n+1)!!}{2^{3n+1}(n!)^2}$$ This last series became the impetus to try and and get to the Golden ratio. Through trial and error I stumbled upon
$$\frac{\sqrt{5}}{11}=\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}$$
| We know that the approximates by partial continued fractions of the golden ratios are the quotients of successive Fibonacci numbers $1,1,2,3,5,8,13,...$. Taking
$$
\phi=\frac{1+\sqrt{5}}2\approx \frac{8}5\implies \sqrt5\approx \frac{11}{5}
$$
we get a lower approximation for $\sqrt5$. To the remainder of the square root after factoring out the approximation we can apply the the Newton/binomial series for the inverse square root
$$
\sqrt5=\frac{11}{5}\sqrt{\frac{125}{121}}=\frac{11}{5}\left(1-4\frac1{5^3}\right)^{-1/2}
=\frac{11}5\sum_{n=0}^\infty\binom{2n}{n}\frac1{5^{3n}}
$$
which is the series you got.
Exploring the same method for one place further in the Fibonacci sequence
$$
\phi=\frac{1+\sqrt{5}}2\approx \frac{13}8\implies \sqrt5\approx \frac{9}{4}
$$
gives an upper approximation of $\sqrt5$ and thus an alternating series,
$$
\sqrt5=\frac{9}{4}\sqrt{\frac{80}{81}}=\frac{9}{4}\left(1+4\frac1{320}\right)^{-1/2}
=\frac{9}4\sum_{n=0}^\infty\binom{2n}{n}\frac{(-1)^n}{320^{n}}
$$
Why that series?
The general binomial series reads as
$$(1+x)^α=\sum_{n=0}^\infty\binom{α}n.$$ For $α=-1/2$ the binomial coefficient can be transformed as
$$
\binom{-1/2}{n}=\frac{(-1/2)(-1/2-1)...(-1/2-n+1)}{n!}=(-1/2)^n\frac{(2n-1)(2n-3)...3\cdot 1}{n!}=(-1/4)^n\frac{(2n)!}{(n!)^2},
$$
resulting in the "simplified" formula
$$\frac1{\sqrt{1-4x}}=\sum_{n=0}^\infty\binom{2n}nx^n.$$
| {
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"url": "https://math.stackexchange.com/questions/3056890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
How to take every third element in a series? $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = 1.644934$ or $\frac{\pi^2}{6}$
What if we take every 3rd term and add them up?
A = $ \frac{1}{3^2} + \frac{1}{6^2} + \frac{1}{9^2} + \cdots = ??$
How to take every 3rd-1 term and add them up?
B = $ \frac{1}{2^2} + \frac{1}{5^2} + \frac{1}{8^2} + \cdots = ??$
How to take every 3rd-2 term and add them up?
C = $ \frac{1}{1^2} + \frac{1}{4^2} + \frac{1}{7^2} + \cdots = ??$
I am not sure how to adapt Eulers methods as he used the power series of sin for his arguments: https://en.wikipedia.org/wiki/Basel_problem
| KM101 deleted his hint... not sure why.
$\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{9^2}+\dots=\frac{1}{3^2 1^2}+\frac{1}{3^2 2^2}+\frac{1}{3^2 3^3}+\dots=\frac{1}{9}(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots)= \frac{\pi^2}{54}$
$\frac{1}{2^2}+\frac{1}{5^2}+\frac{1}{8^2}+\dots=\frac{1}{2^2 1^2}+\frac{1}{3^2?? 1^2}+\frac{1}{3^2 ??1^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Fredholm integral - degenerate kernel method I have started answering a fredholm integral equation of the second kind and do not know where to go from here.
The answer has to be written in the form
$$ \sum a_jx^{j-1} $$
The Fredholm integral equation is
$$ x^3+\frac16x^2+\frac15x = g(x) + \int_0^1(x^2y+xy^2) f(y) dy$$.
My method so far:
Let: $$C_1 = \int_0^1yf(y)dy$$ and $$C_2 = \int_0^1y^2f(y)dy$$
Then
$$ x^3+\frac16x^2+\frac15x = (C_1x^2 +C_2x) + g(x)$$.
Eliminating f(y) to get
$$C_1 = (\frac14C_1 + \frac13C_2) + \int_0^1yg(y)dy$$
and
$$C_2 = (\frac15C_1 + \frac14C_2) + \int_0^1y^2g(y)dy$$
I don't know where to go from here to get it into the form
$$ \sum a_jx^{j-1} $$
Do I put it into matrix form, and solve simultaneously, (not sure how to do this)
If I have gotten anything wrong here please let me know.
Or if you need any more information I may be able to provide. (Like how I got to a specific equation)
Any help will be appreciated
Thank you very much
| $$g(x) = x^3 - \frac{38}{1077} x^2 + \frac{58}{1795} x$$
By rearranging the original equation we can see that g(x) = x^3 + Ax^2 + Bx. This is because both of the integrals on the right evaluate to a constant multiple of x^2 or x respectively. By using this formula for g(x) in each of the integrals we get:
The integral from 0 to 1 of
$$x^2y(y^3+Ay^2+By) dy = x^2[\frac{y^5}{5} + \frac{Ay^4}{4} + \frac{By^3}{3}] = x^2(\frac{1}{5} + \frac{A}{4} + \frac{B}{3})$$
and the integral from 0 to 1 of
$$xy^2(y^3+Ay^2+By) dy = x[y^6/6 + \frac{Ay^5}{5} + \frac{By^4}{4}] = x(\frac{1}{6} + \frac{A}{5} + \frac{B}{4})$$
So the equality becomes:
$$x^3 + \frac{1}{6} x^2 + \frac{1}{5} x = (x^3 + Ax^2 + Bx) + x^2(\frac{1}{5} + \frac{A}{4} + \frac{B}{3}) + x(\frac{1}{6} + \frac{A}{5} + \frac{B}{4})$$
Rearranging gives:
$$x^2(\frac{-1}{30} - \frac{5A}{4} - \frac{B}{3}) + x(\frac{1}{30} - \frac{A}{5}- \frac{5B}{4}) = 0$$
So, by comparing coefficients;
$$\frac{-1}{30} - \frac{5A}{4} - \frac{B}{3} = 0$$ and
$$\frac{1}{30} - \frac{A}{5} - \frac{5B}{4} = 0$$
Solving simultaneously gives
$$A = - 38/1077 $$ and
$$B = \frac{58}{1795} $$
$$ g(x)= x^3-\frac{38}{1077} x^2 + \frac{58}{1795}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve this simple equation $\frac{46}{y} + y = 25$? How do I solve this simple equation? $\frac{46}{y} + y = 25$
I know that the answer is $2$, but how do I arrive at that?
| Multiply both sides by $y$ and you get
$$46 + y^2 = 25y$$
$$\Leftrightarrow 0 = y^2 -25y + 46$$
Now calculating the discriminant $D = (-25)^2 -4\cdot 46 = 625-184 = 441 = 21^2$ gives
$$y = \frac{25 \pm \sqrt{21^2}}{2}$$
So $y = \frac{25+21}{2} = 23$ or $y = \frac{25-21}{2} = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Trick for Inverse Hollow Matrix Calculation (Self-Answered) Let $A$ be the hollow matrix :
$$
A=\begin{pmatrix}
0&1&1&1\\
1&0&1&1\\
1&1&0&1\\
1&1&1&0
\end{pmatrix}
$$
Find the inverse matrix $A^{-1}$ without using any elementary row/ column operations.
| Notice that
$$
A=
\begin{pmatrix}
1&1&1&1\\
1&1&1&1\\
1&1&1&1\\
1&1&1&1
\end{pmatrix} - I_{4\times4}
$$
Let
$$
B=
\begin{pmatrix}
1&1&1&1\\
1&1&1&1\\
1&1&1&1\\
1&1&1&1
\end{pmatrix}
$$
then $B=A+I$. Observe that $B^2 = (A+I)^2=4B$. Expanding,
$$
A^2+2A+I = 4(A+I) \implies A^2-2A=3I
$$
Therefore $A^{-1}=\frac{1}{3}(A-2I)$.
More generally, this method gives us an easy way to express the inverse of a hollow matrix $A$ (of the particular stated form) of any order $n$, which is the main reason for posting this question.
| {
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"url": "https://math.stackexchange.com/questions/3062821",
"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{a}{c\sqrt{a^2+1}}+\frac{b}{a\sqrt{b^2+1}}+\frac{c}{b\sqrt{c^2+1}}\ge \frac{3}{2}$ Let $a,b,c\in \Bbb R^+$ such that $a+b+c=abc$. Prove that $$\frac{a}{c\sqrt{a^2+1}}+\frac{b}{a\sqrt{b^2+1}}+\frac{c}{b\sqrt{c^2+1}}\ge \frac{3}{2}$$
Idea 1.From $a+b+c=abc\Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1$. Let $\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\rightarrow \left(x;y;z\right)$
So i need to prove $\frac{z}{\sqrt{1+x^2}}+\frac{y}{\sqrt{z^2+1}}+\frac{x}{\sqrt{y^2+1}}\ge \frac{3}{2}$
By AM-GM $\frac{x}{\sqrt{y^2+1}}=\frac{x}{\sqrt{\left(x+y\right)\left(y+z\right)}}\ge \frac{2x}{x+2y+z}$
$$LHS\ge 2\sum _{cyc}\frac{x}{x+2y+z}=2\sum _{cyc}\frac{x^2}{x^2+2xy+xz}\ge 2\frac{\left(x+y+z\right)^2}{\sum _{cyc}x^2+\sum _{cyc}3xy}$$
Or $4\left(x+y+z\right)^2\ge 3\left(x^2+y^2+z^2+3xy+3yz+3xz\right)$
Or $x^2+y^2+z^2\ge xy+yz+xz$ (true)
Idea 2. By Holder $$\left(\sum _{cyc}\frac{a}{c\sqrt{a^2+1}}\right)\left(\sum _{cyc}\frac{a}{c\sqrt{a^2+1}}\right)\sum _{cyc}\left(c^2\left(a^2+1\right)a\right)\ge \left(\sum _{cyc}a\right)^3$$
I will prove the inequality $\frac{\left(a+b+c\right)^3}{c^2a\left(a^2+1\right)+a^2b\left(b^2+1\right)+b^2c\left(c^2+1\right)}\ge \frac{3}{2}$
Or $\frac{abc\left(a+b+c\right)^3}{\left(a+b+c\right)\left(a^2b^3+b^2c^3+c^2a^3\right)+abc\left(a^2b+b^2c+c^2a\right)}\ge \frac{3}{2}$
I tried $SOS$ but failed help me improve "idea 2" use Holder
| We need to prove that
$$\sum_{cyc}\frac{a}{c\sqrt{a^2+\frac{abc}{a+b+c}}}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\sqrt{\frac{a(a+b+c)}{c^2(a+b)(a+c)}}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\sqrt{\frac{a^2b}{c(a+b)(a+c)}}\geq\frac{3}{2}$$ or
$$\sum_{cyc }\sqrt{a^3b^2(b+c)}\geq\frac{3}{2}\sqrt{abc(a+b)(a+c)(b+c)}$$ or
$$\sum_{cyc }\sqrt{(a^3b^3+a^3b^2c)}\geq\frac{3}{2}\sqrt{abc(a+b)(a+c)(b+c)}.$$
Now, let $ab=z$, $ac=y$ and $bc=x$.
Thus, we need to prove that
$$\sum_{cyc}\sqrt{z^3+z^2x}\geq\frac{3}{2}\sqrt{(x+y)(x+z)(y+z)}.$$
By Holder
$$\left(\sum_{cyc}\sqrt{z^3+z^2x}\right)^2\sum_{cyc}\frac{z}{z+x}\geq(x+y+z)^3.$$
Id est, it's enough to prove that
$$4(x+y+z)^3\geq9\prod_{cyc}(x+y)\sum_{cyc}\frac{z}{z+x}$$ or
$$4(x+y+z)^3\geq9\sum_{cyc}x(x+z)(y+z)$$ or
$$4(x+y+z)^3\geq9\sum_{cyc}(2x^2y+x^2z+xyz).$$
Now, by Rearrangement easy to show that $$x^2y+y^2z+z^2x+xyz\leq\frac{4}{27}(x+y+z)^3.$$
Thus, it's enough to prove that
$$4(x+y+z)^3\geq\frac{4}{3}(x+y+z)^3+9\sum_{cyc}\left(x^2y+x^2z+\frac{2}{3}xyz\right)$$ or
$$\frac{8}{3}(x+y+z)^3\geq9(x+y)(x+z)(y+z)$$ or
$$\left(\frac{x+y+x+z+y+z}{3}\right)^3\geq(x+y)(x+z)(y+z),$$ which is true by AM-GM.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the domain and range of a function The question is: find the domain and range of $f(x) = \sqrt{{(1-x)}/x}$.
For the domain, I took $x ≠ 0$ and $(1-x)/x ≥ 0$, which gave me $1 ≥ x$.
So the domain I found is $(-∞,0)∪(0,1]$. However, the proposed solution says that the domain is just $(0,1],$ without explaining, so I wonder why.
Also, any quick tip to find the range (without taking monotonicity, roots, limits to infinity etc.)? I tried inverting the function and the domain of the inverse is $\Bbb R$, which is clearly not the range of $f$.
Thanks in advance!
| $f(x) = \sqrt{\frac {1-x}x}$ which is undefined if $x =0$ and if $\frac {1-x}x = \frac 1x - 1 < 0$. $x\ne 0$ and $\frac 1x -1 < 0 \implies \frac 1x \ge 1$.
If $x > 0$ then $\frac 1x \ge 1 \implies x \le 1$ and if $x < 0$ implies $x \ge 1$ which is a contradiction so $0 < x \le 1$ and domain is $(0,1]$.
[Alternatively $\frac 1x \ge 1 > 0$ so $\frac {\frac 1x}{\frac 1x} \ge \frac 1{\frac 1x}>\frac 0{\frac 1x} \implies 1 \ge x > 0$.]
As for range.
Now that we know $0 < x \le 1$ then $\frac 1x \ge 1$ and $\frac 1x - 1 \ge 0$ and $\frac {1-x}x = \frac 1x -1 \ge 0$ and $\sqrt{\frac {1-x}x} \ge 0$ with no restrictions.
That implies the range is $[0, \infty)$.
But just to make certain we should test if $\sqrt{\frac {1-x}x} = m$ is solvable for all $m \ge 0$.
$\sqrt{\frac {1-x}x} = m \ge 0 \implies$
$\frac{1-x}x =\frac 1x- 1= m^2 \implies$
$\frac 1x = m^2 + 1$. So $m^2 + 1 \ge 1$ we can conclude
$x = \frac 1{m^2 + 1}$ and that exists for all possible $m\ge 0$
So range is all $[0, \infty)$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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A very strange and difficult hyperbolic integral This integral gave me serious problems, I tried to solve it by parts but it is madness! The calculations are too long and difficult, I do not think we should solve this.
$$\int _{ -\frac { 1 }{ 3 } }^{ \frac { 1 }{ 3 } }{ \sqrt { 36{ x }^{ 4 }-40{ x }^{ 2 }+4 } \cosh { (3x+\tanh ^{ -1 }{ (3x)-\tanh ^{ -1 }{ (x)) } } } }$$
the answer must be $$\frac { 12+4{ e }^{ 2 } }{ 9e }$$
Can some expert help me?
| This integral is more complicated than it looks and only requires comfort with hyperbolic trigonometric definitions. My initial instinct is to look at the expression inside the $\cosh$ to see if I can simplify it. In fact we have by the definition of the hyperbolic trigonometric functions,
$$\tanh^{-1}(x) = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) $$
Now using log laws we have that,
$$\tanh^{-1}(3x) - \tanh^{-1}(x) = \frac{1}{2} \ln \left(\frac{(1+3x)(1-x)}{(1-3x)(1+x)} \right), \ \ (*)$$
We note at this point that,
$$\sqrt{36x^4 - 40x^2 + 4} = 2\sqrt{(1-9x^2)(1-x^2)} = 2\sqrt{(1-3x)(1+3x)(1-x)(1+x)}$$
There is a lot in common between the above two equations which motivates us to press forward, we now use the following identity which we can prove just with the definitions of the hyperbolic trig functions,
$$\cosh(x+y) = \cosh(x) \cosh(y) + \sinh(x) \sinh(y)$$
We use to simplify the $\cosh$ in the integral, we find that,
$$\cosh(3x + (\tanh^{-1}(3x) - \tanh^{-1}(x))) = \cosh(3x) \cosh(\tanh^{-1}(3x) - \tanh^{-1}(x)) + \sinh(3x) \sinh(\tanh^{-1}(3x) - \tanh^{-1}(x)) $$
Using the following definitions of the hyperbolic functions,
$$\cosh(x) = \frac{e^x + e^{-x}}{2} $$
$$\sinh(x) = \frac{e^x - e^{-x}}{2} $$
We find that using $(*)$, leaving the details to you,
$$\cosh(\tanh^{-1}(3x) - \tanh^{-1}(x)) = \cosh\left(\frac{1}{2}\ln \left(\frac{(1+3x)(1-x)}{(1-3x)(1+x)} \right)\right) = \frac{1-3x^2}{\sqrt{(1-9x^2)(1-x^2)}}$$
Similarly, we have
$$\sinh(\tanh^{-1}(3x) - \tanh^{-1}(x)) = \frac{2x}{\sqrt{(1-9x^2)(1-x^2)}} $$
Putting this all together,
$$I = \int_{-1/3}^{1/3} \sqrt{36x^4 - 40x^2 + 4} \cosh(3x + \tanh^{-1}(3x) - \tanh^{-1}(x)) \ \mathrm{d}x $$
$$I = \int_{-1/3}^{1/3} 2\sqrt{(1-9x^2)(1-x^2)} \left(\cosh(3x) \frac{1-3x^2}{\sqrt{(1-9x^2)(1-x^2)}} + \sinh(3x)\frac{2x}{\sqrt{(1-9x^2)(1-x^2)}} \right) \ \mathrm{d}x$$
Finally the integral simplifies to,
$$I = 2 \int_{-1/3}^{1/3} (1-3x^2)\cosh(3x) \ \mathrm{d}x + 2 \int_{-1/3}^{1/3} 2x \sinh(3x) \ \mathrm{d}x $$
Which is a lot easier to compute with IBP and I will leave that to you to complete, this does indeed give the correct answer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Prove whether $xyz =1$ implies that $yzx=1$ or $yxz=1$. Let $x,y,z$ be elements of a group $G$ and $xyz=1$. I am trying to prove whether this implies $yzx=1$ or $yxz=1$.
My proof goes as follows: Let $x^{-1}$ denote the inverse of $x$, then $xx^{-1} = 1 = x(yz)$. By the cancellation property of groups, $x^{-1}= yz$. This implies $(yz)x = x^{-1}x = 1$. Therefore $xyz = 1 \implies yzx = 1$.
By applying the same argument to $y(zx) = 1$ one can prove that $xyz = 1 \implies yxz =1$.
However while trying to my proof online I stumbled upon the following counterexample for the proposition $xyz = 1 \implies yxz =1$. If we take $G$ to be the group of $2\times 2$ matrices and let $x = \left( \begin{array} { c c } { 1 } & { 2 } \\ { 0 } & { 2 } \end{array} \right)$, $y = \left( \begin{array} { l l } { 0 } & { 1 } \\ { 2 } & { 1 } \end{array} \right)$ and $z = \left( \begin{array} { c c } { - 1 / 2 } & { 3 / 4 } \\ { 1 } & { - 1 } \end{array} \right)$. Then $x y z = \left( \begin{array} { c c } { 1 } & { 0 } \\ { 0 } & { 1 } \end{array} \right) = 1$ but $y x z = \left( \begin{array} { c c } { 2 } & { - 2 } \\ { 5 } & { - 9 / 2 } \end{array} \right) \neq 1$.
I don't understand where my proof went wrong.
| Here $xyz=1$ gives
$$\begin{align}
x^{-1}xyz=x^{-1}1 & \iff yz=x^{-1} \\
&\iff yzx=x^{-1}x \\
& \iff yzx=1,
\end{align}$$
which gives
$$\begin{align}
y^{-1}yzx=y^{-1}1 & \iff zx=y^{-1} \\
& \iff zxy=y^{-1}y \\
&\iff zxy=1.
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
$ \int_0^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx \, \, \, (\phi>0) $ Evaluate
$$
\int_0^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx \, \, \, (\phi>0)
$$
See my answer below for a solution using a nice substitution.
| Solution
$$ \small
\int_0^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx = \int_0^{1} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx + \int_1^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx
$$
Notice that with the substitution $y=\frac{1}{x}$,
$$\begin{aligned}
\int_1^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx &= \int_1^0 \frac{1}{\left( 1+\frac{1}{y^2}\right)\left(1+\frac{1}{y^{\phi}} \right)}\cdot \left(\frac{-1}{y^2} \right)\, dy\\
&=\int_0^1\frac{y^{\phi}}{(1+y^2)(1+y^{\phi})}\, dy
\end{aligned}$$
Therefore
$$\begin{aligned}
\int_0^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx &= \int_0^1\frac{1+x^{\phi}}{(1+x^2)(1+x^{\phi})}\, dx \\ &= \int_0^1 \frac{1}{1+x^2}\, dx \\ &= \arctan x|_0^1 \\ &= \frac{\pi}{4}
\end{aligned}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Non-linear partial differential equation of order 1 Solve the PDE $$2(pq+py+qx)+x^2+y^2=0$$
where $\displaystyle p = \frac{\partial z}{\partial x}$, $\,\displaystyle q = \frac{\partial z}{\partial y}$
$f_p=2q+2x\\ f_q=2p+2x\\ f_z=0\\ f_x=2q+2x\\ f_y=2p+2y$
Using Charpit's method
$$\frac{dp}{2q+2x}=\frac{dq}{2p+2y}=\frac{dz}{-4pq-2py-2qx}=\frac{dx}{-2q-2y}=\frac{dy}{-2p-2x}$$
I deduce $$p+q+x+y=a \\ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
I am not sure how to proceed from here. Kinda stuck.
| Using Charpit's method
$$\frac{dp}{2q+2x}=\frac{dq}{2p+2y}=\frac{dz}{-4pq-2py-2qx}=\frac{dx}{-2q-2y}=\frac{dy}{-2p-2x}$$
Adding first two and last two fractions
$$\frac{dp+dq}{2p+2y+2q+2x}=\frac{dx+dy}{-2q-2y-2p-2x}$$
This implies, $a$ being an arbitrary constant $$p+x+q+y=a \tag{1}$$
Now the equation can be rewritten as $$2(p+x)(q+y)+(x-y)^2=0\tag{2}$$
Using $1,2$ $$(p+x)-(q+y)=\sqrt{a^2-2(x-y)^2}$$
Solving for $p+x$ and $q+y$ ,
$$\begin{align}
p+x &= \frac12(a+\sqrt{a^2+2(x-y)^2}) \\
q+y &= \frac12(a-\sqrt{a^2+2(x-y)^2})
\end{align}$$
This gives
$$\begin{align}
p&= -x+ \frac12(a+\sqrt{a^2+2(x-y)^2}) \\
q &= -y+ \frac12(a-\sqrt{a^2+2(x-y)^2})
\end{align}$$
$$dz=pdx+qdy$$
Substituting and integrating gives $$z=-\frac{x^2+y^2}{2}+\frac{a}{2}(x+y) +\frac12\int\sqrt{a^2-2(x-y)^2}d(x-y)$$
So the solution is $$z=-\frac{x^2+y^2}{2}+\frac{a}{2}(x+y)+\dfrac{a^2\ln\left(\left|a\sqrt{2(x-y)^2+a^2}+\sqrt{2}\left|a\right|(x-y)\right|\right)}{2^\frac{5}{2}}+\dfrac{a(x-y)\sqrt{\frac{2(x-y)^2}{a^2}+1}}{2^2}+b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3074155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
find the probability about sum of random variables Let $X_1, X_2, X_3, Y_1, Y_2, Y_3, Z_1, Z_2, Z_3$ be random variables which have uniform distribution between 0 and 1. It means, the average of $X_1 = 0.5$
Let: $X=X_1 + X_2 + X_3,$
$Y=Y_1 + Y_2 + Y_3$, $Z=Z_1 + Z_2 + Z_3$
In this case, the probability of $\{X$ is bigger than $Y$ and $Z$ both$\}$ would be $\dfrac{1}{3}$.
My question is:
What is the probability of "$c+X$ is bigger than $Y$ and $Z$" when $c$ is a constant"?
For example: what is $\mathbb{P}\left[0.2+X >\max\{Y,Z\}\right]$?
| Thank you @gt6989b
At first, the answer of $F_X(x)$ is
$F(x)=\dfrac{1}{6}x^3 $ when $0\leq x<1$
$F(x)=-\dfrac{1}{3}x^3+\dfrac{3}{2}x^2-\dfrac{3}{2}x+\dfrac{1}{2} $ when $1 \leq x <2$
$F(x)=1-\dfrac{1}{6}(3-x)^3$ when $2\leq x<3$
and secondly,
$f(x)=\dfrac{1}{2}x^2$ when $0\leq x<1$
$f(x)=-x^2 +3x-\dfrac{3}{2}$ when $1\leq x <2$
$f(x)=\dfrac{1}{2}(3-x)^2$ when $2\leq x<3$
And $f(y)$ and $f(z)$ follow same way.
Hence,
$$
\begin{split}
\mathbb{P}\left[X >\max\{Y,Z\}\right]
&= \iiint_{[0,3]^3} \mathbb{I}_{[x > \max\{y,z\}]} f(x)f(y)f(z) dxdydz \\
&= \dfrac{1}{3}\
\end{split}
$$
And I think I can go further. Thank you again.
| {
"language": "en",
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"answer_id": 1
} |
For $A_i$ on $y=\sqrt{x}$ and $B_i$ on $x$-axis, with $\triangle B_{i-1}B_iA_i$ equilateral of side $\ell_i$. Find $\ell_1+\cdots+\ell_{300}$.
Let $O$ be the origin, $A_1,A_2,A_3,\ldots$ be distinct points on the curve $y=\sqrt{x}$ and $B_1,B_2,B_3,\cdots$ be points on the positive $X$-axis such that the triangles $OB_1A_1,B_1B_2A_2,B_2B_3A_3,\ldots$ are all equilateral triangles with side lengths $l_1,l_2,l_3,\cdots$ respectively. Find the value of $l_1+l_2+\ldots+l_{300}$.
My Attempt:
We have $OA_1:y=\tan (60^{\circ})x=\sqrt{3}x$ so
$$
A_1=\left(\frac{1}{3},\frac{\sqrt{3}}{3}\right)
\quad \text{and} \quad
B_1=\left(\frac{2}{3},0\right),\
l_1=\frac{2}{3}.
$$
Let $B_i=(x_i,0)$ then
\begin{align*}
B_iA_{i+1}:& y=\sqrt{3}(x-x_i) \\
\iff & \sqrt{3}(x-x_i)=\sqrt{x} \\
\iff & 3x^2-x(6x_i+1)+3x_i^2=0 \\
\iff & x=\frac{6x_i+1+\sqrt{12x_i+1}}{6}.
\end{align*}
and so $A_{i+1}:\left(\frac{6x_i+1+\sqrt{12x_i+1}}{6},\sqrt{\frac{6x_i+1+\sqrt{12x_i+1}}{6}}=\frac{\sqrt{3}(1+\sqrt{12x_i+1})}{6}\right)$ and
\begin{align*}
x_{i+1}
& =2\left(\frac{6x_i+1+\sqrt{12x_i+1}}{6}\right)-x_i \\
& =\frac{3x_i+1+\sqrt{12x_i+1}}{3}
\end{align*}
\begin{align*}
l_{i+1}
& = x_{i+1}-x_i \\
& = \frac{1+\sqrt{12x_i+1}}{3} \\
& = \frac{1+\sqrt{12(\ell_1+\ell_2+\ldots+\ell_i)+1}}{3}.
\end{align*}
We will prove, that $\ell_i=\frac{2i}{3}$
For $i=1$ it is true.
For $i=n+1$:
\begin{align*}
3 \ell_{n+1}
& = 1+\sqrt{12(\ell_1+\ell_2+\ldots+\ell_n)+1} \\
& = 1+\sqrt{8(1+2+\ldots+n)+1} \\
& = 1+\sqrt{(4n^2+4n+1)} \\
& = 2n+2.
\end{align*}
so
$$
\ell_{n+1}=
\frac{2(n+1)}{3}.
$$
Hence
$$
\ell_1+\ldots+\ell_{300}
=\frac{2}{3} (1+\ldots+300)
=30100.
$$
Am I true for this?
| Let $A_i=(x_i, y_i)$ and $B_i=(z_i, 0)$ then because $y_i>0$, $x_i-z_i>0$ and $z_{i+1}-x_i>0$ we must have
$$
\frac{y_i}{x_i-z_i}=\frac{y_i}{z_{i+1}-x_i}=\tan\frac{\pi}{3}=\sqrt{3}
$$
then
$$
z_i=y_i^2-\frac{\sqrt {3}y_i}{3}\\
z_{i+1}=y_i^2+\frac{\sqrt {3}y_i}{3}\\
l_i=\frac{2\sqrt{3}}{3}y_i
$$
Confronting them let $y_{i+1}=y_i+u_i$ with $u_i>0$ it follows
$$
y_{i+1}^2-\frac{\sqrt {3}y_{i+1}}{3}=y_i^2+\frac{\sqrt {3}y_i}{3}\Rightarrow u_i^2+2u_iy_i-\frac{\sqrt {3}u_i}{3}=\frac{2\sqrt{3}}{3}y_i\\
\Rightarrow u_i=-y_i+\frac{\sqrt 3}{6}+\sqrt{y_i^2+\frac{1}{12}-\frac{\sqrt 3}{3}y_i+\frac{2\sqrt{3}}{3}y_i}=-y_i+\frac{\sqrt 3}{6}+y_i+\frac{\sqrt 3}{6}=\frac{\sqrt 3}{3}
$$
then set $x_0=y_0=0$ we have
$$
y_{i+1}=y_i+\frac{\sqrt 3}{3}=\frac{\sqrt 3}{3}(i+1)\Rightarrow l_i=\frac{2\sqrt{3}}{3}\frac{\sqrt 3}{3}i=\frac{2}{3}i
$$
and this is an arithmetic succession and you can easily find its sum.
| {
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"url": "https://math.stackexchange.com/questions/3079161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $ I struggle for a while solving limit of this chain:
$
a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
$
I know from WolframAlpha result will be $ \frac{1}{4\sqrt{2}} $, but step-by-step solution is overcomplicated(28 steps). Usually I solve limits like this by property $ \left(a-b\right)\left(a+b\right)=a^2-b^2 $
I made this far:
$$
\lim _{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4\:}+1}}-\sqrt{2}n\right)\right)
=n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}=\frac{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\right)\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)} = \frac{-n^8+n^6\sqrt{n^4+1}}{n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}}$$
I will appreciate every help. Thank you
| From the numerator and denominator take $-n^{3}$ as common term and then rationalize the numerator again.
Once you do that you will be left with :
$$\frac{n^{3}}{(n^{2}+\sqrt{n^{4}+1})(\sqrt{n^{2}+\sqrt{n^{4}+1}}+n\sqrt{2})}$$
Divide the numerator and denominator by $n^{3}$, and once you divide the denominator by $n^{3}$ for the first factor divide n^{2} and the second factor just n. You will get :
$$\frac{1}{(1+\sqrt{(1/n^{4})+1})(\sqrt{1+\sqrt{(1/n^{4})+1}}+\sqrt{2})}$$
Apply the limit you will get the answer as $\frac{1}{4\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
on the quadratic form $5x^2-y^2$ Consider the following subsets of $\mathbb{Z}$:
$A=\{\frac{5p^2-y^2}{4}\, | \, p, y \,\,\text{odd positive integers}, p\,\, \text{prime}\}$ and
$B=\{\frac{5x^2-y^2}{4}\, | \, x, y \,\,\text{odd positive integers}\}$.
Are there elements in $B$ which are not in $A$?
| Note that $\frac{5\cdot9^2-3^2}{4}=99$, and if $\frac{5p^2-y^2}{4}=99$ for some prime number $p$, then
$$5p^2-y^2=396\equiv0\pmod{3},$$
from which it follows that $p\equiv y\equiv0\pmod{3}$. Because $p$ is prime it follows that $p=3$, and hence
$$y^2=5p^2-396=-351,$$
a contradiction. So $99\in B$ but $99\notin A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solution to Differential equation 2nd order, x*e^x I'm trying to solve the following differential equation:
$y'' + y = x \cdot e^x$
I already have the homogeneous solution: $y_h = c_1 \cdot cos(x) + c_2 \cdot sin(x)$ but I'm struggeling to find the particular solution.
I tried using $y_p = a \cdot x \cdot e^x$ but that didn't quite work, or maybe I did something wrong...
Can somebody explain to me how to solve this?
Thanks.
Edit: Here's the solution:
$y_p = (a \cdot x + b) \cdot e^x$
$y_p' = (a + a \cdot x + b) \cdot e^x$ and $ y_p'' = (2 \cdot a + a \cdot x + b) \cdot e^x$
$y_p'' + y_p = (2 \cdot a \cdot x + 2 \cdot a + 2 \cdot b) \cdot e^x$
$\Rightarrow 2 \cdot a = 1 \Rightarrow a = \frac{1}{2}$ and $2 \cdot a + 2 \cdot b = 0 \Leftrightarrow 2 \cdot a = 1 = - 2 \cdot b \Rightarrow b = - \frac{1}{2}$
$\Rightarrow y_p = (a \cdot x + b) \cdot e^x = \frac{x-1}{2} \cdot e^x$
All in all: $y = y_h + y_p = c_1 \cdot cos(x) + c_2 \cdot sin(x) + \frac{x-1}{2} \cdot e^x$
| Try $y_p = (ax+b)e^x$.
[Additional characters to bypass the 30 characters rule]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the value of $a$ where $√a$ is area of a trapezoid which touches the circle with center $O$ (diameter is 2)?
The sides $AB, BC, CD$ of trapezoid $ABCD$ touches the circle with
center $O$ and they are equal. $AD$, goes through the point O. If
diameter is 2, then the area of the trapezoid is $√a$ . What is
the value of a?
Source: Bangladesh Math Olympiad 2016 Junior Catagory
I tried but I could not find any possible solution. How am I supposed to get $BC + AD$?
| Warning: might be a little convoluted, uses a bit of assumed knowledge, and is probably not the most elegant/efficient solution. I've triple-checked the method and arithmetic though so unless there's a fundamental flaw with my solution, I think this should be correct.
We make a duplicate trapezoid of $ABCD$, with diameter $AD$, which basically inscribes the circle in a regular hexagon:
Since the duplicate trapezoid has the same area, the area of the hexagon is $2\sqrt a$.
The area of a regular hexagon can be shown to be given by $\frac{3 \sqrt 3}{2} s^2$, where $s$ is the length of a side of the hexagon. Thus,
$$2 \sqrt a = \frac{3 \sqrt 3}{2} s^2 \implies a = \frac{27}{16} s^4$$
The area of a regular polygon can also be shown (link above) to be given by $xp/2$, for perimeter $p$ and apothem $x$. Here, $p = 6s$, and $x = 1$, the radius of the circle, clear by the construction above and that the diameter is $2$. Thus, the area is given by $(6s)(1)/2 = 3s$.
Thus,
$$2 \sqrt a = 3s \implies a = \frac{9}{4}s^2$$
Thus,
$$a = \frac{27}{16}s^4 = \frac{9}{4}s^2$$
We subtract the right fraction from the left, and multiply through by $16$ as we begin solving for $s$. We get the equation
$$27s^4 - 36s^2 = 0$$
Let $u = s^2$ for ease of use, then the above equation becomes
$$27u^2 - 36u = 0 \implies 9u(3u - 4) = 0 \implies u = 0, u = \frac{4}{3}$$
$u=0$ is obviously not what we want, but $u=4/3$ is fine. Then, going back through our substitutions,
$$u=\frac{4}{3} \implies s^2 = \frac{4}{3} \implies s = \frac{2}{\sqrt 3} \implies 2 \sqrt a = 3 \cdot \frac{2}{\sqrt 3} \implies a = 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluating $\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)\,dx$
How to prove $$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x) \, dx = \frac43G + \frac13\pi\ln\left(2+\sqrt3\right),$$where $G$ is Catalan's constant?
I have a premonition that this integral is related to $\Im\operatorname{Li}_2\left(2\pm\sqrt3\right)$.
Attempt
$$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x) \, dx \\ =\int_0^\infty\frac{\operatorname{arcsinh}(2x)}{1+x^2} \, dx\\
=2\int_0^\infty\frac{x\cosh x}{4+\sinh^2x} \, dx\\
=2\int_0^\infty\frac{x\cosh x}{3+\cosh^2x} \, dx\\
=2\int_0^\infty\sum_{n=0}^\infty x(-3)^n\cosh^{-2n-1}(x) \, dx$$
I failed to integrate $x\cosh^{-2n-1}(x)$. Mathematica returns a hypergeometric term while integrating it.
| Sorry in a rush, so only a PARTIAL SOLUTION:
Here I will employ Feynman's Trick:
\begin{equation}
I = \int_0^{\infty}\frac{\operatorname{arcsinh(2x)}}{1 + x^2}\:dx
\end{equation}
Let
\begin{equation}
J(t) = \int_0^{\infty}\frac{\operatorname{arcsinh(tx)}}{1 + x^2}\:dx
\end{equation}
We observe that $J(2) = I$ and $J(0) = 0$. Using Leibniz's Integral rule we differentiate with respect to '$t$':
\begin{equation}
J'(t) = \int_0^{\infty}\frac{x}{\sqrt{1 + t^2x^2}}\frac{1}{1 + x^2}\:dx = \left[\frac{1}{\sqrt{t^2 - 1}} \cdot \arctan\left(\sqrt{\frac{1 + t^2x^2}{t^2 - 1}} \right) \right]_0^{\infty} = \frac{1}{\sqrt{t^2 - 1}}\left[\frac{\pi}{2} - \arctan\left(\frac{1}{\sqrt{t^2 -1}} \right) \right]
\end{equation}
Thus,
\begin{align}
J(t) &= \int \frac{1}{\sqrt{t^2 - 1}}\left[\frac{\pi}{2} - \arctan\left(\frac{x}{\sqrt{t^2 -1}} \right) \right]\:dt \\
&= \int \frac{\pi}{2}\cdot \frac{1}{\sqrt{t^2 - 1}}\:dt - \int \frac{1}{\sqrt{t^2 - 1}}\arctan\left(\frac{1}{\sqrt{t^2 -1}} \right)\:dt = I_1 - I_2
\end{align}
For $I_1$:
\begin{equation}
I_1 = \int \frac{\pi}{2}\cdot \frac{1}{\sqrt{t^2 - 1}}\:dt = \frac{\pi}{2}\ln\left| \sqrt{t^2 - 1} + t\right| + C_1
\end{equation}
Where $C_1$ is the constant of integration. Note that $I_1(2) = \ln\left|2 + \sqrt{3} \right|$
For $I_2$:
\begin{equation}
I_2 = \int \frac{1}{\sqrt{t^2 - 1}}\arctan\left(\frac{1}{\sqrt{t^2 -1}} \right)\:dt
\end{equation}
Unfortunately this is not so easy to evaluate. I will first make the substitution $u = \frac{1}{\sqrt{t^2 - 1}}$:
\begin{align}
I_2 &= \int \frac{1}{\sqrt{t^2 - 1}}\arctan\left(\frac{1}{\sqrt{t^2 -1}} \right)\:dt = \int u \cdot \arctan(u) \cdot \frac{-u^4}{\sqrt{1 +u^2}}\:du\\
& = - \int \frac{-u^5}{\sqrt{1 +u^2}} \cdot \arctan(u) \:du
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 1
} |
Find all continuous functions in $0$ that $2f(2x) = f(x) + x $ I need to find all functions that they are continuous in zero and
$$ 2f(2x) = f(x) + x $$
About
I know that there are many examples and that forum but I don't understand one thing in it and I need additional explanation. (Nowhere I see similar problem :( )
My try
I take $ y= 2x$ then
$$f(y) = \frac{1}{2}f\left(\frac{1}{2}y\right) + \frac{1}{4}$$
after induction I get:
$$f(y) = \frac{1}{2^n}f\left(\frac{1}{2^n}y\right) + y\left(\frac{1}{2^2} + \frac{1}{2^4} + ... + \frac{1}{2^{2n}} \right)$$
I take $\lim_{n\rightarrow \infty} $
$$ \lim_{n\rightarrow \infty}f(y) = f(y) = \lim_{n\rightarrow \infty} \frac{1}{2^n}f\left(\frac{1}{2^n}y\right) + y\cdot \lim_{n\rightarrow \infty} \left(\frac{1}{2^2} + \frac{1}{2^4} + ... + \frac{1}{2^{2n}} \right)$$
$$f(y) = \lim_{n\rightarrow \infty} \frac{1}{2^n} \cdot f\left( \lim_{n\rightarrow \infty} \frac{1}{2^n}y \right) + \frac{1}{3}y$$
Ok, there I have question - what I should there after? How do I know that $$f(0) = 0 $$?
I think that it can be related with " continuous functions in $0$ " but
function is continous in $0$ when
$$ \lim_{y\rightarrow 0^+}f(y)=f(0)=\lim_{y\rightarrow 0^-}f(y)$$
And I don't see a reason why $f(0)=0$
edit
*
*Ok, I know why $f(0) =0$ but why I need informations about "Continuity at a point $0$ " ? It comes to
$$\lim_{n\rightarrow \infty}f\left(\frac{1}{2^n}y\right) = f\left( \lim_{n\rightarrow \infty} \frac{1}{2^n}y \right)$$ ?
| This recurrence equation is linear then
$$
f(x) = f_h(x)+f_p(x)
$$
such that
$$
a f_h(a x)-f_h(x) = 0\\
a f_p(a x)-f_p(x) = x
$$
for the homogeneous equation we assume
$$
f_h(x) = \frac Cx
$$
and then for the paticular we assume
$$
f_p(x) = \frac{C(x)}{x}
$$
then
$$
a\frac{C(a x)}{a x}-\frac{C(x)}{x} = x
$$
or
$$
C(a x)-C(x) = x^2
$$
for this last recurrence equation we choose
$$
C(x) = \frac{x^2}{a^2-1}
$$
so the final solution is
$$
f(x) = \frac{C}{x}+\frac{x}{a^2-1}
$$
in our case $a = 2$ then
$$
f(x) = \frac Cx+\frac x3
$$
and to assure continuity at $x=0$ we choose $C = 0$ so the final result is
$$
f(x) = \frac x3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3083632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
convergence of a recursive sequence with parameter a How can you determine if the following recursive sequence converges:
$$x_{n+1}=\frac{1}{2}(a+x_n^2)$$
where $0\le a \le 1$ and $x_1=0$
I know that the limit x (if it exists) satisfies the following equation:
$$x=\frac{1}{2}(a+x^2)$$
since $\lim_{x\rightarrow \infty} x_n = \lim_{x\rightarrow \infty} x_{n+1}$. Therefore $x=1\pm \sqrt{1-a}$
Thank you in advance :)
| By induction we can show that $0\leq x_n<1$.
Now, for $a=1$ we obtain $$x_{n+1}-x_n=\frac{1}{2}(x_n-1)^2\geq0$$ and the rest is smooth.
But for $a<1$ we obtain $0<r=\frac{2-\sqrt{1-a}}{2}<1$ and $$|x_{n+1}-1+\sqrt{1-a}|=\left|\frac{1}{2}x_n^2+\frac{1}{2}a-1+\sqrt{1-a}\right|=$$
$$=\frac{1}{2}\left|x_n^2-\left(1-\sqrt{1-a}\right)^2\right|=\frac{1}{2}\left|x_n-1+\sqrt{1-a}\right|\cdot\left|x_n+1-\sqrt{1-a}\right|\leq$$
$$\leq r\left|x_n-1+\sqrt{1-a}\right|\leq...\leq r^n\left|x_1-1+\sqrt{1-a}\right|\rightarrow0.$$
| {
"language": "en",
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} |
Approximating $\pi$ with arctangent Use the fact that $\frac{\pi}{4} = \text{arctangent}(\frac{1}{2}) + \text{arctangent}(\frac{1}{3})$ to determine the number of terms summed to ensure an approximation to $\pi$ less than $10^{-3}$.
So far I've got
\begin{align*}
\pi = 4\Bigg(\sum_{i=1}^{\infty} (-1)^{i+1}\frac{1}{2^{2i-1}(2i-1)} + \sum_{i=1}^{n} (-1)^{i+1}\frac{1}{3^{2i-1}(2i-1)}\Bigg)
\end{align*}
from the given identity. But I don't know how to proceed.
| Rewrite the equality as:
\begin{array}{}
\pi = 4\left[\sum\limits_{i=0}^{n-1} (-1)^{i}\left(\frac{1}{2^{2i+1}(2i+1)} +\frac{1}{3^{2i+1}(2i+1)}\right)+\sum\limits_{i=n}^{\infty} (-1)^{i}\left(\frac{1}{2^{2i+1}(2i+1)} +\frac{1}{3^{2i+1}(2i+1)}\right)\right]
\end{array}
to obtain the inequality to be solved for $n$:
$$
\begin{array}{}
4\left|\sum\limits_{i=n}^{\infty} (-1)^{i}\left(\frac{1}{2^{2i+1}(2i+1)} +\frac{1}{3^{2i+1}(2i+1)}\right)\right|<\epsilon=10^{-3}.\tag1
\end{array}$$
Observe now that the series (1) is alternating with monotonously decreasing terms, and thus the absolute value of the series is less than that of its first term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Maximum value of sum with integer parts How can I find the max value of $\sum_{i}^{n} \lfloor\frac{i}{k}\rfloor (k-1)$ with $k$ integer $\in [1,100]$? I can express this sum in terms of n and k (it's quite easy) but I'm not able to find its maximum ...
$\sum_{i}^{n} \lfloor\frac{i}{k}\rfloor (k-1) = \frac{m(k-1)(2N+2-mk-k)}{2}$ with m the greatest integer such that $mk \leq 100$
| In order to calculate the sum we use Iverson brackets $[P]$ to cope with the floor function.
We obtain
\begin{align*}
\color{blue}{\sum_{i=1}^n\left\lfloor\frac{i}{k}\right\rfloor}
&=\sum_{i=1}^n\sum_{j=1}^{\lfloor n/k\rfloor}j\left[j=\left\lfloor\frac{i}{k}\right\rfloor\right]\\
&=\sum_{i=1}^n\sum_{j=1}^{\lfloor n/k\rfloor}j\left[j\leq \frac{i}{k}<j+1\right]\\
&=\sum_{j=1}^{\lfloor n/k\rfloor}j\sum_{i=1}^n\left[kj\leq i<k(j+1)\right]\\
&=\sum_{j=1}^{\lfloor n/k\rfloor-1}j\sum_{i=kj}^{k(j+1)-1}1+\left\lfloor\frac{n}{k}\right\rfloor\sum_{i=k\lfloor n/k\rfloor}^{n}1\\
&=k\sum_{j=1}^{\lfloor n/k\rfloor-1}j+\left\lfloor\frac{n}{k}\right\rfloor\left(n-k\left\lfloor\frac{n}{k}\right\rfloor+1\right)\\
&=\frac{k}{2}\left\lfloor\frac{n}{k}\right\rfloor\left(\left\lfloor\frac{n}{k}\right\rfloor-1\right)
+\left\lfloor\frac{n}{k}\right\rfloor\left(n-k\left\lfloor\frac{n}{k}\right\rfloor+1\right)\\
&\,\,\color{blue}{=\left\lfloor\frac{n}{k}\right\rfloor\left(n+1-\frac{k}{2}\left\lfloor\frac{n}{k}\right\rfloor-\frac{k}{2}\right)}\tag{1}
\end{align*}
In order to find the maximum value of $$(k-1)\sum_{i=1}^n\left\lfloor\frac{i}{k}\right\rfloor$$ where $k\in[1\ldots 100]$, we approximate $\left\lfloor\frac{n}{k}\right\rfloor$ with $\frac{n}{k}$.
We calculate using (1)
\begin{align*}
\frac{d}{dk}\left((k-1)\frac{n}{k}\left(\frac{n}{2}+1-\frac{k}{2}\right)\right)=0
\end{align*}
which gives $k=\sqrt{n+2}$. We conclude the maximum value of $k$ is $$\color{blue}{k=\left\lfloor\sqrt{n+2}\right\rfloor}$$
| {
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
The Integral $\int \frac {dx}{(x^2-2ax+b)^n}$ Recently I came across this general integral,
$$\int \frac {dx}{(x^2-2ax+b)^n}$$
Putting $x^2-2ax+b=0$ we have,
$$x = a±\sqrt {a^2-b} = a±\sqrt {∆}$$
Hence the integrand can be written as,
$$
\frac {1}{(x^2-2ax+b)^n}
=
\frac {1}{(x-a-\sqrt ∆)^n(x-a+\sqrt ∆)^n}
$$
Resolving into partial fractions we have,
$$
\frac {1}{(x^2-2ax+b)^n}
=
\sum \frac {A_r}{(x-a-\sqrt ∆)^r} + \sum \frac {B_r}{(x-a+\sqrt ∆)^r}
$$
Putting $-\frac {1}{2\sqrt ∆} = D$ , I could produce a table of the coefficients $A$ and $B$ for different $n$.
\par
For $n=1$,
$$A_1=-D , B_1=D$$
For $n=2$,
$$A_1=2D^3 , B_1=-2D^3$$
$$A_2=D^2 , B_2 = D^2$$
For $n=3$,
$$A_1=-6D^5 , B_1=6D^5$$
$$A_2=-3D^4 , B_2 = -3D^4$$
$$A_3=-D^3, B_3=D^3$$
For $n=4$,
$$A_1=20D^7, B_1=-20D^7$$
$$A_2=10D^6 , B_2 = 10D^6$$
$$A_3=4D^5, B_3=-4D^5$$
$$A_4=D^4, B_4=D^4$$
For $n=5$,
$$A_1=-70D^9, B_1=70D^9$$
$$A_2=-35D^8, B_2 = -35D^8$$
$$A_3=-15D^7, B_3=15D^7$$
$$A_4=-5D^6, B_4=-5D^6$$
$$A_5=-D^5, B_4=D^5$$
Yet I am unable to deduce a general formula for the coefficients. If I have the coefficients, the integral is almost solved , for then I shall have a logarithmic term and a rational function in $x$. More directly, I seek a result of the form,
$$\kappa \log \left( \frac {x-a-\sqrt ∆}{x-a+\sqrt ∆}\right) + \frac {P(x)}{Q(x)}$$
Any help would be greatly appreciated.
Conjecture 1(Proved below)
$$A(n,r)= (-1)^n \binom {2n-r-1}{n-1} D^{2n-r}$$
$$B(n,r)= (-1)^{n-r} \binom {2n-r-1}{n-1} D^{2n-r}$$
| All right, now I've got it.
The easiest way to get all the coefficients? Expand in a Laurent series around one of the roots. Substituting $z=x-a-\sqrt{\Delta}$ and later defining $D=\frac1{2\sqrt{\Delta}}$, we get
\begin{align*}\frac1{(x^2-2ax+b)^n} &= \frac1{(x-a-\sqrt{\Delta})^n(x-a+\sqrt{\Delta})^n}=\frac1{z^n(z+2\sqrt{\Delta})^n}=\frac1{z^n}\cdot\frac{(2\sqrt{\Delta})^{-n}}{(1+\frac{z}{2\sqrt{\Delta}})^n}\\
\frac1{(x^2-2ax+b)^n} &= \frac{(-D)^n}{z^n(1-Dz)^n} = \frac{(-D)^n}{z^n}\sum_{j=0}^{\infty} \binom{n+j-1}{j}D^jz^j\\
&=(-1)^n\sum_{j=0}^{\infty}\binom{n+j-1}{j}D^{n+j}z^{j-n}\end{align*}
We claim that the coefficients $(-1)^n\binom{n+j-1}{j}D^{n+j}$ for $j<n$ are precisely the coefficients of $\frac1{z^{n-j}}$ in the partial fractions expansion of $\frac1{z^n(z+2\sqrt{\delta})^n}$. Why? Subtract the negative-exponent terms of the Laurent series from the partial fractions expansion. The difference is locally bounded, with a nice power series. But then, the only terms in the partial fractions expansion that aren't locally bounded are the $\frac1{z^k}$ terms - so their coefficients all have to match with the terms from the Laurent series.
Let $k=n-j$, and we get $A(n,k)=(-1)^n\binom{2n-k-1}{n-k}D^{2n-k}=(-1)^n\binom{2n-k-1}{n-1}D^{2n-k}$ in the partial fractions expansion
$$\frac1{z^n(z+2\sqrt{\Delta})^n}=\sum_{k=1}^n \frac{A(n,k)}{z^k} +\sum_{k=1}^n \frac{B(n,k)}{(z+2\sqrt{\Delta})^k}=\sum_{k=1}^n \frac{A(n,k)}{(x-a-\sqrt{\Delta})^k} +\sum_{k=1}^n \frac{B(n,k)}{(x-a+\sqrt{\Delta})^k}$$
Oh, yes - in my comment, I didn't actually define my notation, and the update to the question imported that without defining it. The purpose is clear; we're just putting both parameters in the notation instead of just the power $k$ of $\frac1{z-a\pm\sqrt{\Delta}}$. Formally, the definition is the line just above.
That's half of the conjecture. For the other half, we expand around the other root.
\begin{align*}\frac1{(x^2-2ax+b)^n} &= \frac1{(x-a-\sqrt{\Delta})^n(x-a+\sqrt{\Delta})^n}=\frac1{(w-2\sqrt{\Delta})^nw^n}=\frac1{w^n}\cdot\frac{(-2\sqrt{\Delta})^{-n}}{(1-\frac{w}{2\sqrt{\Delta}})^n}\\
\frac1{(x^2-2ax+b)^n} &= \frac{D^n}{w^n(1+Dw)^n} = \frac{D^n}{w^n}\sum_{j=0}^{\infty} \binom{n+j-1}{j}(-D)^jw^j\\
&=\sum_{j=0}^{\infty}(-1)^j\binom{n+j-1}{j}D^{n+j}w^{j-n}\end{align*}
Again, extract the negative-exponent terms to get $B(n,k)=(-1)^{n-k}\binom{2n-k-1}{n-k}D^{2n-k} =(-1)^{n-k}\binom{2n-k-1}{n-1}D^{2n-k}$. The conjecture is confirmed, and we have our general formula.
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve $x^{18} \equiv 7^{99} - 7, \mod 592$ What I tried:
$x^{18} \equiv 7^{99} - 7, \mod 592 \iff \begin{cases} x^{18} \equiv 7^{99}-7 & \mod 7 \\ x^{18} \equiv 7^{99}-7 & \mod 2 \\ x^{18} \equiv 7^{99}-7 & \mod
3\end{cases} \iff x^{18} \equiv 0, \mod 7,2,3. $
I'm not sure how to proceed: is the last step equivalent to saying $x^{18} \equiv 0, \mod 42 (=7 \cdot 2 \cdot 3)$ or $x^{18} \equiv 0, \mod 592$?
| Hint $\bmod 37\!:\,\ x^{\large 18}\equiv \color{#c00}{7^{\large 99}}\!-7\equiv -6\,\overset{\rm square}\Longrightarrow\,x^{\large 36}\equiv -1\,$ contra little Fermat
because: $\ \ 7 \equiv 3^{\large 4}\,\Rightarrow\, \color{#c00}{7^{\large 99}}\equiv (3^{\large 4})^{\large 99}\equiv (3^{\large 36})^{\large 11}\equiv 1^{\large 11}\equiv\color{#c00}{\bf 1}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\sum_ \limits{n=1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1) } $ $$\sum_ \limits{n=1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1) } $$
$$1 \cdot 3 \cdot 5 \cdots (2n+1) = \frac{1 \cdot 2 \cdot 3 \cdots (2n+2)}{2 \cdot 4 \cdot 6 \cdots (2n+2)} = \frac{(2n+2)!}{2^{n+1} \cdot (n+1)!} $$
But the follow up gives me $\infty$. How to approach this type of exercises?
The result should be $1/2$
| HINT:
Note that
$$\begin{align}
\frac{n}{(2n+1)!!}&=\frac12\left( \frac{2n+1-1}{(2n+1)!!}\right)\\\\&=\frac12\left(\frac{1}{(2n-1)!!}-\frac1{(2n+1)!!}\right)
\end{align}$$
Now, telescope.
| {
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 0
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Roots of polynomials and their formulae relating to coefficients Write down the cubic equation given that $\alpha + \beta + \gamma = 4$, $\alpha^2 + \beta^2 + \gamma^2 = 66$, and $\alpha^3 + \beta^3 + \gamma^3 = 280$
Ok so, the sum of roots is given and I'm able to use the sum of the roots and the sum of the roots squared to get the sum of the combination of roots, but I'm unable to get the product of roots, because I can't seem to manipulate the sum of the cubes of roots to resemble the sum of roots and sum of squares of roots.
| Hint, changing to $\;a,b,c\;$ for the roots:
$$(a+b+c)^2=a^2+2a(b+c)+(b+c)^2$$
$$(a+b+c)^3=a^3+3a^2(b+c)+3a(b+c)^2+(b+c)^3$$
Play with the above and the given data...
Added Or the other way around:
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$$
$$a^3+b^3+c^3=(a+b+c)^3-3a^2(b+c)-3a(b+c)^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 3,
"answer_id": 1
} |
Evaluating $\lim_{\epsilon\to 0^{+}}\ \frac{ _2F_1\left( \tfrac{1}{2} - \nu, \tfrac{1}{2} + \nu; \epsilon; y \right) }{\Gamma(\epsilon)}$ For $\nu \in \mathbb{C}$ and negative $y<0$ is there a way to compute the limit
$$
f(\nu,y) \equiv \lim_{\epsilon \to 0^{+}} \ \frac{ _2F_1\left( \tfrac{1}{2} - \nu, \tfrac{1}{2} + \nu; \epsilon; y \right) }{\Gamma(\epsilon)}
$$
in terms of simpler special functions (ideally not hypergeometrics...)? So far, I thought about writing this in terms of the series
$$
f(\nu,y) \ = \ \sum_{n=0}^\infty \frac{\Gamma(\frac{1}{2} - \nu + n )\Gamma(\frac{1}{2} + \nu + n )}{(n-1)! n!} \ y^n
$$
I think that this series converges at least for $-1 < y < 0$, but I am also interested in values of $y \leq -1$.
(EDIT: I know that $_2F_1(a,b;c;z)$ has poles at $c=0,-1,-2,\ldots$, this makes it hard to learn anything about this limit using Mathematica)
| We may express
\begin{equation}
_2F_1\left( \tfrac{1}{2} - \nu, \tfrac{1}{2} + \nu; \epsilon; y \right) =\frac{\Gamma\left(\epsilon\right)}{\Gamma\left(\tfrac{1}{2} - \nu\right)\Gamma%
\left(\tfrac{1}{2} + \nu\right)}\sum_{s=0}^{\infty}\frac{\Gamma\left(\tfrac{1}{2} - \nu+s\right)\Gamma\left(\tfrac{1}{2} + \nu+s%
\right)}{\Gamma\left(\epsilon+s\right)s!}y^{s}
\end{equation}
and thus,
\begin{align}
\lim_{\epsilon \to 0^{+}} \ \frac{ _2F_1\left( \tfrac{1}{2} - \nu, \tfrac{1}{2} + \nu; \epsilon; y \right) }{\Gamma(\epsilon)}&=\frac{1}{\Gamma\left(\tfrac{1}{2} - \nu\right)\Gamma%
\left(\tfrac{1}{2} + \nu\right)}\sum_{s=1}^{\infty}\frac{\Gamma\left(\tfrac{1}{2} - \nu+s\right)\Gamma\left(\tfrac{1}{2} + \nu+s%
\right)}{\Gamma\left(s\right)s!}y^{s}\\
&=\frac{y}{\Gamma\left(\tfrac{1}{2} - \nu\right)\Gamma%
\left(\tfrac{1}{2} + \nu\right)}\sum_{t=0}^{\infty}\frac{\Gamma\left(\tfrac{3}{2} - \nu+t\right)\Gamma\left(\tfrac{3}{2} + \nu+s%
\right)}{\Gamma\left(t+2\right)t!}y^{t}\\
&=\frac{y\Gamma\left(\tfrac{3}{2} - \nu\right)\Gamma%
\left(\tfrac{3}{2} + \nu\right)}{\Gamma(2)\Gamma\left(\tfrac{1}{2} - \nu\right)\Gamma%
\left(\tfrac{1}{2} + \nu\right)} \,_2F_1\left(\tfrac{3}{2} - \nu,\tfrac{3}{2} +\nu;2;y \right)\\
&=y\left( \frac{1}{4}-\nu^2 \right)\,_2F_1\left(\tfrac{3}{2} - \nu,\tfrac{3}{2} +\nu;2;y \right)
\end{align}
From this representation of the associated Legendre function
\begin{equation}
_2F_1\left(a,b;\tfrac{1}{2}(a+b+1);z\right)=\left(-z(1-z)\right)^{%
\frac{(1-a-b)}{4}}\,P^{(1-a-b)/2}_{(a-b-1)/2}\left(1-2z\right)
\end{equation}
with $a=3/2+\nu,b=3/2-\nu$ we can express
\begin{equation}
f(\nu,y)=\left( \frac{1}{4}-\nu^2 \right)\sqrt{\frac{-y}{1-y}}P^{-1}_{\nu-1/2}\left( 1-2y \right)
\end{equation}
where $1-2y>0$.
| {
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What is the value of $x$ in $ABCD$ rectangle where $AE = 4, BE = 6, CE=5$ and $DE = x$?
In the diagram of rectangular $ABCD$, $AE=4, BE= 6, CE = 5$ and
$DE=x$, find the value of $x$
Source: Bangladesh Math Olympiad 2015 Junior Catagory
I can not relate these information with $DE$.
| Another way is:
$\hspace{1cm}$
$$\begin{cases}6^2=a^2+b^2\\ 5^2=b^2+c^2\\ x^2=c^2+d^2\\ 4^2=a^2+d^2\end{cases} \Rightarrow \\
x^2=(a^2+d^2)-(a^2+b^2)+(b^2+c^2)=\\
4^2-6^2+5^2=5 \Rightarrow x=\sqrt{5}.$$
| {
"language": "en",
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"question_score": "2",
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Evaluating $\sin\frac{\pi}{2}\sin\frac{\pi}{2^2}\sin\frac{\pi}{2^3}\cdots\sin\frac{\pi}{2^{11}}\cos \frac{\pi}{2^{12}}$ Evaluate :
$$\sin\frac{\pi}{2} \times \sin \frac{\pi}{2^2} \times \sin \frac{\pi}{2^3} \cdots \times \sin\frac{\pi}{2^{11}} \times \cos \frac{\pi}{2^{12}}$$
I tried to solve it by using double angle formula, by replacing $\sin\displaystyle\frac{\pi}2$ with $2 \sin\displaystyle\frac{\pi}4 \cos\displaystyle\frac{\pi}4$ and similarly replacing $\sin\displaystyle\frac{\pi}{2^2}$ with double angle formula. But I'm not getting anywhere from this, can you please suggest me how can I approach to this problem, especially that cosine term at last is making everything so weird.
The thing which is most important to ask is : although I know every theory which applies over here but then also why I'm unable to solve this question? What is the reason for my failure?
| Whatever you would do, the result would be a very small number very close to $0$
$$\sin\frac{\pi}{2} \times \sin \frac{\pi}{2^2} \times \sin \frac{\pi}{2^3}\times \cdots \times \sin\frac{\pi}{2^{11}} \times \cos \frac{\pi}{2^{12}}$$
We know the value of the sine for some of the angles. The first are
$$\{1 , \frac {\sqrt 2} 2,\frac{\sqrt{2-\sqrt{2}}}{2},\frac{\sqrt{2-\sqrt{2+\sqrt{2}}}}{2},\frac{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2},\frac{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}} }{2} \}$$ and the pattern continue for ever.
For the next one, corresponding to $\sin \frac{\pi}{2^8}$ the exact value is
$$\frac{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}}}{2}\approx0.012271538$$ while
$$\frac{\pi}{2^8} \approx 0.012271846$$ that is to say that you can approximate all the sines by the argument. Concerning the cosine, the argument is so small that you can assume it is almost equal to $1$.
If you take into account the fact that $\sin(x)<x$, an upper bound for the product of sines would be
$$ \frac{ \pi^9 \,\sqrt 2}{2^{64}}\approx 2.285 \times 10^{-15}$$ while it should be $\approx 2.208 \times 10^{-15}$.
Concerning
$$\cos \frac{\pi}{2^{12}}\approx 0.99999970586$$
All of that makes the value of the expression quite close to the machine epsilon.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}$ $$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}$$
My working:
$$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\int_{0}^{1}\frac{\ln x}{x^2-1}dx=-\sum_{r\ge 1}\int_0^{1}x^{2r}\ln x\,dx =\sum_{r\ge 1}\frac {1}{(2r-1)^2}=
\frac{\pi^2}{8}$$
Is there any other approach?
| I will provide two different methods. The first relies on properties of the polygamma function, the second converts the integral to a double integral first before evaluating it.
Let
$$I = \int_1^\infty \frac{\ln x}{x^2 - 1} \, dx.$$
Method 1 - A polygamma approach
By enforcing a substitution of $x \mapsto 1/x$ we see that
$$I = \int_0^1 \frac{\ln x}{x^2 - 1} \, dx.$$
From the following integral representation for the digamma function $\psi (x)$, namely
$$\psi (x + 1) = -\gamma + \int_0^1 \frac{1 - t^x}{1 - t} \, dt,$$
where $\gamma$ is the Euler–Mascheroni constant, on differentiating with respect to $x$ we have
$$\psi^{(1)} (x + 1) = - \int_0^1 \frac{t^x \ln t}{1 - t} \, dt.$$
Here $\psi^{(1)} (z)$ denotes the trigamma function. Substituting $t = u^2$, $dt = 2u \, du$ on finds
$$\psi^{(1)} (x + 1) = 4 \int_0^1 \frac{u^{2x + 1} \ln u}{u^2 - 1} \, du.$$
Setting $x = -1/2$ then yields
$$\int_0^1 \frac{\ln x}{x^2 - 1} \, dx = \frac{1}{4} \psi^{(1)} \left (\frac{1}{2} \right ).$$
Here the dummy variable $u$ has been reverted back to $x$.
To find the value for the trigamma function at $x = 1/2$ we note that for the polygamma function one has (see Eq. (16) here)
$$\psi^{(n)} \left (\frac{1}{2} \right ) = (-1)^{n + 1} n! (2^{n + 1} - 1) \zeta (n + 1).$$
Here $\zeta (z)$ denotes the Riemann zeta function. Setting $n = 1$ yields
$$\psi^{(1)} \left (\frac{1}{2} \right ) = 3 \cdot \zeta (2) = 3 \cdot \frac{\pi^2}{6} = \frac{\pi^2}{2}.$$
Thus
$$\int_1^\infty \frac{\ln x}{x^2 - 1} \, dx = \frac{1}{4} \cdot \frac{\pi^2}{2} = \frac{\pi^2}{8}.$$
Method 2 - Using a double integral
The problem the first method suffers from is its heavy reliance on a knowledge of the polygamma function. In this second approach, knowing any properties for the polygamma and zeta functions are completely avoided altogether.
Note that as
$$\int_0^\infty \frac{\ln x}{x^2 - 1} \, dx = \int_0^1 \frac{\ln x}{x^2 - 1} \, dx + \int_1^\infty \frac{\ln x}{x^2 - 1} \, dx,$$
we have
$$I = \frac{1}{2} \int_0^\infty \frac{\ln x}{x^2 - 1} \, dx = \frac{1}{4} \int_0^\infty \frac{\ln (x^2)}{x^2 - 1} \, dx. \tag1$$
Observing that
$$\ln (x^2) = \int_0^\infty \left (\frac{x^2}{1 + x^2 t} - \frac{1}{1 + t} \right ) \, dt,$$
we can rewrite (1) as
\begin{align}
I &= \frac{1}{4} \int_0^\infty \int_0^\infty \left (\frac{x^2}{1 + x^2 t} - \frac{1}{1 + t} \right ) \frac{1}{x^2 - 1} \, dt \, dx\\
&= \frac{1}{4} \int_0^\infty \frac{1}{1 + t} \int_0^\infty \frac{1}{1 + x^2 t} \, dx \, dt,
\end{align}
after the order of integration has been changed. Evaluating we have
\begin{align}
I &= \frac{1}{4} \int_0^\infty \frac{1}{\sqrt{t} (1 + t)} \big{[} \tan^{-1} (x \sqrt{t}) \big{]}_0^\infty \, dt\\
&= \frac{\pi}{8} \int_0^\infty \frac{dt}{\sqrt{t} (1 + t)}\\
&= \frac{\pi}{4} \int_0^\infty \frac{dy}{1 + y^2} \qquad \text{(let $t = y^2$)}\\
&= \frac{\pi}{4} \big{[} \tan^{-1} y \big{]}_0^\infty\\
&= \frac{\pi}{4} \cdot \frac{\pi}{2}\\
&= \frac{\pi^2}{8},
\end{align}
as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\lim_{n \to \infty} (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$ and $\lim_{n\to\infty} \frac{1}{\ln(n+1)}=0$
Use the definition of the limit of a sequence to prove that $\lim_{n \to \infty} (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$.
We have
$$\begin{align}
\left|\frac{n^2-1}{2n^2+3}-\frac{1}{2}\right| & =\left|\frac{2n^2-2-2n^2-3}{2(2n^2+3)}\right| \\
&= \left|\frac{-5}{2(2n^2+3)}\right|\\
&= \frac{5}{2(2n^2+3)} \\
&<\frac{5}{4n^2},
\end{align}$$
$$\frac{5}{4n^2}<\epsilon \iff \frac{1}{n^2}<\frac{4 \epsilon}{5} \iff n>\sqrt{\frac{5}{4\epsilon}}$$
We choose $n_0=\left[\sqrt{\frac{5}{4\epsilon}} \right]+1$, Then $\lim_{n \to \infty} \left(\frac{n^2-1}{2n^2+3}\right)=\frac{1}{2}$.
Let $(x_n)=\frac{1}{\ln(n+1)}$ for $n \in \mathbb{N}$.
a) Use the definition of the limit to show that $\lim(x_n)=0$.
$|\frac{1}{\ln(n+1)}-0|=\frac{1}{\ln(n+1)}<\epsilon \Leftrightarrow ln(n+1) > \epsilon \Leftrightarrow n> e^{\epsilon} -1$
We choose $n_0=\left[ e^\epsilon -1 \right]+1$, Then $\lim(x_n)=0$.
b) Find specific value of $n_0 (\epsilon)$ as required in definition of limit for $\epsilon=\frac{1}{2}$.
$n_0=\left[\sqrt{e}-1\right]+1$
Is that true, please?
| For a) We have $\lim_{n\rightarrow \infty} \frac{n^2-1}{2n^2+3}=\lim_{n\rightarrow \infty} \frac{n^2(1-\frac{1}{n^2})}{n^2(2+\frac{3}{n^2})}= \lim_{n\rightarrow \infty} \frac{1-\frac{1}{n^2}}{2+\frac{3}{n^2}}=\frac{1}{2}$
For b) basically the same,meaning $ln$ is monotone so for $n\rightarrow \infty$ it follows that $\ln(n)\rightarrow \infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
How to solve $\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ I am stuck on factoring out everything properly. I feel like I am combining these fractions wrong or something because I always have an extra 1.
edit: edit: I am still stuck. Math isn't working out, I am making a mess with the constant edits, I will stop editing and not touch this so people can review the question. Sorry
a) Prove that P(1) is true
$~$
$~$
$$\frac{1}{1*2} = \frac{1}{1+1} = \frac{1}{2}$$
Show that P(k+1) is true as well
$$\frac{1}{(k+1)(k+1+1)} = \frac{k+1}{k+1+1} - \frac{k}{k+1}$$
$~~$
$$ = \frac{k+1}{k+1+1} \frac{k+1}{k+1} - \frac{k}{k+1} \frac{k+1+1}{k+1+1}$$
$~~$
$$ = \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)}$$
$~~$
$$ = \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)}$$
$~~$
$$ = \frac{(k+1)\bigg((k+1) - k(+1)\bigg)}{(k+1)(k+1+1)}$$
$~~$
$$ = \frac{k-k+1}{k+1+1} = \frac{1}{k+1+1} \neq \frac{1}{(k+1)(k+1+1)}$$
| Hint:
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\ldots+\frac{1}{n(n+1)}=\\\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n}-\frac{1}{n+1}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Power series representation of $f(x) = 3x^2 - (x^2 + 1)\ln(1 - x^2) - 2x \ln \left( \frac{1+x}{1-x} \right)$
We consider the power series:
$$ f(x) = \sum_{n=1}^{+ \infty} \frac{1}{n(n+1)(2n+1)}.x^{2n+2} $$
Prove that:
$$ f(x) = 3x^2 - (x^2 + 1)\ln(1 - x^2) - 2x \ln \bigg( \frac{1+x}{1-x} \biggr) $$
Starting from the second expression I get :
$$ f(x) = 3x^2 - (x - 1)^2 \ln(1 - x) - (x + 1)^2 \ln(1 + x) $$
Using power series of $\ln(1 + x)$ and $ \ln(1 - x)$ does not lead to the result wanted.
I do not know how to proceed to get the first expression.
| Hint: $\frac{1}{n(n+1)(2n+1}=\frac{a}{n}+\frac{b}{n+1}+\frac{c}{2n+1}$. Find $a$, $b$ and $c$. Rewrite your series as sum of such series, justify convergence of those series and compute to get the needed function.
Recall: $\sum_{n=1}^{\infty}{\frac{t^n}{n}}=\log(1-t), \forall t, |t|<1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Binomial Theorem expansion and proving an interesting identity? In the identity
$$\frac {n!}{x(x+1)(x+2)...(x+n)} = \sum ^n_{k=0}\frac {A_k}{x+k} $$
Prove that $$A_k =(-1)^{k}\:\cdot\: ^{n}C_k$$
Also from this deduce that,
$$ \;^{n}C_0\frac 1{1.2} - \:^{n}C_1\frac1{2.3} +\; ^{n}C_2\frac1{3.4} \; ... \;{(-1)^n}\; ^{n}C_n\frac1{(n+1)(n+2)}\;=\frac1{(n+2)}$$
So I have to tried to use the binomial theorem on,
$(b-a)^n$, and then multiplied both sides by $a^{x-1}$.
Now I integrated both sides with respect to $a$. This gives me the binomially expanded side as same as the right hand side of the identity that we have to prove when I substitute $a=1$. I dont know how to integrate $a^{x-1}\;(b-a)^n$ with respect to $a$. I am not able to prove the identity and solve the deduction. Can someone please help me out ? Thanks a lot.
| There are many ways to demonstrate such an interesting identity.
a) Induction
I do not know at what level you are, so let's start with what should be the simpler: Induction
Given the thesis
$$
F(x,n) = {{n!} \over {x\left( {x + 1} \right) \cdots \left( {x + n} \right)}} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
\left( { - 1} \right)^{\,k} \binom{n}{k}{1 \over {x + k}}}
$$
*
*for $n=0$ it is true for whichever value of $x$ different from $0$
$$
n = 0\quad \Rightarrow \quad F(x,0) = {1 \over x} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,0}
\binom{0}{k}{1 \over {x + k}}} = {1 \over x}\;:\;TRUE
$$
*for $n+1$ the LHS is
$$
\eqalign{
& F(x,n + 1) = {{\left( {n + 1} \right)!} \over {x\left( {x + 1} \right) \cdots \left( {x + n} \right)\left( {x + n + 1} \right)}} = \cr
& = {{\left( {x + n + 1 - x} \right)n!} \over {x\left( {x + 1} \right) \cdots \left( {x + n} \right)\left( {x + n + 1} \right)}} = \cr
& = {{n!} \over {x\left( {x + 1} \right) \cdots \left( {x + n} \right)}} - {{n!} \over {\left( {x + 1} \right) \cdots \left( {x + n} \right)\left( {x + n + 1} \right)}} = \cr
& = F(x,n) - F(x + 1,n) \cr}
$$
the same as the RHS
$$
\eqalign{
& F(x,n + 1) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n+1}{k}{1 \over {x + k}}} = \cr
& = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}{1 \over {x + k}}}
+ \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k-1}{1 \over {x + k}}} = \cr
& = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}{1 \over {x + k}}}
- \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k - 1} \binom{n}{k-1}{1 \over {x + 1 + k - 1}}} = \cr
& = F(x,n) - F(x + 1,n) \cr}
$$
and the thesis is demonstrated.
b) Finite Difference
Writing the Forward Difference of the function $f(x)$ wrt to the variable $x$ as
$$
\Delta _{\,x} \,f(x) = f(x + 1) - f(x)
$$
its iteration gives
$$
\Delta _{\,x} ^{\,n} \,f(x) = \Delta _{\,x} \,\left( {\Delta _{\,x} ^{\,n - 1} f(x)} \right) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
\left( { - 1} \right)^{\,n - k} \binom{n}{k}f(x + k)}
$$
So
$$
F(x,n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}{1 \over {x + k}}}
= \left( { - 1} \right)^{\,n} \Delta _{\,x} ^{\,n} \,\left( {{1 \over x}} \right)
$$
We can easily see that
$$
\eqalign{
& \Delta _{\,x} ^{\,0} \,\left( {{1 \over x}} \right)\mathop \equiv \limits^{def} {1 \over x} \cr
& \Delta _{\,x} ^{\,1} \,\left( {{1 \over x}} \right) = {1 \over {x + 1}} - {1 \over x} = {{\left( { - 1} \right)} \over {x\left( {x + 1} \right)}} \cr
& \Delta _{\,x} ^{\,2} \,\left( {{1 \over x}} \right) = - {1 \over {\left( {x + 1} \right)\left( {x + 2} \right)}} + {1 \over {x\left( {x + 1} \right)}} = {{\left( { - 1} \right)\left( { - 2} \right)} \over {x\left( {x + 1} \right)\left( {x + 2} \right)}} \cr
& \quad \quad \vdots \cr
& \Delta _{\,x} ^{\,n} \,\left( {{1 \over x}} \right) = {{\left( { - 1} \right)^{\,n} n!} \over {x\left( {x + 1} \right) \cdots \left( {x + n} \right)}} \cr}
$$
and that demontrates the thesis.
c) Falling / Rising Factorials
For a more general approach, it's advisable to resort to
the properties of the Rising and Falling Factorials, in order that we can write
$$
{{n!} \over {x\left( {x + 1} \right) \cdots \left( {x + n} \right)}} = {{n!} \over {x^{\,\overline {\,n + 1\,} } }}
= n!\;\left( {x - 1} \right)^{\,\underline {\, - \,\left( {n + 1} \right)} }
$$
where $x^{\,\underline {\,k\,} } ,\quad x^{\,\overline {\,k\,} } $ represent respectively the Falling and Rising Factorial.
One of the basic properties of the falling factorial is that its Delta resembles the rule of differentiation of "normal" powers
$$
\Delta _{\,x} \;x^{\,\underline {\,n\,} } = \left( {x + 1} \right)^{\,\underline {\,n\,} } - x^{\,\underline {\,n\,} } = nx^{\,\underline {\,n - 1\,} }
$$
Therefore one automatically derives that
$$ \bbox[lightyellow] {
\eqalign{
& F(x,n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}{1 \over {x + k}}}
= \left( { - 1} \right)^{\,n} \Delta _{\,x} ^{\,n} \,\left( {{1 \over x}} \right) = \cr
& = \left( { - 1} \right)^{\,n} \Delta _{\,x} ^{\,n} \,\left( {\left( {x - 1} \right)^{\,\underline {\, - \,1} } } \right) = \left( { - 1} \right)^{\,n} \Delta _{\,x - 1} ^{\,n} \,\left( {\left( {x - 1} \right)^{\,\underline {\, - \,1} } } \right) = \cr
& = \left( { - 1} \right)^{\,n} \left( { - 1} \right)\left( { - 2} \right) \cdots \left( { - n} \right)\left( {x - 1} \right)^{\,\underline {\, - 1 - \,n\;} } = \cr
& = \left( { - 1} \right)^{\,n} \left( { - 1} \right)^{\,\underline {\,\,n\;} } \left( {x - 1} \right)^{\,\underline {\, - 1 - \,n\,} } = 1^{\,\overline {\,n\,} } \left( {x - 1} \right)^{\,\underline {\, - 1 - \,n\,} } = {{n!} \over {x^{\,\overline {\,n + 1\,} } }} \cr}
}$$
Also, the second part of your question is easily solved as
$$
\eqalign{
& G(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
\left( { - 1} \right)^{\,k} \binom{n}{k}{1 \over {\left( {k + 1} \right)\left( {k + 2} \right)}}} = \cr
& = \left. {\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
\left( { - 1} \right)^{\,k} \binom{n}{k} {1 \over {\left( {x + k} \right)\left( {x + 1 + k} \right)}}\;} } \right|_{\,x\, = \,1} = \cr
& = \left. {G(x,n)} \right|_{\,x\, = \,1} \cr}
$$
therefore
$$ \bbox[lightyellow] {
\eqalign{
& G(x,n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}
{1 \over {\left( {x + k} \right)\left( {x + 1 + k} \right)}}\;} = \cr
& = \left( { - 1} \right)^{\,n} \Delta _{\,x} ^{\,n} \left( {{1 \over {\left( x \right)\left( {x + 1} \right)}}} \right)
= \left( { - 1} \right)^{\,n} \Delta _{\,x} ^{\,n} \left( {{1 \over {x^{\,\overline {\,2\,} } }}} \right) = \cr
& = \left( { - 1} \right)^{\,n} \Delta _{\,x} ^{\,n} \left( {x - 1} \right)^{\,\underline {\, - 2\,} }
= \left( { - 1} \right)^{\,n} \left( { - 2} \right)^{\,\underline {\,n\,} } \left( {x - 1} \right)^{\,\underline {\, - 2 - n\,} } = \cr
& = {{2^{\,\overline {\,n\,} } } \over {x^{\,\overline {\,n + 2\,} } }}
= {{\left( {n + 1} \right)!} \over {x^{\,\overline {\,n + 2\,} } }}\quad \Rightarrow \cr
& \Rightarrow \quad G(1,n) = {{\left( {n + 1} \right)!} \over {1^{\,\overline {\,n + 2\,} } }}
= {{\left( {n + 1} \right)!} \over {\left( {n + 2} \right)!}} = {1 \over {\left( {n + 2} \right)}} \cr}
}$$
Finally, it is surely interesting to point out that the Inversion property of the Binomial Convolution
implies
$$
f(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}g(k)}
\quad \Leftrightarrow \quad
g(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}f(k)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3100400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How do I simplify $\sqrt {4(2- \sqrt{3})}$ into $\sqrt{6} - \sqrt{2}$ This might be a stupid question, but how do I get from $$\sqrt {4(2- \sqrt{3})}$$ to $$\sqrt{6} - \sqrt{2}$$ It is obvious if you squared both, they both equal $8 - 4 \sqrt{3}$, but I'm wondering how you can find the answer from the original expression.
| Note that
\begin{align}
4(2-\sqrt{3}) = 8 - 4\sqrt{3} = (\sqrt{6})^2 - 2\sqrt{6}\sqrt{2} + (\sqrt{2})^2
=
(\sqrt{6}-\sqrt{2})^2.
\end{align}
Therefore,
$$
\sqrt{6} -\sqrt{2} = \sqrt{4(2-\sqrt{3})}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\sin (100^\circ-x) \sin 20^\circ =\sin (80^\circ-x)\sin 80^\circ$
Solve for $x$ such that
$$\sin (100^\circ-x) \sin 20^\circ =\sin (80^\circ-x)\sin 80^\circ$$
First, I use the co-function formula:
$$\sin 80^\circ = \cos 10^\circ \tag{1}$$
Also,
$$\sin 20^\circ = 2\sin 10^\circ \cos 10^\circ \tag{2}$$
From these, I got
$$\sin(100^\circ-x)\cdot 2\sin 10^\circ =\sin (80^\circ-x) \tag{3}$$
I thought to use
$$2\sin a \sin b =\cos(a-b)-\cos(a+b) \tag{4}$$
but I'm stuck.
Help me please.
| $$\dfrac{\sin(100^\circ-x)}{\sin(80^\circ-x)}=\dfrac{\sin80^\circ}{\sin20^\circ}$$
Use Componendo et Dividendo,
$$\dfrac{\sin(100^\circ-x)-\sin(80^\circ-x)}{\sin(100^\circ-x)+\sin(80^\circ-x)}=\dfrac{\sin80^\circ-\sin20^\circ}{\sin80^\circ+\sin20^\circ}$$
Use Prosthaphaeresis Formulas
$$\dfrac{\tan10^\circ}{\cot x}=\dfrac{\tan30^\circ}{\tan50^\circ}$$
$$\iff\tan x=\tan30^\circ\tan40^\circ\tan80^\circ$$
Now using Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$,
$$\tan20^\circ\tan(60^\circ-20^\circ)\tan(60^\circ+20^\circ)=\tan(3\cdot20^\circ)$$
$$\implies\tan x=\cot20^\circ$$
The rest should be easy.
More generally we have reached at $$\tan y\tan(60^\circ-y)\tan(60^\circ+y)\tan(90^\circ-3y)=1$$
See also :
$\cot(x+110^\circ)=\cot(x+60^\circ)\cot x\cot(x-60^\circ)$
Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$
How can I find the following product? $ \tan 20^\circ \cdot \tan 40^\circ \cdot \tan 80^\circ.$
Proving:$\tan(20^{\circ})\cdot \tan(30^{\circ}) \cdot \tan(40^{\circ})=\tan(10^{\circ})$
| {
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"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Flux through region : paraboloid and sphere I have this region :
$D=\{(x,y,z)\mid y^2+z^2\le 3|x|,(x-2)^2+y^2+z^2 \le 4\}$
with vector field $\mathbf F=(-2x,-2y,xy)$
I can use the divergence theorem : $\mathrm{div}(\mathbf F)=-4$
Attempt :
Let's see where they intersect :
$(x-2)^2+3x=4 \implies x^2-x=0 \implies x=0,1$
In order to find the radius of integration I can plug $x=1$ into $y^2+z^2\le 3x$. so radius is $r=\sqrt{3}$
Now It's time to get the $x$ range :
$(x-2)^2+y^2+z^2 = 4 \implies x^2-4x+y^2+z^2=0 \implies x_{1,2}=\frac{4\pm\sqrt{16-4(y^2+z^2)}}{2}$
I take the positive one.
$x\in\left[\frac{y^2+z^2}{3},2+2\sqrt{4-y^2-z^2}\right]$
Integration cylindrical coordinates :
$x\in\left[\frac{r^2}{3},2+2\sqrt{4-r^2}\right]$
$\theta\in[0,2\pi]$
$r\in[0,\sqrt{3}]$
The final integral is :
$$\text{Flux} = -4\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}\int_{\frac{r^2}{3}}^{2+2\sqrt{4-r^2}}r\,\mathrm dx\,\mathrm dr\,\mathrm d\theta=...=-\frac{166}{3}\pi$$
Is my Integral set-up right ?
UPDATE :
I think I can also split the region into two parts : the first one is the paraboloid with $x \in [0,1]$ and the second one in the sphere with $x \in [1,4]$
$\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}\int_{\frac{r^2}{3}}^{1}rdxdrd\theta=\frac{3}{2}\pi$
$\int_{1}^{4}(\sqrt{4x-4x^2})^2\pi dx=9\pi$
so Flux = $-4(\frac{3}{2}\pi+9\pi)=-42\pi$
?
| I’ll skip the divergence theorem and concentrate my answer on a computation of $\int_ D (-4) dV$.
It is not clear for me why your solutions of the equalities assure that the proposed cylindrical coordinates of the integration correspond to $D$ (and below we’ll see that the proposed bounds are wrong). But I can provide you a more simple integral setup by the following parametrization of $D$. Introduce cylindrical coordinates $y=r\cos\theta$, $z=r\sin\theta$, $r\ge 0$, $0\le\theta<2\pi$. Then $$D=\{(r,x,\theta): r^2\le\min \{3|x|, 4x-x^2\}\}.$$ If $(r,x,\theta)\in D$ then since $4x-x^2\ge 0$ we have $0\le x\le 4$. Solving inquality $3x\le 4x-x^2$, we see that $3x\le 4x-x^2$ provided $0\le x\le 1$ and $3x\ge 4x-x^2$ provided $1\le x\le 4$. Thus (I am not perfectly sure it the calculations in the following paragraph. To make them rigorous I need to recall to myself the respective analysis theorems. But I expect that they are correct).
$$\int_ D (-4) dV=$$ $$-4\int_0^{2\pi}\left(\int_0^1\int_0^{3x} r dr dx + \int_1^4\int_0^{4x-x^2} r dr dx\right)
d\theta=$$ $$-8\pi\left(\int_0^1\int_0^{3x} r dr dx + \int_1^4\int_0^{4x-x^2} r dr dx\right)=$$
$$-8\pi\left(\int_0^1 \frac 12 r^2\Big|_0^{3x} dx + \int_1^4 \frac 12 r^2\Big|_0^{4x-x^2} dx\right)=$$
$$-8\pi\left(\int_0^1 \frac {9x^2}2 dx + \int_1^4 \frac {(4x-x^2)^2}2 dx\right)=$$ $$\dots=$$ $$-\frac {672}5\pi.$$
The final value I calculated by MathCad.
Now concerning the proposed bounds. The upper bound $r\le \sqrt{3}$ is wrong, because, for instance, when $x=2$ then the bound $r^2\le 3|x|$ becomes $r^2\le 6$ and the bound $r^2\le 4x-x^2$ becomes $r^2\le 4$, so a value $r=2$ is possible. Also the bound $r^2\le 4x-x^2$ implies $|x-2|\le \sqrt{4-r^2}$ that is $2-\sqrt{4-r^2}\le x\le 2+\sqrt{4-r^2}$, but not only a bound $x\le 2+\sqrt{4-r^2}$.
| {
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How to obtain a column of a matrix representing a homogeneous linear system by the value of the adjacent column? A homogeneous linear system of equations has a coefficient matrix A which
is row equivalent to the following matrix R in reduced echelon form:
\begin{bmatrix}
1 & 2 & 0 & 3 & 0 &5\\
0 & 0 & 1 &4&0&2 \\
0 & 0 & 0&0&1 &-3
\end{bmatrix}
(a) Describe the solution set in parametric vector form.
(b) Suppose the first column of A is:
\begin{bmatrix}
2\\
3 \\
5
\end{bmatrix}
What is the second column of A?
(c) If, in addition to the above, the third column of A is:
\begin{bmatrix}
3\\
-1 \\
4
\end{bmatrix}
what is the fourth column of A? Hint: Use special solutions of Ax = 0
obtained by setting one free variable to 1, and the others to 0. Each
solution of Ax = 0 gives a linear combination of the columns of A that
is 0. The special solutions make it easy to solve for the free columns
in terms of the pivot columns.
I already did part (a) but I am unsure on how to get the columns of A. Could I just obtain them through performing row operations on R, since they are row equivalent? And how can I use special solutions to solve for the fourth column?
| In part (a), we find that every solution to $A\vec{x}=\vec{O}$ is of the form
\begin{align*}
\vec{x}
&= \left[\begin{array}{r}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4} \\
x_{5} \\
x_{6}
\end{array}\right]
= \left[\begin{array}{r}
-2 \, x_{2} - 3 \, x_{4} - 5 \, x_{6} \\
x_{2} \\
-4 \, x_{4} - 2 \, x_{6} \\
x_{4} \\
3 \, x_{6} \\
x_{6}
\end{array}\right] \\
&= x_2\left[\begin{array}{r}
-2 \\
1 \\
0 \\
0 \\
0 \\
0
\end{array}\right]+x_4\left[\begin{array}{r}
-3 \\
0 \\
-4 \\
1 \\
0 \\
0
\end{array}\right]+x_6\left[\begin{array}{r}
-5 \\
0 \\
-2 \\
0 \\
3 \\
1
\end{array}\right]
\end{align*}
This means that the three vectors
\begin{align*}
\vec{v}_1 &= \left\langle-2,\,1,\,0,\,0,\,0,\,0\right\rangle &
\vec{v}_2 &= \left\langle-3,\,0,\,-4,\,1,\,0,\,0\right\rangle &
\vec{v}_3 &= \left\langle-3,\,0,\,-4,\,1,\,0,\,0\right\rangle
\end{align*}
form a basis of $\operatorname{Null}(A)$.
In particular, the vector $\vec{v}_1=\left\langle-2,\,1,\,0,\,0,\,0,\,0\right\rangle$ satisfies $A\vec{v}_1=\vec{O}$. This means that
$$
(-2)\cdot\vec{a}_1+1\cdot\vec{a}_2+0\cdot\vec{a}_3+0\cdot\vec{a}_4+0\cdot\vec{a}_5+0\cdot\vec{a}_6=\vec{O}
$$
where $\{\vec{a}_1, \vec{a}_2, \vec{a}_3, \vec{a}_4, \vec{a}_5, \vec{a}_6 \}$ are the columns of $A$. This gives $\vec{a}_2=2\cdot\vec{a}_1$.
So, if the first column of $A$ is $\vec{a}_1=\left\langle2,\,3,\,5\right\rangle$, then the second column of $A$ is $\vec{a}_2=2\cdot\vec{a}_1=\left\langle4,\,6,\,10\right\rangle$.
Also note that the vector $\vec{v}_2=\left\langle-3,\,0,\,-4,\,1,\,0,\,0\right\rangle$ satisfies $A\vec{v}_2=\vec{O}$. This means that
$$
3\cdot\vec{a}_1+0\cdot\vec{a}_2+(-4)\cdot\vec{a}_3+1\cdot\vec{a}_4+0\cdot\vec{a}_5+0\cdot\vec{a}_6=\vec{O}
$$
Solving for $\vec{a}_4$ gives
$$
\vec{a}_4 = -3\cdot\vec{a}_1+4\cdot\vec{a}_3
$$
So, if $\vec{a}_1=\left\langle2,\,3,\,5\right\rangle$ and $\vec{a}_3=\left\langle3,\,-1,\,4\right\rangle$, then
$$
\vec{a}_4
= -3\cdot\vec{a}_1+4\cdot\vec{a}_3
= -3\cdot\left\langle2,\,3,\,5\right\rangle+4\cdot\left\langle3,\,-1,\,4\right\rangle
= \left\langle6,\,-13,\,1\right\rangle
$$
| {
"language": "en",
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"answer_count": 1,
"answer_id": 0
} |
Fermat's last theorem short proof attempt Fermat's last theorem states:
(1) $x^n + y^n = z^n$
has no solutions for x, y, z and n positive coprime integers and n > 2. An open question is whether there exists a simple proof hinted at by Fermat. If you can spot an error in my thoughts below, please point it out.
This can be rewritten as:
(2) $x^n+(x+a)^n=(x+b)^n$ with $y=x+a$ and $z=x+b,$ a and b integers, then,
(3) $x^n+(x^n+{n\choose n-1}x^{n-1}a+...+{n\choose r}x^ra^{n-r}+...+a^n)$=
$(x^n+{n\choose n-1}x^{n-1}b+...+{n\choose r}x^rb^{n-r}+...+b^n)$, or rearranging and cancelling $x^n$ on both sides
(4) $x^n-{n\choose n-1}x^{n-1}(b-a)+...-{n\choose r}x^r(b^{n-r}-a^{n-r})+...-b^n+a^n=0$
So if we think about what this would mean for the geometrically tractable case of n=3 for a cube, so substituting n=3 in (4) gives
(5) $x^3-3x^2(b-a)-3x(b^2-a^2))-b^3+a^3=0$
The cube of side x = the sum of the different cuboids between the 2 largest cubes. These differences are composed of descending terms in $x^r$. By taking these terms all on to the LHS, we are saying that the RHS = $0$.
There is another way to create a cube of volume $0$, starting from a cube of side x. If we start with a cube of side x and imagine removing a length of d until x=d, then that will create a volume of $0$. We can write this as:
(6) $(x-d)^3 = 0$, expanding gives
(7) $x^3-3dx^2+3d^2x-d^3=0$
we can then compare coefficients in ${n\choose r}x^r$ between (5) and (7) as these are both expressions describing the same geometric operation of taking $x^3$ and subtracting volumes until RHS = $0$. So, for a given value of d, we have 2 variables a and b constrained by the 3 independent equations:
(8) $d=b-a$
(9) $d^2=a^2-b^2$
(10) $d^3=b^3-a^3$
Therefore this set of equations is over-constrained and has no solutions for a and b and therefore for x, y, z for n=3. With 2 variables it is possible to define 2 of the 3 sides, but then the 3rd is not definable in the integers. Therefore (1) is impossible for n=3.
We can extend this approach for the general case of n. Equations (8), (9) and (10) will always be produced for n>2 as will
$(-1)^rd^r=a^r-b^r$ (general equation)
$(-1)^nd^n=a^n-b^n$ (final equation)
Generally there will be n equations formed when comparing coefficients but always only 2 variables. 2 of the n-dimensional sides are definable, but then the remaining dimensions have no solutions in integers.
The simultaneous equations formed in comparing coefficients between (5) and (7) will always be over-constrained for n>2. Therefore (1) is impossible.
| The problem is that if two polynomials of degree 3 share a common root, their coefficients do not necessarily have to be equal (as you state in (8)-(10) comparing (5) and (7)). Look at the polynomial $(x-1)(x-2)(x-3) = x^3-6x^2+11x-6$. Its coefficients are neither the same as those of $(x-1)^3$, $(x-2)^3$ nor $(x-3)^3$.
| {
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"source": "stackexchange",
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Find a relation between $I_n$ and $I_{n+1}$, where $I_{n}=\int_a^b(x-a)^n \sqrt{b-x}\;dx$ $n \in \Bbb{N}^*$ and $0<a<b$ and
$$I_{n}=\int_a^b(x-a)^n \sqrt{b-x}\;dx$$
I'm tasked with finding a relation between $I_n$ and $I_{n+1}$ to prove later that
$$I_n=\frac{2^{2n+2}(n+1)!n!}{(2n+3)!}(b-a)^{n+1}\sqrt{b-a}$$
Any ideas?
| Putting
$$
\left\{ \matrix{
y = {{x - a} \over {b - a}}\quad x = a + \left( {b - a} \right)y \hfill \cr
dy = {1 \over {b - a}}dx \hfill \cr} \right.
$$
we get the expression of the integral in terms of the Beta function, so
in terms of the Gamma, or Rising Factorial, etc.
$$
\eqalign{
& \int_{x = a}^{\;b} {\left( {x - a} \right)^{\,n} \sqrt {\left( {b - x} \right)} \,dx} = \cr
& = \left( {b - a} \right)^{\,n + 3/2} \int_{y = 0}^{\;1} {y^{\,n} \left( {1 - y} \right)^{\,1/2} \,dy} = \cr
& = \left( {b - a} \right)^{\,n + 3/2} {\rm B}(n + 1,3/2) = \cr
& = \left( {b - a} \right)^{\,n + 3/2} {{\Gamma \left( {n + 1} \right)\Gamma \left( {3/2} \right)} \over {\Gamma \left( {n + 5/2} \right)}} = \cr
& = {{\sqrt \pi } \over 2}\left( {b - a} \right)^{\,n + 3/2} {1 \over {\left( {n + 1} \right)^{\,\overline {\,3/2\,} } }} = \cdots \cr}
$$
Concerning the recurrence on $n$, you can put
$$
\eqalign{
& I_{\,n + 1} = \int_{x = a}^{\;b} {\left( {x - a} \right)^{\,n + 1} \sqrt {\left( {b - x} \right)} \,dx} = \cr
& = - {2 \over 3}\int_{x = a}^{\;b} {\left( {x - a} \right)^{\,n + 1} d\left( {\left( {b - x} \right)^{\,3/2} } \right)} = \cr
& = - {2 \over 3}\left( {\left. {\left( {x - a} \right)^{\,n + 1} \left( {b - x} \right)^{\,3/2} \,} \right|_{x = a}^b
- \left( {n + 1} \right)\int_{x = a}^{\;b} {\left( {x - a} \right)^{\,n} \left( {b - x} \right)^{\,3/2} dx} } \right) = \cr
& = {2 \over 3}\left( {\left( {n + 1} \right)\int_{x = a}^{\;b} {\left( {b - x} \right)\left( {x - a} \right)^{\,n} \left( {b - x} \right)^{\,1/2} dx} } \right) = \cr
& = {{2\left( {n + 1} \right)} \over 3}\left( {\int_{x = a}^{\;b} {\left( {b - a + a - x} \right)\left( {x - a} \right)^{\,n} \left( {b - x} \right)^{\,1/2} dx} } \right) = \cr
& = {{2\left( {n + 1} \right)} \over 3}\left( {\left( {b - a} \right)\int_{x = a}^{\;b} {\left( {x - a} \right)^{\,n} \left( {b - x} \right)^{\,1/2} dx}
- \int_{x = a}^{\;b} {\left( {x - a} \right)^{\,n + 1} \left( {b - x} \right)^{\,1/2} dx} } \right) = \cr
& = {{2\left( {n + 1} \right)} \over 3}\left( {\left( {b - a} \right)I_{\,n} - I_{\,n + 1} } \right) \cr}
$$
i.e.
$$
I_{\,n + 1} = {{2\left( {n + 1} \right)} \over {3 + 2\left( {n + 1} \right)}}\left( {b - a} \right)I_{\,n}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107381",
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"source": "stackexchange",
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What are the solutions of $x=\cot x$? Need to find intervals in which the function $y=\frac{x}{2}\cdot \cos x$ is increasing and decreasing. I tried to solve it on the way below but don't know how to continue.
$ \\
y=\frac{x}{2}\cdot \cos x,\ x\in (0,2\pi)\\
y=\frac{1}{2}\cdot x\cdot \cos x \\
{y}'=({\frac{1}{2}\cdot x\cdot \cos x})' \\
{y}'=\frac{1}{2}\cdot ({x\cdot \cos x})' \\
{y}'=\frac{1}{2}\cdot ({x}'\cdot \cos x+x\cdot (\cos x)') \\
{y}'=\frac{1}{2}\cdot(\cos x-x\cdot \sin x) \\
{y}'=0 \\
\frac{1}{2}\cdot(\cos x-x\cdot \sin x)=0 \\ $
$\hspace{1cm}$
$\\
\cos x-x\cdot \sin x=0\ /(\cos x) \\
1-x\cdot \tan x=0 \\
-x\cdot \tan x=-1 \\
x\cdot \tan x=1 \\
x=\frac{1}{\tan x} \\
x=\cot x $
| Hint:
The roots are close to the vertical asymptotes of the cotangent, occurring at $x=k\pi$.
To get a first approximation, we can linearize the cotangent close to a root and
$$\cot x-x\approx\frac1{x-k\pi}-x=0.$$
The positive solution is
$$x=\frac{k\pi+\sqrt{k^2\pi^2+4}}{2}.$$
From this, you start Newton's iterations, which will quickly converge.
Though a bit cluttered, the plot below shows you how the hyperbolas match the true curve close to the roots:
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $f:I\to \mathbb{R}$ is convex and interval $I$ is bounded, prove that $f$ is bounded below.
Let $I$ be a bounded interval and $f:I\to \mathbb{R}$ be a convex function. Prove that $f$ is bounded below in $I.$
Attempt. Let $a,~b\in I$, by convexity of $f$ on $[a,b]:$
$$f(x)\leq g(x):=f(a)+\frac{f(b)-f(a)}{b-a}(x-a)$$ for all $x\in [a,b]$. So it is enough to prove that:
*
*$f(x)\geq g(x)$ for $x\in I,~x<a$ or $x>b$,
*$f$ attains a minimum value $m$ on $[a,b]$,
Thanks for the help.
| It is easy to show from your first inequality that the upper bound for $f$ on $[a,b]$ is $M = \max(f(a),f(b))$.
To find a lower bound, write $x = \frac{a+b}{2} + \theta$. Since $\frac{a+b}{2} = \frac{1}{2} \left(\frac{a+b}{2} + \theta \right) + \frac{1}{2} \left(\frac{a+b}{2} - \theta \right) $, we have by convexity
$$f\left(\frac{a+b}{2}\right) \leqslant \frac{1}{2}f \left(\frac{a+b}{2} + \theta \right) + \frac{1}{2}f \left(\frac{a+b}{2} - \theta \right),$$
and for all $x \in [a,b]$,
$$f(x) = f\left(\frac{a+b}{2} + \theta \right) \geqslant 2f\left(\frac{a+b}{2}\right)- f \left(\frac{a+b}{2} - \theta \right)$$
But from the upper bound we have $-f \left(\frac{a+b}{2} - \theta \right) \geqslant -M$, and it follows that
$$f(x) \geqslant 2f\left(\frac{a+b}{2}\right)-M = m,$$
providing a lower bound $m$ for $f$ on $[a,b].$
| {
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"source": "stackexchange",
"question_score": "3",
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Finding derivative of $\sqrt[3]{\sin(2x)}$ using only definition of derivative First post here, so hello everyone.
Here's the problem:
Find the first derivative of: $$\sqrt[3]{\sin(2x)}$$
But, you can only use the difference quotient...
(i.e. the limit of $\frac{f(x+h)-f(x)}{h}$ as $h \rightarrow 0$)
This is a particular problem given by my prof. It's not for credit or even an assignment, she gave it as an interestingly difficult problem with some twist and turns. I've worked on it for several hours now and would like to see how others approach solving it as I'm wrapped around an axel after using double-angle ID and clearing the cubes in the numerator. Any advice is appreciated.
Thanks.
| Are you allowed to use the trigonometric limit identities of
$$\lim \limits_{u \to 0} \frac{\sin u}{u} = 1 \hspace{.5 in} \text{and} \hspace{.5 in} \lim \limits_{u \to 0} \frac{1-\cos u}{u} = 0?$$
If not, the link above gives good hints. You'll need to prove these first. If you are allowed to use them, then let's proceed. The first step is to use some algebra to get rid of those pesky third roots:
\begin{align*}
&\lim \limits_{h \to 0} \frac{\sqrt[3]{\sin(2x+2h)} - \sqrt[3]{\sin(2x)}}{h}\\
=& \lim \limits_{h \to 0} \frac{\sqrt[3]{\sin(2x+2h)} - \sqrt[3]{\sin(2x)}}{h} \cdot \frac{\sin(2x+2h)^{2/3} + \sin(2x+2h)^{1/3}\sin(2x)^{1/3} + \sin(2x)^{2/3}}{\sin(2x+2h)^{2/3} + \sin(2x+2h)^{1/3}\sin(2x)^{1/3} + \sin(2x)^{2/3}}\\
=&\lim \limits_{h \to 0} \frac{\sin(2x+2h) - \sin(2x)}{h(\sin(2x+2h)^{2/3} + \sin(2x+2h)^{1/3}\sin(2x)^{1/3} + \sin(2x)^{2/3})}
\end{align*}
That denominator looks a little messy, so maybe we could call the part in parentheses
$$W(x,h) = \sin(2x+2h)^{2/3} + \sin(2x+2h)^{1/3}\sin(2x)^{1/3} + \sin(2x)^{2/3}.$$
Notice that $W(x,0) = 3\sin(2x)^{2/3}$. That will come in handy later on. Now let's rewrite our limit using our $W$ shorthand and use the angle sum identity to move forward:
\begin{align*}
\lim \limits_{h \to 0} \frac{\sin(2x+2h) - \sin(2x)}{h \cdot W(x,h)} &= \lim \limits_{h \to 0} \frac{\sin(2x)\cos(2h) - \cos(2h)\sin(2x) - \sin(2x)}{h \cdot W(x,h)}\\
&= \lim \limits_{h \to 0} \frac{\cos(2h)-1}{h} \cdot \frac{\sin(2x)}{W(x,h)} + \lim\limits_{h \to 0} \frac{-\sin(2h)}{h} \cdot \frac{\cos(2x)}{W(x,h)}\\
&= \lim \limits_{h \to 0} \frac{1-\cos(2h)}{2h} \cdot \frac{-2\sin(2x)}{W(x,h)} + \lim\limits_{h \to 0} \frac{\sin(2h)}{2h} \cdot \frac{-2\cos(2x)}{W(x,h)}
\end{align*}
Then, by the identities mentioned above, we have
\begin{align*}
\lim \limits_{h \to 0} \frac{\sqrt[3]{\sin(2x+2h)} - \sqrt[3]{\sin(2x)}}{h} &= 0 \cdot \frac{-2\sin(2x)}{W(x,0)} + 1 \cdot \frac{-2\cos(2x)}{W(x,0)}\\
&= \frac{-2\cos(2x)}{3\sin(2x)^{2/3}}
\end{align*}
which is the answer you're looking for.
| {
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Ramanujan's radical and how we define an infinite nested radical I know it is true that we have
$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3$$
The argument is to break the nested radical into something like
$$3 = \sqrt{9}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}=...=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$
However, I am not convinced. I can do something like
$$4 = \sqrt{16}=\sqrt{1+2\sqrt{56.25}}=\sqrt{1+2\sqrt{1+3\sqrt{\frac{48841}{144}}}}=...=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$
Something must be wrong and the reason behind should be a misunderstanding of how we define infinite nested radical in the form of
$$ \sqrt{a_{0}+a_{1}\sqrt{a_{2}+a_{3}\sqrt{a_{4}+a_{5}\sqrt{a_{6}+\cdots}}}} $$
I researched for a while but all I could find was computation tricks but not a strict definition. Really need help here. Thanks.
| As others have said, the rigorous definition of an infinite expression comes from the limit of a sequence of finite terms. The terms need to be well-defined, but in practice, we just try to make sure the pattern is clear from context.
Now let's see what goes wrong with your other example. You wrote:
$$4 = \sqrt{16}=\sqrt{1+2\sqrt{56.25}}=\sqrt{1+2\sqrt{1+3\sqrt{\frac{48841}{144}}}}=...=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$
The problem is that each term in the sequence (e.g. if we stop at $4$) fails to include an "extra" amount (and this amount is not going to zero). So if we look at the partial sums, we see they won't converge to $4$ unless we include the extra amounts we keep pushing to the right. It's the same logical mistake as taking
\begin{align*}
2 &= 1 + 1 \\
&= \frac{1}{2} + \frac{1}{2} + 1 \\
&= \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + 1 \\
&= \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + 1
\end{align*}
And then saying, wait, the partial sums in the last line only converge to $1$ instead of $2$. But this is a mistake because we can't push the extra $1$ "infinitely far" right. Otherwise, the partial sum terms we write down will just look like $\frac{1}{2} + \frac{1}{4} + \cdots$ and will never include the $1$. The same thing is happening (roughly) in your example.
| {
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Suppose $a, b, c \in I$ such that greatest common divisor of $x^2 + ax + b$ and $x^2 + bx + c$ is $(x + 1)$ and the least common multiple...
Suppose $a, b, c \in I$ such that greatest common divisor of $x^2 + ax + b$ and $x^2 + bx + c$ is $(x + 1)$ and the least common multiple of $x^2 + ax + b$ and $x^2 + bx + c$ is $(x^3 - 4x^2 + x + 6)$. Find the value of $|a + b + c|$.
My attempt :
$$x^2 + ax + b = (x + 1)Y$$
$$x^2 + bx + c = (x + 1)Z$$
From here onwards I do not how to continue. Please help. Thank you!
| We know that $x^2+ax+b$ and $x^2+bx+c$ have the same factor $x+1$.
But $$x^3-4x^2+x+6=x^3+x^2-5x^2-5x+6x+6=$$
$$=(x+1)(x^2-5x+6)=(x+1)(x-2)(x-3),$$ which gives that our polynomials they are:
$$(x+1)(x-2)=x^2-x-2$$ and $$(x+1)(x-3)=x^2-2x-3.$$
Id est, $b=-2$, $a=-1$, $c=-3$ and
$$|a+b+c|=6.$$
| {
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Maximum value of expression $a+b+c$
If $a,b,c$ are non negative integers such that $$2(a^3+b^3+c^3)=3(a+b+c)^2.$$
Then maximum value of $a+b+c$ is ?
My Try: Using Jensen Inequality
Let $f(x)=x^3$. Then $f''(x)>0$ for $x>0$ is convex function
So $$\frac{f(a)+f(b)+f(c)}{3}\geq f\bigg(\frac{a+b+c}{3}\bigg)$$
$$\frac{a^3+b^3+c^3}{3}\geq \bigg(\frac{a+b+c}{3}\bigg)^3\cdots (1)$$
From given condition $$\frac{a^3+b^3+c^3}{3}=\frac{(a+b+c)^2}{2}\cdots (2)$$
So we have $$\frac{(a+b+c)^2}{2}\geq \frac{(a+b+c)^3}{27}$$
$$a+b+c\leq \frac{27}{2}=13.5$$
equality hold when $a=b=c=4.5$
but $a,b,c$ are non negative integers
Could some help me to solve it, Thanks
| So you have the maximum possible sum and need to restrict to integers.
What is the maximum possible sum for positive integers (hint has to be less than or equal to the sum for arbitrary reals)? Call this the target sum
Check that the target sum is even (the given condition implies that).
Then is there a solution to the equality with the target sum? If not try the next one down. You have a finite search space and cubes are quite sparse.
| {
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Computing $\int_{-a}^a \int_{-b}^b \frac 1{(x^2 + y^2 +c^2)^{3/2}}\, dxdy$. The question is exactly as in the title:
$$\int_{-a}^a \int_{-b}^b \frac 1{(x^2 + y^2 +c^2)^{3/2}}\, dxdy$$
It's been so long since the last time I tried to calculate something like this. I first thought about polar coordinates but that doesn't go well with the domain of integration.
What kind of substitution do we need for this kind of problem? I am sorry if similar problem has been asked, I just couldn't manage to find it.
| The inner integral is simple even without any substitution because ot the exponent $\frac 32$ in the denominator. Just write
$$\int \frac {dx}{(x^2 + y^2 +c^2)^{3/2}}=\frac {P(x)}{(x^2 + y^2 +c^2)^{1/2}}$$ Differentiate both sides, simplify and identify to get
$$\left(c^2+x^2+y^2\right) P'(x)-x P(x)=1$$ which is separable and the solution is
$$P(x)=\frac{x}{(c^2+y^2)}+C_1 \sqrt{c^2+x^2+y^2}$$ So
$$\int \frac {dx}{(x^2 + y^2 +c^2)^{3/2}}=\frac{1}{(c^2+y^2)}\frac {x}{(x^2 + y^2 +c^2)^{1/2}}$$
$$\int_{-b}^b \frac {dx}{(x^2 + y^2 +c^2)^{3/2}}=\frac{2 b}{\left(c^2+y^2\right) \sqrt{b^2+(c^2+y^2)}}$$ Now, as J.G. commented, a trigonometric change of variable would make the problem quite simple for the outer integral.
| {
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Factor $x^5-x+15$ It's possible to factor $x^5-x+15$. WolframAlpha gives the answer of: $$(x^2+x+3)(x^3-x^2-2x+5)$$
According to the wikipedia article on quintic functions, the general form $x^5-x+a$ is factorable only when $a=±15$, $±22440$, or $±2759640$.
Question: How would one factor such an expression? For me, it seems close to impossible by hand.
Bonus question: Why do those specific values of $a$ make the expression factorable? If there's no simple answer, is there a paper/further reading about it?
| I see there is an answer to 15. I tried, it appears there is no integer root to $x^5 - x \pm 22440.$ As 7 gives 16800 but 8 gives 32760. We arrive at
$$ (x^3 + A x^2 + B x + C)(x^2 + D x + E) = x^5 - x \pm 22440 $$
The point is not to solve the whole system at once, rather do one coefficient at a time and rewrite the system. The degree four term must be 0, so $A+D = 0$
$$ (x^3 + A x^2 + B x + C)(x^2 -A x + E) = x^5 - x \pm 22440 $$
Next the cubed term is zero, so $E-A^2 + B = 0, $ or $E = A^2 - B.$
$$ (x^3 + A x^2 + B x + C)(x^2 -A x + (A^2 - B)) = x^5 - x \pm 22440 $$
Next $x^2$ has 0, or $A^3 - AB -AB + C = 0,$ or $C = 2AB - A^3,$
$$ (x^3 + A x^2 + B x + (2AB -A^3))(x^2 -A x + (A^2 - B)) = x^5 - x \pm 22440 $$
Linear coefficient is $-1,$ so $A^2 B - B^2 - 2 A^2 B + A^4 = -1,$ or $A^4 - A^2 B - B^2 = -1.$
Getting there; taking $x = A^2, y = B,$ we have $x^2 - xy - y^2 = -1,$ meaning that $x,y$ are consecutive Fibonacci numbers... as $(3,2), (8,5), (21,13), (55,34), (144, 89).$ Since $x$ needs to be a square, we will try $A^2 = 144,$ $A = 12,$ $B = \pm 89$
One selection is
$$ (x^3 + 12 x^2 + 89 x + 408)(x^2 - 12 x + 55) = x^5 - x + 22440$$
Just negating $x$ gives
$$ (-x^3 + 12 x^2 - 89 x + 408)(x^2 + 12 x + 55) = -x^5 + x + 22440,$$
$$ (x^3- 12 x^2 + 89 x - 408)(x^2 + 12 x + 55) = x^5 - x - 22440,$$
So there you go. We were able to use the first Fibonacci number that is also a square. Bigger than $1$ I guess.
| {
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$x+x^2+x^3=x^4+x^5+x^6$ implies $x^4=x$ in a ring Let $(A,+,.)$ be a ring s.t. $x+x^2+x^3=x^4+x^5+x^6$ for all $x \in A$. Prove that $x^4=x$ for all $x$ in $A$. Can somebody give me some tips, please?
| Hint 1: Plugging in $x=-1$ shows that $2=0$ in $A$.
Hint 2: The given equation can be rearranged to get
$$(x+x^2+x^3)(1-x^3)=0,$$
or equivalently $(x^4-x)(1+x+x^2)=0$.
Multiplying through by $x-x^2$ shows that
$$(x^4-x)^2=0.$$
| {
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What's maximum value of $x (1-x^2)$ for $0 < x <1$? Since we are being taught about AM-GM inequality, I decided to use the method however I am getting two different answers by slightly different methods.
Method 1
\begin{equation}
v=x (1-x^2)$
\implies v^2=x^2 (1-x^2)^2
\end{equation}
Using the AM-GM-inequality we obtain
\begin{equation}
x^2+(1-x^2)+(1+x)+(1-x)
>4 (v^2)^{\frac{1}{4}}
\implies \frac{9}{16} \ge v.
\end{equation}
Therefore the max value is $\frac{9}{16}$.
Method 2
$$
v=x (1-x^2)
\implies 2v^2= 2x^2 (1-x^2)^2
$$
With the AM-GM-inequality
$$2x^2+(1-x^2)+(1-x^2)
>3 (2v^2)^{\frac{1}{3}}
\implies \frac{2}{(27)^{\frac{1}{2}}} > v.
$$
| Let $x=\frac{a}{\sqrt3}.$
Thus, by AM-GM $$x(1-x^2)=\frac{1}{3\sqrt3}(3a-a^3)=\frac{1}{3\sqrt3}(2-(a^3+2-3a))\leq$$
$$\leq\frac{1}{3\sqrt3}(2-(3\sqrt[3]{a^3\cdot1^2}-3a))=\frac{2}{3\sqrt3}.$$
The equality occurs for $a=1$ or $x=\frac{1}{\sqrt3},$ which says that we got a maximal value.
We can use AM-GM also by the following way.
$$x(1-x^2)=\sqrt{\frac{1}{2}\cdot2x^2(1-x^2)^2}\leq$$
$$\leq\sqrt{\frac{1}{2}\left(\frac{2x^2+1-x^2+1-x^2}{3}\right)^3}=\frac{2}{3\sqrt3}.$$
| {
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How to prove that $\sum\limits_{k=0}^{\infty}\frac{(4k-1)!!}{(2k+1)!\cdot2^{4k+1}}=\frac{\sqrt{3}-1}{\sqrt{2}}$ How to prove that $$\sum\limits_{k=0}^{\infty}\frac{(4k-1)!!}{(2k+1)!\cdot2^{4k+1}}=\frac{\sqrt{3}-1}{\sqrt{2}}$$
I need any hint to start to prove it.
Thanks for any help.
| Solution 1. Using the generating function for the central binomial coefficients
$$\frac{1}{\sqrt{1-4x}} = \sum_{n=0}^{\infty}\binom{2n}{n}x^n,$$
we have
$$ \sum_{n=0}^{\infty} \frac{(4n-1)!!}{(2n+1)!2^{4n+1}}
= 4 \int_{0}^{\frac{1}{8}}\sum_{n=0}^{\infty} \binom{4n}{2n} x^{2n} \, \mathrm{d}x
= 2\int_{0}^{\frac{1}{8}} \left( \frac{1}{\sqrt{1-4x}} + \frac{1}{\sqrt{1+4x}} \right) \, \mathrm{d}x, $$
which computes to the desired answer.
Solution 2. Write
\begin{align*}
\sum_{n=0}^{\infty} \frac{(4n-1)!!}{(2n+1)!2^{4n+1}}
&= \sum_{n=0}^{\infty} \frac{\Gamma(2n+\frac{1}{2})}{(2n+1)!2^{2n+1}\sqrt{\pi}} \\
&= \sum_{n=0}^{\infty} \frac{1}{(2n+1)!2^{2n+1}\sqrt{\pi}} \int_{0}^{\infty} x^{2n-\frac{1}{2}} e^{-x} \, \mathrm{d}x \\
&= \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\sinh(x/2)}{x^{3/2}} e^{-x} \, \mathrm{d}x \\
&= \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} \frac{e^{-s^2} - e^{-3s^2}}{s^2} \, \mathrm{d}s \tag{$x=2s^2$}
\end{align*}
Taking integration by parts,
\begin{align*}
\require{cancel}
&\frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} \frac{e^{-s^2} - e^{-3s^2}}{s^2} \, \mathrm{d}s \\
&\qquad = \frac{1}{\sqrt{2\pi}} \cancelto{0}{\left[ - \frac{e^{-s^2} - e^{-3s^2}}{s} \right]_{0}^{\infty}} + \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} \left( 3e^{-3s^2} - e^{-s^2} \right) \, \mathrm{d}s,
\end{align*}
which computes to the desired answer.
| {
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Inequality proof (strange) Given $a^2 +b^2 +c^2 +d^2 =1$ where $a,b,c,d$ are positive real numbers, prove that $a+b+c+d-1 \geq 16abcd$
How can I prove the inequality ?
My attempts:
By Cauchy-Schwarz : $ (a+b+c+d)^{2} \leq (a^2 +b^2 +c^2 +d^2 )\cdot (1^2 +1^2 +1^2 + 1^2 ) $
Or $ 0\leq a+b+c+d -1 \leq 1 $ .
By AM-GM : $ a^2 + b^2 +c^2 +d^2 \geq 4 \sqrt[4]{(abcd)^2 } $
or $ 16abcd \leq 1 $
What can I do from here ?
| Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$ and $u^2=tv^2$.
Thus, since $u^2\geq v^2$ it's
$$\sum_{sym}(a-b)^2\geq0,$$ we get $t\geq1$ and since by AM-GM
$$ab+ac+bc+ad+bd+cd\geq6\sqrt{abcd},$$ it's enough to prove that
$$(a+b+c+d)\sqrt{(a^2+b^2+c^2+d^2)^3}-(a^2+b^2+c^2+d^2)^2\geq16v^4$$ or
$$2u\sqrt{(4u^2-3v^2)^3}-(4u^2-3v^2)^2\geq v^4$$ or
$$t(4t-3)^3\geq(8t^2-12t+5)^2$$ or
$$(t-1)(48t^2-68t+25)\geq0$$ and we are done!
| {
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Evaluate the surface integral of scalar function $\int_{S} (x + y + z )dS$ where $S$ is the boundary of the unit ball $B$ Evaluate the surface integral of scalar function $\int_{S} (x + y + z )dS$ where $S$ is the boundary of the unit ball $B$
my attempt
$S = \int_{S} (x + y + z) dS$
Unit ball $s: x^2 + y^2 + z^2 = 1 \to z^2 = 1 - x^2 - y^2$
$2z \frac{dz}{dx} = -2x$
$\frac{dz}{dx} = -x/z$
$2z\frac{dz}{dy} = -2y$
$\frac{dz}{dy} = -y/z$
since we know that
$\sqrt{1 + \left(\frac{dz}{dx} \right)^2 + \left( \frac{dz}{dy}\right)^2}dx dy = \sqrt{\frac{x^2 + y^2 + z^2}{z^2}}dxdy$
$ds = \frac{1}{2}dx dy$ (since $x^2 + y^2 + z^2 = 1$)
now set $z = 0$, then the limits are
$x$ goes to $-1$ and $1$
and $y$ goes to $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$m then
$$S = \int \int (x + y + z) \frac{1}{2}dxdy = \int \int \left( \frac{x}{\sqrt{1-x^2-y^2}} + \frac{y}{\sqrt{1-x^2-y^2}} + 1\right)dxdy $$
now set $z = 0$, then the limits are
$x$ goes to $-1$ and $1$
and $y$ goes to $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$m then
$$\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left( \frac{x}{\sqrt{1-(x^2+y^2)}} + \frac{y}{\sqrt{1-(x^2+y^2)}} + 1\right)dxdy $$
since we know that $\int_{-a}^{a} f(x)dx = 0$ when $f(x)$ is a odd number
$$S = \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} [0 + 0 + 1]dxdy = \pi$$
would this be correct?
| By linearity,
$$ \int_{S} (x + y + z) dS=\int_{S} xdS+\int_{S} ydS+\int_{S} zdS$$
Let $S_{x^+}$ the boundary of the unit ball in the half-plane $x\geq 0$ and define in a similar way $S_{x^-}$. Then
$$\int_{S} xdS=\int_{S_{x^+}} xdS+\int_{S_{x^-}} xdS=\int_{S_{x^+}} xdS+\int_{S_{X^+}} (-X)dS=0.$$
where in the second integral we let $X=-x$. In a similar way we prove that $\int_{S} ydS=\int_{S} zdS=0$, and we may conclude that the given integral is zero.
More generally, if $f$ is a integrable over the boundary of a ball $S$ (centered at the origin) and $f(-x,-y,-z)=-f(x,y,z)$ for all $(x,y,z)\in S$, then $\int_Sf(x,y,z)\,dS=0$.
P.S. In your work.
$$\sqrt{1 + \left(\frac{dz}{dx} \right)^2 + \left( \frac{dz}{dy}\right)^2}dx dy = \sqrt{\frac{x^2 + y^2 + z^2}{z^2}}dxdy=\frac{dxdy}{|z|}.$$ Therefore, since $z/|z|=1$ when $z>0$ and $z/|z|=-1$ when $z<0$, we have that
$$\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left( \frac{x}{\sqrt{1-(x^2+y^2)}} + \frac{y}{\sqrt{1-(x^2+y^2)}} + 1\right)dxdy\\+\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left( \frac{x}{\sqrt{1-(x^2+y^2)}} + \frac{y}{\sqrt{1-(x^2+y^2)}} - 1\right)dxdy=\pi-\pi=0$$
| {
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Evaluating indefinite integrals of the form $\int \frac{x^2 \,dx}{a x^5 + b}$
Evaluate the indefinite integral
$$\int \frac{x^2 \,dx}{a x^5 + b},$$
for real parameters $a, b \neq 0$.
No apparent substitutions simplify the expression (if the exponent of $x$ in the denominator were an integral multiple of $3$, the form of the integrand would suggest the substitution $u = x^3$, $du = 3 x^2 \,dx$, but the exponent is not). Applying integration by parts with $u = \frac{1}{a x^5 + b}$, $dv = x^2 dx$ is straightforward, but it produces an integrand with a much larger degree in the denominator and so appears only to make the situation worse. Applying integration by parts instead with $dw = \frac{x^k dx}{a x^5 + b}$ results in integrating $\frac{x^k dx}{a x^5 + b}$, which, except when $k \equiv 4 \pmod 5$ (which doesn't appear immediately fruitful), doesn't appear much easier than the given integral.
| An explicit antiderivative is messy, but here's an outline for evaluating this integral by hand.
First, make a linear substitution $x = \alpha u$ for an appropriate constant $\alpha$, which transforms the integral $$\int \frac{x^2 \,dx}{a x^5 + b}$$ into some constant multiple of
$$\int \frac{u^2 \,du}{1 - u^5} .$$
This is a rational expression, so in principle we can apply partial fractions and solve, but $1 - u^5$ factors over $\Bbb Q$ into $u - 1$ and a quartic polynomial irreducible over $\Bbb Q$. Thus, to factor the denominator into a product of linear and quadratic polynomials we need to resort to irrational coefficients.
Factoring a generic real quartic over $\Bbb R$ is unpleasant, but we can take advantage of the special form of the denominator: The roots of $1 - u^5$ are precisely the $5$th roots of unity, namely $1$ and the paired complex conjugates $e^{\pm 2 \pi i / 5}$ and $e^{\pm 4 \pi i / 5}$. Thus, one real quadratic factor of $1 - u^5$ is
$$(u - e^{2 \pi i / 5}) (u - e^{-2 \pi i / 5}) = u^2 - 2 \cos \left(\frac{2 \pi}{5}\right) u + 1$$
and the other is
$$(u - e^{4 \pi i / 5}) (u - e^{-4 \pi i / 5}) = u^2 + 2 \cos \left(\frac{\pi}{5}\right) u + 1 .$$
Optionally, we can rewrite these expressions using the facts that $2 \cos \frac{\pi}{5} = \phi$, where $\phi := \frac{1}{2}(1 + \sqrt{5})$ is the Golden Ratio, and $2 \cos \frac{2 \pi}{5} = \frac{1}{\phi}$.
Applying the Method of Partial Fractions thus gives a decomposition
$$\frac{u^2}{1 - u^5} = \frac{A}{u - 1} + \frac{B u + C}{u^2 + \phi u + 1} + \frac{D u + E}{u^2 - \frac{1}{\phi} u + 1}$$
for some constants $A, B, C, D, E$, so
$$\int \frac{u^2\, du}{1 - u^5} = A \int \frac{du}{u - 1} + \int \frac{(B u + C) du}{u^2 + \phi u + 1} + \int \frac{(D u + E) du}{u^2 - \frac{1}{\phi} u + 1} .$$
*
*The integral $$\int \frac{du}{u - 1}$$ is elementary.
*We can rewrite the integral of the second term as a linear combination of
$$\int \frac{(2 u + \phi) du}{u^2 + \phi u + 1} \qquad \textrm{and} \qquad \int \frac{du}{u^2 + \phi u + 1} .$$
The left integral can be handled with the substitution $v = u^2 + \phi u + 1, dv = (2 u + \phi) du$, which gives $$\int \frac{(2 u + \phi) du}{u^2 + \phi u + 1} = \int \frac{dv}{v} = \log |v| + K = \log (u^2 + \phi u + 1) + K .$$ A linear substitution $w = \beta u + \gamma, dw = \beta \,du$ transforms the integral on the right into a multiple of $$\int \frac{dw}{w^2 + 1} = \arctan w + K' = \arctan (\beta u + \gamma) + K' .$$
*The third integral can be handled much like the second integral.
With all of the integrals in $u$ now expressed in terms of elementary functions, all that remains is to undo the original substitution, that is, back-substitute $u = \frac{x}{\alpha}$ to produce an antiderivative in $x$.
| {
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"url": "https://math.stackexchange.com/questions/3135455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given three a-triangle-sidelengths $a,b,c$. Prove that $3\left((a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\right)\geqq b(a+b-c)(a-c)(c-b)$ . If you are interested in IMO 1983 please see: Given three a-triangle-sidelengths $a,b,c$. Prove that:
$$3\left ( a^{2}b(a- b)+ b^{2}c(b- c)+ c^{2}a(c- a) \right )\geqq b(a+ b- c)(a- c)(c- b)$$
If $c\neq {\rm mid}\{a, b, c\}$, the inequality is obviously true!
If $c={\rm mid}\{a, b, c\}$, we have $(a- c)(c- b)= 0\Leftrightarrow c= \dfrac{c^2+ ab}{a+ b}$. I tried to prove that:
$$f(c)- f(\frac{c^2+ ab}{a+ b})= (a- c)(c- b)F\geqq 0$$
where $f(c)= 3\left ( a^{2}b(a- b)+ b^{2}c(b- c)+ c^{2}a(c- a) \right )- b(a+ b- c)(a- c)(c- b)$ but without success! I found this inequality by using discriminant and some coefficient skills. Thank you so much
| I found a nice identity to prove this!
$$3\left ( a^{2}b(a- b)+ b^{2}c(b- c)+ c^{2}a(c- a) \right )- b(a+ b- c)(a- c)(c- b)$$
$$=(a + b - c)(a + c)(a - c)^2 + (a + b - c)( c + b-a)(a - b)^2 + ( c + b-a)(2\,a - b + c)( b-c)^2 \geqq 0$$
By the way$,$ with $k=constant, k \in [0,1]$ and $a,b,c$ is three side of the triangle$:$
$$\sum\,\it{a}^{\,\it{2}}\it{b}\it{(}\,\,\it{a}- \it{b}\,\,\it{)}\geqq \it{k}\,.\,\it{b}\it{(}\,\,\it{a}+ \it{b}- \it{c}\,\,\it{)}\it{(}\,\,\it{a}- \it{c}\,\,\it{)}\it{(}\,\,\it{c}- \it{b}\,\,\it{)}$$
Proof: $$\text{LHS}-\text{RHS}=k \left\{ b \left( a+b-c \right) \left( a-c \right) ^{2}+a \left( b+c- a \right) \left( b-c \right) ^{2} \right\} + \left( 1-k \right) \left\{ {a}^{2}b \left( a-b \right) +{b}^{2}c \left( -c+b \right) +{c} ^{2}a \left( -a+c \right) \right\}$$
Where the last inequality$:$ $$ {a}^{2}b \left( a-b \right) +{b}^{2}c \left( -c+b \right) +{c} ^{2}a \left( -a+c \right) \geqq 0$$ is IMO 1983!
You can see also here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.
Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.
*
*If $n$ is odd
$7 \equiv -1 \mod 8$
$7^n \equiv (-1)^n \mod 8$
$7^n \equiv -1 \mod 8$
$7^n +1 \equiv 0 \mod 8$
Therefore, $7^n+1$ is divisible by $8$ if $n$ is odd.
*
*If $n$ is even
$7 \equiv -1 \mod 8$
$7^n \equiv (-1)^n \mod 8$
$7^n \equiv 1 \mod 8$
$7^n +1 \equiv 2 \mod 8$
Therefore, the remainder of the division of $7^n+1$ is $2$.
Is that true, please?
| Yes, you are right. Another way to present it would be like so:
For $n$ odd,
$$7^n+1 \equiv (-1)^n+1 \equiv -1+1 \equiv 0 \pmod 8$$
and for $n$ even,
$$7^n+1 \equiv (-1)^n+1 \equiv 1+1 \equiv 2 \pmod 8$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate $\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ via trig substitution $x = a\sec\theta, dx = \sec\theta \tan\theta$
$\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{(\tan^2\theta)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{\tan^{\frac{7}{2}}\theta}$ =
$\int \frac{\sec\theta \tan\theta}{\tan\theta ^{\frac{7}{2}}}$ =
$\int \tan^{\frac{-5}{2}}\theta \sec\theta$
Here is where I get stuck...I tried converting $\tan\theta$ and $\sec\theta$ in terms of $\cos\theta$ and $\sin\theta$, but that didn't seem to get me anywhere...What is my next move from here? Did I even start this problem correctly? I can't tell :(
Update with more work after initial answers:
$\int \frac{\cos\theta}{\sin^2\theta}$ $u = \sin\theta, du = \cos\theta d\theta$
I found $\sin^{-1}\theta = \frac{\sqrt{x^2-1}}{x}$
$= \int \frac{du}{u^2} = \frac{1}{ \frac{1}{3}u^3} =
\frac{1}{3\sin^3\theta}
= 3 \bigg( \frac{x}{\sqrt{x^2-1}} \bigg)^3$
| Note that $$(\tan^2 x)^{3/2}=\tan^3 x$$
Thus, yous should instead obtain
$$\int\frac1{(x^2-1)^{3/2}}dx
=\int\frac{\sec t\tan t}{\tan^3 t}dt
$$
which simplifies to
$$\int\frac{\cos t}{\sin^2 t}=\int\frac1{\sin ^2 t}d(\sin t)=-\frac1{\sin t}+C=-\csc t+C$$
Now reverse the substitution by the identity $\csc^2 t=\frac1{1-\cos^2 t}=\frac1{1-x^{-2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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S is the part of the cylinder $x^2+y^2=2x$ parametrize $S$ S is the part of the cylinder $x^2+y^2=2x$
with $0 \leq z \leq \sqrt{x^2 + y^2}$
how would I parameterize this ?
$x^2+y^2=2x$ can be made into
$(x-1)^2 + y^2 = 1$
| In cylindrical coordinates the surface and its limits are $r=2\cos\theta$ for $x^2+y^2=2x$ and $0\leq z\leq r$ for $0 \leq z \leq \sqrt{x^2 + y^2}$ This suggest to take $\theta$ and $z$ as parameters. But we have the restriction $0 \leq z \leq 2\cos\theta$ limiting the variation for $z$. We can use instead other parameter $t$ in such a way that given its maximum value makes $z$ to reach its own maximum value for that value of $\theta$. So is we can make $z=tr=t2\cos\theta$ with $0\leq t\leq 1$
$$\begin{align*}
\theta &= s \\
r &= 2\cos s\\
z &= 2t\cos s
\end{align*}
\qquad t \in [0,1], \; s \in [-\pi/2,\pi/2)$$
It can easily be expressed in cartesian coordinates:
$$\begin{align*}
x &= 2\cos^2 s \\
y &= 2\cos s\sin s\\
z &= 2t\cos s
\end{align*}
\qquad t \in [0,1], \; s \in [-\pi/2,\pi/2)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Trig Subsitution When There's No Square Root I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.
I'm currently trying to solve the following question:
$$Ar \int_a^\infty \frac{dx}{(r^2+x^2)^{(3/2)}}$$
Anyway, so far, I have that:
$$x = r\tan \theta$$
$$dx = r\sec^2 \theta$$
$$\sqrt {(r^2+x^2)} = r\sec\theta$$
The triangle I based the above values on:
Given that $(r^2+x^2)^{(3/2)}$ can be rewritten as $ (\sqrt{r^2+x^2})^3$, I begin to solve.
Please pretend I have $\lim \limits_{b \to \infty}$ in front of every line please.
\begin{align}
&= Ar \int_a^b \frac{r\sec^2\theta}{(r\sec\theta)^3}d\theta \\
&= Ar \int_a^b \frac{r\sec^2\theta}{r^3\sec^6\theta}d\theta \\
&= \frac{A}{r} \int_a^b \frac{1}{\sec^4\theta}d\theta \\
&= \frac{A}{r} \int_a^b \cos^4\theta d\theta \\
&= \frac{A}{r} \int_a^b (\cos^2\theta)^2 d\theta \\
&= \frac{A}{r} \int_a^b \left[\ \frac12 1+\cos(2\theta))\ \right]^2d\theta \\
&= \frac{A}{4r} \int_a^b 1 + 2\cos(2\theta) + \cos^2(2\theta)\ d\theta \\
&= \frac{A}{4r} \int_a^b 1 + 2\cos(2\theta)\ d\theta \quad+\quad \frac{A}{4r} \int_a^b \cos^2(2\theta)\ d\theta
\end{align}
And from there it gets really messed up and I end up with a weird semi-final answer of $$\frac{A}{4r}[2\theta+\sin(2\theta)] + \frac{A}{32r} [4\theta+\sin(4\theta)]$$ which is wrong after I make substitutions.
I already know that the final answer is $\dfrac{A}{r}\left(1-\dfrac{a}{\sqrt{r^2+a^2}}\right)$, but I really want to understand this.
| Firstly you made an error in the first line of working
$$(r\sec{(\theta)})^3=r^3\sec^3{(\theta)}$$
Secondly, you need to change the range of integration after performing a substitution. If $\theta=\arctan{(\frac{x}{r})}$ then the limits should change as $x=a \implies \theta=\arctan{(\frac{a}{r})}$ also $x=\infty \implies \theta=\frac{\pi}2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Picture proof that the area of a right triangle is $xy$ I stumbled on the following result by accident:
Let $A, B, C$ be the vertices of a right triangle, with opposite side lengths $a, b, c$ respectively, where $\angle C = 90^\circ$ and $a^2 + b^2 = c^2$.
Draw the incircle, and let $x, y, z$ be the length of the tangent from the vertices $A$, $B$, and $C$ respectively to the incircle. (So $a = y + z$, $b = x + z$, and $c = x + y$.)
Then the area of the triangle is $\boldsymbol{xy}$.
I can prove this algebraically,$^1$ but is there a picture proof of this fact?
What I have in mind is that we cut up the triangle $ABC$ into finitely many pieces, and rearrange them into a rectangle with sides $x$ and $y$.
$^1$ Using $x = \frac{b+c-a}{2}$ and $y = \frac{a+c-b}{2}$, we get $xy = \frac14\left(c - (a-b)\right)\left(c + (a-b)\right) = \frac14\left(c^2 - a^2 - b^2 + 2ab\right)$. From $c^2 = a^2 + b^2$ this reduces to $\frac14 (2ab) = \frac12 ab$, which is the area of the triangle.
| My solution uses a little bit of algebra.
If you superpose the triangles as shown in the figure, the area of the blue square is counted twice, but the green area is not counted.
The green area is (using Pythagoras) $(x-z)(y-z) = 2z^2$. One $z^2$ is the area of the triangles that we haven't count yet, and the other $z^2$ comes from the double counting.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that $A^{-1} + B^{-1}$ nonsingular by showing that $(A^{-1} + B^{-1} )^{-1} = A( A + B )^{-1} B$ Let $A$, $B$, and $A + B$ be nonsingular matrices. Prove that $A^{-1} + B^{-1}$ is nonsingular by showing that
$( A^{-1} + B^{-1} )^{-1} = A( A + B )^{-1} B$
I have done progress to only knowing that $( A^{-1} + B^{-1} )^{-1} = (A + B)$
and from there am lost completely.
Someone please give me pointers
| To show that $A(A + B)^{-1} B$ is the inverse to $A^{-1} + B^{-1}$, you need to verify the definition of inverses. Specifically, we call the matrix $C$ the inverse of $A^{-1} + B^{-1}$ if
$$C(A^{-1} + B^{-1}) = (A^{-1} + B^{-1})C = I.$$
If such a matrix $C$ exists, then it must be unique, hence why we can call it the inverse.
That is, we just need to show,
$$A(A + B)^{-1} B(A^{-1} + B^{-1}) = (A^{-1} + B^{-1})A(A + B)^{-1} B = I.$$
This is a little daunting to do as written. First note that
$$A(A + B)^{-1} B + A(A + B)^{-1} A = A(A + B)^{-1}(A + B) = AI = A,$$
hence
$$A(A + B)^{-1} B = A - A(A + B)^{-1} A.$$
Similarly,
$$A(A + B)^{-1} B = B - B(A + B)^{-1} B.$$
Hence,
\begin{align*}
(A^{-1} + B^{-1})A(A + B)^{-1} B &= A^{-1} (A(A + B)^{-1} B) + B^{-1} (A(A + B)^{-1} B) \\
&= A^{-1}(A - A(A + B)^{-1} A) + B^{-1}(B - B(A + B)^{-1} B) \\
&= I - (A + B)^{-1} A + I - (A + B)^{-1} B \\
&= 2I - (A + B)^{-1} (A + B) \\
&= I.
\end{align*}
See if you can show that
$$A(A + B)^{-1} B(A^{-1} + B^{-1}) = I$$
in a similar manner!
| {
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"timestamp": "2023-03-29T00:00:00",
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If $85_b = 58_c$, what is the smallest possible value of $b$? I have searched and watched videos online and can't find a method to solve this problem:
If $85_b=58_c$ for some positive integer bases $b$ and $c$, what is the smallest possible value for $b$?
| well think it out:
$8b + 5 = 5c + 8$.
So $8b - 5c = 3$.
$b = 1; c= 1$ is a solution but it is too small as we must have $b,c > 8$. But there are an infinite number of solutions:
If $8b - 5c = 3$ then $8(b+ 5k)-(c+8k)= (8b-5c) + 40k - 40k$ will also be a solution.
$8$ and $5$ are relatively prime so all solutions are of the form:
$8(b + 5k) - (c + 8k) = 3$. And as $b=c = 1$ is one solution all solutions are of form:
$8(1 + 5k) - (1 + 8k)$.
So we just need to find the smallest $k$ where $1 + 5k, 1+8k > 8$. That number is clearly $k = 2$ and $b = 11> 8$ and $c = 17> 8$.
And $85_{11} = 58_{17}=93$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146878",
"timestamp": "2023-03-29T00:00:00",
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$\lim\limits_{n\to\infty} \prod\limits_{k=1}^{n} \left( 1 + \tan{\frac{k}{n^2}} \right) $ I want to calculate $$\lim\limits_{n\to\infty} \prod_{k=1}^{n} \left( 1 + \tan{\frac{k}{n^2}} \right) $$
Taking logarithms, it's enough to find
$$\lim\limits_{n\to\infty} \sum_{k=1}^{n} \log\left( 1 + \tan{\frac{k}{n^2}} \right).$$
Since $\lim\limits_{n\to\infty} \tan{\frac{x}{n^2}} = 0$, we can combine the Taylor series near $0$ of $\log(1+x)$ with the taylor series of $\tan{x}$ near $0$ to obtain the limit $e^\frac{1}{2}$.
My question is: is there any nicer way of evaluating this limit?
| Probably not nicer, but still a different way is to use the facts that
$$\lim\limits_{x\rightarrow0}\frac{\tan{x}}{x}=1$$
and, as shown here
$$\lim\limits_{n\rightarrow\infty} \frac{1}{n}\sum\limits_{k=1}^n f\left(\frac{k}{n}\right)= \int\limits_{0}^{1} f(x)dx$$
$$\sum_{k=1}^{n} \log\left( 1 + \tan{\frac{k}{n^2}} \right)=
\sum_{k=1}^{n} \tan{\frac{k}{n^2}} \cdot \log\left( 1 + \tan{\frac{k}{n^2}} \right)^{\frac{1}{\tan{\frac{k}{n^2}} }}=\\
\sum_{k=1}^{n} \frac{k}{n^2}\cdot \color{red}{ \frac{\tan{\frac{k}{n^2}}}{\frac{k}{n^2}} \cdot \log\left( 1 + \tan{\frac{k}{n^2}} \right)^{\frac{1}{\tan{\frac{k}{n^2}} }}}$$
Because the part in red $\rightarrow 1$ when $n\rightarrow\infty$ for any $k=1..n$, using the definition of the limit, $\forall \varepsilon, \exists N(\varepsilon)$ s.t. $\forall n > N(\varepsilon)$
$$(1-\varepsilon)\left(\sum_{k=1}^{n} \frac{k}{n^2}\right)<\sum_{k=1}^{n} \log\left( 1 + \tan{\frac{k}{n^2}} \right)<(1+\varepsilon)\left(\sum_{k=1}^{n} \frac{k}{n^2}\right)$$
leading to
$$\lim\limits_{n\rightarrow\infty}\sum_{k=1}^{n} \log\left( 1 + \tan{\frac{k}{n^2}} \right)=
\lim\limits_{n\rightarrow\infty}\sum_{k=1}^{n} \frac{k}{n^2}$$
But then
$$\lim\limits_{n\rightarrow\infty}\sum_{k=1}^{n} \frac{k}{n^2}=
\lim\limits_{n\rightarrow\infty}\frac{1}{n}\left(\sum_{k=1}^{n} \frac{k}{n}\right)=\int\limits_{0}^{1}x dx =\frac{1}{2}$$
and the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do we compute $\sqrt[3]{x_1} +\sqrt[3]{x_2} $ using the fact that $x_1 + x_2 = 4 , x_1x_2 = -1$? Given quadratic equation
$$x^2 -4x-1 = 0$$
How do we compute $\sqrt[3]{x_1} +\sqrt[3]{x_2} $ using the fact that $x_1 + x_2 = 4 , x_1x_2 = -1$?
Regards
| Using $a^3+b^3= (a+b)(a^2-(ab)+b^2)$:
$(x_1+x_2)=$
$(x_1^{1/3}+x_2^{1/3})(x_1^{2/3} -(x_1x_2)^{1/3}+x_2^{2/3}).$
Set $X = x_1^{1/3}+x_2^{1/3}$
$4=X(x_1^{2/3}+x_2^{2/3}+1).$
Express $x_1^{2/3}+x_2^{2/3}$ in terms of $X$:
$X^2= (x_1^{1/3}+x_2^{1/3})^2=$
$x_1^{2/3}+x_2^{2/3} +2x_1^{1/3}x_2^{1/3}$;
$X^2+2= x_1^{2/3}+x_2^{2/3}.$
Finally:
$4=X(X^2+3)$.
$X^3+3X -4=0.$
$X=1$ is a solution, factoring:
$(X-1)(X^2 +X+4)=0$.
Any other solutions?
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove/disprove that $\frac{x^2 - \sqrt{yz}}{yz - x} + \frac{y^2 - \sqrt{zx}}{zx - y} + \frac{z^2 - \sqrt{xy}}{xy - z} \ge 0$
Prove/disprove that
$$ \frac{x^2 - \sqrt{yz}}{yz - x} + \frac{y^2 - \sqrt{zx}}{zx - y} + \frac{z^2 - \sqrt{xy}}{xy - z} \ge 0$$ with $x$, $y$ and $z$ are positives.
I tried to use the Schur's inequality for this but it didn't help.
If the inequality is not correct, what minor change could have been done to the inequality so that it is correct for $\forall x, y, z \in \mathbb R^+|yz - x, zx - y, xy - z \ne 0$?
|
Prove/disprove that $$\dfrac{x^2 - \sqrt{yz}}{yz - x} + \dfrac{y^2 - \sqrt{zx}}{zx - y} + \dfrac{z^2 - \sqrt{xy}}{xy - z} \ge 0$$with $x$, $y$ and $z$ are positives.
Choosing for example $(x,y,z)=(1,1,4)$ we get $$\frac{1^2-\sqrt{4}}{4-1}+\frac{1^2-\sqrt{4}}{4-1}+\frac{4^2-\sqrt{1}}{1-4}=-\frac{1}{3}-\frac{1}{3}-\frac{15}{3}=-\frac{17}{3}<0,$$
so the inequality does not hold.
If the inequality is not correct, what minor change could have been done to the inequality so that it is correct for $\forall x, y, z \in \mathbb Z^+|yz - x, zx - y, xy - z \ne 0$?
Sure, we can change it for example to
$$\dfrac{x^2 - \sqrt{yz}}{yz + x} + \dfrac{y^2 - \sqrt{zx}}{zx + y} + \dfrac{z^2 - \sqrt{xy}}{xy + z} \ge 0.$$
(Notice that condition $yz - x, zx - y, xy - z \ne 0$ can be now dropped). This holds since for example for $x\geq1$ we have $x^2\geq x$, also without loss of generality let's $x \leq y \leq z$, and so we have
$$
\sum \dfrac{x^2 - \sqrt{yz}}{yz + x} \geq \frac{1}{z^2+z} \sum (x - \sqrt{yz})
$$
and sum on the right is non-negative due to $x+y+z\geq \sqrt{xy}+\sqrt{yz}+\sqrt{zx}$, which is just rearrangement inequality applied on $\sqrt{x} \leq \sqrt{y} \leq \sqrt{z}$.
| {
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"answer_id": 0
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Find $f^{(n)}(0) \text{ for } n = 1,2,3,...$ Find $f^{(n)}(0) \text{ for } n = 1,2,3,...$ where
$$ f(x) = \begin{cases}
\frac{e^x - 1}{x}, & \text{when } x \neq 0 \\
1, & \text{when } x = 0 \\
\end{cases} $$
My approach
I decided to calculate some first $f$ derivatives. In that case I defined:
$$ g(x) = \frac{e^x - 1}{x} \text{ for } x\neq0$$
$$g'(x) = ...= \frac{e^x x-e^x+1}{x^2} $$
$$g''(x) = \text{..a lot of calculus..} =\frac{e^x x^2-2 e^x x+2 e^x-2}{x^3} $$
$$ g^{(3)}(x) = ... = \frac{e^x x^3-3 e^x x^2+6 e^x x-6 e^x+6}{x^4} $$
$$ g^{(4)}(x) = \frac{e^x x^4-4 e^x x^3+12 e^x x^2-24 e^x x+24 e^x-24}{x^5}$$
But I don't see a pattern. Some of last factors are $n!$. Denominator is $x^{n-1}$. But for the rest I haven't got idea.
I know also that for $n>0$
$$ f^{(n)}(x) = \begin{cases}
\frac{e^x - 1}{x}, & \text{when } x \neq 0 \\
0, & \text{when } x = 0 \\
\end{cases} $$
so theoretically the answer is just $ 0 $ but I am not sure if solution can be so simple...
| Since$$f(x)=1+\frac x{2!}+\frac{x^2}{3!}+\cdots+\frac{x^n}{(n+1)!}+\cdots,$$then$$(\forall n\in\mathbb{Z}^+):\frac{f^{(n)}(0)}{n!}=\frac1{(n+1)!}$$and therefore$$(\forall n\in\mathbb{Z}^+):f^{(n)}(0)=\frac1{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove for all $n\in \mathbb{N}$ that $\sum_{i=0}^{n} i\cdot F_{2i} = (n+1)F_{2n + 1} - F_{2n + 2}$. $F_n$ denotes the Fibonacci sequence where $n$ is the term of the Fibonacci number in the sequence. ($F_0=0$, $F_1=1$, $F_2=1$, $F_3=2$, $F_4=3$, ... $F_n=F_{n-1} + F_{n-2}$)
I want to prove this using strong induction and have formulated base cases for when $n=0$ and when $n=1$.
My induction hypothesis is that for all $k\in \mathbb{N}$, $k\geq 1$ we can assume that for $k-1$, $\sum_{i=0}^{k-1} i\cdot F_{2i} = (k)F_{2k - 1} - F_{2k}$ and that for $k$, $\sum_{i=0}^{k} i\cdot F_{2i} = (k + 1)F_{2k + 1} - F_{2k + 2}$.
For my induction step I want to show that for $k+1$, $\sum_{i=0}^{k+1} i\cdot F_{2i} = (k+2)F_{2k + 3} - F_{2k + 4}$. I've been stuck on the equation below for the past couple of hours trying to figure out how I can derive the equation for $k + 1$.
\begin{align*}
\sum_{i=0}^{k+1} i\cdot F_{2i} &= (k + 1)\cdot F_{2k + 2} + \sum_{i=0}^{k} i\cdot F_{2i}\\
&= (k + 1)\cdot F_{2k + 2} + (k + 1)F_{2k + 1} - F_{2k + 2} + \sum_{i=0}^{k - 1} i\cdot F_{2i}\\
&= (k + 1)\cdot F_{2k + 2} + (k + 1)F_{2k + 1} - F_{2k + 2} + (k)F_{2k - 1} - F_{2k}\\
\end{align*}
Any help is appreciated.
| \begin{align}
\sum_{i=0}^{k+1}iF_{2i}&=(k+1)F_{2k+2}+(k+1)F_{2k+1}-F_{2k+2}\\
&=kF_{2k+2}+(k+1)(F_{2k+3}-F_{2k+2})\\
&=(k+1)F_{2k+3}-F_{2k+2}\\
&=(k+1)F_{2k+3}-(F_{2k+4}-F_{2k+3})\\
&=(k+2)F_{2k+3}-F_{2k+4}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find minimum of $4(a^3 + b^3 + c^3) + 15abc$ subject to $a + b + c = 2$
$a$, $b$ and $c$ are three sides of a triangle such that $a + b + c = 2$. Calculate the minimum value of $$\large P = 4(a^3 + b^3 + c^3) + 15abc$$
Every task asking for finding the minimum value of an expression containing the product of all of the variables scares me.
Here what I've done.
Using the AM-GM inequality and the Schur's inequality, we have that
$$a^3 + b^3 + c^3 \ge 3abc \implies P \ge \dfrac{9}{2}(a^3 + b^3 + c^3 + 3abc)$$
$$\ge \dfrac{9}{2}[ab(a + b) + bc(b + c) + ca(c + a)] = \dfrac{9}{2}[ab(2 - c) + bc(2 - a) + ca(2 - b)]$$
$$\ge \dfrac{9}{2}[2(ab + bc + ca) - 3abc] \ge \dfrac{27}{2}[2\sqrt[\frac{3}{2}]{abc} - abc]$$
Let $abc = m \implies m \le \left(\dfrac{a + b + c}{3}\right)^3 = \dfrac{8}{27}$
The problem becomes
Find the minimum value of $P' = 2\sqrt[\frac{3}{2}]{m} - m$ when $ 0 < m \le \dfrac{8}{27}$.
which is invalid because there isn't a minimum with the given condition.
| This is particularly easy to prove by calculation.
With the substitution $a=u$, $b=u+v$, $c=u+v+w$, we get
$$4(a^3+b^3 + c^3)+15 a b c- (a+b+c)^3 = 3(u v^2 + u v w + u w^2 + 2 v w^2 + w^3)$$
$\bf{Added:}$ We have equality if and only if $w=0$ and $u v=0$, that is, if and only if $a=b=c$, or one of the $a$, $b$, $c$ is $0$ and the other two are equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Reaching upon $9=1$ while solving $x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$
$x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$
Substituting $y=x+45^{\circ}$, we get
$$3\tan{(y-60^{\circ})}=\tan{(y-30^{\circ})}$$
$$3\frac{\tan y - \sqrt3}{1+\sqrt3\tan y}=\frac{\tan y - 1/\sqrt3}{1+1/\sqrt3\cdot\tan y}$$
$$3(\tan ^2 y-3)=3\tan ^2-1$$
$$9=1$$
The solution provided by the book $x=n\pi + \pi/4$ fits, so why did i get $9=1$?
| Hint:
Set $\tan y=\dfrac1a$
$$\dfrac{3(1-\sqrt3a)}{a+\sqrt3}=\dfrac{\sqrt3-a}{\sqrt3a+1}$$
$$\iff3-a^2=3(1-3a^2)\iff a=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Soving PDE $u_{xx}-u_{yy} + \frac{4}{x}u_x+\frac{2}{x^2}u=0$ I have some problems with solving PDEs.
*
*\begin{cases}
\ u_{xx}-u_{yy} + \frac{4}{x}u_x+\frac{2}{x^2}u=0 \\[2ex]
u(x,x)=1,\quad u(1,y)=y
\end{cases}
What I've done: $$u(x,y)=\frac{1}{x^2}v(x,y)$$
$$u_{xx}-u_{yy} + \frac{4}{x}u_x+\frac{2}{x^2}u=0 \Rightarrow v_{xx}-v_{yy}=0$$
$$v(x,y)=C_1(x+y)+C_2(x-y)$$
$$u=\frac{1}{x^2}(C_1(x+y)+C_2(x-y))$$
From $u(x,x)=1, u(1,y)=y$ we get
$$\begin{cases}
\ C_1(2x)+C_2(0)=1 \\[2ex]
C_1(1+y)+C_2(1-y)=y
\end{cases}$$
And this is where I stopped. No idea how to solve such a system. Any ideas?
*\begin{cases}
\ u_{xy}-\frac{1}{x-y}(u_x-u_y)=1 \\[2ex]
u(x,x)=0,\quad u(2,y)=2+2y+\frac{1}{2}y^2
\end{cases}
$$u(x,y)=\frac{1}{x-y}v(x,y)$$
$$u_{xy}-\frac{1}{x-y}(u_x-u_y)=1 \Rightarrow v_{xy}=x-y$$
$$v=\frac{1}{2}x^2y-\frac{1}{2}y^2x+C_1(x)+C_2(y)$$
$$u=\frac{1}{x-y} \left(\frac{1}{2}x^2y-\frac{1}{2}y^2x+C_1(x)+C_2(y)\right)$$
$$\begin{cases}
\ C_1(x)+C_2(-x)=x^3 \\[2ex]
C_1(2)+C_2(y)=4-\frac{y^3}{4}
\end{cases}$$
And again the same problem. What's wrong? Thank you.
| The equation $C_1(2x)+C_2(0)=1$ shows that $C_1(2x)$ doesn't vary with $x$, i.e. $C_1$ is a constant function, say $C_1(x) = c.$
Inserting this in $C_1(1+y)+C_2(1-y)=y$ we get $C_2(1-y) = y-c$ i.e. $C_2(y) = (1-y)-c = (1-c)-y.$
Thus,
$$
\begin{cases}
C_1(x) = c \\
C_2(x) = (1-c)-y
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$.
Solve the equation: $$\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$$
Here's what I've done.
Let $\sqrt{x + 2} = a$ and $\sqrt{3 - x} = b$
$\implies
\left\{
\begin{align}
a^2 + b^2 &= 5\\
a^2 - b^2 &= 2x - 1
\end{align}
\right.$.
We have that $a - b = (x - 3)^2 \implies a + b = \dfrac{a^2 - b^2}{a + b} = \dfrac{2x - 1}{(x - 3)^2}$.
$\left\{
\begin{align}
a = \dfrac{(a + b) + (a - b)}{2} = \dfrac{x^4 - 12x^3 + 54x^2 - 106x + 80}{2(x - 3)^2}\\
b = \dfrac{(a + b) - (a - b)}{2} = \dfrac{x^4 - 12x^3 + 54x^2 - 110x + 82}{2(x - 3)^2}
\end{align}
\right.$
| Let $3-x=t^2$, with $t>0$. Then
$$\sqrt{5-t^2}-t=t^4$$ and
$$p(t):=(t^4+t)^2-\left(\sqrt{5-t^2}\right)^2=t^8+2t^5+2t^2-5=0.$$
Now the derivative
$$8t^7+10t^4+4t$$ cancels for $t=0$ (minimum), and $8(t^3)^2+10t^3+4$ has no real root. As $p(0)<0$, the polynomial has exactly one positive and one negative root, which we discard.
Finally, by inspection, $t=1$ is a root and $x=2$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the volume of a solid rotated around the x-axis $y=x^3, y=x, x \geq 0$
$$ \begin{align}
V &= \int_{0}^{1} A(x)dx = \int_{0}^{1} \pi (x-x^3)^2dx = \pi \int_{0}^{1}(x^2-2x^4 + x^6)dx \\
&= \pi \bigg[ \frac{1}{3}x^3 - \frac{2}{5}x^5 + \frac{1}{7}x^7\bigg]_{0}^{1}
\end{align}$$
The book answer says the answer is $\frac{4\pi}{21}$ but I am not getting anywhere close to that.
I think there are three main areas where I checked that I could have messed up, but I am not seeing anything.
1) setup
2) integration
3) getting the volume
Here is a visual of what the graph looks like and the area I am trying to integrate
| The correct setup is the difference of the squares $$\int_{0}^{1} \pi \left( x^2 - (x^3)^2 \right) dx$$
and not the square of the difference $(x-x^3)^2$.
The radius of the larger disk is $x$ and its area is $\pi x^2$. The radius of the small disk is $x^3$ thus area $\pi (x^3)^2$. The ring (annulus) is the large disk minus the small disk thus $\pi\left( x^2 - (x^3)^2 \right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3162501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral with two distinct roots in them, such as: $\int \frac{\sqrt{x}}{x^2(\sqrt{x+1}+\sqrt{x})}dx$ I'm getting familiar with basic indefinite integrals and these are the hardest ones I've met so far:
*
*$\int \frac{\sqrt{x}}{x^2(\sqrt{x+1}+\sqrt{x})}dx$
*$\int \frac{\sqrt[3]{x+2}-\sqrt[3]{x}}{x^2(\sqrt[3]{x+2}+\sqrt[3]{x})}$
Any hints? Please note that the course I am taking does not anticipate usage of hyperbolic functions. I am not familiar with them.
The first integral I attempted:
$\int \frac{\sqrt{x}}{x^2(\sqrt{x+1}+\sqrt{x})}dx = \int \frac{\sqrt{x}(\sqrt{x+1}-\sqrt{x})}{x^2(\sqrt{x+1}+\sqrt{x})(\sqrt{x+1}-\sqrt{x})}dx = \int \frac{\sqrt{x}\sqrt{x+1}-x}{x^2}dx$
Now I can split it into two integrals. Problem is with:
$\int \frac{\sqrt{x}\sqrt{x+1}}{x^2}dx$
and the major problem is that I don't know how to solve integrals that have some distinct roots with different values inside those roots. Second task's integral seems even harder.
If speaking of "different values under roots", I am only familiar with how to solve such integrals:
$\int \frac{(\sqrt{\frac{x+2}{x-1}}-1)^2}{3(\sqrt{\frac{x+2}{x-1}}+2)}dx$
because there's simple algorithm that I can follow to solve it.
Hints, tips, advices appreciated. Thanks.
EDIT: $\int \frac{\sqrt{x}\sqrt{x+1}}{x^2}dx = \int \frac{\sqrt{x^2+x}}{x^2}dx = \int \frac{x^2 + x }{x^2\sqrt{x^2+x}}dx = \int \frac{x+1}{x\sqrt{x^2+x}}dx$
And now with Euler's substitution should work?
| Hint: For the integral $$\int \frac{\sqrt{x^2+x}}{x^2}dx$$ substitute $$\sqrt{x^2+x}=x+t$$ it is the Eulerian substitution.
Then we get by squaring $$x=\frac{t^2}{1-2t}$$ and $$dx=-2\,{\frac {t \left( -1+t \right) }{ \left( -1+2\,t \right) ^{2}}}dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all functions $f:\Bbb{R} \to \Bbb{R}$ such that for all $x,y,z \in \Bbb{R} $ , $f(xf(x)+f(y))=x^2+y$
Find all functions $f:\Bbb{R} \to \Bbb{R}$ such that for all $x,y, \in \Bbb{R} $ , $f(xf(x)+f(y))=x^2+y$
We can easily get a strong condition $f(f(y))=y $ by setting $x=0$ . By this equation we know $f$ is injective and surjective. I got lost from there. By observation I know $f(x)=x $ and $f(x)=-x$ are solution. So I was trying to make $x^2+y=f(xf(x)+f(y))$ close to $f(x)^2+y$ or $x^2+f(y)$. Any hints would be helpful.
| $$ \begin{align}
&\text{As already noted, we have }\qquad\qquad\qquad\qquad\qquad f(f(y)) = y \implies f(x) = f^{-1}(x) \\
&\text {Also, using the substitution } x\to y \text{ we get}\,\,\quad f((y+1)f(y)) = y^2+y \to f(0) = 0\\
&\text{using } f= f^{-1}:\quad f(x*f(x) + f(y)) = x^2+y \implies x*f(x) + f(y) = f(x^2+y)\\
&\text{from above line we then get}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad f(x) = \frac{f(x^2+y) -f(y)}{x}
\end{align}$$
From $f(x) = \frac{f(x^2+y) -f(y)}{x}$ we now obtain $xf(x) = f(x^2)$ by setting $y=0$.
A substitution $x\mapsto f(x)$ now yields $f(x)x = f(f(x)^2)$.
Combining those two equations, we get $f(x^2) = f(f(x)^2) \implies x^2 = f(x)^2 \implies f(x) = \pm x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the remainder when the polynomial $1+x^2+x^4+x^6+....+x^{22}$ is divided by $1+x+x^2+x^3+...+x^{11}$ Find the remainder when the polynomial $$1+x^2+x^4+x^6+....+x^{22}$$ is divided by $$1+x+x^2+x^3+...+x^{11}$$
$1+x^2+x^4+x^6+....+x^{22}=\frac{x^{24}-1}{x^2-1}$
$1+x+x^2+x^3+...+x^{11}=\frac{x^{12}-1}{x-1}$
Now$$\frac{1+x^2+x^4+x^6+....+x^{22}}{1+x+x^2+x^3+...+x^{11}}=\frac{x^{12}+1}{x+1}$$
Dont know how to proceed from here
| Proceeding from where I left off
$$\frac{1+x^2+x^4+x^6+....+x^{22}}{1+x+x^2+x^3+...+x^{11}}=\frac{x^{12}+1}{x+1}$$$$=\frac{x^{12}-1+2}{x+1}=\frac{x^{12}-1}{x+1}+\frac{2}{x+1}=P(x)+\frac{2}{x+1}$$where $P(x)$ is a polynomial since $x^{12}-1$ is divisible by $x+1$.
So $$1+x^2+x^4+x^6+....+x^{22}=P(x)\left(1+x+x^2+x^3+...+x^{11}\right)+\frac{2\left(1+x+x^2+x^3+...+x^{11}\right)}{x+1}$$$$=P(x)\left(1+x+x^2+x^3+...+x^{11}\right)+2\frac{x^{12}-1}{x^2-1}=P(x)\left(1+x+x^2+x^3+...+x^{11}\right)+2Q(x)$$where $Q(x)=1+x^2+x^4+x^6+x^8+x^{10}$
So answer should be $$2\left(1+x^2+x^4+x^6+x^8+x^{10}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A nice Nesbitt inequality from a strange inequality Given $a,\,b,\,c> 0$$,$ prove that$:$
$$\frac{a}{b+ c}+ \frac{b}{c+ a}+ \frac{c}{a+ b}+ \frac{63}{5}\left [ \frac{2\,c^{\,2}}{(\,a+ b\,)^{\,2}}- \frac{c}{a+ b} \right ]\geqq 0$$
See$:$ $\lceil$ https://artofproblemsolving.com/community/c6h354642p1923888 $\rfloor$
The only way I tried is Buffalo Way but the coefficient of$:$ $\text{coef}[\,c^{\,4}\,],\,\text{coef}[\,c^{\,3}\,],\,\text{coef}[\,c^{\,2}\,],\,\text{coef}[\,c\,],\,\text{coef}[\,c^{\,0}\,]$ aren$'$t same non$-$negative$,$ maybe I$'$m wrong because the equality occurs with $a= b$$.$ So I ask$,$ hope to see the best way$!$ Good luck everybody$!$
| Let $\frac{c}{a+b}=\frac{x}{2}$.
Thus, by C-S and AM-GM $$\sum_{cyc}\frac{c}{a+b}+\frac{63}{5}\left(\frac{2c^2}{(a+b)^2}-\frac{c}{a+b}\right)=\frac{a}{b+c}+\frac{b}{a+c}+\frac{x}{2}+\frac{63}{5}\left(\frac{x^2}{2}-\frac{x}{2}\right)=$$
$$=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{x}{2}+\frac{63}{5}\left(\frac{x^2}{2}-\frac{x}{2}\right)\geq\frac{(a+b)^2}{2ab+c(a+b)}+\frac{x}{2}+\frac{63}{5}\left(\frac{x^2}{2}-\frac{x}{2}\right)=$$
$$=\frac{1}{\frac{2ab}{(a+b)^2}+\frac{c}{a+b}}+\frac{x}{2}+\frac{63}{5}\left(\frac{x^2}{2}-\frac{x}{2}\right)\geq\frac{1}{\frac{1}{2}+\frac{x}{2}}+\frac{x}{2}+\frac{63}{5}\left(\frac{x^2}{2}-\frac{x}{2}\right).$$
Id est, it's enough to prove that
$$\frac{1}{\frac{1}{2}+\frac{x}{2}}+\frac{x}{2}+\frac{63}{5}\left(\frac{x^2}{2}-\frac{x}{2}\right)\geq0$$ or $$63x^3+5x^2-58x+20\geq0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A probability question: Poor Alex
Alex remembers all but the last digit of his friend's telephone number. He decides to choose the last digit at random in an attempt to reach him. Given that, Alex has only enough money to make two phone calls, the probability that he dials the right number before running out of money can be expressed as an irreducible fraction p/q.
What I did was--
WLOG let us assume the correct number be 1. Total number of outcomes can be calculated this way
(0,1)(0,2)(0,3)(0,4)(0,5)(0,6)(0,7)(0,8)(0,9)
(1)
(2,0)(2,1)(2,3)(2,4)(2,5)(2,6)(2,7)(2,8)(2,9)
(3,0)(3,1)(3,2)(3,4)(3,5)(3,6)(3,7)(3,8)(3,9)
(4,0)(4,1)(4,2)(4,3)(4,5)(4,6)(4,7)(4,8)(4,9)
(5,0)(5,1)(5,2)(5,3)(5,4)(5,6)(5,7)(5,8)(5,9)
(6,0)(5,1)(5,3)(5,4)(5,6)(5,7)(5,7)(5,8)(5,9)
(7,0)(7,1)(7,2)(7,3)(7,4)(7,5)(7,6)(7,8)(7,9)
(8,0)(8,1)(8,2)(8,3)(8,4)(8,5)(8,6)(8,7)(8,9)
(9,0)(9,1)(9,2)(9,3)(9,4)(9,5)(9,6)(9,7)(9,8)
So total number of outcomes is 82 and total number of favorable outcomes is 10 so my answer is 10/82 or 5/41. But it is not correct. Please help me understand this.....
| So you are choosing twice from a set of ten Without replacement. That means the first digit tried, called it A is taken from 0 through 9. The second digit, digit B, should be different, because A has been eliminated.
Let's say the correct digit is X.
What are the chances that neither A or B is X?
That question can be restated: How many ways are there to pick two digits from nine (the ten digits minus X), divided by the number of ways of picking two digits.
There are 36 ways of picking a combination of two distinct digits none of which is X.
There are 45 ways of choosing two distinct digits from all 10.
The probability of failing to get the right digit is 4/5.
So the probability of success is 1/5=1-4/5.
Alternatively:
There's a 1/10 chance of getting it right on the first try.
There's a 9/10 chance of not. We only make the second try if we failed the first. The chances of succeeding the second time around is is 1/9.
So the chances of success is (1/10)(1/1)+(9/10)(1/9)=2/10=1/5, just as via the other method.
This is all based on choosing two distinct digits. If you repeat the same failed digit, the analysis changes a little.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\sum_{0 \le k } \binom{n+k}{2k} \binom{2k}{k} \frac{(-1)^k}{k+1}$
Calculate $$\sum_{0 \le k } \binom{n+k}{2k} \binom{2k}{k}
\frac{(-1)^k}{k+1}$$
My approach
$$\sum_{0 \le k } \binom{n+k}{2k} \binom{2k}{k}
\frac{(-1)^k}{k+1} = \\
\sum_{0 \le k } \binom{n+k}{k} \binom{n}{k}
\frac{(-1)^k}{k+1} = \\
\frac{1}{n+1}\sum_{0 \le k } \binom{n+k}{k} \binom{n+1}{k+1}(-1)^k = \\
\frac{1}{n+1}\sum_{0 \le k } \binom{k - 1 - n - k}{k} \binom{n+1}{k+1}
$$
But unfortunately I have stucked, I don't know how I can finish that...
The main obstacle which I see is
$$\binom{- 1 - n}{k} $$
is looks so dangerous because $- 1 - n<0$
| Starting from
$$\sum_{k=0}^n {n+k\choose 2k} {2k\choose k} \frac{(-1)^k}{k+1}$$
for a self-contained answer we observe that
$${n+k\choose 2k} {2k\choose k} =
\frac{(n+k)!}{(n-k)! \times k! \times k!}
= {n+k\choose k} {n\choose k}$$
so we find
$$\sum_{k=0}^n {n+k\choose k} {n\choose k} \frac{(-1)^k}{k+1}$$
which is
$$\frac{1}{n+1}
\sum_{k=0}^n {n+k\choose k} {n+1\choose k+1} (-1)^k
\\ = \frac{1}{n+1}
\sum_{k=0}^n {n+k\choose k} {n+1\choose n-k} (-1)^k
\\ = \frac{1}{n+1} [z^n] (1+z)^{n+1}
\sum_{k=0}^n {n+k\choose k} z^k (-1)^k.$$
The coefficient extractor controls the range (with $k\gt n$ we will
always have $[z^n] (1+z)^{n+1} z^k = 0$) and we may continue by
extending $k$ to infinity:
$$\frac{1}{n+1} [z^n] (1+z)^{n+1}
\sum_{k\ge 0} {n+k\choose k} z^k (-1)^k
\\ = \frac{1}{n+1} [z^n] (1+z)^{n+1}
\frac{1}{(1+z)^{n+1}}
= \frac{1}{n+1} [z^n] 1 = [[n=0]].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.