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How do we express the equation corresponding to the intersection of the planes $x + y + z = 1$ and $x + 2y + 2z = 0$? If a system with three unknowns and two equations are such that
$$
\begin{align}
x+y+z=1&\\
x+2y+2z=0
\end{align}
$$
In the answer it says that this system can be represented as
$$
\begin{pmatrix}
x \\ y \\z
\end{pmatrix} = \begin{pmatrix}
2 \\-1\\0
\end{pmatrix}
+t\begin{pmatrix}
0\\-1\\1
\end{pmatrix}, t\in{\\R}
$$
But I am confused on how they came up to this parametric equation. Is there a general way of coming up with this parametric form?
|
The proposed system of equations corresponds to the intersection of two planes which are not parallel, also known as a line. In order to obtain one of its possible parametrizations, subtract the first equation from the second in order to obtain:
\begin{align*}
(x + 2y + 2z) - (x + y + z) = 0 - 1 & \Longleftrightarrow y + z = -1
\end{align*}
Hence we conclude that $x = 2$. Consequently, if we let $z = t\in\mathbb{R}$, it results that $y = -1 - t$.
Finally, one gets the desired equation of the intersecting line:
\begin{align*}
\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} =
\begin{pmatrix}
2\\
-1 - t\\
t
\end{pmatrix} =
\begin{pmatrix}
2\\
-1\\
0
\end{pmatrix} + t
\begin{pmatrix}
0\\
-1\\
1
\end{pmatrix}
\end{align*}
Hopefully this helps!
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4600941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Why are there complex numbers in the exact solution of $\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}$? Knowing that
\begin{align}
\cot(z)=\frac{1}{z}-2z\cdot\sum_{n=1}^{\infty} \dfrac{1}{\pi^2n^2-z^2}
\end{align}
we can easily calculate the value of
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^2+1}
\end{align}
by just plugging in $z=i\pi$. Therefore:
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^2+1}=\frac{1}{2}\cdot\left(\pi \coth(\pi)-1\right)
\end{align}
In this case, the "$i's$" cancel eachother out nicely.
I also wanted to calculate the exact value of
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}
\end{align}
with an equal approach. After partial fraction decomposition:
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}=\frac{1}{2i}\left(\sum_{n=1}^{\infty}\dfrac{1}{n^2-i}-\dfrac{1}{n^2+i}\right)=\frac{1}{2i}\left(\sum_{n=1}^{\infty}\dfrac{1}{n^2-i}-\sum_{n=1}^{\infty}\dfrac{1}{n^2+i}\right)
\end{align}
By plugging in $z=\sqrt{i}\pi$ and $z=i\sqrt{i}\pi$, I arrive at:
\begin{align}
\frac{1}{2i}\left[\sum_{n=1}^{\infty}\dfrac{1}{n^2-i}-\sum_{n=1}^{\infty}\dfrac{1}{n^2+i}\right]&=\frac{1}{2i}\left[\frac{1}{2i}-\frac{\pi}{2\sqrt{i}} \cdot \cot(\sqrt{i}\pi)-\left(-\frac{1}{2i}-\frac{1}{2i\sqrt{i}}\cdot \cot(i\sqrt{i}\pi)\right)\right]=\\&=\frac{1}{4}\left[-2+\pi\sqrt{i}\cot(\sqrt{i}\pi)-\frac{\pi}{\sqrt{i}}\cot(i\sqrt{i}\pi)\right]
\end{align}
In this case I can't get completely rid of the "$i's$" and the exact value of
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}
\end{align}
is a complex number. But the answer must obviously be a real number.
Wolfram Alpha gets:
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}\approx0.57848+0.\times10^{-22}\,i
\end{align}
You now can argument that the imaginary part is negligible and so the value is a real number, but that doesn't satisfy my question, because there always will be a imaginary part in the answer.
If anyone can explain to me, why it is the case, that the exact value for
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}
\end{align}
is a complex number, I would be very glad.
|
Define $t=\pi\sqrt{i}$ and $f(z):=z\cot(z)$.
Now $f(-z)=f(z)$ for all $z$ so $f(z)=g(z^2).$
Your result is
$$ S:=\sum_{n=1}^\infty\frac1{n^4+1} = \frac14\big[-2+f(t)+f(it)\big]
= -1/2 + (g(i\pi^2)+g(-i\pi^2))/4.$$
Rewrite this as $ S = -1/2 + \Re g(i\pi^2)/2 \approx 0.57848 $ where
$\Re z$ is the real part of $z$.
|
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"url": "https://math.stackexchange.com/questions/4602781",
"timestamp": "2023-03-29T00:00:00",
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|
For integer $x \ge 1$, does it follow that $\left(\frac{x^2+2x+1}{x^2+2x}\right)^{x+1} > \frac{x+2}{x+1}$ Would I be correct that the answer here is yes?
Here is my thinking:
*
*For $x \ge 1$, $\dfrac{x+2}{x+1}$ is strictly decreasing.
*For $x \ge 1$, $\left(\dfrac{x^2 + 2x + 1}{x^2 + 2x}\right)^{x+1}$ is strictly increasing
*$\dfrac{16}{9} > \dfrac{3}{2}$
To show that $\left(\dfrac{x^2 + 2x + 1}{x^2 + 2x}\right)^{x+1}$ is strictly increasing I take the derivative which is positive for $x > 0$
|
As commenters have pointed out, the left side is not increasing, but decreasing.
You can see this by showing:
$$\left(\frac{x^2+2x+1}{x^2+2x}\right)^{(x+1)^2}\to e,$$ so the left side of your inequality is in the range of $e^{1/(x+1)},$ and converges to $1.$
Bernoulli's inequality says that if $y\geq -1$ and $n$ is a non-negative integer, then:
$$(1+y)^n\geq 1+ny\tag1$$
We only need it for $y> 0,$ where $(1)$ obviously follows from the binomial theorem.
In your case, $n=x+1, y=\frac{1}{x^2+2x}$ gives you:
$$\left(\frac{x^2+2x+1}{x^2+2x}\right)^{x+1}\geq 1+\frac{x+1}{x^2+2x}>1+\frac{x+1}{x^2+2x+1}=1+\frac1{x+1}=\frac{x+2}{x+1}$$
|
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"question_score": "5",
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|
Solve the equation $\log_x(9x^2)\cdot\log_3^2(x)=4$ Solve the equation $$\log_x(9x^2)\cdot\log_3^2(x)=4$$
The answers are $x_1=\dfrac19$ and $x_2=3$.
For the range we have: $$D_x:\begin{cases}x>0\\x\ne1\\9x^2>0\iff x\ne0\end{cases}\iff x\in(0;1)\cup(1;+\infty)$$
I wrote the equation as follows using $\log_a(b)=\dfrac{1}{\log_b(a)}:$ $$\log_x(9x^2)\cdot\dfrac{1}{\log_x^2(3)}=4\\\iff \log_x(9x^2)=4\log_x^2(3)=\log_x^2(3^4)$$ Is this reasonable and how can one continue?
|
You are on the right track!
Based on the assumption that $x\in(0,1)\cup(1,+\infty)$, we can rewrite the first term from the LHS of the original equation as follows:
\begin{align*}
\log_{x}(9x^{2}) = \log_{x}(9) + 2\log_{x}(x) = \log_{x}(3^{2}) + 2 = 2\log_{x}(3) + 2
\end{align*}
As you have already noticed, the following identity holds:
\begin{align*}
\log_{x}(3) = \frac{1}{\log_{3}(x)}
\end{align*}
Gathering all the previous results and letting $y = \log_{3}(x)$, one arrives at the desired solution set:
\begin{align*}
\left(\frac{2}{y} + 2\right)y^{2} = 4 & \Longleftrightarrow (2 + 2y)y^{2} = 4y \tag{$y\neq 0$}\\
& \Longleftrightarrow y(2y^{2} + 2y - 4) = 0 \tag{$y\neq 0$}\\\\
& \Longleftrightarrow 2y(y^{2} + y - 2) = 0 \tag{$y\neq 0$}\\\\
& \Longleftrightarrow y\in\{-2,1\}\\\\
& \Longleftrightarrow \log_{3}(x)\in\{-2,1\}\\\\
& \Longleftrightarrow x\in\left\{\frac{1}{9},3\right\}.
\end{align*}
Hopefully this helps!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int \frac{\sqrt{1+x^4}}{1-x^4}dx$ How to evaluate
$$\int \frac{\sqrt{1+x^4}}{1-x^4}dx.$$
If we substitute $x=\sqrt{\tan{\theta}}$, it becomes
$$\int \frac{1}{2\cos{\theta}\cos{2\theta}\sqrt{\tan{\theta}}}d\theta.$$
What can I do next ?
Edit
Are these steps correct ?
$$=\int \frac{1}{2\cos{\theta}\cos{2\theta}\sqrt{\frac{\sin\theta}{\cos\theta}}}d\theta$$
$$=\int \frac{1}{2\cos{2\theta}\sqrt{\cos^2 \theta\frac{\sin\theta}{\cos\theta}}}d\theta$$
$$=\int \frac{1}{2\cos{2\theta}\sqrt{\cos^2 \theta\frac{\sin\theta}{\cos\theta}}}d\theta$$
$$=\int \frac{1}{\sqrt{2}\cos{2\theta}\sqrt{\sin{2 \theta}}}d\theta$$
Now If we substitute $t=\sin{2\theta}$, it becomes
$$=\int \frac{1}{\sqrt{2}(1-t^2)\sqrt{t}}dt$$
$$=\int \frac{\frac{1}{t^2}}{\sqrt{2}\frac{(1-t^2)}{t}\sqrt{\frac{t}{t^2}}}dt$$
$$=\int \frac{\frac{1}{t^2}}{\sqrt{2}(\frac{1}{t} -t)\sqrt{\frac{1}{t}}}dt$$
Now If we substitute $u^2=\frac{1}{t}$, it becomes
$$=\int \frac{-2u}{\sqrt{2}(u^2 -\frac{1}{u^2})u}du$$
$$=\int \frac{-u^2}{\sqrt{2}(u^4 -1^2)}du$$
$$=\int \frac{-u^2}{\sqrt{2}(u^2 +1)(u^2-1)}du$$
$$=-\frac{1}{2\sqrt{2}}\int \frac{u^2+1+u^2-1}{(u^2 +1)(u^2-1)}du$$
This can be solved easily now and we will get an elementary solution.
But Wolfram alpha gives a solution in non elementary functions.
Therefore I am confused. I guess the above solution is correct only for some restricted values of x.
|
Alternatively, substitute $t=\frac{\sqrt2 x}{\sqrt{1+x^4}}$. Then
$$\frac{\sqrt{1+x^4}}{1-x^4}=\frac{xt}{\sqrt2(t^2-x^2)}, \>\>\>\>\>dx =\frac{x}{t(1-t^2x^2)}dt$$
and
\begin{align}
&\int \frac{\sqrt{1+x^4}}{1-x^4}dx\\
=&\ \frac1{\sqrt2}\int \frac{x^2}{(t^2-x^2)(1-t^2 x^2)}dt
= \frac1{\sqrt2}\int \frac1{t^2(\frac1{x^2}+x^2)-1-t^4}dt\\
=& \frac1{\sqrt2}\int \frac1{1-t^4}dt= \frac1{2\sqrt2}\int \frac1{1-t^2}+ \frac1{1+t^2}\ dt\\
=& \ \frac1{2\sqrt2}\left(\tanh^{-1}t +\tan^{-1}t\right)+C
\end{align}
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the limit $\lim_{n \rightarrow \infty}\sum_{k=1}^n\frac{1}{x_{1}^{2}x_{2}^2...x_{k}^2}$ We have the sequences $(x_{n})_{n\geq1}$$(y_{n})_{n\geq1}$ with positive real numbers. $x_{1}=\sqrt{2}$, $y_{1}=1$ and $y_{n}=y_{n-1}\cdot x_{n}^{2}-3$ for every $n\geq2$. We know the sequence $(y_{n})_{n\geq1}$ is bounded. Find $$\lim_{n \rightarrow \infty}\sum_{k=1}^n\frac{1}{x_{1}^{2}x_{2}^2...x_{k}^2}$$
I tried rewriting $y_{n}$ only with terms from $(x_{n})_{n\geq1}$ but we get an ugly formula. I don't know exactly how we can we use the fact that $(y_{n})_{n\geq1}$ is bounded.
|
Introduce an auxilary sequence $z_n = \frac{y_n}{y_n+3} \iff y_n = 3\frac{z_n}{1-z_n}$. For $i \ge 2$, we have
$$y_i = y_{i-1}x_i^2 -3
\implies \frac{1}{x_i^2} = \frac{y_{i-1}}{y_i+3} =
\frac{3\frac{z_{i-1}}{1-z_{i-1}}}{
3\frac{z_i}{1-z_i} + 3} = z_{i-1}\frac{1-z_i}{1-z_{i-1}}
$$
Multiply from $i = 2$ to any $k \ge 2$, we get
$$\prod_{i=2}^k \frac{1}{x_i^2}
= \frac{1-z_k}{1-z_1}\prod_{i=1}^{k-1}z_i
= \frac{1}{1-z_1}\left(\prod_{i=1}^{k-1}z_i - \prod_{i=1}^{k}z_i\right)
$$
Summing from $k = 2$ to $n$, we get
$$\sum_{k=2}^n \prod_{i=2}^k \frac{1}{x_k^2}
= \frac1{1-z_1}\left(z_1 - \prod_{i=1}^n z_i\right)
$$
This leads to
$$\sum_{k=1}^n \prod_{i=1}^k \frac{1}{x_k^2}
= \frac{1}{x_1^2}\left( 1 + \sum_{k=2}^n \prod_{i=2}^k \frac{1}{x_k^2}\right) = \frac1{x_1^2(1-z_1)}\left(1 - \prod_{i=1}^n z_i\right)$$
If $y_n$ is bounded from above by $M$, then $z_n$ is bounded from above by $\frac{M}{M+3} < 1$.
As a result, $\prod\limits_{i=1}^n z_i \le \left(\frac{M}{M+3}\right)^n$ and hence converges to $0$ as $n \to \infty$.
From this, we can deduce
$$\lim_{n\to\infty}
\sum_{k=1}^n \prod_{i=1}^k \frac{1}{x_k^2}
= \frac1{x_1^2(1-z_1)} = \frac{y_1+3}{3x_1^2} = \frac23$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do we prove $x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0$? Question
How do we prove the following for all $x \in \mathbb{R}$ :
$$x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0 $$
My Progress
We can factorise the left hand side of the desired inequality as follows:
$$x^6+x^5+4x^4-12x^3+4x^2+x+1=(x-1)^2(x^4+3x^3+9x^2+3x+1)$$
However, after this I was unable to make any further progress in deducing the desired inequality.
I appreciate your help
|
Let $P(x) = x^4 + 3x^3 + 9x^2 + 3x + 1$
$P(x≥0) \;≥\; 1 \;>\; 0\;$
$P(y=-x) \;=\; y^4 - 3y^3 + 9y^2 - 3y + 1 = (y-1)^4 + (y^3+3y^2+y)$
$P(x<0 \;⇒\;y>0) \;>\; 0 \quad\quad $ // RHS first term ≥ 0, second term > 0
$\forall x\in\mathbb {R} \;⇒\; P(x)>0$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\frac {dy}{dx}$ if $y=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}$ taking JDs advice i used $(fg)'=f'g+fg'$ rule
$$f=\frac x{\sqrt{(1-x^2)}}$$ $$f'=\frac 1{\sqrt{(1-x)}^3}$$
$$g=sin^{-1}x$$ $$g'=\frac{1}{\sqrt{1-x^2}}$$
so anyway adding together
we get
$$\frac 1{(\sqrt{(1-x)}^3}*sin^-x+\frac x{\sqrt{(1-x^2)}}*\frac{1}{\sqrt{1-x^2}}$$
which can be simplified
$$\frac {sin^{-1}x}{(1-x)^3}+\frac x{\sqrt{(1-x^2)^2}}$$
i then multiplied $$\frac x{\sqrt{(1-x^2)^2}}$$ with $\sqrt{(1-x^2)}$ on both numerator and denominator
getting
$$\frac {x(\sqrt{(1-x^2)})}{\sqrt{(1-x^2)^3}}$$
and combining them both we get
$$\frac{sin^{-1}x+x(\sqrt{(1-x^2)}}{\sqrt{(1-x^2)^3}}$$
ps
i never used arcsin before and am compeletely unfamilliar with it
|
A faster way is logarithmic differentiation
$$y=\frac{x \sin ^{-1}(x)}{\sqrt{1-x^2}}\implies \log(y)=\log(x)+\log(\sin ^{-1}(x))-\frac 12 \log(1-x^2)$$
$$\frac{y'}y=\frac 1 x+\frac{1 } {\sin ^{-1}(x)\sqrt{1-x^2}}+\frac x{1-x^2}=\frac {x \sqrt{1-x^2} +\sin ^{-1}(x)}{x \left(1-x^2\right) \sin ^{-1}(x) }$$
$$y'=\frac{y'}y \times y=\frac {x \sqrt{1-x^2} +\sin ^{-1}(x)}{x \left(1-x^2\right) \sin ^{-1}(x) } \times \frac{x \sin ^{-1}(x)}{\sqrt{1-x^2}}$$ Just simplify
|
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|
Integrating $\frac{x^3}{\sqrt{x^2 + 4x + 6}}$ Question:
Evaluate $\displaystyle\int \frac{x^3}{\sqrt{x^2 + 4x + 6}}\ dx $.
My attempt:
$\begin{align} \int \frac{x^3}{\sqrt{x^2 + 4x + 6}}\ dx & = \int \frac{x^3}{\sqrt{(x+2)^2 + 2}}\ dx \\& \overset{(1)}= \int \frac{(\sqrt{2} \tan(t) - 2 )^3 \sqrt{2} \sec^2(t)\ }{\sqrt{2\tan^2(t) + 2}}\ dt\\& = \frac{1}{\sqrt{2}}\int\frac{(\sqrt{2} \tan(t) - 2 )^3 \sqrt{2} \sec^2(t)}{\sec(t)}\ dt\\& = \int(\sqrt{2} \tan(t) - 2 )^3 \sec(t)\ dt\\& = \int -8 \sec(t) + 2 \sqrt2 \tan^3(t) \sec(t) - 12\tan^2(t) \sec(t)\\&\qquad + 12 \sqrt2 \tan(t) \sec(t)\ dt\\\\& = -8\ln|\sec(t) + \tan(t)| + 12\sqrt{2} \sec{(t)} \\
&\qquad + 2\sqrt{2} \int \tan^3(t) \sec(t) \ dt - 12\int\tan^2(t) \sec(t) \ dt\end{align}$
Now both of these integrals can be evaluated using some sort of substitution and integration by parts rule.
This would give us,
$$\boxed{-8\ln|\sec(t) + \tan(t)| + 12\sqrt{2} \sec{(t)} + 2\sqrt{2}\left[\frac{\sec^3(t)}{3} - \sec(t)\right] - 12\left[\frac{-1}{2} \ln|\sec t + \tan t| + \frac12 \sec t \tan t \right] + C}$$
$(1)$ Here I've made a substitution $t = \tan^{-1}\left(\frac{x+ 2}{\sqrt 2}\right)$.
Wolframalpha gives answer as $$\boxed{\frac{1}{3}(x^2 - 5x + 18) \sqrt{x^2 +4x + 6} - 2 \sinh^{-1}\left(\frac{x+2}{\sqrt{2}}\right) + C}$$
How would I simplify my answer equal to this? Undoing my substitution $t = \tan^{-1}\left(\frac{x+ 2}{\sqrt 2}\right)$ is also not an easy task I think.
|
By hyperbolic substitution
Letting $x+2=\sqrt 2 \sinh \theta$ transforms the integral into
$$
\begin{aligned}
I= & \int \frac{(\sqrt{2} \sinh \theta-2)^3}{\sqrt{2} \cosh \theta} \cdot \sqrt{2} \cosh \theta d \theta \\
= & \int\left(2 \sqrt{2} \sinh ^3 \theta-12 \sinh ^2 \theta+12 \sqrt{2} \sinh \theta-8\right) d \theta \\
= & 2 \sqrt{2} \int\left(\cosh ^2 \theta-1\right) d(\cosh \theta)-12 \int \frac{\cosh 2 \theta-1}{2} d \theta +12 \sqrt{2} \cosh \theta-8 \theta \\
= &\frac{2 \sqrt{2}\cosh ^3 \theta}{3}-3\sinh 2\theta +10\sqrt{2} \cosh \theta-2 \theta+C \cdots (*)\\=& \frac{2 \sqrt{2}}{3} \cosh \theta\left(15+\cosh ^2 \theta-\frac{9}{\sqrt{2}} \sinh \theta\right)-2 \theta+C
\end{aligned}
$$
Putting back $x+2=\sqrt 2 \sinh \theta$ yields
$$
\begin{aligned}
I & =\frac{2 \sqrt{2}}{3} \sqrt{1+\left(\frac{x+2}{\sqrt{2}}\right)^2}\left[15+1+\left(\frac{x+2}{\sqrt{2}}\right)^2-\frac{9}{\sqrt{2}} \cdot \frac{x+2}{\sqrt{2}}\right] +C\\
& =\frac{1}{3} \sqrt{x^2+4 x+6}\left(x^2-5 x+18\right)-2 \sinh ^{-1}\left(\frac{x+2}{\sqrt{2}}\right)+C
\end{aligned}
$$
By trigonometric substitution
Questioner using $x+2=\sqrt 2\tan t$ got the answer
$$I=-8\ln|\sec(t) + \tan(t)| + 12\sqrt{2} \sec{(t)} + 2\sqrt{2}\left[\frac{\sec^3(t)}{3} - \sec(t)\right] - 12\left[\frac{-1}{2} \ln|\sec t + \tan t| + \frac12 \sec t \tan t \right] + C$$
Using $x+2=\sqrt2\tan t =\sqrt 2\sinh \theta \Leftrightarrow \tan t=\sinh \theta \textrm{ and } \sec t=\cosh \theta$, we get
$$
\begin{aligned}
I=-2 & \ln |\sec t+\tan t|+\frac{2 \sqrt{2}}{3} \sec ^3 t+10 \sqrt{2} \sec t -6 \sec t \tan t+C \\
=- & 2 \ln |\cosh \theta+\sinh \theta|+\frac{2 \sqrt{2}}{3} \cosh ^3 \theta+10\sqrt{2} \cosh \theta -6 \cosh \theta \sinh \theta+C
\end{aligned}
$$
Since
$\ln |\cosh \theta+\sinh \theta |= \ln \left(\frac{e^\theta+e^{-\theta}}{2}+\frac{e^\theta-e^{-\theta}}{2}\right)
= \theta$, therefore $$I= \frac{2 \sqrt{2}\cosh ^3 \theta}{3}-3\sinh 2\theta +10\sqrt{2} \cosh \theta-2 \theta+C \cdots (*)$$
We can now conclude that both substitutions gives the same result.
|
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"timestamp": "2023-03-29T00:00:00",
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|
find all triplets of positive rationals satisfying this property Find all $(x, y, z) \in Q^{+} | (a, b, c) \in Z^{+} $ for $ a = x + y + z, b = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}, c = xyz$.
I have tried using Viete's formulas, setting $x, y, z$ as the roots of some polynomial $p(t) = a_3t^3 + a_2t^2 + a_1t^2 + a_0 = a_3(t-x)(t-y)(t-z)$, but made little progress... one idea I had was to, since $(x, y, z) \in Q^{+}$, using the rational roots theorem (assuming that we can write p(t) with integer coefficients):
Let $x = \frac{s_1}{t_1}, y = \frac{s_2}{t_2}, z = \frac{s_3}{t_3}$ where $(s_1, s_2, s_3, t_1, t_2, t_3) \in Z$. This gives that $s | a_0 \ \forall s \in \{s_1, s_2, s_3\}$ and $t | a_3 \ \forall t \in \{t_1, t_2, t_3\}$. Does anybody have any ideas?
|
You've made good progress. To finish, first use Vieta's formulas or just expand to get
$$\begin{equation}\begin{aligned}
p(t) & = (t-x)(t-y)(t-z) \\
& = t^3 - (x+y+z)t^2 + (yz+xz+xy)t - xyz
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note $bc = yz+xz+xy$ is an integer. Thus, all of the coefficients in \eqref{eq1A} are integers, with the coefficient of $t^3$ (i.e., $1$) and $xyz$ being non-zero, so the Rational root theorem states all of the rational roots of \eqref{eq1A} are of the form $\frac{p}{q}$ where $q$ is a factor of the coefficient of $t^3$, i.e., $1$, so $q=1$. Thus, $x$, $y$ and $z$ are all positive integers.
This automatically means that $a$ and $c$ are integers, so they don't impose any constraints. For dealing with $b$, due to symmetry, WLOG consider $x \le y \le z$.
With $x = 1$, we can have $y = 1$, which means that $z = 1$. If $y = 2$, then $b = 1 + \frac{1}{2} + \frac{1}{z}$, so if $z \ge 3$, this is not an integer, which means $z = 2$. Finally, if $y \ge 3$, then we have $\frac{1}{y}+\frac{1}{z} \le \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$, so $b$ can't be an integer.
With $x = 2$, if $y = 2$, then $\frac{1}{z}$ must be an integer, which is not possible since $z \ge 2$. If $y = 3$, then $z = 6$ is required. Next, if $y = 4$, then only $z = 4$ works. Finally, if $y \ge 5$, then $b \le \frac{1}{2} + 2\left(\frac{1}{5}\right) = \frac{9}{10}$, which is not permitted.
With $x = 3$, only $y = z = 3$ allows $b$ to be an integer (i.e., $1$ in this case).
Finally, $x \ge 4$ means that $b \le 3\left(\frac{1}{4}\right)$, which is not allowed.
Thus, the $(x,y,z)$ are, in some order of vales, the elements of $\{(1,1,1),(1,2,2),(2,3,6),(2,4,4),(3,3,3)\}$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4617474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
The critical point $(0,0)$ is assymptotically stable. Demostrate $a_{11}+a_{22}<0$ and $a_{11}a_{22}-a_{12}a_{21}>0.$ I've got the following linear system:
$$\frac{dx}{dt}=a_{11}x+a_{12}y$$
$$\frac{dy}{dt}=a_{21}x+a_{22}y$$
The critical point $(0,0)$ is an assymptotically stable critical point of the system.
We have to demostrate that $$a_{11}+a_{22}<0$$ and $$a_{11}a_{22}-a_{12}a_{21}>0.$$
I have writted the system like this:
$$\begin{pmatrix}
x' \\
y'
\end{pmatrix}=\begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{pmatrix}\begin{pmatrix}
x \\
y
\end{pmatrix}$$
If we call $A=\begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{pmatrix}$
$Traza(A)=a_{11}+a_{22}$
and
$Det(A)=a_{11}a_{22}-a_{12}a_{21}$
So now, I have to do something with the eigenvalues, true?
How can I follow? I'm a bit lost.
|
Let's take an alternative approach without solving for the eigenvalues. The phrase "asymptotically stable" implies that Lyapunov's method can be used, in which the following conditions must be satisfied:
*
*$ V\left(\mathbf{0}\right) = 0 $
*$ V\left(\mathbf{x}\right) > 0, \quad \forall \; \mathbf{x} \neq \mathbf{0} $
*$ \dot{V}\left(\mathbf{x}\right) < 0, \quad \forall \; \mathbf{x} \neq \mathbf{0} $
Consider the Lyapunov function candidate:
$$ V\left(\mathbf{x}\right) = - \frac{1}{2} a_{21} x^{2} + \frac{1}{2} a_{12} y^{2}. $$
The first condition is satisfied when $(x, y) = (0, 0)$. In order to satisfy the second condition $V\left(\mathbf{x}\right) > 0$, then we must have
$$ a_{12} > 0, $$
$$ a_{21} < 0. $$
Taking the time derivative for $V\left(\mathbf{x}\right)$, we have
$$ \dot{V}\left(\mathbf{x}\right) = - a_{21} x \dot{x} + a_{12} y \dot{y} $$
$$ \dot{V}\left(\mathbf{x}\right) = - a_{21} x \left(a_{11} x + a_{12} y\right) + a_{12} y \left(a_{21} x + a_{22} y\right) $$
$$ \dot{V}\left(\mathbf{x}\right) = - a_{11} a_{21} x^{2} - a_{12} a_{21} x y + a_{12} a_{21} x y + a_{12} a_{22} y^{2}. $$
Cancelling like terms yields
$$ \dot{V}\left(\mathbf{x}\right) = - a_{11} a_{21} x^{2} + a_{12} a_{22} y^{2}. $$
In order to satisfy the third condition $\dot{V}\left(\mathbf{x}\right) < 0$, then we must have
$$ a_{11} a_{21} > 0, $$
$$ a_{12} a_{22} < 0. $$
From $a_{12} > 0$ and $a_{21} < 0$, it must be true that
$$ a_{11} < 0, $$
$$ a_{22} < 0. $$
Since $a_{11} < 0$ and $a_{22} < 0$, then
$$ a_{11} + a_{22} < 0. $$
Similarly, since $a_{11} a_{22} > 0$ and $a_{12} a_{21} < 0$, then
$$ a_{11} a_{22} - a_{12} a_{21} > 0. $$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4617815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find the intersection points of two circles Find the intersection points of the circles $$k_1:(x-4)^2+(y-1)^2=9\\k_2:(x-8)^2+(y+4)^2=100$$
The intersections point (if they exist) will satisfy the equations of both the circles, so we can find their coordinates by solving the system $$\begin{cases}(x-4)^2+(y-1)^2=9\\(x-8)^2+(y+4)^2=100\end{cases}\iff\begin{cases}x^2-8x+y^2-2y=-8\\x^2-16x+y^2+8y=20\end{cases}$$
Substracting these equations, gives $4x-5y=-14\Rightarrow x=\dfrac{5y-14}{4}$. Substituting into the first, I got (if I didn't mess up the calculations) $$41y^2-332y+766=0$$ which has no real solutions (negative discriminant). Therefore the system has no solutions as well, so the circles don't intersect. Is there something else we can use in order to conclude that they do not intersect? Maybe something which requires less calculations? The center of the first is $O_1(4;1)$, the center of the second cirlce is $O_2(8;-4)$ and their radii are $r_1=3$ and $r_2=10$, respectively, if this somehow helps. The distance $O_1O_2=\sqrt{16+25}=\sqrt{41}$ and $r_1+r_2=13=\sqrt{169}$, but I am not sure how to interpret these findings.
|
Any point on the first circle : $P(4+3\cos t,1+3\sin t)$
If $P$ has to be on the second circle as well,
$$100=(4+3\cos t-8)^2+(1+3\sin t+4)^2=-24\cos t+30\sin t+16+25$$
$$\iff5\sin t-4\cos t=\dfrac{41}6$$
Now $|5\sin t-4\cos t|\le\sqrt{4^2+5^2}=\sqrt{41}$ which is $<\dfrac{41}6$ as $6<\sqrt{41}$
So, no real intersection!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4618115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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|
Calculate the values of the serieses Given this function: $f(x)=|\sin x|$, I was supposed to find its cosinuses series, which came out $$\frac{2}{\pi}+\sum_{k=1}^\infty\frac{4}{\pi\left(1-4k^2\right)}\cos(2kx)\,.$$
After that, I was asked to find the values of these two serieses:
$$\sum_{n=1}^\infty\frac{1}{4n^2-1}$$
$$\sum_{n=1}^\infty\frac{n^2}{\left(4n^2-1\right)^2}$$
Using Parseval Theorem, I got to this:
$$\sum_{k=1}^\infty\frac{1}{\left(4k^2-1\right)^2}=\frac{3\pi^2}{16}-\frac{1}{2}$$
but I don’t know what to do next…
|
Since $\;\displaystyle 0=f(0)=\frac{2}{\pi}+\sum_{k=1}^\infty\frac{4}{\pi\left(1-4k^2\right)}\,,\;$ it follows that
$\displaystyle\sum_{k=1}^\infty\frac{4}{\pi\left(4k^2-1\right)}=\frac2\pi\;\;,$
$\displaystyle\sum_{k=1}^\infty\frac1{4k^2-1}=\frac12\;.$
Hence,
$\displaystyle\sum_{n=1}^\infty\frac1{4n^2-1}=\frac12\;.$
Moreover, by using Parseval Theorem, we can get that :
$\displaystyle\sum_{k=1}^\infty\frac{1}{\left(4k^2-1\right)^2}=\frac{\pi^2}{16}-\frac{1}{2}\;.$
Consequently ,
$\displaystyle\sum_{n=1}^\infty\frac{n^2}{\left(4n^2-1\right)^2}= \frac14\left[\sum_{n=1}^\infty\frac{4n^2-1}{\left(4n^2-1\right)^2}+\sum_{n=1}^\infty\frac1{\left(4n^2-1\right)^2}\right]=$
$\displaystyle= \frac14\left[\sum_{n=1}^\infty\frac1{4n^2-1}+\sum_{n=1}^\infty\frac1{\left(4n^2-1\right)^2}\right]=$
$=\dfrac14\left(\dfrac12+\dfrac{\pi^2}{16}-\dfrac12\right)=\dfrac{\pi^2}{64}\;.$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4619555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Acute triangle $\triangle ABC$ has $(\sin A - 2 \sin 2B) = 2 - 2 \cos 2B$ Find the range of $\frac{\sin B + \sin C}{\sin A}$ Angles in acute triangle $\triangle ABC$ has $(\sin A - 2 \sin 2B) \tan A = 2 - 2 \cos 2B$
Find the range of $$\frac{\sin B + \sin C}{\sin A}$$
We can prove that $\frac{a^2}{bc} = 4$ through some tedious trigonometric computations
Let $\frac{b+c}{a} = k$, then
$1 > \cos A = \frac{b^2+c^2-a^2}{2bc}$, and we plug in $\frac{a^2}{bc} = 4$, and we might be able to find the range of $k$ somehow? But I can't finish and the computation seems daunting. I am also not sure if $\cos A <1$ condition is too loose.
|
Looks like you are very close.
You have established $\frac {\sin B + \sin C} {\sin A} = \frac {b+c}a$ using the Law of Sines.
From $\frac {a^2}{bc} = 4$ we obtain $a^2 = 4bc$ and $bc = \frac{a^2}4$.
Now:
\begin{align}
1 &> \cos A\\ &= \frac {b^2 + c^2 - a^2}{2bc}\\
&=\frac {b^2 + c^2 - 4bc}{2bc}\\
&=\frac {b^2 + 2bc + c^2 - 6bc}{2bc}\\
&=\frac {(b+c)^2}{a^2/2}-6\\
&=2k^2-3\\
k^2& < 2
\end{align}
|
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"url": "https://math.stackexchange.com/questions/4625234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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|
prove $(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]$ $(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]$
How to deal with this problem?
Observing that when $x=\pi/2$, the above inequality becomes equality.
Firstly, denote $f(x)=(\sin x)^{-2}-x^{-2}$ and then take derivative of $f(x)$. We have
$$-2(\sin x)^{-3}\cos x+2x^{-3}$$ Next, how to analysis the sign of $f'(x)$?
Any hints are wellcome! Thanks!
|
Let $f(x)=\sin^{-2}x-x^{-2}$.
Note
$$ f'(x)=\frac{2}{x^3}-\frac{2\cos x}{\sin^3x}=\frac{2(\sin^3x-x^3\cos x)}{x^3\sin^3x}=\frac{g(x)}{x^3\sin^3x}. $$
where
$$ g(x)=\sin^3x-x^3\cos x.$$
Now
$$ g'(x)=x^3\sin x-3\cos x(x^2-\sin^2x)\ge0, \forall x\in(0,\frac\pi2] $$
which implies that $g(x)$ is increasing and hence
$$ g(x)> g(0)=0, \forall x\in(0,\frac\pi2]. $$
This also implies that $f(x)$ is increasing. So $x\in[0,\frac\pi2]$
$$ f(x)\le f(\frac\pi2) $$
or
$$(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]. $$
$\bf{Update:}$ Actually we don't need $g'(x)\ge0$. To prove $g(x)\ge 0$ for $x\in(0,\pi/2]$, using
$$ \sin x\ge x-\frac{x^3}6, \cos x\le 1-\frac{x^2}2+\frac{x^4}{24}, $$
one has
\begin{eqnarray}
g(x)&=&x^3\bigg[\bigg(\frac{\sin x}{x}\bigg)^3-\cos x\bigg]\\
&=&x^3\bigg[\bigg(1-\frac{x^2}6\bigg)^3-\bigg(1-\frac{x^2}2+\frac{x^4}{24}\bigg)\bigg]\\
&=&\frac1{216}x^7(9-x^2)>0, x\in(0,\pi/2].
\end{eqnarray}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4626032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
The projective plane as a smooth surface in a 4-dimensional space For $(a, b, c)$ pairwise distinct:
$$
\begin{align}
f: & S^2 & \mapsto & \quad \mathbb{R}^4 \\
& (x, y, z) & \mapsto & \quad (X, Y, Z, W) = (y \cdot z, x \cdot z, x \cdot y, a \cdot x^2 + b \cdot y^2 + c \cdot z^2)
\end{align}
$$
is a smooth double covering map of the real unit sphere $S^2$ to a
surface in the real 4-dimensional space $\mathbb{R}^4$, and opposite
points have the same image, such that it can also be interpreted as an
injective embedding of the real projective plane. In particular, the
image of $f()$ is a surface, that is, each point has a non-degenerate
tangential plane.
The image of $f()$ can be defined in the $(X, Y, Z, W)$ coordinates as follows.
If $X \cdot Y \cdot Z \neq 0$:
$$
(Y \cdot Z)^2 + (X \cdot Z)^2 + (X \cdot Y)^2 = X \cdot Y \cdot Z \\
a \cdot (Y \cdot Z)^2 + b \cdot (X \cdot Z)^2 + c \cdot (X \cdot Y)^2 = X \cdot Y \cdot Z \cdot W
$$
If $X = Y = 0$ and $Z \neq 0$:
$$
(a - b)^2 \cdot (1 - 4 \cdot Z^2) = (2 \cdot W - (a + b))^2
$$
If $X = Z = 0$ and $Y \neq 0$:
$$
(a - c)^2 \cdot (1 - 4 \cdot Z^2) = (2 \cdot W - (a + c))^2
$$
If $Y = Z = 0$ and $X \neq 0$:
$$
(b - c)^2 \cdot (1 - 4 \cdot Z^2) = (2 \cdot W - (b + c))^2
$$
If $X = Y = Z = 0$:
$$
(W - a) \cdot (W - b) \cdot (W - c) = 0
$$
Question 1:
Is the image of $f()$ an algebraic set in the $(X, Y, Z, W)$ coordinates ? Can
it be defined by a single ideal, without case analysis ?
If yes, what is the ideal in question, explicitly ?
If no, how to prove it ? Furthermore, the image of f() consists then of five
algebraic pieces fitting together smoothly, without being, as a whole,
algebraic; I find it surprising, what happens where the pieces meet ?
Question 2:
If $X \cdot Y \cdot Z \neq 0$, or, equivalently, $x \cdot y \cdot z \neq 0$,
then
$$
[x : y : z] = [Y \cdot Z : X \cdot Z : X \cdot Y]
$$
Is there a rational formula for each of the other pieces ?
If yes, what are the respective formulæ, explicitly ?
If no, how to prove it ?
|
Response to question 1
This is a complement to the excellent answer of Federico Fallucca.
The ideal characterizing the image of $f()$ is already generated by
the following four polynomials:
$$
\begin{align}
& (b - c)^2X^3 + (c - a)^2XY^2 + (a - b)^2XZ^2 + (W - b)(W - c)X - (a - b)(a - c)YZ & (C_0) \\
& (c - a)^2Y^3 + (a - b)^2YZ^2 + (b - c)^2YX^2 + (W - c)(W - a)Y - (b - c)(b - a)ZX & (C_1) \\
& (a - b)^2Z^3 + (b - c)^2ZX^2 + (c - a)^2ZY^2 + (W - a)(W - b)Z - (c - a)(c - b)XY & (C_2) \\
& (b - c)^2(W - a)X^2 + (c - a)^2(W - b)Y^2 + (a - b)^2(W - c)Z^2 + (W - a)(W - b)(W - c) & (C_3)
\end{align}
$$
In particular, this ideal contains the following two polynomials
(naturally derived from the constraints $x^2 + y^2 + z^2 = 1$ and
$ax^2 + by^2 + cz^2 = W$, but superseded by the generators
above):
$$
\begin{align}
& Y^2Z^2 + Z^2X^2 + X^2Y^2 - XYZ & (D_0) \\
& aY^2Z^2 + cX^2Y^2 + bZ^2X^2 - XYZW & (D_1)
\end{align}
$$
Indeed:
$$
\begin{align}
D_0 &= \frac{(b - c)YZC_0 + (c - a)ZXC_1 + (a - b)XYC_2}{(a - b)(b - c)(c - a)} \\
D_1 &= \frac{a(b - c)YZC_0 + b(c - a)ZXC_1 + c(a - b)XYC_2}{(a - b)(b - c)(c - a)}
\end{align}
$$
Response to question 2
The real projective hemisphere $S$ (that is, the real unit sphere with
opposite points identified) can be identified with the real projective
plane $P$ by the following map:
$$
\begin{align}
P & \longrightarrow S \\
[\bar{x} : \bar{y} : \bar{z}] & \longrightarrow (x, y, z)
= \frac{1}{\sqrt{\bar{x}^2 + \bar{y}^2 + \bar{z}^2}} \cdot (\bar{x}, \bar{y}, \bar{z})
\end{align}
$$
Thus, for inversing $f()$, it is enough to determine $(x, y, z)$ up to
a non-zero constant scaling factor, that is, to determine
$[x : y : z]$.
Assume $(X, Y, Z, W) = f(x, y, z)$ from now on; in particular:
$$
\begin{align}
& x^2 + y^2 + z^2 = 1 \\
& X = y \cdot z \\
& Y = z \cdot x \\
& Z = x \cdot y \\
& W = a \cdot x^2 + b \cdot y^2 + c \cdot z^2
\end{align}
$$
Our goal is to express $[x : y : z]$ as polynomial functions
$g_i(X, Y, Z, W)$, using the constraints above. For any polynomial
$t_i(x, y, z)$, we have:
$$
[t_i(x, y, z) \cdot x : t_i(x, y, z) \cdot y : t_i(x, y, z) \cdot z]
= [x : y : z]
$$
unless $t_i(x, y, z) = 0$. The left-hand side is said to be determinate
at $(x, y, z)$ if $t_i(x, y, z)$ does not vanish, and indeterminate
otherwise. We have thus to find polynomials $t_i()$ such that the
left-hand side can be expressed in the $(X, Y, Z, W)$ coordinates,
that is:
$$
g_i(X, Y, Z, W) = [x : y : z]
$$
Furthermore, we have to find enough of such polynomials $t_i()$ resp.
$g_i()$ such that, for each $(X, Y, Z, W)$ in the image of $f()$, some of
these polynomials is determinate at $(X, Y, Z, W)$, that is,
$g_i(X, Y, Z, W) \neq [0 : 0 : 0]$.
A possibility would be to choose $t_0(x, y, z) = x \cdot y \cdot z$, which
yields:
$$
\begin{align}
[x : y : z]
&= [t_0(x, y, z) \cdot x : t_0(x, y, z) \cdot y : t_0(x, y, z) \cdot z] \\
&= [x \cdot y \cdot z \cdot x : x \cdot y \cdot z \cdot y : x \cdot y \cdot z \cdot z ] \\
&= [Y \cdot Z : Z \cdot X : X \cdot Y]
\end{align}
$$
and thus:
$$
g_0(X, Y, Z, W) \overset{\underset{\mathrm{def}}{}}{=}
[Y \cdot Z : Z \cdot X : X \cdot Y]
$$
which is indeterminate at $(X, Y, Z, W)$ exactly if
$X \cdot Y \cdot Z = 0$, or, equivalently, $x \cdot y \cdot z = 0$.
But more comprehensive polynomials $t_i()$ can be derived using the
following equality:
$$
\begin{align}
(W - a)
&= a \cdot x^2 + b \cdot y^2 + c \cdot z^2 - a \\
&= a \cdot x^2 + b \cdot y^2 + c \cdot z^2 - a \cdot (x^2 + y^2 + z^2) \\
&= (b - a) \cdot y^2 + (c - a) \cdot z^2
\end{align}
$$
Analogously:
$$
(W - b) = (c - b) \cdot z^2 + (a - b) \cdot x^2 \\
(W - c) = (a - c) \cdot x^2 + (b - c) \cdot y^2
$$
from which follows:
$$
\begin{align}
(W - a) \cdot x^2
&= (b - a) \cdot x^2 \cdot y^2 + (c - a) \cdot z^2 \cdot x^2 \\
&= (b - a) \cdot Z^2 + (c - a) \cdot Y^2
\end{align}
$$
and
$$
\begin{align}
(c - b) \cdot (W - a) \cdot z^2
&= (W - a) \cdot ((W - b) + (b - a) \cdot x^2) \\
&= (W - a) \cdot (W - b) + (b - a)^2 \cdot Z^2 + (b - a) \cdot (c - a) \cdot Y^2
\end{align}
$$
Analogously:
$$
\begin{align}
(W - b) \cdot y^2 &= (c - b) \cdot X^2 + (a - b) \cdot Z^2 \\
(a - c) \cdot (W - b) \cdot x^2 &= (W - b) \cdot (W - c) + (c - b)^2 \cdot X^2 + (c - b) \cdot (a - b) \cdot Z^2 \\
(W - c) \cdot z^2 &= (a - c) \cdot Y^2 + (b - c) \cdot X^2 \\
(b - a) \cdot (W - c) \cdot y^2 &= (W - c) \cdot (W - a) + (a - c)^2 \cdot Y^2 + (a - c) \cdot (b - c) \cdot X^2
\end{align}
$$
Choosing $t_1(x, y, z) = (a - c) \cdot (W - b) \cdot x$ yields:
$$
\begin{align}
[x : y : z]
&= [t_1(x, y, z) \cdot x : t_1(x, y, z) \cdot y : t_1(x, y, z) \cdot z] \\
&= [(a - c) \cdot (W - b) \cdot x^2 : (a - c) \cdot (W - b) \cdot x \cdot y : (a - c) \cdot (W - b) \cdot z \cdot x] \\
&= [(W - b) \cdot (W - c) + (c - b)^2 \cdot X^2 + (c - b) \cdot (a - b) \cdot Z^2 : (a - c) \cdot (W - b) \cdot Z : (a - c) \cdot (W - b) \cdot Y]
\end{align}
$$
And thus:
$$
g_1(X, Y, Z, W) \overset{\underset{\mathrm{def}}{}}{=}
[(W - b) \cdot (W - c) + (c - b)^2 \cdot X^2 + (c - b) \cdot (a - b) \cdot Z^2 : (a - c) \cdot (W - b) \cdot Z : (a - c) \cdot (W - b) \cdot Y]
$$
Choosing $t_2(x, y, z) = (b - a) \cdot (W - c) \cdot y$ resp.
$t_3(x, y, z) = (c - b) \cdot (W - a) \cdot z$ yields, analogously:
$$
g_2(X, Y, Z, W) \overset{\underset{\mathrm{def}}{}}{=}
[(b - a) \cdot (W - c) \cdot Z : (W - c) \cdot (W - a) + (a - c)^2 \cdot Y^2 + (a - c) \cdot (b - c) \cdot X^2 : (b - a) \cdot (W - c) \cdot X]
\\
g_3(X, Y, Z, W) \overset{\underset{\mathrm{def}}{}}{=}
[(c - b) \cdot (W - a) \cdot Y : (c - b) \cdot (W - a) \cdot X : (W - a) \cdot (W - b) + (b - a)^2 \cdot Z^2 + (b - a) \cdot (c - a) \cdot Y^2]
$$
The constraints for these last three maps $g_1()$, $g_2()$ and $g_3()$
to all be indeterminate at $f(x, y, z)$ for some $(x, y, z)$ on the
real projective hemisphere are
$t_1(x, y, z) = t_2(x, y, z) = t_3(x, y, z) = 0$, that is,
$
(a - c) \cdot (W - b) \cdot x
= (b - a) \cdot (W - c) \cdot y
= (c - b) \cdot (W - a) \cdot z = 0
$,
that is, by the equalities established above:
$$
(W - b) \cdot (W - c) + (c - b)^2 \cdot X^2 + (c - b) \cdot (a - b) \cdot Z^2 = 0 \\
(W - c) \cdot (W - a) + (a - c)^2 \cdot Y^2 + (a - c) \cdot (b - c) \cdot X^2 = 0 \\
(W - a) \cdot (W - b) + (b - a)^2 \cdot Z^2 + (b - a) \cdot (c - a) \cdot Y^2 = 0
$$
that is, by expressing (X, Y, Z, W) in terms of (x, y, z) and simplifying:
$$
x \cdot ((c - b) \cdot z^2 + (a - b) \cdot x^2) = 0 \\
y \cdot ((a - c) \cdot x^2 + (b - c) \cdot y^2) = 0 \\
z \cdot ((b - a) \cdot y^2 + (c - a) \cdot z^2) = 0
$$
which have no solution on the real projective hemisphere, such that no
point of the hemisphere has an image $f(x, y, z)$ at which all three
maps $g_1()$, $g_2()$ and $g_3()$ are indeterminate.
In other words, the inverse of $f()$ is at least one of the polynomials
$g_1()$, $g_2()$ and $g_3()$, whichever is determinate, depending their
argument (a point $(X, Y, Z, W)$ in the image of $f()$), and there is no
argument at which all three are indeterminate.
Finally, the function $f()$ can be equivalently defined on the real
projective plane:
$$
\begin{align}
f([x : y : z])
&= f\left(\frac{x}{\sqrt{x^2 + y^2 + z^2}}, \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \frac{z}{\sqrt{x^2 + y^2 + z^2}}\right) \\
&= \frac{1}{x^2 + y^2 + z^2} \cdot (y \cdot z, z \cdot x, x \cdot y, a \cdot x^2 + b \cdot y^2 + c \cdot z^2)
\end{align}
$$
and thus the real projective plane and the image of $f()$ are
birationally equivalent.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Prove $\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b} \geq \frac{9}{2}$ for $a,b,c\geq1$ Prove $\dfrac{ab^2+2}{a+c} +\dfrac{bc^2+2}{b+a} +\dfrac{ca^2+2}{c+b} \geq \dfrac{9}{2}$ for $a,b,c\geq1$.
I tried using the Titu Andreescu form of the Cauchy Schwarz inequality and got to this point:
$\dfrac{(b\sqrt{a}+c\sqrt{b}+a\sqrt{c})^2+18}{a+b+c} \geq 9$.
I don't know what to do next, any help is appreciated.
Thanks!
|
Update: Add proof 2 without using Holder inequality.
Proof 2:
We have
\begin{align*}
&\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b}\\[6pt]
={}& \frac{ab^2}{a+c} +\frac{bc^2}{b+a} +\frac{ca^2}{c+b} + \frac{2}{a+c} + \frac{2}{b+a} + \frac{2}{c+b}\\[6pt]
\ge{}&3\sqrt[3]{\frac{ab^2}{a+c} \cdot\frac{bc^2}{b+a} \cdot\frac{ca^2}{c+b}} + 3\sqrt[3]{\frac{2}{a+c} \cdot \frac{2}{b+a} \cdot \frac{2}{c+b}} \tag{1}\\[6pt]
={}& \frac{3abc + 6}{\sqrt[3]{(a+b)(b+c)(c+a)}}\\[6pt]
\ge{}& \frac{3abc + 6}{\frac{a+b + b+c + c+a}{3}} \tag{2}\\[6pt]
={}& \frac{9abc + 18}{2(a + b + c)}\\[6pt]
={}& \frac92 \cdot \frac{abc + 2}{a + b + c}\\
\ge{}& \frac92.\tag{3}
\end{align*}
Explanations:
(1): AM-GM.
(2): AM-GM.
(3): $$abc + 2 - a - b - c$$
$$=
(a-1)(b-1)(c-1) + (a-1)(b-1) + (b-1)(c-1) + (c-1)(a-1) \ge 0.$$
We are done.
Proof 1:
We have
\begin{align*}
&\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b}\\[6pt]
\ge{}& 3\sqrt[3]{\frac{ab^2+2}{a+c} \cdot \frac{bc^2+2}{b+a} \cdot \frac{ca^2+2}{c+b}} \tag{1}\\[6pt]
\ge{}& 3\sqrt[3]{\frac{(abc + 2)^3}{\left(\frac{a + c + b + a + c + b}{3}\right)^3}} \tag{2}\\[6pt]
={}& \frac92 \cdot \frac{abc + 2}{a + b + c}\\
\ge{}& \frac92. \tag{3}
\end{align*}
Explanations:
(1): AM-GM.
(2): $(ab^2 + 2)(bc^2 + 2)(ca^2 + 2) \ge (\sqrt[3]{ab^2\cdot bc^2 \cdot ca^2} + 2)^3$ by Holder's inequality;
$(a+c)(b+a)(c+b) \le \left(\frac{a + c + b + a + c + b}{3}\right)^3$ by AM-GM.
(3): $$abc + 2 - a - b - c$$
$$=
(a-1)(b-1)(c-1) + (a-1)(b-1) + (b-1)(c-1) + (c-1)(a-1) \ge 0.$$
We are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sine curve that passes through (1,1) (2,2) and (3,3) Question: Find a sine function in the form $a\sin(bx) + c$ that passes through points (1,1) (2,2) and (3,3)
Working so far:
*
*We have three points for three unknown variables in the function, so we can use simultaneous equations to solve for them.
*Simultaneous equations:
*
*$a\sin(b) + c = 1$
*$a\sin(2b) + c = 2$
*$a\sin(3b) + c = 3$
*Equation 1, solving for c:
*
*$c = 1 - a\sin(b)$
*Equation 2: solving for a:
*
*$a\sin(2b) + 1 - a\sin(b) = 2$
*$a\sin(2b) - a\sin(b) = 1$
*$a(\sin(2b) - \sin(b)) = 1$
*$a = \frac{1}{\sin(2b) - \sin(b)}$
*Equation 3: solving for b:
*
*$(\frac{1}{\sin(2b) - \sin(b)})(\sin(3b)) + 1 - (\frac{1}{\sin(2b) - \sin(b)})(\sin(b)) = 3$
*$\frac{\sin(3b)}{\sin(2b) - \sin(b)} - \frac{\sin(b)}{\sin(2b) - \sin(b)} = 2$
*$\frac{\sin(3b) - \sin(b)}{\sin(2b) - \sin(b)} = 2$
*$\sin(3b) - \sin(b) = 2\sin(2b) - 2\sin(b)$
*$\sin(3b) + \sin(b) = 2\sin(2b)$
*$3\sin(b) - 4\sin^{3}(b) + \sin(b) = 4\sin(b)\cos(b)$
*$4\sin(b) - 4\sin^{3}(b) = ±4\sin(b)\sqrt{1 - \sin^{2}(b)}$
*Substitute $u = \sin(b)$
*$4u - 4u^3 = ±4u\sqrt{1-u^2}$
*$u - u^3 = ±u\sqrt{1-u^2}$
*$1 - u^2 = ±\sqrt{1-u^2}$
*$1 - 2u^2 + u^4 = 1 - u^2$
*$u^4 - u^2 = 0$, $u = 0$
*$u^2 - 1 = 0$
*$u^2 = 1$
*$u = 1$, $u = -1$
*For $u = -1$:
*
*$\sin(b) = -1$
*$b = \arcsin(-1) = (\frac{4n - 1}{2})\pi$ and $(\frac{4n + 3}{2})\pi$, n is an integer.
*For $u = 0$:
*
*$b = \arcsin(0) = n\pi$, n is an integer.
*For $u = 1$:
*
*$b = \arcsin(1) = (\frac{4n + 1}{2})\pi$, n is an integer.
*$b = n\pi$ and $b = n\pi - \frac{\pi}{2}$
*Therefore $a = \frac{1}{\sin(2(n\pi)) - \sin((n\pi))}$
*$a = \frac{1}{0 - 0} = \frac{1}{0}$, let's see if the other solution works.
*$a = \frac{1}{\sin(2(n\pi - \frac{\pi}{2})) - \sin(n\pi - \frac{\pi}{2})}$
*$a = \frac{1}{1} = 1$ when n is an even integer and $a = \frac{1}{-1} = -1$ when n is an odd integer.
*$c = 1 - (1)(\sin(n\pi)) = 1$, $c = 1 - (-1)(\sin(n\pi)) = 1$, $c = 1 - (1)(\sin(n\pi - \frac{\pi}{2})) = 2$ when n is an even integer and $c = 1 - (1)(\sin(n\pi - \frac{\pi}{2})) = 0$ when n is an odd integer, $c = 1 - (-1)(\sin(n\pi - \frac{\pi}{2})) = 0$ when n is an even integer and finally $c = 1 - (-1)(\sin(n\pi - \frac{\pi}{2})) = 2$ when n is an odd integer
*With these results, let's try n = 0:
*
*$a = 1$, $b = 0$ and $b = -\frac{\pi}{2}$, $c = 1$, $c = 2$, $c = 0$
*$y = \sin(-\frac{\pi}{2}x) + 1$ touches none of the three points.
A way to get the curve to approximately touch the 3 points would be to equate $a$ to a very large number, and equate $b$ to the reciprocal of $a$. And have $c$ equal to 0, in this case the graph mimics the straight-line function $f(x) = x$ in the desired range. But this doesn't account for the possibility that the graph could pass through several 'waves' between $x = 0$ and $x = 3$ to pass through all three points in those waves.
What could the values of a, b and c be for the sine curve to pass through all three points?
|
A little geometric intuition is helpful here. Your desired points are odd-symmetric around $(2,2)$, so you can imagine shifting your function up 2 steps and then scaling it horizontally to have period $4$ (instead of $2\pi$).
It would be straightforward then to shift the origin to $(2,2)$ and scale $x$ by $\pi/2$, but the resulting form of the expression is $c + \sin(ax+b)$, which is not quite what you asked for. There are tricks for transforming it in the way you need, but once you've drawn the graph another much easier idea is clear anyway; shift up $2$ units and scale $x$ by negative $\pi/2$ to exploit the symmetry of $sin$.
$$ f(x) = 2-\sin(\pi x/2) $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
For what values of h are the three vectors linearly dependent?
For the vectors below find the value(s) of h for which the vectors are linearly dependent.
$$
\vec v_1=\left[\begin{array}{c} 2 \\ -4 \\ 1 \end{array} \right]\ \vec v_2 = \left[\begin{array}{c} -6 \\ 7 \\ -3 \end{array} \right]\ \vec v_3 = \left[\begin{array}{c} 8 \\ h \\ 4 \end{array} \right]
$$
Attempt:
I started by reducing the matrix into echelon form but am confused on where to go from here and how to tell what values of h would make the vectors linearly dependent. Any help is greatly appreciated.
$$
\begin{bmatrix} 2 & -6 & 8 \\ -4 & 7 & h \\ 1 & -3 & 4 \end{bmatrix} \sim> \begin{bmatrix} 1 & 0 & \frac{-3h}{5} - \frac{28}{5} \\ 0 & 1 & \frac{-h}{5} - \frac{16}{5} \\ 0 & 0 & 0 \end{bmatrix}
$$
EDIT: Someone asked for me to show the work I did to get from the LHS to RHS. It is below:
*
*$R_1 = \frac{R_1}{2}$
$$
\begin{bmatrix} 1 & -3 & 4 \\ -4 & 7 & h \\ 1 & -3 & 4 \end{bmatrix}
$$
*$R_2 = R_2 + 4(R_1)$
$$
\begin{bmatrix} 1 & -3 & 4 \\ 0 & -5 & h+16 \\ 1 & -3 & 4 \end{bmatrix}
$$
*$R_3 = R_3 - R_1$
$$
\begin{bmatrix} 1 & -3 & 4 \\ 0 & -5 & h+16 \\ 0 & 0 & 0 \end{bmatrix}
$$
*$R_2 = -\frac{R_2}{5}$
$$
\begin{bmatrix} 1 & -3 & 4 \\ 0 & 1 & -\frac{h}{5} - \frac{16}{5} \\ 0 & 0 & 0 \end{bmatrix}
$$
*$R_1 = R_1 + 3(R_2)$
$$
\begin{bmatrix} 1 & 0 & \frac{-3h}{5} - \frac{28}{5} \\ 0 & 1 & \frac{-h}{5} - \frac{16}{5} \\ 0 & 0 & 0 \end{bmatrix}
$$
|
Your attempt it is correct, we only need to interpret in terms of the definitions what is happening. We always can use the definition: we says that $\{v_1,v_2,v_3\}$ is linearly dependent if there exists scalars $\alpha_1,\alpha_2,\alpha_3$ not all zeros such that $\alpha_{1}v_1+\alpha_{2}v_2+\alpha_{3}v_3=0$. In terms the linear system equation in matrix form we have $$\begin{bmatrix}2&-6&8\\-4&7&h\\1&-3&4 \end{bmatrix}\begin{bmatrix}\alpha_1\\\alpha_2\\\alpha_3\end{bmatrix}=\begin{bmatrix}0\\0\\0 \end{bmatrix}\iff \left[\begin{array}{ccc|c}2&-6&8&0\\-4&7&h&0\\1&-3&4&0 \end{array}\right]$$
But the linear system is homogeneous (i.e., $Ax=0$) so we know that always it is consistent. Moreover if it has only solution we have independence linear but with infinite solutions we have dependence linear. We can also notice that if at least one row after of the elimination to become in zero, then we have dependence linear. So we only need look at the matrix $$\begin{bmatrix}2&-6&8\\-4&7&h\\1&-3&4 \end{bmatrix}$$ and then to make elimination by row to see under what situation a row could become the zero row, which is our study objective. Here comes your work, let's try to clarify this.
\begin{align*}
\begin{bmatrix}2&-6&8\\-4&7&h\\1&-3&4 \end{bmatrix}&\sim \begin{bmatrix}1&-3&4\\-4&7&h\\2&-6&8 \end{bmatrix}\\
&\sim \begin{bmatrix}1&-3&4\\0&-5&h+16\\2&-6&8\end{bmatrix}\\
&\sim \begin{bmatrix}1&-3&4\\0&1&\frac{-h-16}{5}\\0&0&0 \end{bmatrix}
\end{align*}
Now, you can see (or verify) that in no step of the row reduction was a condition/constraint placed on the value of $h$. But we have obtained a row full of zeros, so the set can never be independent set regardless of the value of $h$, that is basically Anne Bauval's solution in the comments. Therefore we conclude that for all value $h$ the set $\{v_1,v_2,v_3\}$ is always dependent.
I think also as it was suggested in the comments that the way using the idea of this concept in terms of determinants can make the work easier. But it looks like you haven't seen the idea yet, so it can't be used yet.
|
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|
Calculate the following integral $ \int\frac{2x+1}{x^{n+2}(x+1)^{n+2}}\ln\left(\frac{2x^2+2x+1}{x^2(x+1)^2}+\frac7{16}\right)dx$ Hello I am trying to solve a pretty complicated integral. It is a from a set of problems, published in a monthly journal for high school students and they are exercises in preparation for a competition.
So the problem is the following: Calculate the integral $ \int \frac {2x+1}{x^{n+2}(x+1)^{n+2}} \ln\left(\frac {2x^2+2x+1}{x^2(x+1)^2} + \frac{7}{16}\right)dx$
I first tried partial integration, integrating $\frac {2x+1}{x^{n+2}(x+1)^{n+2}}$, it is happily easy to do it and we will end up with $\int \frac {2x+1}{x^{n+2}(x+1)^{n+2}} = \frac {-1}{x^{n+1}(x+1)^{n+1}}$.
I also differentiated $\ln\left(\frac {2x^2+2x+1}{x^2(x+1)^2} + \frac{7}{16}\right)$, ending up with $\frac {-64x^3-96x^2-96x-32}{7x^6+21x^5+71x^3+48x^2+16x}$
Now when I try to solve the integral of the product of these two, I am stuck.
I also thought about some kind of recursive relationship in terms of $n$ but I am not sure about it.
I would happily accept any help in solving this problem. Thanks in advance.
|
As said in comments, using $u=x(x+1)$ as @David Quinn suggested the problem reduces to
$$I=\int\frac 1 {u^{n+2}} \log \left(\frac{7 u^2+32 u+16}{16 u^2}\right)\,du$$ which is
$$I=-\frac{2+(n+1) \log \left(16 u^2\right)}{(n+1)^2\, u^{n+1}}+\int \frac 1 {u^{n+2}}\log(7 u^2+32 u+16)\,du $$
$$J=\int \frac 1 {u^{n+2}}\log(7 u^2+32 u+16)\,du $$
$$J=\log(7)\int \frac {du}{u^{n+2}}+\int \frac{\log(u+4)}{u^{n+2}}\,du+\int \frac{\log \left(u+\frac{4}{7}\right)}{u^{n+2}}\,du$$
$$J=-\frac{\log(7)}{(n+1) u^{n+1} }+\int \frac{\log(u+4)}{u^{n+2}}\,du+\int \frac{\log \left(u+\frac{4}{7}\right)}{u^{n+2}}\,du$$
$$K_a=\int \frac{\log \left(u+a\right)}{u^{n+2}}\,du$$
$$K_a=-\frac{u \,\, _2F_1\left(1,-n;1-n;-\frac{u}{a}\right)+a n \log (u+a) } {a n (n+1) u^{n+1} }$$
|
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|
Equation of parabola passes through $4$ distinct points
Equation of axis of parabola which
passes through the point $(0,1)\ , \ (0,2)$
And $(2,0)\ ,\ (2,2)$ is
Let general equation of conic is
$ax^2+2hxy+by^2+2gx+2fy+c=0\cdots (1)$
And it represent parabola if $h^2=ab$
Parabola passes through $(0,1)$
Then put into $(1)$
$b+2f+c=0\cdots(2)$
Also parabola passes through $(0,2)$
Then $4b+4f+c=0\cdots (3)$
Also parabola passes through $(2,0)$
Then $4a+4g+c=0\cdots (4)$
Also parabola passes through $(2,2)$
Then $4a+8h+4b+4g+4f+c\cdots (5)$
$(4a+4g+c)+(4b+4f+c)+8h-c=0$
From $(2)$ and $(3)$, we get $\displaystyle h=\frac {c}{8}$
From $(2)$ and $(3)$
$\displaystyle b+2f=4b+4f\Longrightarrow 2f=-3b $
Put into $(2)$ and $(3)$
$\displaystyle b=\frac{c}{2}$ and $\displaystyle f=-\frac{3c}{4}$
And $\displaystyle h^2=ab\Longrightarrow \frac{c^2}{64}=\frac{ac}{2}\Longrightarrow a=\frac{c}{32}$
Put all into $(4)$
$\displaystyle \frac{4c}{32}+4g+c=0\Longrightarrow g=-\frac{9c}{32}$
Put all values into $(1)$
$\displaystyle \frac{c}{32}x^2+\frac{c}{4}xy+\frac{c}{2}y^2-\frac{9}{16}x-\frac{3c}{2}y+c=0$
$x^2+8xy+16y^2-18x-48y+32=0$
Buti did not know how I find axis of parabola
Please have a look
|
I'm going to show a solution which uses the following claim, and then add a proof of the claim and an important fact which might interest you.
Claim : The equation of a parabola can be written as
$$\bigg(f(x,y)\bigg)^2+g(x,y)=0$$
where
$f(x,y)=0$ is the equation of the axis of symmetry, and $g(x,y)=0$ is the equation of the tangent at vertex. (Note that the axis of symmetry is perpendicular to the tangent at vertex.)
Solution :
You correctly got
$$x^2+8xy+16y^2-18x-48y+32=0$$
which can be written as
$$(x+4y+c)^2+(-2c-18)x+(-8c-48)y-c^2+32=0$$
Since we want to find $c$ such that the line $x+4y+c=0$ is perpendicular to the line $(-2c-18)x+(-8c-48)y-c^2+32=0$, solving
$$1\times (-2c-18)+4\times (-8c-48)=0$$
gives $c=-\dfrac{105}{17}$, and so the equation of the axis of symmetry is $x+4y-\dfrac{105}{17}=0$.
In the following, I'll add a proof of the claim.
Claim : The equation of a parabola can be written as
$$\bigg(f(x,y)\bigg)^2+g(x,y)=0$$
where
$f(x,y)=0$ is the equation of the axis of symmetry, and $g(x,y)=0$ is the equation of the tangent at vertex.
Proof :
The equation
$$ax^2+2bxy+cy^2+2dx+2fy+g=0\tag2$$
represents a parabola iff $ac=b^2$ and
$$\begin{vmatrix}
a & b & d \\
b & c & f \\
d & f & g \\
\end{vmatrix}\not=0\tag3$$
(see here)
Multiplying the both sides of $(2)$ by $a$, and letting $$A=a,C=b,D=2ad,E=2af,F=ag$$
we see that, in general, the equation of a parabola is given by
$$(Ax+Cy)^2+Dx+Ey+F=0\tag4$$
(where $(A,C)\not=(0,0)$, and $(3)\iff CD-AE\not=0$) which can be written as
$$\bigg(f(x,y)\bigg)^2+g(x,y)=0$$
where
$$\begin{align}f(x,y)&=Ax+Cy+\frac {AD+CE}{2(A^2+C^2)}
\\\\g(x,y)&=\frac{CD-AE}{A^2+C^2}(Cx-Ay+G)
\\\\G&=\frac{4F(A^2+C^2)^2-(AD+CE)^2}{4(A^2+C^2)(CD-AE)}\end{align}$$
Note that $f(x,y)=0$ is perpendicular to $g(x,y)=0$.
Now, we can see the followings by some calculations :
*
*The line $g(x,y)=0$ is tangent to the parabola $(4)$ at $P\bigg(H,\dfrac{CH+G}{A}\bigg)$ where $H=\dfrac{-2CG(A^2+C^2)-A(AD+EC)}{2(A^2+C^2)^2}$.
*The line $f(x,y)=0$ intersects the parabola $(4)$ only at $P$.
From these, we can say the followings :
*
*$f(x,y)=0$ is the equation of the axis of symmetry
*$g(x,y)=0$ is the equation of the tangent at the vertex
*$P$ is the vertex of the parabola.$\quad\blacksquare$
Finally, I'll add an important fact which might interest you.
The equation of a parabola
$$(Ax+Cy)^2+Dx+Ey+F=0$$
can be written as
$$\bigg(\frac{f(x,y)}{\sqrt{A^2+C^2}}\bigg)^2=\frac{AE-CD}{(A^2+C^2)^{\frac 32}}\cdot\frac{Cx-Ay+G}{\sqrt{A^2+C^2}}$$
where
*
*$\dfrac{|f(x,y)|}{\sqrt{A^2+C^2}}$ represents the distance from $(x,y)$ to the axis of symmetry
*$\dfrac{|AE-CD|}{(A^2+C^2)^{\frac 32}}$ represents the length of its latus rectum (see here)
*$\dfrac{|Cx-Ay+G|}{\sqrt{A^2+C^2}}$ represents the distance from $(x,y)$ to the tangent at the vertex.
|
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|
Is there any other simple method to evaluate $\int_0^{1} \frac{\ln ^n x}{\left(1-x^2\right)^2}dx $? After reading the fantastic solutions to the integral in the post,
$$\int_0^1 \frac{\ln x}{1 - x^2} \mathrm{d}x=-\frac{\pi^2}8,$$
and found that $$
\int_0^1 \frac{\ln ^2 x}{1-x^2} d x =\frac{7\zeta(3)}{4}. $$
Then I keep on exploring the integral in general and obtain a formula,
$$
\int_0^{1} \frac{\ln ^n x}{\left(1-x^2\right)^2}dx= \frac{(-1)^n n !}{2}\left[\left(1-\frac{1}{2^n}\right)\zeta(n)+\left(1-\frac{1}{2^{n+1}}\right) \zeta(n+1)\right]
$$
where $n\ge 2.$
First of all, we use its partner integral
$$
I(a)=\int_0^1 \frac{x^a}{\left(1-x^2\right)^2} d x
$$
Using the series for $|x|<1,$
$$
\frac{1}{\left(1-x^2\right)^2}=\sum_{k=1}^{\infty} k x^{2 k-2},
$$
we have $$
I(a) =\sum_{k=1}^{\infty} k \int_0^1 x^{2 k-2+a} d x =\sum_{k=1}^{\infty} \frac{k}{2 k-1+a}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln ^n x}{\left(1-x^2\right)^2} d x & =\left.\sum_{k=1}^{\infty} \frac{\partial^n}{d a^n}\left(\frac{k}{2 k-1+a}\right)\right|_{a=0} \\
& =(-1)^n n ! \sum_{k=1}^{\infty} \frac{k}{(2 k-1)^{n+1}} \\
& =\frac{(-1)^n n !}{2} \sum_{k=1}^{\infty}\left[\frac{1}{(2 k-1)^n}+\frac{1}{(2 k-1)^{n+1}}\right] \\
& =\boxed{\frac{(-1)^n n !}{2}\left[\left(1-\frac{1}{2^n}\right)\zeta(n)+\left(1-\frac{1}{2^{n+1}}\right) \zeta(n+1)\right]}
\end{aligned}
$$
For examples,
$$
I_2=\frac{3}{4} \zeta(2)+\frac{7}{8} \zeta(3)= \frac{\pi^2}{8} +\frac{7}{8} \zeta(3)
$$
$$
I_3=-3\left[\frac{7}{8} \zeta(3)+\frac{15}{16} \zeta(4)\right]=-\frac{21}{8} \zeta(3)-\frac{\pi^2}{32}
$$
My question: Is there any other simple method to evaluate $\int_0^{1} \frac{\ln ^n x}{\left(1-x^2\right)^2}dx $?
|
Utilize
\begin{align}
J(m)=\int_0^1 \frac{\ln^m x}{1-x^2}dx
=&\int_0^1 \frac{\ln^m x }{1-x}dx -\int_0^1 {\frac{x\ln^m x }{1-x^2} } \overset{x^2\to x}{dx} \\=& \ \left(1-\frac1{2^{m+1}}\right)\int_0^1\frac{\ln^mx}{1-x}dx\\
=&\ \left(1-\frac1{2^{m+1}}\right)(-1)^m m!\ \zeta(m+1)
\end{align}
to evaluate
\begin{align}
\int_0^{1} \frac{\ln^n x}{\left(1-x^2\right)^2}dx
&=\int_0^{1} \frac{\ln ^n x}{2x}\ d\left(\frac{x^2}{1-x^2}\right)
\overset{ibp}=\frac12J(n)
-\frac n2 J(n-1)\\
&= \frac{(-1)^n n !}{2}\left[\left(1-\frac{1}{2^n}\right)\zeta(n)+\left(1-\frac{1}{2^{n+1}}\right) \zeta(n+1)\right]
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Evaluating $\int_{-\infty}^{\infty}\frac{\ln\left(\frac{1}{2}+x+x^{2}\right)}{1+x^{2}}dx$ (Motivation) This is an integral I made up for fun. WolframAlpha doesn't seem to come up with a closed form for it and I'm surprised there doesn't seem to be a duplicate after using Approach0, but I believe it equals
$$\pi\ln\left(\frac{5}{2}\right)$$
(Question) Since there seems to be a nice-looking closed form, could there be a cooler or more elegant way of solving this other than my attempt below?
If anyone is willing to look over my attempt and provide constructive criticism, I would greatly appreciate it as someone who seeks to expand his skills in complex analysis.
(Attempt) Let $f(z) = \displaystyle\frac{\log(1/2+z+z^2)}{1+z^2}$. Its poles are $z \in \left\{i,-i\right\}$ and its branch points are $\displaystyle z \in \left\{-\frac{1}{2}+\frac{i}{2}, -\frac{1}{2}-\frac{i}{2}\right\}$. For simplicity's sake, let $\displaystyle z_a = -\frac{1}{2}+\frac{i}{2}$ and $\displaystyle z_b = -\frac{1}{2}-\frac{i}{2}$. With these, we can rewrite $f(z)$ as
$$
\begin{align}
\frac{\log\left(\frac{1}{2}+z+z^{2}\right)}{1+z^{2}} &= \frac{\log\left(\left(z-z_{a}\right)\left(z-z_{b}\right)\right)}{1+z^{2}} \\
&= \frac{1}{1+z^{2}}\left(\log\left(\left|z-z_{a}\right|\cdot\left|z-z_{b}\right|\right)+i\operatorname{arg}\left(\left(z-z_{a}\right)\left(z-z_{b}\right)\right)\right) \\
&= \frac{1}{1+z^{2}}\left(\log\left|z-z_{a}\right|+\log\left|z-z_{b}\right|+i\operatorname{arg}\left(z-z_{a}\right)+i\operatorname{arg}\left(z-z_{b}\right)\right).
\end{align}
$$
But for the equalities to hold, define $\displaystyle \operatorname{arg}(z-z_a) \in \left(\frac{3\pi}{4},\frac{11\pi}{4}\right)$ and $\displaystyle \operatorname{arg}(z-z_b) \in \left(-\frac{11\pi}{4},-\frac{3\pi}{4}\right)$. Here is a visual of the keyhole contour.
By Cauchy's Residue Theorem, we get
$$2\pi i \operatorname{Res}(f(z),z=i) = \left(\int_{-R}^{R}+\int_{\Gamma}+\int_{\lambda_1}+\int_{\gamma}+\int_{\lambda_2}\right)f(z)dz.$$
(I forgot to put this in the picture, but $\gamma$ has a small radius $r$ and the gaps between the branch cut and the $\lambda$s have a small length of $\epsilon$).
As $R \to \infty$ and $r \to 0$, it can be proved that $\displaystyle \int_{\Gamma}f(z)dz$ and $\displaystyle \int_{\gamma}f(z)dz$ go to $0$.
For the contour integrals over $\lambda_1$ and $\lambda_2$, let them approach the branch cut $\Lambda$. So
$$
\begin{align}
& \lim_{R \to \infty}\lim_{\lambda_1, \lambda_2 \to \Lambda}\left(\int_{\lambda_1} + \int_{\lambda_2}\right)f(z)dz \\
=& \lim_{R \to \infty}\lim_{\epsilon \to 0}\int_{Rz_{a}}^{z_{a}}f\left(z+i\epsilon\right)d\left(z+i\epsilon\right) + \lim_{R \to \infty}\lim_{\epsilon \to 0}\int_{z_{a}}^{Rz_{a}}f\left(z-i\epsilon\right)d\left(z-i\epsilon\right) \\
=& \lim_{R \to \infty}\lim_{\epsilon \to 0}\int_{Rz_{a}}^{z_{a}}\frac{d\left(z+i\epsilon\right)}{1+\left(z+i\epsilon\right)^{2}}\left(\log\left|z+i\epsilon-z_{a}\right|+\log\left|z+i\epsilon-z_{b}\right|+i\operatorname{arg}\left(z+i\epsilon-z_{a}\right)+i\operatorname{arg}\left(z+i\epsilon-z_{b}\right)\right) \\
&+ \lim_{R \to \infty}\lim_{\epsilon \to 0}\int_{Rz_{a}}^{z_{a}}\frac{d\left(z-i\epsilon\right)}{1+\left(z-i\epsilon\right)^{2}}\left(\log\left|z-i\epsilon-z_{a}\right|+\log\left|z-i\epsilon-z_{b}\right|+i\operatorname{arg}\left(z-i\epsilon-z_{a}\right)+i\operatorname{arg}\left(z-i\epsilon-z_{b}\right)\right) \\
=& -\int_{z_a}^{i\infty}\frac{dz}{1+z^2}\left(\log\left|z-z_{a}\right|+\log\left|z-z_{b}\right|+\frac{11\pi i}{4}+i\operatorname{arg}\left(z-z_{b}\right)\right) \\
&+ \int_{z_a}^{i\infty}\frac{dz}{1+z^2}\left(\log\left|z-z_{a}\right|+\log\left|z-z_{b}\right|+\frac{3\pi i}{4}+i\operatorname{arg}\left(z-z_{b}\right)\right) \\
=& -2\pi i \int_{z_a}^{i\infty}\frac{dz}{1+z^2} \\
&= i\pi\left(\frac{\pi}{2}+\arctan\left(\frac{1}{2}\right)\right)-\frac{\pi}{2}\ln\left(5\right). \\
\end{align}
$$
Next, we will evaluate the residue at the simple pole $z=i$ like this:
$$2\pi i \operatorname{Res}(f(z),z=i) = 2\pi i \lim_{z \to i}\frac{\left(z-i\right)\log\left(\frac{1}{2}+z+z^{2}\right)}{\left(z-i\right)\left(z+i\right)} = \pi\ln\left(\frac{\sqrt{5}}{2}\right)+i\pi\left(\frac{\pi}{2}+\arctan\left(\frac{1}{2}\right)\right).$$
Gathering everything together and taking the appropriate limits, we get
$$\pi\ln\left(\frac{\sqrt{5}}{2}\right)+i\pi\left(\frac{\pi}{2}+\arctan\left(\frac{1}{2}\right)\right) = \int_{-\infty}^{\infty}f\left(x\right)dx + 0 + i\pi\left(\frac{\pi}{2}+\arctan\left(\frac{1}{2}\right)\right)-\frac{\pi}{2}\ln\left(5\right) + 0.$$
In conclusion,
$$\int_{-\infty}^{\infty}\frac{\ln\left(\frac{1}{2}+x+x^{2}\right)}{1+x^{2}}dx = \pi\ln\left(\frac{5}{2}\right).$$
|
There is even an antiderivative.
Write
$$\frac{\log \left(x^2+x+\frac{1}{2}\right)}{x^2+1}=\frac{\log \left((x-a)(x-b)\right)}{(x+i)(x-i)}$$ wtih
$$a=-\frac{1+i}{2} \qquad \text{and} \qquad b=-\frac{1-i}{2}$$
$$\frac 1{(x+i)(x-i)}=\frac i 2\left(\frac{1}{x+i}-\frac{1}{x-i}\right)$$ and face four integrals looking like
$$\int \frac{\log(x-c)}{x-d}=\text{Li}_2\left(-\frac{x-c}{c-d}\right)+\log (x-c) \log \left(\frac{x-d}{c-d}\right)$$
Edit
As @Ninad Munshi suggested in comments, write
$$I(a)=\int_{-\infty}^{+\infty}\frac{\log \left(\left(x+\frac{1}{2}\right)^2+a\right)}{x^2+1}$$
$$I'(a)=\int_{-\infty}^{+\infty}\frac{dx}{\left(x^2+1\right)
\left(\left(x+\frac{1}{2}\right)^2+a\right)}$$
$$I'(a)=\frac{4 \pi \left(1+\frac{1}{\sqrt{a}}\right)}{4 a+8 \sqrt{a}+5}$$
$$\int_0^b I(a)\,da=\pi \log \left(\frac{4 b+8 \sqrt{b}+5}{5}\right) $$ and
$$I(0)= \pi \log \left(\frac{5}{4}\right)$$
So, as a total
$$\int_{-\infty}^{+\infty}\frac{\log \left(\left(x+\frac{1}{2}\right)^2+b\right)}{x^2+1}=\pi \log \left(b+2 \sqrt{b}+\frac{5}{4}\right)$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4642590",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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|
Investigate on the convergence of $I(a)= \int_0^{\infty} \frac{x}{\sqrt{1+x^a}} d x $ for $a>0$ and its exact value in case of convergence. Inspired by the post,
I start to investigate the convergence of $$I(a)= \int_0^{\infty} \frac{x}{\sqrt{1+x^a}} d x $$ for $a>0$.
For any $0\le a\le 4$, if $x\ge 1$, we have
$$
\frac{x}{\sqrt{1+x^a}}\ge\frac{x}{\sqrt{1+x^4}}=\frac{1}{\sqrt{x^2+\frac{1}{x^2}}}\ge \frac{1}{\sqrt{x^2+x^2}} =\frac{1}{x\sqrt2}
$$
Since $\int_0^{\infty} \frac{1}{x \sqrt{2}} d x$ is divergent, therefore $ I(a)$ is divergent for any $0\le a\le 4$.
For $a>4$, let $x=\tan ^{\frac{2}{a}} \theta$, then
$$
\begin{aligned}
I(a)& =\frac{2}{a} \int_0^{\frac{\pi}{2}} \frac{\tan ^{\frac{2}{a}} \theta}{\sec \theta} \cdot \tan ^{\frac{2}{a}-1} \theta \sec ^2 \theta d \theta \\
& =\frac{2}{a} \int_0^{\frac{\pi}{2}} \tan ^{\frac{4}{a}-1} \theta \sec \theta d \theta \\
& =\frac{2}{n} \int_0^{\frac{\pi}{2}} \sin ^{\frac{4}{a}-1} \theta \cos ^{-\frac{4}{a}} \theta d \theta \\
& =\frac{1}{a} B\left(\frac{2}{a}, \frac{1}{2}-\frac{2}{a}\right) \\
&=\frac{\Gamma\left(\frac{2}{a}\right) \Gamma\left(\frac{1}{2}-\frac{2}{a}\right)}{a \sqrt{\pi}}
\end{aligned}
$$
which are convergent $ \Leftrightarrow \frac{1}{2}-\frac{2}{a}>0 \Leftrightarrow a>4$.
Comments and alternative solutions are highly appreciated!
|
The antiderivative
$$\int \frac{x}{\sqrt{1+x^a}} dx=\frac{1}{2} x^2 \,
_2F_1\left(\frac{1}{2},\frac{2}{a};\frac{a+2}{a};-x
^a\right)$$ leads to the same result for the definite integral.
What is interesting to notice is that, if $a=4+\epsilon$, the definite integral can be approximated as
$$\int_0^\infty \frac{x}{\sqrt{1+x^{4+\epsilon}}} dx=\frac{2}{\epsilon }+\frac{\log (2)}{2}+\frac{\pi ^2-6 (2-\log (2))
\log (2)}{96} \epsilon +O\left(\epsilon ^2\right)$$
Making $\epsilon=1$ gives
$$\frac{\pi ^2}{96}+\frac{32+6 \log (2)+\log ^2(2)}{16}= 2.39277$$ while
$$\frac{\Gamma \left(\frac{1}{10}\right) \Gamma
\left(\frac{7}{5}\right)}{2 \sqrt{\pi }}=2.38116$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Why is the volume of a sphere $\frac{4}{3}\pi r^3$? I learned that the volume of a sphere is $\frac{4}{3}\pi r^3$, but why? The $\pi$ kind of makes sense because its round like a circle, and the $r^3$ because it's 3-D, but $\frac{4}{3}$ is so random! How could somebody guess something like this for the formula?
|
Let us consider the following sphere of radius r with an inscribed pyramid with base area G:
We can calculate the volume of the inscribed pyramid as follows:
$V_{pyramid} = \frac{1}{3} \cdot G \cdot r$
Notice that we can now express $V_{sphere}$ as the sum of the volume of infinitely small pyramids over the total surface area of the sphere, whereby $A_{sphere} = 4 \cdot \pi \cdot r^2$. When doing so, we get the following result:
$V_{sphere} = \sum volume\:of\:pyramid \\
= \sum \frac{1}{3} \cdot G \cdot r \\
= \sum \frac{1}{3} \cdot (4\pi r^2) \cdot r \\
= \frac{4}{3} \pi r^3$
So what do we take from this? The factor $\frac{4}{3}$ can be intuitively considered as a composition of the factor $4$ deriving from the area of the sphere and the factor $\frac{1}{3}$ coming from the volume of a pyramid.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 19,
"answer_id": 18
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|
Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$
Does it converge? If so, what is its sum?
|
I want to use the infinite series expansion of the integral of a function to compute the sum. If you repeat the process of integration by parts over and over again:
$$\int f(x) dx = xf(x)-\int xf'(x)dx = xf(x) - \frac{x^2}{2}f'(x)+\int \frac{x^2}{2}f''(x)dx$$
$$\Rightarrow \int f(x) dx = xf(x) - \frac{x^2}{2}f'(x) + \frac{x^3}{6}f''(x)-\int \frac{x^3}{6}f''(x)dx$$
$$\Rightarrow \int f(x) dx = xf(x) - \frac{x^2}{2}f'(x) + \frac{x^3}{6}f''(x)-\frac{x^4}{24}f'''(x)+ \int \frac{x^4}{24}f'''(x)dx$$
Continuing this pattern we can prove using Mathematical Induction (for all $n\ \epsilon\ \mathbb{Z}^+$)that:
$$\int f(x)dx = \sum_{k=0}^{n}\left[\frac{(-1)^kf^{(k)}(x)x^{k+1}}{(k+1)!}\right]+\frac{(-1)^n}{n!}\int x^n f^{(n)}(x)dx \rightarrow (1)$$
Limiting both sides of equation (1) as $n \rightarrow \infty$ we can write the infinite series expansion:
$$\int f(x)dx = \sum_{k=0}^{\infty}\left[\frac{(-1)^kf^{(k)}(x)x^{k+1}}{(k+1)!}\right] + C=C+xf(x) - \frac{x^2}{2}f'(x) + \frac{x^3}{6}f''(x)-+...\rightarrow (2)$$
We can certainly remove the integration constant $C$ from both sides by taking the definite integral:
$$\int_{0}^{x} f(t)dt =xf(x) - \frac{x^2}{2}f'(x) + \frac{x^3}{6}f''(x)-\frac{x^4}{24}f'''(x)+-...\rightarrow (3)$$
We can prove the Taylor series expansion for functions using this idea. Now let $f(x)=\frac{1}{1+x}$ and we get:
$$\ln(1+x)=\frac{1}{1+\frac{1}{x}}+\frac{1/2}{(1+\frac{1}{x})^2}+\frac{1/3}{(1+\frac{1}{x})^3}+\frac{1/4}{(1+\frac{1}{x})^4}+\frac{1/5}{(1+\frac{1}{x})^5}+...\rightarrow (4)$$
$$y=\frac{x}{x+1} \Rightarrow \ln (\frac{1}{1-y})=y+\frac{y^2}{2}+\frac{y^3}{3}+\frac{y^4}{4}+\frac{y^5}{5}+...\rightarrow (5)$$
By letting $y=-1$ and multiplying both sides of equation $(5)$ by $-1$ we get the sum we desired:
$$\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5}-\frac{1}{6}+-...$$
P.S. You can also use $(4)$ to get another interesting expansion for $\ln 2$ by putting $x=1$:
$$\ln 2 = \frac{1}{1 \cdot 2^1}+\frac{1}{2 \cdot 2^2}+\frac{1}{3 \cdot 2^3}+\frac{1}{4 \cdot 2^4}+\frac{1}{5 \cdot 2^5}+...$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Summing the series $ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \cdots \ \text{ad inf}$ How does one sum the given series: $$ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{2n+5} + \frac{ 1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{1}{2n+7} + \cdots \ \text{ad inf}$$
Given, such a series, how does one go about solving it. Getting an Integral Representation seems tough for me.
I thought of going along these lines, Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$ but couldn't succeed.
|
Observe that
$$\displaystyle \frac{1}{4^n} {2n \choose n} = \frac{(2n-1)(2n-3)(2n-5)...}{2n(2n-2)(2n-4)...}$$
and recall that
$$\displaystyle \frac{1}{ \sqrt{1 - x^2} } = \sum_{k \ge 0} \frac{1}{4^k} {2k \choose k} x^{2k}.$$
It follows that the desired quantity is
$$\displaystyle \int_0^1 \frac{x^{2n}}{\sqrt{1 - x^2}} dx.$$
But letting $x = \sin \theta$ this is just
$$\displaystyle \int_0^{ \frac{\pi}{2} } \sin^{2n} \theta d \theta = \frac{\pi}{2} \frac{1}{4^n} {2n \choose n}.$$
This is equivalent to Mariano's closed form via the identities $\Gamma(x+1) = x \Gamma(x)$ and $\Gamma \left( \frac{1}{2} \right) = \sqrt{\pi}$. I should mention that the integral at the end of Moron's answer is quite doable; write it in terms of a cosine and use the substitution $\cos \theta = \frac{1 - t^2}{1 + t^2}$ where $t = \tan \frac{\theta}{2}$ to reduce the problem to the integral of a rational function, and then one can use one of several related methods (partial fractions, contour integration).
Remark: The last integral identity above happens to be one of my favorite identities. I describe a representation-theoretic and combinatorial proof of it in this blog post. Another approach implicitly occurs in this blog post.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do you show that $d\theta = \frac{x dy - y dx }{x^2 + y^2}$? If $(r, \theta)$ are polar coordinates on $\mathbb{R}^2\setminus \{ (0,0)\}$, then how do I show/prove that
\begin{equation*}
d\theta =\dfrac{x dy - y dx}{x^2 + y^2}?
\end{equation*}
|
On $R^2\setminus\{0\}$ we have $\theta=\Im(\ln(x+iy))$ (here $\Im$ stands for "the imaginary part of"). Therefore $$d\theta = \Im\left(\frac1{x+iy}\right)dx+\Im\left(\frac{i}{x+iy}\right)dy=\Im\left(\frac{x-iy}{x^2+y^2}\right)dx+\Im\left(\frac{i(x-iy)}{x^2+y^2}\right)dy=$$$$\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy=\frac{xdy-ydx}{x^2+y^2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/6083",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluating the nested radical $ \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots}}} $. How does one prove the following limit?
$$
\lim_{n \to \infty}
\sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots \sqrt{1 + (n - 1) \sqrt{1 + n}}}}}
= 3.
$$
|
Let me provide a full and simple proof here (6 years later)
Set, for $m<n$
$$
a_{m,n}=\sqrt{1+m\sqrt{1+(m+1)\sqrt{1+\cdots+(n-1)\sqrt{1+n}}}}.
$$
We shall first show that $$a_{m,n}<m+1,\tag{1}$$ in the following way using
Backwards induction. Clearly,
$a_{n,n}=\sqrt{1+n}<1+n$, and if $a_{k+1,n}<k+2$, for some $k<n$, then
$$
a_{k,n}=\sqrt{1+ka_{k+1,n}}<\sqrt{1+k(k+2})=\sqrt{k^2+2k+1}=k+1.
$$
In particular, in the same way we can show that
$$
m+1=\sqrt{1+m\sqrt{1+(m+1)\sqrt{1+\cdots+(n-1)\sqrt{1+{\color{red}{n^2+2n}}}}}}
$$
Next, observe that
$$
0<m+1-a_{m,n}= \\
=\sqrt{1+m\sqrt{1+\cdots+(n-1)\sqrt{1+{\color{red}{n^2+2n}}}}}
-\sqrt{1+m\sqrt{1+\cdots+(n-1)\sqrt{1+{n}}}} \\
=\frac{m\Big(\sqrt{1+(m+1)\sqrt{1+\cdots+(n-1)\sqrt{1+{\color{red}{n^2+2n}}}}}
-\sqrt{1+(m+1)\sqrt{1+\cdots+(n-1)\sqrt{1+{n}}}}\Big)}{\sqrt{1+m\sqrt{1+\cdots+(n-1)\sqrt{1+{\color{red}{n^2+2n}}}}}
+\sqrt{1+m\sqrt{1+\cdots+(n-1)\sqrt{1+{n}}}}} \\
<\frac{m}{m+2}(m+2-a_{m+1,n}) <\cdots < \frac{m(m+1)\cdots (n-1)}{(m+2)(m+3)\cdots(n+1)}(n+1-a_{n,n})=\frac{m(m+1)(n+1-\sqrt{n+1})}{n(n+1)}<\frac{m(m+1)}{n}.
$$
Thus
$$
\lim_{n\to\infty}a_{m,n}=m+1.
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/7204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Does Stirling's formula give the correct number of digits for $n!\phantom{}$? It is known that the number of digits of a natural number $n > 0$, which represent by $d(n)$ is given by:
$d(n)= 1 + \lfloor\log n\rfloor\qquad (\text{I})$
($\log$ indicates $\log$ base $10$)
Well .. the classical approach to the Stirling factorial natural number $n > 1$ is given by:
$$n! \approx f(n) = [(2n\pi) ^{1/2}] [(n / e) ^ n]$$
The number of digits $n!$, according to equality (I), is:
$d(n!) = 1 + \lfloor\log n!\rfloor$
It seems to me that for all natural $n> 1$, $\log n!$ and $\log [f (n)]$ have the same floor:
$$\lfloor\log(n!)\rfloor = \lfloor\log(f(n))\rfloor$$
Here's my big question!
Therefore, we could write:
$$d (n!) = 1 + \lfloor\log(f(n))\rfloor$$
Hope someone has some little time for the theme.
|
Here are some thoughts on the conjecture that may lead one to suspect that it is true. This is not a proof that it is true.
We want to know whether
$$ \left\lfloor \log_{10} n! \right\rfloor = \left\lfloor \log_{10} \lfloor f(n) \rfloor \right\rfloor$$
is true for all $n > 1.$
We note that this would NOT be true if the interval
$I_n = ( \log_{10} n!,\log_{10} \lfloor f(n) \rfloor)$ contains an integer but is true otherwise.
So let's look at the length of $I_n.$
From the Stirling series we have
$$\frac{1}{\log_{e}10} \left( \frac{1}{12n} - \frac{1}{360n^3} \right) <
\log_{10} n! - \log_{10} f(n) <
\frac{1}{\log_{e}10} \frac{1}{12n}.$$
And so taking the integer part of $f(n)$ we have
$$\frac{1}{\log_{e}10} \left( \frac{1}{12n} - \frac{1}{360n^3} \right) <
\log_{10} n! - \log_{10} \lfloor f(n) \rfloor <
\frac{1}{\log_{e}10} \left( \frac{1}{12n} + \frac{1}{ \lfloor f(n) \rfloor } \right),$$
where to achieve the RHS we note that
$$ \log_{10} f(n) - \log_{10} \lfloor f(n) \rfloor < \frac{1}{ \lfloor f(n) \rfloor \log_e 10 }.$$
Hence
$$\text{Length}(I_n) < \frac{1}{\log_{e}10} \left( \frac{1}{360n^3} +
\frac{1}{ \lfloor f(n) \rfloor } \right).$$
We can verify that the conjecture holds for $n=2,3,\ldots,10,$ so summing up the remaining lengths of the $I_n$ we have
$$\sum_{n=11}^\infty \text{Length}(I_n) < \frac{1}{360 \log_e 10}
\left( \sum_{n=11}^\infty \frac{1}{n^3} + \sum_{n=11}^\infty \frac{1}{ \lfloor f(n) \rfloor } \right).$$
Now $ \lfloor f(n) \rfloor \ge (n-1)! $ and so, doing some calculations (replacing the
$ \lfloor f(n) \rfloor $ on the RHS by $ (n-1)! $ , we have
$$\sum_{n=11}^\infty \text{Length}(I_n) < \frac{1}{360 \log_e 10}
\left( 0.00452492 + 0.00000030 \right) < 5.5 \times 10^{-6}.$$
Hence the probability that an integer falls in any of the intervals is very small.
|
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|
Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However, doing the following gets a completely different answer:
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{\sin x}{\sin^2(x)\cos x} dx\\
&=&\int \frac{\sin x}{(1-\cos^2(x))\cos x} dx.
\end{eqnarray*}
let $u=\cos x, du=-\sin x dx$; then
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{-1}{(1-u^2)u} du\\
&=&\int \frac{-1}{(1+u)(1-u)u}du\\
&=&\int \left(\frac{-1}{u} - \frac{1}{2(1-u)} + \frac{1}{2(1+u)}\right) du\\
&=&-\ln|\cos x|+\frac{1}{2}\ln|1-\cos x|+\frac{1}{2}\ln|1+\cos x|+C
\end{eqnarray*}
I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?
|
Another way to integrate would be to let $\displaystyle t = \frac{\sin{x}}{\cos{x}}$, then $\displaystyle \frac{dt}{dx} = \frac{1}{\cos^2{x}} \Rightarrow dx = \cos^2{x}\;{dt}$.
Thus $ \displaystyle I = \int\frac{\cos^2{x}}{\sin{x}}\;{dt} = \int\frac{\cos{x}}{\sin{x}}\;{dt} = \int\frac{1}{t}\;{dt} = \ln{t}+k = \ln\frac{\sin{x}}{\cos{x}}+k$.
|
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|
$n^2 + 3n +5$ is not divisible by $121$ Question:
Show that $n^2 + 3n + 5$ is not divisible by $121$, where $n$ is an integer.
|
Make a contradiction that $n^2 + 3n + 5$ is divisible by $121$
Let $k$ be any positive integer, we can say that
$n^2 + 3n + 5 = 121\cdot k$
$n^2 + 3n + (5 - (121\cdot k)) = 0$
Solve for $n$,
$$\begin{align}
n=&\frac{-3 \pm \sqrt {(3)^2 - 4\cdot1\cdot(5-(121\cdot k))}}{2\cdot1}\\
n=&\frac{-3 \pm \sqrt {(484\cdot k)-11}}{2}
\end{align}$$
Given that $n$ is an integer, so $\sqrt {(484\cdot k)-11}$ should be an integer
We can represent $\sqrt {(484\cdot k)-11}$ as $(\sqrt{11}\cdot \sqrt{(44\cdot k)-1})$, whose value can't be an integer as value of $\sqrt{11}$ is irrational.
So we can say that our assumption is wrong, $n^2 + 3n + 5$ is not divisible by $121$.
|
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|
$x^y = y^x$ for integers $x$ and $y$ We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($x\neq y$) which satisfies the equality $x^y = y^x$?
|
Although this thing has already been answered, here a shorter proof
Because $x^y = y^x $ is symmetric we first demand that $x>y$
Then we proceed simply this way:
$ x^y = y^x $
$ x = y^{\frac x y } $
$ \frac x y = y^{\frac x y -1} $
$ \frac x y -1 = y^{\frac x y -1} - 1 $
Now we expand the rhs into its well-known exponential-series
$ \frac x y -1 = \ln(y)*(\frac x y -1) + \frac {((\ln(y)*(\frac x y -1))^2}{2!} + ... $
Here by the definition x>y the lhs is positive, so if $ \ln(y) $ >=1 we had lhs $\lt$ rhs
Thus $ \ln(y) $ must be smaller than 1, and the only integer y>1 whose log is smaller than 1 is y=2, so there is the only possibility $y = 2$ and we are done.
[update] Well, after having determined $y=2$ the same procedure can be used to show, that after manyally checking $x=3$ (impossible) $x=4$ (possible) no $x>4$ can be chosen.
We ask for $x=4^{1+\delta} ,\delta > 0 $ inserting the value 2 for y:
$ 4^{(1+\delta)*2}=2^{4^{(1+\delta)}} $
Take log to base 2:
$ (1+\delta)*4=4^{(1+\delta)} $
$ \delta =4^{\delta} - 1 $
$ \delta = \ln(4)*\delta + \frac { (\ln(4)*\delta)^2 }{2!} + \ldots $
$ 0 = (\ln(4)-1)*\delta + \frac { (\ln(4)*\delta)^2 }{2!} + \ldots $
Because $ \ln(4)-1 >0 $ this can only be satisfied if $ \delta =0 $
So indeed the only solutions, assuming x>y, is $ (x,y) = (4,2)$ .
[end update]
|
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|
Summing up the series $a_{3k}$ where $\log(1-x+x^2) = \sum a_k x^k$ If $\ln(1-x+x^2) = a_1x+a_2x^2 + \cdots \text{ then } a_3+a_6+a_9+a_{12} + \cdots = $ ?
My approach is to write $1-x+x^2 = \frac{1+x^3}{1+x}$ then expanding the respective logarithms,I got a series (of coefficient) which is nothing but $\frac{2}{3}\ln 2$.But this approach took some time for me (I can't solved it during the test but after the test I solved it)... Any other quick method?
|
Lemma: Let $f(x) = \sum a_n x^n$ be a power series. Then
$$\sum a_{3n} x^{3n} = \frac{f(x) + f(\omega x) + f(\omega^2 x)}{3}$$
where $\omega = e^{ \frac{2\pi i}{3} }$.
Proof. Ignoring convergence it suffices to prove this for a single term, and then it boils down to the identity
$$\frac{1 + \omega^n + \omega^{2n}}{3} = \begin{cases} 1 \text{ if } 3 | n \\\ 0 \text{ otherwise} \end{cases}.$$
This is a special case of the discrete Fourier transform. Applying the lemma, we readily obtain that the desired sum is
$$\frac{\ln 1 + \ln (1 - \omega + \omega^2) + \ln (1 - \omega^2 + \omega)}{3} = \frac{1}{3} \ln (2 \omega^2 \cdot 2 \omega) = \frac{2}{3} \ln 2$$
where we use the fact that $1 + \omega + \omega^2 = 0$.
|
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|
Is this definite integral really independent of a parameter? How can it be shown? I want to find a nice simple expression for the definite integral
$$\int_0^\infty \frac{x^2\,dx}{(x^2-a^2)^2 + x^2}$$
Now, I can numerically compute this integral, and it seems to converge to $\pi/2$ for all real values of $a$. Is this integral actually always equal to $\pi/2$? How can I show this?
Also, why does Wolfram Alpha give me something that appears to depend on $a$? Is there a good reason it doesn't eliminate $a$?
|
Yes it is true!
Let
$$\displaystyle I = \int_{0}^{\infty} \dfrac{x^2}{x^2 + (x^2-a^2)^2} \ \text{dx}$$
Make the substitution $\displaystyle x = \dfrac{a^2}{t}$
We get
$$\displaystyle I = \int_{0}^{\infty} \dfrac{a^6}{t^4\left(\dfrac{a^4}{t^2} + \left(\dfrac{a^4}{t^2} - a^2\right)^2\right)} \ \text{dt} = \int_{0}^{\infty} \dfrac{a^2}{t^2 + (a^2 - t^2)^2} \ \text{dt}$$
i.e.
$$\displaystyle I = \int_{0}^{\infty} \dfrac{a^2}{x^2 + (a^2 - x^2)^2} \ \text{dx}$$
Therefore
$$\displaystyle 2I = \int_{0}^{\infty} \dfrac{x^2}{x^2 + (x^2-a^2)^2} \ \text{dx} + \int_{0}^{\infty} \dfrac{a^2}{x^2 + (a^2 - x^2)^2}\ \text{dx}$$
$$ = \int_{0}^{\infty} \dfrac{x^2 + a^2}{x^2 + (x^2-a^2)^2} \ \text{dx}$$
$$\displaystyle = \int_{0}^{\infty} \dfrac{1 + \dfrac{a^2}{x^2}}{1 + \left(x-\dfrac{a^2}{x}\right)^2} \ \text{dx}$$
Making the substitution $\displaystyle t = x - \dfrac{a^2}{x}$
Gives us
$$\displaystyle 2I = \int_{-\infty}^{\infty} \dfrac{\text{dt}}{1 + t^2} = \pi$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Find the vertical asymptote of a function For an assignment, I was asked to find the vertical asymptote of the function $$g(x)= \frac{\frac{1}{2}x^3-4x^2+6x}{7x^2-56x+84}.$$
According to my text, a reliable method of finding the asymptote is to factor the numerator and denominator, and what left in the denominator that was not cancelled out is the asymptote.
I factored the numerator to $\frac{1}{2}(x^2-6x)(x-2)$, and the denominator factored to $7(x-2)(x-6)$, therefore $(x-2)$ cancelled out, leaving $(x-6)$ in the denominator.
However, 6 was not accepted as the answer, and I would like to know why.
|
You did not finish factoring the numerator. That's where the problem is.
You started out all right, but then you had to keep factoring that $x^2-6x$: it is not degree $1$, and it is not irreducible quadratic, so it can still be factored. In fact, it has a factor of $x$. So you really have:
$$\frac{1}{2}x^3 - 4x^2+6x = \frac{1}{2}\left(x^3 - 8x^2 + 12x\right) = \frac{1}{2}x\left(x^2-8x+12\right) = \frac{1}{2}x(x-6)(x-2).$$
In fact, both $x-2$ and $x-6$ cancel out in $g(x)$.
As a function, $g(x)$ is equal to
$$g(x) = \left\{\begin{array}{ll}
\frac{1}{14}x &\mbox{if $x\neq 2$ and $x\neq 6$,}\\
\mbox{undefined} &\mbox{if $x=2$ or $x=6$.}
\end{array}\right.$$
So it does not have any vertical asymptotes.
|
{
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How to compute the following definite integral Studying some integral table, I came across the following definite integral
$$\int_0^{\pi} \log [ a^2 + b^2 -2 a b \cos \phi ]\,d\phi$$ for $a,b \in \mathbb{R}$. Does somebody know a nice way to get the results?
|
Here is a different way, and is strangely the first thing that came to my mind. It uses generating series, and power series as well as other various techniques:
(Warning!: It is significantly more complicated)
Assume that $a\geq b\geq0$ without loss of generality. Notice that our integral is $$\pi \log(a^2+b^2)+\int_{0}^{\pi}\log\left( 1-\frac{2ab\cos\phi}{a^2+b^2}\right). $$ Since $\log(1-x)=-\sum_{i=1}^{\infty}\frac{x^{i}}{i}$ we have
$$\int_{0}^{\pi}\log\left( 1-\frac{2ab\cos\phi}{a^2+b^2}\right)d\phi=-\int_{0}^{\pi}\sum_{i=1}^{\infty} \frac{1}{i}\left(\frac{2ab\cos\phi}{a^2+b^2}\right)^i d\phi$$
Switch the order or integration and summation to find: (More on why this is legal at the end)
$$-\int_{0}^{\pi}\sum_{i=1}^{\infty} \frac{1}{i}\left(\frac{2ab\cos\phi}{a^2+b^2}\right)^i d\phi=-\sum_{i=1}^{\infty} \frac{1}{i}\left(\frac{2ab}{a^2+b^2}\right)^i \int_{0}^{\pi} \cos^i\phi d\phi$$
For $n$ odd we have $\int_0^\pi \cos^n xdx=0$. For even $n=2r$ we see by integration by parts that $\int_{0}^{\pi}\cos(x)^{2r}dx=\frac{2r-1}{2r}\int_{0}^{\pi}\cos(x)^{2r-2}dx$, so that induction yields $$\int_{0}^{\pi}\cos(x)^{2r}dx=\pi\prod_{i=1}^{r}\frac{2i-1}{2i}=\pi\frac{\left({2r\atop r}\right)}{4^{r}}$$
Thus our sum becomes
$$-\pi \sum_{r=1}^{\infty} \frac{1}{2r}\left(\frac{ab}{a^2+b^2}\right)^{2r} \left({2r\atop r}\right ) $$
Now let $$f(z)=\sum_{r=1}^{\infty} \frac{1}{r} \left({2r\atop r}\right) \frac{z^r}{4^r}$$
Then we need to find $-\frac{\pi}{2} f\left(\left( \frac{2ab}{a^2+b^2} \right)^2\right)$.
Let $Y=\left( \frac{2ab}{a^2+b^2} \right)^2$.
Recall the generating series $$\sum_{n=0}^{\infty}\left({2n\atop n}\right)x^{n}=\frac{1}{\sqrt{1-4x}}. $$ Differentiating yields $1+xf^{'}(x)=\frac{1}{\sqrt{1-x}}$ and hence $f(Y)=\int_{0}^{Y}\frac{1-\sqrt{1-x}}{x\sqrt{1-x}}dx$.
Make the substitution $x=\sin^{2}(u)$, and we get $$f(Y)=\int_{0}^{\theta}\frac{1-\cos(u)}{\sin^{2}(u)\cos(u)}2\sin(u)\cos(u)du=2\int_{0}^{\theta}\csc(u)-\cot(u)du$$ where $\theta$ is in the first quadrant and satisfies $\sin (\theta)=\frac{2ab}{a^2+b^2}.$
Then since $\int\csc x\, dx=-\log\left|\csc x+\cot x\right|$, and $\int\cot x dx=\log\left|\sin x\right|$, we see that $\int\csc(u)-\cot(u)du=-\log|\sin(u)\csc(u)+\sin(u)\cot u|=-\log|1+\cos(u)|$. Thus $$f(Y)=-2\log|1+\cos(u)|\biggr|_{u=0}^{u=\theta}$$
Drawing the triangle with sides $2ab$, $a^2-b^2$ and $a^2+b^2$ tells us that $\cos(\theta)=\frac{a^2-b^2}{a^2+b^2}$, and hence $$f(Y)=2\log 2- 2\log \left(1+\frac{a^2-b^2}{a^2+b^2}\right)$$
Thus the final answer is $$\pi \log(a^2+b^2)-\frac{\pi}{2}f(Y)=\pi\log(a^2+b^2)-\pi \log 2 +\pi\log\left(1+\frac{a^2-b^2}{a^2+b^2}\right)=2\pi\log(a)$$
|
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"timestamp": "2023-03-29T00:00:00",
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$n^3 \equiv n^5 \pmod{12} $? I am proving that
$$5n^3 + 7n^5 \equiv 0 \pmod{12}$$
It would suffice to show
$$n^3 \equiv n^5 \pmod{12}$$
How would I go about doing that?
I suppose I could just go through each $n \equiv r \pmod{12}$ with $r$ from $1$ to $11$ and show that $n^3 \equiv n^5 \pmod{12}$ for each, but that would be tedious. Surely there's a better way.
|
*
*$n^2 = 0,1,4$ or $9 \mod 12$ by checking each number $0,1,2,3,4,5,6$ (we don't have to check $7,8,9,10$ and $11$ since they are negatives and $x^2 = (-x)^2).$
*$n^4 = (n^2)^2 = n^2 \mod 12$ by checking each number 0,1,4 and 9.
*thus $n^3 = n^5.$
|
{
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How do I come up with a function to count a pyramid of apples? My algebra book has a quick practical example at the beginning of the chapter on polynomials and their functions. Unfortunately it just says "this is why polynomial functions are important" and moves on. I'd like to know how to come up with the function (the teacher considers this out of scope). Even suggesting google search terms to find out more information on this sort of thing would be helpful.
The example
Consider a stack of apples, in a pyramid shape. It starts with one apple at the top. The next layer is 2x2, then 3x3, and so on. How many apples are there, given x number of layers?
The polynomial function
$$f(x) = \frac{2x^3 + 3x^2 + x}{6}$$
What I do and don't understand
Thanks to @DJC, I now know this is a standard function to generate Square Pyramidal Numbers, which is part of Faulhaber's formula. Faulhaber's formula appears to be about quickly adding sequential coefficients which all have the same exponent. Very cool. But how does one get from:
$$\sum_{k=1}^{n} k^p$$
to the spiffy function above? If I'm sounding stupid, how do I make the question better?
Fwiw, I'm in intermediate algebra in the USA. The next course would be Trigonometry or Calculus. (to help people place my current knowledge level)
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The first layer has $1^2$ elements; the second has $2^2$ elements; the next has $3^2$ elements, and so on.
So this is a matter of finding the formula for
$$1^2 + 2^2 + 3^2 + \cdots + n^2$$
in terms of $n$.
For squares, this is a well known formula: $1^2+2^2+\cdots +n^2 = \frac{n(n+1)(2n+1)}{6}$.
Note that
$$\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} = \frac{2n^3 + 3n^2 + n}{6} = \frac{n(2n^2+3n+1)}{6} = \frac{n(2n+1)(n+1)}{6}$$
same formula.
Of course, you might be more interested in how to figure out the formulas in the first place. How does one express $1^k + 2^k + 3^k + \cdots + n^k$?
This is known as Faulhaber's formula. For the general case,
$$1^p + 2^p + \cdots + n^p = \frac{1}{p+1}\sum_{j=0}^p(-1)^jB_j\binom{p+1}{j} n^{p+1-j}$$
where $B_m$ is the $m$th Bernoulli number.
For $p=2$, you get the spiffy formula with:
\begin{align*}
1^2+\cdots+n^2 &= \frac{1}{3}\left(\binom{3}{0}B_0n^3 - \binom{3}{1}B_1n^2 + \binom{3}{2}B_2n - \binom{3}{3}B_3\right)\\
&= \frac{1}{3}\left(B_0n^3 - 3B_1n^2 + 3B_2n - B_3\right)\\
&= \frac{1}{3}\left(n^3 - 3\left(-\frac{1}{2}\right)n^2 + 3\left(\frac{1}{6}\right)n - 0\right)\\
&= \frac{1}{3}n^3 +\frac{1}{2}n^2 + \frac{1}{6}n\\
&= \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6},
\end{align*}
using the fact that $B_0 = 1$ $B_1 = -\frac{1}{2}$, $B_2 = \frac{1}{6}$, and $B_3=0$ (in fact, $B_n = 0$ for all odd $n\gt 1$).
|
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Why does the sum of a division applied to individual items not equal the division applied to the sum of those items? When $a_2/a_1 = b_2/b_1$, $a_1 \neq b_1$, we have
$$\frac{a_{1}}{a_{2}/a_{1}}+\dfrac{b_{1}}{b_{2}/b_{1}}=
\frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}.$$
So why when $a_2/a_1 \neq b_2/b_1 , a_1 \neq b_1$ we don't have a similar equality?
$$\frac{a_{1}}{a_{2}/a_{1}}+\dfrac{b_{1}}{b_{2}/b_{1}}\neq \frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}?$$
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I'm not sure where the figures are coming from, but an interpretation is the inequality ($a,b,c > 0$)
$$\frac{a}{a+b} + \frac{a}{a+c} \ge \frac{2a}{a+ (b+c)/2}, \quad (1)$$
which follows from
$$((a+b)-(a+c))^2 \ge 0$$
and so
$$(a+b)^2+(a+c)^2 \ge 2(a+b)(b+c).$$
Therefore, on adding $2(a+b)(b+c)$ to both sides,
$$((a+b)+(a+c))^2 \ge 4(a+b)(b+c)$$
and dividing both sides by $((a+b)+(a+c))(a+b)(b+c)$ and multiplying by $a$ will give us $(1).$
|
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Proving $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^{2}$ for all positive integers $n$
Prove $(2n+1) + (2n+3) + (2n+5) +
\cdots + (4n-1) = 3n^{2}$ for all
positive integers $n$.
So the provided solution avoids induction and makes use of the fact that $1 + 3 + 5 + \cdots + (2n-1) = n^{2}$ however I cannot understand the first step: $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1) = (1 + 3 + 5 + \cdots + (4n-1)) -(1 + 3 + 5 + \cdots + (2n-1))$. Once that has been established I can follow the rest, but I was hoping someone could help me understand why $(1 + 3 + 5 + \cdots + (4n-1))$ which already looks like less than the LHS can be made equal to the LHS by subtracting a positive number.
Additionally, I wanted to prove the equality using induction, but had trouble with that as well. I think I am thrown off by the last term $4n-1$. Even the initial case of $n=1$ is not totally clear to me: does it hold because $2(1)+1 = 3(1)^{2}$ or is it because $4(1)-1 = 3(1)^{2}$?
Either way, my approach was to replace every $n$ on the LHS with $n+1$ which resulted in: $$(2(n+1)+1) + (2(n+1)+3)+ \cdots + (4(n+1)-1)$$ and I am not sure what the second to last term in the sequence would be... I tried simplifying anyway $$(2n+3) + (2n+5) + \cdots + (4n+3)$$ and I this point I thought I could subtract $(2n+1)$ from both sides of the induction assumption resulting in $(2n+3) + (2n+5) + \cdots ? = 3n^{2} - (2n+1) - (4n-1)$ substituting this yields: $$3n^{2} - 2n - 1 - 4n + 1 + 4n + 3 = 3n^{2} + 6n + 3 - 8n + 1 = 3(n+1)^{2} - 8n + 1$$ but I guess I don't want the $-8n + 1$... I also tried substituting $3n^{2} - (2n+1)$ without that last term being subtracted, but that did not work out either. If anyone can help me understand how to do this properly I would really appreciate it. Thanks!
|
The simple solution using $1+3+\ldots+(2n-1)=n^2$ is as follows:
$(2n+1)+(2n+3)+\ldots+(4n-1)=2n\cdot n+ (1+3+\ldots+(2n-1))=3n^2$
The fact that $1+3+\ldots+(2n-1)=n^2$ is true is a simple induction: Assume it's true for $n$. Then for $n+1$ we get $1+3+\ldots+(2n-1)+(2(n+1)-1)=n^2+2n+1=(n+1)^2$
|
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|
How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$? How can I evaluate
$$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$?
I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.
In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$
|
If you want a solution that doesn't require derivatives or integrals, notice that
\begin{eqnarray}
1+2x+3x^2+4x^3+\dots = 1 + x + x^2 + x^3 + \dots \\
+ x + x^2+ x^3 + \dots\\
+ x^2 + x^3 + \dots \\
+x^3 + \dots \\
+ \dots \\
=1 + x + x^2 + x^3+\dots \\
+x(1+x+x^2+\dots) \\
+x^2(1+x+\dots)\\
+x^3(1+\dots)\\
+\dots \\
=(1+x+x^2+x^3+\dots)^2=\frac{1}{(1-x)^2}
\end{eqnarray}
|
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|
Finding the cos angle between two matrices using the euclidean inner product I wanted to know if I did this problem right or not.
$A$ and $B$ are the following matrices:
$A=\begin{bmatrix}2&6\\1&-3\end{bmatrix},$
$B=\begin{bmatrix}3&2\\1&0\end{bmatrix}.$
Then
$$\frac{\langle A,B\rangle}{\|A\|\|B\|}
=\frac{\left\langle\begin{bmatrix}2&6\\1&-3\end{bmatrix}, \begin{bmatrix}3&2\\1&0\end{bmatrix}\right\rangle}{\sqrt{\left\langle\begin{bmatrix}2&6\\1&-3\end{bmatrix},\begin{bmatrix}2&6\\1&-3\end{bmatrix}\right\rangle}\sqrt{\left\langle\begin{bmatrix}3&2\\1&0\end{bmatrix},\begin{bmatrix}3&2\\1&0\end{bmatrix}\right\rangle}}.$$
I ended up getting:
$\arccos(19/184) = 84.07^\circ$
|
$a\cdot b=|a||b|\cos\theta$ so
$$\theta=\arccos\Big(\frac{a\cdot b}{|a||b|}\Big).$$
we have
$a\cdot b=6+12+1=19, |a||b|=\sqrt{4+36+1+9}\sqrt{9+4+1}=10\sqrt{7}$.
so the angle is $\arccos(19/(10\sqrt{7}))$
|
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|
Manifold with minimum surface distance between two points The book "The World is Flat" uses flatness as a metaphor for a global economy. In fact, a spherical world would seem to be better than a flat world in terms of reducing the distances between two random points on the surface of the world. The shorter the distance between any two points, the easier it is for information and objects to travel between different places. While it may be obvious that a spherical world is better than a flat world, it's far from obvious that a spherical world is optimal in this regard, which brings me to the question: what should the book have been titled? More precisely:
Question: Define a world to be a 2-manifold with some fixed surface area S and a metric d that calculates distance on the surface of the manifold. What shaped world minimizes average distance between any two randomly selected points in the world?
Does the answer depend on whether the world can be embedded in $\mathbb{R}^3$?
Does it depend on the specific metric used?
Does it depend on how we define "average distance" or "randomly selected?"
|
The average distance on a sphere of radius $r$ is $\frac{\pi}{2}r$ (which we can get without any integrations because there's as much area at a distance $\frac{\pi}{2}+\theta$ from a given point as there is at $\frac{\pi}{2}-\theta$); so this is
$$\bar{d}=\frac{\pi}{2}\sqrt{\frac{A}{4\pi}}=\frac{\sqrt{\pi}}{4}\sqrt{A}\approx 0.443 \sqrt{A}\;.$$
The average distance on the flat manifold $(\mathbb R / a\mathbb Z)^2$ is $\frac{1}{6}(\sqrt{2}+\sinh^{-1} 1)a$ (the average Euclidean distance between the point $(a/2,a/2)$ and a random point in $[0,a]^2$), and the area of that manifold is $a^2$, so that makes
$$\bar{d}=\frac{1}{6}(\sqrt{2}+\sinh^{-1} 1)\sqrt{A}\approx 0.383 \sqrt{A}\;.$$
So it's not true that a flat manifold has higher average distance than a sphere; it's the other way around. I suspect the comparison you mean is between a sphere and a flat disk; but a disk is a manifold with boundary, not a manifold, and the higher average distance is due to the boundary, not to the flatness.
[Edit:] Here's the calculation of the average distance in the flat case, as requested. [This is a corrected version.]
We'll need an integral of the form $\sqrt{x^2+c^2}$ several times, so I'll do that in general form first, using the substitution $x=c\sinh u$:
$$
\begin{eqnarray}
\int\sqrt{x^2+c^2}\mathrm dx
&=&
\int c\sqrt{1+\sinh^2 u}\;c\cosh u\mathrm du
\\
&=&
c^2\int\cosh^2u\mathrm du
\\
&=&
\frac{c^2}{2}(u+\sinh u\cosh u)
\\
&=&
\frac{c^2}{2}\left(\sinh^{-1}\frac{x}{c}+\frac{x}{c}\sqrt{1+\left(\frac{x}{c}\right)^2}\right)
\\
&=&
\frac{1}{2}\left(c^2\sinh^{-1}\frac{x}{c}+x\sqrt{x^2+c^2}\right)\;.
\end{eqnarray}
$$
Then the integral over the distance in the primitive cell of a square lattice with $a=2$ is
$$
\begin{eqnarray}
\int_{-1}^1\int_{-1}^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy
&=&
4\int_0^1\int_0^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy
\\
&=&
4\int_0^1\left[
\frac{1}{2}\left(y^2\sinh^{-1}\frac{x}{y}+x\sqrt{x^2+y^2}\right)
\right]_0^1 \mathrm dy
\\
&=&
2\int_0^1
\left(y^2\sinh^{-1}\frac{1}{y}+\sqrt{y^2+1}\right)
\mathrm dy\;.
\end{eqnarray}
$$
We can deal with the first term by integrating by parts twice:
$$
\begin{eqnarray}
\int y^2\sinh^{-1}\frac{1}{y}\mathrm dy
&=&
\frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+\int y\frac{1}{\sqrt{1+(1/y)^2}}\mathrm dy\right)
\\
&=&
\frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+\int y\frac{y}{\sqrt{y^2+1}}\mathrm dy\right)
\\
&=&
\frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1}-\int \sqrt{y^2+1}\mathrm dy\right)\;.
\end{eqnarray}
$$
Putting everything together, we get
$$
\begin{eqnarray}
\int_{-1}^1\int_{-1}^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy
&=&
2\left\{\frac{1}{3}\left[
y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1}
\right]_0^1
+\left(1-\frac{1}{3}\right)
\int_0^1\sqrt{y^2+1}\mathrm dy\right\}
\\
&=&
\frac{2}{3}\left[
y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1}
+\sinh^{-1}y+y\sqrt{y^2+1}\right]_0^1
\\
&=&
\frac{4}{3}\left(\sinh^{-1}1+\sqrt{2}\right)\;.
\end{eqnarray}
$$
This has to be divided by the area $2^2$ to get the average distance for $a=2$, and then by $2$ to get the average distance for $a=1$, since the average distance scales linearly with $a$; that leads to the above result.
For the average distance on the quotient of the plane with respect to a hexagonal lattice, we can use symmetry to restrict the calculation to half of one of the $6$ equilateral triangles. The integrals are rather complicated to work out by hand, but can be done analytically; WolframAlpha gives
$$
\int_0^{\sqrt{3}/2} \int_0^{1-y/\sqrt{3}} \sqrt{x^2+y^2}\mathrm dx \mathrm dy
=\frac{4+\log 27}{32\sqrt{3}}
$$
for side length $a=1$. The area of one of these half-triangles is $\frac{\sqrt{3}}{8} a^2$, so the total area of the manifold is $\frac{3\sqrt{3}}{2}a^2$, and the average distance comes out as
$$\bar{d}=\left(\frac{4+\log 27}{32\sqrt{3}}/\frac{\sqrt{3}}{8}\right)a=\frac{4+\log 27}{12}\sqrt{\frac{A}{3\sqrt{3}/2}}\approx 0.370 \sqrt{A}\;,$$
so yasmar's manifold indeed slightly improves on the one using a square lattice.
|
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|
How to prove $\left \lceil \frac{n}{m} \right \rceil = \left \lfloor \frac{n+m-1}{m} \right \rfloor$? everybody, how can I prove that, for natural $m$ and $n$,
$$
\left \lceil \frac{n}{m} \right \rceil = \left \lfloor \frac{n+m-1}{m} \right \rfloor \; ?
$$
Thanks a lot.
|
what about Mathematical induction prove :
n and m is natural number :
\begin{aligned} m=1 \rightarrow \left \lfloor \frac{1+n-1}{1} \right \rfloor = \left \lceil \frac{n}{1} \right \rceil \rightarrow true \end{aligned}
\begin{aligned}m=2 \rightarrow \left \lfloor \frac{2+n-1}{2} \right \rfloor = \left \lceil \frac{n}{2} \right \rceil \rightarrow true \end{aligned}
\begin{aligned} \left \lceil \frac{n}{2} \right \rceil = \begin{Bmatrix}
\frac{n}{2} \ n \ is \ even
\ \
\frac{n+1}{2} \ n \ is \ odd
\end{Bmatrix}
\end{aligned}
\begin{aligned}
if \ m=k \rightarrow \left \lfloor \frac{k+n-1}{k} \right \rfloor = \left \lceil \frac{n}{k} \right \rceil
\end{aligned}
\begin{aligned}
prove \ m=k+1 \rightarrow \left \lfloor \frac{k+1+n-1}{k+1} \right \rfloor = \left \lceil \frac{n}{k+1} \right \rceil
\end{aligned}
\begin{aligned}
\left \lfloor \frac{n-1}{k+1}+1 \right \rfloor = \left \lceil \frac{n}{k+1} \right \rceil
\end{aligned}
\begin{aligned}
\left \lfloor \frac{n-1}{k+1} \right \rfloor+1 = \left \lceil \frac{n}{k+1} \right \rceil
\end{aligned}
... ?
|
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|
Bell Numbers and Equating Coefficients I need help solving the following exercise:
Let B(n) be the Bell - Numbers of $[n]$ and $$\exp(x) := \sum_{n\geq 0} \frac{x^n}{n!}$$ the exponential function.
Prove by equating the coefficients and using
$$ B(n) = \sum\limits_{a_1+\cdots+a_k=n, a_i \geq 1} \frac{1}{k!} \binom{n}{a_1 \cdots a_k}$$ that for $n \geq 0$
$$\sum\limits_{n \geq 0} B(n)/n! \cdot x^n = \exp(\exp(x) - 1)$$
I tried to play around with the equation but haven't gotten anywhere yet.
For the left side I am not sure how to handle the range of the 2nd sum:
$$\sum\limits_{n \geq 0} B(n)/n!\cdot x^n = \sum\limits_{n \geq 0} \frac{\sum\limits_{a_1+\cdots+a_k=n, a_i \geq 1}\frac{1}{k!}\binom n {a_1 \cdots a_k}}{n!}\cdot x^n$$
How do I get this simplified and converted to a polynomial form?
I know that
$$\exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!}
= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$
But how do I proceed for $\exp(\exp(x)-1) = e^{e^x-1}$?
Thank you in advance!
|
We are interested in expansion of the term $\exp(\exp(x)-1)$, now $$\exp(x) - 1 = \sum_{n=1}^\infty \frac{x^n}{n!}$$ and since $\displaystyle \exp(z) = \sum_{m=0}^\infty \frac{z^m}{m!}$ we should start by considering $\displaystyle \left(\sum_{n=1}^\infty \frac{x^n}{n!}\right)^m$ for $m=2,3,\ldots$, here is a table just for $m=2$, extrapolation will give all other $m$:
$$\begin{array}{c|l}
\sum_{n=1}^\infty \frac{x^n}{n!} & \frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\ldots \\
\times \frac{x^1}{1!} & \frac{x^2}{1!1!}+\frac{x^3}{1!2!}+\frac{x^4}{1!3!}+\frac{x^5}{1!4!}+\frac{x^6}{1!5!}+\ldots \\
\times \frac{x^2}{2!} & \frac{x^3}{1!2!}+\frac{x^4}{2!2!}+\frac{x^5}{2!3!}+\frac{x^6}{2!4!}+\frac{x^7}{2!5!}+\ldots \\
\times \frac{x^3}{3!} & \frac{x^4}{1!3!}+\frac{x^5}{2!3!}+\frac{x^6}{3!3!}+\frac{x^7}{3!4!}+\frac{x^8}{3!5!}+\ldots \\
\vdots & \cdots
\end{array}$$
and so we find $$\exp\left(\sum_{n=1}^\infty \frac{x^n}{n!}\right) = \frac{1}{0!} + \sum_{n=1}^\infty \left(\frac{1}{1!} \sum_{a=n}\frac{x^{a}}{a!} + \frac{1}{2!} \sum_{a+b=n}\frac{x^{a+b}}{a!b!} + \frac{1}{3!} \sum_{a+b+c=n}\frac{x^{a+b+c}}{a!b!c!} + \ldots \right)$$
So you can try to see whether this is equal to the strange multinomial coefficient you have written.
Now I will show how to use the above formula to compute Bell numbers,
First the 0th Bell number (coefficient of $x^0$) is just 1/0! = 1. The first Bell number is the coefficient of $x^1$, and the only term which contributes this is 1/1! x^1/1! so B_1 = 1 as well. A more interesting one is B_3:
All contributions to $x^3$ terms come from $\sum_{n=1}^\infty \frac{1}{1!} \sum_{a=n}\frac{x^{a}}{a!} + \frac{1}{2!} \sum_{a+b=n}\frac{x^{a+b}}{a!b!} + \frac{1}{3!} \sum_{a+b+c=n}\frac{x^{a+b+c}}{a!b!c!}$ we could remove the "..." part since a+b+c+d+.. can never be 3 (since all the numbers must be $\ge 1$).
Now consider each sum one at a time,
*
*{a = 3}
*{a = 2,b = 1}, {a = 1, b = 2}
*{a = 1,b = 1,c = 1}
so we have the $x^3$ coefficient being: $1/1! x^3/3! + 1/2! (x^3/1!2! + x^3/2!1!) + 1/3! x^3/1!1!1! = 5/6 x^3$ so the third bell number is $5$.
|
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|
Lower bound over a convex function I will be to grateful if help me find a tight lower bound $g(x)$ over the following convex function:
$$f(x) = \sqrt{1+4x^2} -1 + \log(\sqrt{1+4x^2}-1) - \log(2x^2) \geq g(x),$$
where $g(x)$ is preferably a polynomial with degree of at least $2$.
The taylor expansion around the point ($x = 0$) of this function is given by:
$$f(x) = x^2 - \frac{x^4}{2} + \frac{2x^6}{3} - \frac{5x^8}{4} + \frac{14x^{10}}{5} - \ldots$$
|
For the case $|x|\leq 1$ I propose the following: Write $g(x):= a x^2- b x^4$ and choose $a$ and $b$ such that $f(1)=g(1)$, $f'(1)=g'(1)$. I did this using Mathematica and obtained
$$a=2\Bigl({3\over 1+\sqrt{5}} +\log {2\over 1+\sqrt{5}}\Bigr)\ \qquad b=
{2\over 1+\sqrt{5}} +\log{2\over 1+\sqrt{5}}\ .$$
Plotting the resulting graph of $f-g$ one gets the impression that
$$0\leq f(x)-g(x)\leq 0.0117 \qquad(0\leq x\leq 1)\ .$$
|
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|
Can we make an integral domain with any number of members? Is it true that rings without zero divisors (integral domains) can have any number of members except for 4,6? and if this is true then what would the multiplication operator be?
|
Another way to define field F4 is by defining addition with:
$$ 0 + a = a $$
$$ a + a = 0 $$
$$ 1 + 2 = 3 $$
and multiplication having:
$$ 0 \times a = 0 $$
$$ 1 \times a = a $$
$$ 2 \times 2 = 3 $$
$$
\begin{array}{c|cccc}
{\Large +} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\
\hline
\overline{0} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\
\overline{1} & \overline{1} & \overline{0} & \overline{3} & \overline{2} \\
\overline{2} & \overline{2} & \overline{3} & \overline{0} & \overline{1} \\
\overline{3} & \overline{3} & \overline{2} & \overline{1} & \overline{0}
\end{array}
\hskip0.5in
\begin{array}{c|cccc}
{\Large \times} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\
\hline
\overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} \\
\overline{1} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\
\overline{2} & \overline{0} & \overline{2} & \overline{3} & \overline{1} \\
\overline{3} & \overline{0} & \overline{3} & \overline{1} & \overline{2}
\end{array}$$
|
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|
Continued proportion implies $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$ I am trying to find a tricky way to proof these:
If $a,b,c,d$ are in continued proportion, prove that $$(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2 ,$$ This result could be extended to $$(a^2+b^2+c^2+d^2)(b^2+c^2+d^2+e^2)=(ab+bc+cd+de)^2$$ when $a,b,c,d,e$ are in continued proportion.
The standard way for solving them could be putting $\frac{a}{b}=\frac{c}{d}=k$ then followed by substitution and which is followed by tedious algebraic manipulations,but that is not what I am looking for could these be solved in a less easy way using some other algebraic method/tricks? Please explain.
|
You know that $b=ka$, $c=kb$ etc so the lhs can be rewritten
$$(a^2+b^2+c^2)(k^2a^2+k^2b^2+k^2c^2) = k^2(a^2+b^2+c^2)^2$$
and the rhs can be written
$$(ka^2 + kb^2 + kc^2)^2 = k^2(a^2+b^2+c^2)^2$$
and you're done. The same trick works for the second example.
|
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|
How do I get the square root of a complex number? If I'm given a complex number (say $9 + 4i$), how do I calculate its square root?
|
Write $(a+bi)^2 = 9+4i$. Then $a^2-b^2 +2abi = 9+4i$.
By comparison of coefficients,
$2ab = 4$ and $a^2-b^2 = 9$.
Thus $ab=2$ and $a^2 = 9 +b^2$.
Setting $a = 2/b$ with $b\ne 0$ gives $4/b^2 = 9 + b^2$ and so $4 = 9b^2 + b^4$, i.e., $b^4 + 9b^2 -4 = 0$.
Solve $x^2 + 9x-4=0$, where $x=b^2$. Solutions are
$x_{1,2} = \frac{-9\pm\sqrt{9^2-4\cdot 1\cdot (-4)}}{2\cdot 1} = \frac{-9\pm\sqrt{97}}{2}$. Solutions are $b_{1,2,3,4} = \pm\sqrt{\frac{-9\pm\sqrt{97}}{2}}$. This gives the corresponding values $a_{1,2,3,4}$. All solution pairs are real-valued.
|
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|
Basic question about mod I'm having a tough time understanding why $(a^x \bmod p)^y \bmod p$ is equal to $a^{xy}\bmod p$.
Does this have a mathematical proof? Please advise.
|
It's usually simpler to work with congruences rather than the binary $\bmod$ operator, and then go tot he binary operator at the end.
Given an integer $n$, we say that $a\equiv b\pmod{n}$ ("$a$ is congruent to $b$ modulo $n$") if and only if $n$ divides $b-a$. It is easy to show that "congruent modulo $n$" is an equivalence relation; that is, $a\equiv a \pmod{n}$ for every $a$; if $a\equiv b\pmod{n}$ then $b\equiv a\pmod{n}$; and if $a\equiv b\pmod{n}$ and $b\equiv c\pmod{n}$, then $a\equiv c\pmod{n}$.
Proposition. The following are equivalent:
*
*$a\equiv b\pmod{n}$.
*$n$ divides $a-b$.
*$a = b+kn$ for some integer $k$.
*$a$ and $b$ have the same remainder when divided by $n$.
Proof. (1)$\Leftrightarrow$(2): This is the definition.
(2)$\Rightarrow$(3): Since $n$ divides $a-b$, there exists $k$ such that $nk = a-b$; hence $a=b+nk$, as claimed.
(3)$\Rightarrow$(4). Let $b=qn+r$, with $0\leq r\lt |n|$. Then $a=b+kn = qn+r+kn = (q+k)n+r$, with $0\leq r\lt |n|$. By the uniqueness clause of the division algorithm, it follows that $r$ is also the remainder of $a$ when divided by $n$.
(4)$\Rightarrow$(2). If $a=qn+r$, $b=pn+r$, $0\leq r\lt|n|$, then $a-b = (q-p)n$, so $n$ divides $a-b$. QED
Condition (4) tells you that if you have an equality between expressions involving remainders of numbers, they are equivalent to the corresponding congruence. In your question, this means that the desired equality holds if and only if $a^{xy}$ is congruent to $(a^x)^y$ modulo $n$, which holds because they are in fact equal. But to get it more carefully:
Corollary. Let $n$ be a positive integer. Then every integer $a$ is congruent modulo $n$ to one and only one of $0$, $1$, $2,\ldots,n-1$.
Proof. Dividing $a$ by $n$ we have $a = qn + r$, $0\leq r\lt n$. Then $a\equiv r\pmod{n}$; so every integer is congruent to at least one of $0,\ldots,n-1$. For uniqueness, let $i,j$ be integers, $0\leq i,j\leq n-1$ such that $i\equiv j\pmod{n}$. Without loss of generality, assume $i\leq j$. Then $0\leq j-i\lt n$, and $n$ divides $j-i$; the only possibility is for $j-i=0$, so $j=i$. QED
Definition. Let $a$ be an integer, $n$ a positive integer. We define $a\bmod n$ ("$a$ mod $n$") to be the unique nonnegative integer $i$, $0\leq i\lt n$, such that $a\equiv i\pmod{n}$. That is, $a\equiv (a\bmod n)\pmod{n}$, and $0\leq (a\bmod n)\lt n$.
Theorem. If $a\equiv b\pmod{n}$ and $c\equiv d\pmod{n}$, then $a+c\equiv b+d\pmod{n}$ and $ac\equiv bd\pmod{n}$.
Proof. Our hypothesis are that $n$ divides both $a-b$ and $c-d$. Therefore, it divides their sum, $(a+c)-(b+d)$ (which implies $a+c\equiv b+d\pmod{n}$); it also divides $(a-b)c$ and $b(c-d)$; adding these latter two gives that $ac-bd$ is a multiple of $n$, so $ac\equiv bd\pmod{n}$. QED
Corollary. If $a\equiv b\pmod{n}$, and $r$ is a positive integer, then $a^r\equiv b^r\pmod{n}$.
Now consider $(a^x)^y$ and $a^{xy}$.
Note that $(a^x)^y = a^{xy}$, so $(a^x)^y\equiv a^{xy}\pmod{n}$. Also, since $a^x \equiv (a^x\bmod n)\pmod{n}$, then $(a^x)^y \equiv (a^x\bmod n)^y\pmod{n}$. Since $a^{xy}\equiv (a^x)^y\pmod{n}$, and $(a^x)^y\equiv (a^x\bmod n)^y\pmod{n}$, then $a^{xy}\equiv (a^x\bmod n)^y\pmod{n}$.
And since $a^{xy}\equiv (a^x\bmod n)^y\pmod{n}$, then $a^{xy}$ and $(a^x\bmod n)^y$ have the same remainder when divided by $n$; that is,
$$a^{xy}\bmod n = (a^x\bmod n)^y \bmod n,$$
as desired.
|
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Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?
|
Late to the party, but I hope I am contributing something new to the answers already posted here.
We seek
$$\sum_{i=1}^Ni^2=1^2+2^2+\cdots+N^2$$
which, geometrically speaking, is equivalent to adding the area of $N$ squares with side lengths $1,2,...,N$. A visual depiction when $N=2$ is
By adding the missing tile on the top-left corner, we may form a rectangle as follows,
with area $2(1+2)$. Then our sum, for $N=2$, is
$$S_2=1^2+2^2=\text{Area rectangle}-\text{Area tile} = 2(1+2)-(1\cdot1).$$
To get a better idea, consider the case with $N=3$,
and again, adding the missing tiles so that we form a rectangle, gives
with area $3(1+2+3)$. The desired sum, for $N=3$, is
$$S_3=1^2+2^2+3^3=\text{Area rectangle}-\text{Area tiles} = 3(1+2+3)-[\underbrace{1\cdot1}_\text{blue tile}+\underbrace{1\cdot(1+2)}_\text{red tile}].$$
Drawing a few more, one may notice that the height of each tile is $1$ and the base is the sum of the bases of the previous squares. In general, we have
$$
\begin{aligned}
S_N = \sum_{i=1}^Ni^2&=N\left(\sum_{i=1}^Ni\right)-\left[1\cdot1+1\cdot(1+2)+\cdots+1\cdot(1+2+\cdots+N-1)\right]\\
& =N\left(\sum_{i=1}^Ni\right)-\sum_{i=1}^{N-1}\sum_{j=1}^{i}j=N\left(\frac{N(N+1)}{2}\right)-\sum_{i=1}^{N-1}\frac{i(i+1)}{2}\\
&=\frac{N^2(N+1)}{2}-\frac12\sum_{i=1}^{N-1}i^2-\frac12\sum_{i=1}^{N-1}i=\frac{N^2(N+1)}{2}-\frac12\sum_{i=1}^{N-1}i^2-\frac{(N-1)N}{4}\\
&=\frac{N^2(N+1)}{2}-\frac12\sum_{i=1}^{\color{red}{N}}i^2+\frac{N^2}2-\frac{(N-1)N}{4},
\end{aligned}$$
and now solving for $\sum i^2$ leads to
$$\frac32\sum_{i=1}^Ni^2=\frac{N^2(N+1)}{2}+\frac{N^2}2-\frac{(N-1)N}{4}\implies\sum_{i=1}^Ni^2=\frac{N(2N+1)(N+1)}{6}.$$
|
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|
Expansion concerning the binomial theorem The question goes:
Expand $(1-2x)^{1/2}-(1-3x)^{2/3}$ as far as the 4th term.
Ans: $x + x^2/2 + 5x^3/6 + 41x^4/24$
How should I do it?
|
Note that:
*
*$\displaystyle (1-x)^{\frac{p}{q}} = 1 - \frac{p}{q} \cdot \frac{x}{1!} + \frac{ \frac{p}{q} \cdot \Bigl(\frac{p}{q} -1\Bigr)}{2!} \cdot x^{2} - \cdots $
Using this formula you have $$(1-2x)^{1/2} = 1- \frac{1}{2} \cdot \frac{2x}{1!} + \frac{\frac{1}{2} \cdot \Bigl(-\frac{1}{2}\Bigr)}{2!}\cdot 4x^{2} - \frac{ \frac{1}{2} \cdot \Bigl(\frac{1}{2}-1\Bigr) \cdot \Bigl(\frac{1}{2}-2\Bigr)}{3!} \cdot8x^{3} + \cdots $$ and $$(1-3x)^{2/3} = 1 - \frac{2}{3} \cdot \frac{3x}{1!} + \frac{ \frac{2}{3} \cdot \Bigl(\frac{2}{3}-1\Bigr)}{2!} \cdot 9x^{2} - \frac{\frac{2}{3}\cdot \Bigl(\frac{2}{3}-1\Bigr) \Bigl(\frac{2}{3}-2\Bigr)}{3!} \cdot 27x^{3} + \cdots$$
simplify and add the above two equations to get the answer.
You may want to see this thread as well:
*
*How can I sum the infinite series $\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad$
|
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Proof by induction that $2^n \gt n^k$ How do I prove using induction that if $k$ is a natural number, then $2^n \gt n^k$ for all $n \geq k^2 + 1$, where $n$ is also a natural number?
|
If found only this proof, which is rather cumbersome. I hoppe that I did not make mistake there and that someone will come up with a more elegant solution.
Lemma 1: $2^k\ge k^2$ for any $k\ge 4$.
EDIT: In the comments you can find a nice combinatorial argument provided by Aryabhata which works for $k\ge5$.
Proof by induction: $1^\circ$ For $k=4$ the equality holds.
$2^\circ$ Suppose that the lemma holds for $k$. For $k+1$ we get
$2^{k+1}=2\cdot2^k \ge \left(\frac{k+1}k\right)^2\cdot k^2 = (k+1)^2$
(since $\frac{k+1}k =1+\frac1k \le \frac 54 \le \sqrt 2$)
Lemma 2: $\left(1+\frac1{k^2}\right)^k < 2$ for $k\ge 2$.
We know that $\left(1+\frac1{k^2}\right)^{k^2} < 3$ (see here and
here, you can find this in many introductory calculus textbooks) which means that $\left(1+\frac1{k^2}\right)^k < 3^{1/k} \le
2$.
Claim: If $n=k^2+t$ for some positive integer $t$ and $k\ge 2$
then $2^n>n^k$.
Induction on $t$.
$1^\circ$ For $t=1$. If $k\ge 4$ then we have $2^k\ge k^2$
$\Rightarrow$ $2^{k^2}\ge(k^2)^k$. If we multiply this inequality by
$2>\left(\frac{k^2+1}{k^2}\right)^k$, which we know from Lemma 2, we get $2^n>n^k$. For $k=2,3$ we can verify this by hand.
$2^\circ$ Suppose that $n=k^2+t+1$ and the claim holds for $t$, i.e., we have
$$2^{k^2+t}>(k^2+t)^k$$
We also get $\left(\frac{k^2+t+1}{k^2+t}\right)^k = \left(1+\frac1{k^2+t}\right)^k \le
\left(1+\frac1{k^2}\right)^k < 2$.
By multiplying these two inequatities we get
$$2^{k^2+t+1}>(k^2+t+1)^k$$
$$2^n>n^k$$
The only remaining case is $k=1$, in which case the claim $n\ge 2$ $\Rightarrow$ $2^n>n$ can be shown by induction.
|
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General approach for problem for finding sum from $1$ to $N$ when all $a$'s are replaced by $b$'s The problem is:
Find the sum of all the numbers from $1$ to $100$ when all the $6$'s are replaced by $9$'s.
I need some ideas on how to approach this kind of problems? Please explain your ideas, assuming one digit replacements.
|
Hint:
*
*Can you find the sum $1 + 2 + \dots 100$?
*What are all the values between $1$ and $100$ that have a $6$ in them? List them.
(Edit:) The process is the same for the numbers $1, \dots , 10$. You would add up all the values $1 + \dots + 10 = \frac{10 \cdot 11}{2} = 55$. Then, subtract each number with a $6$ to get $55 - 6 = 49$. Then, change the $6$ to a $9$, and add it back in: $49 + 9 = 58$. The same process works for the numbers $1, \dots , 100$:
*
*$1 + \dots + 100 = 5050$.
*Add together all the digits with a $6$:
$$\begin{align}
&6 + 16 + \dots + 56 + 60 + 61 + \dots + 69 + 76 + 86 + 96
\\
&= (6 + 16 + \dots + 96) + (60 + \dots + 69) - 66
\\
&= (10\cdot 6 + 10 + 20 + \dots + 90) + (10 \cdot 60 + 1 + 2 + 3 + \dots + 9) - 66
\\
&= \left(60 + 10(1 + 2 + \dots + 9)\right) + \left(600 + \frac{9\cdot 10}{2} \right) - 66
\\
&= (60 + 10\cdot 45) + (600 + 45) - 66
\\
&= 1089.
\end{align}$$
EDIT (#$2$):
This process amounts to subtracting all the values with a $6$, and then adding the numbers with a $9$. Perhaps a quicker way to do this is to add to $5050$, the quantity
$$
9-6 + 19-16 + 29-26 + \dots 59-56 + 90-60+ 91-61 + \dots 99-69 + 79-76 + \dots.
$$
|
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$x^4+4$ is composite for $x>1$
$x^4+4$ is composite for $x>1$
I know the Sophie Germain indentity and the get the factorization $$x^4+4 = (x^2+2-2x)(x^2+2+2x)$$
But I am stuck here. I cannot see any general factor here.
|
A version in two variables is
$$ ( x^2 + 2 x y + 2 y^2) (x^2 - 2 x y + 2 y^2) = x^4 + 4 y^4 $$
where both quadratic forms are positive definite, so that there are only a finite number of $(x,y)$ pairs such that either factor is $1.$
|
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|
A "fast" way to ,find the maximum value of $(x^2) \times (y^3)$, if $3x+4y=12$ for $x,y \ge 0$ If $3x+4y=12$ $\forall x,y \ge 0$, the maximum value of $(x^2) \times (y^3)$ is
*
*$6 \times (6/5)^5$
*$3 \times (6/5)^5$
*$ (6/5)^5 $
*$7 \times (6/5)^5$
How to approach this problem? I thought of using the approach for finding maxima-minima for two independent variable but I am not sure about that as the $x$ and $y$ vanishes after the first partial derivative itself.
|
AM-GM gives
$$12 = \frac{3}{2} x + \frac{3}{2} x + \frac{4}{3} y + \frac{4}{3} y + \frac{4}{3} y \ge 5 \left( \frac{16}{3} x^2 y^3 \right)^{1/5}$$
hence
$$x^2 y^3 \le \frac{3}{16} \left( \frac{12}{5} \right)^5 = 6 \cdot \left( \frac{6}{5} \right)^5$$
with equality when $\frac{3}{2} x = \frac{4}{3} y$, which is attainable.
|
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Challenging inequality: $abcde=1$, show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{33}{2(a+b+c+d+e)}\ge{\frac{{83}}{10}}$ Let $a,b,c,d,e$ be positive real numbers which satisfy $abcde=1$. How can one prove that:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} +\frac{1}{e}+ \frac{33}{2(a + b + c + d+e)} \ge{\frac{{83}}{10}}\ \ ?$$
|
This is only a partial solution but I think someone more familiar with such elementary inequalities than myself might be able to finish it. You can replace $a,b,c,d$, and $e$ with their reciprocals and the inequality in question becomes
$$a + b + c + d + e + {33 \over 2}{1 \over ({1 \over a} + {1 \over b} + {1 \over c} + {1 \over d} + {1 \over e})} \geq {83 \over 10}$$
Since still $abcde = 1$, we can rewrite this as
$$a + b + c + d + e + {33 \over 2}{1 \over ({1 \over a} + {1 \over b} + {1 \over c} + {1 \over d} + {1 \over e})} \geq {83 \over 10}(abcde)^{1 \over 5}$$
Some algebra converts this into
$${a + b + c + d + e \over 5} - (abcde)^{1 \over 5} \geq {33 \over 50}(abcde)^{1 \over 5} - {33 \over 50}{5 \over ({1 \over a} + {1 \over b} + {1 \over c} + {1 \over d} + {1 \over e})}$$
In other words, $AM - GM \geq {33 \over 50}(GM - HM)$. This is needed only when $abcde = 1$, but by scaling this should then hold for all $a,b,c,d,$ and $e$. So you inequality experts out there... is this something that follows from well-known inequalities?
|
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In a acute angled triangle, we have $\tan(A)\cdot\tan(B)\cdot\tan(C) \geq 3\sqrt{3}$ How to show:
In a acute angled $\triangle \ ABC$ show that $$\tan(A) \cdot \tan(B)\cdot \tan(C) \geq 3\sqrt{3}$$
Any ideas?
|
$$A+B=\pi-C$$
\begin{align*}
&\tan (A+B)= \tan (\pi-C)\\
&(\tan A+ \tan B)/(1-\tan A \tan B)= (\tan \pi- \tan C)/(1+\tan \pi \tan C)=-\tan C\\
&(\tan A+ \tan B)= -\tan C(1-\tan A \tan B)\\
&\tan A + \tan B= -\tan C+ \tan A \tan B \tan C\\
&\tan A + \tan B+ \tan C= \tan A \tan B \tan C
\end{align*}
|
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Limit of $(x+3)^{1 + 1/x} - x^{1 + 1/(x+3)}$ when $x\to \infty$ Okay, this is the last limit I have to solve but it's not that easy ;)
$$\lim_{x\to \infty} (x+3)^{1 + 1/x} - x^{1 + 1/(x+3)}$$
|
For constants a, b, consider the asymptotics of $(x+a)^{1+1/(x+b)}$ as $x\to\infty$. This is equal to $(x+a)e^{\log(x+a)/(x+b)}$.
Now since $\log(x+a)/(x+b)$ tends to 0, we know that $e^{\log(x+a)/(x+b)}=1+\log(x+a)/(x+b)+\ldots$ where the $\ldots$ term tends to 0 faster than $\log(x+a)^2/(x+b)^2$.
Then $(x+a)^{1+1/(x+b)}=(x+a)+(x+a)\log(x+a)/(x+b)+(x+a)\ldots$, and since $(x+a)\log(x+a)^2/(x+b)^2$ tends to 0, so does $(x+a)\ldots$.
Therefore $(x+a)^{1+1/(x+b)}-(x+a)-(x+a)\log(x+a)/(x+b)\to0$ as $x\to\infty$.
Note that $(x+a)\log(x+a)/(x+b)-\log(x+a)=(a-b)\log(x+a)/(x+b)\to0$ as $x\to\infty$.
Note also that $\log(x+a)-\log x=\log(1+a/x)\to\log 1=0$ as $x\to\infty$.
Adding these three limits together shows that $(x+a)^{1+1/(x+b)}-(x+a)-\log x\to 0$, which we can rewrite as $(x+a)^{1+1/(x+b)}-(x+\log x)\to a$.
Applying with $a=3$ and $b=0$ gives that $(x+3)^{1+1/x}-(x+\log x)\to 3$. Applying with $a=0$ and $b=3$ gives that $x^{1+1/(x+3)}-(x+\log x)\to 0$.
Subtracting one from the other gives $(x+3)^{1+1/x}-x^{1+1/(x+3)}\to 3$.
|
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Finding limit of a quotient I can't find this for some reason. I know I asked about 6 of these before, and I was able to finish my homework but now I went back to review and I can't do a single one of these problems on my own. Even the ones I did figure out on my own. I spent probably a total of 14 hours on the homework, not sure what is wrong with me but I don't know how to factor or do basic algebra.
Anyways I need to find
$$\lim_{x\to -2}\frac{x+2}{x^3+8}.$$
I have spent at least an hour on it and I can't figure it out.
|
From the comments, it appears that it would be useful to give a fairly elementary and pedantic discussion of using the factor theorem (from precalculus or college algebra, in the U.S.).
Factor Theorem: If $r$ is a zero of a polynomial $P(x)$ (i.e. a solution to $P(x) = 0$ is $x=r$), then $x-r$ is a factor of $P(x)$.
Example 1: Factor $x^3 - 8$.
First, solve $x^3 - 8 = 0$ to find a value for $r$. Solving gives $x^3 = 8$, or $x = 2$. Therefore, we can use $r = 2$, which tells us that $x - 2$ is a factor of $x^3 - 8$. Now use long division (or synthetic division) to determine the quotient when $x - 2$ is divided into $x^3 - 8$. The quotient will be $x^2 + 2x + 4$. Therefore, we can factor $x^3 - 8$ as $(x-2)(x^2+2x+4).$
Example 2: Factor $x^3 + 8$.
First, solve $x^3 + 8 = 0$ to find a value for $r$. Solving gives $x^3 = -8$, or $x = -2$. Therefore, we can use $r = -2$, which tells us that $x - (-2)$, or $x+2$ is a factor of $x^3 + 8$. Now use long division (or synthetic division) to determine the quotient when $x + 2$ is divided into $x^3 + 8$. The quotient will be $x^2 - 2x + 4.$ Therefore, we can factor $x^3 + 8$ as $(x+2)(x^2-2x+4).$
Example 3: Factor $x^5 - 100,000$.
First, solve $x^5 - 100,000 = 0$ to find a value for $r$. Solving gives $x^5 = 100,000$, or $x = 10$. Therefore, we can use $r = 10$, which tells us that $x - 10$ is a factor of $x^5 - 100,000$. Now use long division (or synthetic division) to determine the quotient when $x - 10$ is divided into $x^5 + 100,000$. The quotient will be $x^4 + 10x^3 + 100x^2 + 1000x + 10,000$. Therefore, we can factor $x^5 - 100,000$ as $(x-10)(x^4 + 10x^3 + 100x^2 + 1000x + 10,000).$
Example 4: Factor $x^5 + x^4 + x^3 + x^2 + x + 1$.
First, solve $x^5 + x^4 + x^3 + x^2 + x + 1 = 0$ to find a value for $r$. In this case, standard methods don't work (unless you know about cyclotomic equations). However, by using the rational root test (google it), you'll get $x = 1$ and $x = -1$ as possible solutions. You'll find by direct substitution that $x = -1$ is a solution to $x^5 + x^4 + x^3 + x^2 + x + 1 = 0$. Therefore, we can use $r = -1$, which tells us that $x - (-1)$, or $x + 1$ is a factor of $x^5 + x^4 + x^3 + x^2 + x + 1$. Now use long division (or synthetic division) to determine the quotient when $x + 1$ is divided into $x^5 + x^4 + x^3 + x^2 + x + 1.$ The quotient will be $x^4 + x^2 + 1$. Therefore, we can factor $x^5 + x^4 + x^3 + x^2 + x + 1$ as $(x+1)(x^4 + x^2 + 1).$
Here's the really nice thing about your situation. There's almost no work in trying to find a solution to $P(x)= 0$. For example, in your specific limit problem you know that $x^3 + 8 = 0$ when $x=-2$ (because you presumably already tried to find the limit by plugging in $x=-2$). So you already know a value that can be used for $r$ in the factor theorem. In other words, something "tricky" to solve like Example 4 won't come up.
|
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Fourier transform of a wavefunction It would be great if someone would help me check that I have done this right.
I have an infinite square well, s.t.
$V(x) = 0$ for $x\in(0,a)$ and $V(x) = \infty$ otherwise.
Given that $\psi(x,0) = \alpha x (a-x)$, where $\alpha\in\mathbb R$
I have found by normalization that $\alpha=\frac{a^3}{6}$.
If I have to write $\psi(x,0)$ as a Fourier series/a combination of normalized eigenstates, it would take the form $\psi(x,0)=\sum_{n=1}^{\infty} C_n u_n(x)$ where $u_n(x) = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)$. Therefore, $$C_n =\int_0^a u_n^*(x)\psi(x,0) dx$$ $$= \frac{a^3}{6}\sqrt{\frac{2}{a}}\int_0^a\sin\left(\frac{n\pi x}{a}\right)x (a-x)dx$$
$$= \frac{a}{n\pi}\frac{a^3}{6}\sqrt{\frac{2}{a}} \int_0^a(2x-a)\cos\left(\frac{n\pi x}{a}\right)dx$$
$$=-2\left(\frac{a}{n\pi}\right)^2\frac{a^3}{6}\sqrt{\frac{2}{a}}\int_0^a\sin\left(\frac{n\pi x}{a}\right)dx$$
Then for odd $n$, $C_n = 4\left(\frac{a}{n\pi}\right)^3\frac{a^3}{6}\sqrt{\frac{2}{a}}=
\frac{2}{3}\frac{a^6}{n^3\pi^3}\sqrt{\frac{2}{a}}$; and $C_n = 0$ otherwise.
Hence $\psi(x,0)=\sum_{\text{odd n}} \frac{4}{3}\frac{a^5}{n^3\pi^3}\sin\left(\frac{n\pi x}{a}\right)$.
Then $\psi(x,t) = \sum_{\text{odd n}} \frac{4}{3}\frac{a^5}{n^3\pi^3}\sin\left(\frac{n\pi x}{a}\right) e^{-iE_nt/\hbar}$.
Thank you in advance!
Added:
The model answer gives a probability for $E_n$ that is independent of $a$. What has gone wrong? Incidentally, the probability of getting $E_n$ is $\frac{960}{n^3\pi^3}$.
Comment: I think it's just down to my rubbish calculations. Thanks to @Eric's suggestion, I think this is resolved.
|
I think you messed up on the normalization. I think it's $\alpha=6/a^3$.
|
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|
Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
|
Hint: $(1+2+...+n+(n+1))^2 - (1+2+...+n)^2 = 2(1+2+...+n)(n+1) + (n+1)^2 = (n+1)(2(1+2+...+n) + (n+1)) = (n+1)(n(n+1) + n+1) = (n+1)(n+1)^2 = (n+1)^3$
We use here the (slightly easier to prove) result $\sum_{k=1}^{n} k = n(n+1)/2$
|
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|
Finding limits of rational functions as $x\to\infty$
Possible Duplicate:
Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q,P$ are polynomials
$$\displaystyle \lim_{x\to\infty}\frac{(2x^2+1)^2}{(x-1)^2(x^2+x)}.$$
I do not know where to start on this, I tried multiplying it out and that didn't help really. It seems very complicated and I know I have to reduce it somehow but everything I do just makes it more complicated.
|
By substitution.
I think the OP is having some difficulty in dividing by a power of $x$ and simplifying the rational function, in terms of $1/x$. I must admit that the simplification process seems a bit unnatural while working with $1/x$.
So, my suggestion is that, at the very beginning, substitute $y = 1/x$. Then $x \to \infty$ is equivalent to $y \to 0$. (Note that we should be talking about the one-sided limit as $x \to 0^+$, but we will ignore that issue here.)
Plugging this in the numerator, we get:
$$
(2x^2+1)^2 = \left(2\frac{1}{y^2} + 1\right)^2 = \left(\frac{2+y^2}{y^2}\right)^2 = \frac{(2+y^2)^2}{y^4}.
$$
Similarly the denominator becomes:
$$
(x-1)^2(x^2+x) = \left( \frac{1}{y} - 1\right)^2 \left( \frac{1}{y^2} + \frac{1}{y} \right) = \frac{(1-y)^2}{y^2} \times \frac{1+y}{y^2} = \frac{(1-y)^2(1+y)}{y^4}.
$$
Dividing the numerator by the denominator, we get
$$
\frac{(2x^2+1)^2}{(x-1)^2(x^2+x)} = \frac{(2+y^2)^2}{y^4} \times \frac{y^4}{(1-y)^2(1+y)}
= \frac{(2+y^2)^2}{(1-y)^2(1+y)}.
$$
(Notice that the problematic $y^4$ terms nicely cancel.)
Finally, we have
$$
\lim_{x \to \infty} \frac{(2x^2+1)^2}{(x-1)^2(x^2+x)}
=
\lim_{y \to 0} \frac{(2+y^2)^2}{(1-y)^2(1+y)}.
$$
I believe that the latter limit should be easy to solve (essentially by plugging in $0$).
|
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|
The square of an integer has form $3n$ or $3n+1$ Prove that if $n\in\mathbb Z$, then $n^2$ is of the form $3q$ or $3q+1$ for some $q\in\mathbb Z$
I would like to show that 3q+2 is = 3q+1 thus $n^2$ can be of the form of 3q or 3q+1.
Case one
$(3k)^2=(3k)(3k)=9k^2=3(3k^2)$ and is still of the form $3q$
When $q=3k^2$
$3q$
Case two
$(3k+1)^2= (3k+1)(3k+1)= 9k^2+6k+1$
$(9k^2+6k)+1 =3(3k^2+2k)+1$
this is of the form $3q+1$ when $q=3k^2+2k$
$3q+1$
Case three
$(3k+2)^2= (3k+2)(3k+2) =9k^2+12k+4$
$(9k^2+12k+4) = 3(3k^2+4k+1)+1$
This is of the form $3q+1$ when $q=3k^2+4k+1$
$3q+1$
using a direct proof with cases we see that when $n$ is of the form $3k$ it's in the form $3q$ after squaring. Also when n is in the form $3k+1$, $n$ squared is still in the form $3q+1$ after squaring. Lastly we saw that when $n$ was in the form $3k+2$ we could simplify to the form $3q+1$.
|
Proof is easy ($k \in Z$)
1) if $n = 3k$, $n^2 = 9k^2$, QED
2) if $n = 3k + 1$, then $n^2 = 3kn + n = 3kn + 3k + 1$, QED
3) if $n = 3k + 2$, then $n^2 = 3kn + 2n = 3kn + 6k + 3 + 1$, QED
|
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|
Simplify with no calculator $\dfrac{(8^3)(-16)^5}{4(-2)^8}$
$\dfrac{8\cdot8\cdot8\cdot-16\cdot-16\cdot-16\cdot-16\cdot-16}{4\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}$
$\dfrac{8\cdot8\cdot8\cdot-16\cdot-16\cdot-16\cdot-16\cdot-16}{2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}$
$\dfrac{8\cdot8\cdot8\cdot-16\cdot-16\cdot-16\cdot-16\cdot-16}{2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}$
is it equal to... ?
$$
\dfrac{8}{2} \cdot \dfrac{8}{2} \cdot \dfrac{8}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2}
$$
I am bit confused, How can I handle this problem ?
|
When multiplying fractions, cancel BEFORE multiplying.
$$
\dfrac{8}{2} \cdot \dfrac{8}{2} \cdot \dfrac{8}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2}
$$
Wherever you see $\dfrac{8}{2}$, put $4$.
Wherever you see $\dfrac{-16}{2}$, put $-8$.
Then you have
$$
4\cdot4\cdot4\cdot(-8)\cdot(-8)\cdot(-8)\cdot(-8)\cdot(-8).
$$
|
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|
Proving : $ \bigl(1+\frac{1}{n+1}\bigr)^{n+1} \gt (1+\frac{1}{n})^{n} $ How could we prove that this inequality holds
$$ \left(1+\frac{1}{n+1}\right)^{n+1} \gt \left(1+\frac{1}{n} \right)^{n} $$
where $n \in \mathbb{N}$, I think we could use the AM-GM inequality for this but not getting how?
|
This is equivalent to showing that
$$
\left(\frac{n}{n-1}\right)^{n-1}\tag{1}
$$
is an increasing function of $n$. Consider the Taylor expansion of
$$
\begin{align}
(n-1)\log\left(\frac{1}{1-1/n}\right)
&=(n-1)\left(\frac1n+\frac12\frac1{n^2}+\frac13\frac1{n^3}+\dots\right)\\
&=1-\frac1{1\cdot2}\frac1n-\frac1{2\cdot3}\frac1{n^2}-\frac1{3\cdot4}\frac1{n^3}+\dots\tag{2}
\end{align}
$$
$(2)$ is obviously an increasing function of $n$. QED
Another useful case
$$
\left(\frac{n}{n-1}\right)^n\tag{3}
$$
is a decreasing function of $n$.
$$
\begin{align}
n\log\left(\frac{1}{1-1/n}\right)
&=n\left(\frac1n+\frac12\frac1{n^2}+\frac13\frac1{n^3}+\dots\right)\\
&=1+\frac12\frac1n+\frac13\frac1{n^2}+\dots\tag{4}
\end{align}
$$
$(4)$ is obviously a decreasing function of $n$.
A boundary case
$$
\left(\frac{n}{n-1}\right)^{n-1/2}\tag{5}
$$
is a decreasing function of $n$.
$$
\begin{align}
(n-1/2)\log\left(\frac{1}{1-1/n}\right)
&=(n-1/2)\left(\frac1n+\frac12\frac1{n^2}+\frac13\frac1{n^3}+\dots\right)\\
&=1+\frac12\left(\frac1{2\cdot3}\frac1{n^2}+\frac2{3\cdot4}\frac1{n^3}+\frac3{4\cdot5}\frac1{n^4}+\dots\right)\tag{6}
\end{align}
$$
$(6)$ is obviously a decreasing function of $n$.
Comparing $(6)$ to $(2)$ shows that $\left(1+\frac1n\right)^{n+1/2}$ is a lot closer to $e$ than is $\left(1+\frac1n\right)^n$.
|
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|
Proving a trigonometric identity How can one prove the validity of this trigonometric identity?
\begin{equation}
2\arccos\sqrt{x} = \frac{\pi }{2}-\arcsin(2x-1)
\end{equation}
|
EDITED in response to valdo's answer.
Your identity $$
2\arccos \sqrt{x}=\frac{\pi }{2}-\arcsin (2x-1),\qquad 0\le x\le 1\tag{0},
$$ may be rewritten as $$
\arcsin (2x-1)=\frac{\pi }{2}-2\arccos \sqrt{x},\qquad 0\le x\le 1\tag{1}.
$$
For identity $(1)$ to be valid$^1$ it is enough that
$$
\sin \left( \arcsin (2x-1)\right) =\sin \left( \frac{\pi }{2}-2\arccos \sqrt{
x}\right).\tag{2}
$$
The LHS of $(2)$ is $$\sin \left( \arcsin (2x-1)\right) =2x-1,\tag{3}$$ and the RHS,
$$
\sin \left( \frac{\pi }{2}-2\arccos \sqrt{x}\right) =\cos \left( 2\arccos
\sqrt{x}\right) =2\cos ^{2}\left( \arccos \sqrt{x}\right) -1\tag{4}.
$$
And so, it is enough that we have
$$2x-1 =2\cos ^{2}\left( \arccos \sqrt{x}\right) -1
=2\left( \sqrt{x}\right) ^{2}-1
=2x-1,\qquad x\ge 0,\tag{5}
$$
which is indeed an identity. Consequently, all the previous identities are valid e so, also the given identity $(0)$.
--
$^1$ See valdo's detailed explanation in his answer.
|
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|
Residue integral problem Have this Integral.
$$\int_{-\infty }^\infty \frac{\cos(2x)}{(x^2+1)(x^2+4)^2}\,dx$$
Been working on similar problems but the cos bother me in this problem. Can anyone help me get started? What should I do first?
I´m given that i could use
$$\operatorname{Re}{\left ( \int_{-\infty }^\infty \frac{e^{2ix}}{(x^2+1)(x^2+4)^2} \, dx \right )} = \int_{-\infty }^\infty \frac{\cos(2x)}{(x^2+1)(x^2+4)^2}dx$$
How can the real part of this integral help me?
|
Very much like the example II in this wiki page, we write:
$$
\int_C \frac{\mathrm{e}^{2 i z}}{(z^2+1)(z^2+4)^2} \mathrm{d} z =
\int_{-\infty}^\infty \frac{\mathrm{e}^{2 i x}}{(x^2+1)(x^2+4)^2} \mathrm{d} x + \int_{0}^{\pi} \frac{\exp(2 i R \mathrm{e}^{i \varphi} )}{((R \mathrm{e}^{i \varphi})^2+1)((R \mathrm{e}^{i \varphi})^2+4)^2} i R \mathrm{e}^{i \varphi} \mathrm{d} \varphi
$$
The latter integral vanishes as $R \to + \infty$ because
$$ \begin{multline}
\lim_{R \to + \infty} \operatorname{abs}\left( \frac{\exp(2 i R \mathrm{e}^{i \varphi} )}{((R \mathrm{e}^{i \varphi})^2+1)((R \mathrm{e}^{i \varphi})^2+4)^2} i R \mathrm{e}^{i \varphi} \right) =\\
\lim_{R \to + \infty} \frac{R \exp(-2 R \sin(\varphi))}{\sqrt{1+2 R^2 \cos(2 \varphi) + R^4} \left( 16 + 8 R^2 \cos(2 \varphi) + R^4 \right)} = 0
\end{multline}
$$
The contour integral is evaluated by residues. Let $f(z) = \frac{\mathrm{e}^{2 i z}}{(z^2+1)(z^2+4)^2}$. Then
$$
\begin{eqnarray}
\int_C \frac{\mathrm{e}^{2 i z}}{(z^2+1)(z^2+4)^2} \mathrm{d} z &=&
2 \pi i \left( \operatorname{Res}_{z=i} f(z) +
\operatorname{Res}_{z=2 i} f(z) \right ) \\
&=& 2 \pi i \left( \left. (z-i)f(z) \right|_{z=i} + \left. \frac{\mathrm{d}}{\mathrm{d} z} (z-2 i)^2 f(z) \right|_{z=2i} \right) \\
&=& \frac{\pi}{9 \mathrm{e}^2} - \frac{23 \pi}{144 \mathrm{e}^4}
\end{eqnarray}
$$
Since the result is real, it is also the value of the original integral:
$$
\int_{-\infty}^\infty \frac{\cos(2x)}{(x^2+1)(x^2+4)^2} \mathrm{d} x =
\frac{\pi}{9 \mathrm{e}^2} - \frac{23 \pi}{144 \mathrm{e}^4} \approx 0.038
$$
|
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|
How many ordered pairs of positive integers
For a prime integer p, how many ordered pairs of positive integers
(a, b) are there that satisfy $$\frac{1}{a} + \frac{1}{b} =\frac{1}{p}$$
For example, for p = 5, $$\frac{1}{6} + \frac{1}{30}$$ and
$$\frac{1}{30} + \frac{1}{6}$$ are two different ways of getting
$\frac{1}{5}$.
Ok, so I tried this for prime numbers from 2 to 19. It seems like that for any prime number there are only 3 such ordered pairs.
(Question: $\frac{1}{10} + \frac{1}{10}$ is just one ordered pair for p=5, right?)
But I don't know how to go about proving this. I can see that:
After $\frac{1}{p}$, take the next smallest fraction of the form $\frac{1}{a}$. Now we always get a fraction $\frac{1}{a} + \frac{1}{b}$. Also, now b is the upper bound for the numbers we need to check. But there is always just one other integer between a and b which satisfies for this property, this integer is is 2p
The question does not specifically ask for a prove, but they always expect a prove for everything.
|
If we multiply through by $pab$, we find that for $a, b \ne 0$, our equation is equivalent to $pb+pa=ab$, which we can rewrite as
$$(a-p)(b-p)=p^2.$$
We are looking for non-zero solutions of this equation. Conveniently, $p^2$ does not have many factorizations!
We can have $a-p=-1$, $b-p=-p^2$, which yields a negative $b$, or $a-p=-p^2$, $b-p=-1$, which yields a negative $a$.
We could have $a-p=-p$, $b-p=-p$, but that yields the impossible $a=b=0$.
We can have $a-p=1$, $b-p=p^2$, or $a-p=p^2$, $b-p=1$, which respectively yield the solutions $a=p+1$, $b=p^2+p$, and $a=p^2+p$, $b=p+1$.
Finally, we can have $a-p=p$, $b-p=p$, which yields $a=b=2p$.
Comment: The same idea can be used to find all integer solutions of the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$, where $n$ is a given positive integer. The factorizations of $n^2$, with the exception of $n^2=(-n)(-n)$, supply the integer solutions.
|
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|
A pencil approach to find $\sum \limits_{i=1}^{69} \sqrt{\left( 1+\frac{1}{i^2}+\frac{1}{(i+1)^2}\right)}$ What is the fastest, paper-pencil method of finding $$\sum \limits_{i=1}^{69} \sqrt{\left( 1+\frac{1}{i^2}+\frac{1}{(i+1)^2}\right)}?$$
This is actually a quantitative aptitude problem, and hence the solutions should be fast enough and probably under a minute.
Using wolframalpha, the sum seems to be $\frac{4899}{70}$; however, I am not sure how to find this quickly in the paper-pencil way.
|
This is a late response (since I joined three weeks ago) but here is another method. Write
\begin{align}
1 + \frac{1}{i^2} + \frac{1}{(i + 1)^2} &= 1 + \frac{2}{i(i+1)} + \left(\frac{1}{i^2} - \frac{2}{i(i+1)} + \frac{1}{(i+1)^2}\right)\\
&= 1 + 2\left(\frac{1}{i} - \frac{1}{i+1}\right) + \left(\frac{1}{i} - \frac{1}{i+1}\right)^2\\
&= \left(1 + \frac{1}{i} - \frac{1}{i+1}\right)^2.
\end{align}
Then
\begin{equation}
\sum_{i = 1}^{69} \sqrt{1 + \frac{1}{i} + \frac{1}{i^2}} = \sum_{i = 1}^{69} \left(1 + \frac{1}{i} - \frac{1}{i + 1}\right) = 69 + \sum_{i = 1}^{69} \left(\frac{1}{i} - \frac{1}{i+1}\right) = 70 - \frac{1}{70} = \frac{4899}{70}.
\end{equation}
|
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|
General formula to obtain triangular-square numbers I am trying to find a general formula for triangular square numbers. I have calculated some terms of the triangular-square sequence ($TS_n$):
$TS_n=$1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056
1882672131025, 63955431761796, 2172602007770041, 73804512832419600
I have managed to find that:
$m^2 = \frac{n(n+1)}{2}$
where $m$ is the $m^{th}$ term of the square number sequence, and $n$ is the $n^{th}$ term of the triangular numbers sequence.
Would anyone be able to point out anything that could help me derive a formula for the square-triangular sequence?
|
$${t_n} = \sum\limits_{k = 1}^n k = 1 + 2 + 3 + \cdots + n = \frac{{n\left( {n + 1} \right)}}{2} = n-th{\text{ triangular number}}$$
$${s_m} = {\text{m-th square number}} = {m^2}$$
$${s_m} = {t_n} \Rightarrow \frac{1}{2}n\left( {n + 1} \right) = {m^2}$$
$$\frac{1}{2}{\left( {n + \frac{1}{2}} \right)^2} = \frac{1}{2}\left( {{n^2} + n + \frac{1}{4}} \right) = \frac{1}{2}\left( {{n^2} + n} \right) + \frac{1}{8}$$
$$\frac{1}{2}n\left( {n + 1} \right) = \frac{1}{2}\left( {{n^2} + n} \right) = \frac{1}{2}{\left( {n + \frac{1}{2}} \right)^2} - \frac{1}{8} = {m^2}$$
$$\frac{1}{2}{\left( {n + \frac{1}{2}} \right)^2} - {m^2} = \frac{1}{8}$$
$$4{\left( {n + \frac{1}{2}} \right)^2} - 8{m^2} = 1$$
$$2 \cdot \left( {n + \frac{1}{2}} \right) \cdot 2\left( {n + \frac{1}{2}} \right) - 8{m^2} = 1$$
$${\left( {2n + 1} \right)^2} - 8{m^2} = 1$$
$${\left( {2n + 1} \right)^2} - 2 \cdot {\left( {2m} \right)^2} = 1$$
$$\boxed{w \equiv 2n + 1}$$
$$\boxed{z \equiv 2m}$$
$$\boxed{{w^2} - 2{z^2} = 1}$$
Finding numbers that satisfy the last equation above isn't all that simple... so I found the first numbers that are both triangular and square using python
import pandas as pd
import numpy as np
triangluar_and_square = []
for n in np.arange(1,10000):
w = 2*n + 1
for m in np.arange(1,10000):
z = 2*m
if w*w - 2*z*z - 1 == 0:
triangluar_and_square.append([m*m,m,n])
triangular_and_square = pd.DataFrame(triangluar_and_square,
columns = ["Square (m-squared) + Trianglular","m","n"],
index = [1,2,3,4,5,6])
print(triangular_and_square)`
Code output
$$\begin{array}{*{20}{c}}
& {{\rm{Square (}}{m^2}){\rm{ and Triangular}}}&& m&& n& \\
\hline
& 1&& 1&& 1& \\
& {36}&& 6&& 8& \\
& {1225}&& {35}&& {49}& \\
& {41616}&& {204}&& {288}& \\
& {1413721}&& {1189}&& {1681}& \\
& {48024900}&& {6930}&& {9800}&
\end{array}$$
|
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|
Integrate $\int\frac{1}{x^6} \sqrt{(1-x^2)^3} ~ dx$ How to integrate the following?
$$\int\frac{\sqrt{(1-x^2)^3}}{x^6} \;dx .$$
|
Integral,
$$
\begin{align*}
I &= \int \frac{\sqrt{(1-x^2)^{3}}}{x^6} dx
\\ &= \int \left( \frac{1-x^2}{x^2} \right)^{3/2} \frac{1}{x^3}dx
\\ &= \int \left(\frac{1}{x^2}-1 \right)^{3/2} \cdot \frac{1}{x^3} dx
\end{align*}
$$
Make the substitution: $z= \frac{1}{x^2}-1$, so that $dz=\frac{-2}{x^3} ~dx$.
$$
\begin{align*}
\text{Integral}
&= -\frac{1}{2}\int z^{3/2} dz
\\ &= -\frac{1}{2} \cdot \frac{2}{5}z^{5/2}+C
\\ &= -\frac{1}{5} \left(\frac{1}{x^2} - 1 \right)^{5/2}+C
\\ &= - \frac{1}{5} \left(\frac{\sqrt{1-x^2}}{x} \right)^5+C.
\end{align*}
$$
|
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|
Solving $(2y-4)(2y+1) = (2y-2)^2$ I'm getting different answer from answer key.
Solving $$(2y-4)(2y+1) = (2y-2)^2$$
FOIL left side $$4y^2+2y-8y-4 = (2y-2)^2$$
Right side $$4y^2+2y-8y-4 = 4y^2+4 $$
Subtract $4y^2$ from both sides
$$2y-8y-4 = 4 $$
Combine $y$
$$6y-4 = 4$$
add 4 to both sides
$$6y = 8$$
But the answer key has $y=4$
|
You're unfolding the right-hand side wrong -- $(2y-2)^2$ is not $4y^2+4$, but $4y^2+4-8y$.
|
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|
chances of getting three of one kind and four of another out of seven dice There are several questions similar to this one but after reading those,
I am still very confused.
I also did a similar problem of this one and I think I got it, but then I got stuck again.
So if four dice are rolled, the chance of getting three of a kind is:
$ \binom{6}{1} \frac{1}{6}* \binom{5}{1} \frac{1}{6}*\binom{4}{1} \frac{1}{6} *\frac{1}{6}$
so if seven dice are rolled, in my understanding,
the chance of getting three of a kind and four of another would be:
$ \binom{7}{1} \frac{1}{6}*\binom{6}{1} \frac{1}{6}*\binom{5}{1} \frac{1}{6} *\binom{4}{1} \frac{1}{6} *\binom{3}{1} \frac{1}{6} *\binom{2}{1} \frac{1}{6} *\binom{1}{1} \frac{1}{6} $
however, the answer in the book is $ \frac{6 *5*\binom{7}{4} }{6^7} $ and I am totally lost.
Please help!
additional problems
The answers you guys gave kind of make sense to me but they also make me very confused.
can I think of it using the way I did above?
for example, another part of the questions asked about the chance of getting two fours, two fives and three sixes.
I think of it as:
$ \binom{7}{2} \frac{1}{6}^2*\binom{5}{2} \frac{1}{6}^2*\binom{3}{3} \frac{1}{6}^3 $ which matches the solution in the book.
|
There are $6^7$ possible outcomes, listed as 7-tuples. Now, how many 7-tuples would consist of two distinct elements, 4 of one kind, and 3 of other. Such 7-tuples can be ordered into $[a,a,a,a,b,b,b]$ with $a \not= b$. $a$ can be chosen 6 different ways. Once $a$ is chosen, $b$ can be chosen 5 different ways. There are $\binom{7}{4}$ ways to rearrange $[a,a,a,a,b,b,b]$ tuple.
The probability is then the ratio of the number of needed outcomes, over the number of total outcomes:
$$
p = \frac{1}{6^7} \binom{7}{4} 6 \times 5
$$
|
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|
Determinant of a specific circulant matrix, $A_n$ Let
$$A_2 = \left[ \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]$$
$$A_3 = \left[ \begin{array}{ccc} 0 & 1 & 1\\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]$$
$$A_4 = \left[ \begin{array}{cccc} 0 & 1 & 1 & 1\\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0\end{array}\right]$$
and so on for $A_n$.
I was asked to calculate the determinant for $A_1, A_2, A_3, A_4$ and then guess about the determinant for $A_n$ in general. Of course the pattern is clear that
$$ \det A_n = (n-1)(-1)^{n-1} $$
but I was wondering as to what the proof of this is. I tried to be clever with cofactor expansions but I couldn't get anywhere.
Could someone explain it to me please?
|
The determinant of a general circulant matrix is given by
\begin{eqnarray*}
\left| \begin{array} {ccccc}
x_0 & x_{n-1} & \cdots & x_2 &x_1 \\
x_1 & x_0 &\cdots & x_3 & x_2\\
x_2 & x_1 & \cdots & x_4& x_3 \\
\vdots & \vdots & ~ & \vdots & \vdots \\
x_{n-2} & x_{n-3} &\cdots &x_0 & x_{n-1} \\
x_{n-1} & x_{n-2} &\cdots &x_1 & x_0 \end{array} \right|
= \prod_{p=0}^{n-1} \left( x_0 + \omega^p x_1 +\cdots + \omega^{(n-1)p}x_{n-1} \right)
\end{eqnarray*}
where $\omega = e^{2 \pi i /n} $ is an $n$th root of unity; it satisfies the identity $ \sum_{q=1}^{n-1} \omega^{qd} = -1 $ for
$d \neq 0$. So
\begin{eqnarray*}
\left| \begin{array} {ccccc}
0 & 1 & \cdots & 1 &1 \\
1 & 0 &\cdots & 1 & 1\\
1 & 1 & \cdots & 1& 1 \\
\vdots & \vdots & ~ & \vdots & \vdots \\
1 & 1 &\cdots & 0 & 1\\
1 & 1 &\cdots & 1 & 0 \end{array} \right|
= (0+1+\cdots +1) \prod_{p=1}^{n-1} \left( \omega^p +\cdots + \omega^{(n-1)p} \right)= (n-1)\prod_{p=1}^{n-1} (-1)= (n-1)(-1)^{n-1}
\end{eqnarray*}
|
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|
Which is the "fastest" way to compute $\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} $?
I am looking for the "fastest" paper-pencil approach to compute $$\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} $$
This is a quantitative aptitude problem and the correct/required answer is $3.75$
In addition, I am also interested to know how to derive a closed form for an arbitrary $n$ using mathematica I got $$\sum \limits_{i=1}^{n} \frac{10i-5}{2^{i+2}} = \frac{5 \times \left(3 \times 2^n-2 n-3\right)}{2^{n+2}}$$
Thanks,
|
First note that $\displaystyle\frac{10i-5}{2^{i+2}} = \frac{5(2i-1)}{2^{i+2}}$
Secondly, using the sum of a geometric series you can show that $2^0 + 2^1 + 2^2 + ... + 2^k = 2^{k+1}-1$.
$\displaystyle \begin{align*} \text{So } \sum \limits_{i=1}^{n} \frac{10i-5}{2^{i+2}}
&= 5(\frac{2\times1-1}{2^{1+2}}+\frac{2\times2-1}{2^{2+2}}+\frac{2\times3-1}{2^{3+2}}+...+\frac{2\times n-1}{2^{n+2}})\\
&= 5(\frac{2^{n-1}(2\times1-1)+2^{n-2}(2\times2-1)+2^{n-3}(2\times3-1)+...+2^{0}(2\times n-1)}{2^{n+2}}) \\
&= 5(\frac{2(2^{n-1}\times1+2^{n-2}\times2+2^{n-3}\times3+...+2^{0}\times n)-(2^{n-1}+2^{n-2}+2^{n-3}+...+1)}{2^{n+2}})\\
&= 5(\frac{2(2^{n-1}\times1+2^{n-2}\times2+2^{n-3}\times3+...+2^{0}\times n)-(2^n-1)}{2^{n+2}})\\
&= 5(\frac{2((2^{n-1}+2^{n-2}+2^{n-3}+...+1)+(2^{n-2}+2^{n-3}+2^{n-4}+...+1)+...+(2+1)+(1))-(2^n-1)}{2^{n+2}})\\
&= 5(\frac{2((2^{n}-1)+(2^{n-1}-1)+(2^{n-2}-1)+...+(2^{2}-1)+(2-1))-(2^n-1)}{2^{n+2}})\\
&= 5(\frac{2((2^{n}+2^{n-1}+2^{n-2}+...+2) + (-1)\times n)-(2^n-1)}{2^{n+2}})\\
&= 5(\frac{2(2\times (2^n - 1))- n)-(2^n-1)}{2^{n+2}})\\
&= 5(\frac{4 \times 2^n - 4 - 2n -2^n + 1}{2^{n+2}})\\
&= 5(\frac{3 \times 2^n - 2n - 3}{2^{n+2}})\text{ which is the closed form you got from Mathematica}\end{align*}$
|
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$5^{x}+2^{y}=2^{x}+5^{y} =\frac{7}{10}$ Work out the values of $\frac{1}{x+y}$ $5^{x}+2^{y}=2^{x}+5^{y} =\frac{7}{10}$
Work out the values of $\frac{1}{x+y}$
|
Initial Solution:
$5^x+2^y=7/10$
$5^x+2^y=(5+2)/(5*2)=(1/2+1/5)=5^{-1}+2^{-1}$ ------ (1)
$5^y+2^x=(5+2)/(5*2)=(1/5+1/2)=5^{-1}+2^{-1}$ ------ (2)
By examining (1) and (2) we get the solution:
$x=y=-1$
Now we consider the relation:
$5^x+2^y=k$ --------(1)
where k is a constant.
$\frac{dy}{dx}=-\frac{5^x}{2^y}log_{2} 5<0$ for all values of x and y. So y is a decreasing function of x.
We consider another relation:
$5^y+2^x=k$ --------------(2)
where k is a constant.
$\frac{dy}{dx}=-\frac{2^x}{5^y}\frac{1}{log_2 5}<0$
Again , the function considered is a decreasing one.
But the slope of the first function[expressed by relation (1)] is always greater in magnitude than the slope of the second one[expressed by relation (2)] For a particular point on the x-axis the curve of the first function has a steeper slope than the second one. So their curves can intersect at only one point.
Incidentally to show the greater steepness/slope of the first curve we may write,
$\frac{5^x}{2^y}log_{2} 5-\frac{2^x}{5^y}\frac{1}{log_{2} 5}$
$=\frac{5^{x+y}(log_{2} 5)^2-2^{x+y}}{10^y log_{2} 5}>0$
Therefore,
$1/(x+y)=-1/2$
|
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|
Problem based on Range Find $a$ and $b$ such that the inequality $a \le 3 \cos{x} + 5\cos\left(x - \frac{\pi}{6}\right) \le b$ holds good for all x.
|
Remark: In the meantime this approach has been also added by André Nicolas to his answer.
We will show that the sum $3\cos x+5\cos \left( x-\frac{\pi }{6}\right) $
can be written as $C\sin (x+\phi )$. From the difference formula for $\cos
\left( x-\frac{\pi }{6}\right) $ and using the values $\cos \frac{\pi }{6}=
\frac{1}{2}\sqrt{3}$ and $\sin \frac{\pi }{6}=\frac{1}{2}$ we get
$$
3\cos x+5\cos \left( x-\frac{\pi }{6}\right) =\left( 3+\frac{5}{2}\sqrt{3}
\right) \cos x+\frac{5}{2}\sin x.
$$
The general identity
$$
\begin{eqnarray*}
A\cos x+B\sin x &=&C\sin (x+\phi ) \\
&=&C\sin \phi \cos x+C\cos \phi \sin x
\end{eqnarray*}
$$
holds if $C\sin \phi =A$, $C\cos \phi =B$ i.e. $C=\sqrt{A^{2}+B^{2}}$ and $
\phi =\arctan \frac{A}{B}$. For $A=3+\frac{5}{2}\sqrt{3}$ and $B=\frac{5}{2}$
it takes the form
$$
\left( 3+\frac{5}{2}\sqrt{3}\right) \cos x+\frac{5}{2}\sin x=\sqrt{34+15
\sqrt{3}}\sin \left( x+\arctan \left( \frac{6}{5}+\sqrt{3}\right) \right).
$$
For $C>0$ the function $C\sin (x+\phi )\in \left[ -C,C\right] $.
Consequently the best narrow bounds are $b=-a=C=\sqrt{34+15\sqrt{3}}$.
|
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How do you divide a polynomial by a binomial of the form $ax^2+b$, where $a$ and $b$ are greater than one? I came across a question that asked me to divide $-2x^3+4x^2-3x+5$ by $4x^2+5$. Can anyone help me?
|
Nobody seems to have posted the synthetic division route; I'll include it here for completeness.
We first monicize the divisor (i.e., set things up such that the divisor has leading coefficient $1$); thus, consider
$$\frac{-\frac12x^3+x^2-\frac34x+\frac54}{x^2+\frac54}$$
and set up the array
$$\begin{array}{r|cc|cc}
&-\frac12&1&-\frac34&\frac54\\\hline
0&\times& & &\times\\
-\frac54&\times&\times& & \\\hline
& & & &
\end{array}$$
Note that the coefficients in the leftmost column are the negated coefficients of the terms with degree less than the degree of the polynomial. The second dividing line (can't do dashed lines in an array environment, sorry) indicates that we expect the quotient to be linear ($3-2=1$). The $\times$ entries indicate the sections of the array that shouldn't be filled.
This generalized synthetic division proceeds like this:
$$\begin{array}{r|cc|cc}
&-\frac12&1&-\frac34&\frac54\\\hline
0&\times& & &\times\\
-\frac54&\times&\times& & \\\hline
&-\frac12& & &
\end{array}$$
(bring down leading coefficient)
$$\begin{array}{r|cc|cc}
&-\frac12&1&-\frac34&\frac54\\\hline
0&\times&0& &\times\\
-\frac54&\times&\times&\frac58& \\\hline
&-\frac12& & &
\end{array}$$
(multiply the bottom of the first column by the coefficients in the leftmost column, and fill the array diagonally)
$$\begin{array}{r|cc|cc}
&-\frac12&1&-\frac34&\frac54\\\hline
0&\times&0& &\times\\
-\frac54&\times&\times&\frac58& \\\hline
&-\frac12&1& &
\end{array}$$
(add numbers in second column)
$$\begin{array}{r|cc|cc}
&-\frac12&1&-\frac34&\frac54\\\hline
0&\times&0&0&\times\\
-\frac54&\times&\times&\frac58&-\frac54\\\hline
&-\frac12&1& &
\end{array}$$
(multiply bottom of the second column by the coefficients in the leftmost column and fill diagonally)
$$\begin{array}{r|cc|cc}
&-\frac12&1&-\frac34&\frac54\\\hline
0&\times&0&0&\times\\
-\frac54&\times&\times&\frac58&-\frac54\\\hline
&-\frac12&1&-\frac18&0
\end{array}$$
(add numbers in third and fourth columns)
We thus have the result
$$\frac{-\frac12x^3+x^2-\frac34x+\frac54}{x^2+\frac54}=-\frac12 x+1+\frac{-\frac18x+0}{x^2+\frac54}=-\frac12 x+1+\frac{-\frac12x}{4x^2+5}$$
|
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|
Are quartic minimal polynomials over $\mathbb{Q}$ always reducible over $\mathbb{F}_p$? This situation arose while studying biquadratic extensions.
Let $\mathbb{Q}(\alpha)$ is some biquadratic extension, with $m(x)$ the minimal polynomial of $\alpha$. Suppose that $m(x)\in\mathbb{Z}[x]$. Basic field theory and simple definitions tells us that $\deg(m(x))=4$, and $m(x)$ is irreducible over $\mathbb{Q}$. However, is $m(x)$ always reducible over $\mathbb{F}_p$, for $p$ a prime?
The reason I'm interested is that I noticed that the quartic $x^4+1$, known to be irreducible over $\mathbb{Q}$ is actually reducible over $\mathbb{F}_p$ for the first few primes at least. For example, $x^4+1$ can be reduced as $(x+1)^4\pmod{2}$, $(x^2+x-1)(x^2-x-1)\pmod{3}$, $(x^2+2)(x^2-2)\pmod{5}$, $(x^2+3x+1)(x^2-3x+1)\pmod{7}$.
So I'm wondering if $m(x)$ is the minimal polynomial of some algebraic integer of degree $4$, is it actually reducible over $\mathbb{F}_p$ for any prime?
|
An elementary argument, without any explicit reference to Galois theory:
Suppose the biquadratic extension is $K = \mathbb{Q}(\sqrt{B}, \sqrt{C})$, where $B$ and $C$ are integers (and none of $B$, $C$, $BC$ a rational square). Then any element $\alpha$ of $K$ has the form
$\alpha = a + b\sqrt{B} + c\sqrt{C} + d \sqrt{BC}$
for some $a, b, c, d \in \mathbb{Q}$. This $\alpha$ satisfies the polynomial
$$\begin{align*}f(X) =& (X - a - b\sqrt{B} - c\sqrt{C} - d\sqrt{BC})\cdot(X - a + b\sqrt{B} + c\sqrt{C} - d\sqrt{BC})\\
&\qquad{}\cdot (X - a + b\sqrt{B} - c\sqrt{C} + d\sqrt{BC})\cdot(X - a - b\sqrt{B} + c\sqrt{C} + d\sqrt{BC}).\end{align*}$$
Convince yourself that $f(X)$ is in $\mathbb{Q}[X]$. (This is where a bit of Galois theory helps.) If $\alpha$ is integral over $\mathbb Q$, then in fact $f(X) \in \mathbb{Z}[X]$. If $\alpha$ generates $K$ over $\mathbb{Q}$, then $f(X)$ is irreducible and is the minimal polynomial of $\alpha$ over $\mathbb{Q}$. We want to show that if $\alpha$ is integral then $f(X)$ will always factor in $\mathbb{F}_p[X]$.
The key observation is that at least one of $B$, $C$, and $BC$ will be a square modulo $p$. Convince yourself that this is true: the product of two nonsquares modulo $p$ is a square modulo $p$. (One way to do this is note that a nonsquare has to be an odd power of a generator of $\mathbb{F}_p^\times$.)
So let's suppose that $BC \equiv S^2 \pmod{p}$. (The argument will work essentially the same way if we assume that $B$ or $C$ is a square modulo $p$ instead; convince yourself of this at the end.)
So we have
$\begin{align*}f(X) =& \big((X - a - d\sqrt{BC}) - (b\sqrt{B} + c\sqrt{C})\big)\big((X - a - d\sqrt{BC}) + (b\sqrt{B} + c\sqrt{C})\big)\\
&\qquad {} \cdot \big((X - a + d\sqrt{BC}) + (b\sqrt{B} - c\sqrt{C})\big)\big((X - a + d\sqrt{BC}) - (b\sqrt{B} - c\sqrt{C})\big)\\ =& \big((X - a - d\sqrt{BC})^2 - (b\sqrt{B} + c\sqrt{C})^2\big)\\ &\qquad {} \cdot \big((X - a + d\sqrt{BC})^2 - (b\sqrt{B} - c\sqrt{C})^2\big)\\ =& \big((X - a)^2 + d^2BC - 2d(X-a)\sqrt{BC} - b^2B - c^2C - 2bc \sqrt{BC}\big)\\ &\qquad {} \cdot \big((X - a)^2 + d^2BC + 2d(X-a)\sqrt{BC} - b^2B - c^2C + 2bc \sqrt{BC}\big)\\ \equiv& \big((X - a)^2 + d^2BC - 2d(X-a)S - b^2B - c^2C - 2bc S\big)\\ &\qquad {} \cdot \big((X - a)^2 + d^2BC + 2d(X-a)S - b^2B - c^2C + 2bcS\big) \pmod{p},\end{align*}$
which is a product of two quadratics in $\mathbb{F}_p[X]$.
In the case where all three of $B$, $C$, $BC$ are squares modulo $p$, the quadratic factors of $f(X)$ will split further into linear factors in $\mathbb{F}_p[X]$.
As to your other question -- or maybe it was just a typo, but in any case -- no, it's not true that every irreducible monic quartic polynomial in $\mathbb{Z}[X]$ factors over every $\mathbb{F}_p$. For example, $g(X) = X^4 + X + 1$ is irreducible in $\mathbb{F}_2[X]$. To see this, first plug $X= 0$ and $X =1$ into $g(X)$ to see that there are no mod $2$ roots. To see that $g(X)$ doesn't factor into a product of two irreducible quadratics, note that $X^2 + X + 1$ is the only monic irreducible quadratic of $\mathbb{F}_2$ (write them all down and check for roots), and $(X^2 + X + 1)^2 = X^4 + 2X^3 + 3X^2 + 2X + 1 \equiv X^4 + X^2 + 1 \not\equiv g(X)$. Since $g(X)$ is not divisible by any irreducible linears or quadratics, it has to be irreducible.
In fact, you can show that every $\mathbb{F}_p[X]$ has irreducible polynomials of every degree.
|
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|
Simple probability problems This may be very simple, but I need to know correct answers.
Problem 1.
There are $6$ balls in the box - $2$ black, and $4$ white. We take $3$ balls from the box. What is the probability that we took exactly 1 black ball.
Problem 2.
What is the probability to score $7$ points when you throw two dices.
My answers
Problem 1: ${\binom{2}{1}\binom{4}{2} \over \binom{6}{3}}=0.6$
Problem 2: Denote by $(s,c)$ a pair where s - is a score, c - amount of possible outcomes. Then we have $(2,1) (3,1) (4,2) (5,2) (6,3) (7,3) (8,3) (9,2) (10,2) (11,1) (12,1)$. Amount of possible outcomes is 21. Amount of desired outcomes is 3, hence $p=\frac 3 {21}=\frac 17$
|
The critical distinction to make in problem 1 is whether you're sampling with or without replacement. This is, when you pick a ball out of the jar, do you note its color and put in back in, or keep it it. If it's the former case, you're sampling with replacement. In that case, the number of black balls in a sample of three is binomially distributed variable (thanks to Dilip for pointing out this & that my original analysis was incorrect) with $n=3$ and $p = \frac{1}{3}$. Thus, the probability of drawing one black ball is given by:
$$\binom{3}{1}\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^2 = \frac{4}{9}$$
If you're sampling with out replacement, the solution is a little different. Consider, you could pick a black ball, then two whites, a white, then a black, then a white, or a two whites and then a black. The probabilities of those situations will just be the products of the probabilities of each draw. Thus, they will be: $$\frac{2}{6}\frac{4}{5}\frac{3}{4} , \frac{4}{6}\frac{2}{5}\frac{3}{4} , \frac{4}{6}\frac{3}{5}\frac{2}{4}$$ respectively. The sum, then, will be the total probability of drawing two white balls and one black: $$\frac{2}{6}\frac{4}{5}\frac{3}{4} + \frac{4}{6}\frac{2}{5}\frac{3}{4} + \frac{4}{6}\frac{3}{5}\frac{2}{4} = \frac{3(4*3*2)}{6*5*4} = \frac{3}{5}$$ This is equivalent to the solution you posted for this problem.
|
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Conditions for intersection of parabolas? What are the conditions for the existence of real solutions for the following equations:
$$\begin{align}
x^2&=a\cdot y+b\\
y^2&=c\cdot x+d\end{align}$$
where $a,b,c,d $ are real numbers.
These represent two parabolas; how might we find out the conditions for the existence of $0,2,4$ real solutions of the equations?
|
Since these are parabolas, $a$ and $c$ must be nonzero. Let $x = a^{2/3} c^{1/3} X$ and $y = a^{1/3} c^{2/3} Y$. Under this scaling, the equations become
$X^2 = Y + B$ and $Y^2 = X + D$ where $B = b a^{-4/3} c^{-2/3}$ and
$D = d a^{-2/3} c^{-4/3}$. Now substituting $Y = B - X^2$ into $Y^2 = X + D$
we get the fourth-degree equation $X^4 - 2 B X^2 - X + B^2 - D = 0$. The discriminant of this, according to Maple, is $-256\,{B}^{3}+288\,B D -27+256\,{B}^{2}{D}^{2}-256\,{D}
^{3}$. The curve where the discriminant is $0$ separates the $BD$ plane into three regions like this:
We have: no real solution in the red region, one on the red-yellow boundary,
two in the yellow region, three on the yellow-blue boundary (except at the sharp cusp $B=D=3/4$ where there are two), and four in the blue region.
|
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|
Find $ n\geq1 $ such that 7 divides $n^n-3$ Find $ n\geq1 $ such that 7 divides $n^n-3$.
Here is what I found:
$ n\equiv 0 \mod7, n^n\equiv 0 \mod7,n^n-3\equiv -3 \mod7$ no solution.
$ n\equiv 1 \mod7, n^n\equiv 1 \mod7,n^n-3\equiv -2 \mod7 $ no solution.
$ n\equiv 2 \mod7, n^n\equiv 2^n \mod7, n^n-3\equiv 2^n-3 \mod7$. Is $2^n\equiv 3 \mod7 $ possible?
$ n=7k+2,..., 2^{7k+2}\equiv 2^{k+2} \mod7$. Is $2^{k+2}\equiv 3 \mod7$ possible?
Studying $k$ modulo 6: $2^{6q+2}\equiv 4 \mod7, 2^{6q+3}\equiv 1 \mod7, 2^{6q+4}\equiv 2 \mod7, 2^{6q+5}\equiv 4 \mod7, 2^{6q+6}\equiv 1 \mod7, 2^{6q+7}\equiv 2 \mod7 $
The congruence is never 3 so there is no solution for $n\equiv 2 \mod7$.
$ n\equiv -2 \mod7,..., n=42q+5 $
$ n\equiv-1 \mod7, n^n-3\equiv (-1)^n-3 \mod7 $: no solution.
$ n\equiv 3 \mod7, n^n-3 \equiv 3^n-3 \mod7$. Is $3^n\equiv 3 \mod 7$ possible?
$n=7k+3,..., 3^{7k+3}\equiv -3^k \mod7 $. Is $3^k \equiv 4 \mod 7 $ possible?
Studying $k$ modulo 6:
$ 3^{6q} \equiv 1 \mod7, 3^{6q+1} \equiv 3 \mod7, 3^{6q+2} \equiv 2 \mod7, 3^{6q+3} \equiv -1 \mod7, 3^{6q+4} \equiv 4 \mod7 , 3^{6q+5} \equiv -2 \mod7 $
So $ n=7k+3=7(6q+4)+3=42q+31 $ is a solution.
$ n\equiv -3 \mod7 $,..., there is no solution.
$42q+31 \equiv 3 \mod 7, (42q+31)^{42q+31} \equiv 3^{42q+31} \mod 7 \equiv 3 \mod7$ OK
$ 42q+5 \equiv 4 \mod7, (42q+5)^{42q+5} \equiv 5^{42q+5} \mod 7 \equiv 3 \mod7$ OK
So 7 divides $ n^n-3 $ if and only if $ n\equiv 31 \mod 42$ or $ n\equiv 5 \mod 42$
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$n^n - 3$ is divisible by $7$ if and only if $n \equiv 5\mod 42$ or $n \equiv 31 \mod 42$.
If $n \equiv 5 \mod 42$, then $n \equiv 5 \mod 7$. Thus $n^n \equiv 5^n \equiv 5^{n-5} \cdot 5^5 \equiv 5^5 \equiv 3 \mod 7$, because $n-5$ is divisible by $6$.
If $n \equiv 31 \mod 42$, then $n \equiv 31 \equiv 3 \mod 7$. As before, $n^n \equiv 3^n \equiv 3^{n-31} \cdot 3^{31} \equiv 3^{31} \equiv 3 \mod 7$.
For the converse, suppose that $n^n \equiv 3 \mod 7$. It is clear that $n \not\equiv 0 \mod 7$ and $n \not\equiv 1 \mod 7$. Possible powers of $2$ and $4$ modulo $7$ are $1$, $2$ and $4$, so $n \not\equiv 2 \mod 7$ and $n \not\equiv 4 \mod 7$. Similarly the possible powers of $6$ modulo $7$ are $1$ and $6$, so $n \not\equiv 6 \mod 7$.
Thus there are two possibilities: $n \equiv 3 \mod 7$ or $n \equiv 5 \mod 7$.
If $n \equiv 3 \mod 7$, then $n^n \equiv 3^n \mod 7$ and $3^n \equiv 3 \mod 7$. Thus $3^{n-1} \equiv 1 \mod 7$, and so $6$ divides $n-1$. Thus $n \equiv 1 \mod 6$. Then $7n \equiv 7 \mod 42$ and $6n \equiv 18 \mod 42$, and so $n \equiv 31 \mod 42$.
If $n \equiv 5 \mod 7$, then $n^n \equiv 5^n \mod 7$ and $5^n \equiv 3 \equiv 5^5 \mod 7$. Thus $5^{n-5} \equiv 1 \mod 7$, and $6$ divides $n-5$. With the same argument as before, $n \equiv 5 \mod 42$.
|
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|
Two players alternately flip a coin; what is the probability of winning by getting a head?
Two players, $A$ and $B$, alternately and independently flip a coin and the first player to get a head wins. Assume player $A$ flips first. If the coin is fair, what is the probability that $A$ wins?
So $A$ only flips on odd tosses. So the probability of winning would be $$ P =\frac{1}{2}+\left(\frac{1}{2} \right)^{2} \frac{1}{2} + \cdots+ \left(\frac{1}{2} \right)^{2n} \frac{1}{2}$$
Is that right? It seems that if $A$ only flips on odd tosses, this shouldn't matter. Either $A$ can win on his first toss, his second toss, ...., or his $n^{th}$ toss. So the third flip of the coin is actually $A$'s second toss. So shouldn't it be $$P = \frac{1}{2} + \left(\frac{1}{2} \right)^{2} + \left(\frac{1}{2} \right)^{3} + \cdots$$
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The probability that the first head occurs on toss $n$ is $2^{-n}$, so the probability that the first head happens on an odd $n$ is
$$
\sum_{k=0}^{\infty}2^{-(2k+1)}=\frac{1}{2}\frac{1}{1-1/4}=2/3
$$
|
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|
Solving a $1^\infty$ indeterminate form. I'm preparing for my calculus exam and I can't solve this limit:
$$\lim_{x\rightarrow\infty}\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)^x$$
The limit tends to $1^\infty$, which is indeterminate.
I've tried several things and I couldn't solve it.
Any idea? Thanks in advance.
|
Asymptotics ...
$$\begin{align}
\operatorname{tan} \biggl(\frac{1}{x}\biggr) &= \frac{1}{x} + \frac{1}{3 x^{3}} + \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\
1 + \operatorname{tan} \biggl(\frac{1}{x}\biggr) &= 1 + \frac{1}{x} + \frac{1}{3 x^{3}} + \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\
1 - \operatorname{tan} \biggl(\frac{1}{x}\biggr) &= 1 - \frac{1}{x} - \frac{1}{3 x^{3}} - \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\\frac{1 + \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}{1 - \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)} &= 1 + \frac{2}{x} + \frac{2}{x^{2}} + \frac{8}{3 x^{3}} + \frac{10}{3 x^{4}} + \frac{64}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\\left(\frac{1 + \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}{1 - \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}\right)^{x} &= \operatorname{e} ^{2} + \frac{4 \operatorname{e} ^{2}}{3 x^{2}} + \frac{20 \operatorname{e} ^{2}}{9 x^{4}} + O \Bigl(x^{(-5)}\Bigr)
\end{align}$$
|
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|
Which is the easiest way to evaluate $\int \limits_{0}^{\pi/2} (\sqrt{\tan x} +\sqrt{\cot x})$?
Which is the easiest way to evaluate $\int \limits_{0}^{\pi/2} (\sqrt{\tan x} +\sqrt{\cot x})$?
I have reduced this problem to $$ 2\int_0^{\pi/2} \sqrt{\tan x} \ dx$$
but now, evaluating this integral is giving me some problems, simply substituting $u=\tan(x)$
and then $\mathrm{d}u=\sec^2(x)\mathrm{d}x \Rightarrow \frac{\mathrm{d}u}{1+u^2}=\mathrm{d}x$ and which in turn gives something a bit ugly, I was wondering which is the most elegant way to evaluate this?
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$${\int_0^{\frac{\pi}{2}} \sqrt{\tan x}dx + \sqrt{\cot x}dx}$$
$$={\int_0^{\frac{\pi}{2}}\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}}dx = \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\frac{\sqrt{2\sin{x}\cos{x}}}{\sqrt{2}}}dx = \sqrt{2}\int_0^{\frac{\pi}{2}} \frac{ \sin{x} + \cos{x}}{\sqrt{1 - (1 - 2 \sin{x} \cos{x})}}dx}$$
$${=\sqrt{2}\int_0^{\frac{\pi}{2}} \frac{ \sin{x} + \cos{x}}{\sqrt{1 - (\sin{x} - \cos{x})^2}}dx}$$
Let ${t = \sin{x} - \cos{x}}$, $\Large {{\small{dx}} = \frac{dt}{\sin{x} + \cos{x}}}$ $${x \to \frac{\pi}{2} \implies t = (\sin{x} - \cos{x}) \to 1}$$ $${x \to 0 \implies t = (\sin{x} - \cos{x}) \to -1}$$
$$\sqrt{2}\int_{-1}^{1} \frac{1}{\sqrt{1 - t^2}}dt = \sqrt{2}\left[\sin^{-1}{t}\right]_{-1}^{1} = \sqrt{2}\left[\frac{\pi}{2} - \left(- \frac{\pi}{2} \right) \right] = \sqrt{2} \pi $$
I think this might be the simplest approach.
|
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|
Partial derivative of an implicit function I am trying to solve this question (sorry if the translation is a bit vague):
$Z(x,y)$ is an implicit function of $x$ and $y$ given in the form of
$$x^3z^2+\frac{2}{9}y^2\sin(z) = xyz$$
in the neighborhood of $x=2, y=\pi, z=\frac{\pi}{6}$.
Find $Z_x$ and $Z_y$ at $(x,y) = (2,\pi)$
I know how to partially/totally differentiate, and I know how to find the derivative of a single-variable implicit function. How do I combine it to solve this?
I have the solution in front of me but I can't understand it. Thanks!
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Note: This answer was posted before the correction. However, the same idea still works.
I assume the given equation is
$$x^3z^2 + \frac{2}{9}y^2\sin z = \frac{\pi^2}{3}.$$
Let's now regard $z = Z(x,y)$ as a function of $x$ and $y$ and implicitly (partially) differentiate with respect to $x$:
$$\frac{\partial}{\partial x}\left( x^3z^2 \right) + \frac{\partial}{\partial x}\left(\frac{2}{9}y^2\sin z \right) = \frac{\partial}{\partial x}\left(\frac{\pi^2}{3} \right).$$
That is,
$$3x^2 z^2 + x^3 \frac{\partial}{\partial x}(z^2) + \frac{2}{9}y^2\frac{\partial}{\partial x}(\sin z) = 0,$$
so $$3x^2 z^2 + x^3 (2z)\, Z_x + \frac{2}{9}y^2(\cos z)\,Z_x = 0.$$
Rearranging gives
$$Z_x(x,y) = \frac{-3x^2z^2}{x^3(2z) + \frac{2}{9}y^2\cos z}.$$
Now, we know that when $x = 2$ and $y = \pi$, we have $z = \frac{\pi}{6}$. Plugging these values in, we find that
$$\begin{align}
Z_x(2, \pi) & = \frac{-3(2^2)(\frac{\pi}{6})^2}{2^3(\frac{2\pi}{6}) + \frac{2}{9}\pi^2\cos\frac{\pi}{6}} \\
& = \frac{ \frac{-\pi^2}{3} }{ \frac{8\pi}{3} + \frac{\pi^2\sqrt{3}}{9} }.
\end{align}$$
Simplifying this expression gives something like $$Z_x(2,\pi) = \frac{-3\pi}{24 + \pi\sqrt{3}},$$ though I might have made a calculation error along the way.
The $Z_y$ case is similar and is left to you.
|
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|
Basic Epsilon-N proof clarification It's been a couple years since I've done analysis, so I was hoping someone could point out any possible flaws I have in the following proof.
Prove $\lim_{n\to\infty}\frac{2^n}{n!} = 0$.
For $n > 2$, we have: $\lim_{n\to\infty}\frac{2^n}{n!} = \lim_{n\to\infty}\frac{2^2}{2!}\times\frac{2}{3}\times\frac{2}{4}\times\frac{2}{5}\times\cdots\times\frac{2}{n}$.
Need to show that $\frac{2^2}{2!}\times\frac{2}{3}\times\frac{2}{4}\times\frac{2}{5}\times\cdots \to 0$. In other words, need to show that $\lim_{n\to\infty}\frac{2}{n} = 0$:
$\forall\epsilon\in\mathbb{N}^+, \exists N\in\mathbb{N} $ such that $\forall n>N$ we have $\frac{2}{n} < \epsilon$. Take $N = \frac{2}{\epsilon}$. Since $n>N$ we have $n > \frac{2}{\epsilon}$. Thus $\frac{2}{n} < \epsilon$. Therefore $\lim_{n\to\infty}\frac{2}{n} = 0$.
And so $\lim_{n\to\infty}\frac{2^2}{2!}\times\frac{2}{3}\times\frac{2}{4}\times\frac{2}{5}\times...\times\frac{2}{n} = 0$.
Then $\lim_{n\to\infty}\frac{2^n}{n!} = 0$.
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A couple of things: first of all, writing $$\lim_{n \to \infty}\frac{2}{n} = 0 \ \ \Rightarrow \ \lim_{n\to\infty}\frac{2^2}{2!}\times\frac{2}{3}\times\frac{2}{4}\times\frac{2}{5}\times...\times\frac{2}{n} = 0$$ isn't completely rigorous, unless you observe that all the factors before $\frac{2}{n}$ are $< 1$ (except for the first, which is $2$), which implies that their product is too. Thus you can write: $$0<\frac{2^2}{2!}\times\frac{2}{3}\times\frac{2}{4}\times\frac{2}{5}\times...\times\frac{2}{n}< 2 \cdot 1 \cdot \frac2n \rightarrow 0$$
because, as you have proved, $2/n \rightarrow 0$. Whithout this observation the statement isn't justified, since all the accumulating factors could have resulted in an unbounded quantity.
Also another minor consideration: in your proof that $\lim_{n \to \infty}\frac{2}{n} = 0$ your $N$ should be a natural number, such as $N = \left[\frac2{\epsilon} \right] + 1$.
|
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|
find all positive integers satisfying $2x^2 - y^{14} = 1$ The following problem was posted to usenet forum de.rec.denksport two weeks ago and no progress was made.
Find all positive integers $x$,$y$ satisfying the equation
$$2x^2 - y^{14} = 1$$
$(1,1)$ is a solution and i suspect it is the only one.
I tried some things, e.g. I started from
$$2x^2=1+y^{14}$$
and using the identities
$$1+y^{14}=(y^2+1)(y^{12}-y^{10}+y^8-y^6+y^4-y^2+1)$$
and
$$y^{12}-y^{10}+y^8-y^6+y^4-y^2+1=(y^2+1)(y^{10}-2y^8+3y^6-4y^4+5y^2-6)+7$$
it was possible to split the equations in two equations
$$2u^2=y^2+1$$
$$v^2=y^{12}-y^{10}+y^8-y^6+y^4-y^2+1$$
$$\gcd(y^2+1,y^{12}-y^{10}+y^8-y^6+y^4-y^2+1)=1$$
Using this result I could check that there is no solution for small integers ($y$ less than $10^{2000}$) by finding the set of solutions of the first of these equations using http://www.alpertron.com.ar/QUAD.HTM and checking if $y^{12}-y^{10}+y^8-y^6+y^4-y^2+1$ is a square. But I had no clue how to solve the problem.
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The equation $v^2=y^{12}-y^{10}+y^8-y^6+y^4-y^2+1$ has only trivial solutions $y=0,1$.
Define $f(x)=1+x+x^2+x^3+x^4+x^5+x^6$ so that the right hand side is $f(-y^2)$. For $x>5$ we have $$(16x^3+8x^2+6x+5)^2<256 f(x)< (16x^3+8x^2+6x+6)^2,$$ while for $x<-4$ we get $$(16x^3+8x^2+6x+5)^2<256 f(x)< (16x^3+8x^2+6x+4)^2.$$
Checking the values $-4\leq x\leq 5$ by hand, we see that $f(x)$ is only a square when $x=0,-1$.
Ribenboim's book on Catalan's conjecture has a detailed analysis of the Diophantine equation $v^2=1+x+x^2+\cdots +x^{n-1}$. The only non-trivial solutions are $n=5, x=3$ and $n=4, x=7$.
See also this MSE problem.
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|
Evaluate fraction of sum So i have to evaluate this sum:
$\displaystyle \frac{1-2^{-2}+4^{-2}-5^{-2}+7^{-2}-8^{-2}+10^{-2}-11^{-2}+\cdots}{1+2^{-2}-4^{-2}-5^{-2}+7^{-2}+8^{-2}-10^{-2}-11^{-2}+\cdots}$
it has the form :
$\displaystyle \frac{\sum^{\infty}_0 [(3n+1)^{-2}-(3n+2)^{-2}]}{\sum^{\infty}_0 (-1)^n[(3n+1)^{-2}+(3n+2)^{-2}]}$
My current attempt : Trying to convert this into power series
$\displaystyle a(n) = (3n+1)^{-2}-(3n+2)^{-2} ~~~~~~ b(n) = (3n+1)^{-2}+(3n+2)^{-2}$
Can a(n) and b(n) be the definite integral of certain polynomial function f(x) ?
Maybe there is a better direction. Can someone give me a hint ?
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First, let $A = 1-\frac{1}{2^2} + \frac{1}{4^2}-\frac{1}{5^2}+... $ and $B = 1 +\frac{1}{2^2}-\frac{1}{4^2}-\frac{1}{5^2}+... $
Notice that the both A and B converges absolutely (by comparison test). Thus, limit law applies: $B-A=2\times\frac{1}{2^2}-2\times\frac{1}{4^2}-2\times\frac{1}{8^2}-2\times\frac{1}{10^2}+... =\frac{A}{2} $ and $\frac{A}{B}=\frac{2}{3}$.
Lol i was so naive back then.
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|
Compute $\int_{0}^{\frac{\pi}{2}}\frac{\sin^2x\cos x}{\sin x+\cos x}dx$ I'd like your help with computing the following integral:$$ \int_{0}^{\frac{\pi}{2}}\frac{\sin^2x\cos x}{\sin x+\cos x}dx.$$ I used $t=\sin x$ and got that $$\int_{0}^{\frac{\pi}{2}}\frac{\sin^2x\cos x}{\sin x+\cos x}dx=\int_{0}^{1}\frac{t^2}{t+\sqrt{1-t^2}}dt,$$ but I can't see how I it use either. I tried also to use $t=2\arctan x$, without any successes.
What do you think?
Thank you very much!
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The light: Denote $\displaystyle I = \int^{\pi/2}_0 \frac{ \sin^2 x \cos x}{\sin x + \cos x} dx .$ Let $ u= \pi/2 -x $ to find $\displaystyle I = \int^{\pi/2}_0 \frac{\cos^2 x \sin x}{\sin x+\cos x} dx .$ Adding these two gives $\displaystyle 2I = \int^{\pi/2}_0 \frac{ \sin x \cos x (\sin x+ \cos x) }{\sin x + \cos x} dx = \int^{\pi/2}_0 \sin x \cos x dx = \frac{\sin^2 x}{2} \biggr|^{\pi/2}_0 = \frac{1}{2}$ so $ I = \frac{1}{4}.$
The dark path: Applying the identity $ \sin x + \cos x = \sqrt{2} \sin ( x +\pi/4) $ gives $\displaystyle I = \frac{1}{\sqrt{2}} \int^{\pi/2}_0 \frac{\sin^2 x \cos x}{\sin (x+\pi/4)} dx = \frac{1}{\sqrt{2}} \int^{3\pi/4}_{\pi/4} \frac{\sin^2 (u-\pi/4) \cos (u-\pi/4)}{\sin u} du.$ Now apply $\displaystyle \sin^2 t = \frac{1-\cos 2t}{2}$ and the addition theorems, the result will eventually come.
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|
Use mathematical induction to prove that for all integers $n \ge 2$, $2^{3n}-1$ is not prime I had a homework due yesterday with this problem.
The TA did the problem last week in discussion but I didn't understand it.
She pulled out a $7k$ almost immediately, and I have no idea from where.
It was like, it wasn't prime if $2^{3n}-1$ was 7 times some constant. I understand that that would make $2^{3n}-1$ not prime, but I don't understand how she just used "7".
Where did she get that from? I was thinking, maybe it was like $2^3 \cdot 2^n-1$... which is $8 \cdot 2^n-1$... but you can't just do $8-1$. How?
|
\begin{align*}
2^{3n}−1 & =(2^3)^n−1\\
& =8^n−1^n\\
& =(8−1)(8^{n−1} \cdot 1^0+8^{n−2} \cdot 1^1+⋯+8^2 \cdot 1^{n−3}+8^1 \cdot 1^{n−2}+8^0 \cdot 1^{n−1})\\
& =7\sum(8^{n−1}+8^{n−2}+⋯+8^2+8+1)\\
& =7\sum_{k=0}^{n−1} 8^k
\end{align*}
\begin{align*}
2^{3n}−1 & =(2^3)^n−1\\
& = 8^n−1^n\\
& = (8−1)(8^{n−1} \cdot 1^0+8^{n−2}\cdot 1^1+ \cdots +8^2⋅1^{n−3}+ 8^1 \cdot 1^{n−2}+ 8^0⋅1^{n−1})\\
& = 7(8^{n−1}+8^{n−2} + \cdots +8^2+8+1)\\
& = 7\sum_{k=0}^{n−1}8^k
\end{align*}
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|
Solving the exponential equation: $3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ I have this exponential equation that I don't know how to solve:
$3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ with $x \in \mathbb{R}$
I tried to factor out a term, but it does not help. Also, I noticed that:
$2 \cdot 9^{x+1} = 2 \cdot 3^{2x+2}$
and tried to write the polynomial as a binomial square, without success.
I know I should solve it using logarithm, but I don't see how to continue.
EDIT: WolframAlpha factors it as: $(3 \cdot 2^x - 2 \cdot 3^x)(2^{x+2} - 3^{x+2}) = 0$ and then the solution is straightforward. Any hint about how to reach that?
|
$$
\begin{eqnarray}
0 &=& 3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1}
\\&=& 12 \cdot (2^x)^2 - 35 \cdot (2^x3^x) + 18 \cdot (3^x)^2
\\&=& 12 t^2 - 35 st + 18 s^2
\qquad\text{for}\qquad s=3^x,~t=2^x
\\&=& (3t - 2s)(4t - 9s)
\\
\implies&&
s=\frac32t \quad\text{or}\quad
s=\frac49t
\\&& s=\left(\frac23\right)^\delta t
\quad\text{for}
\quad\delta=\frac{1\pm3}{2}=-1~\text{or}~2
\\&& 3^x=\left(\frac23\right)^\delta 2^x
\\&& 1=\left(\frac23\right)^{x+\delta}
\\&& 0=(x+\delta)\log\left(\frac23\right)
\\&& x+\delta=0
\\&& x=-\delta=-\frac{1\pm3}{2}=\boxed{~1~\text{or}~-2~}
\end{eqnarray}
$$
To make the factorization in the fourth line,
from the signs of the coefficients $12,-35,18$,
we can see that we need to find
a combination of the factors
from a row on the left
and a row on the right
$$
\begin{eqnarray}
1 \cdot 12 &\quad& 1 \cdot 18
\\ 2 \cdot 6 &\quad& 2 \cdot 9
\\ 3 \cdot 4 &\quad& 3 \cdot 6
\end{eqnarray}
$$
so that a sum of the products
(either of the first numbers and the second numbers,
or of the first numbers with the second numbers)
is $35$. So taking $3 \cdot 4$ for $12$ and $2 \cdot 9$ for $18$,
we can obtain $2\cdot 4+3\cdot 9=8+27=35$ to get the factors above.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/108447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
}
|
find shortest distance from origin to parabolic form suppose that,we have given following parabolic arc
$\sqrt{x}+\sqrt{y}=\sqrt{a}$
we are trying to find shortest distance from origin to this line, i think that if we rewrite it as $\sqrt{y}=\sqrt{a}-\sqrt{x}$, then
$y=a-2\sqrt{ax}-x$ or if we rearrange terms,we get $y+a-\sqrt{ax}-x=0$
distance from origin $(0,0)$ to line $Ax+by+c=0$ is equal
$D=(A\cdot0+b\cdot0+c)/(\sqrt{A^2+b^2})$ so in my case what would be minimum distance? please help me
|
One asks that the vector $(x,y)$ is orthogonal to the tangent at $(x,y)$ of the curve. Thus, assume that $(x,y)$ and $(x+h,y+k)$ are on the curve and that $h,k\to0$. Then
$$
\sqrt{x}+\sqrt{y}=\sqrt{a}=\sqrt{x+h}+\sqrt{y+k}=\sqrt{x}\sqrt{1+h/x}+\sqrt{y}\sqrt{1+k/y}
$$
hence
$$
\sqrt{x+h}+\sqrt{y+k}=\sqrt{x}+h/(2\sqrt{x})+o(h)+\sqrt{y}+k/(2\sqrt{y})+o(k).
$$
In particular, $h/\sqrt{x}\sim-k/\sqrt{y}$ hence the tangent vector is $(\sqrt{x},-\sqrt{y})$.
The tangent is orthogonal to $(x,y)$ if and only if $x\sqrt{x}-y\sqrt{y}=0$, that is, $x=y$, that is $x=y=a/4$. The shortest distance is $\sqrt{x^2+y^2}=\sqrt2a/4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/110937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Finding domain of $\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$ How can I find the domain of:
$$\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$
I think the hard part will be to find:
$$\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} \ge 0$$
So far I have: not sure how to preceed:
$$\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$
For $\sqrt{g(x)}$ to be valid, $g(x) \ge 0$
For $f(x)$ to be valid, $(x^2-2)(x^2-4)(x^2-6) \ne 0$
Thus, $x \ne \sqrt 2, 2, \sqrt 3$
$$g(x) = { \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)}} \ge 0$$
|
The first thing you should note is that the expression
$$\Phi=\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} $$
is undefined at any zero of the denominator.
So, the points $x=\pm \sqrt 2$, $\pm 2$, and $\pm\sqrt 6$ are
not in the domain of $\sqrt\Phi$.
Now, to find the domain of $\sqrt\Phi$, we need to find when $\Phi$ is nonnegative.
Towards solving $\Phi\ge0$,
the fundamental observation to make is that
the only points "across which" $\Phi$ can change signs are at the zeros of its numerator or at the zeroes of its denominator.
The zeroes of the numerator are: $$\pm 1, \pm \sqrt 3, \pm \sqrt 5$$
and the zeroes of the denominator are
$$
\pm \sqrt 2, \pm 2 \pm\sqrt 6.
$$
As already mentioned, the only points across which $\Phi$ can change sign are at one of the zeroes above.
Let's also note the expression $\Phi$ is "even": if $\Phi(x)\ge 0$, then $\Phi(-x)\ge 0$. So, let's find the $x$ values on the nonnegative $x$-axis that satisfy $\Phi(x)\ge0$. Then by "reflection" we'll obtain the points on the negative $x$-axis where $\Phi(x)\ge0$.
So, draw a number line with the zeroes listed above:
The zeroes subdivide the nonnegative $x$-axis into intervals, the endpoints of which, except the endpoint 0, are the zeroes of either the numerator or the denominator of $\Phi$. In each subinterval, if you pick a point $x_0$ (not an endpoint, save for 0), evaluate $\Phi(x_0)$, and note its sign, then across
that entire subinterval, the expression $\Phi$ will have that sign.
For example in $[0,1)$, picking $x=1/2$ it is easy to see that the sign of $\Phi(1/2)$ is
$$\frac{((1/2)^2-1)((1/2)^2-3)((1/2)^2-5)}{((1/2)^2-2)((1/2)^2-4)((1/2)^2-6)}
={ (-)(-)(-)\over(-)(-)(-)}
\ge 0$$
So on all of $[0,1)$, the expression $\Phi$ is positive. (Note, you only need to find the sign, there is no need to do the actual arithmetic.)
The complete "sign chart" is shown below:
And we see that $\Phi>0$ on
$$
[0, 1)\cup(\sqrt2,\sqrt3)\cup(2,\sqrt5)\cup(\sqrt6,\infty).
$$
By symmetry, $\Phi>0$ on
$$
( -\infty,-\sqrt6)\cup(-\sqrt5,2)\cup (-\sqrt3,-\sqrt2)\cup (-1,0].
$$
So, the domain of $\sqrt\Phi$ is the union of the two sets above, together with
the zeroes of the numerator of $\Phi$ that aren't zeros of the denominator of $\Phi$: $x=\pm1$, $x=\pm \sqrt3$, and $x=\pm \sqrt5$.
Below is a plot of $y=\Phi$ and $y=\sqrt\Phi$. Note that the zeroes of the denominator of $\Phi$ are vertical asymptotes of the graph of $y=\Phi$ and the zeroes of the numerator of $\Phi$ are the zeroes of $\Phi$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/112416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Finding the outward flux through a sphere Problem:
Find the flux of of the field $F$ across the portion of the sphere $x^2 + y^2 + z^2 = a^2$ in the first octant in the direction away from the origin, when $F = zx\hat{i} + zy\hat{j} + z^2\hat{k}$.
|
The way you calculate the flux of $F$ across the surface $S$ is by using a parametrization $r(s,t)$ of $S$ and then
$$
\int\!\!\!\!\int_S F\cdot n\, dS =
\int\!\!\!\!\int_D F(r(s,t))\cdot (r_s\times r_t)\, dsdt,
$$
where the double integral on the right is calculated on the domain $D$ of the parametrization $r$.
In this case, since $S$ is a sphere, you can use spherical coordinates and get the parametrization
$$
r(\theta, \phi)=(a\cos\theta\sin\phi, a\sin\theta\sin\phi, a\cos\phi),\ \ 0\leq\theta\leq\frac\pi2,\ \ 0\leq\phi\leq\frac\pi2.
$$
The "first octant" is chosen by the region where we let $\theta$ and $\phi$ vary (if you think carefully about it you'll see that $\pi/2$ is the right choice above).
Now the partial derivatives:
$$
r_\theta=(-a\sin\theta\sin\phi,a\cos\theta\sin\phi, 0),\ \ \ r_\phi=(a\cos\theta\cos\phi, a\sin\theta\cos\phi, -a\sin\phi).
$$
The normal vector:
\begin{eqnarray}
r_\theta\times r_\phi&=&\left|\begin{matrix}i& j& k\\
-a\sin\theta\sin\phi&a\cos\theta\sin\phi& 0\\ a\cos\theta\cos\phi& a\sin\theta\cos\phi& -a\sin\phi
\end{matrix}\right|
\\ \ \\
&=&(-a^2\cos\theta\sin^2\phi, -a^2\sin\theta\sin^2\phi, -a^2\sin\phi\cos\phi).
\end{eqnarray}
Since we want the direction away from the origin, we need to reverse the signs in the normal vector.
Now
\begin{eqnarray}
F(r(\theta,\phi))\cdot(r_\theta\times r_\phi)&=&
(a^2\cos\theta\sin\phi\cos\phi,a^2\sin\theta\sin\phi\cos\phi,a^2\cos^2\phi) \\
& &\cdot(a^2\cos\theta\sin^2\phi, a^2\sin\theta\sin^2\phi, a^2\sin\phi\cos\phi)
\\ &=&
a^4\cos^2\theta\sin^3\phi\cos\phi+a^4\sin^2\theta\sin^3\phi\cos\phi+a^4\sin\phi\cos^3\phi\\
&=&
a^4\sin\phi\cos\phi(\cos^2\theta\sin^2\phi+\sin^2\theta\sin^2\phi+\cos^2\phi)\\
&=&a^4\sin\phi\cos\phi.
\end{eqnarray}
Finally,
$$
\int\!\!\!\!\int_S F\cdot n\, dS = \int_0^{\pi/2}\!\!\int_0^{\pi/2}a^4\sin\phi\cos\phi\,d\theta d\phi=\frac\pi2\,a^4\left.\frac{\sin^2\phi}2\right|_0^{\pi/2}=\frac{\pi a^4}4
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/113363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Show $4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$ Show
$$4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$$
The hint says "note $\sin{\frac{3\pi}{5}} = \sin{\frac{2\pi}{5}}$" and "use double/triple angle or otherwise"
So I have
$$4 \cos^2{\frac{\pi}{5}} - 2 (2 \cos^2{\frac{\pi}{10}} - 1) - 1$$
$$4 \cos^2{\frac{\pi}{5}} - 4 \cos^2{\frac{\pi}{10}} +1$$
Now what? Theres $\frac{\pi}{5}$ and $\frac{\pi}{10}$ and I havent used the tip on $\sin$ so perhaps I am missing something?
|
The hint says to use the following:
$$\sin 3x = 3 \sin x - 4 \sin^3x$$
and
$$ \sin 2x = 2 \sin x \cos x$$
So if $x = \frac{\pi}{5}$, then we have, using $\sin 3x = \sin 2x$ that
$$3 \sin x - 4 \sin^3 x = 2 \sin x \cos x $$
Since $\sin x \neq 0$, cancel it, and use $\sin^2 x = 1 - \cos ^2x$.
For a geometric way to find $\cos \frac{\pi}{5}$ see: http://www.cut-the-knot.org/pythagoras/cos36.shtml
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/113466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
}
|
limit law and product of matrices Are there two $ n\times n $ matrices $A$ and $B$ such that $ \lim_{m\to\infty} A^m$ and $ \lim_{m\to\infty} B^m $ both exists but $ \lim_{m\to\infty} (A \cdot B)^m $ doesn't ?
|
Consider
$$
A = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{pmatrix} \qquad B = A^T
$$
Then $\forall m \geqslant 3$, $A^m = B^m = 0$, but since
$$
A \cdot B = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 2\end{pmatrix}
$$
$$
\lim_{m \to \infty} (A \cdot B)^m = \lim_{m \to \infty} \begin{pmatrix}
0 & 0 & 0 \\
0 & \phi ^{2-2 m}+\phi ^{2 m-2} & -\phi ^{1-2 m}+\phi ^{2 m-1} \\
0 & -\phi ^{1-2 m}+\phi ^{2 m-1} & \phi ^{-2 m}+\phi ^{2 m} \end{pmatrix} =
\begin{pmatrix} 0 & 0 & 0 \\ 0 & \infty & \infty \\ 0 & \infty & \infty\end{pmatrix}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/113668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
A problem about parametric integral How to solve the following integral.
$I(\theta) = \int_0^{\pi}\ln(1+\theta \cos x)dx$ where $|\theta|<1$
|
Differentiating under the integral sign yields: $$ I'(\theta) = \int^{\pi}_0 \frac{\cos x}{1+ \theta \cos x} dx .$$
So $$\theta I'(\theta) = \int^{\pi}_0 1 - \frac{1}{1+\theta \cos x} dx= \pi - \int^{\pi}_0 \frac{1}{1+\theta \cos x} dx$$
To deal with the last integral, consider $$
\begin{aligned} I & = \int_{0}^{\pi} \frac{1}{a+b\cos{x}}\;{dx} \\& = \int_{0}^{\pi} \frac{1}{a\left(\sin^2\frac{1}{2}x+\cos^2\frac{1}{2}x\right)+b\left(\cos^2\frac{1}{2}x-\sin^2\frac{1}{2}x\right)}\;{dx} \\& =\int_{0}^{\pi}\frac{1}{(a-b)\sin^2\frac{1}{2}x+(a+b)\cos^2\frac{1}{2}x}\;{dx} \\& =\int_{0}^{\pi}\frac{\sec^2{\frac{1}{2}x}}{(a+b)+(a-b)\tan^2\frac{1}{2}x}\;{dx} \\& = 2\int_{0}^{\infty}\frac{1}{(a+b)+(a-b)t^2}\;{dt} \\& = 2\int_{0}^{\infty}\frac{1}{(\sqrt{a+b})^2+(\sqrt{a-b})^2t^2}\;{dt}\\& = \frac{2}{{\sqrt{a^2-b^2}}}\tan^{-1}\bigg(\frac{\sqrt{a-b}}{\sqrt{a+b}}~t \bigg)\bigg|_{0}^{\infty} \\& = \frac{\pi}{\sqrt{a^2-b^2}}.\end{aligned}
$$
Thus, $$ \theta I'(\theta) = \pi \left(1 - \frac{1}{\sqrt{1-\theta^2}} \right).$$
You can now do some integration of your own to find $I'(\theta).$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/114401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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|
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.