Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to find the radius of the smallest circle such that the inner ellipse is tangent to the circle at exactly one point? Consider the ellipse
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$ where $a > b > 0.$ As a function of $a$ and $b,$ find the radius of the smallest circle that contains the ellipse, is centered on the $y$-axis, and which intersects the ellipse only at $(0,b).$
I have no clue how to solve this. The only thing I've thought of is that maybe it somehow involves the fact that for any $x$, the corresponding $y$ value of the circle is bigger than the corresponding $y$ value of the ellipse. But I haven't made much progress with that.
| A solution without calculus.
Eliminating $x$ in the set $\cases{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ x^2+(y-y_0)^2= r^2}$ we obtain
$$
\left(1-\frac{a^2}{b^2}\right)y^2-2y_0y+y_0^2-r^2=0
$$
now solving for $y$ to determine the intersections we have
$$
y = \frac{-y_0\pm\sqrt{y_0^2-(\frac{a^2}{b^2}-1)(r^2-y_0^2-a^2)}}{(\frac{a^2}{b^2}-1)}
$$
but we need tangency so this implies on an unique solution so the condition is
$$
y_0^2-(\frac{a^2}{b^2}-1)(r^2-y_0^2-a^2)=0
$$
solving for $r$ we get
$$
r = \frac{\sqrt{a^4-a^2b^2+a^2y_0^2}}{\sqrt{a^2-b^2}}
$$
but $r=b-y_0$ and thus we obtain $y_0 = \frac{b^2-a^2}{b}$ and $r = \frac{a^2}{b}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
The sequence $(a_n), \ a_1=1, a_{n+1}=2+1/n+3/a_n$ converges The sequence $(a_n), \ a_1:=1, a_{n+1}:=2+1/n+3/a_n$ converges.
Proof.
Let $a_{k-1}\geq 3$ for some positive integer $k$. Then $a_k\leq3+\frac{1}{k-1}=\frac{3k-2}{k-1}$. And $a_{k+1}\geq 2+1/k+\frac{3k-3}{3k-2}$. It is easy to check that $1/k+\frac{3k-3}{3k-2}>1$ for $k>1$. But $a_2=6$, $a_3=3$ and $3\leq a_n$ is proved by induction for positive inthegers greater than $1$.
Now consider a sequence $(x_i)_{i=2}^\infty$, $x_{n+1}:=3+3/n$. Notice that $a_{n+1}=\frac{2n+1}{n}+3/a_n\leq \frac{2n+1}{n}+1$. But $x_{n+1}$ may be rewritten as $\frac{2n+3}{n}+1$. Therefore $x_n\geq a_n$.
And by the sandwich theorem we conclude that $\lim a_n=3$.
P.s. English is not my first language, so it will be excellent to get some hints about grammar, especially articles.
| Clearly $a_n>2$, for $n>1$ and hence $3/a_n<3/2$ and thus $a_{n+1}=2+1/n+3/a_n=4$. Hence
$$
2<a_n \le 4, \quad \text{for}\,\,n>2.
$$
Next
$$
|a_{n+2}-a_{n+1}|=
\left|
\frac{1}{(n+1)(n+2)}
+\frac{3(a_{n}-a_{n+1})}{a_na_{n+1}}\right|\le
\frac{1}{(n+1)(n+2)}
+\frac{3|a_{n}-a_{n+1}|}{a_na_{n+1}} \\
\le
\frac{1}{(n+1)(n+2)}
+\frac{3}{4}|a_{n}-a_{n+1}|
$$
So, if $b_n=|a_{n+1}-a_n|$, then
$$
0\le \frac{b_{n+1}}{\left(\frac{3}{4}\right)^{n+1}}-\frac{b_n}{\left(\frac{3}{4}\right)^{n}}\le \frac{1}{(n+1)(n+2)\left(\frac{3}{4}\right)^{n+1}}
$$
and hence
$$
\frac{b_n}{\left(\frac{3}{4}\right)^{n}}\le \frac{3}{4}|a_1-a_2|
+\sum_{k=1}^{n-1}\frac{1}{k(k+1)\left(\frac{3}{4}\right)^{k}}
\,\,\Longrightarrow\,\, b_n\le \left(\frac{3}{4}\right)^{n+1}b_1+
\sum_{k=1}^{n-1}\frac{\left(\frac{3}{4}\right)^{n-k}}{k(k+1)}
$$
and hence
$$
\sum_{n=1}^\infty b_n\le \sum_{n=1}^\infty \left(\frac{3}{4}\right)^{n+1}b_1+\sum_{n=1}^\infty\sum_{k=1}^{n-1}\frac{\left(\frac{3}{4}\right)^{n-k}}{k(k+1)}=3b_1+\frac{16}{3}\sum_{k=1}^\infty
\frac{1}{k(k+1)}<\infty
$$
Hence $\sum_{n=1}^\infty |a_{n+1}-a_n|<\infty$, which implies that $\{a_n\}$ converges. Let $a_n\to x$. Then
$$
x \leftarrow a_{n+1}=2+\frac{1}{n}+\frac{3}{a_n}\to 2+0+\frac{3}{x}.
$$
Thus $x=2+3/x$ or $x^2-2x-3=0$. This means that $x=3$ or $x=-1$. But $a_n\ge 2$, for all $n>1$, and finally $x=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find a basis and dimension of two subspaces together with their intersection space?
Given a vector $[a,b,c,d]^T$ ,find a basis for the subspace $S$ of $\Bbb{R}^4$ such that $a+c+d = 0$
and for the subspace $T$ such that $a+b = 0$ and $c = 2d$.
What is the dimension of the intersection $S∩T$?
My approach, it is known that to form a basis the vectors must be linearly independent, and this should be a spanning set.
Basis for subspace $S$,
$\begin{align}&\{(a,b,c,d)| a+c+d=0\}\\&=\{(a,b,-a-d,d) | a+c+d=0\}\\&=\{a(1,0,-1,0)\} +\{b(0,1,0,0)\}+\{d(0,0,-1,1)\}\end{align}$
Hence $\dim = 3$ and basis is $\{(1,0,-1,0),(0,1,0,0),(0,0,-1,1)\}$
Basis for subspace $T$,
$\begin{align}&\{(a,b,c,d): a+b=0 \text{ and } c= 2d\}\\&=\{(a,-a,2d,d)\}
\\&=\{a(1,-1,0,0) + d(0,0,2,1)\}\end{align}$
$\dim T= 2$ and basis is $\{(1,-1,0,0),(0,0,2,1)\}$.
Hence
$\dim(S∩T)= 1 $
I'd appreciate corrections or if there is a more technical way to approach this. I am little confused on the transpose part as well, I want to understand the significance of this representation. Might be it wanted to reflect that those are column vectors. I need validation on the answer. thanks!
Note: Here $[a, b, c, d]^T$ means the column vector $\{\begin{pmatrix}a\\b\\c\\d\end{pmatrix}\}$
| $S=\textrm{null}\begin{pmatrix}1&0&1&1\end{pmatrix}$
$\begin{align}\mathcal{B}_S&=\{\begin{pmatrix}0\\1\\0\\0\end{pmatrix},\begin{pmatrix}-1\\0\\1\\0\end{pmatrix},
\begin{pmatrix}-1\\0\\0\\1\end{pmatrix}\}\end{align}$
$\dim S=3$
$T=\textrm{null}\begin{pmatrix}1&1&0&0\\0&0&1&-2\end{pmatrix}$
$\mathcal{B}_T=\{\begin{pmatrix}-1\\1\\0\\0\end{pmatrix},\begin{pmatrix}0\\0\\2\\1\end{pmatrix}\}$
$\dim T=2$
$S\cap T=\textrm{null}\begin{pmatrix}1&0&1&1\\1&1&0&0\\0&0&1&-2\end{pmatrix}$
$\begin{align}\textrm{rref}\begin{pmatrix}1&0&1&1\\1&1&0&0\\0&0&1&-2\end{pmatrix}&=\begin{pmatrix}1&0&0&3\\0&1&0&-3\\0&0&1&-2\end{pmatrix}\end{align}$
$\mathcal{B}_{S\cap T}=\{\begin{pmatrix}-3\\3\\2\\1\end{pmatrix}\}$
$\dim(S\cap T) =1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4515662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\lim\limits_{x \to \infty} \frac{\pi\sqrt{x}}{\sin{(\frac{\pi}{\sqrt{x}} })} - x = \frac{\pi^2}{6}$? As we know, the result of some expressions and series is equal to $\frac{\pi^2}{6} $ that the most important of them is $\zeta(2)$
Now, I have founded an equation whose limit at point $x=\infty$ is equal to $\frac{\pi^2}{6} : $
$$\lim_{x \to \infty} \frac{\pi\sqrt{x}}{\sin{(\frac{\pi}{\sqrt{x}} })} - x = \frac{\pi^2}{6}$$
I want to know how can we prove it?
| $$y=\frac{\pi\sqrt{x}}{\sin \left(\frac{\pi }{\sqrt{x}}\right)} - x $$
$$x=\frac {\pi^2}{v^2} \quad \implies \quad y=\pi^2 \left(\frac {1}{v\sin(v) }-\frac {1}{v^2}\right)$$
By Taylor
$$\sin(v)=v-\frac{v^3}{6}+\frac{v^5}{120}+O\left(v^7\right)$$
$$v\sin(v)=v^2-\frac{v^4}{6}+\frac{v^6}{120}+O\left(v^8\right)$$ Long division
$$\frac {1}{v\sin(v) }=\frac{1}{v^2}+\frac{1}{6}+\frac{7 v^2}{360}+O\left(v^4\right) \quad \implies \quad y=\pi^2 \left(\frac{1}{6}+\frac{7 v^2}{360}+O\left(v^4\right)\right)$$ Back to $x$
$$y=\frac{\pi ^2}{6}+\frac{7 \pi ^4}{360 x}+O\left(\frac{1}{x^2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
how to find maximum value without differentiating let x be positive real number, find max possible value of the expression
$$y = \frac{x^2 + 2 - \sqrt{x^4 + 4}}{x}$$
it can be found by differentiating, but is there no other way of finding it, like using AM $\geq$ GM.
or any other method.
i tried
$$y = x + \frac{2}{x} - \sqrt{x^2 + \frac{4}{x^2}}$$
but it gives nothing
| You are on the right track!
To begin with, suppose that $x > 0$. Then we can rearrange the proposed expression as
\begin{align*}
f(x) = x + \frac{2}{x} - \sqrt{x^{2} + \frac{4}{x^{2}}} = x + \frac{2}{x} - \sqrt{\left(x + \frac{2}{x}\right)^{2} - 4}
\end{align*}
If we make the substitution $u = x + \dfrac{2}{x}$, one gets:
\begin{align*}
f(u) = u - \sqrt{u^{2} - 4}
\end{align*}
Now you can apply the derivative method. Analogous approach applies to the case where $x < 0$.
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a^2, b^2$ and $c^2$ be three distinct numbers in AP. If $ab + bc + ca = 1$ then $(b + c), (c + a)$ and $(a + b)$ are in which series Let $a^2, b^2$ and $c^2$ be three distinct numbers in AP. If $ab + bc + ca = 1$ then $(b + c), (c + a)$ and $(a + b)$ are in
(1)AP
(2) GP
(3) HP
(4) none of these
My approach is as follow
$2{b^2} = {a^2} + {c^2} \Rightarrow 2{b^2} + {b^2} = {a^2} + {c^2} + {b^2} \Rightarrow 3{b^2} = {a^2} + {c^2} + {b^2}$
${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right)$
${\left( {a + b + c} \right)^2} = 3{b^2} + 2 \Rightarrow {\left( {a + b + c} \right)^2} - {b^2} = 2{b^2} + 2$
$\Rightarrow \left( {a + b + c - b} \right)\left( {a + b + c + b} \right) = 2{b^2} + 2$
$\Rightarrow \left( {a + c} \right)\left( {a + c + 2b} \right) = 2{b^2} + 2 \Rightarrow {\left( {a + c} \right)^2} + 2b\left( {a + c} \right) = 2{b^2} + 2$
$ \Rightarrow {\left( {a + c} \right)^2} + 2\left( {ab + bc} \right) = 2{b^2} + 2 \Rightarrow {\left( {a + c} \right)^2} + 2\left( {1 - ac} \right) = 2{b^2} + 2$
$\Rightarrow {\left( {a + c} \right)^2} - 2ac = 2{b^2} \Rightarrow {\left( {a + c} \right)^2} = 2\left( {{b^2} + ac} \right)$
$ \Rightarrow \frac{{a + c}}{{{b^2} + ac}} = \frac{2}{{\left( {a + c} \right)}}$
Not able to proceed further.
| $a^2, b^2, c^2$ are in AP.
Then $a^2+1,b^2+1,c^2+1$ are also in AP.
$a^2+ab+bc+ca, b^2+ab+bc+ca, c^2+ab+bc+ca$ are in AP.
$\begin{align}a^2+ab+bc+ca&=a(a+b) +c(b+a)\\&=(a+c) (a+b) \end{align}$
$\begin{align}b^2+ab+bc+ca&=b(b+a) +c(b+a)\\&=(b+c) (a+b) \end{align}$
$\begin{align}c^2+ab+bc+ca&=c(c+a) +b(c+a)\\&=(b+c) (c+a) \end{align}$
Now divide by $(a+b) (b+c) (c+a) $.
Hence $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in AP.
$(b+c), (c+a), (a+b) $ are in HP.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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In the set {1, 2, 3, ..., 2019}, we pick a, b, and c randomly without any conditions. What is the probability for abc + bc + c to be divisible by 3? 3 integers a, b, and c are randomly taken from the set {1, 2, 3, ..., 2019} without any conditions. What is the probability for abc + bc + c is divisible by 3?
Correct me if I'm wrong please.
abc + bc + c = c(ab + b + 1).
We can conclude that c has to be a multiple of 3, so we have 1/3 probability for c. a and b are any integers from the set.
AND, ab + b + 1 has to be a multiple of 3 in the case that c is not a multiple of 3.
ab + b + 1 = b(a + 1) + 1.
b(a + 1) has to be in the form 3n - 1 for b(a + 1) + 1 to be a multiple of 3.
And this is where I get stuck. New solutions, hints, or suggestions are welcome.
| For a solution to $b(a+1)+1=3n$ where $n$ is arbitrary, we need to have $$(b(a+1)+1) \mod 3=0$$ which means we need $(b(a+1)) \mod 3=2$. That can happen if $a\mod 3=0$ and $b\mod 3=2$ or $a\mod 3=1$ and $b\mod 3=1$.
For example $a=3$, $b=8$ has $a\mod3=0$ and $b\mod3=2$ and $8(3+1)+1=33=3\times11$, and $a=4$, $b=4$ has $a\mod3=b\mod3=1$ and $4(4+1)+1=21=3\times7$.
Since these two sets don't overlap each other, you can just add them, so $\frac292019$ of pairs of $a$ and $b$ match the criterion, in addition to the $c$s which match.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4524568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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finding radius of convergence for $a_n=\frac{n^2-5}{4^n+3n}$. I am asked to find the radius of convergence for the power series
$$\sum_{n=0}^{\infty} a_nz^n$$ with $a_n=\frac{n^2-5}{4^n+3n}$.
I feel like I can use the ratio test but I am unsure because when I tried it I just got a mess.
$\frac{(n+1)^2-5}{4^{n+1}+3(n+1)}$ $\times$ $\frac{4^n+3n}{n^2-5}$
I would appreciate some help with this one.
| Starting where you left off:
$$\frac{(n + 1)^2 - 5}{4^{n + 1} + 3(n + 1)} \cdot \frac{4^n + 3n}{n^2 - 5}$$
$$= \frac{n^2 + 2n + 1 - 5}{4^{n + 1} + 3(n + 1)} \cdot \frac{4^n + 3n}{n^2 - 5}$$
$$=\frac{n^2 + 2n + 1 - 5}{n^2 - 5} \cdot \frac{4^n + 3n}{4^{n + 1} + 3(n + 1)}$$
Taking $n^2$ out of the first part and $4^n$ in the second part:
$$=\frac{1 + \frac{2}{n} - \frac{4}{n^2}}{1 - \frac{5}{n^2}} \cdot \frac{1 + \frac{3n}{4^n}}{4 + \frac{3(n + 1)}{4^n}}$$
So:
$$r = \lim_{n \to \infty}\left(\frac{1 + \frac{2}{n} - \frac{4}{n^2}}{1 - \frac{5}{n^2}} \cdot \frac{1 + \frac{3n}{4^n}}{4 + \frac{3(n + 1)}{4^n}}\right)^{-1}$$
Since $4^n$ increases exponentially faster than $3n$, therefore:
$$\lim_{n \to \infty}\frac{3n}{4^n} = 0$$ and $$lim_{n \to \infty}\frac{3(n + 1)}{4^{n + 1}} = 0$$
Then:
$$r = \lim_{n \to \infty}\left(\frac{1 + \frac{2}{n} - \frac{4}{n^2}}{1 - \frac{5}{n^2}} \cdot \frac{1 + \frac{3n}{4^n}}{4 + \frac{3(n + 1)}{4^n}}\right)^{-1}$$
$$= \left(\frac{1 + 0 + 0}{1 - 0} \cdot \frac{1 + 0}{4 + 0}\right)^{-1}$$
$$= \left(\frac{1}{4}\right)^{-1}$$
$$= 4$$
If I've made a mistake, can someone correct me? I believe I haven't, but as with life, I could be wrong. Muchly appreciated.
Edit: Thanks to @Jens's point-out, I had forgot to do the reciprocation. (My bad)
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Finding the value of $\sin^3x+\cos x$, if $x$ is an acute angle satisfying $2\sin x\sin\left(\frac{x}{2}\right)=1-\sin x$ Given that $x$ is an acute angle satisfying $$2\sin x\sin\left(\frac{x}{2}\right)=1-\sin x$$
Then find the exact value of $$\sin^3 x+\cos x$$
My try:
Letting $x=2t$ we get
$$2\sin(2t)\sin(t)=1-\sin(2t)=(\cos t-\sin t)^2$$
So we get
$$2\sin t\sqrt{\cos t}=\cos t-\sin t$$
But i am stuck here
| If there is no typo, there is no analytical solution to the first equation.
For example, if you let $x=4\tan^{-1}(t)$, you would need to solve for $t$ the sextic polynomial
$$t^6+4 t^5+19 t^4-13 t^2-4 t+1=0$$ which does not factor.
Beside purely numerical methods, we can try series expansions. Graphing or by inspection, for an acute angle, the solution is just above $\frac \pi 5$. Using the first term of the series expansion, an approximation is
$$x=\frac \pi 5+\frac{4-\sqrt{2 \left(5+\sqrt{5}\right)}}{3+2 \sqrt{5}}=0.654519$$ while the solution given by Newton method is $0.654434$ as @Suzu Hirose already reported.
Edit
Using my favored, $1,400$ years old, approximation of the sine function
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ we end with a quartic equation (better than the sextic), namely,
$$108 x^4-324 \pi x^3+111 \pi ^2 x^2+110 \pi ^3 x-25 \pi ^4=0$$
Letting $x=\pi\,y$
$$108 y^4-324 y^3+111 y^2+110 y-25=0$$
which shows four real solutions which express in terms of radicals; the one we look for is the smallest positive root.
Just for the fun, the solution is
$$36\,y=27 -\sqrt{3} \sqrt{169+a+\frac{9649}{a}}+\sqrt{3}
\sqrt{338-a-\frac{1056
\sqrt{3}}{\sqrt{169+a+\frac{9649}{a}}}-\frac{9649}{a}}$$ where
$$a=\sqrt[3]{450793+480 i \sqrt{3017094}}$$ which gives
$$y=0.208264 \quad \implies \quad x=0.654282$$ which is quite good for an approximation (relative error of $0.023$%).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4530338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Proving Big-Theta bound Prove by induction that $$ \frac{n^2}{2} + 3n + \ln(n) \in \Theta (n^2).$$
This my attempt.
Base case ($n=1$):$\frac{7}{2} \in \Theta(1)$ is true.
Inductive step: Assume $\frac{n^2}{2} + 3n + \ln(n) \in \Theta (n^2) $ is true for some $n \in \mathbb N$. Then we have that there is $c$ and $n_0$ s.t. $\frac{n^2}{2}+3n +\ln(n) \leq c n^2$ for $n \geq n_0$. We also have $\frac{(n+1)^2}{2} + 3(n+1) + \ln (n+1) = \frac{n^2}{2} +3n + \ln(n) + n + \frac{7}{2} + \ln(\frac{n+1}{n})\leq cn^2 + n + \ln(\frac{n+1}{n}).$ What else can I do?
| Continuing off from where you were, since $\ln(\frac{n+1}{n}) \leq n$, we have that $cn^2+n+\ln(\frac{n+1}{n})\leq cn^2+2n$. WLOG assume $c>1$ and we have $cn^2+2n\leq cn^2+2cn+c=c(n+1)^2$. Proven as required.
Sorry for the poor formatting I’m on my phone waiting for breakfast at a hotel.
| {
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"timestamp": "2023-03-29T00:00:00",
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$x^2+y^2=4x+8y+5$ intersects $3x-4y=m$ at two distinct points
$x^2+y^2=4x+8y+5$ intersects $3x-4y=m$ at two distinct points. What are $m$'s possible values?
I got that the centre of the circle is $C=(2,4)$ and its radius is $r=5$. If the line $l$ intersects the circle at two points, the distance between the line and $C$ is less than $r$:
$$\frac{|6-16-m|}5<5\implies-35<m<15$$
Is this correct?
| An alternative approach is, solve $3x - 4y = m$ for $y$. Plug that into the other equation:
$$x^2 + \left(\frac{3x-m}{4}\right)^2 = 4x + 8\frac{3x-m}{4} + 5$$
Then solve for $x$ using quadratic:
$$x=\frac{3m+80 \pm 4\sqrt{-m^2-20m+525}}{25}$$
2 solutions will exist when $-m^2-20m+525 = -(m-15)(m+35) > 0$
This will tell you whether your final result is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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How does $\frac {(x^2 + 2x) - (a^2 + 2a)}{x-a}$ reduce to $x + a + 2$? With the given function:
$$f(x) = x^2 + 2x $$
I am trying to evaluate the following expression:
$$\frac {f(x) - f(a)}{x-a} $$
I've been informed that the solution is:
$x + a + 2$, where $x \ne a$
But I don't know how to get there. Can someone help me understand the logic behind the solution and where I'm going wrong? I hope this question is appropriate. It's my first on the site.
I started by replacing the functions with their bodies respecting the given inputs.
$$\frac {(x^2 + 2x) - (a^2 + 2a)}{x-a}$$
Then, it appears to me that one could factor out the X's and the A's.
$$\frac {x(x + 2) - a(a + 2)}{x - a} $$
At this point I'm actually stumped. I can see there's an $x$ and an $a$ with subtraction between them, but I don't see how they're related.
But given the solution, I guess the $x-a$ in the denominator cancels out the $x-a$ in the numerator (somehow?). Then I'd be left with
$$(x+2)+(a+2)$$
Adding this up I get $x+a+4$. I'm not sure what problems like these are called, so I'm not sure what to search to find what I'm missing.
Thank you in advance for helping me.
| We have that
$$\frac {(x^2 + 2x) - (a^2 + 2a)}{x-a} =\frac {x^2 -a^2 + 2x-2a}{x-a}=\frac {(x+a)(x-a) + 2(x-a)}{x-a}=$$
$$=\frac {(x+a)(x-a) }{x-a}+\frac {2(x-a)}{x-a}=x+a+2$$
for $x\neq a$, the key point it to recognize that $x^2-a^2=(x+a)(x-a)$.
As an alternative we can proceed using that
$$x^2 +2x-a^2-2a=0 \implies x=\frac{-2\pm \sqrt{4-4(-a^2-2a)}}{2}=-1\pm(a+1)$$
and therefore, since $u=-a-2$ and $v=a$ are roots for the polynomial
$$x^2 +2x-a^2-2a=(x-u)(x-v)=(x+a+2)(x-a)$$
Another way by long division
\begin{array}{rrr|ll}
x^2 & +2x & -a^2-2a & x - a
\\ -x^2 & +ax & & & x +a+2
\\ \hline & (a+2)x & -a^2-2a
\\&-(a+2)x & a^2+2a & & & &
\\ \hline & 0& 0 \end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Proving discontinuity of a map between $\mathbb{R}^2$ and $\mathbb{R}$, at a single point I am trying to show that, no matter what $c \in \mathbb{R}$ we take in the below, $f$ is not continuous at $0$:
$$f(x)=\left\{\begin{array}{ll}
\frac{x_1^2-x_2^2}{x_1^2+x_2^2} & \text { if } x \neq 0 \\
c & \text { if } x=0
\end{array}\right.
$$
Is my attempt correct?
Consider first some arbitrary $c\in \mathbb{R} \backslash \{-1, 1\}$. Set $\epsilon:= \min\{|c-1|, |c+1|\}$. Consider some arbitrary $\delta>0$. Choose $x:= (x_1, x_2)$ such that $d(x, 0)=\sqrt{x_1^2+x_2^2}<\delta$. Now, note that: $|f(0)-f(x)|=|c-\frac{x_1^2-x_2^2}{x_1^2+x_2^2}|=|\frac{(x_1^2+x_2^2)c-x_1^2+x_2^2}{x_1^2+x_2^2}|=|\frac{x_1^2c+x_2^2c+x_2^2-x_1^2}{x_1^2+x_2^2}|=|\frac{x_1^2(c-1)+x_2^2(c+1)}{x_1^2+x_2^2}|\geq\frac{x_1^2|(c-1)|+x_2^2|(c+1)|}{|x_1^2+x_2^2|} > \frac{(x_1^2+x_2^2)\min\{|c-1|, |c+1|\}}{|x_1^2+x_2^2|}=$ $\min \{|c-1|, |c+1|\}$.
Next, consider $c=1$. Set $\epsilon:=1$. Consider some arbitrary $\delta>0$. Choose $x:= (x_1, x_2)$ such that $x_1=0$, and $x_2=\sqrt\frac{\delta}{2}$. Then $d(x, 0)<\delta$, but $|f(0)-f(x)|=|c-\frac{x_1^2-x_2^2}{x_1^2+x_2^2}|=|\frac{2x_2^2}{x_1^2+x_2^2}|={2}>1.$
Finally, consider $c=-1$. Set $\epsilon:=1$. Consider some arbitrary $\delta>0$. Choose $x:= (x_1, x_2)$ such that $x_2=0$, and $x_1=\sqrt\frac{\delta}{2}$. Then $d(x, 0)<\delta$, but $|f(0)-f(x)|=|c-\frac{x_1^2-x_2^2}{x_1^2+x_2^2}|=|\frac{2x_1^2}{x_1^2+x_2^2}|={2}>1.$
| The problem with your approach lies in the fact that if for real numbers we have $a>b$ then we cannot deduce that $|a|>|b|$. In your case, while it's true that $x_1^2(c-1)+x_2^2(c+1) \geq (x_1^2+x_2^2)\cdot \min\{c-1,c+1\}$, you cannot take absolute values of both sides because you have no guarantee that they are both positive. As an example, for $c=-1$ the inequality gives $-2x_1^2 \geq -2(x_1^2+x_2^2)$ but when you take absolute values you get $x_1^2\geq x_1^2+x_2^2$ which is obviously not correct.
The simplest way to attack this problem is to find two paths that give rise to two different limits which will prove that the limit of $f$ at $(0,0)$ does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the pattern in the powers of $\sqrt{2}-\sqrt{1}$? What is the pattern in this?
$$\begin{align}
\left(\sqrt{2}-\sqrt{1}\right)^1 &= \sqrt{2}-\sqrt{1}\\
\left(\sqrt{2}-\sqrt{1}\right)^2 &= \sqrt{9}-\sqrt{8}\\
\left(\sqrt{2}-\sqrt{1}\right)^3 &= \sqrt{50}-\sqrt{49}\\
\left(\sqrt{2}-\sqrt{1}\right)^4 &= \sqrt{289}-\sqrt{288}\\
\end{align}$$
I thought of applying the binomial theorem
| Given that
$$\left(\sqrt{2}-\sqrt{1}\right)^k = \sqrt{A_k}-\sqrt{A_k-1}$$
then
$$\left(\sqrt{2}-\sqrt{1}\right)^{k+1} = \left(\sqrt{2}-\sqrt{1}\right)(\sqrt{A_k}-\sqrt{A_k-1})=\\=\sqrt 2\sqrt{A_k}-\sqrt 2\sqrt{A_k-1}-\sqrt{A_k}+\sqrt{A_k-1}=$$
$$=\sqrt{A_{k+1}}-\sqrt{A_{k+1}-1}$$
with
*
*$\sqrt{A_{k+1}}=\sqrt 2\sqrt{A_k}+\sqrt{A_k-1} \implies A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)}$
*$\sqrt{A_{k+1}-1}=\sqrt 2\sqrt{A_k-1}+\sqrt{A_k}\implies A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)}$
that is
$$A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)}$$
and $A_1=2$.
Using an approach similar to this one, we have
$$A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)} \\\iff 2A_{k+1}-1=3(2A_{k}-1)+2\sqrt2\sqrt{(2A_k-1)^2-1}$$
and by $2A_{k+1}-1=\frac12\left(t_k+\frac1{t_k}\right) \implies t_1=3+2\sqrt 2$ we obtain
$$t_{k+1}+\frac1{t_{k+1}}=3\left(t_k+\frac1{t_k}\right)+2\sqrt 2\sqrt{\left(t_k+\frac1{t_k}\right)^2-4}$$
$$t_{k+1}+\frac1{t_{k+1}}=3\left(t_k+\frac1{t_k}\right)+2\sqrt 2\left(t_k-\frac1{t_k}\right)$$
$$t_{k+1}+\frac1{t_{k+1}}=(3+2\sqrt 2)t_k+\frac{3-2\sqrt 2}{t_k}$$
$$t_{k+1}+\frac1{t_{k+1}}=(3+2\sqrt 2)t_k+\frac{1}{(3+2\sqrt 2)t_k}$$
$$t_{k}=(3+2\sqrt 2)^k$$
and then
$$A_k=\frac{(3+2\sqrt 2)^k+(3-2\sqrt 2)^k+2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4540918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Derivative of capital Pi product I wanted to find the derivative of this function at $x=6$
$$y= \prod_{i=1}^{10} (x-i) = (x-1)(x-2) \cdots (x-10) $$
without expanding all of the brackets, so I used the product rule to find a pattern. However, the resulting sum tells me that the derivative is zero at every whole number which is obviously not true. I've been over my solution and I can't see how I've gone wrong. Please could someone highlight where I went wrong? Thank you in advance.
\begin{align*}
\frac{\textit{d}y}{dx} &= (x-2)(x-3) \cdots (x-10) + (x-1) \frac{d}{dx} \biggl((x-2) \cdots (x-10) \biggr) \\
&= \prod_{i=2}^{10} (x-i) + (x-1) \frac{d}{dx} \biggl(\prod_{i=2}^{10} (x-i) \biggr) \\
&= \prod_{i=2}^{10} (x-i) + (x-1)\prod_{i=3}^{10} (x-i) + (x-1)(x-2)\frac{d}{dx} \biggl(\prod_{i=3}^{10} (x-i) \biggr) \\
&= \prod_{i=2}^{10} (x-i) + (x-1)\prod_{i=3}^{10} (x-i) + (x-1)(x-2)\biggl(\prod_{i=4}^{10} (x-i) \biggr) + \cdots \\
&= \frac{y}{x-1} + \frac{y}{x-2} + \frac{y}{x-3}+\cdots + \frac{y}{x-10} \\
&= \sum_{i=1}^{10} \biggl(\frac{y}{x-i}\biggr)
\end{align*}
| The pattern is simpler to understand than this. In general suppose $f(x) = (x - a) g(x)$ and we want to compute $f'(a)$. (Here $g(x)$ could be a polynomial or more generally any differentiable function.) Then
$$f'(x) = g(x) + (x - a) g'(x)$$
which gives $\boxed{ f'(a) = g(a) }$. This is particularly easy if you notice that applying the definition of the derivative in this case directly gives
$$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = \lim_{x \to a} g(x) = g(a).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4541689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Sum of absolute values of coefficients of an equation? If $x =√2+√5$ is a root of $kx^4+mx^3+nx^2+l = 0$ where $k, m, n, l$ are integers
find the value of $|k| + |m| + |n| + |l|$.
Now as all coefficients are integers, so conjugate of x is also a root of the above.
So putting $x=√2+√5$ and $x=√2-√5$ in the above eq one by one and then subtracting the two equations we got.
We get $k.52.√10 + m.22.√5+ n.4.√10+l=0$
Now $m=0$ and $l=0$ for it to become integer.
and $k.52.√10+ n.4.√10=0$
$\implies n=-13k$
Now putting all values in the original equation
we get $kx^4-13kx^2=0 \implies x^2(kx^2-13k)=0 $ and for x not equal to 0
we get $kx^2-13k=0\implies k=0$ is the only integer solution
Hence $k=l=m=n=0$
hence final answer becomes $0$
Is this the correct way to deduce the problem?
One other way is below
$x =√2+√5 $
$⇒ x−√2 = √5 $
$⇒ (x−√2)^2 = (√5)^2 $
$⇒ x^2+2−2√2x =5 $
$⇒ (x^2 − 3) = 2√2x $
$⇒ (x^2 − 3)^2 = (2√2x)^2 $
$⇒ x^4 − 6x^2 + 9 = 8x^2 $
$⇒x^4 − 14x^2 + 9 = 0 $
$⇒ k = 1, m = 0, n = −14, l = 9 $
$⇒ |k| + |m| + |n| + |l| =1 + 0 + 14 + 9 = 24
$
| $x=\sqrt{2}+\sqrt{5}$ is a root of the equation
$$Nx^4-14Nx^2+9N=0$$
for any integer $N\neq 0$. Because, $(x^2-7)^2-40= x^4-14x+9=0.$
So, the sum asked can be $24|N|=24M$ for some positive integer $M$.
It can not be zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\lim\limits_{x\to 1} \frac{1-\cos(\sin(x^3-1))}{x^3-1}$ without L'Hospital's Rule? I am asked to solve $\lim\limits_{x\to 1} \frac{1-\cos(\sin(x^3-1))}{x^3-1}$ without using L'Hospital's Rule.
I'm not sure how to go about it.
There is an indeterminate form $(\frac{0}{0})$ at $x=1$ and L'Hospital's Rule seems like the best course of action.
I tried to multiply by $\frac{1+\cos(\sin(x^3-1))}{1+\cos(\sin(x^3-1))}$ to get $\lim\limits_{x\to 1} \frac{1-\cos^2(\sin(x^3-1))}{x^3-1}\cdot\frac{1}{1+\cos(\sin(x^3-1))} = \lim\limits_{x\to 1} \frac{\sin(\sin(x^3-1))}{x^3-1}\cdot \sin(\sin(x^3-1))\cdot\frac{1}{1+\cos(\sin(x^3-1))}$
I'd try to get a limit of the form $\lim\limits_{u\to 0} \frac{\sin(u)}{u}$ because I know that that limit is $1$, but the problem is that sine is composed with itself.
Any help would be appreciated!
| Let $y=x-1$
so $y \to 0$.
Then
$x^3-1
=(y+1)^3-1
=y^3+3y^2+3y
=y(y^2+3y+3)
=3y+O(y^2)
$
so
$\begin{array}\\
\dfrac{1-\cos(\sin(x^3-1))}{x^3-1}
&=\dfrac{1-\cos(\sin(3y+O(y^2)))}{3y+O(y^2)}\\
&=\dfrac{1-\cos(3y+O(y^2))}{3y+O(y^2)}\\
&=\dfrac{1-(1-\frac{(3y+O(y^2))^2}{2}+O(y^2))}{3y+O(y^2)}\\
&=\dfrac{\frac{(3y+O(y^2))^2}{2}+O(y^2))}{3y+O(y^2)}\\
&=\dfrac{O(y^2)}{3y+O(y^2)}\\
&=O(y)\\
&\to 0\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4546269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
To find the doubling map of the elliptic curve $y^2=x^3+1$ I want to verify that the doubling map of the elliptic curve $y^2=x^3+1$ is given by
\begin{align}
P=(x,y) &\to 2P \\
(x,y) &\mapsto \left(\frac{x^4-8x}{4x^3+4}, \frac{2x^6+40x^3}{8y^3} \right)
\end{align}
For calculation flexibility we denote $P=(x',y')$. We will find the tangent line at $P$ and make intersect with the curve to find $2P$.
Differentiating the curve with respect to $x$, we get $$\frac{dy}{dx}=\frac{3x^2}{2y}.$$
So the tangent line at $P$ is given by $$y-y'=\frac{3x'^2}{2y'}(x-x') \Rightarrow y=\frac{3x'^2x-3x'^3+2y'^2}{2y'}.$$
Substituting $y$ in the curve, we get
\begin{align}
&\left(\frac{3x'^2x-3x'^3+2y'^2}{2y'} \right)^2=x^3+1 \\
&\Rightarrow (3x'^2x-3x'^3+2y'^2)^2=4y'^2x^3+4y'^2 \\
&\Rightarrow 9x'^4x^2+9x'^6+4y'^4+18x'^5x+12x'^3y'^2-12x'^2y'^2x=4y'^2x^3+4y'^2 \\
& \Rightarrow 4y'^2x^3-9x'^4x^2+(12x'^2y'^2-18x'^5)x+(4y'^2-9x'^6-4y'^4-12x'^3y'^2)=0, \cdots (1)
\end{align}
The elliptic passes through $(0,1)$, and hence plugging it in $(1)$, we get
$$4y'^2-9x'^6-4y'^4-12x'^3y'^2=0, \cdots (2)$$
But now I am out of way, how to solve for $x', y'$ ?
what is the apropriate way to solve my question ?
| As hinted by @David Lui and @kelalaka, I would like to answer partially.
The elliptic curve is $y^2=x^3+1$. The tangent line at $P=(x',y')$ is $$L: y=\frac{3x'^2x-3x'^3+2y'^2}{2y'}.$$
Substituting $L$ in the elleiptic curve, we get
$$4y'^2x^3-9x'^4x^2+(12x'^2y'^2-18x'^5)x+(4y'^2-9x'^6-4y'^4-12x'^3y'^2)=0,$$
which is a cubic equation with doubale zero $x'$. If $b$ be another zero of this cubic equation, then by theory of equations we have the folowing relation:
\begin{align} &2x'+b=x'+x'+b=\frac{9x'^4}{4y'^2} \\ \Rightarrow &b=\frac{9x'^4}{4y'^2}-2x' \\ \Rightarrow &b=\frac{9x'^4}{4(x'^3+1)}-2x', ~~(\because y'^2=x'^3+1) \\
\Rightarrow &b=\frac{x'^4-8x'}{4x'^3+4}:=x \end{align}
Now substituting $\frac{x'^4-8x'}{4x'^3+4}:=x $ in the tangent line and using $y'^2=x'^3+1$, we get
\begin{align}y=\frac{2x'^6+40x'^3}{8y'^3}. \end{align}
Therefore $2P=\left(\frac{x^4-8x}{4x^3+4}, \frac{2x^6+40x^3}{8y^3} \right)$. This concludes the doubling map.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the number of ordered triplets $x,y,z$
Find the number of ordered triplets $(x,y,z)$ of positive integers less than $13$ such that the product $x\cdot y\cdot z$ is divisible by $20$.
My work:
The possible values of $x\cdot y\cdot z$ are $20,40,60,80,100,120,140,160$. I began to make cases. I made cases for the product to be equal to $20$. I got $27$ cases.
Then for the product to be equal to $40$, I got $36$ cases.
I see that the number of cases are getting bigger. I don't know the best method of doing these types of questions.
Any help is greatly appreciated.
EDIT
Can we do it by the inclusion-exclusion principle$?$ Like counting all the triplets whose product does not divide both $5$ and $4$.
Here, $x,y,z$ have to be different as if it's not the case then the possible number of triplets exceed $216$.
| Here is a generating function approach. We consider the set of integers $\{1,2,3,\ldots,12\}$ and mark the occurrence of prime factors $2$ and $5$ according to their multiplicity with $z$ resp. $w$. We obtain
\begin{align*}
\begin{array}{cccccccccccc}
1&2&3&4&5&6&7&8&9&10&11&12\\
1&\color{blue}{z}&1&\color{blue}{z^2}&\color{blue}{w}&\color{blue}{z}&1&\color{blue}{z^3}&1&\color{blue}{zw}&1&\color{blue}{z^2}\\
\end{array}\tag{1}
\end{align*}
We obtain from (1) the generating function $A(z,w)$ as
\begin{align*}
\color{blue}{A(z,w)=5+\left(2z+2z^2+z^3\right)+(1+z)w}\tag{2}
\end{align*}
Since we want to count
\begin{align*}
|\{(x,y,z)\,:\,1\leq x,y,z\leq 12, 20|xyz\}|
\end{align*}
we calculate $A(z,w)^3$ which corresponds to the sum of products $x\cdot y\cdot z$ with $1\leq x,y,z\leq 12$ and count all terms which contain $z^2w$. These are the terms which correspond to a multiple of $20$.
Denoting with $[z^n]$ the coefficient of $z^n$ of a series we obtain
\begin{align*}
\color{blue}{\sum_{k\geq 2}}&\color{blue}{\sum_{l\geq 1}[z^k][w^l]A(z,w)^3}\\
&=\sum_{k\geq 2}\sum_{l\geq 1}[z^k][w^l]\left(5+\left(2z+2z^2+z^3\right)+(1+z)w\right)^3\tag{3.1}\\
&=\sum_{k\geq 2}[z^k]\left((1+z)^3+3\cdot 5(1+z)+3\cdot 5(1+z)^2\right.\\
&\qquad\quad+3\left(2z+2z^2+z^3\right)^2(1+z)\\
&\qquad\quad+3\left(2z+2z^2+z^3\right)\left(1+z\right)^2\\
&\qquad\quad+\left.6\cdot 5\left(2z+2z^2+z^3\right)(1+z)\right)\tag{3.2}\\
&=\sum_{k\geq 2}[z^k]\left(\left(3z^2+z^3\right)+3\cdot 5\cdot 0+3\cdot 5z^2\right.\\
&\qquad\quad+3(5)^2(2)\\
&\qquad\quad+3\left(\left(2z^2+z^3\right)+2\left(2z^2+2z^3+z^4\right)+\left(2z^3+2z^4+z^5\right)\right)\\
&\qquad\quad\left.+6\cdot 5\left(2z+2z^2+z^3\right)(1+z)\right)\tag{3.3}\\
&=4+0+15+150+3\left(3+2\cdot 5+5\right)+6\cdot 5\left(3+5\right)\tag{3.4}\\
&\,\,\color{blue}{=463}
\end{align*}
in accordance with other answers.
Comment:
*
*In (3.1) we consider a trinomial expansion
\begin{align*}
(a+b+c)^3&=a^3+b^3+c^3+3\left(a^2b+a^2c+ab^2+ac^2+b^2c+bc^2\right)\\
&\qquad+6abc
\end{align*}
*In (3.2) we select all terms which contains a factor $w$, i.e. which contains a factor $c$ in the expansion from $(a+b+c)^3$.
*In (3.3) we simplify terms by skipping constant and linear terms in $z$ and by counting some of the terms which contain a factor $z^k$ with $k\geq 2$.
*In (3.4) we simplify further until we finally derive the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4549873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\sum\limits_{k=0}^{n-a-b}\binom{n-a-b}{k}(a+k-1)!(n-a-k)!$ I'm solving a probability problem, and I've ended up with this sum:
$$\sum\limits_{k=0}^{n-a-b}\binom{n-a-b}{k}(a+k-1)!(n-a-k)!$$
WolframAlpha says I should get the answer $\frac{n!}{a\binom{a+b}{a}}$, but I don't see how to get there. I tried to get to something containing $\binom{a+k-1}{k}$ so that I could use the Hockey Stick Theorem, but I wasn't successful.
So any hints would be very welcome, thanks for any help
| We seek to find a closed form of
$$(n-1)! \sum_{k=0}^{n-a-b}
{n-a-b\choose k} {n-1\choose n-a-k}^{-1}$$
where $n\gt a+b$ and $a,b\ge 1.$
Recall from MSE
4316307 the
following identity which was proved there: with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1}
= [z^n] \log\frac{1}{1-z} (z-1)^{n-k}.$$
We get for our sum
$$(n-1)! \sum_{k=0}^{n-a-b} {n-a-b\choose k}
(n-a-k) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{a-1+k}
\\ = (n-1)! \sum_{k=0}^{n-a-b} {n-a-b\choose k}
(k+b) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-1-b-k}.$$
We get two pieces, the first is
$$b [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-1-b}
\sum_{k=0}^{n-a-b} {n-a-b\choose k} (z-1)^{-k}
\\ = b [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-1-b}
\left[1+\frac{1}{z-1}\right]^{n-a-b}
\\ = b [z^{a+b-1}] \log\frac{1}{1-z} (z-1)^{a-1}
\\ = {a+b-1\choose b}^{-1}.$$
The second is
$$ (n-a-b) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-1-b}
\sum_{k=1}^{n-a-b} {n-a-b-1\choose k-1} (z-1)^{-k}
\\ = (n-a-b) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-2-b}
\sum_{k=0}^{n-a-b-1} {n-a-b-1\choose k} (z-1)^{-k}
\\ = (n-a-b) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-2-b}
\left[1+\frac{1}{z-1}\right]^{n-a-b-1}
\\ = (n-a-b) [z^{a+b}] \log\frac{1}{1-z} (z-1)^{a-1}
\\ = \frac{n-a-b}{b+1} {a+b\choose b+1}^{-1}.$$
Collecting everything we find
$$(n-1)! {a+b\choose b}^{-1}
\left[ \frac{a+b}{a} + \frac{n-a-b}{b+1} \frac{b+1}{a} \right]
\\ = \frac{n!}{a} {a+b\choose b}^{-1}.$$
This is the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Can anyone help me to proof the 2nd expression? We need to proof these expressions when a and b are real numbers
Here is what I have tried so far
| When $a = 0$ and/or $b = 0$, the inequalities clearly hold. Now consider the case $a \neq 0$ and $b \neq 0$.
Both 2) and 3) reduce to 1). For 2), you have
\begin{align}
a^6 + a^4b^2 + a^2b^4 + b^6 &\text{ vs. } a^6 + 2a^3b^3 + b^6, \\
a^4b^2 + a^2b^4 &\text{ vs. } 2a^3b^3,\\
a^2 + b^2 &\text{ vs. } 2ab.
\end{align}
In the final step, you simply divide both sides by $a^2b^2 > 0$.
For 3), you have
\begin{align}
2a^2+2b^2 &\text{ vs. } a^2 + 2ab + b^2, \\
a^2+b^2 &\text{ vs. } 2ab.
\end{align}
Now just do the same as what you did for 1).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $x-\frac{1}{x}$, given $x^3 - \frac{1}{x^3} = 108+76\sqrt{2}$
If $x^3 - \dfrac{1}{x^3} = 108+76\sqrt{2}$, find the value of $x-\dfrac{1}{x}$.
Here's what I've tried so far.
$$\begin{align}
\left(x-\dfrac{1}{x}\right)^3&=x^3-\dfrac{1}{x^3}-3\left(x-\dfrac{1}{x}\right) \\
\rightarrow \quad \left(x-\dfrac{1}{x}\right)^3&=108+76\sqrt{2}-3\left(x-\dfrac{1}{x}\right) \\u:=x-\dfrac{1}{x} \quad\rightarrow \quad u^3+3u-108-76\sqrt{2}&=0
\end{align}$$
Got stuck here since I didn't know how to solve this cubic equation.
I also tried factorizing $x^3-\dfrac{1}{x^3}$.
$$\begin{align}x^3-\dfrac{1}{x^3}&=\left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+1\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+2-1^2\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(\left(x+\dfrac{1}{x}\right)^2-1^2\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}+1\right)\left(x+\dfrac{1}{x}-1\right)
\end{align}$$
Again, I didn't know what I could do with this.
| Here is a suggestion to get an answer systematically - though not fully analytic.
Others have already noted that we need to solve for the desired $u = x- \frac{1}{x}$ in $$f(u) = u^3+3u- 108- 76\sqrt{2} = 0$$
Now observe that any expression of the form $y = a + b \sqrt{2}$, when taken to an integer power, reproduces this form, i.e. $y^n = a_n + b_n \sqrt{2}$. Since the constant part in $f(u)$ is also of this form, we can write $u = a + b \sqrt{2}$, which gives
$$f(u) = a^3 + 6 a b^2 + 3 a -108 + \sqrt{2} (3 a^2 b +2 b^3 + 3 b - 76) = 0 $$
This enables us to solve for the rational and the irrational parts separately, i.e.
$$a^3 + 6 a b^2 + 3 a -108 = 0\\
3 a^2 b +2 b^3 + 3 b - 76 = 0 $$
where $a$ and $b$ are rational numbers. Note that it is not clear beforehand whether $a$ and $b$ are integers, as other answers have tested some integer combinations.
Now this is again two coupled third-degree equations so one might ask what has been won (maybe one of you has a nice idea how to solve this analytically).
I introduce now a way to manipulate these equations to come up with an iterative procedure with works without any knowledge or presupposition of the nature of $a$ and $b$. Also, it does not need any roots (of quadratic or cubic equations) which have been used earlier but which can become cumbersome.
Writing equivalently
$$a^2 + 6 b^2 + 3 -\frac{108}{a} = 0\\
3 a^2 +2 b^2 + 3 - \frac{76}{b} = 0 $$
allows to isolate the terms with $a^2$ and $b^2$ by weighted adding, which gives
$$4 a^2 + 54/a + 3 = 114/b \\
8 b^2 + 38/b + 3 = 162/a$$
Now let $A = 1/a$ and $B = 1/b$ to end with
$$B = \frac{9 A}{19} + \frac{1}{38} + \frac{2}{57 A^2} \\
A = \frac{19 B}{81} + \frac{1}{54} + \frac{4}{81 B^2}$$
One can make this an iterative pair of equations; start with some $A$, obtain directly $B$ from the first equation, plug into the second and obtain a new $A$, repeat ...
Both equations are of the same functional form.
Since there must be a solution, the first one should fall faster with $A$ then the second one with $B$, so this iteration should converge. Here are some values which are obtained when starting with $A=1$:
$$A=1 , B = 0.535, A = 0.3165, B = 0.526, A = 0.320, B = 0.520, A = 0.323, \cdots $$
Convergence is not very fast, but letting this go for a while leads to values which are approximating $$A=\frac13 , B = \frac12$$ and indeed one can check that this solves the equations.
So the result is $u = 3 + 2 \sqrt{2}$, q.e.d. $\qquad \Box$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555856",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Finding anti-derivative of $f(x)= \sin^3 x \cos^2 x $
Finding anti-derivative of $f(x)= \sin^3 x \cos^2 x $
so, integrate $\int \sin^3 x \cos^2 x dx = \int \sin x (1- \cos^2 x) \cos^2 (x) dx $
let $u = \cos x$
$\int -u^2 (1-u^2) du = \int -u^2 dx + \int u^4 dx = \frac{-u^3}{3} + \frac{u^5}{5} + C $
Therefore $ \frac{\cos^3 x}{3}+\frac{\cos^5 x}{5} + C$
Why am I wrong? My teacher said the answer is $ \frac{-\cos x}{8} - \frac{\cos (3x)}{48} + \frac{\cos (5x)}{80} $
| Note that
\begin{align}
\cos^3 x &= \frac34\cos x +\frac14\cos 3x\\
\cos^5 x & = \frac58\cos x +\frac5{16}\cos 3x+\frac1{16}\cos 5x
\end{align}
and
\begin{align}
&-\frac13\cos^3x+\frac15\cos^5x\\
=& -\frac13\left(\frac34\cos x +\frac14\cos 3x\right)+\frac15\left(\frac58\cos x +\frac5{16}\cos 3x+\frac1{16}\cos 5x\right)\\
=&-\frac18\cos x-\frac1{48}\cos 3x +\frac1{80}\cos5x
\end{align}
Thus, the two anti-derivatives are the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $s(x)$ for $0Consider the series
$$
s(x)=\sum_{j=1}^{\infty} \frac{x^j+(1-x)^j}{j^2} .
$$
Evaluate $s(x)$ for $0<x<1$. (You may use that $\sum_1^{\infty} \frac{1}{j^2}=\frac{\pi^2}{6}$.)
I have proved that $s(x)$ is continues and continuously differentiable in $(0,1)$ as well as uniform convergent by M-Test. But how can I derive the answer which is
$s(x)=\frac{\pi^2}{6}-(\ln x)(\ln (1-x))$
I have also done the following according to the comment
$$
S(x)=\sum_{j=1}^{\infty} \frac{x^j+(1-x)^j.}{j^2}=\sum_{j=1}^{\infty} \frac{\frac{1}{1-x}+\frac{1}{1-(1-x)}}{j^2}
$$
$S(x)$ converge unitormly
so $\int s(x) d x=\sum_{j=1}^{\infty} \frac{1}{j^2} \int \frac{1}{(1-x) x} d x$
$$
=\frac{\pi^2}{6}(\operatorname{In}(x)-\operatorname{In}(1-x))+c
$$
Am I doing something wrong here?
Thanks in advance! :D
| Consider the derivative of $\ln(x) \, \ln(1-x)$ which is
$$ \frac{d}{dx} \, \ln(x) \, \ln(1-x) = \frac{\ln(1-x)}{x} - \frac{\ln(x)}{1-x}.$$
Now integrate both sides which gives
$$ \text{Li}_{2}(x) + \text{Li}_{2}(1-x) = c_{0} - \ln(x) \, \ln(1-x),$$
where $\text{Li}_{2}(x)$ is the dilogarithm function.
Setting $x = 0$, or $x=1$ leads to $c_{0} = \text{Li}_{2}(1) = \zeta(2)$ and
$$ \text{Li}_{2}(x) + \text{Li}_{2}(1-x) = \zeta(2) - \ln(x) \, \ln(1-x). $$
Since $$\text{Li}_{2}(x) = \sum_{n=1}^{\infty} \frac{x^n}{n^2}$$
then $$ \sum_{n=1}^{\infty} \frac{x^n + (1-x)^n}{n^2} = \zeta(2) - \ln(x) \, \ln(1-x). $$
Process without using any named functions:
By using
\begin{align}
\frac{d}{dx} \, \ln(x) \, \ln(1-x) &= \frac{\ln(1-x)}{x} - \frac{\ln(x)}{1-x} \\
&= \frac{\ln(1-x)}{x} - \frac{\ln(1 - (1-x))}{1-x} \\
&= - \sum_{n=1}^{\infty} \frac{x^{n-1}}{n} + \sum_{n=1}^{\infty} \frac{(1-x)^{n-1}}{n}
\end{align}
and then integrating both sides gives
$$ \sum_{n=1}^{\infty} \frac{x^n + (1-x)^n}{n^2} = c_{0} - \ln(x) \, \ln(1-x). $$
Setting $x = 0$, or $x=1$, it will be shown that $c_{0} = \zeta(2)$ and leads to the series
$$ \sum_{n=1}^{\infty} \frac{x^n + (1-x)^n}{n^2} = \zeta(2) - \ln(x) \, \ln(1-x). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that for all $n\in \mathbb{N}$ we have that $6$ divides $n^3-n$ I'd like to know if my proof of the following is valid and if there's a shorter proof:
Prove that for all $n\in \mathbb{N}$ we have that $6$ divides $n^3-n$
Proof:
We proceed by induction...
$P(1)=(1)^3-(1)=0$ which $6$ divides
Suppose the proposition is true for $n=k$, that is, there exists $j\in \mathbb{N}$ s.t. $k^3-k=6j$.
We show that $P(k+1)=(k+1)^3-(k+1)=6i$ for some $i$:
$$(k+1)^3-(k+1)=k^3+3k^2+2k$$
$$=(k^3-k)+3k^2+3k$$
$$=\color{red}{6j+3k(k+1)}$$
Now we show that $6$ divides $3k(k+1)$. Either $3k(k+1)$ is even (case 1), or odd (case 2).
(case 1) There exists $l$ s.t. $k=2l\implies3k(k+1)=6l(2l+1)$
(case 2) There exists $m$ s.t. $k=2m+1\implies3k(k+1)=3(2m+1)(2m+1+1)=6(2m^2+3m+1)$
Now, (case 1) and (case 2) $\implies 6$ divides $3k(k+1)\implies 3k(k+1)=\color{blue}{6q}$ for some $q\in\mathbb{N}$
Next, $$P(k+1)=\color{red}{6j+3k(k+1)}=6j+\color{blue}{6q}=6(i)$$
Therefore, by induction $6$ divides $n^3-n$ for all $n\in \mathbb{N}$
| Since $n^3 \equiv n \pmod{2}$, we know that $2$ divides $n^3 - n$.
Now use induction to show that $n^3 - n$ is also divisible by $3$:
Base case: Certainly $3$ divides $n^3 - n$ when $n = 0$.
Inductive step:
Now suppose $3$ divides $n^3 - n$.
Algebra allows us to write
$\quad (n+1)^3 - (n+1) = n^3 + 3n^2 + 2n = n^3 - n + 3n^2 + 3n$
But then
$\quad [(n+1)^3 - (n+1)] \pmod{3}\equiv$
$\quad\quad [n^3 - n] \pmod{3} + [3n^2 + 3n]\pmod{3}\equiv$
$\quad\quad 0 \pmod{3} + 0 \pmod{3}\equiv 0\pmod{3}$
We conclude that $3$ also divides $n^3 -n$.
Since both $2$ and $3$ are prime factors, the product $6$ must also divide $n^3 -n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4559277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Stability of zero of a system of equations Given a system
$$
\begin{cases}
\frac{dx}{dt}=ax-y-x^5\\
\frac{dy}{dt}=ay-z-y^5\\
\frac{dz}{dt}=az-x-z^5\\
\end{cases}
$$
We want to study the stability of the zero solution.
We can linearize the system and get a matrix
$$
\begin{pmatrix}
a & -1 & 0\\
0 & a & -1\\
-1 & 0 & a\\
\end{pmatrix}
$$
The eigenvalues are $a-1$, $a+\frac12+\frac{\sqrt{3}}{2}i$, $a+\frac12-\frac{\sqrt{3}}{2}i$.
If $a>-\frac12$, there is an eigenvalue with positive real part, meaning instability.
If $a<-\frac12$, all eigenvalues have negative real part, meaning asymptotic stability.
If $a=-\frac12$, linearization no longer works. I try to find a Lyapunov function. I tested $V=x^2+y^2+z^2$, but this fails to work.
Question: How to study the stability of zero when $a=-\frac12$?
| I remember that when eigenvalues with real part are equal to $0$, the system is still Lyapunov stable (there are several types of stability). But I took this course longtime ago and almost forget all. This solution should be easier than the one I propose below.
Now we find the closed-form solution of the linearized system for the case $a =-\frac{1}{2}$.
$$
\begin{cases}
\frac{dx}{dt}=-\frac{1}{2}x-y\\
\frac{dy}{dt}=-\frac{1}{2}y-z\\
\frac{dz}{dt}=-\frac{1}{2}z-x\\
\end{cases}
\Longleftrightarrow
\begin{cases}
\left(\frac{dx}{dt}+\frac{1}{2}x\right)e^{\frac{t}{2}}=-e^{\frac{t}{2}}y\\
\left(\frac{dy}{dt}+\frac{1}{2}y\right)e^{\frac{t}{2}}=-e^{\frac{t}{2}}z\\
\left(\frac{dz}{dt}+\frac{1}{2}z\right)e^{\frac{t}{2}}=-e^{\frac{t}{2}}x\\
\end{cases}
\Longleftrightarrow
\begin{cases}
\frac{d}{dt}\left(e^{\frac{t}{2}}x \right)=-e^{\frac{t}{2}}y\\
\frac{d}{dt}\left(e^{\frac{t}{2}}y \right)=-e^{\frac{t}{2}}z\\
\frac{d}{dt}\left(e^{\frac{t}{2}}z \right)=-e^{\frac{t}{2}}x\\
\end{cases}\\ \tag{1}
$$
For the sake of simplicity, let's denote
$$(X(t),Y(t),Z(t)) = \left(e^{\frac{t}{2}}x, e^{\frac{t}{2}}y,e^{\frac{t}{2}}z\right) $$
Then
$$(1)\Longleftrightarrow
\begin{cases}
\frac{dX}{dt}=-Y\\
\frac{dY}{dt}=-Z\\
\frac{dZ}{dt}=-X
\end{cases} \tag{2}$$
We have
$$(2) \Longrightarrow X'''(t) =-X(t) \Longleftrightarrow X(t)=c_1e^{-t}+c_2e^{\frac{t}{2}}\sin\left(\frac{\sqrt{3}}{2}t\right)+c_3e^{\frac{t}{2}}\cos\left(\frac{\sqrt{3}}{2}t\right)$$
We deduce easily $Y(t)$,$Z(t)$ and also the solution $(x(t),y(t),z(t))$
$$
\begin{cases}
x(t)=c_1e^{-\frac{t}{2}}+c_2\sin\left(\frac{\sqrt{3}}{2}t\right)+c_3\cos\left(\frac{\sqrt{3}}{2}t\right)\\
y(t)=c_4e^{-\frac{t}{2}}+c_5\sin\left(\frac{\sqrt{3}}{2}t\right)+c_6\cos\left(\frac{\sqrt{3}}{2}t\right)\\
z(t)=c_7e^{-\frac{t}{2}}+c_8\sin\left(\frac{\sqrt{3}}{2}t\right)+c_9\cos\left(\frac{\sqrt{3}}{2}t\right)\\
\end{cases} \tag{3}$$
With $(c_i)_{i=1,...,9}$ can be determined from initial condition.
It's my memory is still good, $(3)$ is stable as the solution converges to a stable solution $(x^{*}(t),y^{*}(t),z^{*}(t))$ for $t \to +\infty$
$$\begin{cases}
x^{*}(t)=c_2\sin\left(\frac{\sqrt{3}}{2}t\right)+c_3\cos\left(\frac{\sqrt{3}}{2}t\right)\\
y^{*}(t)=c_5\sin\left(\frac{\sqrt{3}}{2}t\right)+c_6\cos\left(\frac{\sqrt{3}}{2}t\right)\\
z^{*}(t)=c_8\sin\left(\frac{\sqrt{3}}{2}t\right)+c_9\cos\left(\frac{\sqrt{3}}{2}t\right)\\
\end{cases} $$
Then the initial system of equations is Lyapunov stable for $a = -\frac{1}{2}$.
| {
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"url": "https://math.stackexchange.com/questions/4568910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of $\sum_{i=1}^{5}x^5_i+\sum_{i=1}^{5}\frac{1}{x^5_i}$
Suppose $x^5+5x^3+1=0$ and $x_i$ denotes all the complex roots. Find the sum of
$$\sum_{i=1}^{5}x^5_i+\sum_{i=1}^{5}\frac{1}{x^5_i}$$
This polynomial is irreducible over $\Bbb Q[x]$.
I used Vieta's general formulas, however I am not sure if my calculations correct. Here are my calculations.
$$x_1x_2+x_1x_3+x_1x_4+x_1x_5+x_2x_3+x_2x_4+x_2x_5+x_3x_4+x_3x_5+x_4x_5=5$$
$$x_1x_2x_3x_4x_5=-1$$
$$x_1+x_2+x_3+x_4+x_5=0$$
$$x_1x_2x_3+x_1x_2x_4+x_1x_2x_5+x_1x_3x_4+x_1x_3x_5+x_1x_4x_5+x_2x_3x_4+x_2x_3x_5+x_2x_4x_5+x_3x_4x_5=0$$
$$x_1x_2x_3x_4+x_1x_2x_3x_5+x_1x_2x_4x_5+x_1x_3x_4x_5+x_2x_3x_4x_5=0$$
This seems so difficult using these formulas. I couldn't make any solution.
How can I simplify this summation?
$$x_1^5+x_2^5+x_3^5+x_4^5+x_5^5$$
I've tried bring the fractions to common denominator and I got
$$\frac {1}{x_1^5}+\frac {1}{x_2^5}+\frac {1}{x_3^5}+\frac {1}{x_4^5}+\frac {1}{x_5^5}=\frac {(x_1x_2x_3x_4)^5+(x_1x_2x_3x_5)^5+(x_1x_2x_4x_5)^5+(x_1x_3x_4x_5)^5+(x_2x_3x_4x_5)^5}{(x_1x_2x_3x_4x_5)^5}=-(x_1x_2x_3x_4)^5-(x_1x_2x_3x_5)^5-(x_1x_2x_4x_5)^5-(x_1x_3x_4x_5)^5-(x_2x_3x_4x_5)^5$$
But, I couldn't make any further progress from here.
| According to Fundamental Theorem of Symmetric Polynomials,
symmetric polynomials can expressed as polynomial of elementary symmetric polynomials, i.e. the LHS of Vieta's formula.
Then by using math software, the calculation is as follow.
The result of $\sum_{i=1}^{5} x_i^5$ is
$$
s_{1}^{5} - 5 s_{1}^{3} s_{2} + 5 s_{1}^{2} s_{3} + 5 s_{1} s_{2}^{2} - 5 s_{1} s_{4} - 5 s_{2} s_{3} + 5 s_{5}$$
The result of $\sum_{i=1}^{5} \frac{1}{x_i^5}$ is
$$ \frac{- 5 s_{1} s_{4} s_{5}^{3} - 5 s_{2} s_{3} s_{5}^{3} + 5 s_{2} s_{4}^{2} s_{5}^{2} + 5 s_{3}^{2} s_{4} s_{5}^{2} - 5 s_{3} s_{4}^{3} s_{5} + s_{4}^{5} + 5 s_{5}^{4}}{s_{5}^{5}}$$
Given the equation $x^5+5x^3+1=0$,
$s_1=0,s_2=5,s_3=0,s_4=0,s_5=-1$.
Hence the final result is $-10$.
| {
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"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
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Show for $n$, $k$ $\in$ N, such that 1 $\leq k \leq n$, $\sum_{k = 1}^{n}\binom{n}{k}\binom{n}{k-1} = \frac{1}{2}\binom{2n+2}{n+1} - \binom{2n}{n}$ We are given a hint to show that both sides are equal to $\binom{2n}{n+1}$
For the right side, I have computed that
$$\frac{1}{2}\binom{2n+2}{n+1} - \binom{2n}{n}$$
$$=> \frac{(2n+2)!}{2((n+1)!)^2} - \frac{(2n)!}{n!^2}$$
$$=> \frac{(2n)!}{n!^2} * \left [ \frac{(2n+2)(2n+1)}{2(n+1)(n+1)} - 1\right ]$$
$$=> \frac{2n!}{(n!)^2}\left [ \frac{2n+1}{n+1} - \frac{n+1}{n+1}\right ]$$
$$=> \frac{2n!}{(n!)^2} * \frac{n}{n+1}$$
$$=> \frac{2n!}{n(n-1)!n!} * \frac{n}{n+1}$$
$$=> \frac{2n!}{n!(n+1)(n-1)!}$$
$$=> \frac{2n!}{(n-1)!(n+1)!}$$
$$=> \binom{2n}{n+1}$$
However, the left side is really giving me some trouble. I am not sure how to deal with the summation. Any help would be appreciated.
| We have
$$\sum_{k=1}^n {n\choose k} {n\choose n+1-k}
= [z^{n+1}] (1+z)^n \sum_{k=1}^n {n\choose k} z^k
\\ = [z^{n+1}] (1+z)^n \sum_{k=0}^n {n\choose k} z^k
= [z^{n+1}] (1+z)^n (1+z)^n
= [z^{n+1}] (1+z)^{2n}
\\ = {2n\choose n+1}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I show $\lim\limits_{x\to\infty}{\frac{x^{3}+x}{2+4x^3}} = \frac{1}{4}$ with the definition of $\lim\limits_{x \to \infty}f(x)$? How do I show $\lim\limits_{x\to\infty}{\frac{x^{3}+x}{2+4x^3}} = \frac{1}{4}$ with the definition of $\lim\limits_{x \to \infty}f(x)$?
What I have done so far:
$\lim\limits_{x\to\infty}{\frac{x^{3}+x}{2+4x^3}} =\lim\limits_{x\to\infty}{\frac{x^{3}(1+\tfrac{1}{x^2})}{x^3(2\tfrac{1}{x^3}+4)}}=\frac{1+0}{0+4}=\frac{1}{4}$
At this point I wanted to use the definition of $\lim\limits_{x \to \infty}f(x)$
Given: $f : D \subset \mathbb{R} \rightarrow \mathbb{R}$
$\exists r \in \mathbb{R}$ with $(r,\infty) \subset D$.
for $\eta \in \mathbb{R}$ define:
$\lim\limits_{x\to\infty}{f(x)}=\eta : \Leftrightarrow \forall \epsilon >0\ \exists \delta > r \forall x > \delta : f(x) \in B_{\epsilon}(\eta) $\
to show: $\lim\limits_{x\to\infty}{\frac{x^{3}+x}{2+4x^3}} = \frac{1}{4}$
let $\epsilon >0$
$\exists \delta>6 \ \forall x > \delta: \left | \frac{x^{3}+x}{2+4x^3}-\frac{1}{4} \right |<\epsilon$
The plan was to go on and find $\delta$ depending on $\epsilon$ but I didn't get it.
| You have that
$$
\left | \frac{x^{3}+x}{2+4x^3}-\frac{1}{4} \right |=\left | \frac{x^{3}+x}{4(\tfrac{1}{2}+x^3)}-\frac{1}{4} \right |=\frac1{4}\left| \frac{x-\tfrac{1}{2}}{x^3+\tfrac{1}{2}} \right|
$$
Then, if you define $M:=\max\{1,\tfrac{1}{2\sqrt{\epsilon }}\}$ then for $x\geqslant M$ we have that
$$
\frac1{4}\left| \frac{x-\tfrac{1}{2}}{x^3+\tfrac{1}{2}} \right|\leqslant \frac1{4}\left| \frac{x}{x^3} \right|\leqslant \frac1{4M^2}\leqslant \epsilon
$$
∎
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4582461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n$ Question:
$$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=?$$
My work:
$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=\lim_{n\to\infty}\left(\dfrac{n^2(1+5/n+3/n^2}{n^2(1+1/n+2/n^2)}\right)^n=\lim_{n\to\infty}\left(\dfrac{1+5/n+3/n^2}{1+1/n+2/n^2}\right)^n$
$\log L=n\log\left(\dfrac{1+5/n+3/n^2}{1+1/n+2/n^2}\right)$
Is this equal to 0? Then the answer would be $e^0=1$.
The answer was given as $e^4$ and I have no idea how to get to that.
| Another way: Note it is easy to show for large enough $n$, (in this case $n>7$):
$$\frac4{n+1} < \frac{4n+1}{n^2+n+2} < \frac4{n}$$
$$\implies \left(1 + \frac4{n+1}\right)^n < \left(\frac{n^2+5n+3}{n^2+n+2} \right)^n < \left(1 + \frac4{n}\right)^n$$
Now squeezing does the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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There are no positive integers such that $a+b^2$, $a^2+b$ are both powers of $5$ Prove that there doesn't exist positive integers $a$ and $b$ such that $a^2+b$ and $b^2+a$ are both powers of $5$.
Assume on the contrary. Let $$a^2+b=5^x, \quad a+b^2=5^y$$
With $x,y\ge 1$. WLOG assume $x\ge y$ which means $a\ge b$ and $$a+b^2\mid a^2+b\implies a^2+b\equiv 0\pmod{a+b^2}$$
Since $a\equiv -b^2\pmod {a+b^2}$ we get $$a^2+b\equiv b^4+b\equiv 0\pmod{a+b^2}$$
Claim $(a,5)=(b,5)=1$. Furthermore $(a,b)=1$.
Proof
Assume $5\mid a$ then immediately $5\mid b$. Let $a=5a_1,$ $ b=5b_1$ plugging this to the original equation $$5a_1^2+b_1=5^{x-1}$$
Now if $x=1$ we get a contradiction. So assume $x\ge 2$ clearly $5\mid b_1$ meaning $b_1=5b_2$. $$a_1^2+b_2=5^{x-2}$$
Again $x=2$ is impossible. So $a_1=5a_2...$ Cleary this can't go on forever hence $a_1=a_2=0$ a contradiction.
If $p\mid (a,b)$ then $p\mid 5$ meaning $p=5$ contradiction.
Back to our result $a+b^2\mid b(b^3+1)$ since $(a+b^2,b)=1$ we get $$a+b^2\mid b^3+1$$
We're pretty much done. we just need the following claim.
Claim $5\mid b+1$.
Proof We know $5\mid b^3+1$ Thus $$b^3\equiv -1\pmod 5\implies b^6\equiv 1\pmod 5$$
$\operatorname{ord}_5b\mid 6$ but $\operatorname{ord}_5b \mid 4$ so $\operatorname{ord}_5b=2$ or $1$. Surely it's not $1$.
$$b^2\equiv 1\pmod 5 \implies b\equiv -1 \pmod 5$$ So $5\mid b+1$.
All our work till now was in order to use the LTE lemma (Lifting the exponent). Since $5\nmid b$ and $5\mid b+1$ we get $$\nu_5(a+b^2)\le \nu_5(b^3+1)=\nu_5(b+1)+\nu_5(3)=\nu_5(b+1)$$
In fact since the LHS is a power of $5$ we get $a+b^2\mid b+1 \implies b+b^2\le a+b^2\le b+1$. Hence $b\le 0$ A contradiction.
I just need a verification and if you see a part where I could've used a better method to do it -especially the first claim- please tell me.
| Let be by contradiction $a^2+b=5^x$ and $a+b^2=5^y$ with $x\lt y$ (one has $x\ne y$ necessarily) so we have $a^2+b\equiv0\pmod{5^k}$ for $k=1,2,\cdots, x$ and similarly $a+b^2\equiv0\pmod{5^k}$.
For $k=1$ we have the table of the function $f(x,y)=x^2+y$ modulo $5$ where one can see that the only $(a,b)$ and $(b,a)$ giving zero is $(4,4)$. Explaining we have $(x,y)=(1,4),(2,1),(3,1)$ makes $0$ modulo $5$ but $(4,1),(1,2),(1,3)$ does not.
$$\begin{array}{|c|c|}\hline &1 & 2 & 3 & 4 &0 \\\hline 1 & 2 & 3 &4&0&1\\\hline 2 &0&1&2&3&4 \\\hline 3&0&1&2&3&4\\\hline4&2&3&4&0&1\\\hline0&1&2&3&4&0\\\hline\end{array}$$
It follows that $a=5A-1$ and $b=5B-1$ (because $4=-1\pmod5$).
Come back to the beginning we have $$\begin{cases}a^2+b=5^x\\a+b^2=5^y\end{cases}\Rightarrow\begin{cases}5^2A^2-10A+5B=5^x\\5^2B^2-10B+5A=5^y\end{cases}\Rightarrow\begin{cases}5A^2-2A+B=5^{x-1}\\5B^2-2B+A=5^{y-1}\end{cases}$$
from which $$(B-A)\left[5(B+A)-3\right]=5^{y-1}-5^{x-1}=5^{x-1}(5^{y-x}-1)$$ so we have $$B-A=5^{x-1}\\5(B+A)=5^{y-x}+2\hspace1cm(*)$$ The second equation $(*)$ leads to the absurd $2\equiv0\pmod5$ or $3\equiv0\pmod5$. This finishes the proof.
| {
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"url": "https://math.stackexchange.com/questions/4585844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Question about rational exponents From a book, the question goes like this:
This is how John simplified $\sqrt[3]{16x^5}$.
$\sqrt[3]{16x^5} = \sqrt[3]{8} \ \cdot \sqrt[3]{2x^5} = 2\sqrt[3]{2x^5}$
Explain the mistake in John's work and then correct it.
My answer: Isn't John correct? I mean,
\begin{align}
\sqrt[3]{16x^5} & = \sqrt[3]{(2\cdot 8)\cdot x^5} \\
& = (2\cdot 8 \cdot x^5)^{1/3}\\
& = 2^{1/3}\cdot8^{1/3}\cdot x^{5/3} \qquad \text{from $(xy)^n = x^ny^n$ }\\
& = 8^{1/3}\cdot2^{1/3}\cdot x^{5/3} \qquad \text{just rearranging}\\
& = \sqrt[3]{8} \cdot (2x^5)^{1/3}\\
& = \sqrt[3]{8} \cdot \sqrt[3]{2x^5} \qquad \text{which is the same with John's answer}\\
& = 2\sqrt[3]{2x^5}
\end{align}
Am I missing something here? Thanks for the help!
| The question here is what does the autor consider to be a simplified expression. John didn't made any mistake, but someone would say that a more simplified expression would be
$\sqrt[3]{16 x^5} = 2^{(\dfrac{4}{3})} \cdot x^{(\dfrac{5}{3})}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4591066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
In any triangle $\triangle ABC$, show that $4R\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})=r$ This is a problem problem I found in a JEE examination prep textbook, it was a "starred" question which I believe implies that it is more challenging than usual. It goes as follows:
In any triangle $\triangle ABC$, show that $$4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)=r$$
Hint: $$2R^2\sin\left(A\right)\sin\left(B\right)\sin\left(C\right)=\Delta$$
Here is my attempt at it. I want to know if this is correct and if there any better alternative approaches to achieve the same result, please do share them!
We know that:
$$\Delta=rs$$
Using the given hint:
$$2R^2\sin\left(A\right)\sin\left(B\right)\sin\left(C\right)=rs$$
$$16R^2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=r\left(\frac{a+b+c}{2}\right)$$
$$16R\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=r\left(\sin\left(A\right)+\sin\left(B\right)+\sin\left(C\right)\right)$$
Now, focusing on the equation of the right hand side for a bit, we know that:
$$A+B+C=\pi$$
$$\frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$$
$$\sin\left(A\right)+\sin\left(B\right)+\sin\left(C\right)=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)+2\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)$$
$$2\cos\left(\frac{C}{2}\right)\left(\cos\left(\frac{A-B}{2}\right)+\cos\left(\frac{A+B}{2}\right)\right)$$
$$4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)$$
Now substituting this back into the original problem:
$$16R\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=\left(4r\right)\left[\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)\right]$$
And that gives us:
$$4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)=r$$
|
In the figure, $O$ is the circumcenter of $\Delta ABC$, $I$ is the in-center and $AI$ is extended to meet the circumcircle at $D$.
$\angle DIC= \frac{A}{2}+\frac{C}{2}=\frac{A+C}{2}=90^o-\frac{B}{2}$
$$\implies \frac{IC}{\sin B}=\frac{CD}{\sin \angle DIC}$$
$$\implies IC=\frac{CD \times \sin B}{\sin (90^o-\frac{B}{2})}$$
$$\implies IC=\frac{CD \times 2\sin \frac{B}{2} \cos \frac{B}{2}}{\cos\frac{B}{2}}$$
$$\implies IC=2CD \sin \frac{B}{2} \tag{1}$$
Note that
$$\frac{CD}{\sin \frac{A}{2}} = 2R $$
$$ \implies CD= 2R \sin \frac{A}{2} \tag{2}$$
and
$$r=AE=IC \sin \frac{C}{2} \tag{3}$$
$(1), (2), (3) \implies$
$$ r=4R \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Not getting the right derivative of $(x-a)\arctan\frac{(y-b)(z-c)}{(x-a)\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}$ with respect to $c$ Derivative of:
$$\left(x-a\right)\arctan\left(\dfrac{\left(y-b\right)\left(z-c\right)}{\left(x-a\right)\sqrt{\left(x-a\right)^2+\left(y-b\right)^2+\left(z-c\right)^2}}\right)$$
w.r.t. $c$ is, according to online calculator:
$$-\dfrac{\left(x-a\right)^2\left(y-b\right)}{\left(\left(c-z\right)^2+\left(x-a\right)^2\right)\sqrt{\left(x-a\right)^2+\left(y-b\right)^2+\left(z-c\right)^2}}$$
But I am getting:
$$- \dfrac{1}{\sqrt{\left(x-a\right)^2+\left(y-b\right)^2+\left(z-c\right)^2}} \left[
\dfrac{\left(x-a\right)^2\left(y-b\right) \left[ \left( x-a \right)^2+\left(y-b\right)^2 \right]}{(x-a)^2 \left[ \left( x-a \right)^2+\left(y-b\right)^2+\left(c-z\right)^2 \right]+ (y-b)^2 (z-c)^2 } \right]$$
Why is this so? Are the both expressions equivalent?
| Make substitutions to make this more manageable.
First, note that the derivative of the inverse tangent of a quotient simplifies a bit:
$$
\biggl[ \arctan \frac{u}{v} \biggr]'
= \frac{\bigl( \frac{u}{v} \bigr)'}{1 + \bigl( \frac{u}{v} \bigr)^2}
= \frac{\frac{u'v - uv'}{v^2}}{1 + \frac{u^2}{v^2}}
= \frac{u'v - uv'}{u^2 + v^2}
$$
Now, set $X = x - a$, $Y = y - b$, and $Z = z - c$, and $R^2 = X^2 + Y^2 + Z^2$. Letting prime notation denote derivative with respect to $c$,
$$
X' = Y' = 0,
\quad
Z' = -1,
\quad\text{and}\quad
R' = -\frac{Z}{R}
$$
You want to find
\begin{align}
\biggl[ X \arctan \frac{YZ}{XR} \biggr]'
&= X \, \frac{(YZ)'(XR) - (YZ)(XR)'}{(YZ)^2 + (XR)^2} \\&
= X \, \frac{-YXR + YZX\frac{Z}{R}}{(YZ)^2 + (XR)^2} \\
&= -\frac{X^2 Y}{R} \, \frac{R^2 - Z^2}{Y^2Z^2 + X^2R^2} \\
&= -\frac{X^2 Y}{R} \, \frac{X^2 + Y^2}{Y^2Z^2 + X^2R^2}.
\end{align}
which agrees with your answer.
To recover the online calculator answer, expand $R^2$ and factor the denominator:
$$
Y^2Z^2 + X^2R^2
= Y^2Z^2 + X^2(X^2 + Y^2 + Z^2)
= (Z^2 + X^2)(X^2 + Y^2),
$$
so the derivative looks like
\begin{align}
&= -\frac{X^2 Y}{R} \, \frac{X^2 + Y^2}{Y^2Z^2 + X^2R^2} \\
&= -\frac{X^2 Y}{R} \, \frac{X^2 + Y^2}{(Z^2 + X^2)(X^2 + Y^2)} \\
&= -\frac{X^2 Y}{(Z^2 + X^2)R}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4598967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find the area of Quadrilateral $ABCD$. A puzzle for 10th graders. As the title suggests, the problem in this post was meant to be a puzzle for 10th graders, so claims the person who posted this on a language exchange platform: The problem is as follows:
Given a Quadrilateral $ABCD$ with internal point $P$, where $AP=1$, $BP=2$ and $CP=3$, and unknown sides $k$ and $2k$, compute the area of this Quadrilateral.
I first tried to inscribe this quadrilateral into a square but that approach did not turn out successful, I was thinking if there are any other ways to solve it, perhaps via setting up a coordinate system, or via a trigonometric method. I will share my own successful approach below as an answer!
| Using $B$ as the origin and $BC$ and $BA$ as the $x$ and $y$ axis resp., we can express $C(2k,0)$ and $A(0,2k)$. Therefore, expressing in two ways the squares of lengths $AP^2,BC^2,BA^2$ we get:
$$\begin{cases}x^2+y^2&=&4\\x^2+(y-2k)^2&=&1\\(x-2k)^2+y^2&=&9\end{cases} \ \iff \ \ \begin{cases}x^2+y^2&=&4\\x^2+y^2-4ky+4k^2&=&1\\x^2+y^2-4kx+4k^2&=&9\end{cases}$$
$$\iff \ \begin{cases}x^2+y^2&=&4\\4ky-4k^2&=&3\\4kx-4k^2&=&-5\end{cases}\tag{1}$$
(Explanations: new row $R_2=$ old row $R_2$ minus row $R_1$ ; similar operation for row $R_3$).
In (1), we can extract $x=\frac{4k^2-5}{4k}$ and $y=\frac{4k^2+3}{4k}$ from the two last equations of (1) ; plugging the results into the first equation, we get:
$$\left(\frac{4k^2-5}{4k}\right)^2+\left(\frac{4k^2+3}{4k}\right)^2=4\tag{2}$$
wich is a biquadratic equation.
Setting $K:=k^2$, (2) becomes the quadratic equation:
$$(-5+4K)^2+(3+4K)^2=64K$$
which amounts to :
$$16K^2-40K+17=0$$
whose roots are $K=\frac14(5 \pm 2 \sqrt{2})$.
Only root $K=\frac14(5 +2 \sqrt{2})$ is compatible with the given lengths, giving finally :
$$area(ABCD)=height \times \frac12 (upperbase+lowerbase)= 2k \times \frac32 k = 3k^2 =3K=\frac34(5 +2 \sqrt{2})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4604319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Rotation matrix from given axis Let $T\colon\mathbb{R^3}\to\mathbb{R^3}$ denotes linear transformation which rotates by $\frac{\pi}{3}$ counter-clockwise along the vector $u=(1,1,1)$.
If $T(0,1,0)=(a,b,c)$, Find $3a^2+b^2+c^2$.
My Attempt
Consider a plane which contains a point $(0,1,0)$ and uses $u$ as normal vector.
Then I got : $x+y+z=1$.
Since $T$ is rotation, $T(0,1,0)=(a,b,c)$ must be on the same plane $x+y+z=1$.
This means $a+b+c=1$.
Again, $T$ preserves norm of vector : $1=|(0,1,0)|=|T(a,b,c)|=a^2+b^2+c^2$.
So, I got two equations :
$$a+b+c=1 \\ a^2+b^2+c^2=1$$
I want to Find $a, b, c$ without finding exact form of $T$ but I need one more equation about $a,b,c$ to solve this system.
Is there any property of rotation which can give me one more equation about $a,b,c$?
| Another ad-hoc approach. Passing to the plane $x+y+z=1$ where the action takes place, we can see that $(0,1,0)$ rotates around $\left(\frac13, \frac13, \frac13\right)$ within the plane by the angle $\frac{\pi}3$ counter-clockwise.
By definition of $T$ we have
*
*The angle $\angle (0,1,0),\left(\frac13, \frac13, \frac13\right),(a,b,c)$ is equal to $\frac{\pi}3$,
*The distances from $\left(\frac13, \frac13, \frac13\right)$ to the points $(a,b,c)$ and $(0,1,0)$ are equal.
$\hspace{2.5cm}$
It follows that the triangle with vertices $(0,1,0),\left(\frac13, \frac13, \frac13\right),(a,b,c)$ is equilateral so in particular
\begin{align}
\left(a-\frac13\right)^2 + \left(b-\frac13\right)^2 + \left(c-\frac13\right)^2 &= d\left(\left(\frac13, \frac13, \frac13\right), (a,b,c)\right)^2 \\
&= d\left((0,1,0),(a,b,c)\right) \\
&= a^2+(b-1)^2+c^2
\end{align}
which is equivalent to $a-2b+c=1$, so this is your third equation as stated by @enzotib.
You can further simplify the system of equations if you show that the triangle with vertices $(0,0,1),\left(\frac13, \frac13, \frac13\right),(a,b,c)$ is also equilateral (and hence congruent to the aforementioned one) so in particular we have
$$(a-1)^2+b^2 + c^2 = d((0,0,1),(a,b,c)) = d((0,1,0),(a,b,c)) = a^2+(b-1)^2+c^2$$
which immediately implies $a=b$, eliminating one of the solutions.
| {
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"url": "https://math.stackexchange.com/questions/4605744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Evaluation of Trigonometric Limit having 5 terms Evaluation of
$\displaystyle \lim_{h\rightarrow 0}\bigg[\frac{\sin(60^\circ+4h)-4\sin(60^\circ+3h)+6\sin(60^\circ+2h)-4\sin(60^\circ+h)+\sin(60^\circ)}{h^4}\bigg]$
Here above limit is in $(0/0)$ form
So we have using D, L Hopital rule
$\displaystyle \lim_{h\rightarrow 0}\bigg[\frac{4\cos(60^\circ+4h)-12\cos(60^\circ+3h)+12\cos(60^\circ+2h)-4\cos(60^\circ+h)+0}{4h^3}\bigg]$
Again above limit is in $(0/0)$ form
So using D, L Hopital rule
$\displaystyle \lim_{h\rightarrow 0}\frac{-16\sin(60^\circ+4h)+36\sin(60^\circ+3h)-24\sin(60^\circ+2h)+4\sin(60^\circ+h)}{12h^2}$
Above limit is in $(0/0)$ form
So agian using D, L Rule
$\displaystyle \lim_{h\rightarrow 0}\frac{-64\cos(60^\circ+4h)+108\cos(60^\circ+3h)-48\cos(60^\circ+2h)+4\cos(60+h)}{24h}$
Again using D, L rule , We get
$\displaystyle \lim_{h\rightarrow 0}\frac{256\sin(60^\circ+4h)-324\sin(60^\circ+3h)+96\sin(60^\circ+2h)-4\sin(60+h)}{24}$
$\displaystyle \lim_{h\rightarrow 0}\frac{24\sin(60^\circ)}{24}=\frac{\sqrt{3}}{2}$
Above is very lengthy way
Please explain me some short way
Thanks
| I prefer radians everywhere and non-ambiguously. Applying
$$\sin\left(\frac\pi3+t\right)=\sin\frac\pi3\cos t+\cos\frac\pi3\sin t=\frac{\sqrt3}2\left(1-\frac{t^2}2+\frac{t^4}{24}\right)+\frac12\left(t-\frac{t^3}6\right)+o(t^4)$$
to $t=4h,$ $3h,$ $2h,$ $h,$ there only remains (after cancellation):
$$\sin\left(\frac\pi3+4h\right)-4\sin\left(\frac\pi3+3h\right)+6\sin\left(\frac\pi3+2h\right)-4\sin\left(\frac\pi3+h\right)+\sin\frac\pi3$$$$=\frac{\sqrt3}2h^4+o(h^4).$$
| {
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"url": "https://math.stackexchange.com/questions/4607061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Solve the following system: $(x^2+1)(y^2+1) = 12xy$ and $(x+1)^2(y+1)^2=30xy$ I have been given the following system to solve:
$$(x^2+1)(y^2+1) = 12xy$$
$$(x+1)^2(y+1)^2=30xy$$
I noticed that the system is symmetric, but any of the methods I know to solve symmetric systems doesn't seem right. I have also tried writing the second equation as $$((x^2+1)+2x)((y^2+1)+2y) = (x^2+1)(y^2+1)+2y(x^2+1)+2x(y^2+1)+4xy = 30xy$$,
but this didn't get me anywhere. I would appreciate any hints towards the right solution.
| Hint
From a formal point of view, you face an octic polynomial.
Use the first equation and solve for $y$
$$y_\pm=\frac{6x\pm\sqrt{-x^4+34 x^2-1}}{x^2+1}$$ Using $y_+$, plug it in the second, group terms and factor; you should have
$$12 x \left(x^2-4 x+1\right) \left(x^2-3 x+1\right)+$$ $$2 \left(x^2-4 x+1\right) \left(x^2-3 x+1\right)\sqrt{-x^4+34 x^2-1}=0$$ that is to say
$$\left(x^2-4 x+1\right) \left(x^2-3 x+1\right)(6x+\sqrt{-x^4+34 x^2-1})=0$$
Now, it is simple.
Do the same using $y_-$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to compute $\lim_{ x\to 0} \frac{\sqrt{x+2}-\sqrt{2-x}}{\sqrt[3]{x+2}-\sqrt[3]{2-x}}$? How to compute the following limit?
$$\lim_{ x\to 0} \frac{\sqrt{x+2}-\sqrt{2-x}}{\sqrt[3]{x+2}-\sqrt[3]{2-x}}$$
In the context of the book I am reading, I must use some change of variables or multiply the expressions by some "conjugate". I tried to multiply it by $\left(\sqrt{2-x}+\sqrt{x+2}\right)$ to obtain:
$$\frac{2 x}{\left(\sqrt[3]{x+2}-\sqrt[3]{2-x}\right)
\left(\sqrt{2-x}+\sqrt{x+2}\right)}$$
And I tried to make the following change of variables $2+x =t^6$ and then I obtained:
$$\frac{t^3-\sqrt{4-t^6}}{t^2-\sqrt[3]{4-t^6}}$$
But I couldn't go much farther than this. Can you give me a hint?
| Hint
$$\frac{\sqrt{x+2}-\sqrt{2-x}}{\sqrt[3]{x+2}-\sqrt[3]{2-x}}$$
$$=\frac{(\sqrt[3]{x+2} )^3-(\sqrt[3]{2-x})^3 }{ \sqrt[3]{x+2}-\sqrt[3]{2-x}}$$
$$=\frac{(\sqrt[3]{x+2}-\sqrt[3]{2-x}) ((\sqrt[3]{x+2} )^2+(\sqrt[3]{2-x})^2+ \sqrt[3]{x+2}\sqrt[3]{2-x}) }{ \sqrt[3]{x+2}-\sqrt[3]{2-x}}$$
$$ =(x+2)^{2/3} +(2-x)^{2/3}+ \sqrt[3]{4-x^2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4616071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find the taylor series of $f(z)=\frac{1}{z^2+z+1}$ I'm trying to find the taylor expansion of the function $f(z)=\frac{1}{z^2+z+1}$ about $z_0=i$ and its radius of convergence. I'm not sure how to do it without differentiating.
| $$\frac{1}{z^2+z+1}=\sum_{n=0}^\infty a_n (z-i)^n$$
Shift $z→z+i$
$$\frac{1}{z^2+(1+2i)z+i}=\sum_{n=0}^\infty a_n z^n$$
$$=\frac{1}{(z-r)(z-s)}=\frac{1}{r-s}\left(\frac{1}{z-r}-\frac{1}{z-s}\right)$$$$=\frac{1}{r-s}\left(\frac{1/s}{1-z/s}-\frac{1/r}{1-z/r}\right)$$$$=\frac{1}{r-s}\sum_{n=0}^\infty\left(\frac{1}{s^{n+1}}-\frac{1}{r^{n+1}}\right)z^n$$
Now that we found $a_n$, we plug it back in, or rather shift $z→z-i$
$$\frac{1}{z^2+z+1}=\frac{1}{r-s}\sum_{n=0}^\infty\left(\frac{1}{s^{n+1}}-\frac{1}{r^{n+1}}\right)z^n$$
$r,s$ are roots of $z^2+(1+2i)z+i$
| {
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"url": "https://math.stackexchange.com/questions/4617717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
If $y=x^{2}$ then what is $\frac{d}{d y}\left(\frac{d y}{d x}\right)$? Here's what I did:
$\begin{array}{l}
y=x^{2} \\
\Rightarrow \frac{d y}{d x}=2 x
\end{array}$
and we know that
$x=\pm \sqrt{y}$, let's take the $+$ for the sake of the example.
so $\frac{d y}{d x}=2 \sqrt{y}$
so $\begin{array}{c}
\frac{d}{d y}\left(\frac{d y}{d x}\right)=
\frac{1}{\sqrt{y}}
\end{array}$
But on Wolfram Alpha it gives $0$ because it switches automatically from $\frac{d} {dy} $ to $\frac{\partial}{\partial y}$.
But still$\begin{array}{c}
\frac{\partial }{\partial y}\left(\frac{d y}{d x}\right) \end{array} = \begin{array}{l}
\frac{\partial}{\partial y}(2 \sqrt{y})
=\frac{1}{\sqrt{y}}
\end{array} $
I'm so confused by when can I take derivatives or partial derivatives.
| By the chain rule,
$$\frac{d\frac{dy}{dx}}{dy} = \frac{d\frac{dy}{dx}}{dx} \cdot \frac{dx}{dy} = 2 \cdot \frac{\operatorname{sgn}(x)}{2\sqrt y}$$
since $y(x)=x^2\implies x(y)=\pm\sqrt y$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Over a polynomial problem Assume that $P$ is a quadratic polynomial such that $P(1+x) = P(3-x)$, and $P(x)\geq 1$. If $P(0) = 13$, how could we find $P(1)$?
If $P(0) = 13$, then $P$ is of the form $P(x) = ax^2+bx+13$. Since $P(1+x) = P(3-x)$, $P(4) = P(0) =13$ and therefore, $16a+4b = 0$, $b = -4a$. And if $P(-\frac{b}{2a}) = 1$, then $4a-8a + 13 = 1$, leading us to $a = 3$ and $P(x) = 3x^2-12x+13$. From which we could conclude that $P(1) = 4$.
| Hint…consider the minimum point$$P(-\frac{b}{2a})=1$$
| {
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Solving a system of $2n-2$ equations, what is the relationship of solutions between successive n? I am trying to solve a system of $2n-2$ equations. The first two rows of the matrix representation are always $$\begin{pmatrix}-8&1&1&0&\ldots& 0\end{pmatrix} \text{and}\begin{pmatrix}-1&-7&2&1&0&\ldots& 0\end{pmatrix}.$$
The bottom rows are always $$\begin{pmatrix}0&\ldots& 0&1&2&-6&2\end{pmatrix} \text{and}\begin{pmatrix}0&\ldots& 0&1&2&-6\end{pmatrix}.$$ The rows inbetween are $\begin{pmatrix}1&2&-6&2&1&0&\ldots& 0\end{pmatrix}$ and shifting over one to the right as you go down. In my particular problem $x_1$ is always $1$ so that's why there's only $2n-2$ variables.
For example, for $n=4$, the matrix equation is:
$$
M_4\bf{x}=\begin{pmatrix}-8 & 1 & 1 & 0 & 0 & 0 \\
1 &-7& 2& 1 & 0 & 0 \\
1 & 2 &-6 & 2 & 1 & 0 \\
0 & 1 & 2 &-6 & 2 & 1 \\
0 & 0 & 1 & 2 &-6 & 2 \\
0 & 0 & 0 & 1 & 2 &-6 \\
\end{pmatrix}_{6\times6}\begin{pmatrix}x_2\\x_3\\x_4\\x_5\\x_6\\x_7\end{pmatrix}=\begin{pmatrix}a\\b\\c\\d\\e\\f\end{pmatrix}
$$
For $n=5$
$$
M_5\bf{x}=\begin{pmatrix}-8 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
1 &-7& 2& 1 & 0 & 0 & 0 & 0 \\
1 & 2 &-6 & 2 & 1 & 0 & 0 & 0 \\
0 & 1 & 2 &-6 & 2 & 1 & 0 & 0 \\
0 & 0 & 1 & 2 &-6 & 2 & 1 & 0\\
0 & 0 & 0 & 1 & 2 &-6 & 2 & 1\\
0 & 0 & 0 & 0 & 1 & 2 &-6 & 2\\
0& 0 & 0 & 0 & 0 & 1 & 2 &-6\\
\end{pmatrix}_{8\times 8}\begin{pmatrix}x_2\\x_3\\x_4\\x_5\\x_6\\x_7\\x_8\\x_9\end{pmatrix}=\begin{pmatrix}a\\b\\c\\d\\e\\f\\g\\h\end{pmatrix}
$$
My particular problem is that I want to solve thousands (possibly millions) of such systems but it is computationally demanding. You'll notice that you can step down - the $6\times6$ is simply the $8\times8$ matrix with the last two rows and columns removed. Similarly, you can step up by extending the previous matrix with two rows and columns.
The core of my question is this: is there some relationship between the inverse of $M_n$ and $M_{n+1}$ that I can utilize to make solving more efficient?
Specifically, I am interested in $x_n$ for each $n$ if it is more efficient to only calculate that.
| Consider the following symmetric "pentadiagonal" Toepliz matrix :
$$T=\begin{pmatrix}-6 & 2 & 1 & 0 & 0 & 0 & 0 & 0 \\
2 &-6& 2& 1 & 0 & 0 & 0 & 0 \\
1 & 2 &-6 & 2 & 1 & 0 & 0 & 0 \\
0 & \ddots&\ddots & \ddots &\ddots & \ddots & 0 & 0 \\
0 & 0 &\ddots & \ddots &\ddots & \ddots & \ddots & 0\\
0 & 0 & 0 & 1 & 2 &-6 & 2 & 1\\
0 & 0 & 0 & 0 & 1 & 2 &-6 & 2\\
0& 0 & 0 & 0 & 0 & 1 & 2 &-6\\
\end{pmatrix}_{n \times n}$$
The matrix you describe in the general case is a so-called "rank-2 perturbation" of matrix $T$, which is
$$M=T-UU^T$$
where
$$U^T=\begin{pmatrix}1&1&0&0&\cdots 0 \\ 1&0&0&0&\cdots 0
\end{pmatrix}$$
Such a perturbation of a matrix gives rise to a similar formula giving the induced perturbation on their inverses in terms of Woodbury Matrix Identity whose general form is
$$\displaystyle{M^{-1}=\left(T+UCV\right)^{-1}=T^{-1}-T^{-1}U\left(C^{-1}+VT^{-1}U\right)^{-1}VT^{-1}}$$
which boils down in our case, taking $C=-I_n$ (identity matrix of order $n$) and $V=U^T$ :
$$M^{-1}=\left(T-UU^T\right)^{-1}=T^{-1}-\underbrace{T^{-1}U\left(-I+U^TT^{-1}U\right)^{-1}U^TT^{-1}}_{\text{perturbation term } P}$$
Dealing with inversion of matrix $T$ itself, see for example this article "An explicit formula for the inverse of a pentadiagonal
Toeplitz matrix".
Edit : In fact, even for moderate values of $n$, $T^{-1}$ is close to $M^{-1}$. Take a look at the following picture displaying $T^{-1}$ on the left and $M^{-1}$ on the right ($50 \times 50$ matrices) under a colored form with a conversion bar on the left. We see that the differences are very tiny, which is confirmed by precisely comparing the different entries (no entry in the perturbation matrix $P$ is larger than $10^{-4}$).
| {
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Prove that $(\sum x^2)^3\ge9\sum x^4yz$
Prove that $\displaystyle\left(x^2+y^2+z^2\right)^3\ge9\left(x^4yz+y^4xz+z^4xy\right)$, for $x$, $y$, $z\in\Bbb R_+$.
The $pqr$ method doesn't seem possible because the power is too high.
$$\iff\left(p^2-2q\right)^2\ge9r\left(p^3-3pq+3r\right).$$
Then expand the expression to get
$$\sum x^6+3\sum\left(x^4y^2+x^2y^4\right)+6x^2y^2z^2\ge9\sum x^4yz.$$
I wanted to use SOS but cannot find the weight of three squares, my progress:
$$3\sum x^4(y-z)^2=3\sum\left(x^4y^2+x^2y^4\right)-6\sum x^4yz.$$
Whats left is $\displaystyle\sum x^6+6x^2y^2z^2-3\sum x^4yz$. I have trouble dealing with it.
| Another way.
By AM-GM we obtain: $$(x^2+y^2+z^2)^3-9xyz(x^3+y^3+z^3)=$$
$$=\sum_{cyc}(x^6+3x^4y^2+3x^4z^2+2x^2y^2z^2-9x^4yz)=$$
$$=\frac{1}{2}\sum_{cyc}(2x^6-x^4y^2-x^4z^2+7x^4y^2+7x^4z^2-14x^4yz-4x^4yz+4x^2y^2z^2)=$$
$$=\frac{1}{2}\sum_{cyc}(x-y)^2((x^2+y^2)(x+y)^2+7z^4-2xyz(x+y+z))=$$
$$=\frac{1}{2}\sum_{cyc}(x-y)^2((z^2-xy)^2+6z^4-2xy(x+y)z+(x^2+y^2)(x+y)^2-x^2y^2)\geq$$
$$\geq\frac{1}{2}\sum_{cyc}(x-y)^2\left(6z^4-\frac{1}{2}(x+y)^3z+\frac{7}{16}(x+y)^4\right)=$$
$$=\frac{1}{2}\sum_{cyc}(x-y)^2\left(6z^4+3\cdot\frac{7}{48}(x+y)^4-\frac{1}{2}(x+y)^3z\right)\geq$$
$$\geq\frac{1}{2}\sum_{cyc}(x-y)^2\left(4\sqrt[4]{6z^4\left(\frac{7}{48}(x+y)^4\right)^3}-\frac{1}{2}(x+y)^3z\right)=$$
$$=\frac{1}{2}\sum_{cyc}(x-y)^2(x+y)^3z\left(4\sqrt[4]{6\left(\frac{7}{48}\right)^3}-\frac{1}{2}\right)\geq0.$$
| {
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"source": "stackexchange",
"question_score": "5",
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$\left|(a + b)(b + c)(c + a)\right| = \left|(a − b)(b − c)(c − a)\right| \implies | \frac{a}{b} + \frac{b}{c} + \frac{c}{a} | \geq 1$ If $a,b,c \in \mathbb{R}$ satisfy $\left|(a + b)(b + c)(c + a)\right| = \left|(a − b)(b − c)(c − a)\right|$ then $\left| \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right| \geq 1$.
Given what we want to prove, additional assumption of $a,b,c \not= 0$ seems to be necessary.
I can only see kind of 'cyclic symmetry', by which I mean that any $2$ variables can be swapped without changing the equation or inequality. Also, if we could assume $a,b,c > 0$ then the result follows simply from the AM-GM inequality:
$ \frac{1}{3} \left( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right) \geq \left(\frac{a}{b} \frac{b}{c} \frac{c}{a}\right)^\frac{1}{3} = 1 \implies \left| \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right| \geq 3 \geq 1$
But it doesn't help. Such assumption definitely doesn't follow from the initial condition since for instance $a=b=1$ and $c=-1$ satisfy both the equation and inequality.
Other than that, I have no idea how to get from that equation to the desired inequality. Any hints will be appreciated.
| For $abc\neq0$ we obtain:
Case1.$$\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)=\sum_{cyc}(a^2c-a^2b),$$ which gives
$$\sum_{cyc}a^2b=-abc$$ or
$$\sum_{cyc}\frac{a}{c}=-1$$
and by AM-GM we obtain: $$\left(\sum_{cyc}\frac{a}{b}\right)^2=\sum_{cyc}\frac{a^2}{b^2}-2\geq3-2=1,$$ which gives
$$|\sum_{cyc}\frac{a}{b}|\geq1.$$
Case2.
$$\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)=-\sum_{cyc}(a^2c-a^2b),$$ which gives
$$\sum_{cyc}a^2c=-abc$$ or
$$\sum_{cyc}\frac{a}{b}=-1,$$ which gives
$$|\sum_{cyc}\frac{a}{b}|=1\geq1.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What can we tell about quadratic residues modulo $an+b$ ($n=1,2,\dots$)? I only know a little about quadratic residues, and I have a question that: What can we say about quadratic residues modulo $an+b$, for example, $30n+1$? $(n=1,2,\dots)$
Of course, $0,1,4,9,16,25$ are quadratic resiudes modulo $30n+1$ for all $n$.
However, $2$ is a quadratic residue mod $31$ while it is not a quadratic residue modulo $61$.
Are there any relations between quadratic residues modulo $30n+1$?
(For example, a statement such as $cn+d$ is a quadratic residue modulo $30n+1$.)
| Comment: we can find a family of x such that :
$x^2\equiv (cn+t)\bmod (3n+1)$
With some conditions. we must have:
$x^2=m(30n +1)+cn+d=n(30m+c)+m+d$
Suppose $m+d=t^2$ we have:
$(x-t)(x+t)=n(30m+c)$
Suppose:
$\begin{cases}x-t=n\\x+t=30m+c\end {cases}$
We get following conditions:
$\begin{cases} m+d=t^2\\x=\frac 12(30m +n+c)\\t=\frac 12(30m+c-n)\end{cases}$
which leads to solving a system of Diophantine equation. For example :
for$m=1, n=2, c=4$ we get:
$x=18$
$30n+1=61$
$18^2\equiv 19\bmod 61$
$cn+d=4\times 2+d=19\rightarrow d=11$
$m+d=1+11=12$
Or another example:
which does not meet the condition $m+d=t^2$
But for $m=4$, $n=2$ and $c=6$ we get:
$x=64$
$64^2\equiv 9\bmod 61$
$9=2\times 6-3\Rightarrow d=-3$
$\Rightarrow m+d =4-3=1=1^2$
So residue is $cn+d=6n-3$
That is for particular magnitudes of n there exist numbers like x such that the quadratic residue is of the form $cn+d$.
To find n, m and c may be a good computer program can give us many residues in form of $cn+d$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Power Inside Trigonometry Derivative with Chain Rule Trying to differentiate the following function with the chain rule, but I'm stumped with the $\cos(2x+3)^3$. Do I power the $(2x+3)$ by $3$?
$${\sinh ^2 (\cos(2x+3)^3)}$$
| There is an alternative way, using the double angle formula for $\sinh^2 x$.
Let $w = \cos (2x+3)^3$. Using the identity $\sinh^2 x = \frac {1}{2} (\cosh 2x -1)$ and differentiating, we get $$\frac {1}{2} (\sinh 2w (2)) = \sinh 2w \ dw$$
Now we need to find $dw$, meaning we differentiate $\cos (2x+3)^3$. Letting $z = 2x + 3$ and $dz = 2$ we have $(\cos^3 z) = -3 \cos^2 z (\sin z) (2)$ or $3 \sin 2z \cos z$ when we factor out $\cos z$ and see that $2 \sin z \cos z = \sin 2z$.
Putting all the pieces together we get
$$\dfrac {dy}{dx}{\sinh ^2 (\cos(2x+3)^3)} = -\sinh 2(2x+3)\cdot 3 \sin 2(2x+3) \cos (2x+3).$$
NOTE: If you didn't want to use the double angle formulas, you can differentiate $\sinh^2 w$ as $2 \sinh w \cosh w$ and (after differentiating the other parts) leave the answer in powers of cosine and sine. The answer would then be $$\dfrac {dy}{dx}{\sinh ^2 (\cos(2x+3)^3)} = -2 \sinh (2x+3) \cosh (2x+3) \cdot 3 \sin (2x+3) \cos^2 (2x+3)$$
| {
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Finding $x^3+y^3$, given $x + y = 5$ and $xy = 1$
Given
$$x + y = 5 \qquad xy = 1$$
Find $x^3 + y^3$.
To solve this, I tried this:
$y = \frac{1} {x}$
$x + \frac{1}{x} = 5$
$x^2 - 5x + 1 = 0$
What formula needs to be used to find the value of $x$?
| Note that $x^3+y^3=(x+y)(x^2-xy+y^3)$. We have $x+y=5$. Then $(x+y)^2=x^2+y^2+2xy=25$. Then, $x^2-xy+y^2=25-3xy=22$. All that is left is to multiply them.
| {
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"timestamp": "2023-03-29T00:00:00",
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Which functions have the same arc length to the origin at every point as the *signed* curvature at the point? Which functions $y\left( x \right)$ have the same arc length $s$ to the origin at every point as the signed curvature $\kappa$ at the point?
We know the formulas for the arc length $s$
$$
s\left( a, b, \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right) = \int_{a}^{b} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x
$$
and signed curvature $\kappa$
$$
\kappa\left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x}, \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}} \right) = \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}} \text{.}\\
$$
Since we are talking about the length of the function $y\left( x \right)$ from the origin to every point, here $a = 0$ and $b = x$, giving us a nonlinear, higher-order ordinal differential equation:
$$
\begin{align*}
s\left( 0, x, \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right) &= \kappa\left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x}, \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}} \right)\\
\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}}\\
\end{align*}
$$
Now comes the hard part: Solving the ODE for y(x).
(From here only my failed attempts follow...)
My first attempt at solving the ODE
Since I can't think of a specific solution, I would try a bit of reshaping first.
$$
\begin{align*}
\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}}\\
\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{3 \cdot \frac{1}{2}}}\\
\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \right)^{3}} \quad\mid\quad u\left( x \right) := \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}}\\
\int_{0}^{x} u\left( x \right) \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{u\left( x \right)^{3}} \quad\mid\quad \cdot u\left( x \right)^{3} \tag{1.}\\
\int_{0}^{x} u\left( x \right) \operatorname{d}x \cdot u\left( x \right)^{3} &= \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}\\
\\
u\left( x \right) &= \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \quad\mid\quad \left( \right)^{2}\\
u\left( x \right)^{2} &= 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \quad\mid\quad -1\\
u\left( x \right)^{2} - 1 &= \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \quad\mid\quad \sqrt{}\\
\sqrt{u\left( x \right)^{2} - 1} &= \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \quad\mid\quad \frac{\operatorname{d}}{\operatorname{d}x}\\
\frac{\frac{\operatorname{d}u\left( x \right)}{\operatorname{d}x}}{2 \cdot \sqrt{u\left( x \right) - 1}} &= \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}} \tag{2.}\\
\\
\int_{0}^{x} u\left( x \right) \operatorname{d}x \cdot u\left( x \right)^{3} &= \frac{\frac{\operatorname{d}u\left( x \right)}{\operatorname{d}x}}{2 \cdot \sqrt{u\left( x \right) - 1}}\\
\end{align*}
$$
And I'm stuck aigan.
My second attempt at solving the ODE
Next, I would try to transform the higher-order nonlinear ODE into a third-order nonlinear ODE, because maybe that will help. Since the substitution simplified the equation in my opinion, I use it again.
$$
\begin{align*}
\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}} \quad\mid\quad \text{useing } \left( 1. \right) \text{ and } \left( 2. \right)\\
\int_{0}^{x} u\left( x \right) \operatorname{d}x &= \frac{\frac{\operatorname{d}u\left( x \right)}{\operatorname{d}x}}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1}} \quad\mid\quad \frac{\operatorname{d}}{\operatorname{d}x}\\
\\
u\left( x \right) &= \dots
\end{align*}
$$
Time to give up... Since it looks harder now.
My numerical attempt
We know from calculus $\frac{\operatorname{d}f\left( x \right)}{\operatorname{d}x} = \frac{f\left( x + h \right) - f\left( x \right)}{h} + \mathcal{O}\left( h \right)$ or $\frac{\operatorname{d}f\left( x \right)}{\operatorname{d}x} \approx \frac{f\left( x + h \right) - f\left( x \right)}{h}, \text{for small } \left| h \right|$ (Euler method) aka:
$$
\begin{align*}
\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}} \quad\mid\quad \text{useing } \left( 1. \right) \text{ and } \left( 2. \right)\\
\int_{0}^{x} u\left( x \right) \operatorname{d}x &= \frac{\frac{\operatorname{d}u\left( x \right)}{\operatorname{d}x}}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1}}\\
\int_{0}^{x} u\left( x \right) \operatorname{d}x &\approx \frac{\frac{u\left( x + h \right) - u\left( x \right)}{h}}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1}}\\
\int_{0}^{x} u\left( x \right) \operatorname{d}x &\approx \frac{u\left( x + h \right) - u\left( x \right)}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h} \quad\mid\quad \frac{\operatorname{d}}{\operatorname{d}x}\\
u\left( x \right) &\approx \frac{\operatorname{d}}{\operatorname{d}x} \frac{u\left( x + h \right) - u\left( x \right)}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h}\\
u\left( x \right) &\approx \frac{\frac{u\left( x + 2 \cdot h \right) - u\left( x + h \right)}{2 \cdot u\left( x + h \right)^{3} \cdot \sqrt{u\left( x + h \right) - 1} \cdot h} - \frac{u\left( x + h \right) - u\left( x \right)}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h}}{h}\\
u\left( x \right) &\approx \frac{u\left( x + 2 \cdot h \right) - u\left( x + h \right)}{2 \cdot u\left( x + h \right)^{3} \cdot \sqrt{u\left( x + h \right) - 1} \cdot h^{2}} - \frac{u\left( x + h \right) - u\left( x \right)}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2}}\\
u\left( x \right) &\approx \frac{u\left( x + 2 \cdot h \right) \cdot u\left( x \right)^{3} \cdot u\left( x \right)^{3} - u\left( x + h \right) \cdot u\left( x \right)^{3}}{2 \cdot u\left( x + h \right)^{3} \cdot \sqrt{u\left( x + h \right) - 1} \cdot h^{2}} - \frac{u\left( x + h \right) \cdot u\left( x + h \right)^{3} - u\left( x \right) \cdot u\left( x + h \right)^{3}}{2 \cdot u\left( x + h \right)^{3} \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2}}\\
u\left( x \right) &\approx \frac{u\left( x + 2 \cdot h \right) \cdot u\left( x \right)^{3} \cdot u\left( x \right)^{3} - u\left( x + h \right) \cdot u\left( x \right)^{3} - \left( u\left( x + h \right) \cdot u\left( x + h \right)^{3} - u\left( x \right) \cdot u\left( x + h \right)^{3} \right)}{2 \cdot u\left( x + h \right)^{3} \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2}} \quad\mid\quad \cdot \left( 2 \cdot u\left( x + h \right)^{3} \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2} \right)\\
u\left( x \right) \cdot 2 \cdot u\left( x + h \right)^{3} \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2} &\approx u\left( x + 2 \cdot h \right) \cdot u\left( x \right)^{3} \cdot u\left( x \right)^{3} - u\left( x + h \right) \cdot u\left( x \right)^{3} - \left( u\left( x + h \right) \cdot u\left( x + h \right)^{3} - u\left( x \right) \cdot u\left( x + h \right)^{3} \right)\\
u\left( x \right)^{4} \cdot 2 \cdot u\left( x + h \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2} &\approx u\left( x + 2 \cdot h \right) \cdot u\left( x \right)^{6} - u\left( x + h \right) \cdot u\left( x \right)^{3} - u\left( x + h \right)^{4} + u\left( x \right) \cdot u\left( x + h \right)^{3}\\
\end{align*}
$$
Time to give up, since this dosn't look solvable for me...
I am grateful for every help, correction and suggestion.
| I can't see a closed-form solution for an explicit function $y(x)$, but we can make some progress.
Differentiating both sides of the equation and rearranging gives the $3$rd order o.d.e.
$$(1 + (y')^2)^3 = y'''(x) \left[(y')^2 + 1\right] - 3 (y'')^2 y' ,$$ and writing $u := y'$ reduces it to a $2$nd order equation,
$$(1 + u^2)^3 = u'' (1 + u^2) - 3 (u')^2 u .$$
Swapping the independent and dependent variables, i.e., regarding $x$ as a function of $u$, yields the equation
$$x''(u) = -(1 + u^2)^2 x'(u)^3 - \frac{3 u x'(u)}{1 + u^2}.$$ We may now reduce to a $1$st order equation in $z := x'$:
$$z' = -(1 + u^2)^2 z^3 - \frac{3 u z}{1 + u^2} .$$
Substituting $z(u) = \frac{1}{(1 + u^2)^{3 / 2} \sqrt{w(u)}}$ transforms the equation to
$$w'(u) = \frac{2}{u^2 + 1},$$
or
$$w(u) = 2 \arctan u + C.$$
Backsubstituting yields the generic solutions
$$z(u) = \pm \frac{1}{(u^2 + 1)^{3 / 2} (\sqrt{2 \arctan u + C})}$$
and thus
$$x(u) = \pm \int_a^u \frac{dt}{(t^2 + 1)^{3 / 2} (\sqrt{2 \arctan t + C})} .$$ The integral on the right-hand side likely has no closed form, but the inverse of this $x(u)$ is $u(x) = y'(x)$, and integrating again gives $y(x)$.
Incidentally, we can compute the truncation of a series solution: Setting $y(0) = 0$, $y'(0) = a$, $y''(0) = b$ gives
$$y(x) = a x + \frac{1}{2} b x^2 + \frac{1}{6} \left[(1 + a^2)^2 + \frac{3 a b^2}{1 + a^2}\right] x^3 + \cdots .$$
Taking $a = b = 0$ yields the solution with the highest order of tangency to the $x$-axis:
$$y(x) = \frac16 x^3 + \frac1{105} x^7 + \frac{293}{237600} x^{11} + \cdots .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4629269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
System of equations using the extreme principle Find the integers $a,b,c,x,y,z$ that verify the system:
$$x^3+2y^3=a^3$$
$$5y^3+3z^3=b^3$$
$$4z^3+6x^3=c^3$$
This is a logical continuation of this another problem:
Find the integers $a,b,x,y$ that verify the system:
$$x^2+6y^2=a^2$$
$$6x^3+y^2=b^2$$
You assume there is a minimal non-trivial solution $(|x|,|y|,|a|,|b|)$, Add the expressions, and because of the modulo classes of perfect squares with 7, it must be that $a,b$ are divisible by 7, so you divide by 7, and you find another even smaller solution, resulting into a contradiction.
That clearly doesn't work for the first problem, since there are some exceptions regarding the modulo classes.
| I think you can also work modulo $9$ here. The cubes modulo $9$ are $-1$, $0$ and $1$. You can always swap $(a,b,c,x,y,z) \to (-a,-b,-c,-x,-y,-z)$, so you have cases $y^3 \equiv 0 \pmod 9$ or $y^3 \equiv 1 \pmod 9$.
In the first case, you get $3z^3 \pmod 9 \in \{-1,0,1\}$, so $z$ is divisible by three, and using the last equation similarly we get $x$ divisible by $3$. Thus we can reduce $(a,b,c,x,y,z) \to (a/3,b/3,c/3,x/3,y/3,z/3)$.
In the second case, $3z^3+5 \pmod 9 \in \{-1,0,1\}$, so $z^3 \equiv 1 \pmod 9$. The last equation then gives $6x^3+4 \pmod 9 \in \{-1,0,1\}$, so $x^3 \equiv 1 \pmod 9$. But then we have a contradiction in the first equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4629687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{x\to 0} [\sin^2(\frac{\pi}{2-px})]^{\sec^2(\frac{\pi}{2-qx})}$. I am attempting to evaluate
$\lim_{x\to 0} [\sin^2(\frac{\pi}{2-px})]^{\sec^2(\frac{\pi}{2-qx})}$
where $p,q\in \Re$
This is the $1^\infty$ form.
We have a general formula for this indeterminate form-
$\lim_{x\to a}f^g$ where $f\to 1$ and $g\to \infty$ is equal to $e^{\lim_{x\to a}g(f-1)}$
On using the formula, the problem becomes
$\large e^{\lim_{x\to 0} [\sin^2(\frac{\pi}{2-px})-1]{\sec^2(\frac{\pi}{2-qx})}}$
$\large e^{\lim_{x\to 0} [-\cos^2(\frac{\pi}{2-px})]{\sec^2(\frac{\pi}{2-qx})}}$
$\large e^{\lim_{x\to 0} \frac{-\cos^2(\frac{\pi}{2-px})}{\cos^2(\frac{\pi}{2-qx})}}$
How do I proceed further? Any other methods to solve such a problem are welcome.
| $$ \begin{align}
L & = \lim_{x \to 0} - \frac{\cos^2(\frac{\pi}{2-px})}{\cos^2(\frac{\pi}{2-qx})} \\
& = \lim_{x \to 0} - \frac{\sin^2(\frac\pi2-\frac{\pi}{2-px})}{\sin^2(\frac\pi2-\frac{\pi}{2-qx})} \\
& = \lim_{x \to 0} - \left(\frac{\sin\left(\frac\pi2-\frac{\pi}{2-px}\right)}{\frac\pi2-\frac{\pi}{2-px}}\right)^2\left(\frac{\frac\pi2-\frac{\pi}{2-qx}}{\sin^2\left(\frac\pi2-\frac{\pi}{2-qx}\right)}\right)^2\left(\frac{\frac\pi2-\frac{\pi}{2-px}}{\frac\pi2-\frac{\pi}{2-qx}}\right)^2 \\
& = \lim_{x \to 0} - \left(\frac{\sin\left(-\frac{\pi p x}{2(2-px)}\right)}{-\frac{\pi p x}{2(2-px)}}\right)^2\left(\frac{-\frac{\pi qx}{2(2-qx)}}{\sin^2\left(-\frac{\pi qx}{2(2-qx)}\right)}\right)^2\left(\frac{p(2-qx)}{q(2-px)}\right)^2 \\
& = -\frac{p^2}{q^2}
\end{align} $$
So,
$$\lim_{x\to 0} \left(\sin^2(\frac{\pi}{2-px})\right)^{\sec^2(\frac{\pi}{2-qx})} = \large e^{\lim_{x\to 0} \frac{-\cos^2(\frac{\pi}{2-px})}{\cos^2(\frac{\pi}{2-qx})}} = \large e^{-\frac{p^2}{q^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4635751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Question from MIT integration Bee 2023 final: Evaluate $\int^1_0 (\sum^\infty_{n=0}\frac{\left\lfloor 2^nx\right\rfloor}{3^n})^2{\rm d}x$ I am trying to evaluate the last question from MIT integration Bee 2023 Final.
$$\int^1_0 \left (\sum^\infty_{n=0}\frac{\left\lfloor 2^nx\right\rfloor}{3^n} \right )^2{\rm d}x$$
My approach is to divide $(0,1)$ into $1/2^n$ intervals and write the general term of the $y$-value. E.g. For $x \in (k/2^n, (k+1)/2^n)$,
$$f(x)=\left (\sum^{n-1}_{k=0}\frac{\left\lfloor k/2^k\right\rfloor}{3^{n-k}}\right)^2$$
I know that the final integral is just summing up the areas of all the infinite rectangles but I can't solve it. Please help. Thank you.
(The final answer of this question is $27/32$. Candidates were allowed to solve it within 4 minutes.)
| Here's an elementary solution:
Define
$$
I_b = \int_0^b \left(\sum_{n=1}^\infty \frac{\lfloor 2^nx\rfloor}{3^n}\right)^2\,dx
$$
The idea is to write $I_1$ in terms of $I_{1/2}$ in two different ways: one using the substitution $x \rightarrow 2x$, the other by splitting the integral range into two and writing both parts in terms of $I_{1/2}$. Then we solve the system of equations for $I_1$.
The substitution is straightforward, you just have to note that $\lfloor 2^1x\rfloor = 0$ for $0\le x<1/2$:
$$
\begin{align*}
I_1 &= 2\int_0^{1/2} \left(\sum_{n=1}^\infty \frac{\lfloor 2^{n+1}x\rfloor}{3^n}\right)^2\,dx = 2\int_0^{1/2} \left(\sum_{n=\color{red}{2}}^\infty \frac{\lfloor 2^nx\rfloor}{3^{n-1}}\right)^2\,dx \\
&= 2\int_0^{1/2} \left(3\sum_{n=\color{red}{1}}^\infty \frac{\lfloor 2^nx\rfloor}{3^n}\right)^2\,dx = 18\,I_{1/2}
\end{align*}
$$
The other way to write $I_1$ is as follows:
$$
\begin{align*}
I_1 &= \int_0^{1/2} \left(\sum_{n=1}^\infty \frac{\lfloor 2^nx\rfloor}{3^n}\right)^2\,dx + \int_{1/2}^1 \left(\sum_{n=1}^\infty \frac{\lfloor 2^nx\rfloor}{3^n}\right)^2\,dx \\
&= I_{1/2} + \int_0^{1/2} \left(\sum_{n=1}^\infty \frac{\lfloor 2^n(x+1/2)\rfloor}{3^n}\right)^2\,dx \\
&= I_{1/2} + \int_0^{1/2} \left(\sum_{n=1}^\infty \frac{\lfloor 2^nx\rfloor}{3^n} + \underbrace{\sum_{n=1}^\infty \frac{2^{n-1}}{3^n}}_{=1}\right)^2\,dx \\
&= 2\,I_{1/2} + 2\int_0^{1/2} \sum_{n=1}^\infty \frac{\lfloor 2^nx\rfloor}{3^n}\,dx + \frac{1}{2}
\end{align*}
$$
Now the integral without the square is much easier to solve:
$$
\begin{align*}
\int_0^{1/2} \sum_{n=1}^\infty \frac{\lfloor 2^nx\rfloor}{3^n}\,dx &= \sum_{n=1}^\infty \frac{1}{3^n} \int_0^{1/2} \lfloor 2^nx\rfloor\,dx \\
&= \sum_{n=1}^\infty \frac{1}{3^n}\cdot\frac{1}{2^n} \int_0^{2^{n-1}} \lfloor x\rfloor\,dx \\
&= \sum_{n=1}^\infty \frac{1}{3^n}\cdot\frac{1}{2^n} \sum_{k=1}^{2^{n-1}-1} k \\
&= \sum_{n=1}^\infty \frac{1}{3^n}\cdot\frac{1}{2^n}\cdot\frac{1}{2}2^{n-1}\left(2^{n-1}-1\right) \\
&= \frac{1}{8}\sum_{n=1}^\infty \left(\frac{2}{3}\right)^n - \frac{1}{4}\sum_{n=1}^\infty \left(\frac{1}{3}\right)^n \\
&= \frac{1}{8}\cdot 2 - \frac{1}{8}\cdot\frac{1}{2} = \frac{1}{8}
\end{align*}
$$
Recalling that $I_{1/2} = I_1/18$, substitute the values and solve for $I_1$:
$$
I_1 = 2\cdot\frac{1}{18}\,I_1 + 2\cdot\frac{1}{8} + \frac{1}{2} \\
\Rightarrow\frac{8}{9}\,I_1 = \frac{3}{4} \\
\Rightarrow I_1 = \frac{27}{32}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4642139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
} |
Finding all the minima and maxima within a range I'm not sure how to find all the maximas and minimas where the range is $1≤x≤18$ and the function is: $$20\sin \left(\fracπ6x-\frac {2\pi}3\right)+22$$ I already found the first derivative which is: $$\frac{10\pi}3\cos\left(\frac{\pi}{6}x-\frac{2\pi}{3}\right)=0$$ where $x$ is $7$ and using $f''(x)$ and subbing in my $x$, I get $-5.48$ which is a maximum.
| One approach might be to note that the maxima of $20\sin(\frac{\pi}{6}x-\frac{2\pi}{3})+22$ correspond to the maximum values of the sine function. The maxima of $sin(x)$ occur when $x=\frac{\pi}{2}+2n\pi$ with $n\in\mathbb{Z}$.
This means $\frac{\pi}{6}x-\frac{2\pi}{3}=\frac{\pi}{2}+2n\pi\implies\frac{1}{6}x-\frac{2}{3}=\frac{1}{2}+2n\implies x=7+12n$. Which yields $x$ of $7$ giving a value of $42$.
By a similar method, the minima are at $x=1$ and $x=13$ with values of 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\int_{1}^{\infty} \frac{1}{\sqrt{x^3 -1}}\,dx $ This integral converges because $$ \int_{1}^{\infty} \frac{1}{\sqrt{x^3 -1}}\,dx < \int_{1}^{\infty} \frac{1}{x\sqrt{x -1}}\,dx $$
$$\int_{1}^{\infty} \frac{1}{x\sqrt{x -1}}\,dx = 2\int_{0}^{\infty} \frac{1}{u^2+1}\,dx =\pi$$
But, how to solve it?
| Using Mellin transform:
$$ \int_{1}^{\infty}\frac{dx}{\sqrt{x^3-1}} = \frac{1}{3} \int_{0}^{\infty}\frac{dx}{\sqrt{u}(u+1)^{\frac{2}{3}}} \quad x^3=u+1 $$
$$\frac{1}{3} \int_{0}^{\infty}\frac{dx}{\sqrt{u}(u+1)^{\frac{2}{3}}} = \frac{1}{3}\mathcal{M}\left (\frac{1}{2}\right)\left[ \frac{1}{(u+1)^\frac{2}{3}}\right ] = \frac{\Gamma \left( \frac{1}{2} \right)\Gamma \left( \frac{1}{6} \right)}{3\Gamma \left( \frac{2}{3} \right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4645752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $|f(-1)|$ where $f(x)=\int\frac{x^7+2}{(x^2+x+1)^2}dx$
Let $$f(x)=\int\frac{x^7+2}{(x^2+x+1)^2}dx$$ subject to $f(0)=\displaystyle\frac{\pi}{3\sqrt3}.$ Find $|f(-1)|.$
Now I rewrote the integral as $$\int\frac{x(x-1)(x^3-1+2)}{x^2+x+1}+\frac{x+2}{(x^2+x+1)^2}dx$$ and then solved it painstakingly. I cannot even write the whole solution as it is too big and I'm somewhat lazy.
I want to know another approach towards this problem that is concised and less lengthy. Any help is greatly appreciated.
| Suggestions: Partial fractionalize the integrand as
$$\frac{x^7+2}{(x^2+x+1)^2}=x^3-2x^2+x+2-\frac{4x+1}{x^2+x+1}+\frac{1-x^2}{(x^2+x+1)^2}
$$
and then integrate piecewise. In particular
\begin{align}
&\int \frac{1-x^2}{(x^2+x+1)^2}dx= \frac x{x^2+x+1}\\
&\int \frac{4x+1}{x^2+x+1}dx= 2\ln(x^2+x+1)-\frac2{\sqrt3}\tan^{-1}\frac{2x+1}{\sqrt3}\\
\end{align}
As a result
$$f(-1)-f(0)=\int_0^{-1}\frac{x^7+2}{(x^2+x+1)^2}dx=-\frac{19}{12}-\frac{2\pi}{3\sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4646233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\int x \sqrt{x^2 - x}\ dx$
Problem is to integrate $ x \sqrt{x^2 - x}$.
My attempt:
I made it ready for a substitution $u = x^2 - x$
$$\begin{aligned} \int x \sqrt{x^2 -x}\ dx &= \int (2x-1)\sqrt{x^2 - x} \ dx - \int(x - 1)\sqrt{x^2 - x}\ dx\\& = \int \sqrt u\ du - \int(x - 1)\sqrt{x^2 - x}\ dx\\& = \frac23(x^2 -x)^{3/2} + C_1 - \int(x-1) \sqrt{x^2 -x}\ dx\end{aligned}$$
I don't know how to continue from here.
Alternatively I tried this:
$$\begin{aligned} \int x \sqrt{x^2 - x}\ dx &= \int x^2 \sqrt{1- \frac{1}{x}}dx\\ & \overset{1- \frac1x = t^2}{=} \int \frac{2t^2}{(1-t^2)^4}\ dt\\& \overset{t =\sin(\theta)}{=} \int \frac{2\sin^2(\theta) \cos(\theta)\ d\theta}{\cos^4(\theta)}\\& = \int 2 \tan(\theta)\tan(\theta) \sec(\theta) \ d\theta\\ & = \int 2\sqrt{\sec^2(\theta) - 1}\tan(\theta) \sec(\theta) \ d\theta\\& \overset{\sec(\theta) = u}{=} \int 2 \sqrt{\sec^2(\theta) - 1}\tan(\theta) \sec(\theta) \ d\theta\\& = \int 2 \sqrt{u^2 - 1}\ du\\& = u \sqrt{u^2 - 1} - \ln|u + \sqrt{u^2- 1}| + C\\& = \sqrt{x^2 - x} - \ln|x + \sqrt{x^2 - x}| + C\end{aligned}$$
This method is very tedious. Is there any easy way to do the original integral?
| Continue with the first approach
\begin{aligned}
I= \int x \sqrt{x^2 -x}\ dx=& \ \frac23(x^2 -x)^{3/2} - I+\int \sqrt{x^2 -x}\ dx\\
=&\ \frac13 (x^2 -x)^{3/2}+ \frac12\int \sqrt{x^2 -x}\ dx\\
\end{aligned}
where
$$\int \sqrt{x^2 -x}\ dx
= \frac12(x-\frac12) \sqrt{x^2 -x}-\frac18\tanh^{-1}\frac{\sqrt{x^2 -x}}{x-\frac12}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4646982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
Help with integrating $\int \frac{dx}{1+\sqrt{\tan(x)}}$ Starting off with subbing $u^2 = \tan(x)$ to remove the square root, I got:
*
*$$\int \frac{2u}{(1+u)(1+u^4)} du$$ (Deriving that $\sec^2(x) = 1+u^4$)
Then by applying the partial fractions method, I get:
*$$\int \frac{-1}{1+u} du + \int \frac{u^3-u^2+u+1}{1+u^4} du$$
The first integral is manageable but for the second one I had to split the individual terms in the numerator into their own fractions to further obtain:
*$$\int \frac{u^3}{1+u^4}du +\int \frac{u}{1+u^4}du + \int \frac{1-u^2}{1+u^4}du $$
Now, the first two I could solve however it is the last one that I am unable to move forward with;
$$ \int \frac{1-u^2}{1+u^4}du $$
| Since the other parts of your integrand are simple, we look at the remaining part as requested. Note that
$$\frac{1-z^2}{1+z^4} = \frac{z^2(z^{-2} - 1)}{z^2(z^{-2} + z^2)} = \frac{-(1 - z^{-2})}{(z + z^{-1})^2 - 2}. \tag{1}$$
Hence the substitution $$v = z + z^{-1}, \quad dv = 1 - z^{-2} \, dz, \tag{2}$$ yields
$$\begin{align}
\int \frac{1-z^2}{1+z^4} \, dz &= -\int \frac{dv}{v^2 - 2} \\
&= \frac{1}{2\sqrt{2}} \int \frac{1}{v + \sqrt{2}} - \frac{1}{v - \sqrt{2}} \, dv \\
&= \frac{1}{2\sqrt{2}} \log \left| \frac{v + \sqrt{2}}{v - \sqrt{2}} \right| + C \\
&= \frac{1}{2\sqrt{2}} \log \left| \frac{z^2 + \sqrt{2}z + 1}{z^2 - \sqrt{2}z + 1} \right| + C.
\tag{3} \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Determine sum of all complex solutions of $x^3+3=0, \ x^4+4=0, \ x^5+5=0$
Determine sum of all complex solutions of $x^3+3=0, \ x^4+4=0, \ x^5+5=0$
This problem is confusing me a lot. Any idea where my logic is wrong?
$$x^3+3 =0\iff x^3 = -3 \Rightarrow x_1=(-3)^\frac{1}{3} \text{ for } x_1\in\mathbb{R} \\ x_1 \text{ is the only real solution, and the remaining two are complex.} \\ \text{From Viete's relations: } x_1+x_2+x_3=0 \Rightarrow x_2+x_3=-x_1=3^\frac{1}{3} \\ x^4 = -4 \text{ has no real solutions, so the sum of its complex solutions is } 0 \text{ (Again, Viete)} \\ \text{For } x^5 =-5, \text{ the sum of its complex roots, applying the same logic from the first equation, is } 5^\frac{1}{5} \\ \text{Thus, the sum of all complex roots is } 3^\frac{1}{3} + 5^\frac{1}{5}.$$
However, my workbook says the sum is $0$, but gives no explanation. Any clues?
| Another way to see that the sum of the roots are $0$:
Consider $$(x^3+3)(x^4+4)(x^5+5)=0$$ Expanding gives a polynomial with no $x$-term, hence by Vietè's formulas the sum of the roots are $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
An explicit form for this differential equation. So I came across this differential equation:
$$xyy'-y^2=(x+y)^2 e^{-y/x}$$
And I managed to simplify it this way:
First I multiplied by $\frac1{xy}$:
$$y'-\frac{y}{x}=\left(\frac{x}{y}+2+\frac{y}{x}\right) e^{-y/x}$$
Then substituted $v = \frac{y}{x}$ sub to get:
$$\begin{split}
v+xv'-v&=\left(\frac1v+2+v\right) e^{-v}\\
xvv'&=(1+2v+v^2) e^{-v}\\
xvv'&=(v+1)^2 e^{-v}
\end{split}$$
Which is separable. I then proceeded to separate and integrate both sides to get
$$\frac{-e^v}{v+1}\ + e^v=\ln(x)+c$$ or $$\frac{e^v}{v+1}\
=\ln(x)+c$$
Is there a way to get a more explicit equation? Is there something wrong that I did?
| I don't see anything wrong with your work, but you can simplify this even further.
You can actually solve the ODE $x \cdot y\left( x \right) \cdot \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} - \left( y\left( x \right) \right)^{2} = \left( x + y\left( x \right) \right)^{2} \cdot e^{-\frac{y\left( x \right)}{x}}$ for $y$ completely explicitly by using the Lambert W-Function $\operatorname{W}_{n}\left( z \right)$ where $\operatorname{W}_{n}\left( z \right) \cdot e^{\operatorname{W}_{n}\left( z \right)} = z$.
First you should substitute $y\left( x \right) ~{:=}~ x \cdot v\left( x \right)$:
$$
\begin{align*}
y\left( x \right) &= x \cdot v\left( x \right)\\
\frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} &= x \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} + v\left( x \right)\\
\end{align*}
$$
So with the substitution you get:
$$
\begin{align*}
x \cdot y\left( x \right) \cdot \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} - \left( y\left( x \right) \right)^{2} &= \left( x + y\left( x \right) \right)^{2} \cdot e^{-\frac{y\left( x \right)}{x}}\\
x \cdot x \cdot v\left( x \right) \cdot x \cdot \left( \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} + v\left( x \right) \right) - \left( x \cdot v\left( x \right) \right)^{2} &= \left( x + x \cdot v\left( x \right) \right)^{2} \cdot e^{-\frac{x \cdot v\left( x \right)}{x}}\\
x^{3} \cdot v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} &= \left( x \cdot \left( 1 + v\left( x \right) \right) \right)^{2} \cdot e^{-v\left( x \right)}\\
x^{3} \cdot v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} &= x^{2} \cdot \left( 1 + v\left( x \right) \right)^{2} \cdot e^{-v\left( x \right)}\\
\end{align*}
$$
Now you can add the terms with the $y\left( x \right)$ from the $x$ and then integrate:
$$
\begin{align*}
x^{3} \cdot v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} &= x^{2} \cdot \left( 1 + v\left( x \right) \right)^{2} \cdot e^{-v\left( x \right)}\\
\frac{v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} \cdot e^{v\left( x \right)}}{\left( 1 + v\left( x \right) \right)^{2}} &= \frac{1}{x}\\
\int \frac{v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} \cdot e^{v\left( x \right)}}{\left( 1 + v\left( x \right) \right)^{2}}\, \operatorname{d}x &= \int \frac{1}{x}\, \operatorname{d}x\\
\int_{0}^{x} \frac{v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} \cdot e^{v\left( x \right)}}{\left( 1 + v\left( x \right) \right)^{2}}\, \operatorname{d}x &= \ln\left( x \right) + c\\
\end{align*}
$$
Using the rule $\frac{\operatorname{d}\frac{e^{v\left( x \right)}}{v\left( x \right) + 1}}{\operatorname{d}x} = \frac{v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} \cdot e^{v\left( x \right)}}{\left( 1 + v\left( x \right) \right)^{2}}$ we get $\int_{0}^{x} \frac{v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} \cdot e^{v\left( x \right)}}{\left( 1 + v\left( x \right) \right)^{2}} \operatorname{d}x = \int_{0}^{x} \frac{\operatorname{d}\frac{e^{v\left( x \right)}}{v\left( x \right) + 1}}{\operatorname{d}x} \operatorname{d}x = \frac{e^{v\left( x \right)}}{v\left( x \right) + 1}$, so we hold:
$$
\begin{align*}
\int_{0}^{x} \frac{v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} \cdot e^{v\left( x \right)}}{\left( 1 + v\left( x \right) \right)^{2}}\, \operatorname{d}x &= \ln\left( x \right) + c\\
\frac{e^{v\left( x \right)}}{v\left( x \right) + 1} &= \ln\left( x \right) + c\\
\left( v\left( x \right) + 1 \right) \cdot e^{-v\left( x \right)} &= \frac{1}{\ln\left( x \right) + c}\\
v\left( x \right) \cdot e^{-v\left( x \right)} + e^{-v\left( x \right)} &= \frac{1}{\ln\left( x \right) + c}\\
\left( -v\left( x \right) + 1 \right) \cdot e^{-v\left( x \right) + 1} &= -\frac{1}{e \cdot \left( \ln\left( x \right) + c \right)}\\
-v\left( x \right) + 1 &= \operatorname{W}_{n}\left( -\frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right)\\
v\left( x \right) + 1 &= -\operatorname{W}_{n}\left( -\frac{1}{e \left( \ln\left( x \right) + c \right)} \right)\\
v\left( x \right) &= -\operatorname{W}_{n}\left( -\frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right) - 1\\
\end{align*}
$$
Now we can substitute $v\left( x \right)$ to $y\left( x \right) ~{:=}~ x \cdot v\left( x \right) \Leftrightarrow v\left( x \right) = \frac{y\left( x \right)}{x}$ and get:
$$
\begin{align*}
v\left( x \right) &= -\operatorname{W}_{n}\left( -\frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right) - 1\\
\frac{y\left( x \right)}{x} &= -\operatorname{W}_{n}\left( -\frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right) - 1\\
y\left( x \right) &= -x \cdot \operatorname{W}_{n}\left( -\frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right) - x\\
\end{align*}
$$
The probe via Wolfram|Alpha confirms the solution, as you can see here.
However, if you don't accept a concatenation with Lambert W-Function $\operatorname{W}_{n}\left( z \right)$ as the solution, then you could also write the solution in terms of the Fox H-Function $\operatorname{H}_{p,\, q}^{m,\, n}\left( \begin{matrix} a_{1}, &\dots, &a_{p}\\ b_{1}, &\dots, &b_{q}\\\end{matrix} \mid z \right)$ for the branch $n = -1$:$^{\left[ 1. \right]}$
$$
\begin{align*}
y\left( x \right) &= -x \cdot \operatorname{W}_{n}\left( -\frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right) - x\\
y\left( x \right) &= \begin{cases} -\overline{x} \cdot \lim_{\beta \to \alpha^{-}} \left[ \overline{\frac{\alpha^{2} \cdot \left( \left( \alpha - \beta \right) \cdot z \right)^{\frac{\alpha}{\beta}}}{\beta} \cdot \operatorname{H}_{1,\, 2}^{1,\, 1} \left( \begin{matrix} \left( \frac{\alpha + \beta}{\beta},\, \frac{\alpha}{\beta} \right)\\ \left( 0,\, 1 \right),\, \left( -\frac{\alpha}{\beta},\, \frac{\alpha - \beta}{\beta} \right)\\\end{matrix} \mid -\left( -\left( \alpha - \beta \right) \cdot \frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right)^{\frac{\alpha}{\beta} - 1} \right) - x} \right],\, \text{for} \left| x \right| < \frac{1}{e \left| \alpha \right|}\\
-\overline{x} \cdot \lim_{\beta \to \alpha^{-}} \left[ \overline{\frac{\alpha^{2} \cdot \left( \left( \alpha - \beta \right) \cdot z \right)^{-\frac{\alpha}{\beta}}}{\beta} \cdot \operatorname{H}_{2,\, 1}^{1,\, 1} \left( \begin{matrix} \left( 1,\, 1 \right),\, \left( \frac{\beta - \alpha}{\beta},\, \frac{\alpha - \beta}{\beta} \right)\\ \left( -\frac{\alpha}{\beta},\, \frac{\alpha}{\beta} \right)\\\end{matrix} \mid -\left( -\left( \alpha - \beta \right) \cdot \frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right)^{1 - \frac{\alpha}{\beta}} \right) - x} \right],\, \text{otherwise}\\ \end{cases}\\
\end{align*}
$$
where $\overline{z}$ is the complex conjugate of $z$ and $\alpha = -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the exact value of $\int_{0}^{2}x[\frac{1}{x}]dx$. Find the exact value of $\int_{0}^{2}x[\frac{1}{x}]dx$.
Let $[x]$ denote $\lceil{x-\frac{1}{2}}\rceil$.
Using Desmos, I got $2.46736022133$ and WolframAlpha does not give me a solution. My intuition tells me that it might be possible to find an exact value using Trapezoidal Reimann Sums but I am not really sure how to go about doing it. After my attempt, I got stuck but I was at a point where I could plug it into WolframAlpha and it gave me $\frac{\pi^2}{4}$. Why did it come out so nicely?
My attempt:
Where $A_n$ denotes the area of the $nth$ trapezoid from the right:
$$A=\frac{h}{2}(a+b)$$
$$A_n=\frac{\frac{2}{2n-1}-\frac{2}{2n+1}}{2}(\frac{2n}{2n-1}+\frac{2n}{2n+1})$$
$$A_n=\frac{\frac{4n+2}{4n^{2}-1}-\frac{4n-2}{4n^{2}-1}}{2}\left(\frac{4n^{2}+2n}{4n^{2}-1}+\frac{4n^{2}-2n}{4n^{2}-1}\right)$$
$$A_n=\frac{2}{4n^{2}-1}\left(\frac{8n^{2}}{4n^{2}-1}\right)$$
$$A_n=\frac{16n^{2}}{\left(4n^{2}-1\right)^{2}}$$
Then:
$$\int_{0}^{2}x[\frac{1}{x}]dx=\sum_{n=1}^{\infty}A_n=\sum_{n=1}^{\infty}\frac{16n^{2}}{\left(4n^{2}-1\right)^{2}}$$
I do not know how to solve this infinite summation so I plugged it into WolframAlpha and it gave me $\frac{\pi^2}{4}$. How did it get to this conclusion? Is there a more efficient way to solve this?
| *
*If $[x]:=n$ iff $n-\tfrac12< x\leq n+\tfrac12$ (i.e. $[x]=\lceil x-\tfrac12\rceil$, which seems to be what the OP had in mind), then
\begin{align}
\int^2_0 x[1/x]\,dx & =\sum^\infty_{n=1}\int_{[\tfrac{2}{2n+1},\tfrac2{2n-1})}nx\,dx\\
&=\sum^\infty_{n=1}\Big(\frac{2n-1+1}{(2n-1)^2}-\frac{2n+1-1}{(2n+1)^2}\Big)\\
&=\sum^\infty_{n=1}\Big(\frac{1}{2n-1}-\frac{1}{2n+1}+\frac{1}{(2n-1)^2}+\frac{1}{(2n+1)^2}\Big)\\
&=1+2\sum^\infty_{n=1}\frac{1}{(2n-1)^2}-1=2\Big(\sum^\infty_{n=1}\frac{1}{n^2}-\frac14\sum^\infty_{n=1}\frac1{n^2}\Big)\\
&=\frac{\pi^2}{4}
\end{align}
*The same estimate holds when $[x]:=n$ iff $n-\tfrac12 \leq x<n+\tfrac12$ (i.e., $[x]=\lfloor x+\tfrac12\rfloor$) since $\lceil x-\tfrac12\rceil=\lfloor x+\tfrac12\rfloor$ a.s.
*
*If by $[\cdot]$ the OP means the ceiling function $\lceil x\rceil=n$ where $n-1<x\leq n$, then
$$\int^2_0 x\lceil 1/x \rceil\,dx=\int^1_0 x\lceil 1/x \rceil\,dx +\int^2_1 x\,dx=\int^1_0 x\lceil 1/x\rceil \,dx + \frac32$$
From
\begin{align}
\int^1_0 x\lceil 1/x \rceil\,dx & =\sum^\infty_{n=2}\int_{[\tfrac{1}{n},\tfrac1{n-1})}nx\,dx=\sum^\infty_{n=2}\frac{n}{2}\Big(\frac{1}{(n-1)^2}-\frac{1}{n^2}\Big)\\
&=\sum^\infty_{n=2}\frac{2n-1}{2n(n-1)^2}=\sum^\infty_{n=2}\frac{1}{(n-1)^2}+\frac{n-1-n}{2n(n-1)^2}\\
&=\frac12\sum^\infty_{n=1}\frac{1}{n^2}+\frac12\sum^\infty_{n=2}\frac{1}{n(n-1)} =\frac{\pi^2}{12}+\frac12
\end{align}
The value of the integral in this case is $\frac{\pi^2}{12}+2\approx 2.822467$
*If the OP means $[\cdot]$ to be the integer part function (floor function) $\lfloor x\rfloor =n$ iff $n\leq x<n+1$, then
$$\int^2_0 x\lfloor 1/x \lfloor\,dx=\int^1_0 x\lfloor 1/x \rfloor\,dx +\int^2_1 0\cdot x\,dx=\int^1_0 x\lfloor 1/x\rfloor \,dx $$
The value of the integral is then given by
\begin{align}
\int^1_0 x\lfloor 1/x \rfloor\,dx & =\sum^\infty_{n=1}\int_{(\tfrac{1}{n+1},\tfrac1n]}nx\,dx=\sum^\infty_{n=1}\frac{n}{2}\Big(\frac{1}{n^2}-\frac{1}{(n+1)^2}\Big)\\
&=\sum^\infty_{n=1}\frac{2n+1}{2n(n+1)^2}=\sum^\infty_{n=1}\frac{1}{(n+1)^2}+\frac{n+1-n}{2n(n+1)^2}\\
&=\frac12\sum^\infty_{n=2}\frac{1}{n^2}-\frac12\sum^\infty_{n=1}\frac{1}{n(n+1)} =\frac{\pi^2}{12}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4650684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
Computing $\lim_{x\to 0}\left(\frac{\cot(x)}{x^3}-\frac{1}{x^4}+\frac{1}{3x^2}\right)$ I am trying to compute the following limit:
$$\lim_{x\to 0}\left(\frac{\cot(x)}{x^3}-\frac{1}{x^4}+\frac{1}{3x^2}\right)$$
I have tried by rewriting it as $\lim_{x\to 0}\left(\frac{3x+(x^3-3)\tan(x)}{3x^4 \tan(x)}\right)$ and applying De l'Hopital's Rule but the expression quickly becomes unmanageable:
$$\lim_{x\to 0}\left(\frac{3x+(x^3-3)\tan(x)}{3x^4 \tan(x)}\right)\overset{H}{=}\lim_{x\to 0}\frac{3+(x^2 - 3) \sec^2(x) + 2 x \tan(x)}{3 x^3 (4 \tan(x) + x \sec^2(x))}$$
so I then tried by using the Maclaurin expression for $\tan(x)$:
\begin{align*}
\lim_{x\to 0}\left(\frac{\cot(x)}{x^3}-\frac{1}{x^4}+\frac{1}{3x^2}\right)&=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{x-\tan(x)}{x^2\tan(x)}+\frac{1}{3}\right)=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{x-\left(x+\frac{x^3}{3}+\frac{2}{15}x^5\right)}{x\left(\frac{x^3}{3}+\frac{2}{15}x^5\right)}+\frac{1}{3}\right)\\
&=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{-\frac{1}{3}-\frac{2}{15}x^2}{1+\frac{x^2}{3}+\frac{2}{15}x^4}+\frac{1}{3}\right)=+\infty\cdot 0
\end{align*}
and I got an indeterminate form.
I am currently out of ideas so I would appreciate some help in figuring this out, thanks.
EDIT: It just occurred to me that
\begin{align*}
\lim_{x\to 0}\frac{1}{x^2}\left(\frac{-\frac{1}{3}-\frac{2}{15}x^2}{1+\frac{x^2}{3}+\frac{2}{15}x^4}+\frac{1}{3}\right)=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{-\frac{1}{3}-\frac{2}{15}x^2+\frac{1}{3}+\frac{1}{9}x^2+\frac{2}{45}x^4}{1+\frac{x^2}{3}+\frac{2}{15}x^4}\right)\\ \lim_{x\to 0} \frac{1}{x^2}\left(\frac{-\frac{1}{45}x^2+\frac{2}{45}x^4}{1+\frac{x^2}{3}+\frac{2}{15}x^4}\right)=\lim_{x\to 0}\frac{-\frac{1}{45}+\frac{2}{45}x^2}{1+\frac{x^2}{3}+\frac{2}{15}x^4}=-\frac{1}{45}.
\end{align*}
| This example is much simpler to solve if one first simplifies to a common denominator...
$$
\lim_{x\to 0}\left(\frac{\cot(x)}{x^3}-\frac{1}{x^4}+\frac{1}{3x^2}\right)=\lim_{x\to 0}\left(\frac{3x\cot(x)-3+x^2}{3x^4}\right)
$$
... and then uses a series expansion for $\cot(x)$ to understand the behavior of the numerator.
$$
\lim_{x\to 0}\left(\frac{3x(\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}-\frac{2x^5}{945}-\dots)-3+x^2}{3x^4}\right)
$$
If you simplify the first two terms, $3x(1/x-x/3)=3-x^2$, then you see that this is cancelled out the two last terms in the denominator.
$$
\lim_{x\to 0}\left(\frac{3x(-\frac{x^3}{45}-\frac{2x^5}{945}-\dots)}{3x^4}\right)=\lim_{x\to 0}\left(\frac{(-\frac{x^3}{45}-\frac{2x^5}{945}-\dots)}{x^3}\right)=\lim_{x\to 0}\left(-\frac{1}{45}-\frac{2x^2}{945}-\dots\right)
$$
which is now easy to solve since all the expanded terms (except the first) go to zero in the limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4651331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int_{|z|=1} \frac{z^4 + 1}{z^2(\overline{z} - a)(b - \overline{z})} dz$ Let there be numbers ,$$0 < |a| < 1 < |b|$$
calculate the integral :
$$\int_{|z|=1} \frac{z^4 + 1}{z^2(\overline{z} - a)(b - \overline{z})} dz$$
I will provide the solution for this problem and I need some explanation if anyone may help.
On the circle $|z|=1$, we note that $z\bar{z} = 1$, so $\bar{z} = \frac{1}{z}$.
Therefore, we have three points, $0$, $\frac{1}{a}$, and $\frac{1}{b}$. However, since $|a|<1$, we note that $z=\frac{1}{a}$ is not in the given domain,Therefore, we are left with two singular points, $z = \frac{1}{b}$ and $z = 0$.
So we calculate the residue for these points. They did not show the calculation, but they said that $\operatorname{Res}(f,0) = 0$ and $\operatorname{Res}(f,\frac{1}{b}) = \frac{1+b^4}{b^4(b-a)}$.
can someone please explain to me how to culculate such thing with $\bar{z}$ ?
solution:
$\int\limits_{|z|=1} \frac{z^4 +1}{(1-az)(bz-1)} \mathrm{d}z$
$\lim\limits_{z\to 0}\frac{\mathrm{d}}{\mathrm{d}z}(z^2f(z))$ = $\lim\limits_{z\to 0}\frac{\mathrm{d}}{\mathrm{d}z}(z^2\cdot\frac{z^4+1}{(1-az)(bz-1)})$
=$$\lim\limits_{z\to 0}\frac{-4abz^7+5az^6+5bz^6-6z^5+az^2+bz^2-2z}{\left(1-az\right)^2\left(bz-1\right)^2}$$=$0$
$\lim\limits_{z\to \frac{1}{b}}((z-\frac{1}{b})\cdot\frac{z^4+1}{(1-az)(bz-1)})$ =
$\frac{b^4+1}{b^3(b-a)}$
| On the unit circle, the inegrand is $$\frac{z^4+1}{(1-az)(bz-1)}=\frac{((z-\frac{1}{b})+\frac{1}{b})^4+1}{(b-a)(1-\frac{ab}{b-a}(z-\frac{1}{b}))(z-\frac{1}{b})}$$ Upon expansion as a Laurent series in $z-\frac{1}{b}$, the coefficient of $(z-\frac{1}{b})^{-1}$ is seen to be $$\frac{\frac{1}{b^4}+1}{b-a}=\frac{1+b^4}{b^4(b-a)}$$
| {
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Prove that $\sum\limits_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$ How can you prove that
$$
\sum_{k=1}^{n-1}\tan^{2}\left(\frac{k \pi}{2n}\right)
=
\frac{\left(n-1\right)\left(2n - 1\right)}{3}
$$
for every integer $n\geq 1$ > ?.
PS: no, it's not a homework... :-)
| I am doing a similar thing for $\cot$ i hope you can reciprocate it.
We have $$(\cot(\theta)-i)^{n} = \frac{ \cos{n\theta} - \sin{n \theta}}{\sin^{n}(\theta)}$$
Equating the real and imaginary parts on both sides we have $$ \frac{\sin(n\theta)}{\sin{\theta}} = \sum\limits_{s} {n \choose 2s+1} (-1)^{s} \cot^{n-2s-1}(\theta)$$
Now take $2n+1$ instead of $n$ we have, $$\sin{(2n+1)\theta} = \sin^{2n+1}(\theta)P_{n}\cot^{2}(\theta)$$
for $\displaystyle 0 < \theta < \frac{\pi}{2}$ and where $P_{n}$ is the polynomial given by $${ 2n+1 \choose 1}T^{n} - {2n+1 \choose 3}T^{n-1} + \cdots$$
Noting that the zeros of $P_{n}$ are precisely $\displaystyle \frac{r\pi}{2n+1},\ n =1,2,..$, we have the first identity from the sum of the roots formula.
From the inequality $\sin{x} < x < \tan{x}$ we have $$\cot^{2m}(x) < \frac{1}{x^{2m}} < (1 + \cot^{2}(x))^{m}$$ we have $$ \sum\limits_{r=1}^{n} \cot^{2m} \frac{r\pi}{2n+1} < \frac{(2n+1)^2m}{\pi^{2m}} \sum\limits_{r=1}^{n} \frac{1}{r^{2m}} < \sum\limits_{r=1}^{n} \Bigl( 1 + \cot^{2} \frac{r\pi}{2n+1} \Bigr)^{m}$$
Therefore, $$\sum\limits_{r=1}^{n} \Bigl(1+\cot^{2} \frac{r\pi}{2n+1} \Bigr)^{m} = \sum\limits_{r=1}^{n} \cot^{2m} \frac{r\pi}{2n+1} + \mathcal{O}(n^{2m-1})$$
In other words to find $c_{2m}$ where $ \sum\limits_{r=1}^{n} \cot^{2m} \frac{r\pi}{2n+1} = c_{2m}n^{2m} + \mathcal{O}(n^{2m-1})$ it suffices to look at the sum, $$ \cot^{2m} \frac{\pi}{2n+1} + \cot^{2m} \frac{2\pi}{2n+1} + \cdots + cot^{2m} \frac{n\pi}{2n+1}$$
which is the sum $s_{m}$ of $m-th$ powers of the roots of the polynomial $P_{n}$. Then you can complete it using newtons formula for the sum of the roots.
| {
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How do I solve inequalities of the form $\left|\frac{f(x)}{g(x)}\right| \geq 1$? I'm stuck trying to solve an inequality of the form $\left|\frac{f(x)}{g(x)}\right| \geq 1$; specifically, $\left|\frac{2x + 5}{x + 1}\right| \geq 1$
I tried the approach that was taught for solving inequalities in general, and solved first for x when the part enclosed in absolute value signs is $\geq 1$, and then solved for when it is $\leq -1$, but plugging in numbers for the resulting domain $[-4, -2]$ back into the original inequality resulted in some answers that were not $\geq 1$.
Where have I gone wrong in trying to solve this inequality? How is this different from solving an inequality like $\left|\frac{3}{x + 1}\right| \geq 1$?
EDIT: Okay, since I'm clearly messing this up horribly(and now realize that I was absent-mindedly taking the intersection and not the union of my results in spite of having not mixed up the two many times before), I'll edit with the specifics of the solution I was trying and which had me confused before I asked the question. So:
$\left|\frac{2x + 5}{x + 1}\right| \geq 1$
Meaning 1: $\frac{2x + 5}{x + 1} \geq 1$ or 2: $\frac{2x + 5}{x + 1} \leq -1$
Tackling the first: multiplied 1 by $x + 1$, forgetting to account for the fact that $x + 1$ could potentially be negative: $2x + 5 \geq x + 1$, then rearrange for $x \geq -4$.
Now the other inequality: multiplied -1 by $x + 1$, again forgetting to consider $x + 1$ being negative: $2x + 5 \leq -x - 1$ , rearrange and divide by 3 for $x \leq -2$.
Domain is the union, so $[-4, \infty] \cup [-\infty, -2]$, so $R - { 1 }$. This is obviously not right, as plugging in -3 shows.
So, in inequalities where the denominator is some function g(x), if I split it and choose to multiply, will I have to solve each part of the inequality twice(once assuming g(x) is positive, a second time assuming it's negative) and then check which solution makes sense by plugging in numbers?
| With the restriction that $x \ne -1$,
$\; \left|\dfrac{2x + 5}{x + 1}\right| \geq 1$ is equivalent to
$\dfrac{2x + 5}{x + 1} \ge 1 \qquad$ or $\qquad \dfrac{2x + 5}{x + 1} \le -1$
If $ x + 1 \gt 0$, then
$\quad (2x + 5 \ge x + 1$ and $x > -1)\qquad$ or
$\qquad (2x + 5 \le -x - 1$ and $x > -1)$
$\quad (x \ge -4$ and $x > -1)\qquad$ or
$\qquad (x \le -2$ and $x > -1)$
$\quad x \in (-1, \infty) \cup \varnothing$
If $ x + 1 \lt 0$, then
$\quad (2x + 5 \le x + 1$ and $x < -1)\qquad$ or
$\qquad (2x + 5 \ge -x - 1$ and $x < -1)$
$\quad (x \le -4$ and $x < -1)\qquad$ or
$\qquad (x \ge -2$ and $x < -1)$
$\quad x \in (-\infty, -4] \cup [-2, -1)$
$x \in (-\infty, -4] \cup [-2, -1) \cup(-1, \infty)$
| {
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Factoring a Cubic Polynomial I've been trying to understand how
${x^3-12x+9}$
factors to
$(x-3) (x^2+3 x-3)$
What factoring rule does this follow? The net result seems to be similar to what is attained through the sum/difference of cubes factoring pattern, but the signs are different.
Additionally, what type of problem is this, so I can make better and more relevant searches for help on future questions. Is it a cubic trinomial?
| One way to see that is that $x=3$ is a root of $x^3 - 12x + 9$. So $(x-3)$ will be a factor (by the Factor Theorem).
Now you can try to get $x-3$ somehow.
One way you can do that is to rewrite
$$x^3-12x + 9 = x^3 - (3^3 - 3^3) - 12(x - 3+3) + 9$$
$$ = x^3 - 27 + 27 - 12(x - 3) - 36 + 9 = x^3 - 27 - 12(x-3) + (9 -36 + 27)$$
$$ = (x-3)(x^2+3x+9) - 12(x-3) = (x-3)(x^2+3x-3) $$
Here we used the fact that $x^3 - a^3 = (x-a)(x^2 + ax + a^2)$
In general, if $r$ is a root of $f(x) = a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{0}$, then $f(x) - f(r) = f(x)$ gives us a way to factorize $f(x)$ as $(x-r)g(x)$.
$$f(x) = a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{0} - (a_{n}r^{n} + a_{n-1}r^{n-1} + \dots +a_{0})$$
$$ = a_{n}(x^n - r^n) + a_{n-1}(x^{n-1} - r^{n-1}) + \dots + a_{1}(x-r)$$
Just like $x^3 - a^2 = (x-a)(x^2 + ax + a^2)$ we have that
$$x^n - r^n = (x-r)(x^{n-1} + rx^{n-1} + \dots + r^{n-1})$$
and so
$$f(x) = (x-r) (a_{n}(x^{n-1} + rx^{n-2} + \dots + r^{n-1}) + a_{n-1}(x^{n-2} + \dots +r^{n-1}) + \dots + a_1)$$
Once we know a root, we can also try using Polynomial Long Division to get the other factor.
For cubics, the roots can be found without the need to guess. Check this out: Cardano's Method.
Does that help?
| {
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How to prove an identity in radicals? (4 / (3 - sqrt(5))) ^ 2 - ((6 - 5 * sqrt(6)) / (5 - sqrt(6))) ^ 2 = 2 * sqrt(61 + 24*sqrt(5))
$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$
How to prove it is right equality?
I come up with $\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}$, but still can't get it to the obvious equality.
Any ideas?
| Simplify both sides to $8+6\sqrt{5}$, as they are both equal to this.
Just square the RHS to see this and rationalise the denominators on the LHS.
After the rationalisation of the denominators on the LHS (which is very quick) you obtain $(3+\sqrt{5})^2-6,$ and hence the $8+6\sqrt{5}.$
| {
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$\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$ converges? Determine whether this series converges or diverges: $$\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$
Thought about using the limit theorem or by comparison but am so stuck. any pointers would be appreciated guys
| We have $$e^{1} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots$$ and $$e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots $$
Subtracting these two we get $$e - e^{-1} = 2 \cdot \Bigl( 1 + \frac{1}{3!} + \frac{1}{5!} + \cdots \Bigr)$$ Therefore the series converges to $$\frac{e-e^{-1}}{2} = \sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$
| {
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
| I saw this proof in an extract of the College Mathematics Journal.
Consider the Integeral : $I = \int_0^{\pi/2}\ln(2\cos x)dx$
From $2\cos(x) = e^{ix} + e^{-ix}$ , we have:
$$\int_0^{\pi/2}\ln\left(e^{ix} + e^{-ix}\right)dx = \int_0^{\pi/2}\ln\left(e^{ix}(1 + e^{-2ix})\right)dx=\int_0^{\pi/2}ixdx + \int_0^{\pi/2}\ln(1 + e^{-2ix})dx$$
The Taylor series expansion of $\ln(1+x)=x -\frac{x^2}{2} +\frac{x^3}{3}-\cdots$
Thus , $\ln(1+e^{-2ix}) = e^{-2ix}- \frac{e^{-4ix}}{2} + \frac{e^{-6ix}}{3} - \cdots $, then for $I$ :
$$I = \frac{i\pi^2}{8}+\left[-\frac{e^{-2ix}}{2i}+\frac{e^{-4ix}}{2\cdot 4i}-\frac{e^{-6ix}}{3\cdot 6i}-\cdots\right]_0^\frac{\pi}{2}$$
$$I = \frac{i\pi^2}{8}-\frac{1}{2i}\left[\frac{e^{-2ix}}{1^2}-\frac{e^{-4ix}}{2^2}+\frac{e^{-6ix}}{3^2}-\cdots\right]_0^\frac{\pi}{2}$$
By evaluating we get something like this..
$$I = \frac{i\pi^2}{8}-\frac{1}{2i}\left[\frac{-2}{1^2}-\frac{0}{2^2}+\frac{-2}{3^2}-\cdots\right]_0^\frac{\pi}{2}$$
Hence
$$\int_0^{\pi/2}\ln(2\cos x)dx=\frac{i\pi^2}{8}-i\sum_{k=0}^\infty \frac{1}{(2k+1)^2}$$
So now we have a real integral equal to an imaginary number, thus the value of the integral should be zero.
Thus, $\sum_{k=0}^\infty \frac{1}{(2k+1)^2}=\frac{\pi^2}{8}$
But let $\sum_{k=0}^\infty \frac{1}{k^2}=E$ .We get $\sum_{k=0}^\infty \frac{1}{(2k+1)^2}=\frac{3}{4} E$
And as a result $$\sum_{k=0}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$$
| {
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Can I find the limit of $\sum_\limits{i=1}^{n^2}\frac{1}{\sqrt{n+i}}$ with the squeeze theorem? Problem: Calculate limit of $\frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + \frac{1}{\sqrt{n+3}} + \cdots + \frac{1}{\sqrt{n+n^2}}$ as $n$ approaches infinity.
Solution: Denote the above some as $X$, then we can bound it:
$$ \infty\longleftarrow\frac{1}{\sqrt{n+n^2}} \lt X \lt \frac{n^2}{\sqrt{n+1}} \lt \frac{n^2}{\sqrt{n}} = \sqrt{\frac{n^4}{n}}\longrightarrow \infty.$$
So, from the Squeeze Principle, $\lim X = \infty$. Am I doing the right thing?
| Yes, you are doing the right thing, assuming you meant $n^2 / \sqrt{n+n^2} < X$ rather than $1 / \sqrt{n+n^2} < X$.
| {
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Determine the matrix relative to a given basis Question: (a) Let $f: V \rightarrow W$ with $ V,W \simeq \mathbb{R}^{3}$ given by: $$f(x_1, x_2, x_3) = (x_1 - x_3, 2x_1 -5x_2 -x_3, x_2 + x_3).$$
Determine the matrix of $f$ relative to the basis $\{(0,2,1),(-1,1,1),(2,-1,1)\}$ of $V$ and $\{(-1,-1,0),(1,-1,2),(0,2,0)\}$ of $W$.
(b) Let $n \in \mathbb{N}$ and $U_n$ the vector space of real polynomials of degree $\leq n$. The linear map $f: U_n \rightarrow U_n$ is given by $f(p) = p'$. Determine the matrix of $f$ relative to the basis $\{1,t,t^{2},...,t^{n}\}$ of $U_n$.
My attempt so far:
(a): First relative to the bases of $W$ I found the coordinates of an arbitrary vector: $\left( \begin{array}{r} a \\ b \\ c \end{array} \right) = x \left( \begin{array}{r} -1 \\ -1 \\ 0 \end{array} \right) + y \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) + z \left( \begin{array}{c} 0 \\ 2 \\ 0 \end{array} \right)$
$\begin{array}{l} a = -x + y \\ b = - x - y + 2z \\ c = 2y \end{array}$ or $\begin{array}{l} x = -a + \frac{1}{2}c \\ z = -\frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c \\ y = \frac{1}{2}c \end{array}$
At this point I believe I have the linear combinations of the given basis in $W$ for an arbitrary vector, so next I take the vectors from $V$ and send them to $W$ using the given function:
$\begin{array}{l} f(v_1) = f(0,2,1) = (-1,-11,3) = (1 + \frac{3}{2})w_1 + \frac{3}{2}w_2 + (\frac{1}{2} - \frac{11}{2} + \frac{3}{2})w_3 \\ f(v_2) = f(-1,1,1) = (-2,-8,2) = (2+1)w_1 + w_2 + (1 - 4 +1)w_3 \\ f(v_3) = f(2,-1,1) = (1,8,0) = w_1 + (-\frac{1}{2} + 4)w_3 \end{array}$
or $\left( \begin{array}{rrc} \frac{5}{2} & 3 & 1 \\ \frac{3}{2} & 1 & 0 \\ -\frac{7}{2} & -2 & \frac{7}{2}\end{array} \right)$
Was I taking the correct steps? I didn't really do anything differently based on the fact that $V,W$ were isometric... Is there a particular significance or interpretation for the resulting matrix?
(b): Not really sure here...
$f(p) = p'$
would it make sense to write something like:
$f(1,t,t^{2},\dots, t^{n}) = (0,1,2t, \dots, nt^{n-1})$?
and if a basis for $(1,t,t^{2},\dots, t^{n})$ would be $A = \left( \begin{array}{ccccc} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\0 & \cdots & \cdots & 0 & 1 \end{array} \right)$
could i write:
$A' = \left( \begin{array}{ccccc} 0 & 0 & 0 & \cdots & 0 \\ 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\0 & 0 & 0 & 1 & 0 \end{array} \right)$?
| Yes, your solution to (a) is correct assuming your computations for how to express the elements is correct. Though you don't need to find how to write an arbitrary vector in terms of the basis of $W$, you only need to find out how to write $f(0,2,1)$, $f(-1,1,1)$, and $f(2,-1,1)$ (the images of the vectors in the given basis for $V$) in terms of the given basis of $W$.
No, you never need to use the fact that $V$ and $W$ are isomorphic (we don't consider a metric here, so "isometric" is not appropriate here).
Your answer for (b), on the other hand, is not well done (or correct).
The basis for your vector space is $\{1,t,\ldots,t^n\}$. To find the matrix representation of $f$ relative to this basis, you need to find the image of each basis vector in the domain, and express it in terms of the basis vectors of the range. But remember: the vectors in $U_n$ are not tuples, they are polynomials. So $(1,t,\ldots,t^n)$ is not an element of $U_n$.
So, you would have to find each of $f(1)$, $f(t)$, $f(t^2),\ldots, f(t^n)$, and then express them in terms of the basis of the range, which happens to again be $\{1,t,t^2,\ldots,t^n\}$. So for instance, $f(t^2) = 2t$, so
$$f(t^2) = 0\cdot 1 + 2\cdot t + 0\cdot t^2 + \cdots + 0\cdot t^n.$$
That means that the third column of the matrix (the one corresponding to the third basis vector, which is $t^2$) will be the transpose of $(0,2,0,\ldots,0)$, that is
$$\left(\begin{array}{c} 0\\2\\0\\ \vdots \\ 0\end{array}\right).$$
Etc.
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Find examples of two series $\sum a_n$ and $\sum b_n$ both of which diverge but for which $\sum \min(a_n, b_n)$ converges Find examples of two series $\sum a_n$ and $\sum b_n$ both of which diverge but for which $\sum \min(a_n, b_n)$ converges. To make it more challenging, produce examples where $a_n$ and $b_n$ are positive and decreasing.
Edit: This problem is taken verbatim from Exercise 2.7.11 on page 68 of Abbott's Understanding Analysis.
| This answer gives a specific example using the ideas proposed in the previous answer:
Let $\sum a_n=\overline{\frac{1}{1^2}}+\frac{1}{2^2}+\overline{\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{3^2}}+\frac{1}{6^2}+\frac{1}{7^2}+\cdots+\frac{1}{14^2}+\overline{\frac{1}{15^2}+\frac{1}{15^2}+\cdots+\frac{1}{15^2}}+\cdots$
and $\sum b_n=\frac{1}{1^2}+\underline{\frac{1}{2^2}}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\underline{\frac{1}{6^2}+\frac{1}{6^2}+\cdots+\frac{1}{6^2}}+\frac{1}{15^2}+\frac{1}{16^2}+\cdots+\frac{1}{71^2}+\cdots$.
Notice that each group of repeated terms has a sum $S\ge\frac{1}{4}$, and that $\sum c_n=\sum\frac{1}{n^2}$.
| {
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Analytic Geometry | Two Planes and a Angle | Two Solutions This is me again, I have another problem which I haven't been able to solve, the
legend goes like this:
Find the equation of the plane that contains the points $P_1(1,0,-1)$, $P_2(2,0,2)$ and forms a 600 ($\theta$) angle with the other plane $2x-2y+z+6=0$ (Two Solutions).
Here's what I've done:
Since we already know two points, P1$(1, 0, -1)$ and P2$(2,0,2)$ they both satisfy a planes equation:
$$Ax+By+Cz+D=0$$
So, we substitute $x,y,z$ respectively for each point, giving use these two equations
*
*$A(1)+B(0)+C(-1)+D=0$
*$A(2)+B(0)+C(2)+D=0$
Simplifying we get:
*
* $A-C+D=0$
* $2A+2C+D=0$
Now, according to the following theorem the angle that is formed by the perpendiculars of to planes is the following:
$$\cos{\theta} = \pm \frac{AA'+BB'+CC'}{\sqrt{A^2+B^2+C^2} \sqrt{(A')^2+(B')^2+(C')^2}}$$ Where $[A,B,C]$ Are the director numbers of the normal in the first plane and $[A',B',C']$ Are the director numbers of the normal in the second plane.
Now, given those and taking the known angle of 60o, we substitute the director numbers of $2x-2y+z+6-0:[A'=2,B'=-2,C'=1]$ and the angle in the last formula, giving us:
$$\cos{60^\circ} = \pm \frac{A(2)-B(-2)+C(1)}{\sqrt{A^2+B^2+C^2} \sqrt{(2)^2+(-2)^2+(1)^2}}$$
Simplifying, we get(positive because $\theta$ is acute):
$$\frac{1}{2}=\frac{2A-2B+C}{3 \sqrt{A^2+B^2+C^2}}$$
Another simplification finally for:
$$3(A^2+B^2+C^2)=(4A-4B+2C)^2$$
Now, we have 3 equations:
*
* $A-C+D=0$
* $2A+2C+D=0$
*$3(A^2+B^2+C^2)=(4A-4B+2C)^2$
If we take 1) and 2) and write them as $D(A,C)$ we get:
*
*$D=C-A$
*$D=-2A-2C$
From which we can obtain A in terms of C
$$-2A-2C=C-A$$
$$A=-3C$$
Now we substitute back in 1):
$$(-3C)-C+D=0$$
$$C=\frac{1}{4}D$$
Since I want every director number in terms of D we substitute again but in A since A was in function of C.
$$A=-3C$$
$$A=-3(\frac{1}{4}D)$$
$$A=\frac{-3}{4}D$$
Now, we have found $A(D)$, $C(D)$ we have yet to find $B(D)$ I'll explain why I think and want everything in terms of D, if we take the general form of a planes equation $Ax+By+Cz+D=0$ and substitute $A$ and $C$ with our known values we get:
$$(\frac{-3}{4}D)x+By+(\frac{1}{4}D)z+D=0$$
if we divide by $D \neq 0$ and then multiply by 4 to eliminate fractions:
$$\frac{-3}{4}x+\frac{B}{D}y+\frac{1}{4}z+1=0$$
$$-3x+\frac{4B}{D}y+z+4=0$$
From here it's clear that if we find B in terms of D and B is linear we have the planes equation, so, in 3) $3(A^2+B^2+C^2)=(4A-4B+2C)^2$ we substitute A and C:
$$3((\frac{-3}{4}D)^2+B^2+(\frac{1}{4}D)^2)=(4(\frac{-3}{4})-4B+2(\frac{1}{4}D))^2$$
$$\vdots$$
$$104B^2+80DB+35D^2=0$$
Here is were I get stuck, I get imaginary parts, since I have to use the quadratic formula, so any help is really appreciated, the problem states that there are two solutions, but this is what I have thought of if anyone can supply any solution I'll be very grateful.
Regards.. Tristian
| To flesh out Robin's answer a little: The thing that you really want to find is the unit normal vector $n=(A,B,C)^t$ to the sought plane. Once you have that, it's easy to find the equation of the plane. This vector has to satisfy three conditions:
*
*Length one: $A^2+B^2+C^2=1$.
*Orthogonal to the vector $\overrightarrow{P_1 P_2} = (1,0,3)^t$.
*60 degrees to the normal vector of the other plane. If we normalize that vector too, this condition becomes $\frac13 (2,-2,1)^t \cdot (A,B,C) = \cos 60^{\circ}$.
Conditions 2 and 3 give a linear system with two equations and three unknowns $A$, $B$, $C$; solve it and you will get a family of solutions depending on one parameter $t$. Condition 1 then gives a quadratic eqation for determining the two possible values of $t$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit, solution in unusual way I have a problem with solution of this limit:
$$\lim_{x\to 0}{\frac{\tan{x}-x}{x^2}}$$
Of course, it's a very easy to solve, using (twice) L'Hôpital's rule, but I need to find out, how to do this without this rule.
I stuck in this point:
$$\lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}$$
Everything I need to know is how to eliminate $\frac{\sin^2{x}}{x^2}$, because - as my tutor said - I can't simply substitute $1$ for this expression.
Thanks for help.
PS: It's not a homework. My tutor showed this problem as a puzzle and said, that it would be a good exercise to solve this without L'Hôpital's rule.
EDIT: Here is a way I got to the point $\lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}$:
\begin{align*}
\lim_{x->0}{\frac{\tan{x}-x}{x^2}} &= \lim_{x->0}{\frac{(\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}-x)\cdot (\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}+x)}{x^2(\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}+x)}}\\
&= \lim_{x->0}{\frac{\frac{\sin^2{x}}{x^2\cos^2{x}}-1}{x(\frac{\sin{x}}{x}\frac{1}{\cos{x}}+1)}}= \lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}
\end{align*}
| Here is an attempt at a geometric proof.
(Figure thanks to J.M)
Consider $\triangle ABC$ such that $\angle{BCA} = x$. Let $BC=1$ and so $AB = \tan x$.
Let $BE$ be the arc of radius 1 and angle $x$ drawn with $C$ as the center (note that $E$ is on AC, between $A$ and $D$ and is kind of hidden in the brown region). Note that $CE = 1$.
Now the area of the gray region is $\dfrac{\tan x}{2} - \dfrac{x}{2}$ (area of $\triangle ABC$ - area of the sector $CBE$).
Let $D$ be the perpendicular on the hypotenuse $AC$ from $B$. It can be seen that $CD = \cos x$ and thus distance from $D$ to $C$ is less than distance from $E$ to $C$ (which is $1$).
Thus the area of the gray region is less than the area of $\triangle BAD$ (gray + brown).
Now $AD = \dfrac{\sin^2 x}{\cos x}$ and thus we have that
$$ 0 < \dfrac{\tan x - x}{2} \lt \dfrac{\sin^3 x}{2\cos x}$$
And so
$$ 0 < \dfrac{\tan x - x}{2x^2} \lt \dfrac{\sin^3 x}{2x^2 \cos x}$$
Since we know that $\lim_{x \to 0+} \dfrac{\sin x}{x} = 1$, and that $\dfrac{\tan x - x}{x^2}$ is an odd function, that the limit is $0$, follows.
Previous answer, which was a feeble attempt at being pedantic:
For a way to find the limit without using more advanced concepts like McLaurin series etc...
Consider
$$\dfrac{\tan(2x) - 2x}{(2x)^2} = \dfrac{ \frac{2\tan x}{1-\tan^2 x} - 2x}{4x^2}$$
$$ = \dfrac{(2\tan x - 2x) + 2x \tan^2 x}{4x^2(1 - \tan^2 x)} = (\dfrac{\tan x - x}{2x^2} + \dfrac{x\tan^2 x}{2x^2}) \dfrac{1}{1-\tan^2 x}$$
Therefore, taking limits as $\displaystyle x \to 0$
$$ L = (\dfrac{L}{2} + 0)\dfrac{1}{1-0}$$
Thus
$$L = 0$$
There is one problem with the above, though. Can you tell what that is?
(Or rather more simply, replace $\displaystyle x$ with $\displaystyle -x$)
| {
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} |
How can I find $ \int{(1 + \cos x)^3\mathrm dx} $? My question is ; How can I solve the following integral question?
$$\int{(1 + \cos x)^3\mathrm dx}$$
Thanks in advance,
| Expand $(1+\cos(x))^3$. And rewrite $\cos^n(x)$ in terms of $\cos(nx)$ and other similar terms.
$(1+\cos(x))^3 = 1 + 3 \cos(x) + 3 \cos^2(x) + \cos^3(x) = \frac{\cos(3x)+6 \cos(2x)+15 \cos(x)+10}{4}$.
Now, integrate term by term to get the desired answer.
$\int \cos(nx) = \frac{\sin(nx)}{n}$.
Hence, $\int (1+\cos(x))^3 = \frac{\frac{\sin(3x)}{3}+6\frac{\sin(2x)}{2}+15 \sin(x) + 10x}{4} + c = \frac{\sin(3x)}{12} + \frac{3\sin(2x)}{4} + \frac{15 \sin(x)}{4} + \frac{5x}{2} + c$.
| {
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Verifying Carmichael numbers I'm trying to understand a solution I was given in a tutorial regarding a problem with Carmichael numbers and I was wondering if you guys can help clarify things:
A composite number $m$ is called a Carmichael number if the congruence $a^{m-1} \equiv 1 \pmod{m}$ is true for every number a with $\gcd(a,m) = 1$.
Verify that $m = 561 = 3 \times 11 \times 17$ is a Carmichael number.
Solution given:
Apply Fermat's Little Theorem to each prime divisor of $m$:
\begin{align*}
a^2 &\equiv 1 \pmod{3}\\
a^{10} &\equiv 1 \pmod{11}\\
a^{16} &\equiv 1 \pmod{17}
\end{align*}
This somehow then implies that $a^{80} \equiv 1 \pmod{561}$ then accordingly $a^{560} \equiv 1 \pmod{561}$.
I am lost as to how the 3 congruences imply $a^{80} \equiv 1 \pmod{561}$ ($80 = \mathrm{LCM}(2,10,16)$).
Can somebody clarify this for me?
Thanks!
| Note that $80 = \mathrm{lcm}(2,10,16)$. So you can write $a^{80} = (a^2)^{40} = (a^{10})^{8} = (a^{16})^5$. So,
\begin{align*}
a^{80}= (a^2)^{40} &\equiv 1^{40} = 1\pmod{3},\\
a^{80}= (a^{10})^{8} &\equiv 1^8 = 1 \pmod{11},\\
a^{80}= (a^{16})^5 &\equiv 1^5 = 1\pmod{17}.
\end{align*}
By the Chinese Remainder Theorem, the system of congruences
\begin{align*}
x&\equiv 1\pmod{3}\\
x&\equiv 1\pmod{11}\\
x&\equiv 1\pmod{17}
\end{align*}
has a unique solution modulo $3\times 11\times 17 = 561$. But both $x=1$ and $x=a^{80}$ are solutions. Since the solution is unique modulo $561$, then
the two solutions we found must be congruent. That is,
$$a^{80}\equiv 1\pmod{561}.$$
(Added. Or, more simply, as Andres points out, since $3$, $11$, and $17$ each divide $a^{80}-1$, and are pairwise relatively prime, then their product divides $a^{80}-1$).
Once you have that $a^{80}\equiv 1\pmod{561}$, then any power of $a^{80}$ is also congruent to $1$ modulo $561$. In particular,
$$a^{560} = (a^{80})^{7} \equiv 1^7 = 1 \pmod{561}$$
as desired.
| {
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Is it possible to simplify $\frac{2^{2x-1} - 2^{x-1}}{2^{2x-1}}$? Is it possible to simplify this expression?
$\frac{2^{2x-1} - 2^{x-1}}{2^{2x-1}}$
| $$\begin{align}\frac{2^{2x-1} - 2^{x-1}}{2^{2x-1}}
&=\frac{2^{2x-1}}{2^{2x-1}}-\frac{2^{x-1}}{2^{2x-1}}
\\
&=1-2^{(x-1)-(2x-1)}
\\
&=1-2^{-x}
\end{align}$$
| {
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Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$. For example, see Robin Chapman's list or the answers to the question "Different methods to compute $\sum_{n=1}^{\infty} \frac{1}{n^2}$?"
Are there any nice ways to prove that $$\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}?$$
I already know some proofs that give all values of $\zeta(n)$ for positive even integers $n$ (like #7 on Robin Chapman's list or Qiaochu Yuan's answer in the linked question). I'm not so much interested in those kinds of proofs as I am those that are specifically for $\zeta(4)$.
I would be particularly interested in a proof that isn't an adaption of one that $\zeta(2) = \pi^2/6$.
| If you are specially interested only in $\zeta(4)$, the following proof would work but this is an adaptation Euler's idea. The idea is just to mimic Euler's proof for the Basel problem. Euler looks at the function whose zeros are at $\pm \pi, \pm 2 \pi, \pm 3 \pi, \ldots$
To evaluate $\zeta(4)$, we can mimic Euler's idea and look at roots at $\pm \pi, \pm i \pi,\pm 2 \pi, \pm 2 i \pi,\pm 3 \pi, \pm 3 i \pi$.
Let $$p(z) = \left(1 - \left(\frac{z}{i \pi}\right)^4 \right) \times \left(1 - \left(\frac{z}{2 i \pi}\right)^4 \right) \times \left(1 - \left(\frac{z}{3 i \pi}\right)^4 \right) \times \cdots$$
It is not hard to guess that $p(z)$ is same as $$\frac{i \sin(z) \times \sin \left( \frac{z}{i} \right)}{z^2} = \left(1-\frac{z^2}{3!} + \frac{z^4}{5!} -\cdots \right) \times \left(1+\frac{z^2}{3!} + \frac{z^4}{5!} + \cdots \right)$$
Compare the coefficient of $z^4$ to get $$\zeta(4) = \frac{\pi^4}{90}$$
This proof could be extended for any even number to give that $$\zeta(2n) = (-1)^{n+1} \frac{B_{2n} 2^{2n}}{2(2n)!} \pi^{2n} $$
As expected for odd numbers, this doesn't work. For instance for $3$, if you try to work out by looking at $$p(z) = \left(1 - \left(\frac{z}{\omega \pi}\right)^3 \right) \times \left(1 - \left(\frac{z}{2 \omega \pi}\right)^3 \right) \times \left(1 - \left(\frac{z}{3 \omega \pi}\right)^3 \right) \times \cdots$$
where $\omega^3 = 1$ there is an asymmetry since
$$\sin(z) \sin \left( \frac{z}{\omega}\right) \sin \left( \frac{z}{\omega^2}\right)$$ extends on both sides and the non-zero roots are at $$\pm \pi,\pm \omega \pi,\pm \omega^2 \pi,\pm 2 \pi,\pm 2 \omega \pi,\pm 2 \omega^2 \pi,\pm 3 \pi,\pm 3 \omega \pi,\pm 3 \omega^2 \pi,\ldots$$ and hence the $\zeta(3)$ terms nicely hides by canceling out and the resulting expression only gives $\zeta(6)$.
| {
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How does law of quadratic reciprocity work? From the book,
Suppose $p \equiv 1 \pmod{4}$, then by law of quadratic reciprocity, we have:
$$\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right) $$
Next, if $p \equiv 2 \pmod{3}$, then $p \equiv 5 \pmod{12}$
Hence, $$\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right) = -1$$
How do they get those Legendre fraction equal to $-1$?
From my understanding:
$$\left(\frac{q}{p}\right) \cdot \left(\frac{p}{q}\right) = (-1)^{\frac{p-1}{2}\cdot\frac{q-1}{2}}$$
So $q = 3 \implies \frac{q-1}{2} = \frac{3-1}{2} = 1$
For $p$, I take $p \equiv 5 \pmod{12} \implies p = 5 + 12k$, for some integers k.
Hence, $\frac{p-1}{2} = \frac{12k + 5 - 1}{2} = \frac{12k + 4}{2} = 6k + 2.$
And this $6k + 2$ is even :( ! How does $(-1)^{even} = -1$?
Any idea? I think I made some logic mistakes somewhere, but I couldn't find where.
Update
The problem was from Elementary Number Theory and Its Application - Kenneth H.Rosen 5th Edition.
Problem
Using the law of quadratic reciprocity, show that if $p$ is an odd prime, then
$$\left(\frac{3}{p}\right) = 1 \text{ if } p \equiv \pm 1 \pmod{12}$$
$$\left(\frac{3}{p}\right) = -1 \text{ if } p \equiv \pm 5 \pmod{12}$$
Solution
Thanks,
Now I'm even more confused :(!
Consider two cases:
Case 1
$$p \equiv 1 \pmod{4} \text{ and } p \equiv 1 \pmod{3}$$
Then,
$$p \equiv 1 \pmod{12} \implies p = 12k + 1$$
Hence,
$$\frac{p - 1}{2} \cdot \frac{3 - 1}{2} = \frac{12k}{2} = 6k = \text{ even }$$
Which implies
$$\left(\frac{3}{p}\right) \cdot \left(\frac{p}{3}\right) = 1$$
So both must be $1$ or $-1$.
Case 2
$$p \equiv 1 \pmod{4} \text{ and } p \equiv 2 \pmod{3}$$
Then,
$$p \equiv 5 \pmod{12} \implies p = 12k + 5$$
Hence,
$$\frac{p - 1}{2} \cdot \frac{3 - 1}{2} = \frac{12k + 4}{2} = 6k + 2 = 2(3k + 1) = \text{ even }$$
Which implies
$$\left(\frac{3}{p}\right) \cdot \left(\frac{p}{3}\right) = 1$$
So both must be $1$ or $-1$.
I don't see how these arguments can be deduced to the solution. Any suggestion?
| Personally I find that a huge step in understanding things is knowing motivation behind the theorem. Douglas and yunone have both probably explained the meaning behind those equations pretty well but take a look at this Wikipedia article for more historical context: http://en.wikipedia.org/wiki/Quadratic_reciprocity#History_and_alternative_statements
| {
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How does this Taylor Polynomial work? The Taylor Polynomial is defined as following:
$$P_n(x) = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \cdots + (-1)^n \dfrac{1.3.5 \cdots (2n - 3)}{2.4.6 \cdots 2n}x^n$$
If $n = 4$, then the last term in the numerator expansion would be $2 \cdot 4 - 3 = 5$.
So we will have 5 terms totally, running from $x^0$ to $x^4$.
But then how was the numerator generated?
First term = ?
Second term = 1
Third term = 1.3
Fourth term = 1.3.5
Fifth term = ?
I don't understand how they have $1$ for the first term as well as the fifth term. It did not make any sense to me. Any idea?
Edit
The basic function for this polynomial is $f(x) = \sqrt{x + 1}$ at $x = 0$.
Thanks,
| It looks like there is an offset of 1 somewhere. $-\frac{1}{8}x^2$ looks like $(-1)^1\frac{1}{2\cdot 4}x^2$. Then the next term would be $(-1)^2\frac{1\cdot 3}{2 \cdot 4 \cdot 6}x^3=+\frac{3}{48}x^3$ followed by $(-1)^3\frac{1\cdot 3\cdot 5}{2 \cdot 4 \cdot 6\cdot 8}x^4=-\frac{15}{384}x^4$. What basic function is this the Taylor Polynomial for? Also please refer to the terms by the power of $x$ in them-is the first term the constant $1$ or the $\frac{1}{2}x$ term?
Added: Wolfram Alpha agrees with these signsfor $\sqrt{1+x}$, so the exponent of $-1$ must be $n+1$ or $n-1$.
| {
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Find a number $b$ such that $a\cdot b\equiv 1\mod m$ Given two integers $a$ and $m$, such that $a\mathop\bot m,$ how can I find an integer $b$ such that $a\cdot b\equiv 1\mod m?$
| I will show by example, $a = 7$ and $m = 17$.
I want to solve $7b + 17y = 1$ because reducing that equation mod $m$ gives $7 \cdot b \equiv 1 \pmod {17}$.
To solve $7b + 17y = 1$ I will reduce it to an easier equation $7b + (7 + 10)y = 7(b+y) + 10y = 7(b+2y) + 3y = 1$, so now we are in a similar situation.
We want to solve $7x + 3y = 1$ and we can do the same kind of reduction, $7x+3y=4x+3(x+y)=x+3(2x+y)=1$ so again we are in the same situation.
Now we want to solve $x + 3z = 1$, that is easy, set $x = 4$ and $z = -1$.. but now we have to go backwards through this to get $b$. $-1 = z = 2x+y$ and $x=4$ so $y = -9$, further $4 = x = b+2y = b-18$ so $b = 22$.
Since this is a very long calculation is it useful to check that this is the right answer: $7 \cdot 22 \equiv 7 \cdot 9 \equiv 1 \pmod {17}$.
Once this idea is very clear it is possible to optimize it so that both forward and backward stages are done simultaneously.
| {
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$\int \frac{dx}{(x^4 + 1)^2}$ What would be a relatively simple method for computing the indefinite integral below?
$\displaystyle \int \frac{dx}{(x^4+1)^2}$
Furthermore, how would one evaluate the following, possibly by detouring the computation of the indefinite integral?
$\displaystyle\int\nolimits_{-\infty}^\infty \frac{dx}{(x^4+1)^2}$
Note: The resolved value of the definite integral (according to wolfram alpha) is $\displaystyle \frac{3\pi}{4\sqrt{2}}$
| The definite integral can be done using residues. Note that the zeros of $(z^4+1)^2$ in the upper half plane are at $\frac{\pm 1 + i}{\sqrt{2}}$, both with multiplicity 2. The residues are $\frac{3 \sqrt{2}}{32}(1-i)$ at $\frac{-1+i}{\sqrt{2}}$ and $\frac{3 \sqrt{2}}{32}(-1-i)$ at $\frac{1+i}{\sqrt{2}}$ (note that the residue of $\frac{1}{((x-r) f(x))^2}$ at $x=r$, where $f$ is analytic at $r$ with $f(r) \ne 0$, is $\frac{-2 f'(r)}{f(r)^3}$). So the integral is $2 \pi i \frac{3 \sqrt{2}}{32} (1-i -1 - i) = \frac{3 \pi \sqrt{2}}{8}$.
| {
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Solving $\sqrt{x+5} = x - 1$ I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:
$$\sqrt{x+5} = x - 1$$
So I used this logic:
$$
\begin{align}
\sqrt{x+5} &= x - 1 \\
x + 5 &= (x-1)^2 \\
x + 5 &= (x-1)(x-1) \\
x + 5 &= x^2 - 2x + 1 \\
0 &= x^2 - 3x - 4 \\
0 &= (x-4)(x+1) \\
\end{align}
$$
Therefore, $x = -1$ and $x = 4$ satisfy the equation $0 = (x-4)(x+1)$.
But then I tried to plug them in the original problem $\sqrt{x+5}=x-1$:
$$
\begin{align}
\sqrt{4 + 5} &= 4 - 1 \\
\sqrt{9} &= 3 \\
3 &= 3
\end{align}
$$
So using 4 works as expected, but when using $-1$:
$$
\begin{align}
\sqrt{-1 + 5} &= -1 - 1 \\
\sqrt{4} &= -2 \\
2 &\ne -2
\end{align}
$$
At what stage am I going wrong?
And according the WolframAlpha, the solution is $x = 4$.
| Work backwards. Starting with $x=-1$ and $x=4$, when you work backwards through your steps and rewrite everything to get to the very beginning, you end by taking a square-root. When you take that square-root, you have two possible solutions: $+\sqrt{x+5}$ and $-\sqrt{x+5}\quad$. $x=-1$ solves one, $x=4$ solves the other.
| {
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Can one show that $\sum_{n=1}^N\frac{1}{n} -\log N - \gamma \leqslant \frac{1}{2N}$ without using the Euler-Maclaurin formula? I would like to prove that
$$
\sum_{n=1}^N\frac{1}{n} -\log N - \gamma \leqslant \frac{1}{2N}
$$
without using the Euler-Maclaurin summation formula. The motivation for this is that I have come very close to doing so (see the answer provided below) but annoyingly have not actually proved the above.
Some may ask why I don't just use the formula. I'm writing a set of analytic number theory notes for my own use and it seems an unwieldy result to introduce and prove, given that the above inequality is all I need, and given that I have gotten so close without using Euler-Maclaurin!
| One can check that $S(N):=\sum_{n=1}^N\frac{1}{n} -\log N - \gamma = \int_N^\infty \frac{x-[x]}{x^2} \: dx$, where $[x]$ is the integer part of $x$. Moreover
$$
\int_n^{n+1} \frac{x-[x]}{x^2}\: dx = \log\Big(\frac{n+1}{n}\Big) - \frac{1}{n+1} < \frac{1}{n} - \frac{1}{n+1} -\frac{1}{2n^2} +\frac{1}{3n^3}, \qquad (1)
$$
by the Taylor series for $\log(1+x)$. But we have that
$$
n(n+1)(3n-1) = 3n^3 + 2n^2 -n > 3n^3
$$
so
$$
\frac{1}{n(n+1)} < \frac{3n-1}{3n^3} = \frac{1}{n^2} - \frac{1}{3n^3}.
$$
Therefore, from equation $(1)$ we find
$$
S(N) < \sum_{n=N}^\infty \frac{1}{n(n+1)} -\frac{1}{2n^2} +\frac{1}{3n^3} < \sum_{n=N}^\infty \frac{1}{n^2} - \frac{1}{3n^3} -\frac{1}{2n^2} +\frac{1}{3n^3},
$$
and so finally,
$$
S(N) < \frac{1}{2}\sum_{n=N}^\infty \frac{1}{n^2} < \frac{1}{2(N-1)},
$$
for all $N \in \mathbb{N}$, by a standard approximation for $\sum \frac{1}{n^2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
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} |
Linear Algebra: different determinant answers I'm having a problem verifying my answer to this question:
Solve for x:
$$\left| \begin{array}{cc}
x+3 & 2 \\
1 & x+2 \end{array} \right| = 0$$
I get:
$(x+3)(x+2)-2=0$
$(x+3)(x+2)=2$
thus:
$x+3=2$ and $x+2=2$
$x=-1$ and $x=0$
The book says that $x=-1$ and $x=-4$ is the correct answer.
I tried doing it a different way by expanding and got totally different answers:
$x^2+5x=-4$
$x(x+5)=-4$
$x=-4$ and $x=-9$
What is going on here?
| The step in your reasoning
$$(x+3)(x+2)=2 $$
$$\Longrightarrow$$
$$x+3=2\text{ and }x+2=2$$
is incorrect - if $x+2=2$, then $x=0$, hence $x+3=0+3=3$, hence $(x+3)(x+2)=3\cdot 2=6$, not $2$. Also the step in your second approach
$$x^2+5x=-4$$
$$\Longrightarrow$$
$$x(x+5)=-2$$
is incorrect. $x^2+5x$ and $x(x+5)$ are the same thing, so it can't equal both $-4$ and $-2$.
Your second approach is on the right track though; you are correct in multiplying $(x+3)$ and $(x+2)$ together, but now bring everything over to one side, leaving $0$ on the other side:
$$x^2+5x+4=0$$
Now use the quadratic formula to get that the solutions are $x=-1$ and $x=-4$.
| {
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"source": "stackexchange",
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} |
Finding power series representation of $ \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt {1 - k^2\sin^2{x}}}\;{dx}$ I want to show that $\displaystyle \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt {1 - k^2\sin^2{x}}}\;{dx} = \frac{\pi}{2}\sum_{n \ge 0}k^{2n}\left({\frac{{1 \cdot 3 \cdots \left( {2n - 1} \right)}}
{{2 \cdot 4 \cdots \cdot 2n}}} \right)$, where $ -1 < k < 1$.
Here is what I did:
$$\displaystyle \begin{aligned}\int_0^{\frac{\pi }{2}} \frac{1}{{\sqrt {1 - k^2\sin^2{x}}}}\;{dx} & = \int_{0}^{\pi/2}\sum_{n \ge 0} \frac{k^{2n}}{2^{2n}}\binom{2n}{n}\sin^{2n}{x}\;{dx} \\& = \sum_{n \ge 0}\int_{0}^{\pi/2} \frac{k^{2n}}{2^{2n}}\binom{2n}{n}\sin^{2n}{x}\;{dx} \\& = \frac{\pi}{2} \sum_{n \ge 0} ~ k^{2n} \bigg(\frac{1}{2^{2n}}\binom{2n}{n}\prod_{1 \le r \le n}\frac{2r-1}{2r} \bigg) \\& = \frac{\pi}{2} \sum_{n \ge 0} ~ k^{2n} \bigg(\prod_{1 \le r \le n}\frac{2r-1}{2r} \cdot \prod_{1 \le r \le n}\frac{2r-1}{2r} \bigg) \\& = \frac{\pi}{2}\sum_{n \ge 0}k^{2n}\bigg(\prod_{1 \le r \le n}\frac{2r-1}{2r}\bigg)^2.\end{aligned}$$
However, there no power on the coefficients in the given series, so they obviously don't match, and I couldn't whatsoever discern a mistake in my calculations. Thanks in advance.
| In the very first step, $(2n$ choose $n)$ should be $(-1/2$ choose $n)$ because the exponent is $1/2$ and you're apparently using a standard Taylor expansion for $(1+X)^M$ for $M=-1/2$ and $X=-k^2\sin^2 x$. You should also add $(-1)^n$.
The combinatorial factor $(-1/2$ choose $n)$ instantly produces the product of odd numbers over the product of even numbers. It is equal to $(-1/2)(-3/2)\dots (-n+1/2) / n!$. Absorbing $(-1)^n$, you get $(1/2)(3/2)\dots (n-1/2)/n!$.
Oh, your description is actually equivalent.
Indeed, you will get another copy of $(1/2)(3/2)\dots (n-1/2)/n!$, together with the $\pi/2$ factor, from the integral $\int_0^{\pi/2} dx\,\sin^2 x$, so your result is correct, and the original formula is only missing the second power of the big ratio and needs to be corrected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/44478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Calculating overall Profit/Loss I was reading a text book and came across the following interesting short-cut:
If two items are sold, each at $X$, one at a gain of $P\%$ and the other at a loss of $P\%$, then overall loss percentage $= P^2/100 \%$.
Can anyone please explain the underlying logic on the basis of which this shortcut works?
Thanks in advance!
| A gain of P% that results in an item cost of $X$ means that the original price, $C_1$, was such that
$$\left(1 + \frac{P}{100}\right)C_1 = X.$$
A loss of P% that results in an item cost of $X$ means that the original price, $C_2$, was such that
$$\left(1 - \frac{P}{100}\right)C_2 = X.$$
In other words, your costs were
$$C_1 + C_2 = \frac {X}{1 + \frac{P}{100}} + \frac{X}{1-\frac{P}{100}} = \frac{100X}{100+P} + \frac{100x}{100-P}.$$
On the other hand, you received $X+X = 2X$. So what is the total profit/loss? It's equal to the amount received minus the amount spent:
$$\begin{align*}
\text{Profit} &= \text{Revenue} - \text{Cost}\\
&= 2X - \frac{100X}{100+P} + \frac{100X}{100-P} \\
&= 2X- \frac{100X(100-P)+100X(100+P)}{(100+P)(100-P)} \\
&= 2X - 100X\left(\frac{100-P+100+P}{(100+P)(100-P)}\right)\\
&= 2x - 100X\left(\frac{200}{(100+P)(100-P)}\right)\\
&= 2x\left( 1 - \frac{10000}{(100+P)(100-P)}\right)\\
&= 2X\left(1 - \frac{10000}{10000 - P^2}\right)\\
&= 2X\left(\frac{10000-P^2 - 10000}{1000-P^2}\right)\\
&= 2X\left(-\frac{P^2}{10000}\right)\\
&= 2X\left( - \frac{(P^2/100)}{100}\right).
\end{align*}$$
So your total loss is $(P^2/100)$%.
| {
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Prove that $6|2n^3+3n^2+n$ My attempt at it: $\displaystyle 2n^3+3n^2+n= n(n+1)(2n+1) = 6\sum_nn^2$
This however reduces to proving the summation result by induction, which I am trying to avoid as it provides little insight.
| You have $2n^3+3n^2+n=n(n+1)(2n+1)$, and $2\mid n(n+1)$. If $3\mid n(n+1)$, then you're done. Otherwise, $n\not\equiv 0\pmod{3}$ and $n\not\equiv -1\pmod{3}$, so $n\equiv 1\pmod{3}$. Then $2n+1\equiv 3\equiv 0\pmod{3}$, so $3\mid 2n+1$ and you get the result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to take the inverse of this signal? I have a little problem computing the inverse of this signal:
$$X(z) = \frac{(z-1)(z+\frac{3}{2})}{(z+\frac{j}{2})(z-\frac{j}{2})(z-\frac{1}{2})}$$
$$X(z^{-1}) = ?$$
I know how to take the inverse, $z=z^{-1}$ and then multiply the brackets and so on...
But my problem is the numbers in the brackets, $1, -\frac{3}{2}, \frac{j}{2}$, are poles and zeros of a Pole-zero plot of the sequence $X(z)$.
And I think I loose that information, don't I?
| if $X(z)$ has a zero/pole at $z=w$ then $X(z^{-1}$ will have a zero at $z=w^{-1}$, so you don't lose any information about the poles and zeros. You end up with:
$$\begin{align}
X(z^{-1})
& = \frac{(z^{-1}-1)(z^{-1}+\tfrac{3}{2})}{(z^{-1}+\tfrac{j}{2})(z^{-1}-\tfrac{j}{2})(z^{-1}-\tfrac{1}{2})} \\
& = \frac{z(1-z)(1+\tfrac{3}{2}z)}{(1+\tfrac{j}{2}z)(1-\tfrac{j}{2}z)(1-\tfrac{1}{2}z)} \\
& = \frac{-\tfrac{3}{2}z(z-1)(z+\tfrac{2}{3})}{\tfrac{j^2}{8}(z + \tfrac{2}{j})(z - \tfrac{2}{j})(z-2)} \\
& = -\frac{12}{j^2} \frac{z(z-1)(z+\tfrac{2}{3})}{(z + \tfrac{2}{j})(z - \tfrac{2}{j})(z-2)} \\
\end{align}$$
Note that $X(z)$ had a zero at infinity, and hence $X(z^{-1})$ has a zero at zero.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Will any value of a free variable satisfy a system of equation? Say I have a reduced row echelon form matrix like this:
$$A=\begin{bmatrix}
1 & 0 & \frac{1}{2} & 0\\
0 & 1 & -\frac{1}{3} & 0\\
0 & 0 & 0 & 1
\end{bmatrix}$$
The number of unknowns is more than the number of known equations. So I can expect an infinite number of solutions.
And $Ax=b$ is like this:
$$
\begin{bmatrix}
1 & 0 & \frac{1}{2} & 0\\
0 & 1 & -\frac{1}{3} & 0\\
0 & 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}
\end{bmatrix}
=
\begin{bmatrix}
b_{1}\\
b_{2}\\
b_{3}
\end{bmatrix}
$$
Then I can say that my $x$ is like this with $x_{3}$ being a free variable in the equation:
$$\begin{bmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}
\end{bmatrix}
=
x_{3}\begin{bmatrix}
\frac{1}{2}\\
\frac{1}{3}\\
1\\
0
\end{bmatrix}+
\begin{bmatrix}
b_{1}\\
b_{2}\\
0\\
b_{3}
\end{bmatrix}
$$
Now, if I let $\; \begin{bmatrix}
b_{1}\\
b_{2}\\
b_{3}
\end{bmatrix}=\begin{bmatrix}
5\\
2\\
7
\end{bmatrix}$, then...
$$\begin{bmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}
\end{bmatrix}
=
x_{3}\begin{bmatrix}
\frac{1}{2}\\
\frac{1}{3}\\
1\\
0
\end{bmatrix}+
\begin{bmatrix}
5\\
2\\
0\\
7
\end{bmatrix}
$$
At this stage, I can say that for any value that I put into the variable $x_{3}$, I would get an answer that is equals to $b$, right?
So assume I just randomly throw a value into $x_{3}=2$, then...
$$\begin{bmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}
\end{bmatrix}
=
\begin{bmatrix}
1\\
\frac{2}{3}\\
2\\
0
\end{bmatrix}+
\begin{bmatrix}
5\\
2\\
0\\
7
\end{bmatrix}=
\begin{bmatrix}
6\\
\frac{8}{3}\\
2\\
7
\end{bmatrix}
$$
From here, I need to tally if the equation really gets back my intended values of $b$, which is $\begin{bmatrix}
b_{1}\\
b_{2}\\
b_{3}
\end{bmatrix}=\begin{bmatrix}
5\\
2\\
7
\end{bmatrix}$. So,
$x_{4} = b_{3} =7$
This is correct.
$b_{2}=x_{2}-\frac{1}{3}x_{3}$
$b_{2}=\frac{8}{3}-\frac{2}{3}=2$
This is also correct.
BUT NOW,
$b_{1}=x_{1}+\frac{1}{2}x_{3}$
$b_{1}=6+1=7$
$b_{1}$ suppose to be 6 but somehow, why am I getting 7 instead? This is weird. Where did I go wrong?
Thanks for any help!
| You simply made a sign error in your third matrix equation, you should have
$x_1 = \frac{-1}{2} x_3 + b_1$ instead of $x_1 = \frac{1}{2} x_3 + b_1$.
That will give you the right answer which is $b_1=5$ by the way.
| {
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Inequality: $(x + y + z)^3 \geq 27 xyz$ Edit: $a,b,c$ and $x,y,z$ are positive, real numbers.
Since $(a-b)^2 \geq 0~$, $a^2 + b^2 - 2ab\geq0~$ and $a^2 + b^2 \geq 2ab~$. Similarly, $a^2 + c^2 \geq 2ac~$ and $b^2 + c^2 \geq 2bc~$.
Adding these inequalities together, $2(a^2 +b^2 + c^2) \geq 2(ab + ac +bc)~$ and accordingly, $a^2 +b^2 + c^2 \geq ab + ac +bc~$
Multiplying both sides by $(a + b + c)$:
$(a^2 +b^2 + c^2)(a+b +c) \geq (ab + ac +bc)(a + b + c)~~$ and simplifying this, $ a^3 + b^3 + c^3 + \Sigma a^2 b \geq 3abc + \Sigma a^2 b $
Therefore, it follows that $a^3 + b^3 + c^3 \geq 3abc~$, and letting $a^3 = x~$, $b^3 = y~$, $c^3 = z~$: $x + y + z\geq 3\sqrt[3]{xyz}$
Cubing both sides, $(x + y + z)^3 \geq 27 xyz~$ which was to be proven.
I was wondering if there are alternative approaches to solve this problem (possibly using higher-level maths), and is my proof entirely correct?
| Look up "arithmetic-geometric mean inequality". Your proof is fine, if you assume the variables $\ge 0$, except that your notation $\Sigma a^2 b$ is nonstandard.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Ways of Turning Sequences into Infinite Series Is writing a sequence as a telescoping sum the only way to turn a sequence into an infinite series? In particular:
Let $s_n = 1 +\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}- \ln(n+1), \ n \geq 1$. Convert this into an infinite series.
So let $a_n = s_n-s_{n-1}$. Then $a_n = \frac{1}{n}-\ln(n+1)+ \ln n$. This is equivalent to $$\sum_{n=1}^{\infty} \left(\frac{1}{n}- \ln \frac{n+1}{n} \right)$$
What does the inside term represent?
| It may be worth noting the following.
$$
s_n = \sum\limits_{k = 1}^n {\frac{1}{k}} - \int_1^{n + 1} {\frac{1}{x}\,dx} = \sum\limits_{k = 1}^n {\frac{1}{k}} - \sum\limits_{k = 1}^n {\int_k^{k + 1} {\frac{1}{x}\,dx} } = \sum\limits_{k = 1}^n {\bigg(\frac{1}{k} - \int_k^{k + 1} {\frac{1}{x}\,dx} \bigg)} .
$$
Hence $s_n = \sum\nolimits_{k = 1}^n {a_k } $ with $a_n \,(= s_n - s_{n-1})$ given by
$$
a_n = \frac{1}{n} - \int_n^{n + 1} {\frac{1}{x}\,dx} = \frac{1}{n} - \ln \frac{{n + 1}}{n}.
$$
Now, from the simple fact that, for any $x > 0$,
$$
0 < x - \ln (1 + x) < x^2
$$
it follows that, for any $n \in \mathbb{N}$,
$$
0 < \frac{1}{n} - \ln \bigg(1 + \frac{1}{n}\bigg) < \frac{1}{{n^2 }},
$$
so
$$
0 < a_n < \frac{1}{{n^2 }}.
$$
Hence
$$
\sum\limits_{n = 1}^\infty {\bigg(\frac{1}{n} - \ln \frac{{n + 1}}{n}\bigg)} = \sum\limits_{n = 1}^\infty {a_n } < \infty .
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
Factoring $X^{16}+X$ over $\mathbb{F}_2$ I just asked wolframalpha to factor $X^{16}+X$ over $\mathbb{F}_2$. The normal factorization is
$$
X(X+1)(X^2-X+1)(X^4-X^3+X^2-X+1)(X^8+X^7-X^5-X^4-X^3+X+1)
$$
and over $GF(2)$ it is
$$
X(X+1)(X^2+X+1)(X^4+X+1)(X^4+X^3+1)(X^4+X^3+X^2+X+1).
$$
Does the second form follow from the first, or is there a different way to factor over $\mathbb{F}_2$? I noticed that simply replacing the $-$ signs with $+$ signs in the first factorization doesn't yield the second one.
| The second form does follow from the first. Note that over $\mathbb F_2$, the pair of polynomials
$$
X^2 - X + 1 \quad \text{ and } \quad X^2 + X + 1
$$
and the other pair
$$
X^4 - X^3 + X^2 - X + 1 \quad \text{ and } \quad X^4 + X^3 + X^2 + X + 1
$$
are the same. The only difference that comes up in $\mathbb F_2$ is that the polynomial of degree $8$ factors.
Hope that helps,
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Number of Ways of Writing $n$ as Sum I have a general question about numbers:
How many ways can we write $n$ as the sum of the numbers $1$,$2$ and $3$?
I know that we start with the following functions: $$\frac{1}{1-z} = 1+z+z^2+ \dots$$ $$\frac{1}{1-z^2} = 1+z^2+ z^4+ z^6 + \dots$$ $$\frac{1}{1-z^3} = 1+ z^3+ z^{6}+ z^{9} + \dots$$
Multiplying these together we get $$\sum C(n) z^n = \frac{1}{(1-z)(1-z^2)(1-z^3)}$$ where $C(n)$ is our answer. In other words, by multiplying out the polynomials and looking at the coefficients, I can verify that I get the answer. But how do I know that this gives me the answer? How would I get an explicit formula for $C(n)$? I think I have to do something with partial fractions. So $$\frac{1}{(1-z)(1-z^2)(1-z^3)} = \frac{A}{1-z} + \frac{B}{1-z^2}+ \frac{C}{1-z^3}$$ and we have to solve for $A$, $B$ and $C$?
| Since I have it worked out, and it hasn't been posted, here is the explicit decomposition of the partial fractions. We have
$$\frac{1}{(1-z)(1-z^{2})(1-z^{3})}$$ $$=\frac{z+2}{9\left(z^{2}+z+1\right)}-\frac{17}{72}\frac{1}{\left(z-1\right)}+\frac{1}{8(z+1)}+\frac{1}{4(z-1)^{2}}-\frac{1}{6(z-1)^{3}}.$$
Personally, I don't know the generating series for $\frac{1}{z^{2}+z+1}$, but I do know it for $\frac{1}{\omega-z}$ and $\frac{1}{\overline{\omega}-z}$, so lets go one step further. We can derive
$$\frac{z+2}{9\left(z^{2}+z+1\right)}=\frac{1}{9}\left(\frac{\omega}{\omega-z}+\frac{\overline{\omega}}{\overline{\omega}-z}\right).$$
Since we know the expansion
$$\frac{1}{\left(a-z\right)^{k+1}}=\frac{1}{a^{k+1}\left(1-\frac{z}{a}\right)^{k+1}}=\frac{1}{a^{k+1}}\sum_{n=0}^{\infty}\tbinom{n+k}{k}\frac{z^{n}}{a^{n}},$$
we can sum all of these to get the explicit formula for $C_{n}$.
Hope that helps,
| {
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The number is a perfect square if and only if $k=n$ Let $k$ and $n$ be positive integers. Show that $$(k+1)^2k^2(n+1)^4-2k(k+1)n(n+1)^2(2kn+k+1)+n^2(k+1)^2$$ is a perfect square if and only if $k=n$.
| For $k=3$ and $n=792$,
$$
(k+1)^2k^2(n+1)^4-2\,k\,(k+1)\,n\,(n+1)^2(2\,k\,n+k+1)+n^2(k+1)^2=309\,396^2.
$$
This is the only perfect square for $1\le k\le2\,000$, $1\le n\le10\,000$ and $k\ne n$.
| {
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"source": "stackexchange",
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Determinant of a special matrix The question is related to the eigenvalue problem. Using MAPLE, I calculated the following determinants:
$$\begin{align}
\begin{vmatrix}
-\lambda & 1 \\
1 & -\lambda \\
\end{vmatrix}&=\lambda^2-1\\
\begin{vmatrix}
-\lambda & 1 & 0 \\
1& -\lambda & 1 \\
0& 1 & -\lambda \\
\end{vmatrix}&=-\lambda^3+2\lambda\\
\begin{vmatrix}
-\lambda & 1 &0 &1 \\
1& -\lambda & 1 &0 \\
0& 1& -\lambda & 1 \\
1& 0& 1& -\lambda \\
\end{vmatrix}&=\lambda^4-4\lambda^2\\
\begin{vmatrix}
-\lambda &1 &0 &1 &0 \\
1& -\lambda &1 &0 &1 \\
0& 1& -\lambda &1 &0 \\
1& 0& 1& -\lambda &1 \\
0& 1& 0& 1& -\lambda \\
\end{vmatrix}&=-\lambda^5+6\lambda^3\\
\begin{vmatrix}
-\lambda &1 &0 &1 &0 &1 \\
1& -\lambda &1 &0 &1 &0 \\
0& 1& -\lambda &1 &0 &1 \\
1& 0& 1& -\lambda &1 &0 \\
0& 1& 0& 1& -\lambda &1 \\
1& 0& 1& 0&1 & -\lambda \\
\end{vmatrix}&=-9\lambda^4+\lambda^6\\
\end{align}
$$
But I have no idea how to calculate the determinants quickly by hand. Here is my question:
What is the determinant in the $n$ by $n$ case?
| Let us denote the matrix in $n$-dimensions by $\Lambda_n$ and the two non vanishing eigenvalues by $a$ and $b$. We have
$a+b = tr(\Lambda_n) = 0$
$a^2+b^2 = tr(\Lambda_n^2) = 2\lceil \frac{n}{2}\rceil\lfloor\frac{n}{2}\rfloor$
which implies that:
$a ^2= b^2 =-ab = \lceil \frac{n}{2}\rceil \lfloor\frac{n}{2}\rfloor$
Thus the two nonvanishing eigenvalues contributtion to the characteristic polynomial:
$\lambda^2-\lceil \frac{n}{2}\rceil \lfloor\frac{n}{2}\rfloor$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/54109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 0
} |
Which integers are the areas of squares with vertices in the 3d integer lattice? For which integers $n$ does there exist a square of area $n$ with vertices in the 3d integer lattice $\mathbb{Z}^3$?
A sufficient condition is that $n$ is a square or the sum of two squares, and I have verified that the condition is also necessary when $n < 10^5$.
Edit: This question was posed by James Tanton on Twitter. I thought it was very interesting, so I took the liberty of posting it here.
Edit 2: I have extended the search to $n < 10^6$ without finding any counterexamples.
| Translate one of the vertices to the origin, then the two adjacent vertices of the square are $(x,y,z)$ and $(u,v,w)$ where $x^2 + y^2 + z^2 = u^2 + v^2 + w^2 = s$ and $xu + yv + zw = 0$, and the area of the square is $s$. Now consider
$$
\begin{align}
(xw-uz)^2 + (yw-vz)^2
&=x^2w^2 - 2xwuz + u^2z^2 + y^2w^2 - 2ywvz + v^2z^2\\
&=(x^2 + y^2)w^2 + (u^2 +v^2)z^2 - 2(xu+yv)wz\\
&=(x^2 + y^2)w^2 + (u^2 +v^2)z^2 + 2(zw)wz\\
&=(x^2 + y^2 + z^2)w^2 + (u^2 +v^2+w^2)z^2\\
&=s(w^2+z^2)
\end{align}
$$
Since a number is a sum of two squares if and only if each prime factor of that number that is equal to $3\pmod{4}$ occurs with even exponent, all prime factors of $(xw-uz)^2 + (yw-vz)^2$ and $w^2+z^2$ equal to $3\pmod{4}$ occur with even exponent, thus each prime factor of $s$ equal to $3\pmod{4}$ must also occur with even exponent. Therefore, $s$ is a sum of two squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/54547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 0
} |
Showing $\tan\frac{2\pi}{13}\tan\frac{5\pi}{13}\tan\frac{6\pi}{13}=\sqrt{65+18\sqrt{13}}$ How can one show :
$$\tan\frac{2\pi}{13}\tan\frac{5\pi}{13}\tan\frac{6\pi}{13}=\sqrt{65+18\sqrt{13}}?$$
| I delete my comments and sum them up as a possible method which can lead to a solution.
Here are my attempts:
$\textbf{Attempt 1}$. Observe that if: $A+B+C = \pi$, then $$ \tan(\pi - A) = \tan(B+C) = \frac{\tan{B}+\tan{C}}{1-\tan{B} \cdot \tan{C}}$$ So you get $$\tan{A} + \tan{B} + \tan{C} = \tan{A} \cdot \tan{B} \cdot \tan{C}$$ So now you want to find the value of $$\tan\frac{2\pi}{13} + \tan\frac{5\pi}{13} + \tan\frac{6 \pi}{13}$$ which $\mathsf{I \ don't \ know}$. You want to see this post:
*
*Proving $\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$
As I say in the post you have to look for some equations for which $\tan$ appears as a root and then perhaps look at sum of the roots.
$\textbf{Attempt 2.}$ Consider the equation $x^{2n}-1=0$. The roots of the equation are, $$1,-1, \cos\frac{\pi}{n}+i\cdot \sin\frac{\pi}{n},\cdots, \cos\frac{(2n-1)\pi}{n}+ i \cdot \sin\frac{(2n-1)\pi}{n}$$ Therefore we can write $$ \small x^{2n}-1 = (x-1) \cdot (x+1) \cdot \biggl(x -\cos\frac{\pi}{n}-i\sin\frac{\pi}{n}\biggr) \cdots \biggl(x - \cos\frac{(n-1)\pi}{n}-i\sin\frac{(n-1)\pi}{n}\biggr)$$ Also we have $$\cos\frac{(2n-k)\pi}{n} = \cos\frac{k\pi}{n} \ \text{and} \ \sin\frac{(2n-k)\pi}{n} = -\sin\frac{k\pi}{n}$$
Using this we can pair up $$\biggl(x-\cos\frac{\pi}{n}-i\sin\frac{\pi}{n}\biggr)\cdot \biggl(x-\cos\frac{(2n-1)\pi}{n}-i\sin\frac{(2n-1)\pi}{n}\biggr)$$ $$ = x^{2} - 2\cdot x \cdot \cos\frac{\pi}{n} + 1$$
Similarly $$\biggl(x-\cos\frac{2\pi}{n}-i\sin\frac{2\pi}{n}\biggr) \cdot (x - \cos\frac{(2n-2)\pi}{n}-i \sin\frac{(2n-2)\pi}{n}$$ $$= x^{2} -2 \cdot x \cdot \cos\frac{2\pi}{n} + 1$$
Continuing in this way, we get
\begin{align*}
\frac{x^{2n}-1}{x^{2}-1} &= 1 + x^{2} + x^{4} + \cdots + x^{2n-2} \\\ &= \small\biggl(x^{2} - 2x \cos\frac{\pi}{n}+1\biggr) \cdot \biggl(x^{2}-2x\cos\frac{2\pi}{n}+1\biggr) \cdots \biggl(x^{2} - 2x \cos\frac{(n-1)\pi}{n}+1\biggr)
\end{align*}
Now putting $x=1$ and using the fact that $1-\cos{x} = 2 \sin^{2}\frac{x}{2}$ we get $$n = 4^{n-1} \cdot \sin^{2}\Bigl(\frac{\pi}{2n}\Bigr) \cdot \sin^{2}\Bigl(\frac{2\pi}{2n}\Bigr) \cdots \sin^{2}\Bigl(\frac{(n-1)\pi}{2n}\Bigr)$$ From this we get $$\prod\limits_{k=1}^{n-1} \sin\biggl(\frac{\pi k}{2n}\biggr) = \frac{\sqrt{n}}{2^{n-1}}$$
Following the similar procedure gives $$ \prod\limits_{k=1}^{n}\sin\biggl(\frac{k\pi}{2n+1}\biggr) = \frac{\sqrt{2n+1}}{{2^n}}, \qquad \prod\limits_{k=1}^{n}\cos\biggl(\frac{k\pi}{2n+1}\biggr) = \frac{\sqrt{n}}{{2^{n-1}}}$$
On dividing the above you get $$\prod\limits_{k=1}^{n} \tan\Bigl(\frac{k\pi}{2n+1}\Bigr) = \frac{\sqrt{2n+1}}{2^{n}} \times \frac{2^{n-1}}{\sqrt{n}} = \frac{\sqrt{2n+1}}{2 \cdot \sqrt{n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/55120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 0
} |
How to express an irrational number as generalized continued fraction? With simple continued fraction, i.e.
$$a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 \ldots}}}$$
I can use this formula:
$$a_k = \lfloor \alpha_k \rfloor$$
$$\alpha_{k+1} = \dfrac{1}{\alpha_k - a_k}$$
I wonder is there a formula to express the "generalized continued fraction" of the form:
$$a_0 + \cfrac{b_0}{a_1 + \cfrac{b_1}{a_2 + \cfrac{b_2}{a_3 \ldots}}}$$
?
Thank you,
| @Chan, Here is an example for $\pi$:
$$
\pi=\textstyle \cfrac{4}{1+\textstyle \frac{1^2}{2+\textstyle \frac{3^2}{2+\textstyle \frac{5^2}{2+\textstyle \frac{7^2}{2+\textstyle \frac{9^2}{2+\ddots}}}}}}
=3+\textstyle \frac{1^2}{6+\textstyle \frac{3^2}{6+\textstyle \frac{5^2}{6+\textstyle \frac{7^2}{6+\textstyle \frac{9^2}{6+\ddots}}}}}
=\textstyle \cfrac{4}{1+\textstyle \frac{1^2}{3+\textstyle \frac{2^2}{5+\textstyle \frac{3^2}{7+\textstyle \frac{4^2}{9+\ddots}}}}}
$$
Each continued fraction converges to $\pi$, but at greatly different rates.
The first is horrifically slow, requiring roughly $3\times 10^n$ terms for n decimal digits.
The second starts off nicely but later requires nearly 50 terms for five decimal digits, 120 for six.
The third is the best, requiring just four terms for each three decimal digits.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/56164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that there isn't $A \in M_2(\mathbb R)$ such that $A^{2004}=\begin{pmatrix} -1 &0 \\ 0&-2 \end{pmatrix}$? What's the easiest and simplest way to show that there isn't $A \in M_2(\mathbb R)$ such that $A^{2004}=\begin{pmatrix}
-1 &0 \\
0&-2
\end{pmatrix}$?
Thank you.
| Suppose $A^{2004}=\begin{pmatrix}
-1 &0 \\
0&-2
\end{pmatrix}.$
Then $B=A^{1002}$ is a real 2x2 matrix such that
$B^{2}=\begin{pmatrix}
-1 &0 \\
0&-2
\end{pmatrix}.$
We show that this is not possible.
Let $\quad B=\begin{pmatrix}
a &b \\
c&d
\end{pmatrix}.$
Then
$B^{2}=\begin{pmatrix}
a^{2}+bc &ab+bd \\
ac+cd&d^{2}+bc
\end{pmatrix}=\begin{pmatrix}
-1 &0 \\
0&-2
\end{pmatrix}\quad \quad \quad (1)$
Considering the (1,2) entry in equation (1) yields
$$ab+bd=0.$$If $b \neq 0 $ then we must have
$a=-d$
but then the diagonal entries of $B^{2}$ would be equal contradicting (1). If $b=0$ then by considering the (1,1) entry in equation (1) we see that $a^{2}=-1$, which is not possible for $a$ real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/58460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Integrate square of the log-sine integral: $\int_0^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx$ $\displaystyle \int_{0}^{\frac{\pi}{2}} \ln(\sin(x))dx=-\frac{\pi}{2}\ln(2)$ is an integral that is common.
But, how can we show $\displaystyle\int_{0}^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx=\frac{{\pi}^{3}}{24}+\frac{\pi}{2}\ln^{2}(2)$?.
Does anyone have any ideas on how to approach $\displaystyle\int_{0}^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx$?.
Thank you very much.
| Perform the change of variable \begin{align}y=\dfrac{\pi}{2}-x\end{align}
\begin{align}J&=\int_0^{\frac{\pi}{2}}\ln^2\left(\sin x\right) \,dx\\
&=\int_0^{\frac{\pi}{2}}\ln^2\left(\cos x\right)\,dx\\\end{align}
Consider the integrals,
\begin{align}A&=\int_0^{\frac{\pi}{2}}\ln^2\left(\tan x\right)\,dx\\
B&=\int_0^{\frac{\pi}{2}}\ln^2(\sin x\cos x)\,dx\end{align}
observe that,
\begin{align}A+B=4J\end{align}
Perform the change of variable \begin{align}y=\tan x\end{align}
\begin{align}A&=\int_0^\infty \frac{\ln^2 x}{1+x^2}\,dx\end{align}
Consider the double integral
\begin{align} K&=\int_0^\infty \int_0^\infty \frac{\ln^2(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\\
&=2\int_0^\infty \int_0^\infty \frac{\ln^2 x}{(1+x^2)(1+y^2)}\,dx\,dy\\
&=\pi A
\end{align}
since,
\begin{align} \int_0^\infty \frac{\ln x}{1+x^2}\,dx=0\end{align}
On the other hand, perform the change of variable $u=xy$,
\begin{align}K&=\int_0^\infty \int_0^\infty \frac{y\ln^2 u}{(u^2+y^2)(1+y^2)}\,du\,dy\\\end{align}
Perform the change of variable $v=y^2$,
\begin{align}K&=\frac{1}{2}\int_0^\infty \int_0^\infty \frac{\ln^2 u}{(u^2+v)(1+v)}\,du\,dv\\
&=\frac{1}{2}\int_0^\infty\left[\frac{\ln\left(\frac{v+1}{v+u^2}\right)}{u^2-1}\right]_{v=0}^{\infty}\ln^2 u\,du\\
&=\int_0^\infty\frac{\ln^3 u}{u^2-1}\,du\\
&=\int_0^1\frac{\ln^3 u}{u^2-1}\,du+\int_1^\infty\frac{\ln^3 u}{u^2-1}\,du\\
\end{align}
In the latter integral perform the change of variable $z=\dfrac{1}{u}$,
\begin{align}K&=2\int_0^1\frac{\ln^3 u}{u^2-1}\,du\\
&=2\int_0^1\frac{\ln^3 u}{u-1}\,du-\int_0^1\frac{2u\ln^3 u}{u^2-1}\,du
\end{align}
In the latter integral perform the change of variable $z=u^2$,
\begin{align}K&=\left(2-\frac{1}{8}\right)\int_0^1\frac{\ln^3 u}{u-1}\,du\\
&=-\frac{15}{8}\int_0^1\left( \sum_{n=0}^\infty u^{n}\right)\ln^3 u\,du\\
&=-\frac{15}{8}\sum_{n=0}^\infty\int_0^1 u^{n}\ln^3 u\,du\\
&=\frac{45}{4} \sum_{n=0}^\infty\frac{1}{(n+1)^4}\\
&=\frac{45}{4}\zeta(4)\\
\end{align}
therefore,
\begin{align}A&=\frac{K}{\pi}\\
&=\frac{45}{4\pi}\zeta(4)\end{align}
\begin{align}
B&=\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{\sin(2x)}{2}\right)\,dx
\end{align}
Perform the change of variable $y=2x$,
\begin{align}
B&=\frac{1}{2}\int_0^{\pi}\ln^2\left(\frac{\sin x}{2}\right)\,dx\\
&=\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{\sin x}{2}\right)\,dx+\frac{1}{2}\int_{\frac{\pi}{2}}^{\pi}\ln^2\left(\frac{\sin x}{2}\right)\,dx\\
\end{align}
In the latter integral perform the change of variable $y=\pi-x$,
\begin{align}
B&=\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{\sin x}{2}\right)\,dx\\
&=\int_0^{\frac{\pi}{2}}\ln^2\left(\sin x\right)\,dx-2\ln 2\int_0^{\frac{\pi}{2}}\ln(\sin x)\,dx+\frac{\pi}{2}\ln^2 2\\
&=J+\frac{3\pi}{2}\ln^2 2
\end{align}
Therefore,
\begin{align}\frac{45}{4}\zeta(4)+J+\frac{3\pi}{2}\ln^2 2=4J
\end{align}
Thus,
\begin{align}J&=\frac{15}{4\pi}\zeta(4)+\frac{1}{2}\pi\ln^2 2
\end{align}
If you know that,
\begin{align}\zeta(4)=\frac{\pi^4}{90}\end{align}
therefore,
\begin{align}\boxed{J=\frac{\pi^3}{24}+\frac{1}{2}\pi\ln^2 2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/58654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 6,
"answer_id": 1
} |
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.