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Finding difference of angles in triangle . If $\sin A+\sin B+\sin C=0$, $\cos A+\cos B+\cos C=0$, then prove that $$A-B=B-C=C-A=\dfrac{2\pi}3$$
| Squaring and adding we get $$\cos(A-B)+\cos(B-C)+\cos(C-A)=-\frac32$$
Again, we have $$\sin A+\sin B=-\sin C\text{ and }\cos A+\cos B=-\cos C$$
Squaring and adding we get $$2+2\cos(A-B)=1\implies \cos(A-B)=-\frac12=\cos\frac{2\pi}3$$
$$\implies A-B=2n\pi\pm\frac{2\pi}3$$ where $n$ is any integer
Case $A:$ Taking the '+' sign,
$$\implies\sin A=\sin\left(B+2n\pi+\frac{2\pi}3\right)=\sin\left(B+\frac{2\pi}3\right)=-\frac12\sin B+\frac{\sqrt3}2\cos B $$
and similarly, $$\cos A=\cos\left(B+2n\pi+\frac{2\pi}3\right)=\cos\left(B+\frac{2\pi}3\right)=-\frac12\cos B-\frac{\sqrt3}2\sin B $$
So, we need $$\sin C=-\frac12\sin B-\frac{\sqrt3}2\cos B\ \ \ \ (1)\text{ and } \cos C=-\frac12\cos B+\frac{\sqrt3}2\cos B\ \ \ \ (2)$$
$$(1)\implies\sin\left(B+\frac\pi3\right)=-\sin C=\sin(-C)\implies B+\frac\pi3=r\pi+(-1)^r(-C) $$
If $r$ is even $\displaystyle=2m,$(say) $\displaystyle B+\frac\pi3=2m\pi-C\implies B+C=2m\pi-\frac\pi3\ \ \ \ (3)$
If $r$ is odd $=2m+1,$(say) $\displaystyle B+\frac\pi3=(2m+1)\pi+C\implies B-C=2m\pi+\frac{2\pi}3 \ \ \ \ (4)$
$$(2)\implies\cos\left(B-\frac{2\pi}3\right)=\cos C\implies B+\frac{2\pi}3=2a\pi\pm C $$
Taking the '+' sign, $$ B-\frac{2\pi}3=2a\pi+C\implies B-C=2a\pi+\frac{2\pi}3 \ \ \ \ (5)$$
Taking the '-' sign, $$ B-\frac{2\pi}3=2a\pi-C\implies B+C=2a\pi+\frac{2\pi}3 \ \ \ \ (6)$$
From $(3),(4),(5),(6);$ $$B-C=2a\pi+\frac{2\pi}3 $$ with $A-B=2n\pi+\frac{2\pi}3$ will satisfy the given condition (where $a,n$) are integers
$\displaystyle\implies C-A=-(A-B)-(B-C)$
$\displaystyle=-2n\pi-\frac{2\pi}3-2a\pi-\frac{2\pi}3=2(-a-n-1)\pi+\frac{2\pi}3$
Case $B:$ I leave this as an exercise for you to reach the other condition as Case $A$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/623765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$ I solved the following equation, and I just want to be sure I did it right.
This is the procedure:
$$
\sqrt{x-4}-\sqrt{x-5}+1=0\\
\sqrt{x-4}=\sqrt{x-5}-1\\
\text{squaring both sides gives me:}\\
(\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\
x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\
x-4=x-5-\sqrt{x-5}+1\\
x-4=x-4-\sqrt{x-5}\\
\text{substracting x, and adding 4 to both sides}\\
0=-\sqrt(x-5)\\
\text{switching both sides}\\
\sqrt{x-5}=0\\
\text{sqaring both sides}\\
x-5=0\\
x=5\\
\text{When I place 5 in the equation, I get:}\\
\sqrt{5-4}-\sqrt{5-5}+1=0\\
\sqrt{1}-\sqrt{0}+1=0\\
1-0+1=0\\
2=0\\
\text{this means that the equation dosent have any solution, right??}\\
$$
Any advice and suggestion is helpful.
Thanks!!!
| The square root function is increasing on $[0,\infty)$, so
$$\sqrt{x-4} - \sqrt{x-5} \ge 0 $$
for all $x\ge 5.$ Hence the expression can never equal -1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute $\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )^2+\cdots+\left (\frac{1}{n-1}+\frac{1}{n} \right )^2+\left (\frac{1}{n} \right )^2$ Compute the value of the following expression
$$\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )^2+\left ( \frac{1}{2}+\cdots + \frac{1}{n}\right )^2+\cdots+\left (\frac{1}{n-1}+\frac{1}{n} \right )^2+\left (\frac{1}{n} \right )^2$$
The answer is $\boxed{2n-\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )}$.
I've been trying to do it but I've been failed. Any ideas ?
Wolfram's test
| $$S_n - S_{n-1} = n\times \left(\frac{1}{n}\right)^2 + 2\frac{1}{n}\left(1 \times \frac11 +2 \times \frac12+\cdots + (n-1)\times \frac{1}{n-1} \right) = 2-\frac1n$$ and $S_0$ can be taken to be $0$, so just add up $(S_n - S_{n-1})+(S_{n-1} - S_{n-2})+\cdots + (S_{1} - S_{0})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/625939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
| Here's an alternative argument using the fact that a set of $n$ elements has $2^n$ subsets. Consider subsets $S$ of the set $\{0,1,2,\dots,9,10\}$. Since this set has 11 elements, it has $2^{11}$ subsets. Now ask, for each $n$ in the range $0\leq n\leq 10$, how many of these subsets $S$ have $n$ as their largest element. Well, any such $S$ consists of the element $n$ together with some subset $S'$ of $\{0,1,\dots,n-1\}$. So there are $2^n$ choices for $S'$. Thus, $n$ is the largest element of exactly $2^n$ subsets $S$ of $\{0,1,2,\dots,9,10\}$. Therefore, the sum $2^{10}+2^9+\dots+2^2+2^1$ counts all the subsets $S$ of $\{0,1,2,\dots,9,10\}$ whose largest element is 10 or 9 or $\dots$ or 2 or 1. That means it counts all of the $2^{11}$ subsets $S$ of $\{0,1,2,\dots,9,10\}$ except for two, namely the empty set and the set $\{0\}$. So the sum is $2^{11}-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/628501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is $x_n =\sqrt{n} (\sqrt{n+1}− \sqrt{n})$ , $n \ge 1$ convergent? $$x_n =\sqrt{n} (\sqrt{n+1}− \sqrt{n}) ,\quad n \ge 1$$
My work is:
First i analyzed the convergence of this sequence in 2 parts:
a) $\lim \sqrt{n+1}− \sqrt{n} = \lim \frac{(\sqrt{n+1}− \sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}} = \lim \frac{n+1-n}{\sqrt{n+1}+\sqrt{n}} = \lim \frac{1}{ \infty} = 0$ when $n \rightarrow \infty$.
But now how can I conclude that $\lim_{n \rightarrow \infty}x_n = 0$
| Is it correct this?
$\begin{align}
x_n
&=\sqrt{n} (\sqrt{n+1}− \sqrt{n})\\
&=\sqrt{n} (\sqrt{n+1}− \sqrt{n})\dfrac{\sqrt{n+1}+ \sqrt{n}}{\sqrt{n+1}+ \sqrt{n}}\\
&=\sqrt{n} \dfrac{(\sqrt{n+1}+ \sqrt{n})(\sqrt{n+1}− \sqrt{n})}{\sqrt{n+1}+ \sqrt{n}}\\
&=\sqrt{n} \dfrac{(n+1)-n}{\sqrt{n+1}+ \sqrt{n}}\\
&= \dfrac{\sqrt{n}}{\sqrt{n+1}+ \sqrt{n}}\\
\end{align}
$
so
$x_n
> \dfrac{\sqrt{n}}{2\sqrt{n+1}}
$
and
$x_n
< \dfrac{\sqrt{n}}{2\sqrt{n}}
=\dfrac12
$.
Doing the same thing to bound
the difference between these two bounds,
$\begin{align}
\dfrac12-\dfrac{\sqrt{n}}{2\sqrt{n+1}}
&=\dfrac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n+1}}\\
&=\dfrac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n+1}}\dfrac{\sqrt{n+1}+ \sqrt{n}}{\sqrt{n+1}+ \sqrt{n}}\\
&=\dfrac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+ \sqrt{n})}{2\sqrt{n+1}(\sqrt{n+1}+ \sqrt{n})}\\
&=\dfrac{1}{2\sqrt{n+1}(\sqrt{n+1}+ \sqrt{n})}\\
&<\dfrac{1}{4n}\\
\end{align}
$
so
$\dfrac12 -\dfrac1{{4n}}
< x_n < \dfrac12$
so $\lim_{n \to \infty} x_n = \dfrac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/629413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Calculate the eigenvectors We calculate the eigenvectors for the matrix
$$
\begin{equation*}
\mathbf{A} = \left(
\begin{array}{ccc}
2 & 0 & 0 \\
1 & 1 & 1 \\
1 & -1 & 3 \\
\end{array}
\right)
\end{equation*}
$$
First, I calculted the eigenvalue polynomial $\det(\mathbf{A}-\lambda \mathbf{I})=0$, and got triple overlapping eigenvalues $\lambda=2$
$$\begin{equation*}
\mathbf{A} - \lambda \mathbf{I} = \left(
\begin{array}{ccc}
0 & 0 & 0 \\
1 & -1 & 1 \\
1 & -1 & 1 \\
\end{array}
\right)
\end{equation*}$$
So, I got only two eigenvectors $x_1=\left(
\begin{array}{ccc}
1 \\
1 \\
0 \\
\end{array} \right)$
and $x_2=\left(
\begin{array}{ccc}
0 \\
1 \\
1 \\
\end{array} \right)$. So the matrix $\mathbf{A}$ is a degenerate matrix.
Did I calculate it correctly?
| Your results are correct, but that does not mean that $A$ is a degenerate matrix, just that $A$ is not diagonalizable.
Just one little thing:
$x_1=\left(
\begin{array}{ccc}
1 \\
1 \\
0 \\
\end{array} \right)$
and $x_2=\left(
\begin{array}{ccc}
0 \\
1 \\
1 \\
\end{array} \right)$ are not the only eigenvectors but every vector in the whole subspace $\operatorname{span}\left(\left(
\begin{array}{ccc}
1 \\
1 \\
0 \\
\end{array} \right),\left(
\begin{array}{ccc}
0 \\
1 \\
1 \\
\end{array} \right)\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/632385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim\limits_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right)=0$
Prove that:
$$\lim_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}
\,\right)=0$$
Epsilon>0. According to the Archimedean Property of reals, we have n1 element N with epsilon*n1>11. (Why 11? Seems so random...)
n0:=max{3,n1} (What's the point of that? It doesn't appear anywhere in the proof...).
For each n element N with n>=n0 we have:
\begin{align}
0&\lt\sqrt{n^4+n^2+20n+7}-\sqrt{n^4+n^2+1} \\\,\\
&=\dfrac{\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}\right)\cdot\left(\sqrt{n^4+n^2+20n+7}+\sqrt{n^4+n^2+1}\right)}{\sqrt{n^4+n^2+20n+7}\,+\,\sqrt{n^4+n^2+1}} \\\,\\
&=\dfrac{20n+6}{\sqrt{n^4+n^2+20n+7}\,+\,\sqrt{n^4+n^2+1}}\leqslant\dfrac{20n+2n}{\sqrt{n^4}+\sqrt{n^4}}=\dfrac{11}n\leqslant\dfrac{11}{n_1}\lt\epsilon
\end{align}
(I don't get the circled part. Why 20n+2n, why "cut off" the roots? And why should it equal 11/n?)
From this we get
$$\left|\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right|\lt\epsilon$$
, thus the proof is complete.
Thanks for the clarifications!
| Extarct n^4 from each radical. So, what is left inside the remaining radicals is (1 + 1/ n^2 + ...) and, since "n" is large, each remaining radical radical is almost 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/632553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Prove that: $\sin{\frac{\pi}{n}} \sin{\frac{2\pi}{n}} ...\sin{\frac{(n-1)\pi}{n}} =\frac{n}{2^{n-1}}$ Using that:
$$
x^{n - 1} + x^{n - 2} + \cdots + x + 1
=
\left(x - w\right)\left(x - w^{2}\right)\ldots\left(x - w^{n - 1}\right)
$$
Prove that:
$$
\sin\left(\pi \over n\right)\sin\left(2\pi \over n\right)\ldots
\sin\left(\left[n-1\right]\pi \over n\right) = {n \over 2^{n - 1}}
$$
| Notice that
$$
w=e^{i\frac{2\pi}{n}}=z^2,
$$
with
$$
z=e^{i\frac{\pi}{n}}
$$
and
\begin{eqnarray}
n&=&(1-w)\ldots(1-w^{n-1})=(|z|^2-z^2)\ldots(|z|^{2n-2}-z^{2n-2})\\
&=&(-1)^{n-1}z^{1+2+\ldots+(n-1)}(z-\bar{z})\ldots(z^{n-1}-\bar{z}^{n-1})\\
&=&(-1)^{n-1}z^{\frac{n(n-1)}{2}}2i\left(\sin\frac{\pi}{n}\right)\ldots2i\sin\left(\frac{(n-1)\pi}{n}\right)\\
&=&(-1)^{n-1}(2i)^{n-1}e^{i\frac{(n-1)\pi}{2}}\left(\sin\frac{\pi}{n}\right)\ldots\sin\left(\frac{(n-1)\pi}{n}\right)\\
&=&2^{n-1}\sin\left(\frac{\pi}{n}\right)\ldots\sin\left(\frac{(n-1)\pi}{n}\right).
\end{eqnarray}
Thus
$$
\sin\left(\frac{\pi}{n}\right)\ldots\sin\left(\frac{(n-1)\pi}{n}\right)=\frac{n}{2^{n-1}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/633321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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To find $R\circ R^{-1}$ in Discrete mathematics Today I came across a question in DMS which says:
If $R$ is the relation “Less Than” from $A = \{1, 2, 3, 4\}$ to $B = \{1,3,5\}$ then find $R\circ R^{-1}$.
Now what is $R\circ R^{-1}$?
I know how to find $R\circ R$,
like in this question firstly we will find $R$ where
$$R= \{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)\}$$
and then $R\circ R$ would be:
$$\{(1,5),(2,5),(3,5),(4,5)\}.$$
Please correct me if I'm wrong.
And also please explain how to find $R\circ R^{- 1}$?
Thanks in advance :-)
| I wanna introduce the domain and the range of $R$, which are defined in Elements of Set Theory by Herbert B. Enderton:
\begin{align}
x \in \mathrm{dom} R &\Leftrightarrow \exists y :\: \left( x,\, y \right) \in R,\\
x \in \mathrm{ran} R &\Leftrightarrow \exists t :\: \left( t,\, y \right) \in R.
\end{align}
It'll help you think.
Because $ R = \left\{ \left(1,\,3\right),\;\left(1,\,5\right),\;\left(2,\,3\right),\;\left(2,\,5\right),\;\left(3,\,5\right),\;\left(4,\,5\right) \right\} $.
We can get $\mathrm{dom} R = A = \left\{1,\,2,\,3,\,4\right\}$, and $\mathrm{ran} R = \left\{3,\,5\right\}$.
To get $R \circ R$, evaluate $\mathrm{ran}R \cap \mathrm{dom}R = \left\{ 3,\, 5 \right\} \cap \left\{ 1,\, 2,\, 3,\, 4 \right\} = \left\{ 3 \right\}$, use the elements of it in the second $R$. It's obvious that the only possible routes are $1 \rightarrow 3 \rightarrow 5$ and $2 \rightarrow 3 \rightarrow 5$, i.e. $R \circ R = \left\{ \left(1,5 \right),\, \left( 2,\, 5 \right) \right\}$.
By definition, $R^{-1} = \left\{ (y,\,x) \mid xRy \right\}$. So $R^{-1} = \left\{ \left(3,\,1\right),\;\left(5,\,1\right),\;\left(3,\,2\right),\;\left(5,\,2\right),\;\left(5,\,3\right),\;\left(5,\,4\right) \right\} $.
$\mathrm{dom}R^{-1} = \mathrm{ran}R = \left\{ 3,\,5 \right\}$.
Similarly, to get $R \circ R^{-1}$, evaluate $\mathrm{ran}R \cap \mathrm{dom}R^{-1} = \left\{ 3,\,5 \right\}$. Therefore we can get $R \circ R^{-1} = \left\{ \left(1,\,1\right),\;\left(2,\,2\right),\;\left(3,\,3\right),\;\left(4,\,4\right),\;\left(1,\,2\right),\;\left(2,\,1\right),\;\left(2,\,3\right),\;\left(3,\,2\right),\;\left(3,\,4\right),\;\left(4,\,3\right),\;\left(1,\,3\right),\;\left(3,\,1\right),\;\left(2,\,4\right),\;\left(4,\,2\right),\;\left(1,\,4\right),\;\left(4,\,1\right) \right\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/634983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Limit of $x^x$ as $x$ tends to $0$ I am trying to solve the following limit:
$$\lim \limits_{x\to0} x^x$$
The only thing that comes to mind is to write $x^x$ as $e^{x\ln{x}}$ and getting the right sided limit would be easy but I don't see how I could get the left sided one seeing that the $\ln$ is not defined for negative numbers.
Is there something I am missing or is there another way to go about it?
P.S.:I don't know anything about derivatives so please keep it to the limits.
| L'Hospital's rule is quickest. I show that other approaches are possible:
For $\ x\in \left[\ \frac{1}{7},\ \frac{1}{6}\ \right),$
$$ \left( \frac{1}{7} \right)^{\frac{1}{6}}<x^x<1.$$
Now using Newton's Binomial expansion,
\begin{align} \left(1-\alpha\right)^\frac{1}{6} = 1 + \frac{1}{6}(-\alpha) + \frac{\left(\frac{1}{6}\right)\left(-\frac{5}{6}\right)}{2!}(-\alpha)^2 + \frac{\left(\frac{1}{6}\right)\left(-\frac{5}{6}\right)\left(-\frac{11}{6}\right)}{3!}(-\alpha)^3+\ldots \\
\\
= 1 - \frac{1}{6}\alpha - \frac{\left(\frac{1}{6}\right)}{2}\frac{\left(\frac{5}{6}\right)}{1} \alpha^2 - \frac{\left(\frac{1}{6}\right)}{3} \frac{\left(\frac{11}{6}\right)}{2} \frac{\left(\frac{5}{6}\right)}{1} \alpha^3 - \ldots\ \\
\\
> 1 - \frac{1}{6}\left(\alpha + \frac{1}{2} \alpha^2 + \frac{1}{3} \alpha^3 + \ldots \right)\\
\\
= 1 - \frac{1}{6}(\ -\ln(1-\alpha)\ ) = 1 + \frac{1}{6}(\ \ln(1-\alpha)\ ).\\
\\
\end{align}
Substituting $\ \alpha = \frac{6}{7},\ $ into the above, we see that
$$ 1+ \frac{1}{6}\ln\left(\frac{1}{7}\right) = 1 - \frac{1}{6}\ln(7) < \left( \frac{1}{7} \right)^{\frac{1}{6}}<x^x<1\quad \forall x\in \left[\ \frac{1}{7},\ \frac{1}{6}\ \right).$$
More generally, for any $\ k \in\mathbb{N},\ $ we have,
$$ 1 - \frac{\ln(k+1)}{k} < \left( \frac{1}{k+1} \right)^{\frac{1}{k}} < x^x < 1 \quad \forall x\in \left[\ \frac{1}{k+1},\ \frac{1}{k}\ \right).$$
Letting $\ k\to\infty\ $ yields the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/637401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 2
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mixed numbers subtraction vertically In the following subtraction we are subtracting $2$ mixed numbers vertically. I know how it works except the last step.
$$ 7 \frac{1}{3} - 4 \frac{1}{2} = 3 + \frac{-1}{6} = 2 + \frac{5}{6} = 2 \frac{5}{6}$$
I am confused about this step: $ 3 + \frac{-1}{6} = 2 + \frac{5}{6}$
How does this work?
| For "normal" subtraction, we all learned in elementary school to "borrow" a 1 from the next highest place if we needed to.
In this case, we're sort of doing the same thing. When we have $3 + -1/6$, we "borrow" a 1 to make $2 + 6/6 + -1/6$, which then becomes $2 + 5/6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/638931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $4^{2n+1}+3^{n+2} : \forall n\in\mathbb{N}$ is a multiple of $13$ How to prove that $\forall n\in\mathbb{N},\exists k\in\mathbb{Z}:4^{2n+1}+3^{n+2}=13\cdot k$
I've tried to do it by induction. For $n=0$ it's trivial.
Now for the general case, I decided to throw the idea of working with $13\cdot k$ and try to prove it via congruences. So I'd need to prove that $4^{2n+1}+3^{n+2}\equiv0\pmod{13} \longrightarrow 4^{2n+3}+3^{n+3}\equiv0\pmod{13}$, that is, $4^{2n+1}+3^{n+2}\equiv4^{2n+3}+3^{n+3}\pmod{13}$
But I have no clue how to do is. Any help?
| $$\begin{gather}4^{2n+3}+3^{n+3}=4^{2n+1}\cdot{4^2}+3^{n+2}\cdot{3}=\\
=4^{2n+1}\cdot{4^2}+3^{n+2}\cdot{4^2}-3^{n+2}\cdot{4^2}+3^{n+2}\cdot{3}=\\
=16\cdot(4^{2n+1}+3^{n+2})-3^{n+2}\cdot(16-3)=\\
=16\cdot(4^{2n+1}+3^{n+2})-13\cdot3^{n+2}\end{gather}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/640859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Simple Question on Triangles... What times the sum of the squares of the sides of a triangle is equal to the sum of the squares of the medians of the triangle.
| Hint:
$\Large 2\cdot m_a^2+\frac{a^2}{2}=b^2+c^2$
$\Large 2\cdot m_b^2+\frac{b^2}{2}=c^2+a^2$
$\Large 2\cdot m_c^2+\frac{c^2}{2}=a^2+b^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/644070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of function with $\frac00$ I am unable to solve following limit:
$$\lim_{x \rightarrow 0_+}\frac{\sin \sqrt{x}}{x^2}\left(\sqrt{x+2x^2}-\sqrt{2\sqrt{1+x}-2}\right)$$
I keep getting $ \frac{0}{0}$. I admit I haven't tried to use l'Hospital rule multiple times as the square root is not so nice to derive more than one time. Is it possible to solve this limit without using l'Hospital rule/Taylor series (which I haven't learned yet)?
Thank you
| Note that $\sqrt{2\sqrt{1+x}-2}\sim\sqrt{x+2x^2}\sim\sqrt x$, while
\begin{align}
(x+2x^2)-(2\sqrt{1+x}-2)
&=2x^2+x+2-2\sqrt{1+x}\\
&=\frac{(2x^2+x+2)^2-4(1+x)}{2x^2+x+2+2\sqrt{1+x}}\\
&\sim\frac 94x^2
\end{align}
thus
\begin{align}
\frac{\sin \sqrt{x}}{x^2}(\sqrt{x+2x^2}-\sqrt{2\sqrt{1+x}-2})
&=\frac{\sin \sqrt{x}}{x^2}\frac{(x+2x^2)-(2\sqrt{1+x}-2)}{\sqrt{x+2x^2}+\sqrt{2\sqrt{1+x}-2}}\\
&\sim \frac{\sqrt x}{x^2}\frac{\frac 94x^2}{2\sqrt x}\\
&=\frac 98
\end{align}
| {
"language": "en",
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} |
How to integrate this integral? I had a problem with this specific integral:
$$
\int{\sqrt{\frac{x^3 -3}{x^{11}}} \ dx}
$$
Thank you ahead of time.
| $$ \int \sqrt{\frac{x^3-3}{x^{11}}}\,dx = \int \frac{1}{x^4} \sqrt{\frac{x^3-3}{x^3}} \,dx
= \int \frac{1}{x^4} \sqrt{1-\frac{3}{x^3}} \,dx $$
Let
$$ u = \frac{1}{x^3}, \; du = -\frac{3}{x^4} dx $$
The above integral becomes
$$ -\frac{1}{3} \int \sqrt{1-3u}\:du = -\frac{1}{3} \cdot -\frac{1}{3} \cdot \frac{2}{3}
(1-3u)^{3/2} + C = \frac{2}{27} \left(1 - \frac{3}{x^3}\right)^{3/2} + C $$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Complex Numbers and exponential form and roots The roots of $z^7 = -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$ are $\text{cis } \theta_1, \text{cis } \theta_2, \dots, \text{cis } \theta_7,$ where $ 0^\circ \le \theta_k < 360^\circ $for all $ 1 \le k \le 7$. Find $\theta_1 + \theta_2 + \dots + \theta_7$. Give your answer in degrees.
In exponential form this is $z^7 = e^ \left(5 \pi i/4 \right)$. How should I simplify? Thanks
| $\def\cis{\operatorname{cis}}$Let's consider a more general problem. We have $a=r\cis\alpha$ and we write its $n$th roots as
\begin{gather}
\sqrt[n]{r}\cis\frac{\beta}{n}\\
\sqrt[n]{r}\cis\left(\frac{\beta}{n}+\frac{2\pi}{n}\right)\\
\sqrt[n]{r}\cis\left(\frac{\beta}{n}+2\frac{2\pi}{n}\right)\\
\dots\\
\sqrt[n]{r}\cis\left(\frac{\beta}{n}+(n-1)\frac{2\pi}{n}\right)
\end{gather}
Can you sum up all those angles? Use $360$ instead of $2\pi$ if you so prefer or are forced to.
| {
"language": "en",
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"source": "stackexchange",
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} |
Solve the error in the simultaneous equation. The question is:
$$\begin{align}
\tag{1} 2x - 3y &= 3\\
\tag{2} 4x^2 - 9y^2 &= 3
\end{align}$$
From equation (1):
$$\tag{3} 2x = 3 - 3y.$$
Substitute equation (3) in (2):
$$\begin{align}
4x^2 - 9y^2 &= 3\\
(2x)^2 - 9y^2 &= 3\\
(3 - 3y)^2 - 9y^2 &= 3\\
[(3)^2 - 2\cdot 3\cdot 3y + (-3y)^2] - 9y^2 - 3 &= 0\\
9 +18y + 9y^2 - 9y^2 - 3 &= 0\\
-18y + 6 &= 0\\
-18y &= -6\\
y &= \frac 1 3
\end{align}$$
But the answer for y in my book is $ - \frac 1 3\ $
Which one is right?
| From $(3)$ you get $x$, so your solution is
$$ \begin{eqnarray}
x&=&1 \\
y&=&\frac{1}{3}
\end{eqnarray}
$$
Substitute these value in your equation then you see that all the equalities hold except the equality $(1)$. Here we get $1$ for the left-hand side and $3$ for the right-hand side. So apparently $(3)$ is erroneous.
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 1
} |
Evaluate $\iint_D\sin(xy)dA$ where $D$ is bounded by $y=\frac 1x, y=\frac2x, y=x, y=2x$ in the first quadrant. Evaluate $\iint_D\sin(xy)dA$ where $D$ is bounded by $y=\frac 1x, y=\frac2x, y=x, y=2x$ in the first quadrant.
By subbing numbers into the equation, I see that $1\leq x\leq 2, 1\leq y\leq 2.$
Without solving the equation, can someone tell me if this is correct? The final answer I got was $-\frac14\sin(4)+\frac 54\sin(2)-\sin(1)$
| Okay, here's what I got. Evaluating this double integral seems to be much harder on usual cartesian coordinates, since the region is pretty ugly.
Courtesy of Mathematica. If we perform the coordinate change
$$
\begin{cases}
xy & = u \\
\frac{y}{x} & = v,
\end{cases}
$$
we get the much neater region that is a square.
What we have to do now is find the Jacobian and integrate. We have
$$dA = dx \, dy = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \, du \, dv.$$
From our definitions we find
$$
\begin{align}
x & = \sqrt{\frac{u}{v}} \\
y & = \sqrt{uv}.
\end{align}
$$
Then
$$
\begin{align}
\mathcal{J} & = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \\
& = \begin{vmatrix} \frac{1}{2} \frac{1}{\sqrt{uv}} & - \frac{1}{2} \frac{\sqrt{u}}{v} \\ \frac{1}{2} \sqrt{\frac{v}{u}} & \frac{1}{2} \sqrt{\frac{u}{v}} \end{vmatrix} \\
& = \frac{1}{4} \left[ \frac{1}{v} + \sqrt{v} \right].
\end{align}
$$
This is nonzero in the region considered, thus a valid coordinate change. Rewriting the integral we get
$$I = \iint\limits_{D} \sin (xy) \, dA = \int_1^2 \hspace{-5pt} \int_1^2 \sin (u) \frac{1}{4} \left[ \frac{1}{v} + \sqrt{v} \right] \, du \, dv.$$
Evaluating this I found
$$I = \frac{1}{4} \left( (\cos (1) - \cos(2)) \left( \ln (2) + \frac{2}{3} \left( 2^{3/2} -1 \right) \right) \right).$$
Using Mathematica I numerically evaluated this, and $I \approx 0,457206$. This seems correct to me.
Note: We must take the absolute value of the Jacobian, but in this case it is positive so the absolute value is unnecessary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/651315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Limits at infinity I'm working with limits at infinity and stumbled upon this exercise where I want to evaluate the indicated limit:
$$\lim_{x \to \infty} \frac{1}{\sqrt{x^2-2x}-x}$$
I tried to solve it by doing the following:
$$\lim_{x \to \infty} \frac{1}{\sqrt{x^2-2x}-x} = \lim_{x \to \infty} \frac{1}{\sqrt{x^2} \sqrt{1-\frac{2}{x}}-x} = \lim_{x \to \infty} \frac{1}{x \sqrt{1-\frac{2}{x}}-x} = \lim_{x \to \infty} \frac{\frac{1}{x}}{\sqrt{1-\frac{2}{x}}-1}$$
But the answer should be $-1$, so what I did must be wrong. How do you evaluate this limit the best way possible?
| The points that we can add to Adi's post are:
$$\frac{1}{\sqrt{x^2-2x}-x}\times\frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x}=\frac{\sqrt{x^2-2x}+x}{(x^2-2x)-x^2}=\color{blue}{\frac{|x|\sqrt{1-2/x}+x}{-2x}}$$ Now if $x\to+\infty$ so $|x|=x$ and so $$\color{blue}{\frac{|x|\sqrt{1-2/x}+x}{-2x}}=\frac{x\sqrt{1-2/x}+x}{-2x}=\frac{x\times \sqrt{1-0}+x}{-2x}\longrightarrow -1$$ And if $x\to-\infty$ so $|x|=-x$ and so $$\color{blue}{\frac{|x|\sqrt{1-2/x}+x}{-2x}}\longrightarrow 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/651969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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} |
How find this value $A+B+C+AB+BC+AC+ABC$ let
$$A=\dfrac{2\sqrt{3}}{3}i\cdot\cos{\left(\dfrac{\pi}{6}+\dfrac{i}{3}\text{arcsinh}{(\dfrac{3\sqrt{3}}{2})}\right)}$$
$$B=\dfrac{2\sqrt{3}}{3}i\cdot\cos{\left(\dfrac{5\pi}{6}+\dfrac{i}{3}\text{arcsinh}{(\dfrac{3\sqrt{3}}{2})}\right)}$$
$$C=\dfrac{2\sqrt{3}}{3}i\cdot\cos{\left(\dfrac{3\pi}{2}+\dfrac{i}{3}\text{arcsinh}{(\dfrac{3\sqrt{3}}{2})}\right)}$$
Find $$A+B+C+AB+BC+AC+ABC$$
I think we must find $A+B+C=?$
$AB+BC+AC=?$
$ABC=?$
| Observe that $$\cos3\left(\frac\pi6+\frac i3\text{arcsinh} \frac{3\sqrt3}2
\right)=\cos\left(\frac\pi2+i\text{arcsinh} \frac{3\sqrt3}2\right)$$
$$=-\sin \left(i\cdot\text{arcsinh} \frac{3\sqrt3}2\right)$$
Now as, $\displaystyle \sin(ix)=i\sinh(x)$
$$-\sin \left(i\cdot\text{arcsinh} \frac{3\sqrt3}2\right)=-i\sinh \left(\text{arcsinh} \frac{3\sqrt3}2\right)=-i \frac{3\sqrt3}2$$
As $\displaystyle\cos3y=4\cos^3y-3\cos y $
So, $\displaystyle\frac A{\dfrac{2\sqrt3}3i}=\cos\left(\frac\pi6+\frac i3\text{arcsinh} \frac{3\sqrt3}2\right)$ is a root of $\displaystyle4t^3-3t=-i \frac{3\sqrt3}2$
If $\displaystyle\cos3P=\cos\left(\frac\pi2+3Q\right),$
$\displaystyle3P=2n\pi\pm \left(\frac\pi2+3Q\right)$ where $n$ is any integer
Taking the '+' sign and $\displaystyle3P=2n\pi+\left(\frac\pi2+3Q\right)=3Q+\frac{(4n+1)\pi}2$
$\displaystyle\implies P=Q+\frac{(4n+1)\pi}6$
Setting $n=0,1,2$ we can see that
$$\displaystyle\frac A{\dfrac{2\sqrt3}3i},\frac B{\dfrac{2\sqrt3}3i}, \frac C{\dfrac{2\sqrt3}3i}$$ are the roots of $$\displaystyle4t^3-3t=-i \frac{3\sqrt3}2\iff8t^3-6t+i3\sqrt3=0\ \ \ \ (1)$$
Now respond to the invitation of Vieta's formulas
Method $1:$
$$\displaystyle\frac A{\dfrac{2\sqrt3}3i}+\frac B{\dfrac{2\sqrt3}3i}+ \frac C{\dfrac{2\sqrt3}3i}=0\iff A+B+C=0$$
$$\displaystyle\sum\frac A{\dfrac{2\sqrt3}3i}\cdot\frac B{\dfrac{2\sqrt3}3i}=-\frac68\iff \sum AB=-\frac68\left(\dfrac{2\sqrt3}3i\right)^2=1$$
$$\displaystyle\frac A{\dfrac{2\sqrt3}3i}\cdot\frac B{\dfrac{2\sqrt3}3i}\cdot\frac C{\dfrac{2\sqrt3}3i}=-\frac{i3\sqrt3}8\iff ABC=-\frac{i3\sqrt3}8\left(\dfrac{2\sqrt3}3i\right)^3$$
$$=-i\frac{3\sqrt3}8\frac{8\cdot3\sqrt3i^3}{27}=-i^4=-1$$
Method $2:$
$\displaystyle A+B+C+AB+BC+CA+ABC=-1+\prod(1+A)$
Let $\displaystyle y=1+A\iff \frac{y-1}{\dfrac{2\sqrt3}3i}=\frac A{\dfrac{2\sqrt3}3i}$ which is a root of $(1)$
So, put the value of $\frac A{\dfrac{2\sqrt3}3i}$ in $(1)$ and simplify to form a Cubic equation in $y$
Now, we need $\displaystyle A+B+C+AB+BC+CA+ABC=-1+\prod(1+A)=-1+\prod y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/655819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving Integral that contain exponential and Power I have an integral of this form:
$$\int_0^\infty e^{-\frac{x}{a}-\frac{z^2}{bx}-\frac{z}{bx}}\left(\frac{c}{c+x+z}\right)^K~dx$$
where $K$ is a positive integer.
$a$ , $b$ and $c$ are reals and $>0$
finally,$z$ is also reals.
My question is that; is it possible to find the solution of this integral?
| Special case of $z=-c$ :
$\int_0^\infty e^{-\frac{x}{a}-\frac{c^2}{bx}+\frac{c}{bx}}\left(\dfrac{c}{x}\right)^K~dx$
$=c^K\int_0^\infty\dfrac{e^{-\frac{x}{a}-\frac{c^2-c}{bx}}}{x^K}dx$
$=c^K\int_0^\infty\dfrac{e^{-\frac{\sqrt{\frac{a(c^2-c)}{b}}x}{a}-\frac{c^2-c}{b\sqrt{\frac{a(c^2-c)}{b}}x}}}{\left(\sqrt{\dfrac{a(c^2-c)}{b}}x\right)^K}d\left(\sqrt{\dfrac{a(c^2-c)}{b}}x\right)$
$=\dfrac{b^\frac{K-1}{2}c^K}{a^\frac{K-1}{2}(c^2-c)^\frac{K-1}{2}}\int_0^\infty\dfrac{e^{-\sqrt{\frac{c^2-c}{ab}}\left(x+\frac{1}{x}\right)}}{x^K}dx$
$=\dfrac{b^\frac{K-1}{2}c^\frac{K+1}{2}}{a^\frac{K-1}{2}(c-1)^\frac{K-1}{2}}\int_{-\infty}^\infty\dfrac{e^{-\sqrt{\frac{c^2-c}{ab}}\left(e^x+\frac{1}{e^x}\right)}}{(e^x)^K}d(e^x)$
$=\dfrac{b^\frac{K-1}{2}c^\frac{K+1}{2}}{a^\frac{K-1}{2}(c-1)^\frac{K-1}{2}}\int_{-\infty}^\infty e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh x}e^{(1-K)x}~dx$
$=\dfrac{b^\frac{K-1}{2}c^\frac{K+1}{2}}{a^\frac{K-1}{2}(c-1)^\frac{K-1}{2}}\left(\int_{-\infty}^0e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh x}e^{(1-K)x}~dx+\int_0^\infty e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh x}e^{(1-K)x}~dx\right)$
$=\dfrac{b^\frac{K-1}{2}c^\frac{K+1}{2}}{a^\frac{K-1}{2}(c-1)^\frac{K-1}{2}}\left(\int_\infty^0e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh(-x)}e^{(1-K)(-x)}~d(-x)+\int_0^\infty e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh x}e^{(1-K)x}~dx\right)$
$=\dfrac{b^\frac{K-1}{2}c^\frac{K+1}{2}}{a^\frac{K-1}{2}(c-1)^\frac{K-1}{2}}\left(\int_0^\infty e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh x}e^{(K-1)x}~dx+\int_0^\infty e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh x}e^{-(K-1)x}~dx\right)$
$=\dfrac{2b^\frac{K-1}{2}c^\frac{K+1}{2}}{a^\frac{K-1}{2}(c-1)^\frac{K-1}{2}}\int_0^\infty e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh x}\cosh((K-1)x)~dx$
$=\dfrac{2b^\frac{K-1}{2}c^\frac{K+1}{2}}{a^\frac{K-1}{2}(c-1)^\frac{K-1}{2}}K_{K-1}\left(2\sqrt{\dfrac{c^2-c}{ab}}\right)$
| {
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Evaluating $\int_{0}^{\infty} \frac{x^{3}- \sin^{3}(x)}{x^{5}} \ dx $ using contour integration EDIT: Instead of expressing the integral as the imaginary part of another integral, I instead expanded $\sin^{3}(x)$ in terms of complex exponentials and I don't run into problems anymore.
\begin{align} \int_{0}^{\infty} \frac{x^{3}-\sin^{3}(x)}{x^{5}} \ dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^{3}-\sin^{3}(x)}{x^{5}} \ dx \\ &= \frac{1}{2} \ \int_{-\infty}^{\infty} \frac{x^{3}+\frac{1}{8i}(e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix})}{x^{5}} \ dx \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \ \int_{-\infty}^{\infty} \frac{x^{3}+\frac{1}{8i}(e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix})}{(x-i \epsilon)^{5}} \ dx \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{x^{3}+\frac{1}{8i} (e^{3ix}-3e^{ix})}{(x-i \epsilon)^{5}} + \frac{1}{16i} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{3e^{-ix}-e^{-3ix}}{(x-i \epsilon)^{5}} \ dx \end{align}
Then I integrated $ f(z) = \frac{z^{3}+ \frac{1}{8i}(e^{3iz}-3e^{iz})}{(z-i \epsilon)^{5}}$ around the upper half of $|z|=R$ and $ g(z) = \frac{3e^{-iz}-e^{-3iz}}{(z-i \epsilon)^{5}}$ around the lower half of $|z|=R$ and applied Jordan's lemma.
\begin{align} \int_{0}^{\infty} \frac{x^{3}-\sin^{3}x}{x^{5}} \ dx &= \frac{1}{2} \lim_{\epsilon \to 0^{+}}2 \pi i \ \text{Res}[f(z),i \epsilon] + \frac{1}{16i} \lim_{\epsilon \to 0^{+}} 2 \pi i (0) \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \frac{2 \pi i}{4!} \lim_{z \to i \epsilon} \frac{d^{4}}{dz^{4}} \Big(z^{3}+\frac{1}{8i}e^{3iz}-\frac{3}{8i}e^{iz} \Big) \\ &= \frac{\pi i}{24} \lim_{\epsilon \to 0^{+}} \ \lim_{z \to i \epsilon}\Big( \frac{1}{8i}(3i)^{4}e^{3iz}- \frac{3}{8i} (i)^{4} e^{iz} \Big) \\ &= \frac{\pi i}{24} \lim_{\epsilon \to 0^{+}} \Big( \frac{81}{8i}e^{- 3\epsilon} - \frac{3}{8i}e^{- \epsilon} \Big) \\ &= \frac{\pi i}{24} \Big(\frac{81}{8i}-\frac{3}{8i} \Big) \\ &= \frac{13 \pi}{32} \end{align}
| I like to calculate this integral as follows:
Let us note that
$$\frac{1}{x^5}=\frac{1}{4!}\int_0^\infty t^4e^{-xt}dt$$ So
$$I=\frac{1}{4!}\int_{0}^{\infty}(x^{3}-\sin^{3}x)\int_0^\infty t^4e^{-xt}\;dt\;dx$$
$$=\frac{1}{4!}\int_{0}^{\infty}t^4\int_{0}^{\infty}(x^{3}-\sin^{3}x)e^{-xt}\;dx\;dt$$
$$=\frac{1}{4!} \int_{0}^{\infty}t^4\left [\frac{6}{t^4}-\frac{6}{(t^2+1)(t^2+9)}\right ]dt$$
$$=\frac{1}{4}\int_{0}^{\infty}\frac{10t^2+9}{(t^2+1)(t^2+9)}dt=\frac{13\pi}{32}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/656757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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Linear Algebra Complex Numbers The solutions to the equation $z^2-2z+2=0$ are $(a+i)$ and $(b-i)$ where $a$ and $b$ are integers. What is $a+b$?
I simplified and got $(z+1)(z+1) = -1$ and now I'm not sure where to go from there. I did this but I'm not sure.
$(a+i)^2 = a^2 - 1$
$(b-i)^2 = b^2 + 1$
$a+b=(a+i)+(b-i)=(a^2-1)+(b^2+1)=a^2+b^2$
| Since this is a polynomial with real coefficients, $z$ is a root if and only if $\overline z$, the complex conjugate of $z$, is. Since $\overline{a + i} = a - i$, we see that $a = b$ necessarily.
Then using that $a + i$ and $a - i$ are roots, we can factor the polynomial as
\begin{align*}
z^2 + 2z + 2 &= \Big(z - (a + i)\Big)\Big(z - (a - i)\Big) \\
&= (z - a - i)(z - a + i) \\
&= (z - a)^2 - (i)^2 \\
&= z^2 - 2az + a^2 - 1
\end{align*}
Comparing coefficients, we see that $2 = -2a$ so that $a = -1$.
Fast way: By expanding $(z - \lambda)(z - \mu)$, we see that the product of the roots is the constant coefficient, and the sum of the roots is negative of the $z$-coefficient. Hence
$$a + b = \Big(a + i\Big) + \Big(b - i\Big) = -2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If$ a+b-1=1+\frac{ln(2^a-1)}{ln4}+\frac{ln(2^b-1)}{ln4}$ then $a=b$? If $$a+b-1=1+\frac{ln(2^a-1)}{ln4}+\frac{ln(2^b-1)}{ln4}$$ where $a,b>0$ are real numbers and ln is $log_e$, then is a=b?
| Let $f(x) = x - \log_4 \left(2^{x+2}-4\right)$. Then the given relation is $f(a)+f(b) = 0$.
Now $f(x)$ has a minimum at $x=1$, which is $f(1) = 0$. So $f(x) > 0$ for all other values of $x>0$. Hence for the relation to be satisfied, $a=b=1$.
Updated: to show the minimum without calculus, note that
$f(x) \ge 0 \iff x \ge \log_4(2^{x+2}-4) \iff 4^{x} \ge 2^{2+x}-4 \iff y^2\ge 4y - 4 \iff (y-2)^2 \ge 0$
where $y = 2^x$
and this also
$\implies f(x) = 0 \iff 2^x=2 \iff x = 1$
| {
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"timestamp": "2023-03-29T00:00:00",
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$(3^x-2^x)^2+3^x-2^x\leq 5^x$ Solve the following inequality
$$(3^x-2^x)^2+3^x-2^x\leq 5^x$$
I'm having difficulty in solving this inequality because there are multiple exponential with different basis. I thought, however, the replacement:
$$y=3^x-2^x$$
and so:
$$y^2+y\leq 5^x$$
treating $5^x$ as the known term of the inequality of the second degree in y.
What do you think? thank you very much
| Your approach will flesh out as follows:
$$\boxed{y^2+y\le5^x}\iff y^2+y-5^x\le0\iff y\in\Big[y_{_1},y_{_2}\Big],\qquad y_{_{1,2}}=\frac{-1\pm\sqrt{5^x+1}}2$$
$$-\sqrt{5^x+1}\le2y+1\le\sqrt{5^x+1}\iff|2y+1|\le|\sqrt{5^x+1}|\iff(2y+1)^2\le5^x+1$$
$$4y^2+4y\le5^x\iff\boxed{y^2+y\le\dfrac{5^x}{4\ \ }}\iff y\in\Big[y'_{_1},y'_{_2}\Big],\qquad y'_{_{1,2}}=\frac{-1\pm\sqrt{\dfrac{5^x}{4\ \ }+1}}2$$
$$4y^2+4y\le\dfrac{5^x}{4\ \ }\iff\boxed{y^2+y\le\dfrac{5^x}{4^2}}\iff y\in\Big[y^"_{_1},y^"_{_2}\Big],\qquad y^"_{_{1,2}}=\frac{-1\pm\sqrt{\frac{5^x}{4^2}+1}}2$$
Traveling down this road, we deduce that $y\in\Big[Y_{_1},Y_{_2}\Big]$, where $Y_{_{1,2}}=\displaystyle\lim_{n\to\infty}\dfrac{-1\pm\sqrt{\frac{5^x}{4^n}+1}}2=\pm1.$
So $-1\le3^x-2^x\le1\iff2^x-1\le3^x\le2^x+1$. Since $\displaystyle\lim_{t\to-\infty}a^t=0\iff x\in(-\infty,x_{_0})$, for some numerically determined value of $x_{_0}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How does $\frac1{n^4} =\frac1{(2n-1)^4} + \frac1{(2n)^4}$ I need to find the sum of $\displaystyle\sum_{n=1}^\infty \frac{1}{n^4}.$
In the solution document it written:
$$\displaystyle\sum_{n=1}^\infty \frac1{n^4} = \sum_{n=1}^\infty \frac1{(2n-1)^4} + \sum_{n=1}^\infty \frac1{(2n)^4}.$$
My question is, how did he get that $\displaystyle\sum_{n=1}^\infty \frac1{n^4} = \sum_{n=1}^\infty \frac1{(2n-1)^4} + \sum_{n=1}^\infty \frac1{(2n)^4} $?
Thanks in advance.
| \begin{align*}
\sum_{n=1}^\infty \frac{1}{n^4}
&= \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \frac{1}{5^4} + \frac{1}{6^4} + \cdots \\
&= \left( \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots \right)
+ \left( \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \cdots \right) \\
&= \sum_{n=1}^\infty \frac{1}{(2n-1)^4} + \sum_{n=1}^\infty \frac{1}{(2n)^4}
\end{align*}
Note that $n$ is just a dummy variable, so it need not be true that $\displaystyle \frac{1}{n^4} = \frac{1}{(2n-1)^4} + \frac{1}{(2n)^4}$.
| {
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Prove that $\left(\frac12(x+y)\right)^2 \le \frac12(x^2 + y^2)$ Prove that $$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
I've gotten that
$$\left(\frac12(x+y)\right)^2 \ge 0 $$
but stumped on where to go from here...
| This substitution is a nice way to do this:
$$x=r\sin\theta , y=r\cos\theta$$
So, our identity:
$$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
$$\frac{1}{4} (x^2+y^2+2xy) \leq \frac12(x^2 + y^2)$$
Applying our substitutions:
$$\frac{1}{4} (r^2+2\cdot r\sin\theta\cdot r\cos\theta) \leq \frac12(r^2)$$
$$\frac{r^2}{2}\cdot \frac{1+\sin2\theta}{2} \leq \frac{r^2}{2}$$
Now, if you note that the maximum value of $\sin 2\theta$ is $1$, then it becomes obvious.
Added: It could have also been done it this way:
$$\left(\frac{1}{2}(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
$$\frac{1}{2}\cdot\frac{1}{2}(r\cos\theta+r\sin\theta)^2 \leq \frac{r^2}{2}$$
$$\frac{r^2}{2}\cdot\frac{1}{2}(\cos\theta+\sin\theta)^2 \leq \frac{r^2}{2}$$
Now, note that $-\sqrt{2} \leq \cos\theta+\sin\theta = \sqrt{2}\cos(x+45^{\circ}) \leq \sqrt{2}$. Thus the maximum value of $(\cos\theta+\sin\theta)^2$ is $2$, and the result follow.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show if $a^2+b^2 \le 2$ then $a+b \le 2$ If $a^2+b^2 \le 2$ then show that $a+b \le2$
I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result
| Hint Use Cauchy-Schwarz.
Second solution
$$(a+b)^2=a^2+b^2+2ab \leq 2+2ab$$
You got that far, you are almost there:
By AM-GM
$$\sqrt{a^2b^2} \leq \frac{a^2+b^2}{2}$$
which implies $$2ab \leq a^2+b^2 \leq 2$$
| {
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Why does a conjugate of a root remain a root of a polynomial? About quadratic equations, I have the follow question,
What is the polynomial, with rational coefficients, knowing that one root is $ x_1 = 1 + \sqrt 3$ ?
So, to solve the problem, I must know that the other root is $x_2 = 1-\sqrt3$. I see the resolution, then I use $x_2 = 1-\sqrt3$. But I don't know why.
I solved in this way,
$\ \ \ \ \, $ Root Sum $= S = (1+\sqrt3) + (1-\sqrt 3)\, =\, 2$
Root Product $ = P = (1+\sqrt3) * (1-\sqrt3) = -2$
Vieta Formula is $\,x^2 - S\, x + P = 0$ so the polynomial is $\,x^2 - 2\,x - 2 = 0$.
Why is $\,x_2 = 1-\sqrt 3\, $ if $\,x_1 = 1+\sqrt3$?
Why is $\,x_2 = 5+\sqrt 3\, $ if $\,x_1 = 5-\sqrt3$?
Why does this happen (switching the operator of the root) when the coefficients are rational?
| You can see this answer as a supplement to Bill Dubuque's, but I am avoiding any technical language.
The key to understanding why if $1+\sqrt{5}$ is a root of a quadratic equation with rational coefficients, then $1-\sqrt{5}$ is also a root, is understanding the symmetry between $\sqrt{5}$ and $-\sqrt{5}$. Here is what I mean:
If you write down any expressions with rational numbers and $\sqrt{5}$, and you do math with them, then all the math is dictated by the fact that $\sqrt{5}$'s square is 5. Example:
$$(1+\sqrt{5})(3-2\sqrt{5}) = 3 - 2\sqrt{5} + 3\sqrt{5} - 2\cdot (\sqrt{5})^2=3+\sqrt{5} - 10 = -7 + \sqrt{5}$$
You get the answer by doing purely rational arithmetic except occasionally whenever you see $(\sqrt{5})^2$, you can write $5$. In other words, the mathematical behavior of $\sqrt{5}$ in this context is completely determined by the fact that $\sqrt{5}$ satisfies the equation $x^2=5$.
Now the key point is this: $-\sqrt{5}$'s square is also 5. In other words, it also satisfies the equation $x^2=5$. This means that $-\sqrt{5}$ behaves in the exact same way as $\sqrt{5}$ does. If we replace $\sqrt{5}$ with $-\sqrt{5}$ throughout the calculation above, everything will work the exact same way, and $-\sqrt{5}$ will also replace $\sqrt{5}$ in the answer. Look:
$$(1-\sqrt{5})(3-2(-\sqrt{5})) = 3 -2(-\sqrt{5}) + 3(-\sqrt{5}) - 2\cdot(-\sqrt{5})^2 $$
$$= 3 - \sqrt{5} - 10 = -7-\sqrt{5}$$
Replacing $\sqrt{5}$ with $-\sqrt{5}$ in the problem just replaced it as well throughout the calculation and in particular in the answer.
So, to your original question, if $1+\sqrt{5}$ is a root of a quadratic equation $ax^2+bx+c=0$ with rational coefficients, that means that
$$a(1+\sqrt{5})^2 + b(1+\sqrt{5})+c=0$$
But by the above train of thought, if you replace $\sqrt{5}$ with $-\sqrt{5}$ throughout the calculation, everything should still work the same. Since you are assuming that $a,b,c$ are rational, then they don't have any $\sqrt{5}$ in them, so there is nothing to replace, and it looks like
$$a(1-\sqrt{5})^2 + b(1-\sqrt{5}) + c = 0$$
So $1-\sqrt{5}$ is also a root.
| {
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Max. and Min. value of $f(x,y) = \frac{x-y}{x^4+y^4+6}$ [1] Calculation of Max. and Min. value of $\displaystyle f(x) = \sqrt{x^3-6x^2+21x+18}$, where $\displaystyle -\frac{1}{2}\leq x\leq 1$
[2] Calculation of Max. and Min. value of $\displaystyle f(x,y) = \frac{x-y}{x^4+y^4+6}\;,$ where $x,y\in \mathbb{R}$
$\bf{My\; Try::(1)}$Let $g(x) = x^3-6x^2+21x+18$, Now Using Derivative Test.
$g^{'}(x)=3x^2-12x+21=3(x^2-4x+7)=3\left\{(x-2)^2+3\right\}>0\;\forall x\in \mathbb{R}$
So $g(x)$ is Strictly Increasing function.
So $g(x)$ is Min. when $\displaystyle x = -\frac{1}{2}$ and $g(x)$ is Max. when $x=1$
$\bf{My\; Try::(2)}$ Let $\displaystyle f(x,y) = z=\frac{x-y}{x^4+y^4+6}.$
Using $\bf{A.M\geq G.M},\;\; $
$x^4+1\geq 2x^2$ and $y^4+1\geq 2y^2\;,$ we get $x^4+y^4+2\geq 2(x^2+y^2)\Rightarrow x^4+y^4+6\geq (x^2+y^2)+4$
Now I did not understand How can I solve it.
Help Required
Thanks.
| If you want Max, $x-y>0$, you want Min,$x-y<0$
for Max:
$x-y \le |x-y| \le|x|+|y|$
your process can go on: $x^2+1\ge 2|x|,y^2+1\ge2|y| \implies x^4+y^4+6\ge 4(|x|+|y|)\implies \dfrac{x-y}{x^4+y^4+6} \le \dfrac{1}{4}$ when $x=-y=1$
you can get Min in same way.
| {
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Optimization with a few Variables (AMC 12 question) In the 2013 AMC 12B, question 17 says:
Let $a$,$b$, and $c$ be real numbers such that
$a+b+c=2$, and $a^2+b^2+c^2=12$
What is the difference between the maximum and minimum possible values of $c$?
I was wondering if there is a quick and easy solution using multivariable calculus for this problem. (I've only taken single variable)
The only solution I've seen uses the Cauchy-Schwarz inequality.
| Here is an geometric interpretation: the locus of points $(a,b,c)$ such that $a+b+c = 2$ and $a^2+b^2+c^2 = 12$ is the intersection of a plane and a sphere in $\mathbb R^3$; i.e., it is a circle. The plane is perpendicular to the line $a = b = c$, and a little visualization leads us to conclude that the extrema of $c$ occur when $a = b$. In such a case, it becomes straightforward to compute the extrema explicitly: we have $2a + c = 2$, $2a^2 + c^2 = 12$, hence the difference is $\frac{16}{3}$.
A slightly more rigorous argument involves the coordinate transformation $x = (a+b)/\sqrt{2}$, $y = (a-b)/\sqrt{2}$. Then our system becomes $c+x\sqrt{2} = 2$, $x^2+y^2+c^2 = 12$, hence $$0 = \frac{3}{2}c^2 - 2c - 10 + y^2.$$ Solving for $c$, we obtain $$c = \frac{2 \pm \sqrt{64-6y^2}}{3},$$ and we observe that these values are greatest/least when $y = 0$. So the difference is $2\frac{\sqrt{64}}{3} = \frac{16}{3}$.
| {
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Trigonometry identities sum and substraction of cosine Someone can help me showing this identity?
$\frac{cos(a-b)}{cos(a+b)}=\frac{1+tan(a)tan(b)}{1-tan(a)tan(b)}$
| Hint:
\begin{align*}
\frac{\cos(a-b)}{\cos(a+b)}&=\frac{\cos a\cos b+ \sin a\sin b}{\cos a\cos b- \sin a\sin b}\\
&=\frac{\displaystyle\frac{\cos a\cos b+ \sin a\sin b}{\cos a\cos b}}{\displaystyle\frac{\cos a\cos b- \sin a\sin b}{\cos a\cos b}}\\
&=\; ?
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$? Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$?
So far I have:
$a^2+(a+2)^2+(a+4)^2+1$
$=a^2+a^2+4a+4+a^2+8a+16+1 $
$=3a^2+12a+21$
$=3(a^2+4a+7) $
where do I go from here.. the solution I have is divisible by $3$ not $12$...
| $$a^2+(a+2)^2+(a+4)^2+1=3a^2+12a+21\equiv3(a^2-1)\pmod{12}$$
Now, if $a$ is odd $a\pm1$ are even $\implies a^2-1$ will be divisible by $4$
In fact, $\displaystyle(2b+1)^2=4b^2+4b+1=8\frac{b(b+1)}2+1\equiv1\pmod8$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Ways of writing $n=2a+b$ with $a$ and $b$ are non-negative integers For a non-negative number $n$, let $r_n$ be the number of ways of writing
$n = a + 2b$, where $a$ and $b$ are non-negative integers.
For example,
$5 = 1 \cdot 5 + 2 \cdot 0 = 3 \cdot 1 + 1 \cdot 2 = 1 \cdot 1 + 2 \cdot 2$,
so $r_5 = 3$.
My question is I have found the rule that:
$r_0=r_1=1$,
$r_2=r_3=2$,
$r_4=r_5=3$,
$r_6=r_7=4$,
......
$r_{n-1}=r_{n}=\frac{n+1}{2}$ where $n$ is odd.
But how can we prove this rule?
| Hint: Let $n=2k$. Then we can use $0$ or $1$ or $2$ or $\dots$ or $k$ two-dollar coins, a total of $k$ choices.
Let $n=2k+1$. Then we can use $0$ or $1$ or $2$ or $\dots$ or $k$ two-dollar coins, a total of $k$ choices.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Real number multiplicative inverses expressed in another form I've been asked to express the multiplicative inverse of $3 + \sqrt{5}$ in the form $c + d\sqrt{5}$, where $c,d$ are rational numbers.
I understand that for some rational numbers $c,d$ we must have:
$$1 = (3 + \sqrt{5})(c + d\sqrt{5}).$$
I was able to answer for the multiplicative inverse of $2 +\sqrt{3}$. We find that $1 = (2 +\sqrt{3})(c + d\sqrt{3})$ where $c = 2$ and $d = -1$. However this seems to be related to the original $2 +\sqrt{3}$; and in the problem at hand this is not the case. So I'm rather confused.
How could i go about solving this? I'd like steps without the answer IF possible; moreover, how could i prepare for more general questions in this form?
| \begin{align}
Answer &= \frac{1}{3+\sqrt{5}}\\
\mbox{Multiply and Divide by Conjugate}&\\
&= \frac{1}{3+\sqrt{5}}\cdot \frac{3-\sqrt{5}}{3-\sqrt{5}}\\
&= \frac{3-\sqrt{5}}{(3)^2-(\sqrt{5})^2}\\
\end{align}
Hidden Answer:
$$\frac{3}{4}-\frac{\sqrt{5}}{4}$$
| {
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Two-variable limit question $\lim\limits_{(x,y)\rightarrow (0,0)} \dfrac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}$
How to solve this two-variable limit? Thanks :D
| Let
\begin{equation}
f(x,y)=\frac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}
\end{equation}
If $y=0$, $\lim_{(x,y)\rightarrow(0,0)}f(x,y)=0$. If $y\neq0$, say approach the $(0,0)$ do not along with x-axis, we have:
\begin{equation}
f(x,y)=\frac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}\leq\frac{x^2y^2}{2|x|y^2\sqrt{x^2+y^2}}=\frac{|x|}{2\sqrt{x^2+y^2}}\\
0\leq\lim_{(x,y)\rightarrow(0,0)}f(x,y)\leq\lim_{(x,y)\rightarrow(0,0)}\frac{|x|}{2\sqrt{x^2+y^2}}=0
\end{equation}
In summary, $\lim_{(x,y)\rightarrow(0,0)}f(x,y)=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factoring quartic equation $$x^4+6x^2+25=0$$
How could I factor it into $(x^2-2x+5)(x^2+2x+5)=0$?
I got the result looking into horrible formulas on wikipedia, but I suppose there's a much easier way. Could you help me?
| Notice that
$$
x^4+6x^2+25=(x^4+6x^2+9)+16=(x+3)^2+4^2=(x^2+3+4i)(x^2+3-4i) \quad \forall x\in \mathbb{C}.
$$
Let $u \in \mathbb{C}$ such that $u^2=3+4i$. It is easy to show that $u\in\{2+i,-2-i\}$. We have
\begin{eqnarray}
x^4+6x^2+25&=&(x^2+u^2)(x^2+\bar{u}^2)=(x+iu)(x-iu)(x+i\bar{u})(x-i\bar{u})\\
&=&(x+iu)\overline{(x+iu)}(x-iu)\overline{(x-iu)}\\
&=&|x+iu|^2|x-iu|^2=|x-1+2i|^2|x+1-2i|^2\\
&=&[(x-1)^2+4][(x+1)^2+4]=(x^2-2x+5)(x^2+2x+5).
\end{eqnarray}
| {
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Express $\;f(x)=\frac{x − 1}{x + 1}\;$ as the sum of an even and an odd function. In homework there is such problem:
Express $\;f(x)=\dfrac{x − 1}{x + 1}\;$ as the sum of an even and an odd function.
(Simplify as much as possible.)
This function is not even and neither odd. Also if we take it as division of 2 functions, neither $x - 1$ nor $x + 1$ are odd or even... so I'm confused...
| I know it's late, but you could also simply multiply both numerator and denominator by $(x-1)$.
$$ \begin{align}
\frac{x-1}{x+1} &= \frac{x-1}{x+1} \cdot \frac{x-1}{x-1} \\
&= \frac{x^2-2x+1}{x^2-1} \\
&= \frac{x^2+1}{x^2-1} + \frac{2x}{1-x^2} \\
\end{align} $$
The left-hand term is even since it is of the form $even/even$, and the right-hand term is odd since it has the form $odd/even$.
| {
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Integer solutions of $ x^3+y^3+z^3=(x+y+z)^3 $ Consider the equation
$$ x^3+y^3+z^3=(x+y+z)^3 $$
for triples of integers $(x, y, z) $.
I noticed that this has infinitely many solutions: $ x, y $ arbitrary and $ z=-y $.
Are there more solutions?
| Too long for a comment. Interestingly, for $k>3$, there are non-trivial solutions to,
$k=4;\quad x_i = -2634, 955, 1770, 5400$: (Jacobi-Madden equation)
$$x_1^4+x_2^4+x_3^4+x_4^4 = (x_1+x_2+x_3+x_4)^4$$
$k=5;\quad x_i = - 3, - 54,24,28,67 $:
$$x_1^5+x_2^5+\dots+x_5^5 = (x_1+x_2+\dots+x_5)^5$$
$k=6;\quad x_i = -4170, -3187, -888, 1854, 3300, 3936, 4230$:
$$x_1^6+x_2^6+\dots+x_7^6 = (x_1+x_2+\dots+x_7)^6$$
$k=7;\quad x_i = -230, -353, -625, -673, 184, 443, 556$:
$$x_1^7+x_2^7+\dots+x_7^7 = (x_1+x_2+\dots+x_7)^7$$
and longer ones for $k = 8,9,10$ (See April 6 update).
P.S. The cases $k=4,5$ involve elliptic curves, hence there are an infinite number of co-prime solutions.
| {
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Factoring a difference of 2 cubes I am trying to factorize the expression $(a - 2)^3 - (a + 1)^3$ and obviously I would want to put it in the form of $(a - b)(a^2 + ab + b^2)$
So I start off with the first $(a - b)$ and I get $(a - 2) - (a + 1)$ which I simplify from $(a^2 + a -2a -2)$ to $(a^2 -3a -2)$
Now I'm up to $(a^2 + ab + b^2)$ and I $a^2$ would equal to $(a^2 - 4)$, $ab$ would be $(a - 2) * (a + 1)$ which is $(a^2 + a -2a -2)$ and $b^2$ would be $(a^2 + 1)$..
Then we get $(a^2 - 3a - 2)((a^2 - 4) (a^2 + a - 4a)(a^2 + 1)) $
At this point I get confused because I'm not sure if I did anything correct, and I don't know how to continue this. Any help is much appreciated
Regards,
| you have $$(a - 2)^3 - (a + 1)^3=[(a - 2) - (a + 1)][(a - 2)^2+(a - 2)(a + 1) + (a + 1)^2]
\\=[-3][a^2-4a +4+a^2 - a -2 + a^2 +2a+ 1]
\\=-3(3a^2-3a+3)
\\=-9(a^2-a+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/689671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find all triples $(p; q; r)$ of primes such that $pq = r+ 1$ and $2(p ^ 2+q ^ 2) =r ^ 2 + 1$. We have to find all triples $(p; q; r)$ of primes such that $pq = r+ 1$ and $2(p ^ 2+q ^ 2) =r ^ 2 + 1$. This question was asked in the 2013 mumbai region RMO but i could not find a solution to it. Can you please help me out with this?
| Well, $r^2+1=2(p^2+q^2)$ is even, so $r^2$ is odd, and so $r$ is odd. But then $pq=r-1$ is even. Thus, $p$ and/or $q$ must be an even prime.
What happens next depends on whether you are considering negative numbers as potential primes. I will proceed as though you are. Let's assume that $p$ is an even prime, so that $p^2=4.$ Squaring the first equation then shows us that $$4q^2=r^2+2r+1,$$ while multiplying the second equation by $2$ gives us $$16+4q^2=2r^2+2,$$ meaning $$4q^2=2r^2-14.$$ Thus, $$2r^2-14=r^2+2r+1\\r^2-2r-15=0\\(r-5)(r+3)=0$$ and so either $r=5$ or $r=-3.$ But we cannot have $r=-3,$, though, since then we would have $$4q^2=r^2+2r-1=9-6+1=4,$$ so $q^2=1,$ which is impossible, since $q$ is prime. Thus, we must have $r=5.$ So, our original equations become $$pq=6$$ and $$8+2q^2=26.$$ Hence, we find the triples $(2,3,5)$ and $(-2,-3,5).$
Assuming that $q$ is an even prime likewise gets us the triples $(3,2,5)$ and $(-3,-2,5).$
From the work above, we can also see that there are only two viable triples if we are not considering negative numbers as potential primes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/689884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How to solve simple trigonometry equation. So we are learning trigonometry in school and I would like to ask for a little help with these. I would really appreciate if somebody can explain me how I can solve such equations :)
*
*$\sin 3x \cdot \cos 3x = \sin 2x$
*$2( 1 + \sin^6 x + \cos^6 x ) - 3(\sin^4 x + \cos^4 x) - \cos x = 0$
*$3 \sin^2 x - 4 \sin x \cdot \cos x + 5 \cos^2 x = 2$
*$\sin^2 x - \sin^4 x + \cos^4 x = 1$
In our students book they're poorly explained with 2 pages, I tried to find a solution on the web, but still couldn't find similar examples. All we got from our teacher was paper with few formulas and we basically have no idea when to use them. I would show what I've tried, but the problem is that I have no idea to even start solving such equations.
| Are your formulae things like $\sin^2 x + \cos^2 x = 1$?
If so then you need to spot where you can apply them.
For example $$\sin^2 x - \sin^4 x + \cos^4 x = 1$$
$$\sin^2 x (1- \sin^2 x) + \cos^4 x = 1$$
$$\sin^2 x \cos^2 x + \cos^4 x = 1 $$
$$(\sin^2 x + \cos^2 x) \cos^2 x = 1 $$
$$ \cos^2 x = 1 $$
$$ \cos x = \pm 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/690465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluate $\int_0^{\pi/2}\log\cos(x)\,\mathrm{d}x$ How can you evaluate $$\int\limits_0^{\pi/2}\log\cos(x)\,\mathrm{d}x\;?$$
| Here is another solution: As we have $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}=\frac{e^{2ix}+1}{2e^{ix}}$ we get:
$$I=\int_0^{\frac{\pi}{2}} \ln(\cos(x))dx=\int_0^{\frac{\pi}{2}} \ln(e^{2ix}+1)dx-\int_0^{\frac{\pi}{2}} \ln(2)dx-\int_0^{\frac{\pi}{2}} ixdx$$
By using the series expansion of $\ln(x)$ and by calculating the two simple integrals we obtain:
$$I=\int_0^{\frac{\pi}{2}} \sum_{k=0}^\infty \frac{(-1)^k}{k+1}\cdot e^{2ix(k+1)}dx-\frac{\pi}{2}\cdot \ln(2)-i\frac{\pi^2}{8}$$
By swapping the integral and the sum, the integral of the infinite sum can be calculated as follows:
$$\int_0^{\frac{\pi}{2}} \sum_{k=0}^\infty \frac{(-1)^k}{k+1}\cdot e^{2ix(k+1)}dx=\sum_{k=0}^\infty \frac{(-1)^k}{k+1}\cdot \int_0^{\frac{\pi}{2}}e^{2ix(k+1)}dx=\sum_{k=0}^\infty \frac{(-1)^k}{k+1}\cdot \left(\frac{e^{i\pi(k+1)}-1}{2i(k+1)}\right)=\frac{i}{2}\cdot\sum_{k=0}^\infty \frac{(-1)^k}{k+1}\cdot \left(\frac{1-(-1)^{k+1}}{(k+1)}\right)$$
Here I used, that $\frac{1}{i}=-i$. Now, when $k$ is uneven, the expression after the sigma is equal to $0$, this yields:
$$\frac{i}{2}\cdot\sum_{k=0}^\infty \frac{(-1)^k}{k+1}\cdot \left(\frac{1-(-1)^{k+1}}{(k+1)}\right)=\frac{i}{2}\cdot\sum_{k=0}^\infty \frac{2\cdot(-1)^{2k}}{(2k+1)^2}=i\cdot\sum_{k=0}^\infty \frac{1}{(2k+1)^2}=i\cdot\left(\sum_{k=1}^\infty \frac{1}{k^2}-\sum_{k=1}^\infty \frac{1}{(2k)^2}\right)=i\cdot\frac{3}{4}\cdot\zeta(2)=i\frac{\pi^2}{8}$$
Therefore, $I$ can be expressed as follows:
$$I=i\frac{\pi^2}{8}-\frac{\pi}{2}\cdot \ln(2)-i\frac{\pi^2}{8}=-\frac{\pi}{2}\cdot \ln(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/690644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 9,
"answer_id": 1
} |
Show that the gcd of $(n^2, n^2 + n + 1) = 1$ We know $(a,\;b)$ = $(a-kb,\;b)$, with $(x,\;y)$ meaning the greatest common divisor of x and y.
$(n^2,\;n^2 + n + 1)$ = $(n^2,\;(n+1)^2 - n)$
Let $d\;|\;n^2$ and let $d\;|\;((n+1)^2 -\;n)$
Then $d\;|\;n+1$, since by linear combination theorem $d\;|\;n^2$ and $d\;|\;(n^2+n+1)$ implies by subtraction that $d\;|\;(n+1)$
Also, since $d\;|\;n^2$, then $d\;|\;n$
Since $d\;|\;n$ then again by linear combination theorem $d\;|\;(n+1)^2$
$d\;|\;(n+1)^2$ implies $d\;|\;(n+1)$
$d\;|\;n$ and $d\;|\;(n+1)$ implies $(d,n) = 1$
Is this proof correct? Seems legit to me. This is my first proof typed out in MathJAX directly, and it started as my asking where to go from the "Also, since ..." but then I managed to complete it!
| Use the Euclidean Algorithm for Greatest Common Divisor:
$$n^2 + n + 1 = n^2 + (n+1)$$
$$n^2 = (n-1)(n+1) + 1$$
Hence the greatest common divisor is $1$.
Another way is to assume $d \mid n^2$ and $d \mid n^2 + n + 1$, where $d$ is prime number. Now $d$ divides their difference so we have:
$$d \mid n^2 + n + 1 - n^2 \implies d \mid n+1$$
Now since $d \mid n^2 \implies d \mid n$, we get that $d$ divides consecutive numbers, which is possible only for $d = 1$. Hence the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/692116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove for any $k$ then have $a^2_{1}+a^2_{2}+a^2_{3}+\cdots+a^2_{k}=m^3$ for any positive integer $k$,there exsit $m\in N$ and $a_{i}\in N,i=1,2,\cdots,k$,such
(1): $a_{i}\neq a_{j},i\forall i\neq j$,
(2):
$$a^2_{1}+a^2_{2}+a^2_{3}+\cdots+a^2_{k}=m^3$$
My idea: if $k=1$,then we let $a_{1}=1$,then $$a^2_{1}=1^3=m^3$$
if
$k=2$, then we let $a_{1}=5,a_{2}=10$,then we have
$$a^2_{1}+a^2_{2}=5^2+10^2=125=5^3=m^3$$
if $k=3$, note
$$3^2+4^2+10^2=125=5^3$$
then let $a_{1}=3,a_{2}=4,a_{3}=10,m=5$
But follow I can't find it and How prove this problem ,Thank you
| Let $1^2+2^2+\cdots+k^2=m$ and multiply through by $m^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/694713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Pre Calculus Expression The questions is:
$$\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}$$
My answer is: $$\dfrac{3(x+2)^2 + 6x^2-4}{(x-3)^2}$$
Am I right? If not, where have I failed?
| Trusty maxima tells me:
factor(ratsimp((3*(x + 2)^2 * (x - 3)^2 - 2 * (x + 2)^3 * (x - 3))/(x - 3)^4));
is:
$$
\frac{(x - 13) (x + 2)^2}{(x - 3)^3}
$$
A trick that helps catch silly errors while simplifying is to replace some simple values, like $x = 0$ and $x = \pm 1$ in the expresssions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/695222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Prove an equation about binomial coefficients Could we prove:
$ \sum_{k} \binom{2k}{k}\binom{n+k}{m+2k} \frac{(-1)^k}{k+1} = \binom{n-1}{m-1}$ when $m,n \in N$
| The generating function for the Central Binomial Coefficients is
$$
\sum_{k=0}^\infty\binom{2k}{k}x^k=(1-4x)^{-1/2}\tag{1}
$$
Integrating $(1)$ and dividing by $x$ yields
$$
\sum_{k=0}^\infty\binom{2k}{k}\frac{x^k}{k+1}=\frac{1-\sqrt{1-4x}}{2x}\tag{2}
$$
Sum the formula against $x^m$:
$$
\begin{align}
&\sum_{m=-\infty}^n\left(\sum_{k=0}^{n-m}\binom{2k}{k}\binom{n+k}{m+2k}\frac{(-1)^k}{k+1}\right)x^m\tag{3}\\
&=\sum_{k=0}^\infty\sum_{m=-2k}^{n-k}\binom{2k}{k}\binom{n+k}{m+2k}\frac{(-1)^k}{k+1}x^m\tag{4}\\
&=\sum_{k=0}^\infty x^{-2k}\sum_{m=0}^{n+k}\binom{2k}{k}\binom{n+k}{m}\frac{(-1)^k}{k+1}x^m\tag{5}\\
&=\sum_{k=0}^\infty x^{-2k}\binom{2k}{k}\frac{(-1)^k}{k+1}(x+1)^{n+k}\tag{6}\\
&=(x+1)^n\sum_{k=0}^\infty\binom{2k}{k}\frac{(-1)^k}{k+1}\left(\frac{x+1}{x^2}\right)^{k}\tag{7}\\
&=(x+1)^n\frac{-1+\sqrt{1+4\frac{x+1}{x^2}}}{2\frac{x+1}{x^2}}\tag{8}\\
&=x(x+1)^{n-1}\frac{-x+(x+2)}{2}\tag{9}\\[9pt]
&=x(x+1)^{n-1}\tag{10}\\[6pt]
&=\sum_{m=1}^n\binom{n-1}{m-1}x^m\tag{11}
\end{align}
$$
Justification:
$\:\ (3)$: sum against $x^m$
$\:\ (4)$: change order of summation
$\:\ (5)$: substitute $m\mapsto m-2k$
$\:\ (6)$: apply the Binomial Theorem
$\:\ (7)$: reorganize
$\:\ (8)$: apply $(2)$
$\:\ (9)$: simplify
$(10)$: simplify
$(11)$: apply the Binomial Theorem
Equating coefficients of $x^m$ yields
$$
\sum_{k=0}^{n-m}\binom{2k}{k}\binom{n+k}{m+2k}\frac{(-1)^k}{k+1}=\binom{n-1}{m-1}\tag{12}
$$
Note that the sum has non-zero terms for $m\lt1$, but these cancel; e.g. $n=3,m=-1$:
$$
\begin{align}
&\binom{0}{0}\binom{3}{-1}\frac11-\binom{2}{1}\binom{4}{1}\frac12+\binom{4}{2}\binom{5}{3}\frac13-\binom{6}{3}\binom{6}{5}\frac14+\binom{8}{4}\binom{7}{7}\frac15\\[6pt]
&=0-4+20-30+14\\[9pt]
&=0
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/697170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$\gcd(x^2+1,x^2+4x+5)$ Is there anything I can tell about $\gcd(x^2+1,x^2+4x+5)$ for any given integer $x$? I believe I've seen similar questions in the past, though I don't remember any details or what to search for. I did write a computer program to check the first $50$ values which seem to alternate between $1$ and $2$, but how do I prove this?
| If $d$ divides both
$d$ will divide $x^2+4x+4-(x^2+1)=4(x+1)$
$d$ will divide $4(x^2+1)-4(x+1)(x-1)=8$
Now if $x$ is even $x^2+1$ is odd $\implies(x^2+1,x^2+4x+5)=1$
If $x$ odd, $x^2\equiv1\pmod8\iff x^2+1\equiv2\implies(x^2+1,x^2+4x+5)=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/697369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Fourier Series for $|\cos(x)|$ I'm having trouble figuring out the Fourier series of $|\cos(x)|$ from $-\pi$ to $\pi$.
I understand its an even function, so all the $b_n$s are $0$
$$a_0 = \frac 2 \pi \int_0^\pi |\cos(x)|\,dx = 0$$
$$a_n = \frac 2 \pi \int _0^\pi |\cos(x)| \cos(nx) \, dx = \frac 2 \pi \int_0^\pi \cos^2(x)\,dx.$$
since for all $j,k$ not equal the integral is zero.
so only $a_1$ remains. is this correct?
How would I evaluate $\sum_{n=1}^\infty (-1)^{n-1} /(4n^2 - 1)\ {}$?
| You must breakup the integral into three intervals:
$\left[-\pi \cdots -\frac{\pi}{2} \right]$,
$\left[-\frac{\pi}{2} \cdots \frac{\pi}{2} \right]$, and
$\left[\frac{\pi}{2} \cdots \pi \right]$
Which represent the regions where the sign of $\cos x$ changes.
\begin{equation*}
\left\vert \cos x\right\vert =
\begin{cases}
-\cos x & -\pi \le x \le -\frac{\pi}{2} \\
\cos x & \frac{\pi}{2} \le x \le \frac{\pi}{2} \\
-\cos x & \frac{\pi}{2} \le x \le \pi
\end{cases}
\end{equation*}
When I plugged the integral over the three regions into Maple I got:
\begin{align*}
a_n &=\frac{1}{2 \pi} \int\limits_{t=-\pi}^{\pi} \left\vert \cos(t) \right\vert \cos(nt) \\
&= \frac{1}{2 \pi} \int\limits_{t=-\pi}^{-\frac{\pi}{2}}(-\cos(t)) \cos(nt) \\
&+ \frac{1}{2 \pi} \int\limits_{t=-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(t) \cos(nt) \\
&+ \frac{1}{2 \pi} \int\limits_{t=\frac{\pi}{2}}^{\pi}(-\cos(t)) \cos(nt) \\
&=-4{\frac {\cos \left( \frac{1}{2} \pi n \right) }{ \left( -1+{n}^{2} \right) \pi }}
\end{align*}
Since, $\left\vert \cos t \right\vert$ is even you could break the integral in two and find $a_n$ as in Julián's answer.
\begin{align*}
a_n &=\frac{2}{\pi} \int\limits_{t=0}^{\pi} \left\vert \cos(t) \right\vert \cos(nt) \\
&= \frac{2}{\pi} \int\limits_{t=0}^{\frac{\pi}{2}}\cos(t) \cos(nt) \\
&+ \frac{2}{\pi} \int\limits_{t=\frac{\pi}{2}}^{\pi}(-\cos(t)) \cos(nt) \\
&=-4{\frac {\cos \left( \frac{1}{2} \pi n \right) }{ \left( -1+{n}^{2} \right) \pi }}
\end{align*}
for $1 < n$ and $a_0 = \frac{2}{\pi}$
My answer is simply an amplification of Julián's answer, but, I hope it helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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A formula for bit strings with no $k$ consecutive $1$'s using generating functions The following problem is taken from the book generatingfunctionology (P.28) by Herbert S. Wilf.
Let $f(n,m,k)$ be the number of strings of $n$ $\,0$'s and $1$'s that contain exactly $m\,$ $1$'s, no $k$ of which are consecutive.
(a) Find a recurrence formula for $f$. It should have $f(n,m,k)$ on the left side, and exactly three terms on the right.
(b) Find, in simple closed form, the generating functions
$$
F_k(x,y)=\sum_{n,m\ge0}f(n,m,k)x^ny^m \,\,\,\,\,\,\,\,\ (k=1,2,...).
$$
(c) Find an explicit formula for $f(n,m,k)$ from the genrating function (this should involve only a single summation, of an expression that involves a few factorials).
I was able to solve (a) and (b) and found that the generating functions is
$$
F_k = \frac{1-x^ky^k}{1-x-xy-x^{k+1}y^k}
$$
However, I cannot seem to be able to extract the sequence from this function. Any help will be appreciated.
Edit:
I've added the way I solved (a) and (b) to allow any criticism in case I got the function wrong.
My solution to (a) goes as follows:
Given $m,n,k$ such that $0\le k\le m \le n$ and $k\lt n$, we look at all of the bit string of length $n$ that contain exactly $m$ $1$'s, no $k$ of which are consecutive. We split these strings into two sets. The first composed of all the strings that end with $0$. This set has $f(n-1,m, k)$ strings since by removing the $0$ from the end of the string we obtain a one to one correspondence between the strings in our set and the strings $f(n-1,m, k)$ counts. The other set is composed of all the strings that end with $1$. By removing the last bit we obtain a one to one correspondence between our set and all strings counted by $f(n-1,m-1, k)$, except those that have $k-1$ consecutive $1$'s at the end. To count the latter we first note that since $k\lt n$ there must be another bit, which must be $0$, before the last $k-1$ $\,1$'s. By removing the last $k$ bits ($10\cdots0$) we obtain a one to one correspondence with the strings that are counted by $f(n-k-1,\,m-k,\,k)$. With that we obtain the recursion: $$f(n,m,k) = f(n-1,m,k) + f(n-1,m-1,k) - f(n-k-1,\,m-k,\,k)$$
In addition, we note the if $0\le m \le n$ and $k > m$, then $f(n,m,k) = \binom{n}m$
, since we can't have more consecutive $1$'s than there are $1$'s, so the problem is reduced to counting the number of strings with length $n$ that have $m$ $1$'s. For all other cases we write $f(n,m,k) = 0$ which completes the definition.
To Solve (b) I did the following:
$$
\begin{align}
F_k(x,y) & = \sum_{n,m\ge0}f(n,m,k) x^n y^m \\
& = \sum_{n=0}^{\infty}\sum_{m=0}^{n}f(n,m,k) x^n y^m \\
& = \sum_{n=0}^{k-1}\sum_{m=0}^{n}f(n,m,k) x^n y^m + \sum_{n=k}^{\infty}\sum_{m=0}^{k-1}f(n,m,k) x^n y^m + \sum_{n=k}^{\infty}\sum_{m=k}^{n}f(n,m,k) x^n y^m \\
& = \sum_{n=0}^{k-1}\sum_{m=0}^{n}\binom{n}m x^n y^m + \sum_{n=k}^{\infty}\sum_{m=0}^{k-1}\binom{n}m x^n y^m + \sum_{n=k}^{\infty}\sum_{m=k}^{n}f(n,m,k) x^n y^m
\end{align}
$$
By using the convention that $\binom{n}m = 0$ when $m\gt n$, and noting that when $m=n\ge k$ we have $f(n,m,k)=0$ we can write the above equation as follows:
$$
\begin{align}
& \sum_{m=0}^{k-1}\sum_{n=0}^{\infty}\binom{n}m x^n y^m + \sum_{n=k+1}^{\infty}\sum_{m=k}^{n-1}f(n,m,k) x^n y^m \\
=& \sum_{m=0}^{k-1} \frac{x^m}{(1-x)^{m+1}} y^m + \sum_{n=k+1}^{\infty}\sum_{m=k}^{n-1}f(n,m,k) x^n y^m \\
=& \frac{1}{1-x} \cdot \omega(x,y) + \sum_{n=k+1}^{\infty}\sum_{m=k}^{n-1}f(n,m,k) x^n y^m
\end{align}
$$
where: $$\omega(x,y) := \frac{1 - (\frac{xy}{1-x})^k}{1 - \frac{xy}{1-x}}.$$
Using the recusion formula obtained in (a) we can expand the second summand like so:
$$
\begin{align}
\sum_{n=k+1}^{\infty}\sum_{m=k}^{n-1}f(n,m,k) x^n y^m & = A + B - C
\end{align}
$$
where:
$$
\begin{align}
A & =\sum_{n=k+1}^{\infty}\sum_{m=k}^{n-1}f(n-1,m,k) x^n y^m \\
& = x\sum_{n=k}^{\infty}\sum_{m=k}^{n}f(n,m,k) x^n y^m \\
& = x\left(\sum_{n=0}^{\infty}\sum_{m=0}^{n}f(n,m,k) x^n y^m -
\sum_{n=0}^{k-1}\sum_{m=0}^{n}f(n,m,k) x^n y^m - \sum_{n=k}^{\infty}\sum_{m=0}^{k-1}f(n,m,k) x^n y^m \right)\\
& = x\left(F_k(x,y) -
\sum_{n=0}^{k-1}\sum_{m=0}^{n}\binom{n}m x^n y^m - \sum_{n=k}^{\infty}\sum_{m=0}^{k-1}\binom{n}m x^n y^m \right)\\
& = x\left(F_k(x,y) -
\sum_{m=0}^{k-1}\sum_{n=0}^{\infty}\binom{n}m x^n y^m\right)\\
& = xF_k(x,y) - \frac{x}{1-x} \cdot \omega(x,y)\\
\\
B & =\sum_{n=k+1}^{\infty}\sum_{m=k}^{n-1}f(n-1,m-1,k) x^n y^m \\
& = xy\sum_{n=k}^{\infty}\sum_{m=k-1}^{n-1}f(n,m,k) x^n y^m \\
& = xy\left(\sum_{n=0}^{\infty}\sum_{m=0}^{n}f(n,m,k) x^n y^m -
\sum_{n=0}^{k-1}\sum_{m=0}^{n}f(n,m,k) x^n y^m - \sum_{n=k}^{\infty}\sum_{m=0}^{k-2}f(n,m,k) x^n y^m \right)\\
& = xy\left(F_k(x,y) -
\sum_{n=0}^{k-1}\sum_{m=0}^{n}\binom{n}m x^n y^m - \sum_{n=k}^{\infty}\sum_{m=0}^{k-2}\binom{n}m x^n y^m \right)\\
& = xy\left(F_k(x,y) -
\sum_{m=0}^{k-2}\sum_{n=0}^{\infty}\binom{n}m x^n y^m - x^{k-1}y^{k-1}\right)\\
& = xyF_k(x,y) -
xy\sum_{m=0}^{k-2}\sum_{n=0}^{\infty}\binom{n}m x^n y^m - x^ky^k\\
& = xyF_k(x,y) -
xy\sum_{m=0}^{k-2}\frac{x^m}{(1-x)^{m+1}} y^m - x^ky^k\\
& = xyF_k(x,y) + 1 - \omega(x,y) - x^ky^k\\
\\
C & =\sum_{n=k+1}^{\infty}\sum_{m=k}^{n-1}f(n-k-1,\,m-k,\,k) x^n y^m = x^{k+1}y^kF_k(x,y)
\end{align}
$$
Summing all the values we get:
$$
F_k = \frac{1}{1-x} \cdot \omega + xF_k - \frac{x}{1-x} \cdot \omega + xyF_k + 1 - \omega - x^ky^k - x^{k+1}y^kF_k
$$
And so (note that $\omega$ disappears):
$$
F_k = \frac{1-x^ky^k}{1-x-xy-x^{k+1}y^k}
$$
| First of all, some unsolicited advice: your derivation for (b) need not be so messy. It greatly simplifies things to observe that
$$
f(n,m,k) = f(n-1,m,k) + f(n-1,m-1,k) - f(n-k-1,m-k,k) \tag{1}
$$
holds not just when $0 \le k \le m \le n$, $k < n$. In fact, it holds for any $n \ge 0, m \ge 0, k \ge 1$, except in the two cases $m = n = k$, and $m = n = 0$. This of courses assumes that $f(n,m,k) = 0$ if $n < 0$ or $m < 0$.
Proof:
*
*If $m > n$, then the LHS and RHS of (1) are both zero, so the equation holds.
*If $m \le n$ and $k > m$, then the LHS is ${n \choose m}$, while the RHS is
${n-1 \choose m} + {n-1 \choose m-1} + 0$, so (1) holds, unless $n = 0$ in which case we have the first exception $m = n = 0$.
*If $m \le n$ and $k \le m$ and $k < n$, then (1) holds by the counting argument you gave.
*If $m \le n$ and $k \le m$ and $k \ge n$, i.e. $k = m = n$, then the LHS is $0$ while the RHS is $0 + 1 - 0 = 1$, so we have the second exception.
This covers all cases. So then you can just write
\begin{align*}
F_k(x,y) & = \sum_{n,m\ge0}f(n,m,k) x^n y^m \\
& = 1 - x^k y^k \\
&\quad + \sum_{n,m\ge0}
\big[ f(n-1,m,k) + f(n-1,m-1,k) - f(n-k-1, m-k,k) \big] x^n y^m \\
&= 1 - x^k y^k + (x + xy - x^{k+1} y^k) F_k(x,y) \\
F_k(x,y) &=
\frac{1 - x^k y^k}{1 - x - xy + x^{k+1}{y^k}}.
\end{align*}
Then you avoid all the gross introduction of $\omega$, etc. ;)
For part (c)
Write as follows:
\begin{align*}
F_k(x,y)
&= \frac{1 - x^k y^k}{1 - x - xy + x^{k+1}{y^k}} \\
&= \frac{1 - x^k y^k}{1 - xy - x(1 - x^ky^k)} \\
&= \frac{1}{x} \left( \frac{x(1 - x^k y^k)}{1 - xy - x(1 - x^ky^k)} \right) \\
&= \frac{1}{x} \left( \frac{1 - xy - \left[1 - xy - x(1 - x^k y^k)\right]}{1 - xy - x(1 - x^ky^k)} \right) \\
&= \frac{1}{x} \left( \frac{1 - xy}{1 - xy - x(1 - x^ky^k)} - 1 \right) \\
&= \frac{1}{x} \left( \frac{1}{1 - x(1 + xy + x^2y^2 + \cdots + x^{k-1}y^{k-1})} - 1 \right) \\
\end{align*}
Now,
\begin{align*}
&\; \frac{1}{1 - x(1 + xy + x^2y^2 + \cdots + x^{k-1}y^{k-1})} \\
&= \sum_{N \ge 0} x^N \left(1 + xy + x^2y^2 + \cdots + x^{k-1}y^{k-1}\right)^N \\
\end{align*}
We want the coefficient of $x^{n+1}y^m$ in this series (because then this will multiply by $\frac{1}{x}$ to give us the coefficient of $x^n y^m$, $f(n,m,k)$.)
For the coefficient of $x^{n+1} y^m$, we need $N = n + 1 -m$. So we want the coefficient of $x^m y^m$ in
$$
\left(1 + xy + x^2y^2 + \cdots + x^{k-1}y^{k-1}\right)^{n+1-m}
=
\frac{(1 - x^k y^k)^{n+1-m}}{(1 - xy)^{n+1-m}}
$$
Now write out the series for $\frac{1}{(1 - xy)^{n+1-m}}$ explicitly, and expand $(1 - x^k y^k)^{n+1-m}$ with the binomial formula, and you should be able to find a finite summation yielding the coefficient of $x^{n+1} y^m$, which will equal $f(m,n,k)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/701923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find all $x$ and $y$ such that $10xy^2=-x^3+4x$ and $13x^2y=-2y^3+4y$
Find all $x$ and $y$ such that $10xy^2=-x^3+4x$ and $13x^2y=-2y^3+4y$.
Well $(0,0)$ is a solution. And we can simplify as:
$10y^2=-x^2+4$ and $13x^2=-2y^2+4$
this is equal to:
$10y^2+x^2-4=0$ and $13x^2+2y^2-4=0$
and therefore
$10y^2+x^2-4=13x^2+2y^2-4$
which gives
$8y^2-12x^2=0$
or
$2y^2-3x^2=0$
But I feel like I'm doing something wrong.
| $\displaystyle10xy^2=-x^3+4x\iff x(10y^2+x^2-4)=0$
If $\displaystyle x=0,0=-2y^3+4y\iff2y(y^2-2)=0\implies y=0$ or $y^2=2$
Else $10y^2+x^2-4=0\ \ \ \ (1)$
Similarly start from $\displaystyle13x^2y=-2y^3+4y\iff y(13x^2+2y^2-4)=0$
Finally solve for $x^2,y^2$ from $(1)$ and $13x^2+2y^2-4=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/704007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
$\epsilon$-$\delta$ limit proof that $\lim_{n\to \infty}\frac{n^2-n+2}{3n^2+2n-4}=\frac{1}{3}$ I need to prove that $$\lim_{n\to \infty}\frac{n^2-n+2}{3n^2+2n-4}=\frac{1}{3}$$ using the epsilon definition.
I'm having specific trouble understanding how to make it less than epsilon once I've simplified the equation.
| Note that
$$
\begin{align}
\left|\frac{n^2-n+2}{3n^2+2n-4} - \frac{1}{3}\right|&=
\frac{1}{3}\left|\frac{3n^2-3n+6}{3n^2+2n-4} - 1\right|\\&=
\left|\frac{-5n+10}{3(3n^2+2n-4)}\right|
\end{align}
$$
Given an $\epsilon>0$, we need to find an $n$ such that this expression is less than $\epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/704311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Primes $p$ such that $p^2$ divides $x^2 + y^2 + 1$ Call a prime $p$ awesome if there exist positive integers $x$ and $y$ such that $p^2$ divides $x^2+y^2+1$.
Observation: $2$ is not awesome, because $x^2+y^2+1\not\equiv 0$ (mod $4$). But $3$ is awesome, because $9$ divides $27=5^{2}+1^{2}+1$. So my question is:
Are there infinitely many awesome primes? Can we find all awesome primes?
Motivation: It is true that for every prime $p$, there exists positive integer $x$ and $y$ such that $p$ divides $x^2+y^2+1$. The proof can be found here. (This is actually a nice result; for example, it is used in a proof of Lagrange's $4$-square theorem).
If this is too trivial, what can we say if $p^2$ is replaced by $p^{k}$? :)
| Well time factor here is probably only 3.
Written in the ideal equation: $X^2+Y^2+1=3Z^3$
Solutions can be written using the following equation Pell. $p^2-2(k^2+k+1)s^2=1$
Then the solutions are of the form:
$Y=2p^2+2(k+2)ps-1$
$X=2p^2+2(2k+1)ps-1$
$Z=2p^2+2(k+1)ps-1$
$k$ - what some integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/705611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
$\prod_{k=0}^{\infty} \cos(x \cdot 2^{-k}).$ Task is to find $$\prod_{k=0}^{\infty} \cos(x \cdot 2^{-k}).$$
I tried to make it with double-angle formula:
$\prod_{k=0}^{\infty} \cos(x \cdot 2^{-k}) = \frac{\prod_{k=0}^\infty \sin(x2^{1-k})}{2^\infty \cdot \prod_{k=0}^\infty \sin(x \cdot 2^{-k})} $
I'm sad to admit, that I'm stuck with that one.
So I would appreciate any help, such as pointing the right direction.
| For the partial products we have
$$\begin{align}
\prod_{k = 0}^N \cos \frac{x}{2^k} &= \frac{\sin \frac{x}{2^N}\cos\frac{x}{2^N}}{\sin \frac{x}{2^N}} \prod_{k=0}^{N-1} \cos \frac{x}{2^k}\\
&= \frac{\sin \frac{x}{2^{N-1}}}{2\sin \frac{x}{2^N}} \prod_{k=0}^{N-1} \cos \frac{x}{2^k}\\
&= \frac{\sin \frac{x}{2^{N-1}}\cos \frac{x}{2^{N-1}}}{2\sin \frac{x}{2^N}} \prod_{k=0}^{N-2} \cos \frac{x}{2^k}\\
&= \frac{\sin \frac{x}{2^{N-2}}}{2^2\sin \frac{x}{2^N}} \prod_{k=0}^{N-2} \cos \frac{x}{2^k}\\
&\qquad\qquad \vdots\\
&= \frac{\sin \frac{x}{2^{N-N}}}{2^N\sin \frac{x}{2^N}}\cos \frac{x}{2^0}\\
&= \frac{\sin x\cos x}{2^N\sin (x\cdot 2^{-N})}.
\end{align}$$
From that, the limit is easily found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/706749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is the exact coefficient of $x^{12}$ in $(2+x^3+x^6)^{10}$? What is the coefficient of $x^{12}$ in $(2+x^3+x^6)^{10}$?
I figure you need to pick $x^3$ 4 times so $C(10,4)$...but what happens with the other numbers/variables???
Can someone explain to me how this is done properly?
Thanks.
EDIT:
$(x + y)^n = C(n,k) \cdot x^{n-k} \cdot y^k$
EX: Find the term for $x^5$ in $(5-2x)^8$
Answer: $C(8,5) \cdot (-2)^5 \cdot 5^3$
How can I use this info to solve a polynomial based question such as the featured?
Answer:
To sum up all information provided by everyone (Thanks!!!):
$(C(10,4) * 2^6) + (C(10,2) * 2^8) + (C(10,1) * C(9,2) + 2^7) = 71040 $
| In combinatorial terms, you're looking for the number of partitions of $12$ that use only $0, 3$, and $6$ and have $10$ parts. You then weight these partitions by multinomial coefficients. There are $3$ such partitions:
\begin{align*}
12 &= 6 + 6 + 0 + \cdots + 0\\
&= 6 + 3 + 3 + 0 + \cdots + 0\\
&= 3 + 3 + 3 + 3 + 0 + \cdots + 0 \, .
\end{align*}
These partitions correspond to the terms
\begin{align*}
x^6 * x^6 * 2^8 &= 2^8 x^{12}\\
x^6 * x^3 * x^3 * 2^7 &= 2^7 x^{12}\\
x^3 * x^3 * x^3 * x^3 * 2^6 &= 2^6 x^{12} \, .
\end{align*}
Now we observe that the number of ways each of these can be chosen is $\binom{10}{8,2}$, $\binom{10}{7,2,1}$, $\binom{10}{6,4}$ ways. So it looks to me that the coefficient of $x^{12}$ is
$$
\binom{10}{8,2} 2^8 + \binom{10}{7,2,1} 2^7 + \binom{10}{6,4} 2^6 \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/708897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
An integral identity for $\frac{x^{a-1}}{x^b-1}$ via. partial fractions Can somebody please confirm or correct the following? If $a$ and $b$ are both positive integers such that $a<b$ and $b$ is even then we can write
$$\frac{x^{a-1}}{x^{b}-1}=\frac{1}{b\left(x-1\right)}+\frac{\left(-1\right)^a}{b\left(x+1\right)}+\frac{2}{b}\sum\limits_{k=1}^{\frac{b-2}{2}}\frac{\cos\frac{2ak\pi}{b}\left(x-\cos\frac{2k\pi}{n}\right)-\sin^2\frac{2ak\pi}{b}}{x^2-2x\cos\frac{2k\pi}{b}+1},$$ and upon integration, I think we get
$$\int\frac{x^{a-1}}{x^{b}-1}dx=\frac{1}{b}\log\left(x-1\right)+\frac{\left(-1\right)^a}{b}\log\left(x+1\right)+\frac{1}{b}\sum\limits_{k=1}^{\frac{b-2}{2}}\cos\left(\frac{2ak\pi}{b}\right)\log\left(x^2-2x\cos\frac{2k\pi}{b}+1\right)-\frac
{2}{b}\sum\limits_{k=1}^{\frac{b-2}{2}}\frac{\sin^2\frac{2ak\pi}{b}}{\sin\frac{2k\pi}{b}}\tan^{-1}\frac{x-\cos\frac{2k\pi}{nb}}{\sin\frac{2k\pi}{b}}.$$
Context: I don't know my mistake at the moment, but I would like to get this instead $$\int\frac{x^{a-1}}{x^b-1}dx=\cdots-\frac{2}{b}\sum\limits_{k=1}^{\frac{b-2}{2}}\sin\left(\frac{2ak\pi}{b}\right)\tan^{-1}\frac{x-\cos\frac{2k\pi}{b}}{\sin\frac{2k\pi}{b}}.$$ That way, the formula could be used to derive another identity using the beta function which is my goal.
Sorry for length of the formula, I wasn't sure about appropriate question length.
| Mistake found
The rest of the partial fractions should have been found like so:
$$\frac{e^{\frac{2ak\pi}{b}i}}{x-e^{\frac{2k\pi}{b}i}}+\frac{e^{-\frac{2ak\pi}{b}i}}{x-e^{\frac{-2k\pi}{b}i}}=\frac{2\cos\frac{2ak\pi}{b}\left(x-\cos\frac{2k\pi}{b}\right)-2\sin\frac{2ak\pi}{b}\sin\frac{2k\pi}{b}}{x^2-2x\cos\frac{2k\pi}{b}+1},$$ which corrects the original formula for the fraction to
$$\frac{x^{a-1}}{x^{b}-1}=\frac{1}{b\left(x-1\right)}+\frac{\left(-1\right)^a}{b\left(x+1\right)}+\frac{2}{b}\sum\limits_{k=1}^{\frac{b-2}{2}}\frac{\cos\frac{2ak\pi}{b}\left(x-\cos\frac{2k\pi}{b}\right)-\sin\frac{2ak\pi}{b}\sin\frac{2k\pi}{b}}{x^2-2x\cos\frac{2k\pi}{b}+1},$$ which, upon integration, yields the formula
$$\int\frac{x^{a-1}}{x^{b}-1}dx=\frac{1}{b}\log\left(x-1\right)+\frac{\left(-1\right)^a}{b}\log\left(x+1\right)+\frac{1}{b}\sum\limits_{k=1}^{\frac{b-2}{2}}\cos\left(\frac{2ak\pi}{b}\right)\log\left(x^2-2x\cos\frac{2k\pi}{b}+1\right)-\frac
{2}{b}\sum\limits_{k=1}^{\frac{b-2}{2}}\sin\left(\frac{2ak\pi}{b}\right)\tan^{-1}\frac{x-\cos\frac{2k\pi}{b}}{\sin\frac{2k\pi}{b}}$$ which is the correct formula. The error was a simple error in expansion of the exponential terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/712078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
How to find the inverse of a symmetric tridiagonal Toeplitz matrix? Let $a, b>0$, and the matrix $A_{n \times n}$ and such
$$A=\begin{bmatrix}
a&b&0&\cdots&0&0\\
b&a&b&\cdots&0&0\\
0&b&a&\cdots&0&0\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
\cdots&\cdots&\cdots&b&a&b\\
\cdots&\cdots&\cdots&\cdots&b&a
\end{bmatrix}$$
Find the inverse $A^{-1}$.
My idea: $$A^{-1}=\dfrac{A^{*}}{|A|}$$
and let $|A|=D_{n}$, then we have
$$D_{n}=aD_{n-1}-b^2D_{n-2}$$
so
$$D_{n}=\begin{cases}
(n+1)\left(\dfrac{a}{2}\right)^n,a^2=4b^2\\
\dfrac{(a+\sqrt{a^2-4b^2})^{n+1}-(a-\sqrt{a^2-4b^2})^{n+1}}{2^{n-1}\sqrt{a^2-4b^2}}&a^2\neq 4b^2
\end{cases}$$
and then I fell very ugly, do you have other methods?
Thank you
| Notice that
$$A=\begin{bmatrix}
a&b&0&\cdots&0&0\\
b&a&b&\cdots&0&0\\
0&b&a&\cdots&0&0\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
\cdots&\cdots&\cdots&b&a&b\\
\cdots&\cdots&\cdots&\cdots&b&a
\end{bmatrix} = a \begin{bmatrix}
1&b/a&0&\cdots&0&0\\
b/a&1&b/a&\cdots&0&0\\
0&b/a&1&\cdots&0&0\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
\cdots&\cdots&\cdots&b/a&1&b/a\\
\cdots&\cdots&\cdots&\cdots&b/a&1
\end{bmatrix} =: aX.$$
So, $A^{-1} = \frac{1}{a} X^{-1}$.
To find $X^{-1}$, apply this answer. You can also find some references here.
In case you want to search further, your matrix is tridiagonal (a special kind of band) Toeplitz matrix.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I solve this question? Compute the value of the following improper integral. If it is divergent, type "Diverges" or "D".
$$\int_0^2 \frac{dx}{\sqrt{4-x^2}}$$
Do I make $u= 4-x^2$ then $du= -2x \, dx$
Not exactly sure..
| The limit of integration for which our integrand is undefined is $x = 2$. So, what we really need to compute is:
\begin{align}
\lim \limits_{t \to 2} \int_0^t\dfrac{dx}{\sqrt{4-x^2}}
\end{align}
First, let's worry about the integral. Consider the substitution $x=2\sin(\theta)$ (do a bit of trigonometry to see why this might be useful). So, $dx=2\cos(\theta)d\theta$. So, the integral we're dealing with is:
\begin{align}
\int \dfrac{2\cos(\theta)d\theta}{\sqrt{4-4\sin^2(\theta)}}\\
= \int \dfrac{2\cos(\theta)d\theta}{2\sqrt{1-\sin^2(\theta)}}\\
= \int \dfrac{\cos(\theta)d\theta}{\cos(\theta)}\\
= \int d\theta\\
= \theta + C
\end{align}
Now, we need to put things back in terms of $x$. Solving for $\theta$, we see that this is equal to $\sin^{-1}(\tfrac{x}{2}) + C$.
Now, back to what we were originally dealing with. We now have:
\begin{align}
\lim \limits_{t \to 2} \left(\sin^{-1}(\tfrac{x}{2}) \Big|_0^t\right)\\
= \sin^{-1}(\tfrac{2}{2}) - \sin^{-1}(\tfrac{0}{2})\\
= \tfrac{\pi}{2} - 0 = \boxed{\tfrac{\pi}{2}}
\end{align}
We now see that this improper integral converges to the value $\tfrac{\pi}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/716435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to divide polynomials? The title says it all : How to divide polynomials?
I don't understand the method below taught in my school.
Would any of you mind explaining it or even better, suggest an alternative way to solve this?
| This is an easier way to do polynomial division, at least I think it is easier.
Basically when we multiply polynomials (or any numbers really) we can sometimes do it in a table form, say if we wanted to multiply $x + 2$ and $x^2 + 3x + 4$ we might write a table like this:
$$
\\ \begin{array} {c|c|c|c}
multiply&x^2 & 3x &4 \\ \hline
x &x^3 &3x^2 &4x \\ \hline
2 &2x^2 & 6x & 8
\end{array}
$$
And from this we get that the product of $x + 2$ and $x^2 + 3x + 4$ is $x^3 + 5x^2 + 10x + 8$. Now we want to do this in reverse.
We know that the highest power term in the product ($2x^2$ in this case) must be the product of the highest powers in the factors. This might seem a bit strangely worded, but basically what I mean is that we know our quadratic (not trinomial) is the product of two linear (not binomial) factors, so our $2x^2$ term only comes from the product of the $x$ terms. So, we know when we do the table multiplication as above we have:
$$
\\ \begin{array} {c|c|c}
divide& * & \\ \hline
x &\color{Red}{2x^2} & \\ \hline
1 &** &
\end{array}
$$
And now we know what the entry in $*$ must be: $2x$, and then we get $**$ also so we have
$$
\\ \begin{array} {c|c|c}
divide& \color{Red}{2x} & ** \\ \hline
x &{2x^2} & * \\ \hline
1 & \color{Red}{2x}&
\end{array}
$$
And now, because in our product $2x^2 + 5x + 3$ we have $5x$, and the power of the term in $*$ is x, we must have that $2x + * = 5x$ so we get $ * = 3x$ so then $** = 3$ so our table is :
$$
\\ \begin{array} {c|c|c}
divide& {2x} & \color{red}{3} \\ \hline
x &{2x^2} & \color{red}{3x} \\ \hline
1 & {2x}& *
\end{array}
$$
And lastle we see * = 3 which is what we want, so we get that $(x+1)(2x+3) = 2x^2 + 5x + 3$ so $\frac{2x^2 + 5x + 3} {x + 1} = 2x + 3$.
This works in general for any order polynomial, and if one polynomial doesn't divide the other, you can easily get the remainder from this algorithm too.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Solve two equations for $a$ and $b$ \begin{cases}
c_2=\dfrac{c_1}{a} \left( \left(\dfrac{c_3}{b}\right)^3 - 1 \right) \\[2ex]
b^2 = a^2 + c_3^2 + 2(a)\, (c_3)\, (c_4) \\
\end{cases}
I am stuck at this point. Not sure on how to move forward. ( A small change made)
| The system \begin{cases}
c_2=\dfrac{c_1}{a} \left( \left(\dfrac{c_3}{b}\right)^3 - 1 \right) \\[2ex]
b^2 = a^2 + c_3^2 + 2ac_3c_4 \\
\end{cases}
is equivalent to
\begin{cases}
a=c_{1}\dfrac{c_{3}^{3}-x^{3}}{c_{2}x^{3}}\\[2ex]
b=x ,
\end{cases}
where $x$ is a solution of the following septic equation I've obtained in SWP:
\begin{eqnarray*}
0 &=&c_{2}^{2}x^{7}+c_{3}c_{2}^{2}x^{6}+\left(
-c_{1}^{2}+2c_{1}c_{3}c_{4}c_{2}\right) x^{5}+\left(
2c_{3}^{2}c_{1}c_{4}c_{2}-c_{3}c_{1}^{2}\right) x^{4} \\[2ex]
&&+\left( 2c_{3}^{3}c_{1}c_{4}c_{2}-c_{1}^{2}c_{3}^{2}\right)
x^{3}+c_{1}^{2}c_{3}^{3}x^{2}+c_{1}^{2}c_{3}^{4}x+c_{1}^{2}c_{3}^{5}.
\end{eqnarray*}
I am stuck at this point.
The general septic equation cannot be solved algebraically.
Note: In the present form of the system $a$ should be different from $0$. So $b=c_{3},a=0$
is no longer a solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
evaluation of $\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)dx$
Compute the indefinite integral
$$
\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\,dx
$$
My Attempt:
First, convert
$$
\frac{\cos x+\sin x}{\cos x-\sin x} = \frac{1+\tan x}{1-\tan x} = \tan \left(\frac{\pi}{4}+x\right)
$$
This changes the integral to
$$
\int \cos (2x)\cdot \ln \left(\tan \left(\frac{\pi}{4}+x\right)\right)\,dx
$$
Now let $t=\left(\frac{\pi}{4}+x\right)$ such that $dx = dt$. Then the integral with changed variables becomes
$$
\begin{align}
\int \cos \left(2t-\frac{\pi}{2}\right)\cdot \ln (\tan t)dt &= \int \sin (2t)\cdot \ln (\tan t)dt\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \frac{\sec^2(t)}{\tan t}\cdot \cos (2t)\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \cot (2t)dt\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\ln \left|\sin (2t)\right|
\end{align}
$$
where $t=\displaystyle \left(\frac{\pi}{4}+x\right)$.
Is this solution correct? Is there another method for finding the solution?
| \begin{array}{l}
\displaystyle \quad \int \cos 2 x \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right) d x
\\=\displaystyle \int \cos 2 x \ln \left(\tan \left(x+\frac{\pi}{4}\right)\right) d x \\
\displaystyle =\frac{1}{2} \int \ln \left(\tan \left(x+\frac{\pi}{4}\right)\right) d(\sin 2 x) \\
\displaystyle =\frac{1}{2} \sin 2 x \ln \left(\tan \left(x+\frac{\pi}{4}\right)\right)-\frac{1}{2} \int \frac{\sin 2 x \sec ^{2}\left(x+\frac{\pi}{4}\right) d x}{\tan \left(x+\frac{\pi}{4}\right)} \\
\displaystyle =\frac{1}{2} \sin 2 x \ln (\tan \left(x+\frac{\pi}{4}\right))-\frac{1}{2} \int \frac{\sin 2 x d x}{\sin \left(x+\frac{\pi}{4}\right) \cos \left(x+\frac{\pi}{4}\right)} \\
\displaystyle =\frac{1}{2} \sin 2 x \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)-\int \frac{\sin 2 x}{\sin \left(2 x+\frac{\pi}{2}\right)} d x\\\displaystyle =\frac{1}{2} \sin 2 x \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)+\frac{1}{2} \int \frac{d(\cos 2 x)}{\cos 2 x} \\
\displaystyle =\frac{1}{2} \sin 2 x \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)+\frac{1}{2} \ln |\cos 2 x|+C
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/718719",
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"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 1
} |
Proving inequality $x^xy^y \geq (\frac{x+y}{2})^{x+y}$ Prove that for all $x,y>0$ the following inequality $x^xy^y \geq (\frac{x+y}{2})^{x+y}$ is true.
It smells like Jensen inequality, but all I can get is that $\frac{x+y}{2}ln(x) + \frac{x+y}{2} ln(y) \geq xln(\frac{x+y}{2})+yln(\frac{x+y}{2})$
| This proof won't use Jensen. Multiply both sides by $x^y y^x$ and regroup:
$$
\begin{alignat*}{}
&\Leftrightarrow &\ x^{x+y} y^{x+y} &\geq{} \left(\frac{x+y}{2}\right)^{x+y} x^y y^x\\
&\Leftrightarrow &\ \left(\frac{2xy}{x+y}\right)^{x+y} &\geq{} x^y y^x \\
&\Leftrightarrow &\ \frac{2xy}{x+y} &\geq{} x^{y/(x+y)} y^{x/(x+y)}
\end{alignat*}
$$
Now apply weighted AM-GM inequality to get:
RHS $\leq \frac{xy}{x+y} + \frac{yx}{x+y} = \frac{2xy}{x+y} =$ LHS. Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/722722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Problem with divisibility Proof of the following without induction :
$13| 3^{n+2} + 4^{2n+1}$
for every $n \in \mathbb{Z}^+$
Any help is apprciated.
| First, rewrite as:
$$\begin{align}3^{n+2} + 4^{2n+1} &= 9\cdot3^n + 4\cdot4^{2n}\\
&= 9\cdot3^n + 4\cdot16^{n}\end{align}$$
Then, use modular arithmetic to prove divisibility by $13$:
$$\begin{align}9\cdot3^n + 4\cdot16^{n}&= 9\cdot3^n + 4\cdot(13+3)^n\\
&\equiv 9\cdot3^n + 4\cdot3^n\pmod{13}\\
&= 3^n\cdot(9 + 4)\pmod{13}\\
&= 3^n\cdot13\pmod{13}\\
&\equiv 0 \pmod{13}\end{align}$$
Hence, $3^{n+2} + 4^{2n+1} \equiv 0 \pmod{13}$, i.e. $13 | (3^{n+2} + 4^{2n+1})$ for all non-negative integers $n$.
| {
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"source": "stackexchange",
"question_score": "1",
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How can I prove this combinatorial identity without using Wilf-Zeilberger? I am trying to prove the following identity without using W-Z algorithm:
\begin{equation}
\sum\limits_{j=0}^n (-1)^{(n-j)} \frac{(n+j+1)!}{(n-j)!(j)!(j+1)!} = 1
\end{equation}
| As observed in the comments, your sum is equal to
$$
\sum_{j=0}^n (-1)^{n-j} \binom{n}{n-j} \binom{n+j+1}{j+1} \, .
$$
Given a (formal) power series $f(x)$, let $[x^m] f(x)$ denote the coefficient of $x^m$. Note that
$$
(1 - x)^n = \sum_{j \geq 0} (-1)^j \binom{n}{j} x^j
$$
and
$$
\frac{1}{(1 - x)^{n+1}} = \sum_{j \geq 0} \binom{n+j}{j} x^j \, .
$$
Then
$$
\sum_{j \geq 0} \binom{n+j+1}{j+1} x^j = \frac{1}{x}\left(\frac{1}{(1 - x)^{n+1}} - 1\right) \, .
$$
By the convolution formula for ordinary generating functions, then
\begin{align*}
\sum_{j=0}^n (-1)^{n-j} \binom{n}{n-j} \binom{n+j+1}{j+1} &= [x^n] \left(\sum_{j \geq 0} (-1)^j \binom{n}{j} x^j \right)\left(\sum_{j \geq 0} \binom{n+j+1}{j+1} x^j\right)\\
&= [x^n] (1 - x)^n \cdot \frac{1}{x}\left(\frac{1}{(1 - x)^{n+1}} - 1\right)\\
&= [x^{n+1}] \left(\frac{1}{1 - x} - (1 - x)^n \right) = 1 - (-1)^{n+1} \binom{n}{n+1}\\
&= 1
\end{align*}
as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find constants of function I have this equality :
$$f(x)=\frac{9}{(x-1)(x+2)^2}$$
I am required to find the constants A, B and C so that,
$$f(x) = \frac{A}{(x-1)} + \frac{B}{(x+2)} + \frac{C}{(x+2)^{2}} $$
How do we go about solving such a question?
I am not sure on how to solve such questions. Approach and Hints to solve these kinds of questions are welcomed. :)
Thank you!
Edit:
Wow, saw all the answers! Didn't know that there were a lot of different ways to solve this question. Math is such a fascinating thing!
| $$\dfrac{9}{(x-1)(x+2)^2}=\dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}$$
Multiplying by the GCD(which means multiplying all the terms by $(x-1)(x+2)^2$)
$$9=A(x+2)^2+B(x-1)(x+2)+C(x-1)$$
Let $x=-2$;
$$9=A(-2+2)^2+B(-2-1)(-2+2)+C(-2-1)$$
$$9=A(0)+B(-3)(0)+C(-3)$$
$$9=-3C$$
$$C=-3$$
Now that we have C,we let x=1;
$$9=A(1+2)^2+B(1-1)(1+2)+C(1-1)$$
$$9=A(3)^2+B(0)(3)+C(0)$$
$$9=9A$$
$$\implies A=1$$
then we let x=3 and substitute back A and C
$$9=(3+2)^2+B(2)(5)-3(2)$$
$$9=25+10B-5$$
$$9=20+10B$$
$$-11=10B$$
$$B=\dfrac{-11}{10}$$
So,now you have the values of A,B and C. This is how you go about doing it. Try it yourself. There might be a small arithmetic error in this solution. Solve it yourself to find it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
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} |
Find the value of $k$ such that $p(x)= kx^3 + 4x^2 + 3x - 4$ and $q(x)= x^3 - 4x + k$ , leave the same remainder when divided by $(x – 3)$. $p(x)= kx^3 + 4x^2 + 3x - 4$ and $q(x)= x^3 - 4x + k$ , leave the same remainder when divided by $(x – 3)$.
(a) -1 (b) 1 (c) 2 (d) -2
I am getting the value of k: $-17/29$ after equating the remainders.
$p(x)= kx^3 + 4x^2 + 3x - 4/(x – 3)$: remainder= $(30k+32)$
$q(x)= x^3 - 4x + k/(x – 3)$: remainder= $(k+15)$
So,
$30k+32=k+15$
=> $k=(-17/29)$
Any help would be appreciated. :)
| Find the value of $k$ if $x^3-4x^2+5x-k$ is completely divisible by $x-4$.
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Cesàro sum of $\sum\limits_{n = 0}^\infty {\cos n}=\dfrac{1}{2}$. Please check my work Thanks to the formula
http://functions.wolfram.com/ElementaryFunctions/Cos/23/01/0001/
Partial sums
$$s_m=\sum\limits_{k = 0}^m {\cos k = } \frac{{\sin \left( {\frac{1}{2}\left( {m + 1} \right)} \right)\cos \frac{m}{2}}}{{\sin \frac{1}{2}}}+1$$
some rearrangement
$$\frac{{\frac{1}{2}\left( {\sin \left( {\frac{m}{2} + \frac{1}{2} + \frac{m}{2}} \right) + \sin \left( {\frac{m}{2} + \frac{1}{2} - \frac{m}{2}} \right)} \right)}}{{\sin \frac{1}{2}}} = \frac{{\frac{1}{2}\left( {\sin \left( {m + \frac{1}{2}} \right) + \sin \frac{1}{2}} \right)}}{{\sin \frac{1}{2}}}$$
and we get
$$s_m=\sum\limits_{k = 0}^m {\cos k = } \frac{{\sin \left( {m + \frac{1}{2}} \right)}}{{2\sin \frac{1}{2}}} + \frac{1}{2}$$
Let's now define
$${a_n} = \sum\limits_{m = 0}^n {\left[ {\frac{{\sin \left( {m + \frac{1}{2}} \right)}}{{2\sin \frac{1}{2}}} + \frac{1}{2}} \right]}$$
which is
$${a_n} = \frac{1}{{2\sin \frac{1}{2}}}\sum\limits_{m = 0}^n {\sin \left( {m + \frac{1}{2}} \right)} + \sum\limits_{m = 0}^n {\frac{1}{2}} $$
using the formula that I found here
http://functions.wolfram.com/ElementaryFunctions/Sin/23/01/0003/
$$\sum\limits_{m = 0}^n {\sin \left( {m + \frac{1}{2}} \right)} = \frac{{{{\sin }^2}\frac{{n + 1}}{2}}}{{\sin \frac{1}{2}}} = \frac{{1 - \cos \left( {n + 1} \right)}}{{2\sin \frac{1}{2}}}$$
I got
$${a_n} = \frac{{1 - \cos \left( {n + 1} \right)}}{{{{\left( {2\sin \frac{1}{2}} \right)}^2}}} + \frac{{\left( {n + 1} \right)}}{2} = \frac{{1 - \cos \left( {n + 1} \right)}}{{2\left( {1 - \cos 1} \right)}} + \frac{{\left( {n + 1} \right)}}{2}$$
and finally
$${a_n} = \frac{{\cos \left( {n + 1} \right) + \left( {n + 1} \right)\cos 1 - n - 2}}{{2\left( {\cos 1 - 1} \right)}}$$
Cesàro sum is the following limit
$$
\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{n} = \mathop {\lim }\limits_{n \to \infty } \frac{{\cos \left( {n + 1} \right) + n\left( {\cos 1 - 1} \right) + \cos 1 - 2}}{{2\left( {\cos 1 - 1} \right)}} $$
split into the sum of two limits
$$\mathop {\lim }\limits_{n \to \infty } \frac{{\cos \left( {n + 1} \right) + \cos 1 - 2}}{{2n\left( {\cos 1 - 1} \right)}} + \mathop {\lim }\limits_{n \to \infty } \frac{{n\left( {\cos 1 - 1} \right)}}{{2n\left( {\cos 1 - 1} \right)}} $$
the first is $0$ because numerator is limited and denominator goes to infinity, the second is $\dfrac{1}{2}$
therefore Cesàro sum is
$$\sum\limits_{n = 0}^\infty {\cos n = } \frac{1}{2}$$
Thank you in advance for your attention
Any comment would be greatly appreciated
| I haven't check your work, but it coincides with other results. Particularly, it is what you get when you do Ramanujan's summation of the series:
Let $f(x)=\cos x$. Then the various expressions for Ramanujan's summation are:
$$\sum _{x\ge0}f(x)=\sum _{n=1}^{\infty }{\frac {f^{(n-1)}(0)}{n!}}B_{n}(1)$$
$$\sum _{x\ge0}f(x)=-\sum _{k=1}^{\infty }{\frac {\Delta ^{k-1}f(x)}{k!}}(-1)_{k}$$
$$\sum _{x\ge0}f(x)=\int _{0}^{x}f(t)dt-{\frac {1}{2}}f(x)+\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}f^{(2k-1)}(1)$$
$$\sum _{x\ge0}f(x)=\int _{0}^{x}f(t)dt+\sum _{k=1}^{\infty }{\frac {c_{k}\Delta ^{k-1}f(1)}{k!}}$$
(where $c_{k}=\int _{0}^{1}{\frac {\Gamma (x+1)}{\Gamma (x-k+1)}}dx$)
In all these cases you also get 1/2.
| {
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"url": "https://math.stackexchange.com/questions/733766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Given the length of two altitudes and one side , find the area of triangle. Segments $BE$ and $CF$ are the altitudes in $\triangle ABC$.
$E$ is on line $AC$ and $F$ is on line $AB$.
$BC = 65$, $BE = 60$ and $CF = 56$.
Find $A(\triangle ABC)/100$.
By the Pythagorean theorem , $CE=25$ , and $BF= 33$.
If the length of altitude from $A$ to B$C$ can be calculated then the area of $\triangle ABC$ can be calculated since the length of $BC$ is known.
But I'm stuck here , so any hints are apreciated .
| We have that $\cot C=25/60$ and $\cot B=33/56$.
Now use the area formula:
$$\text{Area}=\frac{a^2}{2(\cot B + \cot C)}$$
Thus:
$$\frac{(ABC)}{100}=\frac{65^2}{200(\frac{25}{60} + \frac{33}{56})}=\frac{13^2}{8(\frac{25}{60} + \frac{33}{56})}=\frac{13^2}{\frac{10}{3} + \frac{33}{7}}=\frac{13^2}{\frac{70+99}{21}}=21$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove the ring $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ has inverse and is a field How can I prove that $\frac{1}{a+b\sqrt[3]{2}+c\sqrt[3]{4}}$ is of the form $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ (i.e. that $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ is a field) for all rational $a,b,c$ and $a+b\sqrt[3]{2}+c\sqrt[3]{4} \neq 0$?
| This is very standard. Let $\phi(a,b,c)=a+b\sqrt[3]{2}+c\sqrt[3]{4}$. Then
you have
$$
\phi(a,b,c)+\phi(a',b',c')=\phi(a+a',b+b',c+c')
$$
$$
\phi(a,b,c)\phi(a',b',c')=\phi(aa'+2bc'+2cb',ab'+ba'+2cc',ac'+bb'+ca')
$$
$$
\phi(a,b,c)^{-1}=\phi(\frac{a^2-2bc}{D},\frac{2c^2-ab}{D},\frac{a^2-2bc}{D}), \
D=a^3+2b^3+4c^3-6abc
$$
Note that $a^3+2b^3+4c^3-6abc \neq 0$ whenever one of $a,b$ or $c$ is nonzero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/734636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $a^n - b^n + c^n - d^n \ge (a-b+c-d)^n$ Following on from an earlier question, and in search of a conceptual insight, I asked myself:
Given real numbers $a \ge b \ge c \ge d \ge 0 \tag{1}$
Prove that
$a^n - b^n + c^n - d^n \ge (a - b + c - d)^n \text{ for all } \underline{n \in \mathbb{R}} \tag{2}$
First, I wrote a toy program to do a quick test, and it hinted that the conjecture may indeed be true as long as $n \ge 1 \tag{3}$
Next, I proceeded to prove it for $n \in \mathbb{N}$ by using a simple substitution technique given in this answer. $\begin{aligned}x = a-b && y = b-c && z = c-d\end{aligned} \tag{S}$
Transforms $(1)$ into: $x, y, z, d \ge 0 \tag{4}$
Transforms $(2)$ into: $(x+y+z+d)^n - (y+z+d)^n + (z+d)^n - d^n \ge (x+z)^n \tag{5}$
which is true since:
$\displaystyle \begin{aligned} & (x+y+z+d)^n - (y+z+d)^n + (z+d)^n - d^n - (x+z)^n \\ = & \sum_{r=1}^n \binom{n}{r}(x+z)^{n-r}(y+d)^r - \sum_{r=1}^n \binom{n}{r}z^{n-r}(y+d)^r + \sum_{r=1}^{n-1}d^{n-r}z^r \\= &\sum_{r=1}^n \binom{n}{r}(y+d)^r\left((x+z)^{n-r} - z^{n-r} \right)+ \sum_{r=1}^{n-1}d^{n-r}z^r \\= & \sum_{r=1}^n \binom{n}{r}(y+d)^r\left(\sum_{q=1}^{n-r} \binom{n-r}{q}x^q z^{n-r-q}\right)+ \sum_{r=1}^{n-1}d^{n-r}z^r \\ =& \text{sum of non-negative terms}\\\ge & 0\end{aligned}$
And this is where I got stuck, as I don't know how to prove it for any real $n \ge 1$.
I would appreciate any help in this direction.
| As you already noted, the problem is equivalent to proving
$$(a+b+c+d)^\alpha-(b+c+d)^\alpha+(c+d)^\alpha-d^\alpha\ge(a+c)^\alpha$$
for $a,b,c,d\ge0$.
Lemma: If $\alpha\ge1$, $x\ge y\ge0$ and $r\ge0$, then $(x+r)^\alpha-(y+r)^\alpha\ge x^\alpha-y^\alpha$.
Proof: It suffices to show that $(x+r)^\alpha-x^\alpha$ is an increasing function in $x$. Taking the derivative we have that $\alpha(x+r)^{\alpha-1}\!-\alpha x^{\alpha-1}=\alpha((x+r)^{\alpha-1}\!-x^{\alpha-1})\ge0$.
Therefore
$$(a+c+(d+b))^\alpha\!-\!(c+(d+b))^\alpha+(c+d)^\alpha\!-\!(0+d)^\alpha\ge(a+c)^\alpha\!-\!c^\alpha+c^\alpha\!-\!0^\alpha=(a+c)^\alpha$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Inequality with trigonometric functions Find all values for $a$ such that the following inequality holds:
$$\sin^6x + \cos^6x + a\sin x \cos x \ge 0$$
To be fair, I didn't manage to get anything helpful wiht my calculations. I tried to cube the basic trigonometric identity $\sin^2x + \cos^2x = 1$ in order to get a replacement for $\sin^6x + \cos^6x$. Also I tried to check all four quadrants separately, but again I didn't manage to get something.
| It is enough to find the minimum value of
$$
f_a(x):=\cos^6x+\sin^6x+a\cos x\sin x
$$
We have
\begin{eqnarray}
f_a(x)&=&\cos^6x+\sin^6x+a\cos x\sin x\\
&=&(\cos^2x+\sin^2x)^3-3(\cos^4x\sin^2x+\cos^2x\sin^4x)+a\cos x\sin x\\
&=&1-3\cos^2x\sin^2x(\cos^2x+\sin^2x)+a\cos x\sin x\\
&=&1-3\cos^2\sin^2x+a\cos x\sin x\\
&=&1+\frac{a}{2}\sin(2x)-\frac{3}{4}\sin^2(2x)\\
&=&P_a(\sin(2x)),
\end{eqnarray}
where
$$
P_a(t)=1+\frac{a}{2}t-\frac{3}{4}t^2, t \in [-1,1]
$$
It follows that
\begin{eqnarray}
\min_{x\in \mathbb{R}}f_a(x)&=&\min_{t\in [-1,1]}P_a(t)=\min P_a([-1,1])=\min\{\frac{1-2a}{4},\frac{1+2a}{4}\}\\
&=&\frac14\min\{1-2a,1+2a\}=\frac{1-2a+1+2a-|1-2a-(1+2a)|}{8}\\
&=&\frac{2-4|a|}{8}=\frac{1-2|a|}{4}.
\end{eqnarray}
We have
$$
f_a \ge 0 \iff \min f_a \ge 0 \iff 1-2|a|\ge 0 \iff |a|\le \frac12,
$$
i.e. $a \in [-1/2,1/2]$.
| {
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How to compute $\int_0^a \sin(\tan^{-1}(b\sin\theta))\ d\theta$ How to compute
$$\int_0^a \sin(\tan^{-1}(b\sin\theta))\ d\theta$$
I've tried to rewrite it as
$$\int_0^a\frac{b\sin\theta}{\sqrt{b^2\sin^2\theta+1}}d\theta$$
But I'm still stuck here.
| The change of variables
$$
\cos\theta=\frac{b+1}{b}x
$$
gives
$$
\int\frac{b\sin\theta}{\sqrt{b^2\sin^2\theta+1}}d\theta=-\sqrt{b^2+1}\int\frac{dx}{\sqrt{1-x^2}}.
$$
Let's check it:
$$
\sin\theta\,d\theta=-\frac{b+1}{b}dx,
$$
$$
b^2\sin^2\theta+1=b^2(1-\cos^2\theta)+1=b^2(1-\frac{(b+1)^2}{b^2}x^2)+1=(b^2+1)(1-x^2).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/745467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the fuction $g$. If $f:x \mapsto x^2 + 3$, find function $g$ such that $gf:x \mapsto 2x^2 + 3$.
I don't know how to do it, there is no such example in my book.
Help?
| $$f(x)=x^2+3,\quad g(f(x))=2x^2+3\\
\implies g(x^2+3)=2x^2+3=2(x^2+3)-3\\
\implies g(x)\text{ could be }g(x):=2x-3$$
If $gf=g(x)\times f(x)$, then
$$g(x)\times f(x)=2x^2+3\\
\implies g(x)=\dfrac{2x^2+3}{x^2+3}=1+\dfrac{x^2}{x^2+3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/747179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show that if $p$ is a prime number $> 3$ then $24 \mid p^2-1$ Hi guys can someone help me with this ?(Without using Modular arithmetic)
Show that if $p$ is a prime number $>3$ then $24$ $\mid$ $p^2-1$
| If $p$ is a prime greater than $3$, than it is equal to $1, 3, 5, 7$ $\pmod 8$
THis implies $p^2 \equiv 1 \pmod 8$ (because $1^2 \equiv 1$, $3^2 \equiv 1, 5^2 \equiv 1, 7^2 \equiv 1 \pmod 8$)
Also, $p \equiv 1, 2 \pmod 3 \Rightarrow p^2 \equiv 1 \pmod 3$
THis implies that $p^2 - 1 $ is divisible by both $8$ and $3$, and so it is divisible by $24$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/749605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How prove this $\cos{x}+\cos{y}+\cos{z}=1$ Question:
let $x,y,z\in R$ and such $x+y+z=\pi$,and such
$$\tan{\dfrac{y+z-x}{4}}+\tan{\dfrac{x+z-y}{4}}+\tan{\dfrac{x+y-z}{4}}=1$$
show that
$$\cos{x}+\cos{y}+\cos{z}=1$$
My idea: let $$x+y-z=a,x+z-y=b,y+z-x=c$$
then
$$a+b+c=\pi$$
and
$$\tan{\dfrac{a}{4}}+\tan{\dfrac{b}{4}}+\tan{\dfrac{c}{4}}=1$$
we only prove
$$\cos{\dfrac{b+c}{2}}+\cos{\dfrac{a+c}{2}}+\cos{\dfrac{a+b}{2}}=1$$
Use
$$\cos{\dfrac{\pi-x}{2}}=\sin{\dfrac{x}{2}}$$
$$\Longleftrightarrow \sin{\dfrac{a}{2}}+\sin{\dfrac{b}{2}}+\sin{\dfrac{c}{2}}=1$$
let
$$\tan{\dfrac{a}{4}}=A,\tan{\dfrac{b}{4}}=B,\tan{\dfrac{\pi}{4}}=C$$
then
$$A+B+C=1$$
and use $$\sin{2x}=\dfrac{2\tan{x}}{1+\tan^2{x}}$$
so we only prove
$$\dfrac{2A}{1+A^2}+\dfrac{2B}{1+B^2}+\dfrac{2C}{1+C^2}=1$$
other idea:let
$$\dfrac{y+z-x}{4}=a,\dfrac{x+z-y}{4}=b,\dfrac{x+y-z}{4}=c$$
then we have
$$a+b+c=\dfrac{\pi}{4},\tan{a}+\tan{b}+\tan{c}=1$$
we only prove
$$\cos{(2(b+c)}+\cos{2(a+c)}+\cos{2(a+b)}=\sin{(2a)}+\sin{(2b)}+\sin{(2c)}=1$$
then I fell very ugly, can you some can help?
Thank you very much!
|
An interesting view of the surface $ \cos x + \cos y + \cos z = 1 $ Three lines ( in red ) are added for clarity. We can see that even along the plane $ x + y + z = \pi $ we must add a network of lines to encompass the simple periodicity condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/749758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How find this system $a^2+b^2=3,a^2+c^2+ac=4,b^2+c^2+\sqrt{3}bc=7$ Find the this system real solution
$$\begin{cases}
a^2+b^2=3\\
a^2+c^2+ac=4\\
b^2+c^2+\sqrt{3}bc=7
\end{cases}$$
I think that one can use Geometry to solve this system. Maybe there exist an algebraic method.
$$a^2+b^2=\sqrt{3}^2$$
$$a^2+c^2-2ac\cos{(120°)}=2^2$$
$$b^2+c^2-2bc\cos{150°}=\sqrt{7}$$
and note $$150°+120°+90°=360°$$
This is Mathematics competition of Zhejiang province today.
| Here is an algebraic solution that yields all the real answers.
Let us introduce the complex numbers:
$$x=\frac{-b+i a}{\sqrt{3}},\quad y=\frac{2a+c}{4}-i\frac{\sqrt{3}}{4}c.$$
The first two equations are equivalent to the statement: $\vert x\vert=\vert y\vert=1$, and the third equation tells us that
$$\vert \sqrt{3} x-2i y\vert^2=\left\vert b+\frac{\sqrt{3}+i}{2} c\right\vert^2
=b^2+c^2+\sqrt{3}bc=7$$
On the other hand, since $\vert x\vert=\vert y\vert=1$ we see that
$\vert \sqrt{3} x-2i y\vert^2=3+4+4\sqrt{3}\,\Re(ix\bar{y})$. Thus, we have
$\Re(ix\bar{y})=0$ that is $x\bar{y}$ is a real number of modulus $1$, or $y=\pm x$.
*
*If $y=x$, we conclude comparing real parts, and imaginary parts that
$$b=-\frac{1}{2\sqrt{3}}a,\qquad c=-\frac{4}{3}a,$$
and replacing in the first equation we get $b=\epsilon\sqrt{3/13}$ with $\epsilon\in\{+1,-1\}$, and this yields the couple of solutions:
$$
a=-\epsilon\frac{6}{\sqrt{13}},~b=\epsilon\frac{\sqrt{3}}{\sqrt{13}},~
c=\epsilon\frac{ 8}{\sqrt{13}},\quad\hbox{with }\epsilon\in\{+1,-1\}$$
*If $y=-x$, we conclude similarly that
$$b=\frac{5}{2\sqrt{3}}a,\qquad c=\frac{4}{3}a,$$
and Replacing in the first equation we obtain the second couple of solutions:
$$
a= \epsilon\frac{6}{\sqrt{37}},~b=\epsilon\frac{5\sqrt{3}}{\sqrt{37}},~
c=\epsilon\frac{ 8}{\sqrt{37}},\quad\hbox{with }\epsilon\in\{+1,-1\}$$
and we are done, with four solutions in total.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/751807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
If all eigenvalues are 1 or -1, is then $A^{12}=I$? True or false: If all the eigenvalues of A are either $\lambda=1$ or $\lambda = -1$ then $A^{12}$= I
If we have a matrix $$\mathbf A = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$$ this has eigenvalues of 1 or -1 then $$\mathbf A = \begin{pmatrix}\lambda-1&0\\0&\lambda+1\end{pmatrix}$$ and if we plug in $\lambda=1 $ $$\mathbf A = \begin{pmatrix}0&0\\0&2\end{pmatrix}$$ So $A^{12}$ $\neq$ I
Is this a correct way of proving this is false?
| Compute powers of the matrix $A$, where
$$A=\begin{pmatrix} 1 & 1\\0&1\end{pmatrix}.$$
You will find that
$$A^n=\begin{pmatrix} 1 & n\\0&1\end{pmatrix}.$$
Remark: If a $2\times 2$ matrix $A$ has both $1$ and $-1$ as eigenvalues, then we will have $A^2=I$. But we can find a $3\times 3$ matrix that has both $1$ and $-1$ as eigenvalues such that $A^n\ne I$ for all positive integers $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/752613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Integral $\frac{1}{\pi}\int_0^{\pi/3}\log\big( \mu(\theta)+\sqrt{\mu^2(\theta)-1} \big)\ d\theta, \quad \mu(\theta)=\frac{1+2\cos\theta}{2}.$ Hi I am trying to calculate this integral:
$$
I=\frac{1}{\pi}\int_0^{\pi/3}\log\left( \frac{1+2\cos\theta}{2}+\sqrt{\bigg( \frac{1+2\cos\theta}{2} \bigg)^2-1} \right)\ d\theta.
$$
The integral evaluation is related to Mahler measures. You may also recognize that it is related to the evaluation of log-sine integrals at $\dfrac{\pi}{3}$.
This integral $I$ is somewhat related to
$$
\int_0^1\log\big|2a+2b\cos (2\pi \theta)\big|\ d\theta=\log \big( |a|+\sqrt{a^2-b^2} \big),
$$
for $a,b\in\mathbb{R}$ with $|a|\geq|b|> 0$. This can be seen in Gradstein and Ryzhik's tables of integrals, but I am not sure how to use that to help me to solve $I$.
Thanks!
| The integral can be expressed in terms of a series of Gauss hypergeometric functions. It is doubtfull that it would be possible to go further on this way.
$I = \frac{1}{\pi}\int_0^{\pi/3}\ln\left(\mu(\theta)+\sqrt{\mu^2(\theta)-1}\right)d\theta = \frac{1}{\pi}\int_0^{\pi/3}\cosh^{-1}(\mu(\theta))$, where $\mu(\theta) = \frac{1}{2}+\cos(\theta)$.
$$\cosh^{-1}(\mu)=\sum_{k=0}^\infty \frac{(-1)^k\Gamma(k+1/2)}{2^{k-1/2}(2k+1)k!\sqrt{\pi}}(\mu-1)^{k+1/2}$$
so
$$\mu(\theta)=\frac{1}{2}+\cos(\theta)\to I = \frac{1}{\pi^{3/2}}\sum_{k=0}^\infty \frac{(-1)^k\Gamma(k+1/2)}{2^{k-1/2}(2k+1)k!}(\cos(\theta)-1/2)^{k+1/2}$$
Let $I_k = \int(cos(\theta) - 1/2)^{k+1/2}d\theta$. Then
$$I_k = -\frac{1}{(2k+3)2^k}\sqrt{\frac{2}{3}}(2\cos(\theta)-1)^{k+3/2}F_1\left(k+\frac{3}{2};\frac{1}{2},\frac{1}{2};k+\frac{5}{2};\frac{1}{3}(1-2\cos(\theta)),2\cos(\theta)-1\right)+C$$
where $C$ is some constant and $F_1$ is the Appell hypergeometric function of two variables.
$2\cos(0)-1=1$, $2\cos(\pi/3)-1=0$, and $F_1(a;b,b;c;-1/3,1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}_2F_1(a,b;c-b;-1/3)$, so
$$I_k=\frac{1}{(2k+3)2^k}\sqrt{\frac{2}{3}}\frac{\Gamma(k+5/2)\Gamma(1/2)}{\Gamma(1)\Gamma(k+2)} {}_2F_1\left(k+\frac{3}{2},\frac{1}{2};k+2;-\frac{1}{3}\right)$$
$$I_k=\sqrt{\frac{2\pi}{3}}\frac{(2k+1)\Gamma(k+1/2)}{2^{k+2}(k+1)!} {}_2F_1\left(k+\frac{3}{2},\frac{1}{2};k+2;-\frac{1}{3}\right)$$
Finally, from $I = \frac{1}{\pi^{3/2}}\sum_{k=0}^\infty \frac{(-1)^k\Gamma(k+1/2)}{2^{k-1/2}(2k+1)k!}I_k$, we have
$$\frac{1}{\pi}\int_0^{\pi/3}\ln\left(\mu(\theta)+\sqrt{\mu^2(\theta)-1}\right)d\theta=\boxed{\frac{1}{\pi\sqrt{3}}\sum_{k=0}^\infty \frac{(-1)^k(\Gamma(k+1/2))^2}{2^{2k+1}(2k+1)k!(k+1)!} {}_2F_1\left(k+\frac{3}{2},\frac{1}{2};k+2;-\frac{1}{3}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/755477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
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} |
Find the first two iteration of the Jacobi method for the following linear system, using $x^{(0)} = 0$ $$3x_{1} - x_{2} + x_{3} = 1,$$
$$3x_{1} + 6x_{2} + 2x_{3} = 0,$$
$$3x_{1}+3x_{2}+7x_{3} = 4$$
So, from this I got
T = \begin{bmatrix}
0 & \frac{-1}{3} & \frac{1}{3} \\[0.3em]
\frac{1}{2} & 0 & \frac{1}{3} \\[0.3em]
\frac{3}{7} & \frac{3}{7} & 0
\end{bmatrix}
$C= [-1/3, 0, -4/7]$
Then, I got $x_{1}^1 = 0.33$, $x_{2}^1 = 0, x_{3}^1 = -0.57$ and
$x_{1}^2 = 0.33, x_{2}^2 = 0.833, x_{3}^2 = 0.142$
But, the answer in the book is $(0.1428571, -0.3571429, 0.4285714)$. Did I do something wrong? I am confused here. Can someone please help me understand this? Thanks.
| First iteration:
$$0.333333, \text{ }0.,\text{ }0.571429$$
Second iteration:
$$0.142857,\text{ }-0.357143\text{ },0.428571$$
So, it looks like somehow the sign of your $x_3$ on the first iteration got corrupted.
We rewrite the system as:
$$x_1 = \dfrac{1}{3} x_2 -\dfrac{1}{3}x_3 + \dfrac{1}{3} \\ x_2 = -\dfrac{1}{2}x_1 + 0 x_2 - \dfrac{1}{3}x_3 \\ x_3 = - \dfrac{3}{7}x_1 - \dfrac{3}{7}x_2 + \dfrac{4}{7}$$
Initial start point: $x^{(0)} = (x_1, x_2, x_3) = (0,0,0)$.
First iteration (substitute $x^{(0)}$ into the system to find the next set of values):
$$x_1 = 0 + 0 + \dfrac{1}{3}, x_2 = 0 + 0 + 0, x_ 3 = 0 + 0 +\dfrac{4}{7}$$
Now, you try the second iteration, just substitute the first iterations values into the three equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/755589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$
I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from solve this question...
| Cauchy-Schwarz inequality allows us to conclude $$((a-1)^{2}+(b-1)^{2})(1+1) \geq (a-1+b-1)^{2}$$ and hence $4\geq (a+b-2)^{2}$ and finally $a+b \leq 4$ with equality holding only when $a=b=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/756213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Explanation of recursive function Given is a function $f(n)$ with:
$f(0) = 0$
$f(1) = 1$
$f(n) = 3f(n-1) + 2f(n-2)$ $\forall n≥2$
I was wondering if there's also a non-recursive way to describe the same function.
WolframAlpha tells me there is one:
$$g(n) = \frac{(\frac{1}{2}(3 + \sqrt{17}))^n - (\frac{1}{2}(3 - \sqrt{17}))^n}{\sqrt{17}}$$
However, I have absolutely no clue how to determine this function, especially the $\sqrt{17}$ makes no sense to me.
Could anyone maybe explain why $f(n)$ and $g(n)$ are the same?
| $$\begin{align} f(n) &= 3\,f(n-1) + 2\,f(n-2) \\
f(n-1) &= 1\,f(n-1) + 0\,f(n-2)\end{align}$$
$$\begin{bmatrix} f(n) \\ f(n-1) \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} f(n - 1) \\ f(n - 2) \end{bmatrix}$$
$$\begin{bmatrix} f(n) \\ f(n-1) \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 1 & 0 \end{bmatrix}^{n - 1} \begin{bmatrix} f(1) \\ f(0) \end{bmatrix}$$
$$\begin{bmatrix} f(n + 1) \\ f(n) \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 1 & 0 \end{bmatrix}^{n} \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$
Diagonalize (find eigen values / eigen vectors) and you have your closed form.
$$\begin{bmatrix} 3 & 2 \\ 1 & 0 \end{bmatrix}
=
\begin{bmatrix} 1 & 1 \\ \frac{2}{3 - \sqrt{17}} & \frac{2}{3 + \sqrt{17}} \end{bmatrix}
\begin{bmatrix} \frac{3 - \sqrt{17}}{2} & 0 \\ 0 & \frac{3 + \sqrt{17}}{2} \end{bmatrix}
\begin{bmatrix} 1 & 1 \\ \frac{2}{3 - \sqrt{17}} & \frac{2}{3 + \sqrt{17}} \end{bmatrix}^{-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/756931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Trigonometric substitution Been out of touch with trigonometry for some time now. Need help proving this expression.
$$\sin^{2}\left(\frac{x}{2}\right) = \frac{1}{2}(1-\cos\left(x\right))$$
Any help will be appreciated. Thanks.
| From the identity $\sin^2\theta+\cos^2\theta=1$ and $\cos 2\theta=\cos^2\theta-\sin^2\theta$, rewrite the LHS as
$$
\begin{align}
\frac{1}{2}(1-\cos x)&=\frac{1}{2}\left(\sin^2\left(\frac x2\right)+\cos^2\left(\frac x2\right)-\left(\cos^2\left(\frac x2\right)-\sin^2\left(\frac x2\right)\right)\right)\\
&=\frac{1}{2}\left(\sin^2\left(\frac x2\right)+\cos^2\left(\frac x2\right)-\cos^2\left(\frac x2\right)+\sin^2\left(\frac x2\right)\right)\\
&=\frac{1}{2}\left(2\sin^2\left(\frac x2\right)\right)\\
&=\sin^2\left(\frac x2\right)
\end{align}
$$
$$\color{blue}{\Large\text{# }\mathbb{Q.E.D.}\text{ #}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/757893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to prove $x^3$ is strictly increasing I am trying to use $f(x)=x^3$ as a counterexample to the following statement.
If $f(x)$ is strictly increasing over $[a,b]$ then for any $x\in (a,b), f'(x)>0$.
But how can I show that $f(x)=x^3$ is strictly increasing?
| Suppose $a<b$. We want to prove that $a^3 < b^3$.
Well, $a<b, \ \therefore \exists \delta>0$ such that $b = a+\delta$. So:
$b^3 = (a+ \delta)^3 = a^3 + 3a^2\delta + 3\delta^2a + \delta^3 = a^3 + (3a\delta(a+\delta)+\delta^3)$.
Now we consider five different cases:
*
*$\ a<-\delta$
*$\ a = -\delta$
*$\ -\delta<a<0$
*$\ a=0$
*$\ a > 0$.
In cases $1, 2, 4$ and $5,$ it is fairly easy to see that $3a\delta(a+\delta)+\delta^3 > 0,\ $ so all that remains is case $3$:
$-\delta<a<0 \implies \exists k$ with $0<k<1,\ $ such that $a = -k\delta.\ $ Then, $\ a\delta(a+\delta) = -k\delta^2(1-k)\delta = -k(1-k)\delta^3$, and using the fact that $k(1-k) \leq \frac{1}{4},\ $ we see that $-k(1-k)\delta^3 \geq -\frac{1}{4}\delta^3,\ $ and so $\ 3a\delta(a+\delta)+\delta^3 \geq (-\frac{3}{4}\delta^3 + \delta^3) = \frac{1}{4}\delta^3 > 0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
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Solve $z^2 - iz = |z - i|$ I have the equation:
$z^2 - iz = |z - i|$
The solutions are $i$, $-\sqrt3/2 + i/2$, $\sqrt3/2 + i/2$
Can someone please walk me through or give me a hint...
| Let $z=x+iy$, $z^2-iz=|z-i|$ implies $0=\Im (z^2-iz)=2xy-x$ then $x=0$ or $y=\frac{1}{2}$.
If $x=0$, then we have $-y^2+y=|y-1|$ i.e. $-y(y-1)=|y-1|$ then $y=1$. So $z=i$ satisfies the equation.
Note that $z^2-iz=|z-i|$ implies $|z||z-i|=|z^2-iz|=|z-i|$, then $|z|=1$. So, if $y=\frac{1}{2}$, then $x=\pm\sqrt{1-\left(\frac{1}{2}\right)^2}=\pm\frac{\sqrt{3}}{2}$. If we put $z=\pm \frac{\sqrt{3}}{2}+i\frac{1}{2}$ we get
\begin{align}
z^2-iz & =z(z-i) \\
& =\left(\pm \frac{\sqrt{3}}{2}+i\frac{1}{2}\right)\left(\pm \frac{\sqrt{3}}{2}-i\frac{1}{2}\right) \\
& =\frac{3}{4}+\frac{1}{4} \\
& = 1 \\
\end{align}
We can see easily $|z-i|=\left|\pm \frac{\sqrt{3}}{2}-i\frac{1}{2}\right|=1$.
Therefore, $i$, $\pm \frac{\sqrt{3}}{2}+i\frac{1}{2}$ are the roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/758872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Definite integral (mixture of functions) I have problem with this integral and I generally don't know how to approach it:
$$\int_{-2}^2 (x^4+4x+\cos(x))\cdot \arctan\left(\frac{x}{2}\right)dx$$
I know that I probably have to make some substitution, but really don't know which.
Thanks for your help.
| Using lab bhattacharjee's result. The integral can be solven by using integration by parts.
Let
\begin{align}
u=\arctan\frac{x}{2}\qquad\rightarrow\qquad du=\frac{2\,dx}{x^2+4}\qquad\text{and}\qquad dv=x\,dx\qquad\rightarrow\qquad v=\frac{1}{2}x^2
\end{align}
Then
\begin{align}
8\int_0^2x\arctan\frac {x}{2}\, dx&=\left.4x^2\arctan\frac{x}{2}\right|_0^2-8\int_0^2 \frac{x^2}{x^2+4}\,dx\\
&=4\pi-8\int_0^2 \frac{x^2+4-4}{x^2+4}\,dx\\
&=4\pi-8\int_0^2 \,dx+16\int_0^2 \frac{2}{x^2+4}\,dx\\
&=4\pi-16+16\, \left.\arctan\frac{x}{2}\right|_0^2\\
&=4\pi-16+4\pi\\
&=8\pi-16
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate this expression: $∫\frac{\sin2x}{(1 - \cos2x)^4}\:\mathrm{d}x$?
How to evaluate: $$∫\frac{\sin2x}{(1 - \cos2x)^4}\:\mathrm{d}x$$ If anyone knows the answer, please help.
| Use $1-\cos 2x=2\sin^2 x$ and $\sin 2x=2\sin x\cos x$, then u will get
$$\int \frac{\sin 2x}{(1-\cos 2x)^4}dx=\int \frac{\cos x}{2^3 \sin ^7x }dx$$
Now letting $u=\sin x\Rightarrow du=\cos x dx$
$$\int \frac{\cos x}{2^3 \sin ^7x }dx= \frac{1}{2^3}\int\frac{du}{u^7}=\frac{1}{2^3}.\frac{-1}{6u^6}+C=\frac{1}{2^3}.\frac{-1}{6\sin^6x}+C=\frac{-1}{2^4.3\sin^6x}+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\sin A = \cfrac{3}{5}$ with $A$ in QII, find $\sec2A$. If $\sin A = \cfrac{3}{5}$ with $A$ in QII, find $\sec2A$.
I'm getting $\sec2A=\cfrac{25}{7}$. Is that correct?
| \begin{align}
\sec2A&=\frac{1}{\cos2A}\\
&=\frac{1}{1-2\sin^2A}\\
&=\frac{1}{1-2\left(\frac{3}{5}\right)^2}\\
&=\frac{1}{1-2\left(\frac{9}{25}\right)}\\
&=\frac{1}{1-\frac{18}{25}}\\
&=\frac{1}{\frac{7}{25}}\\
&=\frac{25}{7}
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/763231",
"timestamp": "2023-03-29T00:00:00",
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No. of real solutions of the equation $2 \cos (\frac{x^2 + x}{6}) = 2^x + 2^{-x} $ How many real solutions are there of the equation $2 \cos (\frac{x^2 + x}{6}) = 2^x + 2^{-x} $?
Please illustrate.
| AM-GM inequality says: $2^x + 2^{-x} \geq 2 \geq 2cos\left(\dfrac{x^2 + x}{6}\right)$. so equation occurs when: $2^x = 2^{-x}$ or $x = 0$, and this is the solution of the equation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When is the sum of consecutive squares a prime? For what integers $x$ do there exist $x$ consecutives integers, the sum of whose squares is prime?
I tried use $$1^2+2^2+...+n^2=\frac {n(n+1)(2n+1)}{6}$$
| The sum of $x$ consecutive squares starting from $n$ is $$n^2+(n+1)^2+...+(n+x-1)^2=xn^2+2n\sum_{k=1}^{x-1} k +\sum_{k=1}^{x-1} k^2 $$
$$\Rightarrow n^2+(n+1)^2+...+(n+x-1)^2=xn^2+nx(x-1)+\frac{(x-1)x(2x-1)}{6}$$
If $x\equiv 1,5\ (mod\ 6)\ $ then we have $(x-1)(2x-1)\equiv0\ (mod\ 6)$
which means that $\frac{(x-1)(2x-1)}{6}\ $ is an integer so the sum becomes:
$$n^2+(n+1)^2+...+(n+x-1)^2=x(n^2+nx(x-1)+\frac{(x-1)(2x-1)}{6})$$ meaning that if $x\equiv 1,5\ (mod\ 6)\ $ the sum of the square of x consectuive integers will never be a prime.
If $x\equiv 0\pmod6\ \text{and}\ x>6$ then $\frac {x}{6}$ will be an integer greater than 1
so we will have: $$n^2+(n+1)^2+...+(n+x-1)^2=\frac {x}{6}(6n^2+6nx(x-1)+(x-1)(2x-1))$$ and again no pirme. In the same fashion if $x\equiv2\ (mod\ 6)$ and $x>2\ $ we take $\frac{x}{2}$ as a factor, and if $x\equiv4\pmod6$ we take $\frac{x}{2}$ again as a factor and finally if $x\equiv3\pmod6\ $ and $x>3\ $ we take $\frac{x}{3}$ as a factor.
Now we need to consider tha cases in which x = 2, 3, 6:
if $x = 2$ then we have $1^2+2^2=5$ which is a prime
if $x = 3 $ then we have $2^2+3^2+4^2=29$ which is prime
and if $x = 6$ then we have $2^2+3^2+4^2+5^2+6^2+7^2 = 139$ which is again a prime.
So the answer for this problem will be 2, 3 and 6
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
A Horrible looking limit I have the following limit question:
$$\lim_{x \rightarrow 1 }\frac {({\rm log} (1+x)-{\rm log}\space 2)(3\times4^{x-1}-3x)}{[(7+x)^{1/3}-(1+3x)^{1/2}]{\rm sin}\space \pi x}$$
This has the form $\frac {0}{0}$ and hence I can apply L'Hospital's Rule. But, this is too large and that way it can go terribly wrong. Is there any clever way to do this?
I found it in a MCQ test and it has the following four alternatives:
A. $\frac{9}{\pi}{\rm log}\space \frac{4}{e}$
B. 1
C. $\frac{3}{\pi}{\rm log} \space \frac{2}{e}$
D. $\frac{1}{\pi}$.
| Put $x=y+1$ and take the limit $y\to 0$, so that we can use some standard limits and simplify the expression. The given limit will be:
\begin{align*}
& \lim_{y\to 0} - \frac{\displaystyle \log\left(1+\frac{y}{2}\right)\Big(3\, \left(4^y-1-y\right)\Big)}{\left(\left(8+y\right)^{1/3}-\left(4+3\, y\right)^{1/2}\right)\, \sin\left(\pi\, y\right)}\tag 1\\
\end{align*}
Divide the numerator and denominator by $y^2$ and we will take four separate limits:
\begin{align*}
\lim_{y\to 0} \frac{\displaystyle \log\left(1+\frac{y}{2}\right)}{\displaystyle \frac{y}{2}\cdot 2} &= \frac{1}{2}\\\\
\lim_{y\to 0} \frac{4^y-1}{y}-1 &= \log{4}-1\\\\
\lim_{y\to 0} \frac{\sin\left(\pi\, y\right)}{\pi\, y}\cdot \pi &= \pi
\end{align*}
and for the remaining one we can use L'Hôpital's rule:
\begin{align*}
\lim_{y\to 0} \frac{\left(8+y\right)^{1/3}-\left(4+3\, y\right)^{1/2}}{y} &= \lim_{y\to 0} \frac{1}{3}\left(8+y\right)^{-2/3}-\frac{3}{2}\left(4+3\, y\right)^{-1/2}\\
&= \frac{1}{3\cdot 4}-\frac{3}{4} = -\frac{2}{3}\\
\end{align*}
Combining all these in $(1)$, we see that
\begin{align*}
\lim_{x \to 1 }\frac {({\log} (1+x)-{\log}\space 2)(3\times4^{x-1}-3x)}{[(7+x)^{1/3}-(1+3x)^{1/2}]{\sin}\space \pi x} = \frac{9}{4\, \pi}\log\left(\frac{4}{e}\right)\approx 0.276662956773403
\end{align*}
which means none of the options are correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove the inequality $\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}\ge 3$
If $a,b,c\in\mathbb R^+$ prove that:
$$\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}\ge 3$$
| By C-S and AM-GM we obtain
$$\sum_{cyc}\frac{a^2+1}{b+c}=\sum_{cyc}\frac{a^2}{b+c}+\sum_{cyc}\frac{1}{b+c}\geq$$
$$\geq\frac{(a+b+c)^2}{2(a+b+c)}+\frac{9}{2(a+b+c)}=\frac{1}{2}\left(a+b+c+\frac{9}{a+b+c}\right)\geq$$
$$\geq\frac{1}{2}\cdot2\sqrt{\frac{9(a+b+c)}{a+b+c}}=3.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
} |
Evaluating the integral $\int\frac{3x^{2}-x+2}{x-1}\;dx$ As the title suggests, the following integral has been given to me
$$\int\frac{3x^{2}-x+2}{x-1}\;dx$$
Yet I still get the wrong answer every time.
Can someone calculate it step-by-step so I can compare it to my own answer?
| Another hint :
\begin{align}
\frac{3x^{2}-x+2}{x-1}&=\frac{3x^{2}-3x+2x+2}{x-1}\\
&=\frac{3x^{2}-3x}{x-1}+\frac{2x}{x-1}+\frac{2}{x-1}\\
&=\frac{3x(x-1)}{x-1}+2\left(\frac{x-1+1}{x-1}\right)+\frac{2}{x-1}\\
&=3x+2\left(1+\frac{1}{x-1}\right)+\frac{2}{x-1}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving $\cos 36° > \tan 36° $ How do we prove that $\cos 36° > \tan 36° $ ? Please help . Thank you.
| $\cos36 > \tan36 \iff \cos36 > \dfrac{\sin36}{\cos36} \iff \cos^2{36} > \sin36 \iff 1 - \sin^2{36} > \sin36 \iff \sin^2{36} + \sin36 - 1 < 0 \iff \sin36 < \dfrac{\sqrt{5} - 1}{2} (*)$. We prove $(*)$ true. let $x = \sin18$, then using $\cos36 = \sin54$ gives: $\cos(2\cdot 18) = \sin(3\cdot 18) \iff 1 - 2x^2 = 3x - 4x^3 \iff (x - 1)(4x^2 + 2x -1) = 0$. Since $0 < x < 1$, this equation gives only one solution: $x = \dfrac{\sqrt{5} - 1}{4}$. So $\cos36 = 1- 2x^2 = \dfrac{\sqrt{5} + 1}{4}$. So $\sin36 = \sqrt{1 - \cos^2{36}} = \dfrac{\sqrt{10 - 2\sqrt{5}}}{4} < \dfrac{\sqrt{5} - 1}{2} \iff \dfrac{10 - 2\sqrt{5}}{16} < \dfrac{6 - 2\sqrt{5}}{4} \iff 3\sqrt{5} < 7 \iff 45 < 49$. The last inequality is obviously true. This implies $(*)$ is true, and we're done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Maximum value of $abc$ for $a, b, c > 0$ and $ab + bc + ca = 12$ $a,b,c$ are three positive real numbers such that $ab+bc+ca=12$.
Then find the maximum value of $abc$
| Or, the same answer also comes by intuition. since a,b,c are positive numbers, $(ab).(bc).(ca)$ is maximum only when $ab=bc=ca=\frac{12}{3}=4$. therefore, $ abc={((ab)(bc)(ca))}^\frac{1}{2}=4^\frac{3}{2}=8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/769901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Solve the following recursive relation by using generating functions $a_n - 9a_{n-1} + 26a_{n-2} - 24a_{n-3} = 0, n \ge 3, a_0 = 0, a_1 = 1,a_2 = 10$
I have tried solving it by the normal way, but I have no idea how to solve it by generating functions. Please give me a detailed answer.
| Check Wilf's "generatingfunctionology". Write the recurrence without subtraction in indices:
$$
a_{n + 3} - 9 a_{n + 2} + 26 a_{n + 1} - 24 a_n = 0
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply by $z^n$, sum over $n \ge 0$ and recognize sums like:
$$
\sum_{n \ge 0} a_{n + m} z^n
= \frac{A(z) - a_0 - a_1 z - \ldots - a_{r - 1} z^{r - 1}}{z^{r - 1}}
$$
to get:
$$
\frac{A(z) - z - 10 z^2}{z^3}
- 9 \frac{A(z) - z}{z^2}
+ 26 \frac{A(z)}{z}
- 24 A(z)
= 0
$$
Solving for $A(z)$, written as partial fractions:
$$
A(z)
= \frac{5}{2} \cdot \frac{1}{1 - 4 z}
- \frac{4}{3} \cdot \frac{1}{1 - 3 z}
+ \frac{3}{2} \cdot \frac{1}{1 - 2 z}
$$
Everything in sight is a geometric series:
$$
a_n = \frac{5}{2} \cdot 4^n - \frac{4}{3} \cdot 3^n + \frac{3}{2} \cdot 2^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{2^{4n + 2} + 1}{5}$ is composed for all $n \ge 2$ The problem is to prove that $$\frac{2^{4n + 2} +1 }{5}$$ is composed for every $n \ge 2$.
I've tried this way: write $2^{4n + 2} + 1$ as $1 - (-4)^{2n + 1}$ and $5 = 1 - (-4)$, then $$\frac{2^{4n + 2} + 1}{5} = \frac{1 - (-4)^{2n + 1}}{1 - (-4)} = 1 - 4 + 16 - 64 + \cdots + (-4)^{2n}$$
How to proceed from this?
Thanks!
| Check you can understand/prove the following
$$\frac{2^{4n+2}+1}5=\frac{3\cdot\left(2^{4n+1}-2^{4n}+\ldots-2^2+2\right)}5$$
But $\;5\nmid 6\;$ , so $\;5\;$ must divide the other factor as the whole thing is an integer since
$$4\cdot16^n+1=0\pmod 5$$
and thus the expression equals $\;6k\;,\;\;k\in\Bbb N\;$ ...composite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrating a top heavy function How do you integrate this function?
$$\int\frac{x^3}{(x+5)^2}dx$$
I have tried it myself by substitution but I can't seem to get rid of the $x$s.
| Long division:
$$
\begin{array}{cccccc}
& & x & - & 10 \\ \\
x^2+10x+25 & ) & x^3 \\
& & x^3 & + & 10x^2 & + & 25x \\ \\
& & & & -10x^2 & - & 25x \\
& & & & -10x^2 & - & 100x \\ \\
& & & & & & 75 x
\end{array}
$$
So we have
$$
\frac{x^3}{x^2+10x+25} = x - 10 + \frac{75x}{(x+5)^2} = x - 10 + \frac{A}{x-5} + \frac{B}{(x+5)^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Why a linear numerator for fractions with irreducible denominators? For example: (2x^3+5x+1)/((x^2+4)(x^2+x+2)) breaks down to (ax+b/(x^2+4))+(cx+d/(x^2+x+2)). I have been told that since the denominators are irreducible, the numerators will be either linear or constant. Now my question is for something like (2x^3+5x+1/(x^2-4)) you would make it equal (A/(x+2))+(B/(x-2)), why do assume that the numerators are constant? Why couldn't the numerators be linear like the irreducible one?
| The reason that we always make a partial fraction's numerator at least $1$ degree less than that of the denominator is that, if the numerator's degree is greater than or equal to that of the denominator, the highest degree can be factored out as a non-fraction term.
Taking your $\displaystyle \frac{2x^3+5x+1}{x^2-4}$ for example, we get
$$\frac{2x^3+5x+1}{x^2-4} = \frac{2x^3 - 8x + 13x + 1}{x^2-4} = \frac{2x (x^2 - 4) + 13x + 1}{x^2-4} = 2x + \frac{13x + 1}{x^2-4}$$
$$= 2x + \frac{A(x+2)+B(x-2)}{(x-2)(x+2)} = 2x + \frac{A}{x-2} + \frac{B}{x+2}$$
So $13x+1 = A(x+2) + B(x-2)$. I leave the math now for you to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/773661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $f(x)=\sqrt{2x-6}$ is continuous at $x=4$ by using precise definition. Please help me get the answer to this question.
Prove $f(x)=\sqrt{2x-6}$ is continuous at $x=4$ by using precise definition. ($\epsilon-\delta$ definition of limits.)
| We have $|x-4|< \delta$.
$|f(x)-f(4)|=|\sqrt {2x-6}-\sqrt {2}|=|\sqrt{2}(\sqrt{x-3}-1)|$
Now, multiplying the numerator and denominator by $(\sqrt{x-3}+1)$,
$=\sqrt{2} |\frac{x-4}{\sqrt{x-3}+1}$|
Now, ${\sqrt{x-3}+1}$ is always greater than or equal to $1$. Thus, $\frac{x-4}{\sqrt{x-3}+1}\leqslant (x-4)$.
Therefore, $|f(x)-f(4)|\leqslant \sqrt{2}|(x-4)| < \sqrt{2} \delta=\epsilon$
Thus, $f(x)$ is continuous at $x=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/775265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
$\int_0^{\pi\over2}{\cos x\sin x\over(x+1)}dx$ = ${1\over2}({1\over2}+{1\over\pi+2}$- $\int_0^\pi{\cos x\over(x+2)^2}dx)$ $$A=\int_0^\pi{\cos x\over(x+2)^2}dx$$
Then prove that, $\displaystyle\int_0^{\Large\pi\over2}{\cos x\sin x\over(x+1)}dx={1\over2}\left({1\over2}+{1\over\pi+2}-A\right)$
| Rewrite:
$$
\int_0^{\Large\pi\over2}{\cos x\sin x\over x+1}dx=\frac12\int_0^{\Large\pi\over2}{\sin 2x\over (x+1)}dx
$$
Now, use IBP by taking $u=\dfrac1{x+1}\;\Rightarrow\;du=-\dfrac{dx}{(x+1)^2}$ and
$$
dv=\sin2x\ dx\quad\Rightarrow\quad v=-\frac12\cos2x
$$
Hence
$$
\begin{align}
\int_0^{\Large\pi\over2}{\sin 2x\over (x+1)}dx
&=-\left.\dfrac{\cos2x}{2(x+1)}\right|_0^{\Large\pi\over2}-\frac12\int_0^{\Large\pi\over2}\dfrac{\cos2x}{(x+1)^2}dx\\
&=\frac{1}{\pi+2}+\frac{1}{2}-\frac12\int_0^{\Large\pi\over2}\dfrac{\cos2x}{(x+1)^2}dx.
\end{align}
$$
Now, let $y=2x\;\Rightarrow\;dx=\dfrac{dy}2$ and $0<x<\dfrac\pi2$ is corresponding to $0<y<\pi$. Thus
$$
\int_0^{\Large\pi\over2}{\cos x\sin x\over x+1}dx=\frac12\left(\frac{1}{\pi+2}+\frac{1}{2}-\int_0^{\Large\pi}\dfrac{\cos y}{(y+2)^2}dy\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/776702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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