Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
for a $3 \times 3$ matrix A ,value of $ A^{50} $ is I f
$$A= \begin{pmatrix}1& 0 & 0 \\
1 & 0 & 1\\
0 & 1 & 0 \end{pmatrix}$$
then $ A^{50} $ is
*
*$$ \begin{pmatrix}1& 0 & 0 \\
50 & 1 & 0\\
50 & 0 & 1 \end{pmatrix}$$
*$$\begin{pmatrix}1& 0 & 0 \\
48 & 1 & 0\\
48 & 0 & 1 \end{pmatrix}$$
*$$\begin{pmatrix}1& 0... | You should learn BenjaLim's answer, which provides a general method for dealing with this kind of problems. However, here is a simple answer just for fun. Note that
$$
A^2=
\begin{pmatrix}
1&0&0\\
1&1&0\\
1&0&1\end{pmatrix}
=I+\underbrace{\begin{pmatrix}
0&0&0\\
1&0&0\\
1&0&0\end{pmatrix}}_{L}
$$
and $L^2=0$. Theref... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/267492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Help with counting factors Can someone help me prove that the product below has $(\large\frac{10^{100}}{6}-\frac{17}{3})$ factors of $7$?
$$\large\prod_{{k}={1}}^{{{10^{100}}}}({4k}+1)$$
| Below is a partial answer.
Let $S_n = \displaystyle \prod_{k=1}^n(4k+1)$.
Number of multiples of $7$ in $S_n$ is $\left \lfloor \dfrac{n +2}7 \right\rfloor$
Number of multiples of $7^2$ in $S_n$ is $\left \lfloor \dfrac{n +37}{49} \right\rfloor$
Number of multiples of $7^3$ in $S_n$ is $\left \lfloor \dfrac{n +86}{343}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/268506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many ways to write one million as a product of three positive integers? In how many ways can the number 1;000;000 (one million) be written as the product
of three positive integers $a, b, c,$ where $a \le b \le c$?
(A) 139
(B) 196
(C) 219
(D) 784
(E) None of the above
This is my working out so far:
$1000000 = 10^{6... | As we can factorise one million as $5^6 \times 2^6 $, it becomes a matter of how many ways we can split up a set of size 12 into three disjoint subsets. This is more of a combinatronics problem, really.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/273137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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I am trying to solve the inequality $\log_{\log{\sqrt{9-x^2}}} x^2 <0$ I am trying to solve the inequality
$$\log_{\log{\sqrt{9-x^2}}} x^2 <0.$$
I got $\mathrm{S.S}=(-\sqrt8 ,-1)\cup( 1,\sqrt8)$, but a friend got $\mathrm{S.S}=(-1,1)- \{0\}$.
Please, what is true?
| $$\log_{\log_c(\sqrt{9-x^2})}x^2=\frac{\log_b x^2}{\log_b(\log_c\sqrt{9-x^2})}<0 $$
Without any loss of generality, we can take base $b>1$
(i)If $\log_b x^2<0 \iff x^2<1$
then we need
$\log_b(\log_c\sqrt{9-x^2})>0\implies \log_c\sqrt{9-x^2}>1\implies x^2<9-c^2$
$\implies x^2<min(1,9-c^2)$
Here observe that for real $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculate $\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$ Please help me calculate this:
$$\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$$
Here I've tried multiplying by $\sqrt[4]{x+9}+2$ and few other method.
Thanks in advance for solution / hints us... | Hint: Use that $\frac{a^4-b^4}{a-b} = a^3 + a^2b + ab^2 + b^3$. Setting $a=\sqrt[4]{x+9}$ and $b=2$, you see $a^4-b^4 = x-7$, and you get $$\frac{1}{\sqrt[4]{x+9}-2} = \frac{a^3 + a^2b + ab^2 + b^3}{x-7}$$
Similarly you can write $\sqrt{x+2}-3$ as:
$$\sqrt{x+2}-3 = \frac{x-7}{\sqrt{x+2}+3}$$
And a similar but uglier r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
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Finding polynomial function's zero values not native English speaker so I may get some terms wrong and so on.
On to the question:
I have as an assignment to find a polynomial function $f(x)$ with the coefficients $a$, $b$ and $c$ (which are all integers) which has one root at $x = \sqrt{a} + \sqrt{b} + \sqrt{c}$.
I've ... | Subtract $\sqrt a$ from both sides, square both sides: now you have $\sqrt a$ on one side, $\sqrt{bc}$ on the other. Solve for $\sqrt a$, square both sides: now you have only $\sqrt{bc}$. Solve for $\sqrt{bc}$, square both sides, voila! all square roots gone.
If you need to know the other zeros, the full set is $\pm\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/276111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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How to show that $\frac{x^2}{x-1}$ simplifies to $x + \frac{1}{x-1} +1$ How does $\frac{x^2}{(x-1)}$ simplify to $x + \frac{1}{x-1} +1$?
The second expression would be much easier to work with, but I cant figure out how to get there.
Thanks
| $\textbf{Hint:}$ Do you know about polynomial long divison?
A simpler way of dealing with this problem is noticing that for all $x\neq 1$,
$$\frac{x^2}{x-1}=\frac{x^2-1+1}{x-1}=\frac{x^2-1}{x-1}+\frac{1}{x-1}=\frac{(x-1)(x+1)}{x-1}+\frac{1}{x-1}=x+1+\frac{1}{x-1}.$$
It's not always this easy, though. So you should lea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/278481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
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Finding $\frac{1}{S_1}+\frac{1}{S_2}+\frac{1}{S_3}+\cdots+\frac{1}{S_{2013}}$ Assume $S_1=1 ,S_2=1+2,S=1+2+3+,\ldots,S_n=1+2+3+\cdots+n$
How to find :
$$\frac{1}{S_1}+\frac{1}{S_2}+\frac{1}{S_3}+\cdots+\frac{1}{S_{2013}}$$
| you can see $ s_{n}=1+2+\cdots=\dfrac{n(n+1)}{2}$,
so
$\dfrac{1}{s_{n}}=\dfrac{2}{n(n+1)}=2\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$
then
$\dfrac{1}{s_{1}}+\dfrac{1}{s_{2}}+\cdots+\dfrac{1}{s_{2013}}=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\cdots+\dfrac{1}{2013}-\dfrac{1}{2014}\right)=\dfrac{2013}{1007}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/281318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve the inequality on the number line? How would I solve the following inequality.
$x^2+10x \gt-24$
How would I solve it and put it in a number line?
| $x^2+10x>−24$
iff $x^2-10x+25 > 1$
iff $(x-5)^2 > 1$.
(Note: "iff" means "if and only if".)
Because of the magical property of $1$ being its own square root
(funny how this happens, eh?)
this is true iff $|x-5| > 1$
which is true
iff $x<4$ or $x > 6$.
By completing the square like this,
you are implicitly solving the e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/282950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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How to find the solution of the differential equation
Find the solution of the differential equation
$$\frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)}, y(0)=1$$
Trial: $$\begin{align} \frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)} \\ \implies \frac{dy}{dx}=-\frac{1+(y/x)^2-10/x^2}{(y/x)(1+(y/x)^2+5/x^2)} \\ ... | Hint: It is an exact equation. Assume there is an differentiable function $f(x,y)$ such that $$f_x=x^3+xy^2-10x,~~~f_y=x^2y+y^3+5y$$ and then find the function. The solution is as $$f(x,y)=C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Consider a function $f(x) = x^4+x^3+x^2+x+1$, where x is an integer, $x\gt 1$. What will be the remainder when $f(x^5)$ is divided by $f(x)$? Consider a function $f(x) = x^4+x^3+x^2+x+1$, where x is an integer, $x\gt 1$. What will be the remainder when $f(x^5)$ is divided by $f(x)$ ?
$f(x)=x^4+x^3+x^2+x+1$
$f(x^5)=x^{2... | Observe that $f(x^5) = f(x)(x^{16}-x^{15}+2x^{11}-2x^{10}+3x^6-3x^5+4x-4) + 5$.
So the remainder will be $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
What is the formula for $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$ How can I find the formula for the following equation?
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$$
More importantly, how would you approach finding the formula? I have fo... | Observe that $$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ $\therefore$ The given series can be written as $$1-\frac12+\frac12-\frac13+\frac13+\cdots -\frac{1}{n}+\frac1n-\frac{1}{n+1}$$ $$=1-\frac1{n+1}$$ $$=\frac{n}{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/286024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Calculation of $x$ in $\tan^{-1}\left(\frac{x}{1-x^2}\right)+\tan^{-1}\left(\frac{1}{x^3}\right) = \frac{3\pi}{4}$ The no. of real values of $x$ satisfying $\displaystyle \tan^{-1}\left(\frac{x}{1-x^2}\right)+\tan^{-1}\left(\frac{1}{x^3}\right) = \frac{3\pi}{4}$
options ::
(a) $0$
(b) $1$
(c) $2$
(d) Infinitely many
M... | *
*If $|x|<1$, then $x^4+1>1>x$. If $|x|\geq 1$, then $x^4+1>x^4\geq x$. In both cases
$$
x^4-x+1>0
$$
so the equation $x^4-x+1=0$ have no real solutions.
*Note that $x=1$ is not a solution of original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/287220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find the value of $h(99)$ in the function? If $$h(x) + h(x+1) = 2x^2$$
and $$h(33) = 99$$
What will be the value of $h(99)$?
| From your definition,
$$h(34)=2\times 33^2-h(33)$$
$$h(35)=2\times 34^2-h(34)=2\times 34^2-2\times 33^2+h(33)$$
$$\dots$$
$$h(99)=2\times 98^2 - 2\times 97^2 + 2\times 96^2 -\cdots + 2\times 34^2 - 2\times 33^2 +h(33)\\=2(98^2-97^2+96^2-\cdots+34^2-33^2)+99\\
=2(98+97+\cdots +34+33)+99\\
=8745$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/287521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to show $AB^{-1}A=A$
Let $$A^{n \times n}=\begin{pmatrix} a & b &b & \dots & b \\ b & a &b & \dots & b \\ b & b & a & \dots & b \\ \vdots & \vdots & \vdots & & \vdots \\ b & b & b & \dots &a\end{pmatrix}$$
where $a \neq b$ and $a + (n - 1)b = 0$.
Suppose $B=A+\frac{11'}{n}$ , where $1=(1,1,\dots,1)'$ is an $n ... | Here is another answer of mine. Let $E={\mathbf 1}{\mathbf 1}^T$. By Sherman-Morrison formula,
\begin{align*}
&B = (a-b)I + \left(b + \frac1n\right){\mathbf 1}{\mathbf 1}^T\\
\Rightarrow&B^{-1} = \frac{I}{a-b} + cE
\end{align*}
for some constant $c$. We will see that the exact value of $c$ is unimportant. Now, note tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/287917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Compute the limit $\displaystyle \lim_{x\rightarrow 0}\frac{n!.x^n-\sin (x).\sin (2x).\sin (3x).......\sin (nx)}{x^{n+2}}\;\;,$ How can i calculate the Given limit
$\displaystyle \lim_{x\rightarrow 0}\frac{n!x^n-\sin (x)\sin (2x)\sin (3x)\dots\sin (nx)}{x^{n+2}}\;\;,$ where $n\in\mathbb{N}$
| $$\dfrac{\sin(kx)}{kx} = \left(1- \dfrac{k^2x^2}{3!} + \mathcal{O}(x^4)\right)$$
Hence,
$$\prod_{k=1}^n \dfrac{\sin(kx)}{kx} = \prod_{k=1}^n\left(1- \dfrac{k^2x^2}{3!} + \mathcal{O}(x^4)\right) = 1 - \dfrac{\displaystyle \sum_{k=1}^n k^2}6x^2 + \mathcal{O}(x^4)\\ = 1 - \dfrac{n(n+1)(2n+1)}{36}x^2 + \mathcal{O}(x^4)$$
H... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/291087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Stochastic Calc (a) Consider the process
$$
\mathrm d\sqrt{v} = (\alpha - \beta\sqrt{v})\mathrm dt + \delta \mathrm dW
$$
Here $\alpha, \beta,$ and $\delta$ are constants. Using Ito's Lemma show that
$$
\mathrm dv = (\delta^2 + 2\alpha\sqrt{v} - 2\beta v)\mathrm dt + 2\delta\sqrt{v}\mathrm dW.
$$
(b) Using Ito's Lemma... | (a) Notice that $v = f(\sqrt{v})$ for $f(x)=x^2$. We have $f'(x)=2x$ and $f''(x)=2$.
Ito's lemma yields:
$$
dv = f'(\sqrt{v})\,d\sqrt{v} + \frac{1}{2}f''(\sqrt{v})\,d\langle\sqrt{v}\rangle.
$$
Hence
\begin{align}
dv &= 2\sqrt{v}\times\left((\alpha-\beta\sqrt{v})\,dt + \delta \,dW\right) + \frac{1}{2}\times 2\times\del... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/291747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number theory - congruence
Let $n$ be an even integer not divisible by $10$, what digit is in the $10$s place for $n^{20}$ and the hundreds place for $n^{200}$, can you generalise this?
I know that for $n^{20}$ they end in $76$ and for $n^{200}$ they all end in $376$
I even went on too see that $n^{2000}$ always ends... | Let us first do the case $n^{20} \pmod{100}$.
You have been advised to split this into two problems
*
*$n^{20} \pmod{4}$. Since $n$ is even, this yields $n^{20} \equiv 0 \pmod{4}$ here.
*$n^{20} \pmod{25}$. Since $n$ is not divisible by $5$, we have $(n, 25) = 1$. Since $\varphi(25) = 20$, we obtain $n^{20} \equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/295361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$ Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical
$$
\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
$$
Taking a cue from Ramanujan's solution method, I defi... | This is not an answer to your question, but you may be interested in the following two similar evaluations:
$$\sqrt{1+\sqrt{4+\sqrt{16+\sqrt{64+...}}}}=2\tag{1}$$
$$\sqrt{1+2^{-1}\sqrt{1+2^{-2}\sqrt{1+2^{-3}\sqrt{1+...}}}}=\frac{5}{4}\tag{2}$$
To prove $(1)$, one may use the fact that
$$2^n+1=\sqrt{4^n+2^{n+1}+1}$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/300299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 6,
"answer_id": 4
} |
Solving Recurrence $T(n) = T(n − 3) + 1/2$; I have to solve the following recurrence.
$$\begin{gather}
T(n) = T(n − 3) + 1/2\\
T(0) = T(1) = T(2) = 1.
\end{gather}$$
I tried solving it using the forward iteration.
$$\begin{align}
T(3) &= 1 + 1/2\\
T(4) &= 1 + 1/2\\
T(5) &= 1 + 1/2\\
T(6) &= 1 + 1/2 + 1/2 = 2\\
T(7) ... | The crucial observation is that the sequence occurs in blocks of 3, so for each $n$ we need to find out "which block of 3 is $n$ in". So using $\lfloor n/3\rfloor$ or $\lceil n/3\rceil$ would be good.
Observe the pattern:
$$\begin{array}{c}
n & T(n) & \lceil n/3\rceil\\\hline
0 & 2/2 & 1\\\hline
1 & 2/2 & 1\\\hline
2 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/300934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Prove $\frac{a}{ab+2c}+\frac{b}{bc+2a}+\frac{c}{ca+2b} \ge \frac 98$ $a,b,c \in \mathbb{R^+} \text{such that }a+b+c=2$. Prove inequality $$\frac a{ab+2c}+\frac b{bc+2a}+\frac c{ca+2b} \ge \frac 98$$
I tried
*
*$$LHS = \sum \frac{1}{b+2\cdot c/a} \ge \frac 9 {2+2(\sum c/a)} \longrightarrow failed$$
*$$\frac a {ab+2... | \begin{align}
\sum_{cyc}{\frac{a}{ab+2c}}=\sum_{cyc}{\frac{a}{ab+(a+b+c)c}}=\sum_{cyc}{\frac{a}{(a+c)(b+c)}}& =\frac{\sum_{cyc}{a(a+b)}}{(a+b)(a+c)(b+c)} \\
& =\frac{a^2+b^2+c^2+ab+ac+bc}{(a+b)(a+c)(b+c)} \\
& \geq \frac{\frac{2}{3}(a+b+c)^2}{(a+b)(a+c)(b+c)} \\
& \geq \frac{\frac{2}{3}(a+b+c)^2}{\left(\frac{(a+b)+(a+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/304777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Integral of type $\displaystyle \int\frac{1}{\sqrt[4]{x^4+1}}dx$ How can I solve integral of types
(1) $\displaystyle \int\dfrac{1}{\sqrt[4]{x^4+1}}dx$
(2) $\displaystyle \int\dfrac{1}{\sqrt[4]{x^4-1}}dx$
| (1) Put $x=\sqrt{\tan t}\implies dx=\frac{\sec^2tdt}{2\sqrt{\tan t}}$ and $1+x^4=1+\tan^2t=\sec^2t$
So, $$\int\frac{dx}{(1+x^4)^\frac14}=\int\frac{\sec^2tdt}{2\sqrt{\tan t}\sqrt{\sec t}}=\int\frac{dt}{2\cos t\sqrt{\sin t}}=\int\frac{\cos tdt}{2\cos^2t\sqrt{\sin t}}$$
Put $\sin t=y^2\implies \cos tdt=2ydy $
$$\int\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/306027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Solve Trignometric Equation
Solve the equation: $\sin^25x+\sin^23x = 1+\cos(8x)$.
I tried : $1+\cos(8x) = 2\cos^2(4x)$ which gives :
$$\begin{align*}
\sin^25x+\sin^23x &= 2\cos^2(4x)\\
&= 2(1-\sin^2(4x))\\
&= 2-2\sin^2(4x)\\
\end{align*}$$
| As $\cos(A-B)\cos(A+B)=\cos^2A-\sin^2B$
$\cos8x=\sin^25x+\sin^23x-1=-(\cos^23x-\sin^25x)=-\cos(5x-3x)\cos(5x+3x)=-\cos8x\cos2x$
$\cos8x(1+\cos2x)=0$
If $\cos8x=0,8x=(2n+1)\frac\pi2,x=(2n+1)\frac\pi{16}$
If $1+\cos2x=0\implies \cos2x=-1=\cos\pi,2x=(2m+1)\pi,x=(2m+1)\frac\pi2$ where $m,n$ are any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to compute Nullspace on maple? I have the matrix
$$A := \begin{bmatrix}6& 9& 15\\-5& -10& -21\\ 2& 5& 11\end{bmatrix}.$$ Can anyone please tell me how to both find the eigenspaces by hand and also by using the Nullspace command on maple? Thanks.
| Given the matrix
$$A = \left(\begin{matrix}6& 9& 15\\-5& -10& -21\\ 2& 5& 11\end{matrix}\right).$$
Find the Eigensystem by hand.
First, lets find the eigenvalues by solving $det(A - \lambda I) = 0$, so we have:
$$det(A - \lambda I) = \left|\begin{matrix}6 - \lambda & 9& 15\\-5& -10 - \lambda & -21\\ 2& 5& 11 - \lambd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/308117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Determine whether $x^3$ is $O(g(x))$ for certain functions $g(x)$. a) $g(x) = x^2$
b) $g(x) = x^3$
c) $g(x) = x^2 + x^3$
d) $g(x) = x^2 + x^4$
e) $g(x) = 3^x$
f) $g(x) = (x^3)/2$
Do you guys have any ideas? Thanks!
| Since you asked in a comment about (e), let's discuss that.
Saying "$x^3$ is $O\bigl(3^x\bigr)$" means the following:
There is some constant $c$, such that for all sufficiently large $x$, $$x^3 < c\cdot3^x.$$
Consider the functions $x^3$ and $3^x$. Clearly, $3^x$ increases much faster than $x^3$. For $x=10$, $3^x$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/309008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
proving $\csc^2 \left( \frac{\pi}{7}\right)+\csc^2 \left( \frac{2\pi}{7}\right)+\csc^2 \left( \frac{4\pi}{7}\right)=8$ How can I prove the following identity using complex variables
$$
\begin{align*}
1) & \csc^2 \left( \frac{\pi}{7}\right)+\csc^2 \left( \frac{2\pi}{7}\right)+\csc^2 \left( \frac{4\pi}{7}\right)=8 \\
2)... | I am going to answer the question 2 by grouping terms.
Firstly, we are going to evaluate the sum $\displaystyle \sum_{k=1}^{7} \tan ^{2}\left(\frac{k \pi}{16}\right). $
$\begin{aligned}\because \tan \frac{(8-k) \pi}{16} &=\tan \left(\frac{\pi}{2}-\frac{k \pi}{16}\right) =\frac{1}{\tan \frac{k\pi}{16}} \\\therefore \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/309271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Implicit differentiation 2 questions? Hello everyone I have two questions on implicit differentiation.
My first one is express $\frac{dy}{dx}$ in terms of $x$ and $y$ if $x^2-4xy^3+8x^2y=20$
what I did is this
$2x-4x(3y^2)(\frac{dy}{dx})+4y^3+8x^2\frac{dy}{dx}+16xy=0$
$-4x(3y^2\frac{dy}{dx})+8x^2\frac{dy}{dx}=-2x-4y^... | Very well done, Fernando. You indeed found the equation for the line tangent to the second equation at the point $(2, 1)$, and your first solution look like you differentiated properly, too.
You could simplify just a tad: $$\frac{dy}{dx}=\frac{2x-4y^3-16}{-4(x)(3y^2)+8x^2} = \frac{2(x-2y^3 - 8)}{4(-3xy^2 +2)} = \frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/310408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Formula for this sequence? $P_n= \prod^n_k_=_2 \frac{k^2-1}{k^2}$ for $n \ge 2$
I calculated $P_1 to P_3$ . I have been trying to come up with a formula but I can't really see any pattern.
$P_2 = \frac{3}{4} , P_3 = \frac{2}{3}, P_4 = \frac{5}{8}$
| $$P_n = \prod_{k=2}^n \dfrac{k^2-1}{k^2} = \prod_{k=2}^n \dfrac{(k-1)(k+1)}{k^2}$$
Hence,
$$P_n = \dfrac{1 \times \color{red}3}{2 \times \color{blue}2} \cdot \dfrac{\color{blue}2 \times \color{green}4}{\color{red}3 \times \color{orange}3} \cdot \dfrac{\color{orange}3 \times \color{magenta}5}{\color{green}4 \times \colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/311745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Maclaurin series for $\frac{x}{e^x-1}$ Maclaurin series for
$$\frac{x}{e^x-1}$$
The answer is
$$1-\frac x2 + \frac {x^2}{12} - \frac {x^4}{720} + \cdots$$
How can i get that answer?
| This is not a straightforward solution, but I added this to show that we have other ways if we know some properties of the function.
Method 2. Using the Taylor series of the logarithm, we have
\begin{align*}
\frac{x}{e^x - 1}
&= \frac{\log(1+(e^x - 1))}{e^x - 1} \\
&= \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} (e^x - 1)^n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/311817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 1
} |
Proof of inequality involving logarithms How could we show that
$$\left|\log\left( \left({1 + \frac{1}{n}}\right)^{n + \frac{1}{2}}\cdot \frac{1}{e}\right)\right| \leq \left|\log\left( \left({1 - \frac{1}{n}}\right)^{n - \frac{1}{2}}\cdot \frac{1}{e}\right)\right| ,\; \forall n \text{ sufficiently large?} $$
I already... | It is enough to prove that
$$\lim_{n\rightarrow \infty }\log \left( \left( 1+\frac{1}{n}\right) ^{n+1/2}
\frac{1}{e}\right) =0$$
and
$$\lim_{n\rightarrow \infty }\log \left( \left( 1-\frac{1}{n}\right) ^{n-1/2}\frac{1}{e}\right) =-2.$$
The first limit can be evaluated as follows:
$$\begin{eqnarray*}
\lim_{n\rightarrow ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/312123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Orthogonal trajectories for a given family of curves Find the orthogonal trajectories for the given family of curves:
$$y^2=Cx^3-2$$
Derivative with respect to x, and finding the value of C:
$$2yy' = C3x^2 $$
$$C=\frac{y^2+2}{x^3}$$
Replacing C, and solving for y':
$$2yy' = \frac{3y^2+6}{x}$$
$$y'=\frac{3y^2+6}{2xy}$$
... | That's what I got
$$
y^2=Cx^3-2 \\
2yy'= 3Cx^2 \\
y'=\frac{3Cx^2}{2y}
$$
Orthogonal curve
$$
y'=-\frac {2y}{3Cx^2} \\
\frac {y'}{y} = -\frac 2{3Cx^2} \\
\ln |y| = \frac 2{3Cx} \\
|y| = \exp \left (\frac 2{3Cx} \right)
$$
and here some plots
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/315317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Simple AM-GM inequality Let $a,b,c$ positive real numbers such that $a+b+c=3$, using only AM-GM inequalities show that
$$
a+b+c \geq ab+bc+ca
$$
I was able to prove that
$$
\begin{align}
a^2+b^2+c^2 &=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{a^2+c^2}{2} \geq \\
&\ge \frac{2\sqrt{a^2b^2}}{2}+\frac{2\sqrt{b^2c^2}}{2}+\... | Reformulate first $a+b+c=S$ and $a=M+d \qquad b=M-d $ then
$$S \ge M^2-d^2 + (S-2M)2M$$
reorganize
$$ 3M^2-2SM+S +d^2 \ge 0 $$
Now make use of the given definition that $S=3$. We get
$$3(M^2-2M+1) +d^2 \ge 0 $$
$$3(M-1)^2 +d^2 \ge 0 $$
which is always true.
Well, this focuses "when and how" it makes sense to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/315699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Finding the limit of $f(x)$ tends to infinity How do you find this limit?
$$\lim_{x \rightarrow \infty} \sqrt[5]{x^5-x^4} -x$$
I was given a clue to use L'Hospital's rule.
I did it this way:
UPDATE 1:
$$
\begin{align*}
\lim_{x \rightarrow \infty} \sqrt[5]{x^5-x^4} -x
&= \lim_{x \rightarrow \infty} x\begin{pmatrix}\sq... | If we are not compelled to use L'Hospital's Rule,
$$\lim_{x \rightarrow \infty} \sqrt[5]{x^5-x^4} -x$$
$$=\lim_{y\to0}\frac{(1-y)^\frac15-1}y$$
$$=\lim_{y\to0}\frac{(1-y)-1}{y\{(1-y)^\frac45+(1-y)^\frac35+(1-y)^\frac25+(1-y)^\frac15+1\}}$$ as $ a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1)$
$$=\frac{-1}{1+1+1+1+1}\text { as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/316865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
The series $\sum_{n=1}^{+\infty}\frac{1}{1^2+2^2+\cdots+n^2}.$ How to justify the convergence and calculate the sum of the series:
$$\sum_{n=1}^{+\infty}\frac{1}{1^2+2^2+\cdots+n^2}.$$
| $$\begin{array}{lcl}
\sum_{n=1}^\infty \frac{1}{1^2+2^2+\cdots+n^2}&=& \sum_{n=1}^\infty\frac{6}{n(n+1)(2n+1)} \\ &=& 6\sum_{n=1}^\infty \frac{1}{2n+1} \left( \frac{1}{n}-\frac{1}{n+1}\right) \\ &=& 12\sum_{n=1}^\infty \frac{1}{2n(2n+1)} -12\sum_{n=1}^\infty \frac{1}{(2n+1)(2n+2)} \\ &=& 12\sum_{n=1}^\infty \left[ \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/317219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 0
} |
Solve $x^{11}+x^8+5\equiv 0\pmod{49}$ Solve $x^{11}+x^8+5\pmod{49}$
My work
$f(x)=x^{11}+x^8+5$
consider the polynomial congruence $f(x) \equiv 0 \pmod {49}$
Prime factorization of $49 = 7^2$
we have $f(x) \equiv 0 \mod 7^2$
Test the value $x\equiv0,1,2,3,4,5,6$ for $x^{11}+x^8+5 \equiv 0\pmod 7$
It works for $x\equiv... | Hint $\rm\ mod\ 49\!:\ 5+(1\!+\!7n)^8\!+(1\!+\!7n)^{11}\!\equiv 5 + (1\!+\!56n) + (1\!+\!77n) \equiv 7 - 14 n\equiv 7(1\!-\!2n)$
Thus $\rm\ 49\mid 7\,(1\!-\!2n)\iff 7\mid 1\!-\!2n\iff n\equiv 4\,\ (mod\ 7),\ $ so $\rm\ x \equiv 1+7n\equiv 29\,\ (mod\ 49).$
Alternatively $ $ we may compute $\rm\:f(1\!+\!7n)\:$ by Taylor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/317458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$2005|(a^3+b^3) , 2005|(a^4+b^4 ) \implies2005|a^5+b^5$ How can I show that if $$2005|a^3+b^3 , 2005|a^4+b^4$$ then $$2005|a^5+b^5$$
I'm trying to solve them from $a^{2k+1} + b^{2k+1}=...$ but I'm not getting anywhere.
Can you please point in me the correct direction?
Thanks in advance
| If prime $p|(a^3+b^3), p|(a+b)(a^3+b^3) \implies p| \{(a^4+b^4)+ab(a^2+b^2)\}$
If $p|(a^4+b^4), p$ must divide $ab(a^2+b^2)$
If $p|a,$ $p$ must divide $b$ as $p|(a^3+b^3)$
If $p|a$ and $p|b,$ then $p|(a^n+b^n)$ for integer $n\ge1$
Else $p\not\mid ab $ and $p|(a^2+b^2)\implies p|(a+b)(a^2+b^2) \implies p| \{(a^3+b^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/319248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Show that that $\lim_{n\to\infty}\sqrt[n]{\binom{2n}{n}} = 4$ I know that
$$
\lim_{n\to\infty}{{2n}\choose{n}}^\frac{1}{n} = 4
$$
but I have no Idea how to show that; I think it has something to do with reducing ${n}!$ to $n^n$ in the limit, but don't know how to get there. How might I prove that the limit is four?
| Hint:
$$
\begin{align}
\binom{2n}{n}
&=\frac{2n(2n-1)}{n^2}\frac{(2n-2)(2n-3)}{(n-1)^2}\frac{(2n-4)(2n-5)}{(n-2)^2}\cdots\frac{4\cdot3}{2^2}\frac{2\cdot1}{1^2}\\
&=2^n\frac{2n-1}{n}\frac{2n-3}{n-1}\frac{2n-5}{n-2}\cdots\frac{3}{2}\frac{1}{1}\\
&=4^n\frac{n-1/2}{n}\frac{n-3/2}{n-1}\frac{n-5/2}{n-2}\cdots\frac{3/2}{2}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/320846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 0
} |
Differential problem, find the maximum and minimum value Find the maximum, minimum value and inflection/saddle point of the following function
*
*$f(x)=12x^5-45x^4+40x^3+6$
*$f(x)=x+\frac{1}{x}$
*$f(x)=(2x+4) (x^2-1)$
Give a little explanation or procedural details if possible
| Many questions!
$1.$ We have $f'(x)=60x^4-180x^3+120x^2=60x^2(x^2-3x+2)=60x^2(x-1)(x-2)$.
Note that $(x-1)(x-2)\gt 0$ if $x\lt 1$ or $x\gt 2$. Ao $f(x)$ is increasing in $(-\infty,1]$, then decreasing in $[-1,2]$, then increasing in $[2,\infty)$. (It hesitates slightly at $x=0$, since the derivative is $0$ there, but ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/321819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $z^3 + (1+i)z - 3 + i = 0$ does not have any roots in the unit circle $|z|\leq 1$. I need help with showing that $z^3 + (1+i)z - 3 + i = 0$ does not have any roots in the unit circle $|z|\leq 1$?
My approach so far has been to try to develop the expression further.
$$ z^3 +(1+i)z-3+i = z(z^2+i+1)-3+i$$
$$z(z^... | $z^3 + (1+i)z - 3 + i = 0\iff z^3+(1+i)z=3-i$
Now, If $|z|\leq 1$, then $|z^3+(1+i)z|\leq |z|^3+|1+i||z|\leq1+\sqrt2$
As $|3-i|=\sqrt{10}\gt 1+\sqrt{2}$
Therefore, $z^3+(1+i)z\neq 3-i$ for any $z\in \Bbb C, |z|\leq 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
prove $ \frac{1}{13}<\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \cdot \frac{99}{100}<\frac{1}{12} $ Without the aid of a computer，how to prove
$$ \frac{1}{13}<\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \cdot \frac{99}{100}<\frac{1}{12} $$
| When you multiply up and down in the product by $2 \cdot 4 \cdot 6 \ldots 50$, the product becomes
$$\frac{1}{2^{100}} \binom{100}{50} = \frac{1}{2^{100}} \frac{100!}{(50!)^2}$$
Use Stirling:
$$n! \sim \sqrt{2 \pi n} n^n e^{-n} \left ( 1+ \frac{1}{12 n}\right )$$
for large $n$. We will see what that means: plug in thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/322224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Equation of one branch of a hyperbola in general position Given a generic expression of a conic:
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F=0$$
is there a way to write an expression for one of the branches as a function of the coefficients? I tried using the quadratic formula to get an expression for $y$:
$$y=\frac{-(Bx+E)\pm \... | Following Will Jagy's suggestion, here are some examples:
Example #1
Consider
$$x^2 - y^2 -1 = 0$$
($A=1$, $B=0$, $C=-1$, $D=0$, $E=0$, $F=-1$). From
$$p_c = \begin{pmatrix}x_c\\y_c\end{pmatrix} = \begin{pmatrix}\frac{BE-2CD}{4AC-B^2}\\ \frac{DB-2AE}{4AC-B^2}\end{pmatrix}$$
we have
$$p_c = \begin{pmatrix}0\\0\end{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/324048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Can the Basel problem be solved by Leibniz today? It is well known that Leibniz derived the series
$$\begin{align}
\frac{\pi}{4}&=\sum_{i=0}^\infty \frac{(-1)^i}{2i+1},\tag{1}
\end{align}$$
but apparently he did not prove that
$$\begin{align}
\frac{\pi^2}{6}&=\sum_{i=1}^\infty \frac{1}{i^2}.\tag{2}
\end{align}$$
Euler ... | One of the best way without leibnizSince $\int_0^1 \frac{dx}{1+x^2}=\frac{\pi}{4}$, we have
$$\frac{\pi^2}{16}=\int_0^1\int_0^1\frac{dydx}{(1+x^2)(1+y^2)}\overset{t=xy}{=}\int_0^1\int_0^x\frac{dtdx}{x(1+x^2)(1+t^2/x^2)}$$
$$=\frac12\int_0^1\int_t^1\frac{dxdt}{x(1+x^2)(1+t^2/x^2)}\overset{x^2\to x}{=}\frac12\int_0^1\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/324249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
If all the signs are negative in an $(a + b + c)^2$ bracket, can I just make them all positive? I have to do the expansion
$$(-y - z - x^2 - y^2 - z^2)^2$$
Can I say that this is
$$(y + z + x^2 + y^2 + z^2)^2$$
as all the signs are the same inside the brackets and so multiplying two negatives together will always give... | Yes.
$$(-a -b -c)^2 = ((-1)(a + b + c))^2 = \underbrace{(-1)^2}_{=+1}(a+b+c)^2 = (a+b+c)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/324681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Need help proving this integration If $a>b>0$, prove that :
$$\int_0^{2\pi} \frac{\sin^2\theta}{a+b\cos\theta}\ d\theta = \frac{2\pi}{b^2} \left(a-\sqrt{a^2-b^2} \right) $$
| $$I = \int_0^{2 \pi} \dfrac{\sin^2(x)}{a+b \cos(x)} dx \implies aI = \int_0^{2 \pi} \dfrac{\sin^2(x)}{1+\dfrac{b \cos(x)}a}dx$$
\begin{align}
aI & = \sum_{k=0}^{\infty}\left(\dfrac{(-b)^k}{a^k} \int_0^{2 \pi}\sin^2(x) \cos^k(x) dx \right) = \sum_{k=0}^{\infty}\left(\dfrac{b^{2k}}{a^{2k}} \int_0^{2 \pi}\sin^2(x) \cos^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/326714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Half-symmetric, homogeneous inequality Let $x,y,z$ be three positive numbers. Can anybode prove the follwing inequality :
$(x^2y^2+z^4)^3 \leq (x^3+y^3+z^3)^4$ (or find a counterexample, or find a reference ...)
| Note that we have $(x^3+y^3+z^3)^2\geq (x^3+y^3)^2+z^6\geq 4x^3y^3+z^6,$ and also $x^3+y^3\geq 2\sqrt{x^3y^3},$ so that it suffices to check that
$$\left(2\sqrt{x^3y^3}+z^3\right)^2(4x^3y^3+z^6)\geq (x^2y^2+z^4)^3.$$
Using the Holder's inequality, we can get a sharper bound:
$$\begin{aligned}\left(2\sqrt{x^3y^3}+z^3\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/331427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to solve this system of equations? How to solve this system of equations?
$$\begin{cases}
1+\sqrt{2 x+y+1}=4 (2
x+y)^2+\sqrt{6 x+3 y},\\
(x+1) \sqrt{2 x^2-x+4}+8 x^2+4
x y=4.
\end{cases}$$
| You have two equations with two variables.
$$1+\sqrt{2x+y+1}-4(2x+y)^2-\sqrt{6x+3y}=0$$
$$(x+1)\sqrt{2x^2-x+4}+8x^2+4xy-4=0$$
Solve it with any root finding algortihm.
If you are looking for real solutions
$$x=0.5\qquad y=-0.5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/332059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Simplify $ \frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}} $ Please help me find the sum
$$
\frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}}
$$
| They're arranged rather nicely. Proceed from left to right and keep using the formula $(a-b)(a+b)=a^2-b^2$ to rewrite the two terms you're about to add with a common (unfactored) denominator. You'll get a good deal of additive cancellation in the numerators, so it won't be all that messy at any stage.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/332191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
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If $p$ is prime, $a \in \Bbb Z$, $ord^a_p=3$. Then how to find $ord^{a+1}_p=?$ If $p$ is prime, $a \in \Bbb Z$, $ord^a_p=3$. Then how to find $ord^{a+1}_p=?$
about $ord_n^a$ we know that is $(a,n)=1$ and smallest integer number as $d$ such that $a^d \equiv 1$ so $d=ord_n^a$
also we have: if $(a,n)=1 $, $a\equiv b \pmod... | Hint: Note that since $(a+1)^6\equiv 1\pmod{p}$, the order of $a+1$ divides $6$. It follows that the only candidates to be eliminated are $1$, $2$, and $3$. The numbers $4$ and $5$ are not in the game.
Added: The fact that the order of $a+1$ is not $1$ is easy to prove, but should be proved. It comes down to the fact ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/332325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Logarithm simplification Simplify: $\log_4(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})$
Can we use the formula to solve this: $\sqrt{a+\sqrt{b}}= \sqrt{\frac{{a+\sqrt{a^2-b}}}{2}}$
Therefore first term will become: $\sqrt{\frac{3}{2}}$ + $\sqrt{\frac{1}{2}}$
$\log_4$ can be written as $\frac{1}{2}\log_2$
Please guide further.... | $$\sqrt{2+\sqrt3}=\sqrt{\frac{4+2\sqrt3}2}=\sqrt{\frac{(\sqrt3)^2+1^2+2\cdot\sqrt3\cdot1}2}=\sqrt{\frac{(\sqrt3+1)^2}2}$$
$$\text{So,}\sqrt{2+\sqrt3}=\frac{\sqrt3+1}{\sqrt2}$$
$$\text{Similarly, }\sqrt{2-\sqrt3}=\frac{\sqrt3-1}{\sqrt2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/332636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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The rational points on the curve: $y^2=ax^4+bx^2+c$. I wonder how to find the rational points on the curve: $y^2=ax^4+bx^2+c$.
Is there infinite rational points on this curve?
For example:$y^2=x^4+3x^2+1.$If we set $y=x^2+k$,then $2kx^2+k^2=3x^2+1$, Can one turn the equation to the form :$y^2=ax^3+bx^2+cx+d$?
Thanks in... | You can turn $y^2 = a x^4 + b x^2 + c$ into $y^2 = x^3 + px + q$ assuming you can find one rational point on $y^2 = a x^4 + b x^2 +c$. The easiest case is when $a$ is square. I do an example of this computation here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/334680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Variation of Pythagorean triplets: $x^2+y^2 = z^3$ I need to prove that the equation $x^2 + y^2 = z^3$ has infinitely many solutions for positive $x, y$ and $z$.
I got to as far as $4^3 = 8^2$ but that seems to be of no help.
Can some one help me with it?
| Take any Pythagorean triplet $(a,b,c)$.
$$\begin{align*}
a^2+b^2 &=c^2\\
a^2\cdot c^4+b^2\cdot c^4&=(c^2)^3\\
(ac^2)^2+(bc^2)^2 &=(c^2)^3
\end{align*}$$
Multiplying $c^{6k-2}$, where $k$ is a natural number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/334839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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"answer_id": 2
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How do I show $\sin^2(x+y)−\sin^2(x−y)≡\sin(2x)\sin(2y)$? I really don't know where I'm going wrong, I use the sum to product formula but always end up far from $\sin(2x)\sin(2y)$. Any help is appreciated, thanks.
| \begin{align}
\Big(\sin(x+y)\Big)^2 - \Big(\sin(x-y)\Big)^2 & = \Big( \sin x\cos y+\cos x\sin y \Big)^2 - \Big( \sin x\cos y+\cos x\sin y \Big)^2 \\[6pt]
& = \Big( \sin^2\cos^2y + 2\sin x\cos y\cos x\sin y + \cos^2x\sin^2y \Big) \\[6pt]
&\phantom{{}=} {}- \Big( \sin^2\cos^2y - 2\sin x\cos y\cos x\sin y + \cos^2x\sin^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/334897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
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How does $\int_{z=-R+0i}^{R+0i} \frac{e^{2iz}-1-2iz}{z^2}\ dx$ become $\int_{-R}^R \frac{\sin^2x}{x^2}\ dx$? While trying to compute $\int_0^\infty \frac{\sin^2 x}{x^2}\ dx$, the author of this book suggests computing $\int_{C_R} \frac{e^{2iz}-1-2iz}{z^2}\ dz$ on a semi-circular contour in the upper half-plane.
The sin... | $$
e^{2iz}-1-2iz=1+2iz+\frac {(2iz)^2}2+\frac{(2iz)^3}6+\ldots\frac {(2iz)^n}{n!}+\ldots-1-2iz=\\
=\frac{(2iz)^2}2+\frac{(2iz)^3}{3!}+\ldots+\frac{(2iz)^{n+2}}{(n+2)!}+\ldots
$$
If you integrate it over $(-R, R)$, obviously all odd powers will drop out since their antiderivatives will be even. So let's consider even po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/335160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find a closed form of the series $\sum_{n=0}^{\infty} n^2x^n$ The question I've been given is this:
Using both sides of this equation:
$$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$$
Find an expression for $$\sum_{n=0}^{\infty} n^2x^n$$
Then use that to find an expression for
$$\sum_{n=0}^{\infty}\frac{n^2}{2^n}$$
This is a... | $\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$
Differentiating (and multiplying with $x$)we have,
$\displaystyle \frac{x}{(1-x)^2}=\sum_{n=0}^{\infty}nx^n$
Differentiating(and multiplying with $x$) we have,
$\displaystyle \frac{[(1-x)^2(1)-(x)2(1-x)(-1)]x}{(1-x)^4}= \frac{x^2+x}{(1-x)^3}=
\sum_{n=0}^{\infty}n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/338852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
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Prove $ax^2+bx+c=0$ has no rational roots if $a,b,c$ are odd If $a,b,c$ are odd, how can we prove that $ax^2+bx+c=0$ has no rational roots?
I was unable to proceed beyond this: Roots are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
and rational numbers are of the form $\frac pq$.
| Suppose that $a,b,c$ are odd. $ax^2+bx+c=0$ has rational roots iff the discriminant is the square of an integer. That is, there is an integer $d$ so that $d^2=b^2-4ac$. Since $a,b,c$ are odd, $d$ must also be odd.
Note that the right hand side of
$$
(b-d)(b+d)=4ac\tag{1}
$$
has exactly two factors of $2$. However, sinc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/339605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 11,
"answer_id": 1
} |
$ (n-7)^2$ is $\Theta(n^2) $ Prove if it's true $$ (n-7)^2 \, \text{is} \, \Theta(n^2) $$
Is this correct?
So far I have:
$ (n-7)^2 \, \text{is} \, O(n^2) \\
n^2 -14n +49 \, \text{is} \, O(n^2) \\
\begin{align} n^2 -14n +49 & \le \, C \cdot n^2 \, , \, n \ge 1 \\
& \le 50n^2 \end{align} \\ $
$ (n-7)^2 \, \text{is... | Easier: $(n-7)^2 > (\frac{n}{2})^2=\Omega(n^2), \ (n-7)^2 < (2n)^2=O(n^2)$ for $n$ large enough, hence $(n-7)^2 = \Theta(n^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/341792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Recursion Formulas of $\int x^{\alpha}\ln x \ \text{dx}$ and $\int\frac{\ln^{\beta}x}{x} \ \text{dx}$ Suppose that I have recursion formulas of $\int x^{\alpha}\ln x \ \text{dx}$ and $\int\frac{\ln^{\beta}x}{x} \ \text{dx}$, and suppose that i found them(integration by parts),
$$\int x^{\alpha}\ln x \ \text{dx}=\frac{x... | For $\alpha=-1$, we have
\begin{align}
\int x^{-1} \ln(x) dx & = \overbrace{\int \dfrac{\ln(x)}x dx = \int t dt}^{t = \ln(x)} = \dfrac{t^2}2 + \text{constant}\\
& = \dfrac{\ln^2(x)}2 + \text{constant} = \dfrac{\ln^{1+1}(x)}{1+1} + \text{constant}
\end{align}
The expression you have is
\begin{align}
\int x^{\alpha} \ln(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ $$\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$$
I have attempted this question by expanding the left side using the cosine sum and difference formulas and then multiplying, and then simplifying till I replicated the identity on the right. I am not s... | Another way of looking at it:
$$\begin{align*}
\cos(A+B)\cos(A-B) &= \frac14\left(e^{i(A+B)}+\frac{1}{e^{i(A+B)}}\right)\left(e^{i(A-B)}+\frac{1}{e^{i(A-B)}}\right) \\
&= \frac14\left[(e^{iA})^2 + \frac{1}{(e^{iA})^2} + (e^{iB})^2 + \frac{1}{(e^{iB})^2}\right] \\
&= \frac14\left[(e^{iA})^2 + \frac{1}{(e^{iA})^2} + 2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/345703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
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Simplifying $\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt {5 +\cdots}}}}$ How to simplify the expression:
$$\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\cdots}}}}.$$
If I could at least know what kind of reference there is that would explain these type of expressions that would be very helpful.
Thank you.
| Let $x=\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}$.
Then $x^2=5+2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}$.
So, $x^2-5=2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}$
Remember that $x=\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}$
So, $2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}$. So, $2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/346303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
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Proving $\tan \left(\frac{\pi }{4} - x\right) = \frac{{1 - \sin 2x}}{{\cos 2x}}$ How do I prove the identity:
$$\tan \left(\frac{\pi }{4} - x\right) = \frac{{1 - \sin 2x}}{{\cos 2x}}$$
Any common strategies on solving other identities would also be appreciated.
I chose to expand the left hand side of the equation and g... | $$
\begin{aligned}
& \frac{1-\sin 2 x}{\cos 2 x} \\
=& \frac{(\cos x-\sin x)^{2}}{\cos ^{2} x-\sin ^{2} x} \\
=&\frac{(\cos x-\sin x)^{2}}{(\cos x+\sin x)(\cos x-\sin x)} \\
=& \frac{\cos x-\sin x}{\cos x+\sin x} \\
=& \frac{1-\tan x}{1+\tan x} \\
=& \tan \left(\frac{\pi}{4}-x\right)
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/347248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Infinite product simplification I found the following identity $\prod_{k=0}^{\infty}(1+\frac{1}{2^{2^k}-1})=\frac{1}{2}+\sum_{k=0}^{\infty}\frac{1}{\prod_{j=0}^{k-1}(2^{2^j}-1)}$
My first thought was to use eulers identity somehow $\prod_{k=1}^{\infty}(1+z^k)=\prod_{k=1}^{\infty}(1-z^{2k-1})^{-1}$ but it does not help ... | To prove this identity, observe that the left-hand side is
\begin{eqnarray*}
\prod_{k\ge 0} (1+\frac{1}{2^{2^k}-1})
&=&
\prod_{k\ge 0} \frac{2^{2^k}}{2^{2^k}-1}\\
&=& \prod_{k\ge 0} \frac{1}{1-2^{-2^k}}\\
&=& \prod_{k\ge 0} (1 + 2^{-2^k} + 2^{-2\cdot 2^k} + 2^{-3\cdot 2^k} + \cdots)
\end{eqnarray*}
and then, partially ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/349404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Integral solutions of hyperboloid $x^2+y^2-z^2=1$ Are there integral solutions to the equation $x^2+y^2-z^2=1$?
| Recall the Brahmagupta formula
$$(ad-bc)^2 + (ac+bd)^2 = (a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2$$
Now find $a,b,c,d$ such that say $ad-bc = 1$. Then set $z = ac+bd$, $x=ac-bd$ and $y = ad+bc$ to get a solution.
For instance, one such one parameter family solution following the above procedure is
$$x = 2t, y = 2t^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/351491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 1
} |
Integrating $\int \sqrt{(2+\sin{3x})\cos{3x}}{dx}$ Integrating $$\int \sqrt{(2+\sin{3x})\cos{3x}}\mathrm{d} x$$
Let, $\sqrt{2+\sin{3x}}=t$
then, $\frac{3\cos{3x}}{2\sqrt{2+\sin{3x}}}\mathrm{d} x=\mathrm{d} t$
Integral = $\frac{2}{3}\int\frac{t^2}{\sqrt{1-(t^2-2)^2}}\mathrm{d} t$
Integral = $\frac{2}{3}\int\frac{t^2}{\s... | Some idea: by parts
$$u:=t\;\;,\;\;u'=1\\v'=\frac{t}{\sqrt{1-(t^2-2)^2}}\;\;,\;\;v=\frac{1}{2}\arcsin(t^2-2)$$
so
$$\frac{2}{3}\int\frac{t^2}{\sqrt{1-(t^2-2)^2}}dt=-\frac{1}{3}t\arcsin(t^2-2)-\frac{1}{3}\int\arcsin(t^2-2) dt\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/351721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Triangle proof using law of sines
In triangle $ABC$, suppose that angle $C$ is twice angle $A$. Use the law of sines to show that $ab= c^2 - a^2$.
| A common way to invoke the Law of Sines is to note what its ratios equal:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
where $d$ is the circum-diameter of $\triangle ABC$. Thus,
$$a = d \sin A \qquad b = d \sin B \qquad c = d \sin C$$
Consequently, the identity in question, upon division by $d^2$, is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/354452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Least square solution of a matrix Determine the least squares solution to the system
$$\begin{pmatrix} 1 & 2 & 1 \\ 1&3&2\\2&5&3\\2&0&1 \\ 3 &1&1\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}-1\\2\\0\\1\\-2\end{pmatrix}$$
So the previous equation was Ax = B
To find the least square regression we need ... | You have computed $AA^T$, not $A^TA$. The correct $A^TA$ should be:
$$\left(\begin{array}\\
19 & 18 & 14 \\
18 & 39 & 24 \\
13 & 24 & 16
\end{array}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/355486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Quadratic equation with tricky conditions. Need to prove resulting inequalities. The roots of the quadratic equation $ax^ 2-bx+c=0,$ $a>0$, both lie within the interval $[2,\frac{12}{5}]$. Prove that:
(a) $a \leq b \leq c <a+b$.
(b) $\frac{a}{a+c}+\frac{b}{b+a}>\frac{c}{c+b}$
So we can use the quadractic formula and ob... | As shown in other answers, let $\alpha,\beta$ be the two real roots of $ax^2-bx+c=0$, i.e. $$ax^2-bx+c=a(x-\alpha)(x-\beta)\iff \alpha+\beta=\frac{b}{a},\ \alpha\beta=\frac{c}{a}.$$
Since $\alpha,\beta\in[2,2.4]$ and $a>0$,
$$\frac{b}{a}=\alpha+\beta\ge 4\Rightarrow a\le \frac{b}{4}<b;$$
$$\frac{b}{c}=\frac{\alpha+\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/356594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$ \int_{C}^{} \frac{ydx-(x+1)dy}{x^2+y^2+2x+1} $ where $C$ is $ |x|+|y|=4 $ Calculate
$$
\int_{C}^{} \frac{ydx-(x+1)dy}{x^2+y^2+2x+1}
$$
where C is the curve
$$
|x|+|y|=4
$$
counterclock-wise, a full revolution.
Answer: $$-2\pi$$
So, I've tried to figure this out for a while now. I've tried Green's formula, that g... | I think we could exploit the fact that the integrand is a gradient of the fundamental function for a Poisson equation.
Notice:
$$
\nabla \big(\ln((x+1)^2+y^2)\big) = 2\left(\frac{x+1}{(x+1)^2+y^2},
\frac{y}{(x+1)^2+y^2}\right)
$$
Denote the square as $\Omega$, the infinitesimal tangent of the curve is $(dx,dy)$, then r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/360668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Summation of Arithmetic-Geometric Series I've been working through my homework paper, and I've come across this question. Now I'm confident in what I have done for the most part, but I am stuck at the end.
I have this recurrence relation, that I am supposed to solve using backward substitution.
$$
T(n) = 2T(n/2) + n +... | Well $$\sum_{j=0}^{k-1} 2^j = 2^k-1$$ since it's just a geometric sum of ratio 2.
But let's check your substitution:
$$T(2^k) = 2 T(2^{k-1}) + 2^k + 11 = 2 \left(2 T(2^{k-2}) + 2^{k-1} + 11\right) + 2^k + 11$$ or
$$2^2T(2^{k-2}) + 2\times 2^k + (2 + 1)\times 11$$
Adding another term gives
$$2^2\left(2T(2^{k-3}) + 2^{k-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/362656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Finding Pi variables from matrix. From PageRank Algorithm. $$\pmatrix{\pi_1 & \pi_2 & \pi_3} = \pmatrix{\pi_1 & \pi_2 & \pi_3}\pmatrix{\frac{1}{6} & \frac{4}{6} & \frac{1}{6} \\ \frac{5}{12} & \frac{2}{12} & \frac{5}{12} \\ \frac{1}{6} & \frac{4}{6} & \frac{1}{6} \\}$$
Answer: $$\pi_1=\frac{5}{18}$$ $$\pi_2=\frac{8}{1... | The key terms are the Perron-Frobenius theorem and the stable state of a Markov chain. In practice, this vector $\pi$ is found as the eigenvector for the eigenvalue $1$, and is normalized so that the sum of its components is $1$.
Subtract $1$ along the diagonal to get
$$B=\pmatrix{-\frac{5}{6} & \frac{4}{6} & \frac{1}... | {
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"url": "https://math.stackexchange.com/questions/365443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to expand this taylor series and find radius of convergence f(x)= √(1-x) at x=0
How do you find the taylor series and radius of convergence?
| $$(1-x)^\frac{1}{2}=1+\sum_{k=1}^\infty(-1)^k\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k+1)}{k!}x^k$$
and
\begin{align}\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k+1)}{k!}&=\frac{(-1)(-3)\cdots(3-2k)}{2^kk!}=(-1)^{k-1}\frac{1\times3\times\cdots\times(2k-3)}{2^kk!}\\&=(-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/365789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove this equation(18)? my book have this:let $f(x)=\sqrt{nx},n\in N,0<x<1$, then use Taylor we have
$$f(x)=1+f'(x_{0})(x-x_{0})-\dfrac{g(x)}{2}$$
where $x_{0}=\dfrac{1}{n},g(x)=(\sqrt{nx}-1)^2$
my question: $g(x)=(\sqrt{nx}-1)^2$? why?
| $$f(x)=\sqrt{nx}, \quad f'(x)=\frac{1}{2}\cdot\frac{1}{\sqrt{nx}}\cdot n = \frac{n}{2\sqrt{nx}}$$
$$x_0=1/n, \quad f(x_0)=\sqrt{n\cdot(1/n)}=1, \quad f'(x_0)=\frac{n}{2\sqrt{n\cdot(1/n)}}=\frac{n}{2}$$
$$\sqrt{nx} = 1 + \frac{n}{2} \cdot (x-1/n) - \frac{g(x)}{2}$$
$$2\sqrt{nx} = 2 + (nx-1) - g(x)$$
$$g(x) = 2 + (nx-1) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/366222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Symmetry properties of $\sin$ and $\cos$. Why does $\cos\left(\frac{3\pi}{2} - x\right) = \cos\left(-\frac{\pi}{2} - x\right)$? For a question such as:
If $\sin(x) = 0.34$, find the value of $\cos\left(\frac{3\pi}{2} - x\right)$.
The solution says that:
\begin{align*}
\cos\left(\frac{3\pi}{2} - x\right) &= \cos\left... | HINT : as $\cos(2\pi+y)=\cos2\pi\cos y-\sin2\pi\sin y=\cos y$
As $$\left(\frac{3\pi}2-x\right)-\left(-\frac\pi2-x\right)=2\pi$$
$$\implies \left(\frac{3\pi}2-x\right)=2\pi+\left(-\frac\pi2-x\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/367938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to estimate the following integral: $\int_0^1 \frac{1-\cos x}{x}\,dx$
How to estimate the following integral?
$$
\int_0^1 \frac{1-\cos x}{x}\,dx
$$
| Since $\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$,
$1-\cos(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n}}{(2n)!}$,
so
$\frac{1-\cos(x)}{x} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n-1}}{(2n)!}$.
Therefore
$\begin{align}
\int_0^1 \frac{1-\cos(x)}{x} dx
&= \int_0^1 dx \sum_{n=1}^{\infty} \frac{(-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Expand into power series $f(x)=\log(x+\sqrt{1+x^2})$ As in the topic, I am also supposed to find the radius of convergence.
My solution: $$\log(x+\sqrt{1+x^2})=\log \left ( x(1+\sqrt{\frac{1}{x^2}+1})\right )=\log(x)+\log(1+\sqrt{\frac{1}{x^2}+1})$$Now I tried to use expansion for $\log(1+x)$ as $x\rightarrow0:\log(1+... | Hint:
$$\dfrac{d}{dx}\ln\left(x+\sqrt{1+x^2}\right)=\dfrac{1}{\sqrt{1+x^2}}=(1+x^2)^{-\frac{1}{2}}.$$
Radius of convergence for the $\ln\left(x+\sqrt{1+x^2}\right)$ will be the same as for $(1+x^2)^{-\frac{1}{2}}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/369733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving recursive sequence using generating functions $$a_{0} = 0$$
$$a_{1} = 1$$
$$a_{n} = a_{n-1} - a_{n-2}$$
I have to find the solution of this equation ($a_{n} = ...$, non-recursive, you know what I mean...). So let's pretend that:
$$ A(x) = \sum_{n=0}a_{n}x^{n}$$
Using this formula and the recursive equation I'm ... | The trick is to forget about convergence. To encourage self-study, I will feature a slightly different series (the Fibonacci numbers), but the technique is the same.
$$a_n=a_{n-1}+a_{n-2} \implies a_nx^n=xa_{n-1}x^{n-1}+x^2a_{n-2}x^{n-2}\quad \text{for n}=2,3,\dots$$
Let $A(x)$ be the generating function of the sequenc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/371714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Finding a coefficient using generating functions I need to find a coefficient of $x^{21}$ inside the following expression:
$$(x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{8}$$
I think the only way (using generating functions) is to express the parentheses content with a generating function.
The generating function for the $... | $(x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{8} = x^{16}(1+x+x^2+x^3+x^4)^8$, so you should find the coefficient of $x^5$ in $(1+x+x^2+x^3+x^4)^8$. If you write $5$ as sum of $8$ terms $4\geq k_i\geq 0$ in all possible ways, you'll to see in how many ways you can get $x^5$. It is also known what is the number of sums of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/371914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Prove That $x=y=z$ If $x, y,z \in \mathbb{R}$,
and if
$$ \left ( \frac{x}{y} \right )^2+\left ( \frac{y}{z} \right )^2+\left ( \frac{z}{x} \right )^2=\left ( \frac{x}{y} \right )+\left ( \frac{y}{z} \right )+\left ( \frac{z}{x} \right ) $$
Prove that $$x=y=z$$
| Let $$ a=\frac{x}{y} $$ $$ b=\frac{y}{z} $$ $$ c=\frac{z}{x} $$
Then $$abc=1$$
$$ a^2+b^2+c^2=a+b+c$$
which implies
$$ (a-0.5)^2+(b-0.5)^2+(c-0.5)^2=0.75$$
I was struck up here, i would appreciate if any one helps me here
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
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Prove the inequality $\frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}\ge{3}$ Let a, b, c be the three side lengths of a triangle. Prove that
$$\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\geq 3$$
Under what conditions is equality obtained?
| solution 1:
let $$b+c-a=x,a+c-b=y,a+b-c=z$$,
then
$$a=\dfrac{y+z}{2},b=\dfrac{x+z}{2},c=\dfrac{x+y}{2}$$
then
$$\sum\dfrac{b}{a+c-b}=\dfrac{1}{2}\left[\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{y}{z}+\dfrac{z}{y}\right)+\left(\dfrac{x}{z}+\dfrac{z}{x}\right)\right]\ge 3$$
solution 2:
by cauchy-Schwarz inequali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/374373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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(USAJMO)Find the integer solutions:$ab^5+3=x^3,a^5b+3=y^3$ Find the integer solutions:
$$a·b^5+3=x^3,a^5·b+3=y^3$$
This is the first problem of today's USAJMO (has finished),I only find a trival result that $x\equiv y \pmod6$ and $abxy≠0 \pmod 3$.
Thanks in advance!
| If $3 \mid a$, then $3\| (a^5b+3)=y^3$, a contradiction. Thus $3\nmid a$. Similarly $3 \nmid b$, so $3 \nmid x, y$. Note that if $3 \nmid n$, then $n^3 \equiv \pm 1 \pmod{9}$. Thus $x^3-3, y^3-3 \equiv 5, 7 \pmod{9}$, so
$$1\equiv(ab)^6 \equiv (x^3-3)(y^3-3) \equiv 4, 7, 8 \pmod{9}$$
We get a contradiction, so there ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/377800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
recursive sequence Given $0<a<b$, $\forall n$ define $x_n$ as $x_1=a$, $x_2=b$, $x_{n+2}=\frac{x_n+x_{n+1}}{2}$. Show that $(x_n)$ converges and find the limit.
In order to prove the convergence, I claim that $|x_{n+2}-x_{n+1}|\le\lambda|x_{n+1}-x_{n}|$, $0<\lambda<1$. In fact, $|x_{n+2}-x_{n+1}|\le|\frac{x_{n+1}+x{n}}... | Simpler: For $A(z) = \sum_{n \ge 0} x_{n + 1} z^n$ the recurrence directly translates into
$$
\frac{A(z) - x_1 - x_2 z}{z^2}
= \frac{1}{2} \frac{A(z) - x_1}{z} + \frac{A(z)}{2}
$$
This gives:
$$
A(z) = \frac{2 a + (2 b - a) z}{2 - z - z^2}
= \frac{2 a - 2 b}{3} \frac{1}{1 + z / 2} + \frac{a + 2 b}{3} \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/378835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Expanding $\frac{1}{1-z-z^2}$ to a power series. How would you expand the analytic function $$\frac{1}{1-z-z^2}$$ to a series of the form $$\sum_{k=0}^\infty a_k z^k \, \, ?$$
| Three ways:
*
*Write as partial fractions:
$$
\frac{1}{1 - z - z^2}
= \frac{1}{(1 - \tau z) (1 - \overline{\tau} z)}
= \frac{\tau}{\sqrt{5} (1 - \tau z)}
- \frac{\overline{\tau}}{\sqrt{5} (1 -\overline{\tau} z)}
$$
Here $\tau$ is the positive root of $r^2 - r - 1 = 0$, $\overline{\tau}$ the negative one (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/379483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding the kth term of an iterated sequence The sequence $x_0, x_1, \dots$ is defined through $x_0 =3, x_1 = 18$ and $x_{n+2} = 6x_{n+1}-9x_n$ for $n=0,1,2,\dots\;$. What is the smallest $k$ such that $x_k$ is divisible by $2013$?
| HINT:
Using Characteristic equation, $$r^2-6r+9=0$$
So, $r=3,3$
$$\text{So,}x_n=(An+B)3^n$$ where $A,B$ are arbitrary constants
$3=x_0=B\implies B=3$ and $18=(A+3)\cdot3\implies A=3\implies x_n=(n+1)3^{n+1}$
Now, $2013=3\cdot11\cdot 61$
As $(3,61)=(3,11)=1\implies (n+1)$ must be divisible by $11\cdot 61=671$
and $(n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/383381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int {\sin 2\theta \over 1 + \cos \theta} \, d\theta $, using the substitution $u = 1 + \cos \theta $
Evaluate $$\int_0^{\pi \over 2} {{\sin 2\theta } \over {1 + \cos \theta }} \, d\theta$$ using the substitution $u = 1 + \cos \theta $
Using
$$\begin{align}
u &= 1 + \cos \theta \\
\frac{du}{d\theta} &= -\s... | Even easier $(t=\cos x$):
$$
\int_{a}^{b}\frac{\sin 2x \, dx}{1+\cos x} =-2 \int_{a}^{b} \frac{\cos x \, d \cos x}{1+\cos x}=-2\int _{\varphi(a)}^{\varphi(b)}\frac{(t+1-1) \, dt}{1+t}
$$
Can you handle from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/387635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What have I done wrong in solving this problem with indices rules? The question asks to simplify:
$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}}.$$
So I used $(a^m)^n=a^{mn}$ to get
$$\dfrac{25}{4}x^{-2} = \dfrac{25}{4} \times \dfrac{1}{x^2} = \frac{25}{4x^2} = \frac{25}{4}x^{-2}$$
However, this isn't the answer, and ... | Just to point out the "rule" you forgot:
It is correct that $$(a^m)^n=a^{mn}$$
But you have more than $(x^4)^{-1/2}$ to consider. You need to apply the fact that when a product and/or quotient of numbers/variables, enclosed in parentheses, is raised to a power, you need to distribute that power across the product in pa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/388748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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An integral involving Fresnel integrals $\int_0^\infty \left(\left(2\ S(x)-1\right)^2+\left(2\ C(x)-1\right)^2\right)^2 x\ \mathrm dx,$ I need to calculate the following integral:
$$\int_0^\infty \left(\left(2\ S(x)-1\right)^2+\left(2\ C(x)-1\right)^2\right)^2 x\ \mathrm dx,$$
where
$$S(x)=\int_0^x\sin\frac{\pi z^2}{2... | Following on from Ron...
Make the substitutions: $x=\sqrt{y}$, $p=1/2+1/2\,\sqrt {1+4\,r}$, to get:
$\displaystyle \dfrac{64}{{\pi }^{2}}\,\int _{0}^{\infty }\!x \left( \int _{1}^{\infty }\!{\frac {\sin
\left( 2\,\pi \,{x}^{2}p \left( p-1 \right) \right) }{p}}{dp}
\right) ^{2}{dx}$,
$\displaystyle=\dfrac{32}{{\pi }^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/390847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "46",
"answer_count": 2,
"answer_id": 0
} |
Reducibility of $x^{2n} + x^{2n-2} + \cdots + x^{2} + 1$ Just for fun I am experimenting with irreducibility of certain polynomials over the integers. Since $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$, I thought perhaps $x^6+x^4+x^2+1$ is also reducible. Indeed:
$$x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$
Let $f_n(x)=x^{2n}+x^{2n-2}+\cdots + ... | Let $g_n(x) = x^n + x^{n-1} + \ldots + x + 1$. Then $g_n(x) = \frac{x^{n+1}-1}{x-1}$. Your polynomial $f_n(x) = g_n(x^2)$. The factorization of $x^n-1$ over $\mathbb{Q}$ is known:
$$
x^n-1 = \prod_{d\mid n}{\Phi_d(x)},
$$
where $\Phi_d(x)$ is the $d$th cyclotomic polynomial. Thus, the factorization of $g_n(x)$ will lea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/391086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
} |
Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$ Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me?
Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$
Isolate one of the square roots: $\sqrt{(2x-5... | $\sqrt{2x-5} - \sqrt{x-1} = 1$
Let $\sqrt{2x-5} + \sqrt{x-1} = y$
Multiplying, we get
$(2x-5) - (x-1) = y$
$y = x - 4$
\begin{align}
\sqrt{2x-5} + \sqrt{x-1} &= x - 4 \\
\sqrt{2x-5} - \sqrt{x-1} &= 1 & \text{subtract}\\
\hline
2\sqrt{x-1} &= x-5 \\
4x-4 &= x^2 - 10x + 25 \\
x^2 -14x + 29 &= 0 \\
x &= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/392308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Simplifying fractions - Ending up with wrong sign I've been trying to simplify this
$$
1-\frac{1}{n+2}+\frac{1}{(n+2) (n+3)}
$$
to get it to that
$$
1-\frac{(n+3)-1}{(n+2)(n+3)}
$$
but I always end up with this
$$
1-\frac{(n+3)+1}{(n+2)(n+3)}
$$
Any ideas of where I'm going wrong?
Wolfram Alpha gets it to correct form ... | Just in case you want to see a full simplification:
\begin{align*}
1-\frac{1}{n+2}+\frac{1}{(n+2)(n+3)} &= \frac{(n+2)(n+3)}{(n+2)(n+3)} - \frac{(n+3)}{(n+2)(n+3)}+\frac{1}{(n+2)(n+3)} \\
&= \frac{(n+2)(n+3)-(n+3)+1}{(n+2)(n+3)}\\
&= \frac{(n^2+5n+6) -n-2 }{(n+2)(n+3)} \\
&= \frac{n^2+4n+4}{(n+2)(n+3)} \\
&= \frac{(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/393530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Knowing that $n= 3598057$ is a product of two different prime numbers and that 20779 a square root of $1$ mod $n$, find prime factorization of $n$.
Knowing that $n= 3598057$ is a product of two different prime numbers and that 20779 is a square root of $1$ mod $n$, find prime factorization of $n$.
What I have done so... | From the condition
(x-1)(x+1)=0 mod n
you can conclude (a) n divides (x-1), or (b) n divides (x+1), or (c) some factor of n divides (x-1) and some other factor of n divides (x+1)
Since (a) and (b) are ruled out (x is not plus or minus 1) we get two factors as
f1 = gcd(x-1,n)
f2 = gcd(x+1,n)
If n is not known to be a pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Partial fraction expansion two variables How to expand
$$\frac{y}{(x-y)(y-1)}$$
by partial fraction expansion.
| $$\text{Plot3D}\left[\left\{\frac{y^4}{(x-1) (y-1)}-\frac{y \left(x^2 y^2+x^2 y+x^2+x y^2+x y+y^2\right)}{x^3},\frac{y}{(y-1) (x-y)},x \left(y+\frac{1}{y}+1\right)+y+1\right\},\{x,-2,2\},\{y,-2,2\},\text{PlotLegends}\to \text{Automatic}\right]$$
use this
$$\frac{y^4}{(x-1) (y-1)}-\frac{y \left(x^2 y^2+x^2 y+x^2+x y^2+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Integrate ${\sec 4x}$ How do I go about doing this? I try doing it by parts, but it seems to work out wrong:
$\eqalign{
& \int {\sec 4xdx} \cr
& u = \sec 4x \cr
& {{du} \over {dx}} = 4\sec 4x\tan 4x \cr
& {{dv} \over {dx}} = 1 \cr
& v = x \cr
& \int {\sec 4xdx} = x\sec 4x - \int {4x\sec 4x\tan 4... | Use two substitutions. The first substitution transforms the integrand into $\sec \theta$, whose evaluation was asked here. The second substitution is the Weirstrass substitution. (See comment below). In the present case the second integral becomes an easy table integral:
$$\begin{eqnarray*}
\int \sec 4x\,dx &=&\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
A question about an inequality How to get the solution set of the inequality $$\left ( \frac{\pi}{2} \right )^{(x-1)^2}\leq \left ( \frac{2}{\pi} \right )^{x^2-5x-5}?$$
| Multiplying either sides by $\left(\frac\pi2\right)^{(x^2-5x-5)}$
$$\left(\frac\pi2\right)^{(x-1)^2+(x^2-5x-5)}\le 1\iff \left(\frac\pi2\right)^{2x^2-7x-4}\le 1=\left(\frac\pi2\right)^0$$
$\implies 2x^2-7x-4\le0$ as $\pi>2\iff \frac\pi2>1,$
Now, if $(x-a)(x-b)\le0$ where $a\le b,$ $a\le x\le b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/395424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
A limit on binomial coefficients Let $$x_n=\frac{1}{n^2}\sum_{k=0}^n \ln\left(n\atop k\right).$$ Find the limit of $x_n$.
What I can do is just use Stolz formula. But I could not proceed.
| The limit is $\frac{1}{2}$.
We have
$$\begin{eqnarray*}
\sum_{k=0}^n \log {n\choose k}
&=& \sum_{k=0}^n \log n! - \sum_{k=0}^n \log k! - \sum_{k=0}^n \log (n-k)! \\
&=& (n+1)\log n! - 2\sum_{k=1}^n \log k!.
\end{eqnarray*}$$
But
$$\begin{eqnarray*}
\sum_{k=1}^n \log k!
&=& \sum_{k=1}^n \sum_{j=1}^k \log j \\
&=& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/395507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 0
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Using generating functions, Find a closed formula to next expression: $\sum_{k=0}^m{k(k+2)}$ Using generating functions, Find a closed formula to next expression:
$\sum_{k=0}^m{k(k+2)}$
If i use calculus power series rules, The question is fairly simple. But how can i find the proper relation with generating functions... | Let the desired sum be given by
\begin{align}
S_{m} = \sum_{k=0}^{m} k(k+2).
\end{align}
Using the generating function method it is seen that the expression to calculate is
\begin{align}
\sum_{m=0}^{\infty} S_{m} \, t^{m}.
\end{align}
The calculation of this series is as follows.
\begin{align}
\sum_{m=0}^{\infty} S_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Uniform convergence of $\sum_{n=1}^{\infty} \frac{\sin(n x) \sin(n^2 x)}{n+x^2}$ I'm not sure wether or not the following sum uniformly converge on $\mathbb{R}$ :
$$\sum_{n=1}^{\infty} \frac{\sin(n x) \sin(n^2 x)}{n+x^2}$$
Can someone help me with it? (I can't use Dirichlet' because of the areas where $x$ is close to $... | The series does converge uniformly. For the proof, put $S_n(x) = \sum_{k = 0}^n \sin{(kx)}\sin{(k^2 x)}$ for $n\geq 0$. The general idea is to use summation by parts to reduce ourselves to showing that $S_n(x)$ is bounded uniformly, and then to prove that by giving a closed form for $S_n(x)$.
First, the summation by pa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/397097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 1,
"answer_id": 0
} |
Laurent expansion problem Expand the function $$f(z)=\frac{z^2 -2z +5}{(z-2)(z^2+1)} $$ on the ring $$ 1 < |z| < 2 $$
I used partial fractions to get the following $$f(z)=\frac{1}{(z-2)} +\frac{-2}{(z^2+1)} $$
then
$$ \frac{1}{z-2} = \frac{-1}{2(1-z/2)} = \frac{-1}{2} \left[1+z/2 + (z/2)^2 + (z/2)^3 +\cdots\right]
... | Hint:
$$\frac{-2}{1+z^2} = \sum_{n=0}^\infty (-(-i)^n-i^n) z^n,\quad\mid z\mid<1$$
and:
$$\frac{-2}{1+z^2} = \sum_{n=0}^\infty (-1+z)^n (-1+i) (2^{-1-n} ((-1-i)^n+i (-1+i)^n))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/397696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
linear Transformation of polynomial with degrees less than or equal to 2 I would like to determine if the following map $T$ is a linear transformation:
\begin{align*}
T: P_{2} &\to P_{2}\\
A_{0} + A_{1}x + A_{2}x^{2} &\mapsto A_{0} + A_{1}(x+1) + A_{2}(x+1)^{2}
\end{align*}
My attempt at solving:
\begin{align}
T(p + q)... | The idea is right but you are starting from what you have to proof
Hint : $T(p+q)=T(A_0+B_0+(A_1+B_1)x+(A_2+B_2)x^2)=...=T(p)+T(q)$.
I see that you had the right idea maybe this could make you start
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/397748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integration of $\displaystyle \int\frac{1}{1+x^8}\,dx$
Compute the indefinite integral
$$
\int\frac{1}{1+x^8}\,dx
$$
My Attempt:
First we will factor $1+x^8$
$$
\begin{align}
1+x^8 &= 1^2+(x^4)^2+2x^4-2x^4\\
&= (1+x^4)^2-(\sqrt{2}x^2)^2\\
&= (x^4+\sqrt{2}x^2+1)(x^4-\sqrt{2}x^2+1)
\end{align}
$$
Then we can rewrite ... | Why not splitting up in fractions until you have first degree polynomials in the nominators?
$$\frac{1}{1+x^8}=\frac{A}{x-e^{i\pi/8}}+\frac{B}{x-e^{-i\pi/8}}+\frac{C}{x-e^{i3\pi/8}}+\frac{D}{x-e^{-i3\pi/8}}+\frac{E}{x-e^{i5\pi/8}}+\frac{F}{x-e^{-i5\pi/8}}+\frac{G}{x-e^{i7\pi/8}}+\frac{H}{x-e^{-i7\pi/8}}$$
or if you pre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Solve equations $\sqrt{t +9} - \sqrt{t} = 1$ Solve equation: $\sqrt{t +9} - \sqrt{t} = 1$
I moved - √t to the left side of the equation $\sqrt{t +9} = 1 -\sqrt{t}$
I squared both sides $(\sqrt{t+9})^2 = (1)^2 (\sqrt{t})^2$
Then I got $t + 9 = 1+ t$
Can't figure it out after that point.
The answer is $16$
| $$\sqrt{t +9} - \sqrt{t} = 1$$
Multiplying by $\sqrt{t +9} + \sqrt{t}$ you get
$$9=\sqrt{t +9} +\sqrt{t} $$
Now adding
$$\sqrt{t +9} + \sqrt{t} =9$$
$$\sqrt{t +9} - \sqrt{t} = 1$$
you get
$$\sqrt{t+9}=5 \Rightarrow t=25-9 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/399199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 2
} |
Finding the sum of a Taylor expansion I want to find the following sum:
$$
\sum\limits_{k=0}^\infty (-1)^k \frac{(\ln{4})^k}{k!}
$$
I decided to substitute $x = \ln{4}$:
$$
\sum\limits_{k=0}^\infty (-1)^k \frac{x^k}{k!}
$$
The first thing I noticed is that this looks an awful lot like the series expansion of $e^x$:
$$
... | You started fine, but then you got sidetracked:
$$\sum\limits_{k\ge 0}(-1)^k \frac{(\ln{4})^k}{k!}=\sum_{k\ge 0}\frac{(-\ln 4)^k}{k!}=e^{-\ln 4}=\frac1{e^{\ln 4}}=\frac14\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.