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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
*
*$$ \begin{pmatrix}1& 0 & 0 \\
50 & 1 & 0\\
50 & 0 & 1 \end{pmatrix}$$
*$$\begin{pmatrix}1& 0 & 0 \\
48 & 1 & 0\\
48 & 0 & 1 \end{pmatrix}$$
*$$\begin{pmatrix}1& 0 & 0 \\
25 & 1 & 0\\
25 & 0 & 1 \end{pmatrix}$$
*$$\begin{pmatrix}1& 0 & 0 \\
24 & 1 & 0\\
24 & 0 & 1\end{pmatrix}$$
I am stuck on this problem. Can anyone help me please...............
| 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 $L^2=0$. Therefore
$$
A^{50} = (I+L)^{25} = I+{25\choose 1}L+\sum_{k=2}^{25}{25\choose k}L^k=I+25L
$$
and hence the answer is 3.
| {
"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 $\left \lfloor \dfrac{n +86}{343} \right\rfloor$
Number of multiples of $7^{2k}$ in $S_n$ is $$\left \lfloor \dfrac{n + 7^{2k} - (7^{2k}-1)/4}{7^{2k}} \right\rfloor = \left \lfloor \dfrac{4n + 3\cdot 7^{2k} + 1}{4 \cdot 7^{2k}} \right\rfloor$$
Number of multiples of $7^{2k+1}$ in $S_n$ is $$\left \lfloor \dfrac{n + 7^{2k+1} - (3 \cdot 7^{2k+1}-1)/4}{7^{2k+1}} \right\rfloor = \left \lfloor \dfrac{4n + 7^{2k+1}+1}{4 \cdot 7^{2k+1}} \right\rfloor$$
Hence, the highest power of $7$ dividing $S_n$ is
$$\sum_{k=1}^{\infty} \left(\left \lfloor \dfrac{4n + 7^{2k-1}+1}{4 \cdot 7^{2k-1}} \right\rfloor + \left \lfloor \dfrac{4n + 3\cdot 7^{2k} + 1}{4 \cdot 7^{2k}} \right\rfloor \right)$$
I computed the sum above using WolframAlpha and it agrees with $\dfrac{10^{100}}6 - \dfrac{17}3$
| {
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"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:
$1000000 = 10^{6} = 2^{6} \cdot 5^{6}$ and then to consider this as the product of three factors i.e.
$10^6 = 2^6 \cdot 5^6$
$= 2^a 5^p \cdot 2^b5^q \cdot 2^c 5^r$ (where $a+b+c = 6$ and $p+q+r = 6$).
However there are repetitions here because $2^3\,5^3 \cdot 2^2\,5^2 \cdot 2^1\,5^1$ is the product of the same three factors as $2^2\,5^2 \cdot 2^3\,5^3 \cdot 2^1\,5^1$.
I think there are 139 such factors.
| 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",
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"question_score": "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 $x, min(1,9-c^2)>x^2>0\implies 9-c^2>0\implies c^2<9$ else there will be no solution.
(ii)If $\log_b x^2>0 \iff x^2>1$
then we need $\log_b(\log_c\sqrt{9-x^2})<0\implies\log_c\sqrt{9-x^2}<1\implies x^2>9-c^2$
$\implies x^2>max(1,9-c^2)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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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 using simple methods.
Edit
Please don't use l'Hosplital rule. We are before derivatives, don't know how to use it correctly yet. Thanks!
| 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 result for $\sqrt[3]{x+20}-3$ using $u-v=\frac{u^3-v^3}{u^2+uv+v^2}$ with $u=\sqrt[3]{x+20}$ and $v=3$, that gives $u^3-v^3 = x-7$.
Note then that the $x-7$s cancel out, and you get an expression where none of the numerators or denominators approach zero as $x\to 7$, so you can finally just plug in $x=7$ in that expression.
Cancelling out the $x-7$ terms, you get:
$$(a^3 + a^2b + ab^2 + b^3)\left(\frac{1}{\sqrt{x+2}+3}-\frac{1}{u^2+uv+v^2}\right)$$
But as $x\to 7$ $a\to b=2$ and $u\to v=3$.
So the limit is:
$$(4\cdot2^3)\left(\frac{1}{6}-\frac{1}{3\cdot 3^2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/275990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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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 done this with $f(x) = 0$ where $x = \sqrt{a} + \sqrt{b}$ through an iterative method which looks like this (forgive me for my pseudo code):
*
*var x = sqrt(a) + sqrt(b)
*Set x to x multiplied by the conjugate of x
*Repeat from step 2 until all square roots are gone
The full calculation looks like this (the exponent signs disappeared, sorry about that):
$(x-(\sqrt{a}+\sqrt{b}))(x+(\sqrt{a}+\sqrt{b}))
= x^2-(\sqrt{a} +\sqrt{b})^2 \\
= x^2-(a+2\sqrt{a}\sqrt{b} +b) \\
= x^2-a-b-2\sqrt{a}\sqrt{b} $
$((x^2-a-b)-(2\sqrt{a}\sqrt{b} ))(( x^2-a-b)+(2\sqrt{a}\sqrt{b} ))
= x^4-2ax^2-2bx^2+2ab+a^2+b^2-4ab \\
= x^4-2ax^2-2bx^2-2ab+a^2+b^2 \\
= x^4-2x^2(a+b)-2ab+a^2+b^2$
$p(x) = x^4-2x^2(a+b)-2ab+a^2+b^2$
And then I used the factor theorem to calculate the remaining roots, which gave the following results:
$x = (\sqrt{a} +\sqrt{b} )$, $x = -(\sqrt{a} +\sqrt{b} )$, $x = (\sqrt{a} -\sqrt{b} )$ and $x = -(\sqrt{a} - \sqrt{b})$.
When I try the same method on $x = \sqrt{a} + \sqrt{b} + \sqrt{c}$, the calculations just become absurd. Any kind of help would be enormously appreciated!
| 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 $\pm\sqrt a\pm\sqrt b\pm\sqrt c$ where the signs are to be taken independently of each other (making 8 zeros in all).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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 learn polynomial long divison.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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",
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"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 equation.
If you start with the inequality
$x^2-2bx > c$
($b=5$ and $c=-24$ in your case)
this becomes
$(x-b)^2 > c + b^2$.
If $c + b^2 < 0$,
all $x$ satisfy this;
otherwise this can be rewritten as
$|x-b| > \sqrt{c+b^2}$
for which the solutions are
$x < b-\sqrt{c+b^2}$
and
$x > b+\sqrt{c+b^2}$
If you start with the inequality
$x^2-2bx < c$
this becomes
$(x-b)^2 < c + b^2$.
If $c + b^2 < 0$,
no $x$ satisfies this;
otherwise this can be rewritten as
$|x-b| < \sqrt{c+b^2}$
for which the solutions are
$b-\sqrt{c+b^2}
< x < b+\sqrt{c+b^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/282950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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: $$\begin{align} \frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)} \\ \implies \frac{dy}{dx}=-\frac{1+(y/x)^2-10/x^2}{(y/x)(1+(y/x)^2+5/x^2)} \\ \implies v+x\frac{dv}{dx}=-\frac{1+v^2-10/x^2}{v(1+v^2+5/x^2)} \end{align}$$ I can't seperate $x$ and $v$.
| 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",
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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 $f(x)$ ?
$f(x)=x^4+x^3+x^2+x+1$
$f(x^5)=x^{20}+x^{15}+x^{10}+x^5+1$
| 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",
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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 found that every time, the denominator number seems to go up by $n+2$, but that's about as far as I have been able to get:
$\frac12 + \frac16 + \frac1{12} + \frac1{20} + \frac1{30}...$ the denominator increases by $4,6,8,10,12,\ldots$ etc.
So how should I approach finding the formula? Thanks!
| 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",
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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
My Try:: Using The formula $\tan^{-1}A+\tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$
So $\displaystyle \tan^{-1}\left(\frac{\frac{x}{1-x^2}+\frac{1}{x^3}}{1-\frac{x}{x^3.(1-x^2)}}\right) = \frac{3\pi}{4}$
So $\displaystyle \frac{x^4+1-x^2}{x^3-x^5-x}= -1$
So $(x-1)(x^4-x+1) = 0$
$x = 1$ and $x^4 -x+1 = 0$
So Iam getting only one value of $x$ which is $x=1$ and How can i calculate any real value of $x$ exists from $x^4-x+1 = 0$
OR any other method by using we can solve this Question
Thanks
| *
*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",
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"timestamp": "2023-03-29T00:00:00",
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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",
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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 $n \times 1$ vector. Show that $AB^{-1}A=A$
Trial: Here $B^{-1}=\frac{1}{\alpha-\beta}I_n-\frac{\beta~11'}{(\alpha-\beta)(\alpha+(n-1)\beta)}$ Where $(\alpha-\beta)=(a+\frac{1}{n}-b-\frac{1}{n})=(a-b)$ and $\alpha+(n-1)\beta=1$ So, $B^{-1}=\frac{1}{a-b}[I_n-(b+\frac{1}{n})11']$. Then pre and post multipling $A$ I can't reach to the desire result . Please help.
| 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 that
\begin{align*}
&A = (a-b)I + bE\\
\Rightarrow&AE = (a-b)E + bE^2 = (a-b)E + nbE = 0.
\end{align*}
Therefore,
\begin{align*}
AB^{-1}A
&= A\left(\frac{I}{a-b} + cE\right)[(a-b)I + bE]\\
&= \left(\frac{A}{a-b} + cAE\right)[(a-b)I + bE]\\
&= \frac{A}{a-b}[(a-b)I + bE]\\
&= A.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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,
$$\prod_{k=1}^n \dfrac{\sin(kx)}{kx} = \prod_{k=1}^n\left(1- \dfrac{k^2x^2}{3!} + \mathcal{O}(x^4)\right) = 1 - \dfrac{\displaystyle \sum_{k=1}^n k^2}6x^2 + \mathcal{O}(x^4)\\ = 1 - \dfrac{n(n+1)(2n+1)}{36}x^2 + \mathcal{O}(x^4)$$
Hence, the limit you have is
$$\lim_{x \to 0} \dfrac{n!x^n - \displaystyle \prod_{k=1}^n \sin(kx)}{x^{n+2}} = n!\left(\lim_{x \to 0} \dfrac{1 - \displaystyle \prod_{k=1}^n \dfrac{\sin(kx)}{kx}}{x^{2}} \right) = \dfrac{n(2n+1)(n+1)!}{36}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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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 to find the SDE satisfied by $U$ given that $U =\ln(Y)$ and $Y$ satisfies
$$
\mathrm dY = \frac{1}{2Y}\mathrm dt + \mathrm dW \\
Y(0) = Y_0.
$$
| (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\delta^2\,dt\\
&=(\delta^2 + 2\alpha\sqrt{v} - 2\beta v)\,dt + 2\delta \sqrt{v}\, dW.
\end{align}
(b) Take $f(y)=\ln y$. We have $f'(y)=\dfrac{1}{y}$ and $f''(y)=-\dfrac{1}{y^2}$. Hence $U(0)=\ln Y(0)$ and
\begin{align}
dU &= df(Y) = \frac{1}{Y}\left(\frac{1}{2Y}dt+dW\right) - \frac{1}{2}\frac{1}{Y^2}dt\\
& = \frac{1}{Y}dW\\
& = e^{-U}dW
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/291747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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 in $9376$.
I was given a clue that $76$ is the only number that is divisible by $4$ that gives $1$ when working $\bmod{25}$, I'm not sure how to relate that and generalise
When working $\bmod{100}$ for $n^{20}$ I noticed, $100 = 4 \times 25 = 2^2 \times 5^2$ , so we looked at a number divisible by $4$ that gave $1$ when working in $\bmod{25}$.
Similarly for $n^{200}$, $1000 = 2^3 \times 5^3$, for which $376$ is divisible by $8$ and gives $1$ when working with $\bmod{125}$?
Am I looking along the right lines?
Any help would be greatly appreciated.
Thanks!
| 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 $n^{20} \equiv 1 \pmod{25}$.
(Here $\varphi$ is Euler's totient function.)
Now you solve the system of congruences
$$
\begin{cases}
x \equiv 0 \pmod{4}\\
x \equiv 1 \pmod{25}
\end{cases}
$$
which has the solution(s) $x \equiv 76 \pmod{100}$.
In the general case $k \ge 2$ you have $n^{2 \cdot 10^{k-1}} \pmod{10^{k}}$. Again, split it into two problems
*
*$n^{2 \cdot 10^{k-1}} \equiv 0 \pmod{2^k}$, as above.
*$n^{2 \cdot 10^{k-1}} \pmod{5^k}$. Since $n$ is not divisible by $5$, we have $(n, 5^k) = 1$. Since $\varphi(5^k) = 5^k - 5^{k-1} = 4 \cdot 5^{k-1}$, we obtain $n^{2 \cdot 10^{k-1}} \equiv 1 \pmod{5^k}$.
Now you solve the system of congruences
$$
\begin{cases}
x \equiv 0 \pmod{2^k}\\
x \equiv 1 \pmod{5^k}.
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/295361",
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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 defined a function $f(x)$ such that
$$
f(x) = \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}}
$$
We can see that
$$\begin{align}
f(0) &= \sqrt{2^0 + \sqrt{2^1 + \sqrt{2^2 + \sqrt{2^3 + \ldots}}}} \\
&= \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
\end{align}$$
And we begin solving by
$$\begin{align}
f(x) &= \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} \\
f(x)^2 &= 2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}} \\
&= 2^x + f(x + 1) \\
f(x + 1) &= f(x)^2 - 2^x
\end{align}$$
At this point I find myself stuck, as I have little experience with recurrence relations.
How would this recurrence relation be solved? Would the method extend easily to
$$\begin{align}
f_n(x) &= \sqrt{n^x + \sqrt{n^{x+1} + \sqrt{n^{x+2} + \sqrt{n^{x+3} + \ldots}}}} \\
f_n(x)^2 &= n^x + f_n(x + 1)~\text ?
\end{align}$$
| 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 $(2)$ can be obtained from $(1)$ by multiplying it by $1/2$ and inverting the first radical.
| {
"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.
$$\begin{align}
T(3) &= 1 + 1/2\\
T(4) &= 1 + 1/2\\
T(5) &= 1 + 1/2\\
T(6) &= 1 + 1/2 + 1/2 = 2\\
T(7) &= 1 + 1/2 + 1/2 = 2\\
T(8) &= 1 + 1/2 + 1/2 = 2\\
T(9) &= 2 + 1/2
\end{align}$$
I couldnt find any sequence here. can anyone help!
| 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:
$$\begin{array}{c}
n & T(n) & \lceil n/3\rceil\\\hline
0 & 2/2 & 1\\\hline
1 & 2/2 & 1\\\hline
2 & 2/2 & 1\\\hline
3 & 3/2 & 2\\\hline
4 & 3/2 & 2\\\hline
5 & 3/2 & 2\\\hline
6 & 4/2 & 3\\\hline
7 & 4/2 & 3\\\hline
8 & 4/2 & 3\\\hline
\end{array}$$
| {
"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
} |
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
*
*$$LHS = \sum \frac{1}{b+2\cdot c/a} \ge \frac 9 {2+2(\sum c/a)} \longrightarrow failed$$
*$$\frac a {ab+2c} \ge \frac 9{16}a \longrightarrow failed$$
| \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)+(b+c)}{3}\right)^3} \\
& =\frac{9}{8}
\end{align}
| {
"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
} |
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 $
$$\int\frac{\cos tdt}{2\cos^2t\sqrt{\sin t}}=\int\frac{2ydy}{2(1-y^4)y}=\frac12\left(\int\frac{dy}{1-y^2}+\int\frac{dy}{1+y^2}\right)$$
$$=\frac12\left(\frac12\ln\left|\frac{1+y}{1-y}\right|+\arctan y\right)+C$$
where $y^4=\sin^2t=\frac1{1+\cot^2t}=\frac1{1+\frac1{x^4}}=\frac{x^4}{1+x^4}$
(2) should be handled similarly by putting $x=\sqrt{\sec t}$
| {
"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
} |
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
} |
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:
$$det(A - \lambda I) = \left|\begin{matrix}6 - \lambda & 9& 15\\-5& -10 - \lambda & -21\\ 2& 5& 11 - \lambda\end{matrix}\right| = 0.$$
This gives us the characteristic polynomial:
$$-\lambda^3 + 7\lambda^2 - 16\lambda + 12 = 0$$
From this we get two eigenvalues (one is repeated) as: $\lambda_1 = 3, ~ \lambda_{2,3} = 2$
Next, we want to find the eigenvector for the eigenvalue $\lambda_1$, by solving the equation $(A - \lambda_1) v_1 = 0$.
$(A-\lambda_1)v_1 = (A-3)v_1 = \left(\begin{matrix}3 & 9& 15\\-5& -13 & -21\\ 2& 5& 8\end{matrix}\right)v_1 = 0.$
Using the row-reduced-echelon-form, this leads to $v_1 = (1,-2,1).$
Next, we want to find the eigenvector for the eigenvalue $\lambda_2$, by solving the equation $(A - \lambda_2) v_2 = 0.$
$(A-\lambda_2)v_2 = (A-2)v_2 = \left(\begin{matrix}4 & 9& 15\\-5& -12 & -21\\ 2& 5& 9\end{matrix}\right)v_2 = 0.$
Using the row-reduced-echelon-form, this leads to $v_2 = (3,-3,1).$
Since we have a repeated eigenvalue, care needs to be taken using algebraic and geometric multiplicities (know what those are), if the matrix is diagonalizable (you can work these details).
*
*The algebraic multiplicity of an eigenvalue is the number of times it is a root.
*The geometric multiplicity of an eigenvalue is the number of linearly independent eigenvectors for the eigenvalue.
To find the eigenvector for $\lambda_3$, we will solve $(A - \lambda_3 I)v_3 = v_2$ (you must have learned why this is in class), so we have (shown in augmented form):
$\left(\begin{array}{@{}ccc|c@{}}
4 & 9& 15 & 3\\-5& -13 & -21 & -3\\ 2& 5& 8 & 1
\end{array}
\right)v_3 = 0$
Using RREF, this results in $v_3 = (3, -1, 0)$.
Thus, we have:
$$\lambda_1 = 3, v_1 = (1, -2, 1)$$
$$\lambda_2 = 2, v_2 = (3, -3, 1)$$
$$\lambda_2 = 2, v_3 = (3,-1, 0)$$
Do you know how to use the information above to write the diagonal form, otherwise known as the Jordan Normal Form?
$$A = P J P^{-1} = \begin{bmatrix} 3 & 3 & 1 \\ -3 & -1 & -2 \\ 1 & 0 & 1\end{bmatrix} \cdot \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \cdot \begin{bmatrix} -1 & -3 & -5 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{bmatrix}$$
Babak Sorouh already told you the nullspace and Mhenni Benghorbal showed the Maple commands, so no need to repeat that.
Regards
| {
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"url": "https://math.stackexchange.com/questions/308117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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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 $x=10$, $3^x$ is already 59,049 and $x^3$ is only 1,000. And $3^x$ triples every time $x$ becomes $x+1$, whereas $x^3$ does not triple so easily; to triple $x^3$ you have to multiply $x$ by something.
So there is no trouble finding the constant $c$ that we want; $c=1$ will do. And indeed, for all $x>10$, it is the case that $x^3 < 1\cdot 3^x$. So $x^3$ is $ O\bigl(3^x\bigr)$.
Now on the other hand, $x^3$ is not $O\bigl(x^2\bigr)$. Why not? Well, if it were, then
There is some constant $c$, such that for all sufficiently large $x$, $$x^3 < c\cdot x^2$$
But clearly that's not true, since no matter what $c$ is, $x^3 > c\cdot x^2$ whenever $x>c$. So $x^3$ is not $O\bigl(x^2\bigr)$.
Does that help?
| {
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"timestamp": "2023-03-29T00:00:00",
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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
$$
\begin{align*}
1) & \csc^2 \left( \frac{\pi}{7}\right)+\csc^2 \left( \frac{2\pi}{7}\right)+\csc^2 \left( \frac{4\pi}{7}\right)=8 \\
2) & \tan^2 \left( \frac{\pi}{16}\right) + \tan^2\left( \frac{3\pi}{16}\right) + \tan^2\left( \frac{5\pi}{16}\right)+ \tan^2\left( \frac{7\pi}{16}\right) = 28
\end{align*}
$$
On earlier problem, I was given, $\displaystyle (z+a)^{2m}-(z-a)^{2m}=4maz \prod_{k=1}^{m-1} \left(z^2+a^2 \cot^2 \left(\frac{k\pi}{2m} \right )\right ) $ for integer $m>1$. I am not sure if I can use this is helpful. I am stumped please help.
| 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 \tan ^{2} \frac{k \pi}{16}+\tan ^{2} \frac{(8-k) \pi}{16}&=\tan ^{2} \frac{k \pi}{16}+\frac{1}{\tan ^{2} \frac{k \pi}{16}} \\&=\frac{\sin ^{4} \frac{k \pi}{16}+\cos ^{4} \frac{k \pi}{16}}{\cos ^{2} \frac{k \pi}{16} \cdot \sin ^{2} \frac{k \pi}{16}} \\&=\frac{1-2 \sin ^{2} \frac{k \pi}{16} \cos ^{2} \frac{k \pi}{16}}{\cos ^{2} \frac{k \pi}{16} \cdot \sin ^{2} \frac{k \pi}{16}} \\&=\frac{4}{\sin ^2\left(\frac{k \pi}{8}\right)}-2\end{aligned} \tag*{} $
$\begin{aligned} \sum_{k=1}^{7} \tan ^{2} \frac{k \pi}{16}=& \sum_{k=1}^{3}\left(\tan ^{2} \frac{k \pi}{16}+\tan ^{2} \frac{(8-k) \pi}{16}\right)+\tan ^{2} \frac{\pi}{4} \\=& \sum_{k=1}^{3}\left(\frac{4}{\sin ^{2} \frac{k \pi}{8}}-2\right)+1 \\=& 4 \sum_{k=1}^{3} \frac{1}{\sin ^{2} \frac{k \pi}{8}}-5\end{aligned} \tag*{} $
Now let’s evaluate the three terms one by one and get
$\displaystyle \sin ^{2}\left(\frac{\pi}{8}\right) =\frac{1-\cos \frac{\pi}{4}}{2} =\frac{1-\frac{\sqrt{2}}{2}}{2} =\frac{2-\sqrt{2}}{4} \tag*{} $
$\displaystyle \sin ^{2}\left(\frac{2 \pi}{8}\right)=\frac{1}{2} \tag*{} $
$\displaystyle \sin ^{2}\left(\frac{3 \pi}{8}\right) =\sin ^{2}\left(\frac{\pi}{2}-\frac{\pi}{8}\right) =\cos ^{2} \frac{\pi}{8} =1-\frac{2-\sqrt{2}}{4} =\frac{2+\sqrt{2}}{4} \tag*{} $
Summing them up yields
$\displaystyle \sum_{k=1}^{7} \tan ^{2}\left(\frac{k \pi}{16}\right)=4\left(\frac{4}{2-\sqrt{2}}+2+\frac{4}{2+\sqrt{2}}\right)-5 =35 \tag*{} $
Moreover,
$$\tan \left(\frac{2\pi}{16}\right)=\sqrt{2}-1 \Rightarrow \tan ^{2}\left(\frac{2\pi}{16}\right)=3-2 \sqrt{2} $$
$$\begin{aligned} \tan \left(\frac{6 \pi}{16}\right) &=\tan \left(\frac{\pi}{2}-\frac{\pi}{8}\right) =\frac{1}{\tan \frac{\pi}{8}} =\sqrt{2}+1 \end{aligned}\Rightarrow \tan ^{2}\left(\frac{6 \pi}{16}\right)=3+2 \sqrt{2}$$
Now we can conclude that
$$ \displaystyle \tan ^{2}\left(\frac{\pi}{16}\right)+\tan ^{2}\left(\frac{3 \pi}{16}\right)+\tan ^{2}\left(\frac{5 \pi}{16}\right)+\tan ^{2}\left(\frac{7 \pi}{16}\right)=35-(3-2\sqrt 2)-1-(3+2\sqrt 2)=28$$
| {
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"url": "https://math.stackexchange.com/questions/309271",
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"source": "stackexchange",
"question_score": "4",
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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
$2x-4x(3y^2)(\frac{dy}{dx})+4y^3+8x^2\frac{dy}{dx}+16xy=0$
$-4x(3y^2\frac{dy}{dx})+8x^2\frac{dy}{dx}=-2x-4y^3-16$
Finally I got $\frac{dy}{dx}=\frac{2x-4y^3-16}{-4(x)(3y^2)+8x^2}$
My second question is
Find an equation for the tangent line to the graph of $2x^2-5y^2+xy=5$ at the point $(2,1)$
Anyway I simplified the equation to
$\frac{dy}{dx}=\frac{-4x-1}{-10y+x}$ plugging in x and y I got $\frac{9}{8}$
so my line is $y-1=\frac{9}{8}(x-2)$ but I am unsure if if I did this correctly.
| 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: $$\frac{dy}{dx}=\frac{2x-4y^3-16}{-4(x)(3y^2)+8x^2} = \frac{2(x-2y^3 - 8)}{4(-3xy^2 +2)} = \frac{x - 2y^3 - 8}{2(-3xy^2 + 2)}$$
| {
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"source": "stackexchange",
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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,
$$P_n = \dfrac{1 \times \color{red}3}{2 \times \color{blue}2} \cdot \dfrac{\color{blue}2 \times \color{green}4}{\color{red}3 \times \color{orange}3} \cdot \dfrac{\color{orange}3 \times \color{magenta}5}{\color{green}4 \times \color{brown}4} \cdot \dfrac{\color{brown}4 \times 6}{\color{magenta}5 \times 5} \cdots \dfrac{(n-2) \times \color{darkgreen}n}{(n-1) \times \color{purple}{(n-1)}} \cdot \dfrac{\color{purple}{(n-1)} \times (n+1)}{\color{darkgreen}n \times n}$$
Rearranging we get,
$$P_n = \dfrac12 \cdot \dfrac{3 \times 2}{2 \times 3} \cdot \dfrac{4 \times 3}{3 \times 4} \cdot \dfrac{5 \times 4}{4 \times 5} \cdots \dfrac{n \times (n-1)}{(n-1) \times n} \cdot \dfrac{n+1}n$$
All the terms in the middle are $1$, except the two end terms.
Hence, we get that
$$P_n = \dfrac12 \cdot \dfrac{n+1}{n}$$
| {
"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 \\
&= \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} x^n \left( \frac{e^x - 1}{x} \right)^n .
\end{align*}
Since we only want to extract terms up to degree 4, we can focus on the following expansion:
\begin{align*}
\frac{x}{e^x - 1}
&= 1 - \frac{x}{2} \left( 1 + \frac{x}{2} + \frac{x^2}{6} + \frac{x^3}{24} + \cdots \right) + \frac{x^2}{3} \left( 1 + \frac{x}{2} + \frac{x^2}{6} + \cdots \right)^2 \\
&\qquad - \frac{x^3}{4} \left(1 + \frac{x}{2} + \cdots \right)^3 + \frac{x^4}{5} \left(1 + \cdots \right)^4
\end{align*}
Method 3. We decompose the function into the odd part and the even part:
\begin{align*}
\frac{x}{e^x - 1}
&= \color{red}{\frac{1}{2}\left( \frac{x}{e^x - 1} - \frac{(-x)}{e^{-x} - 1} \right) } + \color{blue}{\frac{1}{2}\left( \frac{x}{e^x - 1} + \frac{(-x)}{e^{-x} - 1} \right)} \\
&= \color{red}{-\frac{x}{2}} + \color{blue}{\frac{x}{2} \cdot \frac{e^x + 1}{e^x - 1}} \\
&= \color{red}{-\frac{x}{2}} + \color{blue}{\frac{\cosh (x/2)}{\left(\frac{\sinh(x/2)}{x/2}\right)}}
\end{align*}
Expanding both the numerator and the denominator of the blue-colored term,
$$ \cosh(x/2) = 1 + \frac{x^2}{8} + \frac{x^4}{384} + \cdots, \qquad \frac{\sinh (x/2)}{x/2} = 1 + \frac{x^2}{24} + \frac{x^4}{1920} + \cdots. $$
Thus using the same trick as other answers we find that
$$ \frac{x}{e^x - 1} = -\frac{x}{2} + \left( 1 + \frac{x^2}{8} + \frac{x^4}{384} + \cdots \right)\left[ 1 - \left( \frac{x^2}{24} + \frac{x^4}{1920} + \cdots \right) + \left( \frac{x^2}{24} + \cdots \right)^2 - \cdots \right]. $$
So the burden of calculation reduces greatly.
| {
"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 calculated in wolfram the limit of the quotient of the logs when $n \rightarrow \infty$. And it is zero. However, I can't prove it.
| 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:
$$\begin{eqnarray*}
\lim_{n\rightarrow \infty }\log \left( \left( 1+\frac{1}{n}\right) ^{n+1/2}\frac{1}{e}\right) &=&\lim_{n\rightarrow \infty }\left(n+\frac{1}{2}\right)\log \left( 1+\frac{1}{n}\right) -1 \\
&=&\lim_{n\rightarrow \infty }n\log \left( 1+\frac{1}{n}\right)+\frac{1}{2}\lim_{n\rightarrow \infty }\log \left( 1+\frac{1}{n}\right) -1 \\
&=&\lim_{n\rightarrow \infty }\frac{\log \left( 1+\frac{1}{n}\right) }{\frac{1}{n}}+0-1 \\
&=&1-1=0;
\end{eqnarray*}$$
and the second:
\begin{eqnarray*}
\lim_{n\rightarrow \infty }\log \left( \left( 1-\frac{1}{n}\right) ^{n-1/2}\frac{1}{e}\right) &=&\lim_{n\rightarrow \infty }\left(n-\frac{1}{2}\right)\log \left( 1-\frac{1}{n}\right) -1 \\
&=&\lim_{n\rightarrow \infty }n\log \left( 1-\frac{1}{n}\right) -\frac{1}{2}%
\lim_{n\rightarrow \infty }\log \left( 1-\frac{1}{n}\right) -1 \\
&=&\lim_{n\rightarrow \infty }\frac{\log \left( 1-\frac{1}{n}\right) }{\frac{1}{n}}-0-1 \\
&=&-1-1=-2.
\end{eqnarray*}
| {
"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':
$$2yy' = \frac{3y^2+6}{x}$$
$$y'=\frac{3y^2+6}{2xy}$$
Finding the orthogonal and separating terms:
$$y' = -\frac{2xy}{3y^2+6}$$
$$\frac{(3y^2+6)}{y}y'=-2x$$
$$\frac{(3y^2+6)}{y}dy=-2xdx$$
Integrating:
$$\int{\frac{(3y^2+6)}{y}dy}=\int{-2xdx}$$
$$\frac{3y^2}{2}+6log(y)=-x^2+C$$
$$C=\frac{3y^2}{2}+6log(y)+x^2$$
Was I close? We weren't given solutions, so I'd appreciate a check on my answer. :)
| 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
$$
\begin{align}
a^2+b^2+c^2 &=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{a^2+c^2}{2} \geq \\
&\ge \frac{2\sqrt{a^2b^2}}{2}+\frac{2\sqrt{b^2c^2}}{2}+\frac{2\sqrt{a^2c^2}}{2} \\
&= ab+bc+ca
\end{align}
$$
but now I am stuck. I don't know how to use the fact that $a+b+c=3$ to prove the inequality. Anybody can give me a hint?
| 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 introduce the condition that $S=3$. Unfortunately the step with the AM-GM-inequality is lost. But maybe you can combine your steps with this derivations?
| {
"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:
$$
\begin{align*}
\lim_{x \rightarrow \infty} \sqrt[5]{x^5-x^4} -x
&= \lim_{x \rightarrow \infty} x\begin{pmatrix}\sqrt[5]{1-\frac 1 x} -1\end{pmatrix}\\
&= \lim_{x \rightarrow \infty} \frac{\sqrt[5]{1-\frac 1 x} -1}{\frac1x}
\end{align*}
$$
Applying L' Hospital's,
$$
\begin{align*}
\lim_{x \rightarrow \infty} \frac{\sqrt[5]{1-\frac 1 x} -1}{\frac1x}&=
\lim_{x \rightarrow \infty} \frac{0.2\begin{pmatrix}1-\frac 1 x\end{pmatrix}^{-0.8}\begin{pmatrix}-x^{-2}\end{pmatrix}(-1)} {\begin{pmatrix}-x^{-2}\end{pmatrix}}\\
&= -0.2
\end{align*}
$$
However the answer is $0.2$, so I would like to clarify the correct use of L'Hospital's
| 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)$
$$=\frac{-1}{1+1+1+1+1}\text { as } y\to0\implies y\ne0$$
| {
"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:
$$\sum_{n=1}^{+\infty}\frac{1}{1^2+2^2+\cdots+n^2}.$$
| $$\begin{array}{lcl}
\sum_{n=1}^\infty \frac{1}{1^2+2^2+\cdots+n^2}&=& \sum_{n=1}^\infty\frac{6}{n(n+1)(2n+1)} \\ &=& 6\sum_{n=1}^\infty \frac{1}{2n+1} \left( \frac{1}{n}-\frac{1}{n+1}\right) \\ &=& 12\sum_{n=1}^\infty \frac{1}{2n(2n+1)} -12\sum_{n=1}^\infty \frac{1}{(2n+1)(2n+2)} \\ &=& 12\sum_{n=1}^\infty \left[ \frac{1}{2n}-\frac{1}{2n+1} \right] - 12\sum_{n=1}^\infty \left[ \frac{1}{2n+1}-\frac{1}{2n+2} \right]\\
&=& 12(1-\ln 2)- 12\left(\ln 2-\frac{1}{2}\right)\\ &=& 18-24\ln 2
\end{array}
$$
| {
"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 $x\equiv1$, $x\equiv1\pmod7$ is a solution.
We proceed to lift the solution $\mod 7^2$
$f'(x) = 11x^{10} + 8x^7 +5$, we have $f(1) = 7$ , $f'(x) = 24$
Since $7$ can not divide $f'(1)$, we need to solve $24t \equiv 0 \mod 7$
we get $t \equiv 0 \pmod 5$
and then ..................... ?????
| 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's formula (vs. Binomial Theorem above)
$$\rm mod\ 49\!:\,\ g(n) = f(1\!+\!7n) = g(0) + g'(0)\, n + \cdots\, \equiv\, f(1) + 7\, f'(1)\,n$$
Thus $\rm\ 7\mid f(1)\:\Rightarrow\:49\mid f(1)+7\,f'(1)\,n\!\iff\! 7\mid f(1)/7+f'(1)\,n\!\iff\! n\equiv\, -\dfrac{f(1)/7}{f'(1)}\:\ (mod\ 7)$
So $\rm\,\ mod\ 7\!:\ n \equiv -\dfrac{f(1)/7}{f'(1)} \equiv \dfrac{-1}{11+8}\equiv \dfrac{6}{-2}\equiv -3\equiv 4.\:$ This is equivalent to using Hensel's Lemma.
It is instructive to compare the two approaches.
| {
"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 $p|(a^2+b^2)\implies p|(a+b)(a^2+b^2) \implies p| \{(a^3+b^3)+ab(a+b)\}$
As $p|(a^3+b^3),p|ab(a+b)\implies p|(a+b)$ as $p\not\mid ab$
As $p|(a+b)$ then $p|(a^m+b^m)$ for odd integer $m\ge1$
Now put $p=5,p=401$ separately and use lcm$[5,401]=2005$
| {
"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:
$$
\begin{align}
\binom{2n}{n}
&=\frac{2n(2n-1)}{n^2}\frac{(2n-2)(2n-3)}{(n-1)^2}\frac{(2n-4)(2n-5)}{(n-2)^2}\cdots\frac{4\cdot3}{2^2}\frac{2\cdot1}{1^2}\\
&=2^n\frac{2n-1}{n}\frac{2n-3}{n-1}\frac{2n-5}{n-2}\cdots\frac{3}{2}\frac{1}{1}\\
&=4^n\frac{n-1/2}{n}\frac{n-3/2}{n-1}\frac{n-5/2}{n-2}\cdots\frac{3/2}{2}\frac{1/2}{1}\tag{1}\\
&\ge4^n\frac{n-1}{n}\frac{n-2}{n-1}\frac{n-3}{n-2}\cdots\frac{1}{2}\cdot1/2\\
&=4^n\frac1{2n}\tag{2}
\end{align}
$$
$(1)$ and $(2)$ show that
$$
\frac1{2n}4^n\le\binom{2n}{n}\le4^n\tag{3}
$$
| {
"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 then decides to keep on increasing for a while.)
So there is a local (relative) maximum at $x=1$, and a local minimum at $x=2$. There is no global maximum, since $f(x)$ is large when $x$ is large positive or negative. But the local minimum at $x=2$ is also a global minimum.
Note that $f''(x)=60(4x^3-9x^2+4x)=60x(4x^2-9x+4)$. Set this equal to $0$. The solutions are $x=0$ and (by the Quadratic Fomula) $x=\frac{9\pm\sqrt{17}}{8}$.
Note that $f''(x)$ is negative for $x\lt 0$, positive between $0$ and the first root of the quadratic, then negative between the two roots of the quadratic, and finally positive. So there is a change of concavity at each of the $3$ roots, and therefore there are $3$ inflection points.
$2.$ This has been done by anorton. Please note that there is no (absolute) maximum, since $x+\frac{1}{x}$ blows up as we approach $0$ from the right. There is also no absolute minimum, for $x+\frac{1}{x}$ becomes very large negative as we approach $0$ from the left.
There is one local maximum, and one local minimum. We have $f'(x)=0$ at $x=\pm 1$. Note also (very importantly) the singularity at $x=0$. So there are $3$ "critical points," $-1$, $0$ and $1$. We examine the behaviour of the function in the four regions determined by the critical points.
For example, note that $f'(x) \gt 0$ if $x\lt -1$, and $f'(x)\lt 0$ if $-1\lt x\lt 0$. So $f(x)$ is increasing in $(-\infty,-1]$ and decreasing in $[-1,0)$. It follows that there is a local maximum at $x=-1$.
$3.$ This is less interesting. We have $f'(x)=6x^2+8x-2$, so we need to use the quadratic Formula to find the critical points.
| {
"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.
$$ z^3 +(1+i)z-3+i = z(z^2+i+1)-3+i$$
$$z(z^2+i+1) = 3 - i \longrightarrow |z(z^2+i+1)| = |3 - i|$$
This gives me the expression:
$z((z^2+1)^2+(1)^2) = \sqrt{10}$
Which can be written as:
$z(z^4 +2z^2 +2) = \sqrt{10}$
But how do I move on from here? Or am I attempting the wrong solution?
Thank you for your help!
| $z^3 + (1+i)z - 3 + i = 0\iff z^3+(1+i)z=3-i$
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 this approximation into the binomial expression above for $n=50$; the result is
$$\frac{1}{2^{100}} \binom{100}{50} \approx \frac{1}{\sqrt{50 \pi}} \left ( 1- \frac{1}{400}\right )$$
Note that the product being about $\frac{1}{\sqrt{50 \pi}}$ is accurate to $1$ part in $400$. Now, it is clear that $50 \pi \approx 157$ to within that margin of error. As $157$ falls between $12^2=144$ and $13^2=169$, the assertion is true.
| {
"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 $y$:
$$y=\frac{-(Bx+E)\pm \sqrt{(Bx+E)^2 - 4(C)(Ax^2 + Dx + F)}}{2C}$$
but this doesn't always work. Consider:
$$xy=1$$
Here, $A=0, B=1, C=0, D=0, E=0, F=0$, so $y=\frac{\cdot}{0}$, which isn't particularly helpful. In other cases it is not as bad, but still not what I'm looking for. E.g.
$$x^2 - y^2 - 1=0$$
Using the formula above, we get
$$y=\pm \sqrt{x^2-1}$$
which seems nice, but $y=\sqrt{x^2-1}$ it is actually one half of each branch rather than one entire branch, as can be seen here.
http://www.wolframalpha.com/input/?i=plot%28x^2+-+y^2+-+1%3D0%29
http://www.wolframalpha.com/input/?i=plot%28y%3Dsqrt%28x^2-1%29%29
I am trying to draw one of these branches, so I need an ordered set of points along a predefined "grid" of either of the variables. Is it possible to do this?
| 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
$$p_c = \begin{pmatrix}0\\0\end{pmatrix}$$
Now, the eigenvalues/vectors of the matrix $$\begin{pmatrix}A & \frac{B}{2}\\\frac{B}{2} & C\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}$$ are $\lambda_1 = -1$, $\lambda_2 = 1$, $v_1 = \begin{pmatrix}0\\1\end{pmatrix}$, $v_2 = \begin{pmatrix}1\\0\end{pmatrix}$
From the parametric form
$$g(t)=p_c + v_1 cosh(t) + v_2 sinh(t)$$
we have
$$g(t) = \begin{pmatrix}0\\1\end{pmatrix} cosh(t) + \begin{pmatrix}1\\0\end{pmatrix} sinh(t)$$
which looks like
Now how would we draw the other branch? That is, how do you determine the values of the parameter that constitute each branch?
Example #2
Consider
$$7x^2 - 3y^2 - 25 = 0$$
($A=7$, $B=0$, $C=-3$, $D=0$, $E=0$, $F=-25$). 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
$$p_c = \begin{pmatrix}0\\\frac{-25}{6}\end{pmatrix}$$
Now, the eigenvalues/vectors of the matrix $$\begin{pmatrix}A & \frac{B}{2}\\\frac{B}{2} & C\end{pmatrix} = \begin{pmatrix}7 & 0 \\ 0 & -3 \end{pmatrix}$$ are $\lambda_1 = -3$, $\lambda_2 = 7$, $v_1 = \begin{pmatrix}0\\1\end{pmatrix}$, $v_2 = \begin{pmatrix}1\\0\end{pmatrix}$
From the parametric form
$$g(t)=p_c + v_1 cosh(t) + v_2 sinh(t)$$
we have
$$g(t) = \begin{pmatrix}0\\ \frac{-25}{6}\end{pmatrix} + \begin{pmatrix}0\\1\end{pmatrix} cosh(t) + \begin{pmatrix}1\\0\end{pmatrix} sinh(t)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/324048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 did, in 1741 (unfortunately, after the demise of Leibniz). Note that this was also before the time of Fourier.
My question: do we now have the tools to prove (2) using solely (1) as the definition of $\pi$? Any positive/negative results would be much appreciated. Thanks!
Clarification: I am not looking for a full-fledged rigorous proof of (1)$\Rightarrow$(2). An estimate that (2) should hold, given (1), would qualify as an answer.
| One of the best way without leibnizSince $\int_0^1 \frac{dx}{1+x^2}=\frac{\pi}{4}$, we have
$$\frac{\pi^2}{16}=\int_0^1\int_0^1\frac{dydx}{(1+x^2)(1+y^2)}\overset{t=xy}{=}\int_0^1\int_0^x\frac{dtdx}{x(1+x^2)(1+t^2/x^2)}$$
$$=\frac12\int_0^1\int_t^1\frac{dxdt}{x(1+x^2)(1+t^2/x^2)}\overset{x^2\to x}{=}\frac12\int_0^1\left(\int_{t^2}^1\frac{dx}{(1+x)(x+t^2)}\right)dt$$
$$=-\frac12\int_0^1\frac{\ln\left(\frac{4t^2}{(1+t^2)^2}\right)}{1-t^2}dt\overset{t=\frac{1-x}{1+x}}{=}-\frac12\int_0^1\frac{\ln\left(\frac{1-x^2}{1+x^2}\right)}{x}dx$$
$$\overset{x^2\to x}{=}-\frac14\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{x}dx=-\frac14\int_0^1\frac{\ln\left(\frac{(1-x)^2}{1-x^2}\right)}{x}dx$$
$$=-\frac12\int_0^1\frac{\ln(1-x)}{x}dx+\frac14\underbrace{\int_0^1\frac{\ln(1-x^2)}{x}dx}_{x^2\to x}$$
$$=-\frac38\int_0^1\frac{\ln(1-x)}{x}dx\Longrightarrow \int_0^1\frac{-\ln(1-x)}{x}dx=\frac{\pi^2}{6}$$
Remark:
This solution can be considered a proof that $\zeta(2)=\frac{\pi^2}{6}$ as we have $\int_0^1\frac{-\ln(1-x)}{x}dx=\text{Li}_2(x)|_0^1=\text{Li}_2(1)=\zeta(2)$
| {
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"source": "stackexchange",
"question_score": "15",
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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 me a positive?
Or if I wanted to show it algebraically, I could do
$$(-y - z - x^2 - y^2 - z^2)^2 = [(-1)(y + z + x^2 + y^2 +z^2)]^2$$
$$ = (-1)^2(y + z + x^2 + y^2 +z^2)^2 = (y + z + x^2 + y^2 +z^2)^2$$
EDIT: Ok, lets say just one of those terms in that bracket was positive, could I still do the $(-1)$ trick and make just one term negative and so its easier to work out, or would I need to leave it as it is and expand it?
| Yes.
$$(-a -b -c)^2 = ((-1)(a + b + c))^2 = \underbrace{(-1)^2}_{=+1}(a+b+c)^2 = (a+b+c)^2$$
| {
"language": "en",
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"source": "stackexchange",
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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^{2k}(x) dx \right)
\end{align}
Note that we have thrown away the odd terms since for $k$ odd, the integral $\displaystyle \int_0^{2 \pi}\sin^2(x) \cos^k(x) dx$ is zero.
\begin{align}
\dfrac{\displaystyle \int_0^{2 \pi}\sin^2(x) \cos^{2k}(x) dx}4 & = \int_0^{\pi/2}\sin^2(x) \cos^{2k}(x) dx\\
&
\int_0^{\pi/2}\cos^{2k}(x) dx - \int_0^{\pi/2}\cos^{2k+2}(x) dx\\
& = \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2 - \dfrac{2k+1}{2k+2} \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2\\
& = \dfrac1{2k+2} \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2\\
& = \dfrac{\pi}{2^{k+2}} \dfrac{(2k-1)(2k-3)\cdots3 \cdot 1}{(k+1)!} = \dfrac{\pi}{2^{2k+2}} \dfrac{(2k)!}{k! (k+1)!}
\end{align}
Hence,
$$\dfrac{aI}{\pi} = \sum_{k=0}^{\infty} \left(\dfrac{b}{2a} \right)^{2k} \underbrace{\dfrac{(2k)!}{k! (k+1)!}}_{\text{Catalan numbers}}$$
Now $$\sum_{k=0}^{\infty} \dfrac{\dbinom{2k}k x^{k}}{k+1} = \dfrac{1-\sqrt{1-4x}}{2x} \,\,\,\,\,\, \forall \vert x \vert < \dfrac14$$ This is the generating function for the Catalan numbers. Hence, in our case, we get that
$$\sum_{k=0}^{\infty} \left(\dfrac{b}{2a} \right)^{2k} \underbrace{\dfrac{(2k)!}{k! (k+1)!}}_{\text{Catalan numbers}} = \dfrac{1-\sqrt{1-4 \cdot \left(\dfrac{b}{2a} \right)^2}}{2 \cdot \left(\dfrac{b}{2a} \right)^2} \,\,\,\,\,\,\,\,\forall \dfrac{b}{2a} < \dfrac12$$
Hence,
$$\dfrac{aI}{\pi} = \dfrac{1-\sqrt{1-\left(\dfrac{b}a\right)^2}}{\dfrac{b^2}{2a^2}} = \dfrac{a-\sqrt{a^2-b^2}}{\dfrac{b^2}{2a}}$$
Hence,
$$I = \dfrac{2\pi}{b^2} (a-\sqrt{a^2-b^2})$$
| {
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"url": "https://math.stackexchange.com/questions/326714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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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:
$$\begin{aligned}\left(2\sqrt{x^3y^3}+z^3\right)^2(4x^3y^3+z^6)&\geq \left(\sqrt[3]{16x^6y^6}+z^4\right)^3\\&=\left(2\sqrt[3]2x^2y^2+z^4\right)^3.\end{aligned}$$
In this stronger version, equality holds for $(x,y,z)=(t,t,\sqrt[3]{2t}).$
For the original inequality, if nonnegative reals are allowed, then equality holds if and only if $x=y=0.$ Otherwise there is no equality.
$\Box$
| {
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"source": "stackexchange",
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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
$$x=0.5\qquad y=-0.5$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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",
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"source": "stackexchange",
"question_score": "6",
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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 $, $a\equiv b \pmod n$then $gcd(b,n)=1$, $ord_n^b=ord_n^a$
if $k \in \Bbb N$ , $a^k \equiv 1 \pmod n$ iff $ord_n^a|k$
$a^{k_1} \equiv a^{k_2} \pmod n$ iff $k_1 \equiv k_2 \pmod { ord_n^a}$
$ord_n^a| \phi(n)$
it's my trying :
$a^3\equiv 1 \pmod p$ so $(a-1)(a^2 +a+1) \equiv 0 \pmod p$ so $a \equiv1 \pmod p$ that is impossible. so $a^2+a+1 \equiv 0 \pmod p$ so $a+1 \equiv -a^2 \pmod p$ how to find smallest $d$ such that $gcd(p,a+1)=1$ and $(a+1)^d \equiv 1 \pmod p$
also we have: $(-(a+1))^d \equiv (a^2)^d \equiv 1 $ also $ord^a_p=ord^{a^2}_p$so
$d=3$, $(a+1)^3 \equiv -1 $ so $(a+1)^6 \equiv 1$ the problem is : Is $6$ smallest?
how to prove for $2,4,5$ that is not ? in fact how to prove :
$(a+1)^i \not \equiv 0 \pmod p$, $i=2,4,5$
| 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 that the order of $a$ is $\ne 2$.
To show $a+1$ does not have order $2$, suppose that it does. Then from $(a+1)^2\equiv 1\pmod{p}$ we get that $a(a+2)\equiv 0\pmod{p}$. Now show that we cannot have $a\equiv -2\pmod{p}$.
To show that the order of $a+1$ is not $3$, suppose it is. Then from $(a+1)^3\equiv 1\pmod{p}$ we obtain $3a^2+3a+1\equiv 0\pmod{p}$. But $p^2+p+1\equiv 0\pmod{p}$. From this one can quickly obtain a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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",
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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 advance.
| 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.
| {
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"question_score": "4",
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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.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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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^2y \Big) \\[6pt]
& = 4\sin x\cos y\cos x\sin y \\[6pt]
& = (2\sin x\cos x)(2\sin y\cos y) \\[6pt]
& = \sin(2x)\sin(2y).
\end{align}
| {
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"question_score": "3",
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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 singularity at $z=0$ is removable, so the function is entire, so the integral becomes zero. He then makes a huge jump and says, "Thus, $-2\int_{-R}^R \frac{\sin^2 x}{x^2}\ dx -2i\int_{\Gamma_R} \frac{dz}{z} + \int_{\Gamma_R} \frac{e^{2i z}-1}{z^2}\ dz = 0$."
Here, $\Gamma_R$ denote the "arc" part of the semi-circular contour.
I get where he gets the second terms from. I don't get where he gets the first.
How do we go from
$$\int_{-R}^R \frac{e^{2ix}-1-2ix}{x^2}\ dx$$
to
$$\int_{-R}^R \frac{\sin^2 x}{x^2}\ dx?$$
| $$
e^{2iz}-1-2iz=1+2iz+\frac {(2iz)^2}2+\frac{(2iz)^3}6+\ldots\frac {(2iz)^n}{n!}+\ldots-1-2iz=\\
=\frac{(2iz)^2}2+\frac{(2iz)^3}{3!}+\ldots+\frac{(2iz)^{n+2}}{(n+2)!}+\ldots
$$
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 powers only
$$
e^{2iz}-1-2iz \stackrel{\int}{\equiv} \frac {(2ix)^2}{2!} + \frac {(2ix)^4}{4!} + \frac {(2ix)^6}{6!} + \ldots + \frac {(2ix)^{2k+2}}{(2k+2)!} + \ldots = \\
= -\frac {(2x)^2}{2!} + \frac {(2x)^4}{4!} - \frac {(2x)^6}{6!} + \ldots + (-1)^{k+1} \frac {(2x)^{2k+2}}{(2k+2)!} + \ldots = \\
= 1 -\frac {(2x)^2}{2!} + \frac {(2x)^4}{4!} - \frac {(2x)^6}{6!} + \ldots + (-1)^{k+1} \frac {(2x)^{2k+2}}{(2k+2)!} + \ldots - 1 = \\
= \cos 2x-1 = -2\sin^2x
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"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 as close as I've gotten:
\begin{align*}
\frac{1}{1-x} & = \sum_{n=0}^{\infty} x^n \\
\frac{-2}{(x-1)^3} & = \frac{d^2}{dx^2} \sum_{n=0}^{\infty} x^n \\
\frac{-2}{(x-1)^3} & = \sum_{n=2}^{\infty} n(n-1)x^{n-2} \\
\frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} n(n-1)\frac{x^n}{x}(x+1) \\
\frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} (n^2x + n^2 - nx - n)\frac{x^n}{x} \\
\frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} n^2x^n + n^2\frac{x^n}{x} - nx^n - n\frac{x^n}{x} \\
\end{align*}
Any help is appreciated, thanks :)
| $\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$
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,
$\displaystyle \frac{[(1-x)^2(1)-(x)2(1-x)(-1)]x}{(1-x)^4}= \frac{x^2+x}{(1-x)^3}=
\sum_{n=0}^{\infty}n^2x^n$
| {
"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
} |
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, since $b$ and $d$ are both odd, $2d\equiv2\pmod{4}$ and so one of $b-d$ and $b+d$ is $0\bmod{4}$ and the other is $2\bmod{4}$. Thus, the left hand side of $(1)$ has at least three factors of $2$. Contradiction.
| {
"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} \, \Omega(n^2) \\
n^2 -14n +49 \, \text{is} \, \Omega(n^2) \\
\begin{align} n^2 -14n +49 \, & \ge \, C \cdot (n^2) \; ,n \ge 1 \\
& \ge -14 n^2 \end{align}$
This proves it true right? Must both proofs have the same constant? I'm very new with Asymptotics.
| 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
} |
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),
$$\int x^{\alpha}\ln x \ \text{dx}=\frac{x^{{\alpha} + 1}}{{\alpha} + 1} \big[\ln x - \frac{1}{{\alpha} + 1}\big] + C_1$$
and,
$$\int\frac{\ln^{\beta}x}{x} \ \text{dx}=\frac{\ln^{\beta + 1}x }{\beta + 1}+C_2$$
I have trouble showing that for $\alpha=-1$ and $\beta=1$ they have the same value(I cannot put $\alpha=-1$)
What to do?
| 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(x) dx & = \dfrac{x^{\alpha+1} \ln(x)}{1+\alpha} - \dfrac{x^{\alpha+1}}{(1+\alpha)^2} + \text{constant}
\end{align}
You cannot directly take the limit as $\alpha \to -1$. But note that you can draw some constants out from the constant term to help you.
\begin{align}
\int x^{\alpha} \ln(x) dx & = \dfrac{(x^{\alpha+1} - 1) \ln(x)}{1+\alpha} + \dfrac{\ln(x)}{1+\alpha}+ \dfrac{1-x^{\alpha+1}}{(1+\alpha)^2} \underbrace{- \dfrac1{(1+\alpha)^2}+ \text{constant}}_{\text{new constant}}\\
& = \dfrac{(x^{\alpha+1} - 1) \ln(x)}{1+\alpha} + \dfrac{\ln(x)}{1+\alpha}+ \dfrac{1-x^{\alpha+1}}{(1+\alpha)^2} + \text{constant} \,\,\,\, (\spadesuit)
\end{align}
Now note that
$$x^{1+\alpha} = \exp((1+\alpha) \ln(x)) = 1 + (1+\alpha) \ln(x) + \dfrac{(1+\alpha)^2}2 \ln^2(x) + \mathcal{O}((1+\alpha)^3)$$
Hence,
\begin{align}
\dfrac{1- x^{1+\alpha}}{1+\alpha} & = - \ln(x) - \dfrac{1+\alpha}2 \ln^2(x) + \mathcal{O}((1+\alpha)^2)\\
\ln(x) + \dfrac{1- x^{1+\alpha}}{1+\alpha} & = - \dfrac{1+\alpha}2 \ln^2(x) + \mathcal{O}((1+\alpha)^2)\\
\dfrac{\ln(x)}{1+\alpha} + \dfrac{1- x^{1+\alpha}}{(1+\alpha)^2} & = - \dfrac{\ln^2(x)}2 + \mathcal{O}((1+\alpha))
\end{align}
Plug this in $(\spadesuit)$ to get
$$\int x^{\alpha} \ln(x) = \ln^2(x) + \dfrac{1+\alpha}2 ln^3x + ln x\,\mathcal{O}((1+\alpha)) - \dfrac{\ln^2(x)}2 + \mathcal{O}((1+\alpha)) = \dfrac{\ln^2(x)}2 + \dfrac{1+\alpha}2 ln^3x + (ln x + 1)\,\mathcal{O}((1+\alpha))$$
Now letting $\alpha \to -1$, we get that
$$\lim_{\alpha \to -1} \int x^{\alpha} \ln(x) = \lim_{\alpha \to -1} \dfrac{\ln^2(x)}2 + \mathcal{O}((1+\alpha)) = \dfrac{\ln^2(x)}2$$
which matches with our original integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/344041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"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 stuck. What's bothering me is that the way I went about this question seemed like a rather "clunky" method. I'm just curious if I've missed some underlying pattern that could have made it easier to reproduce the identity on the right.
The way I did it:
$$\begin{array}{l} \cos (A + B)\cos (A - B)\\ \equiv (\cos A\cos B - \sin A\sin B)(\cos A\cos B + \sin A\sin B)\\ \equiv {\cos ^2}A{\cos ^2}B - {\sin ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A(1 - {\sin ^2}B) - (1 - {\cos ^2}A){\sin ^2}B\\ \equiv {\cos ^2}A - {\cos ^2}A{\sin ^2}B - {\sin ^2}B + {\cos ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A - {\sin ^2}B\end{array}$$
| Another way of looking at it:
$$\begin{align*}
\cos(A+B)\cos(A-B) &= \frac14\left(e^{i(A+B)}+\frac{1}{e^{i(A+B)}}\right)\left(e^{i(A-B)}+\frac{1}{e^{i(A-B)}}\right) \\
&= \frac14\left[(e^{iA})^2 + \frac{1}{(e^{iA})^2} + (e^{iB})^2 + \frac{1}{(e^{iB})^2}\right] \\
&= \frac14\left[(e^{iA})^2 + \frac{1}{(e^{iA})^2} + 2 - 2 + (e^{iB})^2 + \frac{1}{(e^{iB})^2}\right] \\
&= \frac14\left[(e^{iA})^2 + \frac{1}{(e^{iA})^2} + 2\right] + \frac14\left[ - 2 + (e^{iB})^2 + \frac{1}{(e^{iB})^2}\right] \\
&= \left(\frac{e^{iA} + e^{-iA}}{2}\right)^2 - \left( \frac{e^{iB}-e^{-iB}}{2i}\right)^2 \\
&= \cos^2 A - \sin^2 B.
\end{align*}
$$
| {
"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
} |
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, $2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}=2x$
This automatically makes our equation to be $x^2-5=2x$. Just solve the quadratic equation, and take the positive root. Why only positive? Think about it: how can $\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}$ equal a negative number? It must be positive! So that's why we only take the positive root.
$$x^2-2x-5=0$$
$$x=\dfrac{2\pm \sqrt{(-2)^2-4(1)(-5)}}{2(1)}$$
$$x=\dfrac{2\pm \sqrt{4+20}}{2}$$
$$x=\dfrac{2\pm \sqrt{24}}{2}$$
$$x=\dfrac{2\pm 2\sqrt{6}}{2}$$
$$x=1\pm \sqrt{6}$$
Since $1-\sqrt{6}$ equals a negative number, we reject that root. So, $x=1+\sqrt{6}$, which is also the answer to the original problem.
Answer:
$$\displaystyle \boxed{\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}=1+\sqrt{6}}$$
| {
"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
} |
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 got stuck here:
$$\frac{\cos x-\sin x}{\cos x+\sin x}$$
| $$
\begin{aligned}
& \frac{1-\sin 2 x}{\cos 2 x} \\
=& \frac{(\cos x-\sin x)^{2}}{\cos ^{2} x-\sin ^{2} x} \\
=&\frac{(\cos x-\sin x)^{2}}{(\cos x+\sin x)(\cos x-\sin x)} \\
=& \frac{\cos x-\sin x}{\cos x+\sin x} \\
=& \frac{1-\tan x}{1+\tan x} \\
=& \tan \left(\frac{\pi}{4}-x\right)
\end{aligned}
$$
| {
"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
} |
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 me.
If you have an idea or know a helpful identity to prove this result I would really appreciate it.
| 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 expanding the infinite product,
\begin{eqnarray*}
&\ & \prod_{k\ge 0} (1 + 2^{-2^k} + 2^{-2\cdot 2^k} + 2^{-3\cdot 2^k} + \cdots)\\
&=& 1+\sum_{k\ge 0} (2^{-2^k} + 2^{-2\cdot 2^k}+ 2^{-3\cdot 2^k} + \cdots)\prod_{0\le j<k} (1 + 2^{-2^j} + 2^{-2\cdot 2^j} + 2^{-3\cdot 2^j} + \cdots)\\
&=& 1+\sum_{k\ge 0} \left(2^{-2^k} + \frac{2^{-2\cdot 2^k}}{1-2^{-2^k}}\right)
\prod_{0\le j<k} (1-2^{-2^j})^{-1}\\
&=& 1+\sum_{k\ge 0} \left(2^{-2^k} \prod_{0\le j<k} (1-2^{-2^j})^{-1}+ 2^{-2^{k+1}} \prod_{0\le j\le k} (1-2^{-2^j})^{-1}\right)\\
&=& 1+\sum_{k\ge 0} 2^{-2^k} \prod_{0\le j<k} (1-2^{-2^j})^{-1}
+\sum_{\ell\ge 1} 2^{-2^\ell} \prod_{0\le j<\ell} (1-2^{-2^j})^{-1}, \ \ \text{setting } \ell=k+1\\
&=& 1-\frac{1}{2}+2\sum_{k\ge 0} 2^{-2^k} \prod_{0\le j<k} (1-2^{-2^j})^{-1}, \qquad \text{lumping the two sums into one}\\
&=& \frac12+\sum_{k\ge 0} 2^{-(2^0+\cdots+2^{k-1})} \prod_{0\le j<k} (1-2^{-2^j})^{-1}\\
&=& \frac12 + \sum_{k\ge 0} \prod_{0\le j\le k-1} \frac{1}{2^{2^j}-1},
\\
\end{eqnarray*}
which is the right-hand side.
| {
"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
$$x = 2t, y = 2t^2-1, z = 2t^2$$
where $a = d = t, b = (t+1), c = (t-1)$ since $t^2 - (t^2-1) = 1$.
| {
"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 = $\frac{2}{3}\int\frac{t^2}{\sqrt{1-(t^2-2)}\sqrt{1+(t^2-2)}}\mathrm{d} t$
| 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
} |
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 equivalent to
$$\sin A \sin B \stackrel{?}{=} \sin^2 C - \sin^2 A$$
Now, $C=2A$ implies $B=\pi-3A$, so that $\sin B = \sin 3A$, and the identity to be proven becomes
$$\begin{align}
\sin A \sin 3A &\stackrel{?}{=} \sin^2 2A - \sin^2 A \\[6pt]
&= \left( \sin 2A + \sin A \right) \left( \sin 2 A - \sin A \right) \\[6pt]
&= 2 \sin\frac{3A}{2}\cos\frac{A}{2} \cdot 2 \sin\frac{A}{2} \cos\frac{3A}{2} \\[6pt]
&= 2 \sin\frac{A}{2}\cos\frac{A}{2} \cdot 2 \sin\frac{3A}{2} \cos\frac{3A}{2} \\[6pt]
&= \sin A \; \sin 3A
\end{align}$$
In the above, I use the sum-to-product identity $\sin\theta \pm \sin\phi = 2 \sin\frac{\theta\pm \phi}{2} \cos\frac{\theta\mp \phi}{2}$, which is familiar to me. Someone familiar with multiple-angle identities may have proceeded thusly:
$$\begin{align}
\sin A \sin 3A &\stackrel{?}{=} \sin^2 2A - \sin^2 A \\[6pt]
&= 4\sin^2 A\cos^2 A - \sin^2 A \\[6pt]
&= \sin A \cdot \sin A \left( 4 \cos^2 A - 1 \right) \\[6pt]
&= \sin A \cdot \sin A \left( 3 - 4\sin^2 A \right) \\[6pt]
&= \sin A \; \sin 3A
\end{align}$$
| {
"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
} |
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 $(A^tA)^{-1}A^t B = x$ wherein the regression line is $ y=x_i+x_{i+1}+...+x_{n}$
The problem is:
A=
$\begin{pmatrix} 1 & 2 & 1 \\ 1&3&2\\2&5&3\\2&0&1 \\ 3 &1&1\end{pmatrix}$
$A^{t}$ =
$\begin{pmatrix} 1 & 1 & 2&2&3 \\ 2&3&5&0&1\\1&2&3&1&1\end{pmatrix}$
$(A^{t}A)$=$\begin{pmatrix} 6 & 9 & 15&3&6 \\ 9&14&23&4&8\\15&23&38&7&14\\3&4&7&5&7\\6&8&14&7&11\end{pmatrix}$
This is the problem:
$(A^{t}A)^{-1}$ = singular
| 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",
"answer_count": 1,
"answer_id": 0
} |
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 obtain
$2 \leq \frac{b \pm \sqrt{b^2-4ac}}{2a} \leq \frac{12}{5}$
and as $a>0$, this implies that
$4a \leq b \pm \sqrt{b^2-4ac} \leq \frac{24a}{5}$.
Then we can divide this into two cases (plus or minus) --- But how can we manipulate it to yield the required result?
Another idea is to make use of the fact that $b^2-4ac$ is non-negative
| 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 $a>0$,
$$\frac{b}{a}=\alpha+\beta\ge 4\Rightarrow a\le \frac{b}{4}<b;$$
$$\frac{b}{c}=\frac{\alpha+\beta}{\alpha\beta}=\frac{1}{\alpha}+\frac{1}{\beta}\le 1\Rightarrow b\le c;$$
and
$$(\alpha-1)(\beta-1)<2\Rightarrow 1+\frac{b}{a}=\alpha+\beta+1>\alpha\beta=\frac{c}{a}\Rightarrow c< a+b.$$
Therefore,
$$\frac{a}{a+c}+\frac{b}{a+b}\ge \frac{a}{a+c}+\frac{b}{a+c}=\frac{a+b}{a+c}>\frac{c}{a+c},$$
where the first inequality is due to $a,b>0$ and $0\le b\le c$ and the second one is due to $a+b>c$ and $a+c>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/356594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
$ \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 gives me 0 as its answer, which is wrong. I've tried to parametrized it, which gives me nothing useful.
Ideas?
| 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 rotate it clockwise by $\pi/2$ we get the normal, and the normal is $(dy,-dx)$. So the original integral $I$ becomes:
$$
I = -\frac{1}{2}\int_{\partial \Omega} \nabla \big(\ln((x+1)^2+y^2)\big) \cdot \mathbf{n} \,ds
$$
Moreover recall the fundamental solution of Poisson equation in 2D is $\displaystyle \phi = \frac{1}{2\pi}\ln|\mathbf{x}-\mathbf{x}_0|$ which solves:
$$
\Delta \phi = \delta(\mathbf{x}-\mathbf{x}_0)
$$
in $\mathbb{R}^2$. Let $\mathbf{x}_0 = (-1,0)$, then $u =\ln\big((x+1)^2+y^2\big) = 2\ln|\mathbf{x}-\mathbf{x}_0|$, and $u$ solves:
$$
\Delta u = 4\pi \delta(\mathbf{x}-\mathbf{x}_0)
$$
Now uses Divergence theorem:
$$
I = -\frac{1}{2}\int_{\partial \Omega} \nabla u \cdot \mathbf{n} \,ds = -\frac{1}{2} \int_{\Omega} \Delta u = -\frac{1}{2} \int_{\Omega} 4\pi \delta(\mathbf{x}-\mathbf{x}_0) = -2\pi
$$
Above method looks like a little bit cheating though.
EDIT: Guess this is a standard Calculus III problem, so we shouldn't use either Dirac delta nor Cauchy integral formula. Thinking of the Cauchy integral formula reminded me a trick we used in Calculus III (maybe it was too long and slipped from my mind, now I recollected how we did this kind of problem):
When the enclosed $\Omega$ by the curve $C$ has a singularity, i.e., $\displaystyle \frac{\partial Q}{\partial x}$ or $\displaystyle \frac{\partial P}{\partial y}$ is not continuous, we could just cut a hole with a small radius $r$ centered at this singularity , in this example it is $(-1,0)$, such that outside this small disk of radius $r$ there is no singularity (See the picture below).
Denote $C'$ the boundary of this disk $B$, rotating clockwisely. Then:
$$
\oint_{C} \frac{ydx-(x+1)dy}{x^2+y^2+2x+1} + \oint_{C'} \frac{ydx-(x+1)dy}{x^2+y^2+2x+1} =
\int_{\Omega\backslash B}\left\{ -\frac{\partial}{\partial x}\left(\frac{x+1}{(x+1)^2+y^2}\right) - \frac{\partial}{\partial y}\left(\frac{y}{(x+1)^2+y^2}\right)\right\} dx dy = 0
$$
Therefore:
$$
\oint_{C} \frac{ydx-(x+1)dy}{x^2+y^2+2x+1} =- \oint_{C'} \frac{ydx-(x+1)dy}{x^2+y^2+2x+1}
$$
Now we could just parametrize the curve $C'$ using $t$, by letting $x+1 = r\cos t$, $y = r\sin t$, so $dx = -r\sin t\, dt$, $dy = r \cos t \,dt$. The integral becomes:
$$
\oint_{C'}\frac{y }{(x+1)^2+y^2}dx-\frac{(x+1)}{(x+1)^2+y^2}dy
= \int^{2\pi}_0 \left(\frac{r\sin t\, }{r^2\cos^2 t + r^2\sin^2 t} (-r\sin t)
- \frac{r\cos t\,}{r^2\cos^2 t + r^2\sin^2 t}r \cos t \right)dt = -2\pi
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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.
$$
T(n) = 2T(n/2) + n + 11
$$
Where n is some power of 2, for all n>1 and T(1) = 0.
I started out by some backward substitution and eventually arrived at this:
$$
2^3T(n/2^3) + 2^2(n) + 2^2(11) + 2(n) + 2(11) + n + 11
$$
And I reasoned that the general case was:
$$
2^kT(n/2^k) + 2^{k-1}(n) + 2^{k-1}(11) + 2^{k-2}(n) + 2^{k-2}(11)...2^0(n) + 2^0(11)
$$
n is a power of two, meaning that I can resolve this further:
$$
T(n) = 2^{k-1}(n) + 2^{k-1}(11) + 2^{k-2}(n) + 2^{k-2}(11)...2^0(n) + 2^0(11)
$$
Now I know I am supposed to summate this in some way, but I really don't know how! Can you guys point me in the right direction please?
Second attempt
I started from the beginning again, and this is how I went. I can see my answer coming closer to yours now, but I'm just struggling with one bit.
$$
T(n) = 2T(n/2) + n + 11
$$
$$
T(n) = 2^2T(n/4) + n/2 + n + 22 + 11
$$
$$
T(n) = 2^3T(n/8) + n/4 + n/2 + n + 44 + 22 + 11
$$
Factorized the 11..
$$
T(n) = 2^3T(n/8) + n/4 + n/2 + n + 11(4 + 2 + 1)
$$
I'm struggling to get from this point to your answer. Sorry to be such a pain!
| 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
$$2^2\left(2T(2^{k-3}) + 2^{k-2} + 11\right) + 2\times 2^k + (2 + 1)\times 11$$
or
$$2^3T(2^{k-3}) + 3\times 2^k + (4 + 2 + 1)\times 11$$
Edit: Note that you accidentally substituted in $T\left(\frac{n}{2}\right) = 2T\left(\frac{n}{4}\right) + \boxed{n} + 11$. The boxed term should be $\frac{n}{2}$. Then you accidentally left out the $\boxed{2} \times 2^{k-1}$ and the $\boxed{2\times 2}\times2^{k-2}$ etc.
In general, we get
$$2^k T(1) + k 2^k + (1 + 2 + \cdots + 2^{k-1})\times 11 = k 2^k+11(2^k-1)$$
or
$$T(2^k) = (k+11)2^k-11$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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: $$\pi_1=\frac{5}{18}$$ $$\pi_2=\frac{8}{18}$$ $$\pi_3=\frac{5}{18}$$
Could someone please explain how to find these variables.
| 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
$$B=\pmatrix{-\frac{5}{6} & \frac{4}{6} & \frac{1}{6} \\ \frac{5}{12} & -\frac{10}{12} & \frac{5}{12} \\ \frac{1}{6} & \frac{4}{6} & -\frac{5}{6} }$$
Since we multiply by the matrix on the right, the desired vector is in the kernel of $B^T$. The Scilab command v=kernel(B')' outputs
v = 0.4682929 0.7492686 0.4682929
This is normalized to have length $1$, but we want the sum of components to be $1$. The command v/sum(v) does the job:
0.2777778 0.4444444 0.2777778
which is your vector. Of course, with sufficient patience one can do it by hand as well.
| {
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"timestamp": "2023-03-29T00:00:00",
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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)^{k-1}\frac{(2k-2)!}{2^{2k-1}k!(k-1)!}\end{align}
hence we have
$$(1-x)^\frac{1}{2}=1-\sum_{k=1}^\infty \frac{(2k-2)!}{2^{2k-1}k!(k-1)!}x^k=1-\sum_{k=1}^\infty a_kx^k$$
Finaly by the D'Alembert criterion we have
$$\frac{1}{R}=\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\to\infty}\frac{2k(2k+1)}{2^2(k+1)k}=1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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?
| $$f(x)=\sqrt{nx}, \quad f'(x)=\frac{1}{2}\cdot\frac{1}{\sqrt{nx}}\cdot n = \frac{n}{2\sqrt{nx}}$$
$$x_0=1/n, \quad f(x_0)=\sqrt{n\cdot(1/n)}=1, \quad f'(x_0)=\frac{n}{2\sqrt{n\cdot(1/n)}}=\frac{n}{2}$$
$$\sqrt{nx} = 1 + \frac{n}{2} \cdot (x-1/n) - \frac{g(x)}{2}$$
$$2\sqrt{nx} = 2 + (nx-1) - g(x)$$
$$g(x) = 2 + (nx-1) - 2\sqrt{nx} = nx - 2 \sqrt{nx} - 1 = (\sqrt{nx}-1)^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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(-\frac{\pi}{2} - x\right)\\
&= \cos\left(\frac{\pi}{2} + x\right)\\
&= -\sin(x)
\end{align*}
Why does $\cos\left(\frac{3\pi}{2} - x\right) = \cos\left(-\frac{\pi}{2} - x\right)$?
Similarly, if $\cos(x) = 0.6$, find $\sin\left(\frac{3\pi}{2} + x\right)$, the solution says $\sin\left(\frac{3\pi}{2} + x\right)$ is equal to $\cos(\pi+x)$. How did they get $\cos(\pi+x)$?
| HINT : as $\cos(2\pi+y)=\cos2\pi\cos y-\sin2\pi\sin y=\cos y$
As $$\left(\frac{3\pi}2-x\right)-\left(-\frac\pi2-x\right)=2\pi$$
$$\implies \left(\frac{3\pi}2-x\right)=2\pi+\left(-\frac\pi2-x\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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)^{n+1} x^{2n-1}}{(2n)!}\\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{(2n)!} \int_0^1 x^{2n-1} dx\\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{(2n)!} \frac1{2n} \\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{2n(2n)!}\\
\end{align}
$
This is an alternating series with terms decreasing in absolute value,
so its sum is between any two consecutive terms.
The first two terms are
$\frac1{4}$ and $\frac{-1}{4\cdot 4!} = \frac{-1}{96}$,
so the result is slightly less than $\frac1{4}$.
| {
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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+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$$
\log(1+(x-1))=(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\cdots=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}(x-1)^k $$ $$\log(1+\sqrt{\frac{1}{x^2}+1})= \sqrt{\frac{1}{x^2}+1}-\frac{(\sqrt{\frac{1}{x^2}+1})^2}{2}+\frac{(\sqrt{\frac{1}{x^2}+1})^3}{3}-\cdots=$$$$=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}(\sqrt{\frac{1}{x^2}+1})^k$$Here I end up with nasty identity $$f(x)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}(x-1)^k+\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}(\sqrt{\frac{1}{x^2}+1})^k$$and I don't know how to evaluate it into one serie. ($|x|<1$). Any hints? Thanks in advance
| 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 $(1+x^2)^{-\frac{1}{2}}.$
| {
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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 getting:
$$A(x) = xA(x) - x^{2}A(x)$$
Substituting $t = A(x)$, solving simple quadratic equation, and I'm getting two solutions:
$t = A(x) = \frac{1 - i\sqrt{3}}{2}$ or $t = A(x) = \frac{1 + i\sqrt{3}}{2}$
So actually this should be the right side of the generating function $A(x)$, it also has no variable so it already is a coefficient - the job is done.
However, the book shows different results, and they differ a lot. Let me write it:
$a_{n} = -\frac{i\sqrt{3}}{3}(\frac{1+i\sqrt{3}}{2})^{n mod6}$ or $a_{n} = \frac{i\sqrt{3}}{3}(\frac{1-i\sqrt{3}}{2})^{n mod6}$
What did I do wrong?
| 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 sequence, $A(x):=\sum_{n=0}^{\infty}a_nx^n$.
$$A(x)-a_1x-a_0=x(A(x)-a_0)+x^2A(x)$$
Note that we subtract some terms from the sum. Next, plug in the initial conditions $a_0=0$ and $a_1=1$.
$$A(x)-1\cdot x-0=x(A(x)-0)+x^2A(x) \implies A(x)=\frac{x}{1-x-x^2}$$
Up to now, the only difference between the sequences was a sign. You should have arrived at $A(x)=\frac{x}{1-x+x^2}$.
Unfortunately, the following part is specific to the Fibonacci series. Now, we use the geometric series to obtain a closed formula.
$$\begin{align*}A(x)&=\frac{1}{\sqrt{5}\left(1-\frac{2}{\left(\sqrt{5}-1\right)}x\right)}-\frac{1}{\sqrt{5}\left(1+\frac{2}{\left(\sqrt{5}+1\right)}x\right)}=\\
&=\frac{1}{\sqrt{5}}\sum_{n=0}^{\infty}\left(\frac{2}{\sqrt{5}-1}x\right)^n-\frac{1}{\sqrt{5}}\sum_{n=0}^{\infty}\left(-\frac{2}{\sqrt{5}+1}x\right)^n=\\
&=\frac{1}{\sqrt{5}}\sum_{n=0}^{\infty}\left(\left(\frac{2\left(\sqrt{5}+1\right)}{4}\right)^n-\left(-\frac{2\left(\sqrt{5}-1\right)}{4}\right)^n\right)x^n\end{align*}$$
Finally, let us identify the coefficients of both expansions.
$$a_n=\frac{1}{\sqrt{5}}\left(\left(\frac{\sqrt{5}+1}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right) \implies a_n=0,1,1,2,3,5,8,13,\dots$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $(1, x, x^{2}, x^{3}, ...)$ sequence is equal to $\frac{1}{1-x}$.
However, well...the expression at the top is not infinite, so I can't really express it as a generating function.
How can I do this?
| $(x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{8} = x^{16}(1+x+x^2+x^3+x^4)^8$, 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 $k$ nonegative terms which sum up in $n$: $n+k-1 \choose k-1$. In your case, there are $8$ sums $5=0+0+...+5=0+...+0+5+0=...=5+0+..+0$,which you cannot get in the products, because $k_i\leq 4$. So the answer is ${5+8-1} \choose{8-1}$$-8$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove That $x=y=z$ If $x, y,z \in \mathbb{R}$,
and if
$$ \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$$
I was struck up here, i would appreciate if any one helps me here
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality $\frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}\ge{3}$ Let a, b, c be the three side lengths of a triangle. Prove that
$$\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$$
solution 2:
by cauchy-Schwarz inequality we have
$$\sum\dfrac{a}{b+c-a}=\sum\dfrac{a^2}{a(b+c-a)}\ge\dfrac{(a+b+c)^2}{\sum a(b+c-a)}$$
$$\Longleftrightarrow (a+b+c)^2\ge\sum a(b+c-a)$$
$$\Longleftrightarrow 2a^2+2b^2+2c^2\ge ab+bc+ac$$
It's obivous.
solution 3:
since$a+b-c>0,b+c-a>0,a+c-b>0$
then we have
$$\sum\dfrac{2a}{b+c-a}=\sum\left(\dfrac{a+b-c}{b+c-a}+\dfrac{a-b+c}{b+c-a}\right)=\sum\left(\dfrac{a+b-c}{b+c-a}+\dfrac{b+c-a}{a+b-c}\right)\ge 6$$
| {
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"timestamp": "2023-03-29T00:00:00",
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(USAJMO)Find the integer solutions:$ab^5+3=x^3,a^5b+3=y^3$ Find the integer solutions:
$$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}$$
We get a contradiction, so there are no integer solutions.
| {
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"answer_id": 0
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recursive sequence Given $0<a<b$, $\forall n$ define $x_n$ as $x_1=a$, $x_2=b$, $x_{n+2}=\frac{x_n+x_{n+1}}{2}$. Show that $(x_n)$ converges and find the limit.
In order to prove the convergence, I claim that $|x_{n+2}-x_{n+1}|\le\lambda|x_{n+1}-x_{n}|$, $0<\lambda<1$. In fact, $|x_{n+2}-x_{n+1}|\le|\frac{x_{n+1}+x{n}}{2}-x_{n+1}|=\frac{1}{2}|x_{n+1}-x_n|$.But I have no idea about how to find the limit. I'd like to get some help.
| Simpler: For $A(z) = \sum_{n \ge 0} x_{n + 1} z^n$ the recurrence directly translates into
$$
\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}{1 - z}
$$
These are two geometric series:
$$
x_n = \frac{2 a - 2 b}{3} 2^{-n - 1} + \frac{a + 2 b}{3}
= \frac{a - b}{3} 2^{-n} + \frac{a + 2 b}{3}
$$
Thus $x_n \rightarrow (a + 2 b) / 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/378835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Expanding $\frac{1}{1-z-z^2}$ to a power series. How would you expand the analytic function $$\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)}
$$
Here $\tau$ is the positive root of $r^2 - r - 1 = 0$, $\overline{\tau}$ the negative one ($\tau$s are zeros of the denominator $1 - z - z^2$). That is:
\begin{align}
\tau &= \frac{-1 + \sqrt{5}}{2} \\
\overline{\tau} &= \frac{-1 - \sqrt{5}}{2}
\end{align}
This is a pair of geometric series:
$$
[z^n] \frac{1}{1 - z - z^2}
= \frac{\tau^{n + 1} - \overline{\tau}^{n + 1}}{\sqrt{5}}
$$
*Expand:
\begin{align}
\frac{1}{1 - z(1 + z)}
&= \sum_{r \ge 0} z^r (1 + z)^r \\
&= \sum_{r \ge 0} z^r \sum_{0 \le s \le r} \binom{r}{s} z^s \\
[z^n] \frac{1}{1 - z - z^2}
&= \sum_{r + s = n} \binom{r}{s} \\
&= \sum_{0 \le k \le n} \binom{k}{n - k}
\end{align}
*Recognize the generating function of the Fibonacci numbers:
$$
F_0 = 0, F_1 = 1, F_{n + 2} = F_{n + 1} + F_n
$$
gives:
$$
F(z) = \sum_{n \ge 0} F_n z^n = \frac{z}{1 - z - z^2}
$$
so that:
$$
\frac{1}{1 - z - z^2}
= \frac{F(z) - F_0}{z}
= \sum_{n \ge 0} F_{n + 1} z^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding the kth term of an iterated sequence The sequence $x_0, x_1, \dots$ is defined through $x_0 =3, x_1 = 18$ and $x_{n+2} = 6x_{n+1}-9x_n$ for $n=0,1,2,\dots\;$. What is the smallest $k$ such that $x_k$ is divisible by $2013$?
| HINT:
Using Characteristic equation, $$r^2-6r+9=0$$
So, $r=3,3$
$$\text{So,}x_n=(An+B)3^n$$ where $A,B$ are arbitrary constants
$3=x_0=B\implies B=3$ and $18=(A+3)\cdot3\implies A=3\implies x_n=(n+1)3^{n+1}$
Now, $2013=3\cdot11\cdot 61$
As $(3,61)=(3,11)=1\implies (n+1)$ must be divisible by $11\cdot 61=671$
and $(n+1)3^{n+1}$ must be divisible by $3$ which is true if $n\ge0$
So, $n=671\cdot a-1$ where $a$ is any integer
The minimum positive value of $n$ will be $671\cdot1-1=670$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/383381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int {\sin 2\theta \over 1 + \cos \theta} \, d\theta $, using the substitution $u = 1 + \cos \theta $
Evaluate $$\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} &= -\sin\theta \\
d\theta &= \frac{du}{-\sin\theta}\\
\end{align}$$
$$\begin{align}
\int_0^{\pi \over 2} {{\sin 2\theta } \over {1 + \cos \theta }} &= \int_0^{\pi \over 2} {{\sin 2\theta } \over {1 + \cos \theta }} {{du} \over { - \sin \theta }} \\
& = \int_0^{\pi \over 2} {{2\sin \theta \cos \theta } \over {1 + \cos \theta }} {{du} \over { - \sin \theta }} \\
& = \int_1^2 {{2(u - 1)} \over { - u}}\, du \\
& = \int_1^2 {{2 - 2u} \over u} \,du \\
&= \int_1^2\left( {2 \over u} - 2\right) \,du \\
& = \left[ 2\ln |u| - 2u \right]_1^2 \\
& = (2\ln 2 - 2(2)) - (2\ln1 - 2(1)) \\
& = (2\ln 2 - 4) - (2\ln 1 - 2) \\
& = 2\ln 2 - 2 \\
\end{align} $$
This answer is incorrect, the answer in the book is
$$2 - 2\ln 2$$
Could someone tell me where and how I went wrong? I'm not sure, but I think it may have been when I distributed the minus sign from the denominator to the numerator.
| Even easier $(t=\cos x$):
$$
\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}$$
However, this isn't the answer, and I can't see what I've done wrong.
This is what the mark scheme says:
$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}} = \left[\left(\frac{4}{25x^4}\right)^{\frac{1}{2}} \text{ or } \left(\frac{5x^2}{2}\right)^{-1} \text{ or } \frac{1}{\left(\dfrac{25x^4}{4}\right)^{\frac{1}{2}}}\right] = \frac{2}{5}x^{-2}$$
To me, my answer looks more simple than theirs, and I can't see what I've done wrong.
| Just to point out the "rule" you forgot:
It is correct that $$(a^m)^n=a^{mn}$$
But you have more than $(x^4)^{-1/2}$ to consider. You need to apply the fact that when a product and/or quotient of numbers/variables, enclosed in parentheses, is raised to a power, you need to distribute that power across the product in parentheses:
$$(ab)^n = a^nb^n\quad \text{or given,} \quad \left(\frac{ab}c\right)^n = \dfrac{a^nb^n}{c^n}$$
So in your case, you need to combine the two rules: $$(ab^m)^n = a^n\cdot \left(b^{m}\right)^n = a^n \cdot b^{mn}$$
So for $$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}}$$ I would simplify matters to express this as $$\left(\frac{25}{4} x^4\right)^{-1/2} = \left(\frac{25}{4}\right)^{-1/2} (x^4)^{-1/2} = \left(\frac{5^2}{2^2}\right)^{-1/2} (x^4)^{-2} = \frac{5^{-1}}{2^{-1}}x^{-2} = \frac 25 x^{-2} = \frac 2{5 x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/388748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
An integral involving Fresnel integrals $\int_0^\infty \left(\left(2\ S(x)-1\right)^2+\left(2\ C(x)-1\right)^2\right)^2 x\ \mathrm dx,$ I need to calculate the following integral:
$$\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}\mathrm dz,$$
$$C(x)=\int_0^x\cos\frac{\pi z^2}{2}\mathrm dz$$
are the Fresnel integrals.
Numerical integration gives an approximate result $0.31311841522422385...$ that is close to $\frac{16\log2-8}{\pi^2}$, so it might be the answer.
| 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 }^{2}}\,\int _{0}^{\infty }\! \left( \int _{0
}^{\infty }\!{\frac {2\,\sin \left( 2\,\pi\, y\, r \right) }{\sqrt {1+4\,
r} \left( 1+\sqrt {1+4\,r} \right) }}{dr} \right) ^{2}{dy}
$, see Appendix,
$\displaystyle=\dfrac{16}{{\pi }^{2}}\,\int _{-\infty}^{\infty }\! \left( \int _{0
}^{\infty }\!{\frac {2\,\sin \left( 2\,\pi \,y \,r \right) }{\sqrt {1+4\,
r} \left( 1+\sqrt {1+4\,r} \right) }}{dr} \right) ^{2}{dy}
$,
$\displaystyle=\dfrac{16}{{\pi }^{2}}\,\int _{-\infty}^{\infty }\! \left( \int _{-\infty
}^{\infty }\!{\frac {H(r)\,\sin \left( 2\,\pi \,\,y\,r \right) }{\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right) ^{2}{dy}
$ : $H \left( r \right) =\cases{1&$0\leq x$\cr -1&$x<0$\cr}$.
Note then that:
$\displaystyle\left(\int _{-\infty
}^{\infty }\!{\frac {H(r)\,\sin \left( 2\,\pi \,\,y\,r \right) }{\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right) ^{2}=\displaystyle\left|-\dfrac{i}{2}\int _{-\infty
}^{\infty }\!{\frac {H(r)\,e^{ -i2\pi \,y\,r } }{\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr}+\dfrac{i}{2}\int _{-\infty
}^{\infty }\!{\frac {H(r)\,e^{ i2\pi\, y\,r } }{\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right| ^{2}$
$=\displaystyle\left|\int _{\infty
}^{\infty }\!{\frac {\left(H(r)-H(-r)\right)\,e^{ i2\pi \,y\,r } }{2\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right| ^{2}=\displaystyle\left|\int _{\infty
}^{\infty }\!{\frac {\,e^{ i2\pi \,y\,r } }{\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right| ^{2}.$
We can now use Plancherel's theorem which states:
$\displaystyle \int _{-\infty }^{\infty }\! \left| f \left( y \right)
\right| ^{2}{dy}=\int _{-\infty }^{\infty }\!
\left| F \left( r \right) \right| ^{2}{dr}
$ : $ \displaystyle F \left( r \right) =\int _{-\infty }^{\infty }\!f \left( y \right) {
e^{-i2\pi r y}}{dy}
$,
and thus:
$\displaystyle\dfrac{16}{{\pi }^{2}}\,\int _{-\infty}^{\infty }\! \displaystyle\left|\int _{\infty
}^{\infty }\!{\frac {\,e^{ i2\pi \,y\,r } }{\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right| ^{2}{dy}$
$=\displaystyle\dfrac{16}{{\pi }^{2}}\,\int _{-\infty}^{\infty }\! \displaystyle\left|{\frac {1 }{\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }} \right| ^{2}{dr}$,
$=\dfrac{16}{{\pi }^{2}}\displaystyle\int _{0}^{\infty }\!{\frac {2}{ \left( 1+4\,r \right) \left( 1+
\sqrt {1+4\,r} \right) ^{2}}}{dr}$.
Now if we undo the substitution made earlier by letting $r = p^2-p$:
$\dfrac{16}{{\pi }^{2}}\displaystyle\int _{0}^{\infty }\!{\frac {2}{ \left( 1+4\,r \right) \left( 1+
\sqrt {1+4\,r} \right) ^{2}}}{dr}=\dfrac{16}{{\pi }^{2}}\int _{1}^{\infty }\!{
\frac {1}{2{p}^{2} \left( 2\,p-1 \right) }}{dp}$,
and then make one final substitution $p=\dfrac{1}{q+2}$ we get:
$\displaystyle\dfrac{16}{{\pi }^{2}}\int _{1}^{\infty }\!{
\frac {1}{2{p}^{2} \left( 2\,p-1 \right) }}{dp}=-\dfrac{16}{{\pi }^{2}}\int_{-2}^{-1}\dfrac{1}{2}+\dfrac{1}{q}{dq}=\dfrac{16\ln(2)-8}{\pi^2}$.
Appendix:
Note that the integral: $$\int _{0
}^{\infty }\!{\frac {2\,\sin \left( 2\,\pi \,y \,r \right) }{\sqrt {1+4\,
r} \left( 1+\sqrt {1+4\,r} \right) }}{dr} $$
converges by the Chartier-Dirichlet test because: $$f(r)={\frac {2\, }{\sqrt {1+4\,
r} \left( 1+\sqrt {1+4\,r} \right) }}$$
is monotonic and continuous on $ \mathbb R^+ $ and $f(r)\rightarrow0$ as $r\rightarrow \infty$, and because:
$$\left|\int_{0}^{b}\sin\left(2\pi\,y\,r\right){dr}\right|$$
is bounded as $b\rightarrow\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/390847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "46",
"answer_count": 2,
"answer_id": 0
} |
Reducibility of $x^{2n} + x^{2n-2} + \cdots + x^{2} + 1$ Just for fun I am experimenting with irreducibility of certain polynomials over the integers. Since $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$, I thought perhaps $x^6+x^4+x^2+1$ is also reducible. Indeed:
$$x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$
Let $f_n(x)=x^{2n}+x^{2n-2}+\cdots + x^2+1$. Using Macaulay2 (powerful software package) I checked that:
$$f_4(x) = (x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$$
We get that
$$ f_5(x)=(x^2+1)(x^2-x+1)(x^2+x+1)(x^4-x^2+1)$$
The polynomial is reducible for $n=6, 7, 8, 9$ as far as I checked. I suspect that $f_n(x)$ is reducible over integers for all $n\ge 2$. Is this true?
Thanks!
| Let $g_n(x) = x^n + x^{n-1} + \ldots + x + 1$. Then $g_n(x) = \frac{x^{n+1}-1}{x-1}$. Your polynomial $f_n(x) = g_n(x^2)$. The factorization of $x^n-1$ over $\mathbb{Q}$ is known:
$$
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 lead to a factorization of $f_n(x) = g_n(x^2)$ in $\mathbb{Q}[x^2]$. It is then possible that this factorization may be refined further. So, the only chance of an irreducible $f_n(x)$ would be when $n = p-1$ for some prime: in this case, it is known that $g_{p-1}(x) = x^{p-1} + \ldots + x + 1$ is irreducible over $\mathbb{Q}$. However, for odd primes $p$, the splitting field $\mathbb{Q}(\zeta_p)$ is equal to $\mathbb{Q}(\zeta_{2p})$. Thus, $\zeta_p = \zeta_{2p}^2$, that is, $\zeta_{p}$ has a square root in the same field. So $\zeta_{2p}$ has a minimal polynomial of degree $p-1$ and it is a root of $f_{p-1}(x)$, which has degree $2(p-1)$. So, $f_{p-1}(x)$ must split into two irreducible factors.
Therefore, the answer is yes, $f_n(x)$ is always reducible when $n\geq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/391086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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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)} = 1 + \sqrt{(x-1)}$
Square both sides: $2x-5 = (1 + \sqrt{(x-1)})^{2}$
We have removed one square root.
Expand right hand side: $2x-5 = 1 + 2\sqrt{(x-1)} + (x-1)$-- I don't understand?
Simplify: $2x-5 = 2\sqrt{(x-1)} + x$
Simplify more: $x-5 = 2\sqrt{(x-1)}$
Now do the "square root" thing again:
Isolate the square root: $\sqrt{(x-1)} = \frac{(x-5)}{2}$
Square both sides: $x-1 = (\frac{(x-5)}{2})^{2}$
Square root removed
Thank you in advance for your help
| $\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 &= 7 \pm 2 \sqrt 5
\end{align}
| {
"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 but it doesn't show me the steps (even in pro version)
Thanks
| 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+2)^2}{(n+2)(n+3)} \\
&= \frac{n+2}{n+3} \\
\end{align*}
Provided $n\neq -2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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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 far:
$n = p \cdot q$
$x^2 \equiv 1\pmod{n}$
$x^2 -1 \equiv 0\pmod{n}$
$(x-1)(x+1) \equiv 0\pmod{n}$
$x-1 = 20779 \lor x + 1=20779$
I have also noticed that:
$(x-1)(x+1) \equiv 0\pmod{p \cdot q}$
$(x-1)(x+1) \equiv 0\pmod{p} \land (x-1)(x+1) \equiv 0\pmod{q}$
But I have no idea what to do next. Any hints?
| 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 product of two primes, then one will try to factor f1 and f2 or prove their primality.
| {
"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 y+y^2\right)}{x^3}$$
as you can see the orange (above ) and the blue (your fraction ) are very near is the generalitated laurent series
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394312",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Integrate ${\sec 4x}$ How do I go about doing this? I try doing it by parts, but it seems to work out wrong:
$\eqalign{
& \int {\sec 4xdx} \cr
& u = \sec 4x \cr
& {{du} \over {dx}} = 4\sec 4x\tan 4x \cr
& {{dv} \over {dx}} = 1 \cr
& v = x \cr
& \int {\sec 4xdx} = x\sec 4x - \int {4x\sec 4x\tan 4xdx} \cr
& \int {4x\sec 4x\tan 4xdx} : \cr
& u = 4x \cr
& {{du} \over {dx}} = 4 \cr
& {{dv} \over {dx}} = \sec 4x\tan 4x \cr
& v = {1 \over 4}\sec x \cr
& \int {4x\sec 4x\tan 4xdx} = x\sec 4x - \int {\sec 4x} dx \cr
& \int {\sec 4xdx} = x\sec 4x - \left( {x\sec 4x - \int {\sec 4xdx} } \right) \cr} $
I don't know where to go from here, everything looks like it equals 0, where have I went wrong?
Thank you!
EDIT: Is there an easier way to do this?
| Use two substitutions. The first substitution transforms the integrand into $\sec \theta$, whose evaluation was asked here. The second substitution is the Weirstrass substitution. (See comment below). In the present case the second integral becomes an easy table integral:
$$\begin{eqnarray*}
\int \sec 4x\,dx &=&\frac{1}{4}\int \sec \theta \,d\theta ,\qquad \theta =4x
\\
&=&\frac{1}{4}\int \frac{1}{\frac{1-t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}
\,dt,\qquad t=\tan \frac{\theta }{2} \\
&=&\frac{1}{2}\int \frac{1}{1-t^{2}}dt=\frac{1}{2}\operatorname{arctanh}t+C \\
&=&\frac{1}{2}\operatorname{arctanh}\left( \tan \frac{\theta }{2}\right) +C \\
&=&\frac{1}{2}\operatorname{arctanh}\left( \tan 2x\right) +C.
\end{eqnarray*}$$
This integral can be rewritten as
$$\frac{1}{2}\operatorname{arctanh}\left( \tan 2x\right) =\frac{1}{4}\ln \left\vert
\tan 2x+1\right\vert -\frac{1}{4}\ln \left\vert 1-\tan 2x\right\vert .$$
Comment: The Weierstrass substitution is a universal standard substitution to evaluate an integral of a rational fraction in $\sin \theta,\cos \theta$, i.e. a rational fraction of the form
$$R(\sin \theta,\cos \theta)=\frac{P(\sin \theta,\cos \theta)}{Q(\sin \theta,\cos \theta)},$$
where $P,Q$ are polynomials in $\sin \theta,\cos \theta$
$$
\begin{equation*}
\tan \frac{\theta }{2}=t,\qquad\theta =2\arctan t,\qquad d\theta =\frac{2}{1+t^{2}}dt
\end{equation*},
$$
which converts the integrand into a rational function in $t$. We know from trigonometry (see this answer) that
$$\cos \theta =\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{
\theta}{2}}=\frac{1-t^2}{1+t^2},\qquad \sin \theta =\frac{2\tan \frac{\theta }{2}}{1+\tan ^{2}
\frac{\theta }{2}}=\frac{2t}{1+t^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
A question about an inequality How to get the solution set of the inequality $$\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 \\
&=& \sum_{j=1}^n \sum_{k=j}^n \log j \\
&=& \sum_{j=1}^n (n-j+1)\log j \\
&=& (n+1)\sum_{j=1}^n \log j - \sum_{j=1}^n j\log j \\
&=& (n+1)\log n! - n^2 \frac{1}{n}\sum_{j=1}^n \frac{j}{n}\log \frac{j}{n} - \sum_{j=1}^n j\log n \\
&=& (n+1)\log n! - n^2\int_0^1 dx\ x\log x - \frac{n(n+1)}{2}\log n + O(n\log n) \\
&=& (n+1)\log n! +\frac{n^2}{4} - \frac{n(n+1)}{2}\log n + O(n\log n).
\end{eqnarray*}$$
(The error estimate above can probably be tightened.)
Using Stirling's approximation we find
$$\begin{eqnarray*}
\frac{1}{n^2}\sum_{k=0}^n \log {n\choose k}
&=& -\frac{n+1}{n^2}(\log n! - n\log n) - \frac{1}{2} + O\left(\frac{\log n}{n}\right) \\
&=& -\frac{n+1}{n^2}(-n + O(\log n)) - \frac{1}{2} + O\left(\frac{\log n}{n}\right) \\
&=& \frac{1}{2} + O\left(\frac{\log n}{n}\right).
\end{eqnarray*}$$
| {
"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
} |
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_{m} \, t^{m} &= \sum_{m=0}^{\infty} \sum_{k=0}^{m} k(k+2) t^{m} \nonumber\\
&= \sum_{m=0}^{\infty} \sum_{k=0}^{\infty} k(k+2) t^{m+k} \nonumber\\
&= \sum_{m=0}^{\infty} t^{m} \cdot \sum_{k=0}^{\infty} k(k+2) t^{k}
= \frac{1}{t(1-t)} \sum_{k=0}^{\infty} k(k+2) t^{k+1} \nonumber\\
&= \frac{1}{t(1-t)} \partial_{t} \, \sum_{k=0}^{\infty} k \, t^{k+2}
= \frac{1}{t(1-t)} \partial_{t} \left[ t^{3} \sum_{k=0}^{\infty} k t^{k-1} \right] \\
&= \frac{1}{t(1-t)} \partial_{t} \left[ t^{3} \partial_{t} \sum_{k=0}^{\infty}
t^{k} \right]
= \frac{1}{t(1-t)} \partial_{t} \left[ t^{3} \partial_{t} \left( \frac{1}{1-t} \right) \right] \\
&= \frac{1}{t(1-t)} \left[ \frac{2 t^{3}}{(1-t)^{3}} + \frac{3 t^{2}}{(1-t)^{2}} \right] = \frac{2t^{2}}{(1-t)^{4}} + \frac{3 t}{(1-t)^{3}}. \\
\end{align}
Since $t = 1-(1-t)$ and $t^{2} =1-2(1-t) + (1-t)^{2}$ then the last result becomes
\begin{align}
\sum_{m=0}^{\infty} S_{m} t^{m} = \frac{2}{(1-t)^{4}} - \frac{1}{(1-t)^{3}} - \frac{1}{(1-t)^{2}}.
\end{align}
Now using the standard series
\begin{align}
\frac{1}{(1-t)^{2}} &= \sum_{m=0}^{\infty} (m+1) t^{m} \\
\frac{1}{(1-t)^{3}} &= \frac{1}{2} \sum_{m=0}^{\infty} (m+1)(m+2) t^{m} \\
\frac{1}{(1-t)^{4}} &= \frac{1}{6} \sum_{m=0}^{\infty} (m+1)(m+2)(m+3) t^{m}
\end{align}
then
\begin{align}
\sum_{m=0}^{\infty} S_{m} t^{m} &= \sum_{m=0}^{\infty} (m+1) \left[ \frac{1}{6}
(m+2)(m+3) - \frac{1}{2} (m+2) - 1 \right] t^{m} \\
&= \sum_{m=0}^{\infty} \left[ \frac{m(m+1)(2m+7)}{6} \right] \, t^{m}
\end{align}
and upon equating the coefficients of both sides it is seen that
\begin{align}
S_{m} = \sum_{k=0}^{m} k(k+2) = \frac{m(m+1)(2m+7)}{6} = \frac{1}{6} (2m^{3}+ 9m^{2} +7m).
\end{align}
This result may be verified by combining the two series,
\begin{align}
\sum_{k=0}^{n} k &= \frac{1}{2}(n^{2} +n) \\
\sum_{k=0}^{n} k^{2} &= \frac{1}{6}(2n^{3}+3n^{2}+n),
\end{align}
in the desired way. This provides
\begin{align}
\sum_{k=0}^{n} k(k+2) = \frac{1}{6} ( 2n^{3}+9n^{2}+7n ).
\end{align}
This example shows that the generating function method yields the desired result.
| {
"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}$$
Can someone help me with it? (I can't use Dirichlet' because of the areas where $x$ is close to $0$)
| The series does converge uniformly. For the proof, put $S_n(x) = \sum_{k = 0}^n \sin{(kx)}\sin{(k^2 x)}$ for $n\geq 0$. The general idea is to use summation by parts to reduce ourselves to showing that $S_n(x)$ is bounded uniformly, and then to prove that by giving a closed form for $S_n(x)$.
First, the summation by parts (I write $S_n$ in place of $S_n(x)$ for brevity):
$$
\begin{align}
\sum_{n = 1}^N{\sin{(nx)}\sin{(n^2 x)}\over n + x^2} & = \sum_{n = 1}^N {1\over n+x^2}(S_n - S_{n-1}) \\
& = \sum_{n = 1}^N {S_n\over n+x^2} - \sum_{n = 1}^N {S_{n-1}\over n+x^2} \\
& = \sum_{n = 1}^N {S_n\over n+x^2} - \sum_{n = 0}^{N-1} {S_{n}\over n+1+x^2} \\
& = {S_N\over N+x^2} - {S_0\over 1 + x^2} + \sum_{n = 1}^{N-1} S_n\left({1\over n+x^2} - {1\over n+1+x^2}\right) \\
& = {S_N\over N+x^2} + \sum_{n = 1}^{N-1} {S_n\over (n+x^2)(n+1+x^2)}.
\end{align}
$$
From here it is clear that it is enough to prove that $S_n = S_n(x)$ is uniformly bounded in $x$.
To do this, note that
$$
\begin{align}
2\sin{(kx)}\sin{(k^2 x)} &= \cos{\{(k^2 - k)x\}} - \cos{\{(k^2 + k)x\}} \\
& = \cos{\{k(k - 1)x\}} - \cos{\{(k+1)k)x\}},
\end{align}
$$
and therefore that
$$
\begin{align}
2S_n(x) & = \sum_{k = 0}^n 2\sin{(kx)}\sin{(k^2 x)} \\
& = \sum_{k = 0}^n\left(\cos{\{k(k - 1)x\}} - \cos{\{(k+1)k)x\}}\right)
\end{align}
$$
The last sum telescopes, and we are left with
$$
2S_n(x) = 1 - \cos{\{n(n+1)x\}},
$$
which is plainly bounded uniformly in $x$. So we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/397097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 1,
"answer_id": 0
} |
Laurent expansion problem Expand the function $$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]
$$
but I'm stuck with $$ \frac{-2} {z^2+1} $$ how can I expand it?
| Hint:
$$\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))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/397696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
linear Transformation of polynomial with degrees less than or equal to 2 I would like to determine if the following map $T$ is a linear transformation:
\begin{align*}
T: P_{2} &\to P_{2}\\
A_{0} + A_{1}x + A_{2}x^{2} &\mapsto A_{0} + A_{1}(x+1) + A_{2}(x+1)^{2}
\end{align*}
My attempt at solving:
\begin{align}
T(p + q) &= p(x+1) + q(x+1)\\
&= \left[A_{0} + A_{1}(x+1) + A_{2}(x+1)^2\right] + \left[b_{0} + b_{1}(x+1) + b_{2}(x+1)^2\right]\\
&= \left(A_{0} + b_{0}\right) + \left(A_{1} + b_{1}\right)(x+1) + \left(A_{2} + b_{2}\right)(x+1)^2\\
&= T(p) + T(q)
\end{align}
Is this right so far? If not, what am I doing wrong?
| The idea is right but you are starting from what you have to proof
Hint : $T(p+q)=T(A_0+B_0+(A_1+B_1)x+(A_2+B_2)x^2)=...=T(p)+T(q)$.
I see that you had the right idea maybe this could make you start
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/397748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integration of $\displaystyle \int\frac{1}{1+x^8}\,dx$
Compute the indefinite integral
$$
\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}
$$
Then we can rewrite the integral as
$$
\int\frac{1}{1+x^8}\,dx = \int \frac{1}{(x^4+\sqrt{2}x^2+1)(x^4-\sqrt{2}x^2+1)}\,dx$$
To use partial fractions let $t = x^2$ to get
$$
\frac {1}{(t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)} = \frac{At+B}{t^2+\sqrt{2}t+1}+\frac{Ct+D}{t^2-\sqrt{2}t+1}
$$
This method of solving the problem becomes very complex. Is there a less complex approach to the problem?
| Why not splitting up in fractions until you have first degree polynomials in the nominators?
$$\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}}$$
or if you prefer without the complex numbers
$$\frac{1}{1+x^8}=\frac{ax+b}{x^2-2\cos(\pi/8)x+1}+\frac{cx+d}{x^2-2\cos(3\pi/8)x+1}+\frac{ex+f}{x^2-2\cos(5\pi/8)x+1}+\frac{gx+h}{x^2-2\cos(7\pi/8)x+1} \; .$$
With the complex formula, you can find the coefficients easily as follows
$$A=\lim_{x \to e^{i\pi/8}}\frac{x-e^{i\pi/8}}{1+x^8}\overset{\text{H}}{=}\lim_{x \to e^{i\pi/8}}\frac{1}{8x^7}=\frac{e^{-i7\pi/8}}{8}$$
where I used de l'Hôpital's rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/398878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Solve equations $\sqrt{t +9} - \sqrt{t} = 1$ Solve equation: $\sqrt{t +9} - \sqrt{t} = 1$
I moved - √t to the left side of the equation $\sqrt{t +9} = 1 -\sqrt{t}$
I squared both sides $(\sqrt{t+9})^2 = (1)^2 (\sqrt{t})^2$
Then I got $t + 9 = 1+ t$
Can't figure it out after that point.
The answer is $16$
| $$\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!}
$$
The first thing I noticed is that this looks an awful lot like the series expansion of $e^x$:
$$
e^x = \sum\limits_{k=0}^\infty \frac{x^k}{k!}
$$
The only obstacle is the $(-1)^k$ term. I tried getting rid of it by rewriting:
\begin{align*}
\sum\limits_{k=0}^\infty (-1)^k \frac{x^k}{k!} &= 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \frac{x^5}{5!} + \dots\\
&= (1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots) - (x + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots)\\
&= \sum\limits_{k=0}^\infty \frac{x^{2k}}{(2k)!} - \sum\limits_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}
\end{align*}
These sums look a lot like the series expansions of $\sin(x)$ and $\cos(x)$:
\begin{align*}
\sin(x) &= \sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}\\
\cos(x) &= \sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!}
\end{align*}
However, these sums do have a $(-1)^k$ term, just when I got rid of it! So now I'm stuck. Can someone help me in the right direction?
| You started fine, but then you got sidetracked:
$$\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
} |
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