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How to find the minimum value of this function? How to find the minimum value of
$$\frac{x}{3y^2+3z^2+3yz+1}+\frac{y}{3x^2+3z^2+3xz+1}+\frac{z}{3x^2+3y^2+3xy+1}$$,where $x,y,z\geq 0$ and $x+y+z=1$.
It seems to be hard if we use calculus methods. Are there another method? I have no idea.
Thank you.
| By C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{x}{3y^2+3z^2+3yz+1}=\sum_{cyc}\frac{x}{3y^2+3z^2+3yz+(x+y+z)^2}=$$
$$=\sum_{cyc}\frac{x}{x^2+4y^2+4z^2+2xy+2xz+5yz}=$$
$$=\sum_{cyc}\frac{x^2}{x^3+4xy^2+4xz^2+2x^2y+2x^2z+5xyz}\geq$$
$$\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(x^3+6x^2y+6x^2z+5xyz)}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(2x^3+6x^2y+6x^2z+4xyz)}=$$
$$=\frac{(x+y+z)^2}{2(x+y+z)^3}=\frac{1}{2}.$$
The equality occurs for $x=y=z=\frac{1}{3}$, which says that the answer is $\frac{1}{2}$.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Rationalizing the denominator 3 It is a very difficult question. How can we Rationalizing the denominator?
$$\frac{2^{1/2}}{5+3*(4^{1/3})-7*(2^{1/3})}$$
| $\frac1{A+B\sqrt[3]{n}+C\sqrt[3]{n^2}}=
\frac{ \begin{vmatrix}
1&\sqrt[3]n&\sqrt[3]{n^2}\\
B&A&Cn\\
C&B&A
\end{vmatrix} }{\begin{vmatrix}
A&Cn&Bn\\
B&A&Cn\\
C&B&A
\end{vmatrix}}=
\frac{A^2-BCn+(C^2n-AB)\sqrt[3]{n}+(B^2-AC)\sqrt[3]{n^2}}
{A^3+B^3n+C^3n^2-3ABCn}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/681873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Why we do division in those steps told, and who invented division? i know how to divide but i dont quit understand why we use those steps told in schools.
like for example
____
3/450 150 quotient
3
-----
15
15
------
000
could someone please tell me why we do these steps is there any other method or can someone explain why?
| Taking your example, write
$$
\frac{450}{3}
= \frac{4 \times 100 + 5 \times 10 + 0 \times 1}{3}
= \frac{4}{3} \times 100 + \frac{5}{3} \times 10 + \frac{0}{3} \times 1.
$$
We will arrange the terms so that the standard method is suggested.
First compute $\frac{4}{3} \times 100 = \left(1 + \frac{1}{3} \right) \times 100$ first; next, take the piece $\frac{1}{3} \times 100$ and add it to the piece $\frac{5}{3} \times 10$ to get $\left(\frac{10}{3} + \frac{5}{3}\right)\times 10 = \frac{15}{3} \times 10 = 5 \times 10$. The last piece is just zero. Your conclusion is
$$
\frac{450}{3}
= 1 \times 100 + 5 \times 10 + 0 \times 1
= 150.
$$
I hope this helps!
| {
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Simplfying $\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}}$ I am trying to simplify the expression:
$\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}}$
I tried to square the expression but I can't do that because it is not an equation so I got stuck. Can someone please give me some pointers on how to proceed?
| Here is another way of doing it.
If we set $a=31+8\sqrt {15}, b=31-8\sqrt {15}$ we have
$a+b=62$ and
$ab=31^2-15\cdot 64=31^2-30\cdot32=1$
We know that $ab$ will work out well, because we know that $(x+y)(x-y)=x^2-y^2$
Then we know that $a$ and $b$ are the roots of $x^2-62x+1=0$
We want the square roots $\alpha$ and $\beta$ of $a$ and $b$ so if we set $y^2=x$ we will create an equation which has these as roots - $y^4-62y^2+1=0$
We know that $ab=1$ because we computed that. When we take the square root we get that $\alpha\beta=1$.
We therefore look for a factorisation $(y^2+py+1)(y^2-py+1)=y^4-62y^2+1$ which gives us $2-p^2=-62$, or $p^2=64$.
We know that one of these factors will have roots $\alpha, \beta$ and the other will have roots $-\alpha, -\beta$ so the final step is to work out which of these gives us the answer.
I have glossed over a couple of things because this is homework - but the double square root coming from a quartic without any terms in $y$ or $y^3$ is closely related to the underlying field theory, and is worth noticing.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve : $\lim_{n\rightarrow \infty} \frac{n!}{n\cdot 2^{n}}$ $$\lim_{n\rightarrow \infty} \frac{n!}{n\cdot 2^{n}}$$
I need to solve the limit problem above. I have no idea about what to do. What do you suggest?
Thanks in advance.
| Since
$$n!=1\cdot 2\cdot 3\cdot 4\cdot 5\cdot \cdots \cdot n\ge 2\cdot 3\cdot 4^{n-3}$$
$$\frac{n!}{n\cdot 2^n} \ge 2\cdot 3\cdot \frac{4^{n-4}}{2^n}=\frac{2\cdot3}{4^4} 2^n$$
for $n\ge 4$. Therefore the sequence diverges.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For all square matrices $A$ and $B$ of the same size, it is true that $(A+B)^2 = A^2 + 2AB + B^2$? The below statement is a true/false exercise.
Statement:
For all square matrices A and B of the same size, it is true that
$(A + B)2 = A^2 + 2AB + B^2$.
My thought process: Since it is not a proof, I figure I can show by example and come to a valid conclusion based on such example.
My work:
Come up with a square matrix A and B let both be a 2 by 2 matrix(rows and cols must be same).
Matrix $A$:
$A = \begin{array}{ccc}
3 & 5 \\
4 & 6 \\
\end{array} $
Matrix $B$:
$B = \begin{array}{ccc}
5 & 8 \\
9 & 4 \\
\end{array} $
$A + B = \begin{array}{ccc}
8 & 13 \\
13 & 10 \\
\end{array}$
$(A + B)^2 = \begin{array}{ccc}
233 & 234 \\
234 & 264 \\
\end{array}$
$A^2 = \begin{array}{ccc}
29 & 45 \\
36 & 56 \\
\end{array}$
$(AB) = \begin{array}{ccc}
60 & 44 \\
74 & 56 \\
\end{array}$
$2(AB) = \begin{array}{ccc}
120 & 88 \\
234 & 112 \\
\end{array}$
$B^2 = \begin{array}{ccc}
97 & 72 \\
81 & 88 \\
\end{array}$
$A^2 + 2AB + B^2 = \begin{array}{ccc}
246 & 205 \\
265 & 256 \\
\end{array}$
Based my above work, the answer is false.
Is there another way to approach the problem? It seems like a lot of work needed to be done for a true/false question which raised my suspicion about whether there is a better way to look at the problem.
| A counterexample is all you need, so you're done. You probably could have picked a simpler counterexample, say with most entries of $A$ and $B$ being $0,$ but yours works just fine.
As an alternative, note that if $A$ and $B$ are square matrices of the same size, then $$(A+B)^2=A(A+B)+B(A+B)=A^2+AB+BA+B^2.$$ From this, it follows that $$(A+B)^2=A^2+2AB+B^2$$ if and only if $AB=BA.$ So, any two square matrices $A,B$ of the same size such that $AB\neq BA$ will yield a counterexample.
| {
"language": "en",
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"question_score": "8",
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Suppose that $2^b-1\mid 2^a+1$. Show that $b = 1$ or $2$. I'm stuck with this one. I would appreciate any idea how to prove this.
| Suppose that $2^b-1 \mid 2^a+1$.
Note that $b$ is the order of $2 \pmod{2^b-1}$. Since $2^{2a} \equiv 1 \pmod{2^b-1}$ we have $b \mid 2a$.
If $b$ is odd then we get $b \mid a$, so $2^a \equiv 1 \pmod{2^b-1}$. However we also have $2^a \equiv -1 \pmod{2^b-1}$, whence $-1 \equiv 1 \pmod{2^b-1}$, so $b=1$.
Otherwise $b$ is even, so $3 \mid 2^b-1$, so $3 \mid 2^a+1$, so $a$ is odd. Write $b=2c, c \in \mathbb{Z}^+$, so $b \mid 2a$ becomes $c \mid a$. Write $a=kc$. Since $a$ is odd, both $k$ and $c$ are odd. We now get $2^{2c}-1 \mid 2^{kc}+1$.
Thus $(2^c-1)(2^c+1) \mid (2^c+1)(2^{(k-1)c}-2^{(k-2)c}+ \ldots -2^c+1)$
$$2^c-1 \mid (2^{(k-1)c}-2^{(k-2)c}+ \ldots -2^c+1)=\sum_{j=0}^{k-1}{(-1)^j2^{jc}}$$
Now $$\sum_{j=0}^{k-1}{(-1)^j2^{jc}} \equiv \sum_{j=0}^{k-1}{(-1)^j} \equiv 1 \pmod{2^c-1}$$
Therefore $2^c-1 \mid 1$, whence $c=1$, so $b=2$.
In conclusion, $2^b-1 \mid 2^a+1 \Rightarrow b=1, 2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving an inequality involving factorials: $(\frac{n}{2})^n \ge n! \geq (\frac{n}{3})^n$ For $n \geq 6$, where $n$ is a natural number, prove that $(\frac{n}{2})^n \ge n! \geq (\frac{n}{3})^n$.
I tried using induction but could not do it.
| Re-trying with induction...assume $(\frac{n}{2})^n \ge n!$ is valid for all natural number $k$ such that $n \ge k \ge 6$, then let's prove it's valid also for $n+1$. For the inductive hypothesis we have
$$ \left(\frac{n}{2}\right)^n \ge n! $$
so
$$ (n+1)\cdot \left(\frac{n}{2}\right)^n \ge (n+1)! $$
but since $\left(\frac{n+1}{2}\right)^{(n+1)} = \left(\frac{n+1}{2}\right)^{(n)}\cdot \left(\frac{n+1}{2}\right) \ge (n+1)\cdot \left(\frac{n}{2}\right)^n$ because
$$\frac{1}{2}\left(\frac{n+1}{2}\right)^{(n)} \ge \left(\frac{n}{2}\right)^n$$
$$ \frac{(n+1)^n}{2} \ge n^n$$
$$ \left(\frac{n+1}{n}\right)^n \ge 2 $$
for all positive integers, $$ \left(\frac{n+1}{2}\right)^{(n+1)} \ge (n+1)\cdot \left(\frac{n}{2}\right)^n \ge (n+1)!$$ for all positive integers $\ge 6$ (it's trivial to prove $\left(\frac{6}{2}\right)^6 \ge 6!$).
As for the second part of the inequality you should re-try with induction...
Hint : assume $n! \geq (\frac{n}{3})^n$ true for all integers $k$ such that $n \ge k \ge 6$, then we have $(n+1)! \geq (n+1)\cdot \left(\frac{n}{3}\right)^n$ but, as before
$(n+1)\cdot \left(\frac{n}{3}\right)^n \geq (\frac{n+1}{3})^{n+1}$, because
$$n^n \geq \frac{(n+1)^n}{3}$$
$$ 3 \geq \left( \frac{n+1}{n}\right)^{n}$$
for all positive integers (since is trivial for the smallest positive numbers, and $lim_{n \to \infty} (\frac{n+1}{n})^n = e \lt 3$ )
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac{a+b}{c+b}=\frac{c+d}{a+d}\Rightarrow a=c$ or $a+b+c+d=0$ Please find my mistake or give hint:
If $\frac{c}{a}\geq 1$, then by mediant inequality:
$1\leq\frac{a+b}{c+b}$ and $\frac{c+d}{a+d}\leq 1\Rightarrow c=a$
If $\frac{a}{c}\geq 1$, then:
$\frac{a+b}{c+b}\leq 1$ and $1\leq \frac{c+d}{a+d}\Rightarrow c=a$
thnks
| Note that
$$
\frac{a+b}{c+b} = \frac{c+d}{a+d} \implies
(a+d)(a+b) = (c+d)(c+b)\implies\\
a^2 + ab + ad + db = c^2 + cb + cd + db \implies\\
a^2 - c^2 +ab - cb + ad - cd = 0 \implies\\
(a-c)(a+c) + (a-c)b + (a-c)d = 0 \implies\\
(a-c)(a+b+c+d) = 0
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Center of Mass with two functions
I am having trouble trying to figure out how to go about this problem. I can do problems with single variables but I can not solve this one. I think I would need to subtract the functions from one another but I am not sure which should be subtracted from which. If anyone can help with tips or solutions it would be greatly appreciated. Thank you.
| The formula for Centroid
$$f\left(x\right) = 9\cos \left(x\right) ; g\left(x\right) = 9\sin \left(x\right)$$
$$\bar x =\dfrac{ β«_0^{\frac{{Ο}}{4}} x\left(f\left(x\right)-g\left(x\right)\right) dx}{β«_0^{\frac{{Ο}}{4}} \left(f\left(x\right)-g\left(x\right)\right)dx}$$
$$\bar y= \dfrac{ β«_0^{\frac{{Ο}}{4}} \frac{\left(f\left(x\right)+g\left(x\right)\right)}{2}\left(f\left(x\right)-g\left(x\right)\right) dx}{β«_0^{\frac{{Ο}}{4}}\left(f\left(x\right)-g\left(x\right)\right)dx}$$
To further elaborate:
Denominator = $$9β«_0^{\frac{{Ο}}{4}}\left(\cos \left(x\right) -\sin \left(x\right)\right) dx = 9.\left(2\right).\frac{1}{\sqrt{2}}-9 = 9.\sqrt{2}-9$$
Numerator for $$\bar x = 9β«_0^{\frac{{Ο}}{4}}\left(x\cos \left(x\right) -x\sin \left(x\right)\right) dx$$
Evaluate the integral by integration by parts. I will show the example for one of them:
$$β«_0^{\frac{{Ο}}{4}}x\cos \left(x\right)dx $$
Put u = x, dv = cos(x)dx => du = dx and v = sin(x)
$$β«_0^{\frac{{Ο}}{4}} udv = uv -β«_0^{\frac{{Ο}}{4}}vdu = \left[x\sin \left(x\right) +\cos \left(x\right)\right] |\frac{Ο}{4},0$$
Similarly Put Put u = x, dv = sinxdx => du = dx and v = -cos(x)
$$β«_0^{\frac{{Ο}}{4}} udv = uv -β«_0^{\frac{{Ο}}{4}}vdu = \left[-x\cos \left(x\right) +\sin \left(x\right)\right] |\frac{Ο}{4},0$$
Thus the numerator $$= 9*\left[x\sin \left(x\right) +\cos \left(x\right) - \left[-x\cos \left(x\right) +\sin \left(x\right)\right]\right]= x\left(\sin \left(x\right) + \cos \left(x\right)\right) + \cos \left(x\right) - \sin \left(x\right) = 9\left(\frac{Ο}{4}.\sqrt{2}-1\right)$$
$$\bar x = \dfrac{9\frac{Ο}{4}\sqrt{2} -9}{9\sqrt{2}-9}$$
For the $\bar y$
Numerator =
$$β«_0^{\frac{{Ο}}{4}} \frac{81}{2} \left[\cos \left(x\right) - \sin \left(x\right)\right]\left[\cos \left(x\right) + \sin \left(x\right)\right]$$
$$β«_0^{\frac{{Ο}}{4}} \frac{81}{2}\left[\cos ^{2}\left(x\right) - \sin ^{2}\left(x\right)\right] $$
$$β«_0^{\frac{{Ο}}{4}} \frac{81}{2}\cos \left(2x\right) = \frac{81}{2} \frac{\sin \left(2x\right)}{2}$$
$$\bar y =\frac{81}{4}.1$$
$$\bar y = \frac{81}{4}$$
Centroid = $$\left(\bar x, \bar y\right) = \left(\dfrac{9\frac{Ο}{4}\sqrt{2}-9}{9\sqrt{2}-9}, {\frac{81}{4\left(9\sqrt{2}-9\right)}}\right)$$
Thanks
Satish
| {
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"timestamp": "2023-03-29T00:00:00",
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Determining number of solutions with inclusion-exclusion NOTE: I know there are similar questions to this, but the ones on this website are much more complex, and I'd like to get a basic understanding before moving on to them. Please do not mark this as a duplicate.
On a practice quiz:
Use inc-exc to determine the number of solutions to $a+b+c+d = 15$ with $0 \leq a,b,c,d \leq 6$
Here's what I have so far:
Let $A_1$ be the solution to $a+b+c+d=15$ with $a\geq 7, 0\leq b,c,d$
Let $A_2$ be the solution to $a+b+c+d=15$ with $b\geq 7, 0\leq a,c,d$
Let $A_3$ be the solution to $a+b+c+d=15$ with $c\geq 7, 0\leq a,b,d$
Let $A_4$ be the solution to $a+b+c+d=15$ with $d\geq 7, 0\leq a,b,c$
$|A_1 \cup A_2 \cup A_3 \cup A_4| = C(4,1) \cdot C((4+4-1), 4) - \dotsb $
And this is where I get stuck. I found my first term using the logic that I want to count $|A_1|+|A_2|+|A_3|+|A_4|$. How do I continue this to find the rest of the terms? I know what to do after I find the number of non-solutions, just subtract it from the total non-restricted solutions...but I can't figure out how to find the number of non-solutions!
Any help would be greatly appreciated, thanks.
| Generating Functions
Look at the coefficient of $x^{15}$ in $\left(1+x+x^2+x^3+\dots+x^6\right)^4$:
$$
\begin{align}
\left(\frac{1-x^7}{1-x}\right)^4
&=\sum_{j=0}^4(-1)^j\binom{4}{j}x^{7j}\sum_{k=0}^\infty(-1)^k\binom{-4}{k}x^k\\
&=\sum_{j=0}^4(-1)^j\binom{4}{j}x^{7j}\sum_{k=0}^\infty\binom{k+3}{k}x^k
\end{align}
$$
which is
$$
\sum_{j=0}^2(-1)^j\binom{4}{j}\binom{18-7j}{3}=180
$$
Inclusion-Exclusion
Without restriction on the size of the terms, using the standard $\mid$ and $\circ$ argument ($15$ $\circ$s and $3$ $\mid$s), there are $\binom{15+3}{3}$ ways to choose 4 non-negative integers that sum to $15$.
$$
\text{one sum for each arrangement}\\
2+4+6+3=\circ\,\circ\mid \circ\circ\circ\,\circ\mid \circ\circ\circ\circ\circ\,\circ\mid \circ\circ\circ
$$
Now let's count how many ways there are to have terms greater than $6$. There are $\binom{4}{1}$ ways to choose which $1$ term should be greater than $6$. To count the number of sums with $1$ term at least $7$, that would be $\binom{15-7+3}{3}$.
$$
\text{consider the red group atomic}\\
2+8+4+1=\circ\,\circ\mid\color{#C00000}{\circ\circ\circ\circ\circ\circ\circ}\,\circ\mid\circ\circ\circ\,\circ\mid\circ
$$
There are $\binom{4}{2}$ ways to choose which $2$ terms should be greater than $6$. To count the number of sums with $2$ terms at least $7$, that would be $\binom{15-14+3}{3}$.
$$
7+0+7+1=\color{#C00000}{\circ\circ\circ\circ\circ\circ\,\circ}\mid\mid\color{#C00000}{\circ\circ\circ\circ\circ\circ\circ}\mid\circ
$$
There is no way for $3$ terms to be greater than $6$. Inclusion-Exclusion says there are
$$
\binom{18}{3}-\binom{4}{1}\binom{11}{3}+\binom{4}{2}\binom{4}{3}=180
$$
ways for $4$ terms to sum to $15$ with each term at most $6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability of 5 cards drawn from shuffled deck
Five cards are drawn from a shuffled deck with $52$ cards. Find the probability that
a) four cards are aces
b) four cards are aces and the other is a king
c) three cards are tens and two are jacks
d) at least one card is an ace
My attempt:
a) $\left(13*12*\binom{4}{4}*\binom{4}{1}\right)/\binom{52}{5}$
b) same as (a)?
c) $\left(13 * 12 * \binom{4}{3} * \binom{4}{2}\right)/\binom{52}{5}$
d) $\left(13 * \binom{4}{1}\right)/\binom{52}{5}$
| a) There are $ \binom{52}{5} = 2,598,960 $ ways of choosing 5 cards. There are $\binom{4}{4} = 1 $ way to select the 4 aces. So there are $\binom{48}{1}= 48 $ ways to select the remaining card. Thus there are a total of 48 ways to select 5 cards such that 4 of them are aces, and the probability is:
$\frac{48}{2,598,960} = \frac{1}{54,145}$.
b) There are $\binom{4}{4} = 1 $ way to choose the 4 aces, and there are $\binom{4}{1} = 4 $ ways to choose a king. So there are $1\times4 = 4 $ ways to choose 5 cards such that 4 are aces and the other is a king card. The probability is: $\frac{4}{2,598,960} = \frac{1}{649,740} $.
c) There are $\binom{4}{3} = 4 $ ways to choose 3 ten cards, and there are $\binom{4}{2} = 6 $ ways of choosing 2 jacks. So there are $ 4\times 6 = 24 $ ways to choose 5 cards such that 3 are ten and 2 are jacks. The probability for this case is: $\frac{24}{2,598,960} = \frac{1}{108,290} $.
d) The probability of 5 non-ace cards is: $\frac{\binom{48}{5}}{\binom{52}{5}} = \frac{1,712,304}{2,598,960} = 0.6588 $,
so the probability of getting 5 card at least one ace is: $1 - 0.6588 = 0.34$.
| {
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Indefinite Integral $\int\frac{1}{1+\tan^{-1}x}\,\text{d}x$ I tried to solve this indefinite integral, $$\int\frac{1}{1+\tan^{-1}x}\,\text{d}x.$$
I tried taking the change of variable $u=\tan^{-1}x$ but failed to reach a solution.
Can anyone help me?
Thanks in advance.
| Let $u=\tan^{-1}x$ ,
Then $x=\tan u$
$\therefore\int\dfrac{1}{1+\tan^{-1}x}~dx$
$=\int\dfrac{d(\tan u)}{u+1}$
$=\dfrac{\tan u}{u+1}-\int\tan u~d\left(\dfrac{1}{u+1}\right)$
$=\dfrac{\tan u}{u+1}+\int\dfrac{\tan u}{(u+1)^2}~du$
$=\dfrac{\tan u}{u+1}+\int\sum\limits_{n=0}^\infty\dfrac{8u}{((2n+1)^2\pi^2-4u^2)(u+1)^2}~du$ (use Mittag-Leffler Expansion of tangent)
$=\dfrac{\tan u}{u+1}-\int\sum\limits_{n=0}^\infty\dfrac{8}{((2n+1)\pi-2)^2((2n+1)\pi+2u)}~du+\int\sum\limits_{n=0}^\infty\dfrac{8}{((2n+1)\pi+2)^2((2n+1)\pi-2u)}~du+\int\sum\limits_{n=0}^\infty\dfrac{8((2n+1)^2\pi^2+4)}{((2n+1)^2\pi^2-4)^2(u+1)}~du-\int\sum\limits_{n=0}^\infty\dfrac{8}{((2n+1)^2\pi^2-4)(u+1)^2}~du$
$=\dfrac{\tan u}{u+1}-\sum\limits_{n=0}^\infty\dfrac{4\ln((2n+1)\pi+2u)}{((2n+1)\pi-2)^2}-\sum\limits_{n=0}^\infty\dfrac{4\ln((2n+1)\pi-2u)}{((2n+1)\pi+2)^2}+\sum\limits_{n=0}^\infty\dfrac{8((2n+1)^2\pi^2+4)\ln(u+1)}{((2n+1)^2\pi^2-4)^2}+\sum\limits_{n=0}^\infty\dfrac{8}{((2n+1)^2\pi^2-4)(u+1)}+C$
$=\sec^21\ln(\tan^{-1}x+1)+\dfrac{x+\tan1}{\tan^{-1}x+1}-\sum\limits_{n=0}^\infty\dfrac{4\ln((2n+1)\pi+2\tan^{-1}x)}{((2n+1)\pi-2)^2}-\sum\limits_{n=0}^\infty\dfrac{4\ln((2n+1)\pi-2\tan^{-1}x)}{((2n+1)\pi+2)^2}+C$
Check by Wolfram Alpha https://www.wolframalpha.com/input/?i=sum+ln%28%282n%2B1%29pi%2Ba%29%2F%28%282n%2B1%29pi-2%29%5E2%2Cn%3D0+to+inf and https://www.wolframalpha.com/input/?i=sum+ln%28%282n%2B1%29pi-a%29%2F%28%282n%2B1%29pi%2B2%29%5E2%2Cn%3D0+to+inf, the two infinite series converges.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Cholesky decomposition of the inverse of a matrix I have the Cholesky decomposition of a matrix $M$. However, I need the Cholesky decomposition of the inverse of the matrix, $M^{-1}$. Is there a fast way to do this, without first computing $M^{-1}$? In other words, is there a relationship between the Cholesky decompositions of a matrix and of its inverse?
My matrix is a covariance matrix and, hence, positive-definite.
| To add to previous answers, if we view $X$ as covariance matrix of data, the relationship between two decompositions reduces to relationship between coefficients of "right-to-left" autoregressive model and "left-to-right" one (details)
In general, switching order of "inverse" and "Cholesky" gives different results. IE
$$
\text{X=}\left(
\begin{array}{ccc}
10 & 0 & 4 \\
0 & 10 & -2 \\
4 & -2 & 15 \\
\end{array}
\right)
$$
Using Cholesky and inverting factors, gives this decomposition
$$X^{-1}=
\left(
\begin{array}{ccc}
1 & 0 & -\frac{2}{5} \\
0 & 1 & \frac{1}{5} \\
0 & 0 & 1 \\
\end{array}
\right)\left(
\begin{array}{ccc}
\frac{1}{10} & 0 & 0 \\
0 & \frac{1}{10} & 0 \\
0 & 0 & \frac{1}{13} \\
\end{array}
\right)\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
-\frac{2}{5} & \frac{1}{5} & 1 \\
\end{array}
\right)
$$
While Cholesky on the inverse directly gives
$$X^{-1}=
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
-\frac{4}{73} & 1 & 0 \\
-\frac{20}{73} & \frac{2}{15} & 1 \\
\end{array}
\right),\left(
\begin{array}{ccc}
\frac{73}{650} & 0 & 0 \\
0 & \frac{15}{146} & 0 \\
0 & 0 & \frac{1}{15} \\
\end{array}
\right),\left(
\begin{array}{ccc}
1 & -\frac{4}{73} & -\frac{20}{73} \\
0 & 1 & \frac{2}{15} \\
0 & 0 & 1 \\
\end{array}
\right)
$$
notebook
| {
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"url": "https://math.stackexchange.com/questions/712993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Solve $\tan x =\sqrt{3}$, finding all solutions. My attempt so far:
$$
\tan x = \sqrt3
$$
$$
\frac{\sin x}{\cos x}= \sqrt3
$$
Then I look at the unit circle to find possible solutions. I find two solutions: $$
\frac{\pi}{3} \text{ and }\frac{4\pi}{3}
$$
However, I find the answer to be:
$$
x = \frac{\pi}{3} + \pi k
$$
Why isn't $4\pi/3$ included in the solution?
(edit: I mistakenly wrote $7\pi/6$ instead of $4\pi/3$. Sorry for the confusion.)
| $$\tan\left(\frac {7\pi}{6}\right) = \dfrac{-\frac 12}{-\frac{\sqrt 3}{2}} = \dfrac 1{\sqrt 3}$$
So $x = \dfrac{7\pi}{6}$ does not solve $\tan x = \sqrt 3$.
The second solution $x$ in the interval $[0, 2\pi]$ is $$x = \dfrac{4\pi}3$$
ADDED after EDITED post:
Note that $$\dfrac {4\pi}{3} = \frac{\pi}{3} + 1\cdot \pi = \dfrac{\pi}{3} + k\pi, \;k = 1$$
And we can characterize all solutions $x_i = \pi/3 + k\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/714392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Factoring $x^5 + x^4 + x^3 + x^2 + x + 1$ without using $\frac{x^n - 1}{x-1}$? I was at a math team meet today and one of the problems was to factor $x^5 + x^4 + x^3 + x^2 + x + 1$. It also gave the hint that it decomposes into two trinomials and a binomial.
The solution they gave was based on the fact that $\frac{x^6 - 1}{x-1} = x^5 + x^4 + x^3 + x^2 + x + 1$ and from there the solution is pretty straightforward. However, I was not aware of that factorization. The only ones I have really learned are $x^2 - y^2 = (x-y)(x+y)$ and $x^3 \pm y^3 = (x \pm y)(x^2 \mp xy + y^2)$. Is there any other way I could have solved this factorization without using the ones they used?
| You should be(come) aware of: $\frac{x^n-1}{x-1} = \sum_{k=0}^{n-1} x^k$
It's quite useful.
$$x^5+x^4+x^3+x^2+x+1 \\ = \frac{x^6-1}{x-1} \\ = \frac{(x^3-1)(x^3+1)}{x-1} \\ = (x^2+x+1)(x^3+1) \\ = (x^2+x+1)(x+1)(x^2-x+1)$$
The only other way would be to guess -1 as a root, because you have six terms in ascending polynomial sequence.
$$ x^5+x^4+x^3+x^2+x+1 = (x+1)(x^4+x^2+1)$$
Then split the resulting tetranomial, by solving: $$\exists a, b : x^4 + x^2 + 1 = (x^2+ax+1)(x^2+bx+1) \\ (x^2+a x + 1)(x^2+b x +1) = x^4 + (a+b) x^3 + (2+ab) x^2 + (a+b)x + 1 \\ \therefore a+b=0 \land ab=-1 \\ \therefore a = \pm 1, b=\mp 1 $$
And so:
$$ x^5+x^4+x^3+x^2+x+1 = (x+1)(x^2+x+1)(x^2-x+1)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression using sums $\sum$ and double sums $\sum$$\sum$?
| One way is by using the Euler-Maclaurin summation formula, something you strangely don't find in most standard calculus books.
http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula
Let $f(x) = x^4$.
Then
$$f'(x) = 4x^3, f'''(x) = 24 x, f^{(5)}(x) = f^{(7)}(x) = ... = 0 $$
And according to the formula,
$$ \sum_{k=1}^{n} k^{4} = \sum_{k=0}^{n} k^{4} = \int_{0}^{n} x^{4} \ dx - B_{1} \left( f(n) - f(0) \right) + \frac{B_{2}}{2!} \left(f'(n) - f'(0) \right) + \frac{B_{4}}{4!} \left(f'''(n) - f'''(0) \right)$$
where $B_{i}$ is a Bernoulli number.
http://en.wikipedia.org/wiki/Bernoulli_number
So
$$\sum_{k=1}^{n} k^{4} = \frac{n^{5}}{5} + \frac{1}{2} (n^{4}) + \frac{1}{2} \left( \frac{1}{6} \right) (4n^{3}) + \frac{1}{24} \left(-\frac{1}{30}\right)(24n) $$
$$ = \frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/718939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Prove if $n^3$ is odd, then $n^2 +1$ is even I'm studying for finals and reviewing this question on my midterm. My question is stated above and I can't quite figure out the proof.
On my midterm I used proof by contraposition by stating:
If $n^2 +1$ is odd then $n^3$ is even.
I let $n^2+1 = (2m+1)^2 + 1$
$= (4m^2 + 4m + 1) + 1$
$= 2(2m^2 + 2m + .5) + 1$
Let $2m^2 + 2m + .5 = k$
$n^2 + 1 => 2k + 1$
Therefore proving that $n^2 + 1$ was odd making $n^3$ even.
I know my logic was messed up somewhere..some guidance would be nice.
| If $n^3$ is odd, then $n^3-1$ is even. Now, $n^3-1=(n-1)(n^2+n+1)$, since that one of $(n-1)$, $(n^2+n+1)$ (or both) must be even.
'
Suppose that $n-1$ is even, then $n$ is odd, $n^2$ is odd and $n^2+1$ is even.
If $n-1$ is odd, then $n$ is even and $n^2+1$ is also odd. Therefore, $n-1$ must be even.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this matrix decomposition possible? Given a $2\times2$ matrix $S$ with entries in $\mathbb{Z}$ or $\mathbb{Q}$ , when is it possible to write $S=\frac{1}{3}(ABC+CAB+BCA)$ such that $A+B+C=0$, where $A, B, C$ are matrices over the same ground ring as S. Always? How would I find $A, B, C$?
| To challenge this knot, we have no choice but to replace any specific matrices.
As one of the approaches, The rotation matrix is a part of correct answer. Because the matrix has special characteristics. As reasons, variables and equations can be reduced and commutativity is allowed under any two dimension. For example, the following figure shows that even if rotating operations are exchanged, eventual angles are corresponded as same direction. Therefore, this property can be applied to supply the better solution.
Initially, let replace $\{A,B,C\}$ with rotation matrices $\{R_{a},R_{b},R_{c}\}$. Further, rotation matrix can be handled to be identity with Euler formula:
$$
R_{a}=R(a)=
\begin{bmatrix}
\cos{(a)} & -\sin{(a)} \\
\sin{(a)} & \cos{(a)}
\end{bmatrix}
\equiv e^{ia}
$$
Then the supplied top equation can be expressed by follows:
$$
\begin{aligned}
3S
=& ABC+BCA+CAB \\
=& R_{a}R_{b}R_{c}+R_{b}R_{c}R_{a}+R_{c}R_{a}R_{b} \\
S=& R_{a}R_{b}R_{c} \\
\Leftrightarrow
e^{i\phi}=& e^{ia} e^{ib} e^{ic} =e^{i(a+b+c)} \\
\end{aligned} \\
$$
where, $S\equiv e^{i\phi}$ and $\phi$ is phase degree. Next, the bottom constraint equation is:
$$
O=R_{a}+R_{b}+R_{c} \\
\Leftrightarrow
0=e^{ia} + e^{ib} + e^{ic}
$$
Hence, two equations can be expressed by the following simultaneous equation:
$$
\begin{cases}
a+b+c=\phi \\
e^{ia}+e^{ib}+e^{ic}=0
\end{cases}
$$
We can solve to replace as $\delta_{1}=c-a$ and $\delta_{2}=c-b$. Eventually, the solutions are below (However, these parameters are limited on range of $[0 \ \ 2\pi + \phi]$):
$$
(a,b,c)
=\biggl\{
\cfrac{\phi}{3},
\cfrac{\phi+2\pi}{3},
\cfrac{\phi+4\pi}{3}
\biggr\}
$$
Therefore, the top equation can be calculated as follows:
$$
\begin{aligned}
S
=& R_{a}R_{b}R_{c} \\
=& R_{\phi /3}(R_{\phi /3} R_{120^\circ})(R_{\phi /3} R_{240^\circ}) \\
=& R_{\phi /3}^3R_{(120+240)^\circ} \\
R_{\phi}=& R_{\phi/3}^3
\end{aligned}
$$
This means that end phase is exactly three times of initial phase. This notion can be drawn on unit ring like following figure. Of course, these operations can be exchangeable as above explanation.
$\hspace{3cm}$
Next, we have to determine the matrix $S$. Then, its internal elements must be rational numbers or integers owing to sentence. To estimate these factors, the Pythagorean triple gives good clue. Following descriptions are initial four triples (Of course inverted ordering or sort sequence are also true. However except unit matrix):
$$
\cfrac{1}{5}
\begin{bmatrix}
3 & -4 \\
4 & 3
\end{bmatrix}
,\cfrac{1}{13}
\begin{bmatrix}
5 & -12 \\
12 & 5
\end{bmatrix}
,\cfrac{1}{17}
\begin{bmatrix}
8 & -15 \\
15 & 8
\end{bmatrix}
,\cdots
$$
To satisfy these conditions, each matrices are applied:
$$
\begin{array}
AA= & \cfrac{1}{5}
\begin{bmatrix}
3 & -4 \\
4 & 5
\end{bmatrix}
,&
\cfrac{1}{13}
\begin{bmatrix}
5 & -12 \\
12 & 5
\end{bmatrix}
,&\cdots
\\
B= & \cfrac{1}{10}
\begin{bmatrix}
-(4\sqrt{3}+3) & -(3\sqrt{3}-4) \\
-(3\sqrt{3}-4) & -(4\sqrt{3}+3)
\end{bmatrix}
,&
\cfrac{1}{26}
\begin{bmatrix}
-(12\sqrt{3}+5) & -(5\sqrt{3}-12) \\
5\sqrt{3}-12 & -(12\sqrt{3}+5)
\end{bmatrix}
,&\cdots
\\
C= & \cfrac{1}{10}
\begin{bmatrix}
-(4\sqrt{3}-3) & -(3\sqrt{3}+4) \\
-(3\sqrt{3}+4) & -(4\sqrt{3}-3)
\end{bmatrix}
,&
\cfrac{1}{26}
\begin{bmatrix}
12\sqrt{3}-5 & 5\sqrt{3}+12 \\
-(5\sqrt{3}+12) & 12\sqrt{3}-5
\end{bmatrix}
,&\cdots
\\
S= & \cfrac{1}{125}
\begin{bmatrix}
-117 & -44 \\
44 & -117
\end{bmatrix}
,&
\cfrac{1}{2197}
\begin{bmatrix}
-2035 & 828 \\
-828 & -2035
\end{bmatrix}
,&\cdots
\end{array}
$$
As well known, Pythagorean triples exist endlessly so that these commutative matrices (and whole solutions) also exist infinity.
| {
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"url": "https://math.stackexchange.com/questions/721155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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} |
Evaluate the integral: $\int\tan^5 (4x)\,\mathrm dx$. We have to use a trig identity for this. I think it's $\tan^2 x = \sec^2 x - 1$.
But I'm having difficulty setting up the integral.
| The secret is to use $1+\tan^2 w=\sec^2 w$. Indeed:
$$\int \tan^5 4x dx=\int \tan^3 4x\cdot \tan^2 4x dx=\int \tan^3 4x(\sec^2 4x-1)dx= $$
$$=\frac{1}{4}\int (\tan 4x)^3(\sec^2 4x \cdot 4 dx)-\int \tan^3 4x dx= $$
$$=\frac{1}{4}\frac{(\tan 4x)^4}{4} -\int \tan 4x\cdot \tan^2 4xdx=$$
$$=\frac{1}{16}\tan^4 4x-\int \tan 4x (\sec^2 4x-1)dx= $$
$$= \frac{1}{16}\tan^4 4x- \frac{1}{4}\int (\tan 4x)^1(\sec^2 4x\cdot 4 dx)+\int \tan 4x dx=$$
$$= \frac{1}{16}\tan^4 4x- \frac{1}{4}\frac{(\tan 4x)^2}{2}+\frac{1}{4}\int \frac{(\sin 4x\cdot 4 dx)}{\cos 4x}=$$
$$=\frac{1}{16}\tan^4 4x- \frac{1}{4}\frac{(\tan 4x)^2}{2}+\frac{1}{4} \ln |\sin 4x|+c.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/722460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding one sided limits algebraically I was wondering what the best method was for proving this limit algebraically:
$$\lim_{x \to 1}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}$$
I know the answer to this question is ;
$$\lim_{x \to 1^+}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}={-\infty}$$
$$\lim_{x \to 1^-}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}={+\infty}$$
The only way I can solve this is using the graph of the function as the limit becomes very apparent. What is the best way to answer this algebraically?
| Hints:
$$3x^4-8x^3+5=(x-1)(3x^3-5x^2-5x-5)$$
$$x^3-x^2-x+1=(x-1)(x^2-1)=\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/724626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove this $\ln{(x+\sqrt{x^2+1})}<\frac{x(a^x-1)}{(a^x+1)\log_{a}{(\sqrt{x^2+1}-x)}}$ let $0<a<1,x<0$,show that
$$\ln{(x+\sqrt{x^2+1})}<\dfrac{x(a^x-1)}{(a^x+1)\log_{a}{(\sqrt{x^2+1}-x)}}$$
My idea:
$$\Longleftrightarrow \ln{(\sqrt{x^2+1}+x)}<\dfrac{x(a^x-1)\ln{a}}{(a^x+1)\ln{(\sqrt{x^2+1}-x)}}$$
Then following I fell very ugly.Thank you
| $$\ln{(x+\sqrt{x^2+1})}<\dfrac{x(a^x-1)}{(a^x+1)\log_{a}{(\sqrt{x^2+1}-x)}}$$
Rewrite the inequality in terms of hyperbolic functions:
1.) $\ln{(x+\sqrt{x^2+1})}=\sinh^{-1}x$
2.) $\frac{1}{\log_{a}{(\sqrt{x^2+1}-x)}}=-\frac{\log{a}}{\sinh^{-1}x}$
3.) $\frac{a^x-1}{a^x+1}=\tanh{\left(\frac{x\log{a}}{2}\right)}$
Using the three identities above, the inequality becomes:
$$\sinh^{-1}x<-\frac{x\log{a}}{\sinh^{-1}x}\tanh{\left(\frac{x\log{a}}{2}\right)}$$
For $x<0$ and $0<a<1$,
$$\sinh^{-1}x < 0,\\
0<-x,\\
0<\frac{\log a}{\sinh^{-1}x},\\
0<\tanh{\left(\frac{x\log{a}}{2}\right)}.$$
Hence,
$$\sinh^{-1}x<0<-\frac{x\log{a}}{\sinh^{-1}x}\tanh{\left(\frac{x\log{a}}{2}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/724722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Remainder of the polynomial A polynomial function $f(x)$ with real coefficients leaves the remainder $15$ when divided by $x-3$, and the remainder $2x+1$ when divided by $(x-1)^2$. Then the remainder when $f(x)$ is divided by $(x-3)(x-1)^2$ is?
What I have thought-The remainder must be of the form $ax^2+bx+c$. Now applying the remainder theorem, I am able to find $2$ equations in $a,b,c$ . eg
Let $f(x)$=$(x-3)$$h(x)$+$15$ ...................................(1)
Also let $f(x)=(x-3)(x-1)^2g(x)+ax^2+bx+c$...........................(2)
Put $x=3$ and using (1) we get $15=9a+3b+c$
Similarly I can get another equation using the other information given. But I am only able to get 2 equations in 3 variables. From where do I get the 3rd equation in $a,b,c$ and hence the remainder?
| Note that
$$
\frac{1}{4}(x-1)^2-\frac{1}{4}(x+1)(x-3)=1.
$$
You can get this by dividing $(x-1)^2$ by $x-3$. So take
$$
f(x)=15\frac{1}{4}(x-1)^2-(2x+1)\frac{1}{4}(x+1)(x-3).
$$
Then the remainder when $f(x)$ is divided by $x-3$ is 15 and by $(x-1)^2$ is $2x+1$ since $(x+1)(x-3)=[(x-1)+2][(x-1)-2]=(x-1)^2-4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the Global maximum and minimum values on the closed disk of this function just wondering if I am doing this correctly:
For $f(x,y) = x^3y + xy^3 - xy + 1 $ Find the Global maximum and minimum values on:
$ D = {(x,y) \in R | x^2 + y^2 \le 4} $
I found that the critical points of the function are: $(0,0),(0,1),(0,-1),(\frac{1}{2},\frac{1}{2}),(\frac{1}{2},-\frac{1}{2}),(-\frac{1}{2},\frac{1}{2}),(-\frac{1}{2},-\frac{1}{2})$
Plugging each point into the function, I get:
$f(0,0) = 1, ), f(0,1) = 1, f(0,-1) = 1, f(\frac{1}{2},\frac{1}{2}) = \frac{7}{8}, f(\frac{1}{2},-\frac{1}{2}) = \frac{9}{8}, f(-\frac{1}{2},\frac{1}{2}) = \frac{9}{8}, f(-\frac{1}{2},-\frac{1}{2}) = \frac{7}{8}$
I then check the boundary points of D, by parametrising the variables, letting $x = 2cos\theta$ and $y = 2sin\theta$ for $ 0 \le \theta \le 2\pi$
Plugging these into the function and defining
$g(\theta)= (2cos\theta)^3(2sin\theta)+ (2cos\theta)(2sin\theta)^3-(2cos\theta)(2sin\theta) + 1$
I take the derivative and get:
$g'(\theta) = 12 cos2\theta$, so $g$ has critical points at $\theta = \frac{\pi}{4} and \frac{5\pi}{4}$.
So the corresponding $x,y$ values are: $x=2cos(\frac{\pi}{4}) = \sqrt2, y=2sin(\frac{\pi}{4})=\sqrt2$ and $x=2cos(\frac{5\pi}{4}) = -\sqrt2, y=2sin(\frac{5\pi}{4})=-\sqrt2$
Finally, plugging these into $f$ I get:
$f(\sqrt2,\sqrt2) = (\sqrt2)^3(\sqrt2)+(\sqrt2)(\sqrt2)^3-(\sqrt2)(\sqrt2) +1 = 7$
$f(-\sqrt2,-\sqrt2) = (-\sqrt2)^3(-\sqrt2)+(-\sqrt2)(-\sqrt2)^3-(-\sqrt2)(-\sqrt2) +1 = 7$
So indeed I get max values at these points and min values for $x = \frac{1}{2}, -\frac{1}{2} $ and $y = \frac{1}{2}, -\frac{1}{2}$
Have I done this correctly? Many thanks for the feedback!
| You may have worked too hard on the circle, and made a little error.
Our function is $xy(x^2+y^2)-xy+1$, which on the circle is $3xy+1$. This is $12\cos\theta\sin\theta+1$, which is $6\sin 2\theta+1$, maximum $7$, minimum $-5$.
Note that parametrization (polar coordinates) could also be useful inside the circle, for our function is $\frac{1}{2}(r^4-r^2)\sin 2\theta +1$. The partial derivatives are easy to find.
In finding critical points, rectangular coordinate version, there are some errors. For note the symmetry between $x$ and $y$. It follows that if for example $(0,1)$ is a critical point (which it is), then $(1,0)$ is also a critical point.
But what's happening in the interior of the disk turns out to be irrelevant. The global maximum is $7$, and the global minimum is $-5$. Expressing the function as $xy(x^2+y^2-1)+1$ makes it clear that the global extrema must be on the boundary.
| {
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Recurrence relation converting to explicit formula Let $a_n = -2a_{n-1 }+ 15a_{n-2 }$ with initial conditions $a_1 = 10 $ and $a_2 = 70$.
a)Write the first 5 terms of the recurrence relation.
b)Solve this recurrence relation.
c)Using the explicit formula you found in part b, evaluate $a_5$. You must show that you are using the equation from part b.
a.)$a_1 = 10$;
$a_2 = 70$;
$a_3 = -2(70) + 15(10) = -140 + 150 = 10$;
$a_4 = -2(10) + 15(70) = -20 + 1050 = 1030$;
$a_5 = -2(1030) + 15(10) = -2060 + 150 = -1910$;
b.) Using an Excel generator, I determined: root1 = -5, root2 = 3, u = 1, v = 5
I came up with the explicit formula: $a_n = 1 * (-5)n + 5 * (3)n$ , but this is where I get lost in the sauce. I tried solving for $a_5$ and came up with a number that I care not to share. So I must have gone wrong somewhere, but I just can't find it. Any help or guidance would be greatly appreciated.
| Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence as:
$$
a_{n + 2} = - 2 a_{n + 1} + 15 a_n
$$
Running the recurrence "backwards" gives $a_0 = 6$ (starting at index 0 is nicer all around). Multiply the recurrence by $z^n$, sum over $n \ge 0$, and recognize:
\begin{align}
\sum_{n \ge 0} a_{n + 1} z^n &= \frac{A(z) - a_0}{z} \\
\sum_{n \ge 0} a_{n + 2} z^n &= \frac{A(z) - a_0 - a_1 z}{z^2}
\end{align}
to get:
$$
\frac{A(z) - 6 - 10 z}{z^2} = - 2 \frac{A(z) - 6}{z} + 15 A(z)
$$
Solving for $A(z)$, and writing as partial fractions, gives:
$$
A(z) = \frac{6 + 22 z}{1 + 2 z - 15 z^2}
= \frac{1}{1 + 5 z} + \frac{5}{1 - 3 z}
$$
Two geometric series:
$$
a_n = (-5)^n + 5 \cdot 3^n
$$
This gives $a_5 = -1910$.
| {
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If $a+b+c+d = 2$, then $\frac{a^2}{(a^2+1)^2}+\frac{b^2}{(b^2+1)^2}+\frac{c^2}{(c^2+1)^2}+\frac{d^2}{(d^2+1)^2}\le \frac{16}{25}$ If $a+b+c+d = 2$, prove that
$$\dfrac{a^2}{(a^2+1)^2}+\dfrac{b^2}{(b^2+1)^2}+\dfrac{c^2}{(c^2+1)^2}+\dfrac{d^2}{(d^2+1)^2}\le \dfrac{16}{25}$$
Also $a,b,c,d \ge 0$.
| Let us consider the function
$$
\frac{x^2}{(1+x^2)^2} +
\frac{y^2}{(1+y^2)^2} +
\frac{z^2}{(1+z^2)^2} +
\frac{t^2}{(1+t^2)^2} +
\lambda(x+y+z+t-2)
$$and write the first order conditions:
$$
0 = -\frac{2x(x^2-1)}{(1+x^2)^3} + \lambda,
$$and the same equation for $y,z,t$. In particular,
$$
\frac{x(x^2-1)}{(1+x^2)^3} = \frac{y(y^2-1)}{(1+y^2)^3}
= \frac{z(z^2-1)}{(1+z^2)^3} = \frac{t(t^2-1)}{(1+t^2)^3}
$$
Taking a look at the variations of this function,
we see that if $x\neq y$ then $x,y < 1$ and $x,y,z,t\in \{u,v\}$, with
$$
\begin{cases}
\frac{u(u^2-1)}{(1+u^2)^3} = \frac{v(v^2-1)}{(1+v^2)^3} \\
u+v=1\\
u<v.
\end{cases}
$$
Then an analysis proves that
$(u,v) = (0,1)$:
Eventually, compute the two potential extrema:
$$
\frac{1^2}{(1+1^2)^2} + \frac{1^2}{(1+1^2)^2}+0+0 =\frac 12;\\
\frac{(1/2)^2}{(1+(1/2)^2)^2} + \frac{(1/2)^2}{(1+(1/2)^2)^2}+
\frac{(1/2)^2}{(1+(1/2)^2)^2} + \frac{(1/2)^2}{(1+(1/2)^2)^2} = \frac {16}{25}.
$$
| {
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Proof by induction that $n^3 + (n + 1)^3 + (n + 2)^3$ is a multiple of $9$. Please mark/grade. What do you think about my first induction proof? Please mark/grade.
Theorem
The sum of the cubes of three consecutive natural numbers is a multiple of 9.
Proof
First, introducing a predicate $P$ over $\mathbb{N}$, we rephrase the theorem as follows.
$$\forall n \in \mathbb{N}, P(n)
\quad \text{where} \quad
P(n) \, := \, n^3 + (n + 1)^3 + (n + 2)^3 \text{ is a multiple of 9}$$
We prove the theorem by induction on $n$.
Basis
Below, we show that we have $P(n)$ for $n = 0$.
$$0^3 + 1^3 + 2^3 = 0 + 1 + 8 = 9 = 9 \cdot 1$$
Inductive step
Below, we show that for all $n \in \mathbb{N}$, $P(n) \Rightarrow P(n + 1)$.
Let $k \in \mathbb{N}$. We assume that $P(k)$ holds.
In the following, we use this assumption to show that $P(k + 1)$ holds.
By the assumption,
there is a $i \in \mathbb{N}$ such that
$i \cdot 9 = k^3 + (k + 1)^3 + (k + 2)^3$.
We use this fact in the following equivalent transformation.
The transformation turns the sum of cubes in the first line,
for which we need to show that it is a multiple of 9,
into a product of 9 and another natural number.
$(k + 1)^3 + (k + 2)^3 + (k + 3)^3 \\
= (k + 1)^3 + (k + 2)^3 + k^3 + 9k^2 + 27k + 27 \\
= k^3 + (k + 1)^3 + (k + 2)^3 + 9k^2 + 27k + 27 \quad | \text{ using the induction hypothesis} \\
= 9i + 9k^2 + 27k + 27 \\
= 9 \cdot i + 9 \cdot k^2 + 9 \cdot 3k + 9 \cdot 3 \\
= 9 \cdot (i + k^2 + 3k + 3)$
We see that the above product has precisely two factors: 9 and another natural number.
Thus the product is a multiple of 9.
This completes the induction.
| I do not think that this is a real question but if you were my student i would give you an A.
It is all fine to me.
| {
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"question_score": "19",
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Prove that if $n(a^2+b^2+c^2)=abc$ then $2\mid n$
Is it true that if $n\in\mathbb N$ and the diophantine equation $$n(a^2+b^2+c^2)=abc,\\(a,b)=(b,c)=(c,a)=1\tag1$$
has positive integer solutions $a,b,c$, then $2\mid n$?
I can prove that $3\mid n:$
1) If $3\not\mid abc$ then $3\mid a^2+b^2+c^2,$ a contradiction.
2) If $3\mid abc,$ since $(a,b)=(b,c)=(c,a)=1$, we can assume that $3\mid a$ and $3\not \mid bc,$ then $3\not\mid a^2+b^2+c^2,$ hence $3\mid n.$
I can prove that equation $(1)$ has infinitely many solutions when $n=6,$ in fact, let $c=17,$ then it become a Pell's equation: $(12a-17b)^2-145b^2=-41616.$
I find some solutions to equation $(1)$:
$\{a,b,c,n\}=\{39,20,17,6\}\{52,29,15,6\}\{68,61,45,18\}\{87,80,61,24\}$
However, I cannot prove that $2\mid n.$ Thanks in advance!
| It is indeed true. We need the following well known fact:
If $p\equiv3\pmod4$ is prime and $p\mid x^2+y^2$, then $p\mid x$ and
$p\mid y$.
We will prove that
Theorem. If $a,b,c$ is a solution to $(1)$, then exactly one of $a,b,c$ is divisible by $4$.
Proof.
First suppose $a,b,c$ are all odd. Then $a^2+b^2+c^2\equiv3\pmod 4$, so has a prime divisor $p\equiv3\pmod 4$. Without loss of generality, suppose $p\mid a$.
Then $p\mid b^2+c^2$, so $p\mid b$ and $p\mid c$ which contradicts the condition $(a,b)=(b,c)=(c,a)=1$.
Therefore one of $a,b,c$ is even. Say $2\mid a$ and suppose $4\nmid a$. Then $a^2+b^2+c^2\equiv6\pmod8$, which means it has a prime divisor $p\equiv3\pmod 4$. Again $p\mid b$ and $p\mid c$, contradiction.
So we should have $4\mid a$. $\square$
In this case, $a^2+b^2+c^2\equiv2\pmod 4$, which means $4\nmid a^2+b^2+c^2$. Therefore, $2\mid n$.
| {
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Is limits $\lim_{x\to\infty}{(x-2)^2\over2x+1}=\dfrac{1}{2}$? $$\lim_{x\to\infty}{(x-2)^2\over2x+1}=\dfrac{1}{2}$$
I used an online calculator and it said it was actually $=\infty$
Here's how I calculate it:
$$\lim_{x\to\infty}\dfrac{x^2+4-4x}{2x+1}=\lim_{x\to\infty}\dfrac{({x\over x}-{2\over x})({x\over x}-{2\over x})}{({2x\over x}+{1\over x})}=\dfrac{(1-0)(1-0)}{(2+0)}={1 \over 2}$$
| You have accidentally divided the numerator by $x^2$ and the denominator only by $x$. For the numerator we get
$$(x-2)^2 = \left[x\left(\tfrac{x}{x}-\tfrac{2}{x}\right)\right]\cdot \left[x\left(\tfrac{x}{x}-\tfrac{2}{x}\right)\right] = x^2\left(1-\tfrac{2}{x}\right)^2,$$
so the fraction is
$$\frac{(x-2)^2}{2x+1} = \frac{x^2\left(1-\tfrac{2}{x}\right)^2}{x\left(2+\frac{1}{x}\right)} = \underbrace{x}_{\to\infty}\underbrace{\frac{\left(1-\tfrac{2}{x}\right)^2}{\left(2+\frac{1}{x}\right)}}_{\to \frac{1}{2}}.$$
| {
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How to prove series convergence: $\sum \limits_{n=1}^\infty \left(\frac1n+\sqrt{1+n^2}-\sqrt{2+n^2}\right)^2$ I have this series:
$$\sum \limits_{n=1}^\infty \left(\frac1n+\sqrt{1+n^2}-\sqrt{2+n^2}\right)^2$$
I know that it's convergent (from WolframAlpha) but I need to prove it is convergent. How can I do it?
| You have that $$\begin{align*} \dfrac 1 n+\sqrt{1+n^2}-\sqrt{2+n^2}&=\frac1n+\frac{(\sqrt{1+n^2}-\sqrt{2+n^2})\cdot(\sqrt{1+n^2}+\sqrt{2+n^2})}{\sqrt{1+n^2}+\sqrt{2+n^2}}=\\\\&=\frac1n-\frac{1}{\sqrt{1+n^2}+\sqrt{2+n^2}}\\& \le \frac{1}{n} \end{align*}$$ Then the convergence is established by squaring both sides and using the comparison test.
For the above implication to be correct we also need to show that the middle part is positive. Indeed $$\frac{1}{\sqrt{1+n^2}+\sqrt{2+n^2}}\ < \frac{1}{\sqrt{n^2}+\sqrt{n^2}}=\frac{1}{2n}$$ or equivalently $$-\frac{1}{\sqrt{1+n^2}+\sqrt{2+n^2}}\ >-\frac{1}{2n}$$ which implies that $$\frac1n-\frac{1}{\sqrt{1+n^2}+\sqrt{2+n^2}}>\frac1n-\frac{1}{2n}=\frac1{2n}>0$$
| {
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Prove that the integers $n$ and $n, 2^{2^n} + 1$ are relatively prime. I need help to prove that
$$\gcd(n, 2^{2^n} + 1)=1,\ n = 1,2,\dots$$
I have no idea how start the proof.
| First note the following: For any prime $p$ dividing $2^{2^n}+1$:
$$p|(2^{2^n}+1) \implies 2^{2^n} \equiv_p -1$$
$$\implies \left(2^{2^n}\right)^2 = 2^{2^{n+1}} \equiv_p 1$$
$$2^{2^n} \equiv_p -1; \ 2^{2^{n+1}} \equiv_p 1 \implies 2^{n+1} = Kc,$$
for some odd integer $K$, where $c$ is the smallest positive integer s.t. $2^c \equiv_p 1$ [and so $c$ divides $|(\mathbb{Z}/p\mathbb{Z})^{\times}|$ $= p-1$]. [To see why $K$ must be odd, as the equation $2^{2^n} \not \equiv_p 1$ holds, it follows that $2^n = \frac{2^{n+1}}{2}$ is not an integral multiple of $c$. So as $2^{n+1}=Kc$ with $K$ integral, it follows that $2^{n}= \frac{Kc}{2}$ with $\frac{K}{2}$ is not integral even though $K$ is integral, which gives $K$ odd.]
So as $c$ divides $p-1$, it follows that $Kc$ must divide $K(p-1)$, and so, as $2^{n+1}$ divides $Kc$, it follows that $2^{n+1}$ must divide $K(p-1)$ as well. Then, as $K$ is odd and $(K,2^{n+1})=1$, it follows that $2^{n+1}$ must not only divide $K(p-1)$ but $2^{n+1}$ must also divide $p-1$ itself. This gives $p > 2^{n+1}>n$. So any prime $p$ dividing $2^{2^{n}}+1$ must satisfy $p >2^{n+1}>n$. Thus,
there is no prime $p'$ dividing both $2^{2^n}+1$ and $n$, because all primes dividing $2^{2^n}+1$ are larger than $n$ and thus cannot divide $n$. This gives $(n,2^{2^n}+1)=1$.
| {
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Finding the mean with absolute value This question is out of my field and topic that I am teaching myself now, but I was wondering how would you solve this problem if it had the absolute value of it.
My Question:
$$f(x) = \begin{cases}
2/9(x-1), & \text{for $1<x<4$} \\
0, & \text{elsewhere} \\
\end{cases}
$$
I know what was given is $\mu=3$ and $\sigma=$ $\sqrt{.5}=.7071$
I wanted to find $E[|x-\mu|]$.
My first thoughts would be $\int_1^4|x-\mu|f(x)dx.$ That is all I can think of because I did not learn about this one yet. I was just wondering out of curiousity how to solve this problem and seeing what it would look like. It seems like something good to know for later problems that are like it.
$\int_1^4|x-\mu|f(x)dx$
$=\int_1^3 (2/9)(x-1)(3-x)dx+\int_3^4 (2/9)(x-1)(x-3)dx$
$2/9\int_1^3(-x^2+4x-3)dx+2/9\int_3^4(x^2-4x+3)dx$
$2/9\int_1^3(-x^3/3+2x^2-3x) + 2/9\int_3^4 (x^3/3-2x^2+3x)$
$(2/9)(-x^3/3+2x^2-3x)|_1^3 + (2/9)(x^3/3-2x^2+3x)|_3^4$
$8/27+8/27$
$16/27$
| $\int_1^4|x-\mu|f(x)dx$
$=\int_1^3 (2/9)(x-1)(3-x)dx+\int_3^4 (2/9)(x-1)(x-3)dx$
$2/9\int_1^3(-x^2+4x-3)dx+2/9\int_3^4(x^2-4x+3)dx$
$2/9\int_1^3(-x^3/3+2x^2-3x) + 2/9\int_3^4 (x^3/3-2x^2+3x)$
$(2/9)(-x^3/3+2x^2-3x)|_1^3 + (2/9)(x^3/3-2x^2+3x)|_3^4$
$8/27+8/27$
$16/27$
| {
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Inequality Exercise in Apostol's Calculus I Let p and n denote positive integers. Show that:
$$n^{p} \lt \frac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^{p}$$
Attempt at Solution
Using the identity $b^{p+1}-a^{p+1} = (b-a)\sum_{k=0}^{p}b^{p-k}a^{k}$, let $b = n+1$ and $a = n$. Then:
$$\frac{(n+1)^{p+1} - n^{p+1}}{p+1} = \frac{\sum_{k=0}^{p}(n+1)^{p-k}n^{k}}{p+1} = \frac{(n+1)^{p} + (n+1)^{p-1}n + .... + (n+1)n^{p-1} + n^{p}}{p+1}$$
I'm not sure where to proceed from here. Am I supposed to use a geometric series in here somewhere?
EDIT: I may have figured it out if someone can confirm that $(n+1)^{p-k}n^{k} > n^{p}$, $1 \le k \le p$, $\forall n$
Assuming the above is true, we have:
$$\frac{(n+1)^{p} + (n+1)^{p-1}n + .... + (n+1)n^{p-1} + n^{p}}{p+1} \gt \frac{n^{p} + n^{p} + .... + n^{p} + n^{p}}{p+1} = \frac{(p+1)n^{p}}{(p+1)} = n^{p}$$
Similarly:
$$\frac{(n+1)^{p} + (n+1)^{p-1}n + .... + (n+1)n^{p-1} + n^{p}}{p+1} \lt \frac{(n+1)^{p} + (n+1)^{p} + .... + (n+1)^{p} + (n+1)^{p}}{p+1} = \frac{(p+1)(n+1)^{p}}{(p+1)} = (n+1)^{p}$$
| We can use Bernoulli's to strengthen this to $n,p$ positive real numbers. I will show with the LHS.
$(p+1)n^p < (n+1)^{p+1}-n^{p+1}$
$n^p(p+n+p) < (n+1)^{p+1}$
$( p/n+1/n+1) < (1+1/n)^{p+1}$ which just follows from Bernoulli's Inequality. Similar for the RHS. Just write the solution forwards.
| {
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Does the order I multiply the characteristic equation's factors in the homogeneous solution matter? I've been doing a recurrence relation exercise in my book. Doing some steps and comparing them to the ones taken by the book.
$$T(0) = 1$$
$$T(1) = 2$$
$$T(k) - 7T(k-1)+10T(k-2)=6+8k$$
Characteristic equation:
$$a^2-7a+10$$
$$(a - 5)(a-2)$$
Homogeneous solution:
$$b_0\cdot(5)^n+b_1\cdot(2)^n$$
Wait. My book did something different. Its homogeneous solution is
$$b_0\cdot(2)^n+b_1\cdot(5)^n$$
As you can observe, the book multiplied the constants by the factors in a different order as I did.
My question, then: does it matter? If yes, what determines the order?
| The order doesn't matter.
A general way of solving recurrences like this is as follows. Define the generating function $G(z) = \sum_{n \ge 0} T(n) z^n$, rewrite the recurrence so there aren't subtractions in indices:
$$
T(n + 2) - 7 T(n + 1) + 10 T(n) = 8 n + 22
$$
Multiply the recurrence by $z^n$, sum over $n \ge 0$, and recognize:
\begin{align}
\sum_{n \ge 0} T(n + 1) z^n &= \frac{G(z) - T(0)}{z} \\
\sum_{n \ge 0} T(n + 2) z^n &= \frac{G(z) - T(0) - T(1) z}{z^2} \\
\sum_{n \ge 0} z^n &= \frac{1}{1 - z} \\
\sum_{n \ge 0} n z^n &= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \\
&= \frac{z}{(1 - z)^2}
\end{align}
to get:
$$
\frac{G(z) - z - 2 z}{z^2} - 7 \frac{G(z) - 1}{z} + 10 G(z)
= 8 \frac{z}{(1 - z)^2} + 22 \frac{1}{1 - z}
$$
Solve fo $G(z)$, express the result as partial fractions (this is the part where the factorization of the characteristic equation comes up):
\begin{align}
G(z) &= \frac{1 - 7 z + 33 z^2 - 19 z^3}{1 - 9 z + 25 z^2 - 27 z^3 + 10 z^4} \\
&= \frac{1 - 7 z + 33 z^2 - 19 z^3}{(1 - 5 z) (1 - 2 z) (1 - z)^2} \\
&= \frac{2}{1 - 5 z} - \frac{9}{1 - 2 z} + \frac{6}{1 - z} + \frac{2}{(1 - z)^2}
\end{align}
Now you can read off the coefficients, if you remember the generalized binomial theorem:
$$
(1 - u)^{-m} = \sum_{k \ge 0} \binom{-m}{k} (-u)^k
= \sum_{k \ge 0} \binom{m + k - 1}{m - 1} u^k
$$
In this case:
\begin{align}
T(n) &= 2 \cdot 5^n - 9 \cdot 2^n + 6 + 2 (n + 1) \\
&= 2 \cdot 5^n - 9 \cdot 2^n + 2 n + 8
\end{align}
| {
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Consider the quadratic equation $ax^2-bx+c=0, a,b,c \in N. $ If the given equation has two distinct real root... Problem :
Consider the quadratic equation $~ax^2-bx+c=0, \quad a,b,c \in N. ~$ If the given equation has two distinct real roots belonging to the interval $~(1,2)~ $ then the minimum possible values of $~a~$ is
$(i) \quad -1 $
$(ii)\qquad 5 $
$(iii)~~\quad 2 $
$(iv)\quad -5 $
$(v) \qquad1 $
My approach :
We know the condition that two roots will between the two numbers viz. $(1,2)$
$(1) \quad f(1) >0 ; \qquad $
$(2)\quad f(2)>0\qquad$
$(3) \quad 1 < \frac{-b}{2a} <2\qquad $
$(4) \quad D \geq 0$
By using the above I got the following :
$(1) \quad a-b+c >0$
$(2)\quad 4a-2b+c >0$
$(3)\quad 1 < \frac{b}{2a} <2$
$(4) \quad b^2-4ac \geq 0$
Please guide further how to get the answer given in above five options. Thanks..
| From Vieta's:
$$2<x_1+x_2=\frac ba<4 \Rightarrow 2a<b<4a\\
1<x_1x_2=\frac ca<4 \Rightarrow a<c<4a\\
0<x_2-x_1<1 \Rightarrow 0<\frac{\sqrt{b^2-4ac}}{a}<1 \Rightarrow 0<b^2-4ac<a^2$$
Also:
$$f(1)=a-b+c>0 \Rightarrow c>b-a\\
f(2)=4a-2b+c>0 \Rightarrow c>2b-4a$$
Cases: $a=1$
$$2<b<4 \Rightarrow b=3\\
1<c<4\Rightarrow c=2,3\\
0<9-4c<1 \Rightarrow \emptyset$$
$a=2$:
$$4<b<8\Rightarrow b=5,6,7\\
2<c<8 \Rightarrow 3,4,5,6,7\\
0<b^2-8c<4 \ \text{and} \ c>b-a \ \text{and} \ c>2b-4a \Rightarrow \emptyset$$
So, by method of elimination, the answer is (if there is a single correct answer) $a=5$.
Let's consider $a=5$ as well:
$$10<b<20 \Rightarrow b=11,12,...,19\\
2<c<20 \Rightarrow c=3,4,...,19\\
0<b^2-20c<25 \ \text{and} \ c>b-5 \ \text{and} \ c>2b-20 \Rightarrow \\
(b,c)=(15,11),(18,15) \Rightarrow \\
5x^2-15x+11=0 \Rightarrow x_{1,2}\approx 1.28; 1.72 \ \color{green}{\checkmark}\\
5x^2-18x+15=0 \Rightarrow x_{1,2}\approx 1.31; 2.29 \not\in (1,2).$$
| {
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"url": "https://math.stackexchange.com/questions/748229",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Find the value of $x$ for which $ff=gf$. Functions $f$ and $g$ are defined by $f:x \mapsto \frac{1}{2x+1}$, $x \neq \frac{-1}{2}$ and $g:x \mapsto x+1$. Find the value of $x$ for which $ff=gf$.
So I started in this way:
$f[f(x)]=g[f(x)]$
$f(\frac{1}{2x+1}) = g(\frac{1}{2x+1})$
$\frac{1}{2(\frac{1}{2x+1})+1}=\frac{1}{2x+1}+1$
Then I solve it in simple algebra. Is this right?
I get the wrong answer, my book says $\frac{-5}{6}$ but I get $\frac{-11}{2}$
Helppp
| $\frac{1}{2(\frac{1}{2x+1})+1}=\frac{1}{2x+1}+1 \Leftrightarrow$
$\frac{2x+1}{2x+3}=\frac{2x+2}{2x+1}\Leftrightarrow$
$(2x+1)^2=(2x+2)(2x+3) \Leftrightarrow$
$4x^2+4x+1=4x^2+10x+6 \Leftrightarrow$
$6x+5=0$. So $x=-\frac{5}{6}$. Your book is right.
| {
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"url": "https://math.stackexchange.com/questions/748695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculation of $\int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2 \cos^2 x}}dx$ Calculate
$$ \int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2 \cos^2 x}}dx$$
$\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2\cos^2 x}}dx = \int_{0}^{\frac{\pi}{4}}\frac{\pi}{2}-\int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{\cos 2x}{2\cos^2 x}}dx$
Using The formula $\displaystyle \tan^{-1}(x)+\cot^{-1}(x) = \frac{\pi}{2}\Rightarrow \tan^{-1}(x) = \frac{\pi}{2}-\cot^{-1}(x).$
Now Let $\displaystyle J = \int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{\cos 2x}{2\cos^2 x}}dx = \int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{\cos^2 x-\sin^2 x}{2\cos^2 x}}dx = \int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{1}{2}-\frac{\tan^2 x}{2}}dx$
Now How can I solve after that? Help required.
Thanks
| Answer: $\displaystyle \int_{0}^{\pi/4}\tan^{-1}\sqrt{\frac{\cos 2x}{2\cos^2 x}}\,dx=\frac{\pi^2}{24}$
Proof:
We are making use of $3$ Lemmas which are (quite ) easy to prove:
1. $\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{x^2+2}\left ( x^2+1 \right )}=\frac{\pi}{6} $
Proof:
$$\begin{align*}
\int_{0}^{1}\frac{dx}{\sqrt{x^2+2}\left ( x^2+1 \right )} &\overset{x=\sqrt{2}\sinh t}{=\! =\! =\! =\! =\! =\! =\!}\int_{0}^{a}\frac{dt}{1+2\sinh^2 t} \\
&= \int_{0}^{a}\frac{dt}{\cosh (2t)}=\int_{0}^{a}\frac{\cosh (2t)}{1+\sinh^2 (2t)}\,dt\\
&=\frac{1}{2}\tanh^{-1}\left ( \sinh (2a) \right ) \\
&=\frac{1}{2}\tanh^{-1}\left ( \sqrt{\left ( 1+2\sinh^2 a \right )^2-1} \right ) \\
&= \frac{1}{2}\tanh^{-1}\sqrt{3}=\frac{\pi}{6}\\
\end{align*}$$
where $\displaystyle a=\sinh^{-1}\frac{1}{\sqrt{2}} $.
2. $\displaystyle \int_{0}^{\infty}\frac{dx}{\left ( x^2+a^2 \right )\left ( x^2+\beta^2 \right )}=\frac{\pi}{2a\beta\left ( a+\beta \right )}$
Proof:
$$\begin{align*}
\int_{0}^{\infty}\frac{dx}{\left ( x^2+a^2 \right )\left ( x^2+\beta^2 \right )} &=\int_{0}^{\infty}\frac{1}{\beta^2-a^2}\left ( \frac{1}{x^2+a^2}-\frac{1}{x^2+\beta^2} \right )\,dx \\
&= \frac{1}{\beta^2-a^2}\left ( \frac{\pi}{2a}-\frac{\pi}{2\beta} \right )\\
&= \frac{\pi}{2a\beta\left ( a+\beta \right )}\\
\end{align*}$$
3. It also holds by definition that: $\displaystyle \tan^{-1}a= \int_{0}^{1}\frac{a}{1+a^2x^2}\,dx$.
And now we are ready to evaluate the integral. Successively we have:
$$\begin{align*}
\int_{0}^{\pi/4}\tan^{-1}\sqrt{\frac{\cos 2\theta}{2\cos^2 \theta}}\,d\theta &=\int_{0}^{\pi/4}\int_{0}^{1}\frac{\sqrt{\frac{\cos 2\theta}{2\cos^2 \theta}}}{1+\left ( \frac{\cos 2\theta}{2\cos^2 \theta} \right )x^2}\,dx \,d\theta\\
&= \int_{0}^{1}\int_{0}^{\pi/4}\frac{\sqrt{1-2\sin^2 \theta}}{2-2\sin^2 \theta+\left ( 1-2\sin^2 \theta \right )x^2}\sqrt{2}\cos \theta \,\,d\theta \,dx\\
&=\int_{0}^{1}\int_{0}^{\pi/2}\frac{\sqrt{1-\sin^2 \phi}}{2-\sin^2 \phi+\left ( 1-\sin^2 \phi \right )x^2}\cos \phi \,\,d\phi \,dx \\
&=\int_{0}^{1}\int_{0}^{\pi/2}\frac{\cos^2 \phi}{\sin^2 \phi+\left ( x^2+2 \right )\cos^2 \phi}\,\,d\phi \,dx \\
&=\int_{0}^{1}\int_{0}^{\pi/2}\frac{d\phi \,dx}{\tan^2 \phi+x^2+2}=\int_{0}^{1}\int_{0}^{\infty}\frac{dx\,dx}{\left ( y^2+x^2+2 \right )\left ( y^2+1 \right )} \\
&=\frac{\pi}{2}\int_{0}^{1}\frac{dy}{\left ( 1+\sqrt{2+y^2} \right )\sqrt{2+y^2}} \\
&=\frac{\pi}{2}\left ( \frac{\pi}{4}-\frac{\pi}{6} \right )=\frac{\pi^2}{24}
\end{align*}$$
which checks numerically with the answer given above.
If I have any typos, because I typed it so quickly please let me know so that I correct them.
T:
| {
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"timestamp": "2023-03-29T00:00:00",
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Polynomials division algebra problem Find sum of coefficients of the quotient obtained in:
$$\frac{2x^n+x^{n-1}+x^{n-2}+...+x^2+x+5}{x-\frac{1}{2}}$$
I got "n" as the answer but according to the book is wrong, I don't know what is wrong exactly, but i want to know why the answer is "2n" ._. , thanks.
| $$\frac{2x^n+x^{n-1}+x^{n-2}+...+x^2+x+5}{x-\frac{1}{2}}=$$
$$\frac{(x-\frac{1}{2})(2x^{n-1}+2x^{n-2}+\dots+2x^2+2x+2)+6}{x-\frac{1}{2}}=$$
$$2x^{n-1}+2x^{n-2}+\dots+2x^2+2x+2+\frac{6}{x-\frac{1}{2}}$$
So the quotient is: $2x^{n-1}+2x^{n-2}+\dots+2x^2+2x+2$
The remainder is: $6$
And the sum of coefficients in the quotient is: $\sum\limits_{i=0}^{n-1}2=2n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/751503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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An Inequality Problem with not nice conditions How to show that $\dfrac{a^3}{a^2+b^2} + \dfrac{b^3}{b^2+c^2} + \dfrac{c^3}{c^2+a^2} \ge \dfrac32$, where $a^2+b^2+c^2=3$, and $a,b,c > 0$ ?
| By weighted AM-HM inequality, $\sum\limits_{cyc}\dfrac{\dfrac{a\cdot a^2}{a^2+b^2}}{3}\ge \dfrac{3}{\sum\limits_{cyc}\dfrac{a^2(a^2+b^2)}{a}}$
Here we consider the weights to be $a^2$,$b^2$ and $c^2$ respectively.
From the inequality above, we can have $\sum\limits_{cyc}\dfrac{a\cdot a^2}{a^2+b^2}\ge \dfrac{9}{\sum\limits_{cyc}\dfrac{a^2(a^2+b^2)}{a}}$
Now, if we can show $a(a^2+b^2)+b(b^2+c^2)+c(c^2+a^2)\le 6$ we will be done.
We can have, $a(3-c^2)+b(3-a^2)+c(3-b^2)\le 6 \implies 3(a+b+c)-(ac^2+ba^2+cb^2)$
From Cauchy-Schwarz inequality, we can have, $3(a+b+c)\le 9$
Now, we are left to prove, $ac^2+ba^2+cb^2\ge 3$
Again, using weighted AM-HM inequality,$ac^2+ba^2+cb^2\ge \dfrac{9}{\frac{ac}{c}+\frac{ab}{a}+\frac{bc}{b}}$
Now, $\dfrac{ac}{c}+\dfrac{ab}{a}+\dfrac{bc}{b}= a+b+c\le 3$
Hence, we are done.
I still believe there exists a simpler solution by a further application of Chebyshev Inequality.
| {
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For $1-r<|\theta|<1/2$, $|\frac{2r\sin{2\pi\theta}}{1-2r\cos{2\pi\theta}+r^2}-\frac{\cos\pi\theta}{\sin\pi\theta}|
For $1-r<|\theta|<1/2$ show that
$$|\frac{2r\sin{2\pi\theta}}{1-2r\cos{2\pi\theta}+r^2}-\frac{\cos\pi\theta}{\sin\pi\theta}|<C\frac{(1-r)^2}{|\theta|^3}$$
This inequality shows that the integral of the left side on $[1-r,1/2]$ is bounded independent of $r$.
$|\frac{2r\sin{2\pi\theta}}{1-2r\cos{2\pi\theta}+r^2}-\frac{\cos\pi\theta}{\sin\pi\theta}|
=|\frac{4r\sin^2\pi\theta\cos\pi\theta-\cos\pi\theta+2r\cos2\pi\theta-4r\sin^2\pi\theta\cos\pi\theta-r^2\sin\pi\theta}{(1-2r\cos{2\pi\theta}+r^2)\sin\pi\theta}|
=|\frac{\cos\pi\theta-2r\cos\pi\theta+r^2\sin\pi\theta}{(1-2r\cos{2\pi\theta}+r^2)\sin\pi\theta}|$
| Just a small trick: write $\frac{\cos \pi \theta}{\sin \pi \theta} = \frac{2\cos^2 \pi \theta}{\sin 2 \pi \theta}$ and use $1+ \cos 2 \pi \theta = 2\cos^2 \pi \theta.$ Now proceed by taking l.c.m, and all that.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I solve this definite integral: $\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$? $$\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$$
I have already solved the indefinite integral by transforming $\sin^{4}x + \cos^{4}x$ as follows:
$\sin^{4}x + \cos^{4}x = (\sin^{2}x + \cos^{2}x)^{2} - 2\cdot\sin^{2}x\cdot\cos^{2}x = 1 - \frac{1}{2}\cdot\sin^{2}(2x) = \frac{1 + \cos^{2}(2x)}{2}$,
and then using the $\tan(2x) = t$ substitution. But if I do the same with the definite integral, both bounds of the integral become $0$.
| We have $\displaystyle\cos^4x+\sin^4x=(\cos^2x+\sin^2x)^2-2\cos^2x\sin^2x=1-2\cos^2x\sin^2x$
$\displaystyle=\frac{2-\sin^22x}2=\frac{2(1+\tan^22x)-\tan^22x}{2\sec^22x}=\frac{\tan^22x+2}{2\sec^22x}$
$$\int\frac{dx}{\cos^4x+\sin^4x}=\int\frac{2\sec^22x}{\tan^22x+2}dx$$
Setting $\tan2x=u,$ $$\int\frac{2\sec^22x}{\tan^22x+2}dx=\int\frac{du}{u^2+(\sqrt2)^2}=\frac1{\sqrt2}\arctan\left(\frac u{\sqrt2}\right)+K$$
$$\implies \int\frac{dx}{\cos^4x+\sin^4x}=\frac1{\sqrt2}\arctan\left(\frac{\tan2x}{\sqrt2}\right)+K\ \ \ \ (1)$$
Now $\displaystyle\tan2x=0\iff 2x=n\pi\iff x=\frac{n\pi}2$ where $n$ is any integer
Establish that $$\int_0^{2a}f(x)dx=\begin{cases} 2\int_0^af(x)dx &\mbox{if } f(2a-x)=f(x) \\
0 & \mbox{if } f(2a-x)=-f(x) \end{cases} $$
Setting $2a=2\pi\iff a=\pi$ and $\displaystyle f(x)=\cos^4x+\sin^4x$
$\displaystyle\cos(2\pi-x)=\cos x,\sin(2\pi-x)=-\sin x\implies f(2\pi-x)=f(x)$
$$\implies I=\int_0^{2\pi}\frac{dx}{\cos^4x+\sin^4x}=2\int_0^{\pi}\frac{dx}{\cos^4x+\sin^4x}$$
Again setting $\displaystyle2a=\pi\iff a=\frac\pi2$
$\displaystyle\cos(\pi-x)=-\cos x,\sin(\pi-x)=+\sin x\implies f(\pi-x)=f(x)$
$$\implies I=2\int_0^{\pi}\frac{dx}{\cos^4x+\sin^4x}2=2\cdot2\int_0^{\dfrac\pi2}\frac{dx}{\cos^4x+\sin^4x}$$
Finally set $\displaystyle2a=\frac\pi2\iff a=\frac\pi4$
$\displaystyle\implies\cos\left(\frac\pi2-x\right)=\sin x,\sin\left(\frac\pi2-x\right)=\cos x$
$\displaystyle\implies f(x)=f\left(\frac\pi2-x\right)$
$\displaystyle\implies I=4\cdot2\int_0^{\dfrac\pi4}\frac{dx}{\cos^4x+\sin^4x}$
From $\displaystyle(1),I=8\left[\frac1{\sqrt2}\arctan\left(\frac{\tan2x}{\sqrt2}\right)+K\right]_0^{\frac\pi4}=\frac8{\sqrt2}\left(\frac\pi2-0\right)$
| {
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Equality of integrals: $ \int_{0}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x = 2 \cdot \int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x $ In Street-Fighting Mathematics (page 16), Prof. Sanjoy Mahajan states that $$ \displaystyle\int_{0}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x = 2 \cdot \displaystyle\int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x $$
I realize this is equivalent to saying that
$$\displaystyle\int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x = \displaystyle\int_{1}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x, $$
but why is this true?
| I assume you mean
$$\int_0^{\infty} \dfrac{dx}{1+x^2} = 2 \int_0^1 \dfrac{dx}{1+x^2}$$
This is true because
$$\underbrace{\int_1^{\infty} \dfrac{dx}{1+x^2} = \int_1^0 \dfrac{-du/u^2}{1+1/u^2}}_{x = 1/u} = \int_0^1 \dfrac{du}{u^2+1}$$
Hence,
$$\int_0^{\infty} \dfrac{dx}{1+x^2} = \int_0^{1} \dfrac{dx}{1+x^2} + \int_1^{\infty} \dfrac{dx}{1+x^2} = \int_0^{1} \dfrac{dx}{1+x^2} + \int_0^1 \dfrac{dx}{1+x^2} = 2 \int_0^1 \dfrac{dx}{1+x^2}$$
| {
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Prove this identity: $\sin^4x = \frac{1}{8}(3 - 4\cos2x + \cos4x)$ The problem reads as follows.
Prove this identity: $$\sin^4x = \frac{1}{8}(3 - 4\cos2x + \cos4x)$$
I started with the right side and used double angles identities for $\cos2x$ and a sum and then then double angle identity for the $\cos4x...$ It all got messy and I hit a dead end. He doesn't give any hints and I'm pretty lost.
I'm thinking that I will eventually need a product-sum identity to get the $\dfrac{1}{8}....$ But I'm just confused how to get there...
Thanks in advance to anyone who can help!
| $\sin^4x = (1 - \cos^2x)^2 = \left(1 - \dfrac{1 + \cos2x}{2}\right)^2 = \dfrac{(1 - \cos2x)^2}{4} = \dfrac{1 - 2\cos2x + \cos^22x}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{\cos^22x}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{\dfrac{1 + \cos4x}{2}}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{1 + \cos4x}{8} = \dfrac{3 - 4\cos2x + \cos4x}{8}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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integer solutions to $x^2+y^2+z^2+t^2 = w^2$ Is there a way to find all integer primitive solutions to the equation $x^2+y^2+z^2+t^2 = w^2$? i.e., is there a parametrization which covers all the possible solutions?
| Bradleyβs excellent paper on equal sums of squares gives the complete parameterization of the equation
$$
x_1^2 + x_2^2 + x_3^2 + x_4^2 = y_1^2
$$
as
\begin{align}
x_1 &= (uz+vy+wz)^2 -m^2(m^2+x^2+y^2+z^2-u^2-v^2-w^2) \\
x_2 &= 2m(um^2+uz^2+xvm-ywm+xwz+yvz) \\
x_3 &= 2m(vm^2+vy^2+zwm-xum+zuy+xwy) \\
x_4 &= 2m(wm^2+wx^2+yum-zvm+yvx+zux) \\
y_1 &= (uz+vy+wz)^2 +m^2(m^2+x^2+y^2+z^2+u^2+v^2+w^2),
\end{align}
with the option of multiplying all terms by a common factor $l$. Note that each term is a homogeneous quartic expression in the seven [integer] parameters.
| {
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Prove that $7^{100}+3^{10}=8^{100}$ or $7^{100}+3^{10}<8^{100}$
Prove that $7^{100}+3^{10}=8^{100}$ or $7^{100}+3^{10}<8^{100}$
I tried using some theorems of divisibility, to show that one divides the other, and the other also divides the first, but could not.
| If you list $7^x$ and $8^x$ you will notice that when $x$ is divisible by $4$, $7^x$ ends in $1$ and $8^x$ ends in $6$. $$7^0=8^0=1$$ $$7^{4+j}=7^j*7^4$$ $$7^4 = 2401$$ $2401$ ends in $1$ as does $7^0$ and any number that ends in $1$ times any number than ends in 1 results in a number that also ends in $1$. This works for $8$ as $8^4$ ends in $6$ and $6$ has the same property.
In order for $7^x+y=8^x$ then $y$ must end in $5$ which is impossible for $3^z$ as any number that ends in $5$ is divisible by $5$.
| {
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Evaluate the indefinite integral $\int \frac{\sqrt{9-4x^{2}}}{x}dx$ $$\int \frac{\sqrt{9-4x^{2}}}{x}dx$$
How Can I attack this kind of problem?
| $$(9-4x^2)^{1/2} = \sum_{n=0}^{\infty}\binom{1/2}{n}9^{1/2-n}(-4x^2)^{n}$$
$$\dfrac{(9-4x^2)^{1/2}}{x} = \sum_{n=0}^{\infty}\binom{1/2}{n}(-4)^{n}\left(\dfrac{9}{x^2}\right)^{1/2-n}$$
$$\int \left(\dfrac{9}{x^2}\right)^{1/2-n}(-4)^{n} \;\mathrm{d}x =(-4)^{n}9^{1/2-n}\int {x}^{1-2n}\; \mathrm{d}x = (-4)^{n}9^{1/2-n}\left(\dfrac{x^{2-2n}}{2-2n}\right) + C$$
$$\begin{align}\int \dfrac{(9-4x^2)^{1/2}}{x} \; \mathrm{d}x &= \sum_{n=0}^{\infty}\int\binom{1/2}{n}(-4)^{n}\left(\dfrac{9}{x^2}\right)^{1/2-n} \mathrm{d}x \\ &= \sum_{n=0}^{\infty}\binom{1/2}{n}(-4)^{n}9^{1/2-n}\left(\dfrac{x^{2-2n}}{2-2n}\right)\\ &= \dfrac{1}{3}\sum_{n=0}^{\infty}\binom{1/2}{n}(-4)^{n}3^{2-2n}\left(\dfrac{x^{2-2n}}{2-2n}\right)\\ &= \dfrac{1}{6}\sum_{n=0}^{\infty}\binom{1/2}{n}(-4)^{n}\left(\dfrac{(3x)^{2-2n}}{1-n}\right) + c\end{align}$$
| {
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Maclaurin series: $\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^8}{8!}+\frac{x^{11}}{11!}+\frac{x^{12}}{12!}+...$ The Taylor series of a real or complex-valued function Ζ(x) that is infinitely differentiable at a real or complex number a is the power series
$$f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots. $$
I have a curiosity concerning the following polynomial:
$$\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^8}{8!}+\frac{x^{11}}{11!}+\frac{x^{12}}{12!}+...$$
Is this a known development in Maclaurin series?
| While I've never seen the series you define here, it can be shown to be an elementary function. The process for doing this goes under the name of multisectioning: the key concept is that if $\zeta$ is an $n$th root of unity, then the Maclaurin coefficients of the functions $e^{\zeta^kx}$ for $0\leq k\lt n$ all have $n$ as a period, and the vectors of the first n coefficients are all linearly independent (and so span $\mathbb{C}^n$). This means that any Taylor series $\sum_n\frac{a_nx^n}{n!}$ with a periodic coefficient sequence $\{a_n\}$ can be expressed as a linear combination of (possibly complex) exponentials.
In the case to hand, it's easiest to start by adding $1$ to the function (giving the series $1+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots$ which is exactly periodic); then we have $n=4$, so $\zeta=i$, and the four functions are $e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots$, $e^{ix} = 1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\ldots$, $e^{-x} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\ldots$, and $e^{-ix} = 1-ix-\frac{x^2}{2!}+\frac{ix^3}{3!}+\ldots$ (where in each case, the ellipses designate a periodic continuation of coefficients). The function is thus a linear combination $ae^x+be^{ix}+ce^{-x}+de^{-ix}$, where equating coefficients gives us:
$$\begin{array}
\\
a+b+c+d&=1\\
a+ib-c-id&=0\\
a-b+c-d&=0\\
a-ib-c+id&=1\\
\end{array}
$$
Now, summing the first and third equations gives $a+c=\frac12$, and summing the second and fourth equations gives $a-c=\frac12$, so we have $a=\frac12, c=0$; similarly, differencing the first and third equations gives $b+d=\frac12$, and differencing the second and fourth gives $i(d-b)=\frac12$, or $b-d=\frac i2$, so we get $b=\frac{1+i}4$ and $d=\frac{1-i}4$; finally, this yields:
$$1+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^8}{8!}+\ldots = \frac12e^x+\frac14\left((1+i)e^{ix}+(1-i)e^{-ix}\right)$$
If you know Euler's identities, then you should be able to see how to turn the seemingly-complex terms here into a linear combination of $\sin(x)$ and $\cos(x)$, but I'll leave that for you to do...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$2 \cos^2 x β 2 \cos xβ 1 = 0$ Find the solutions if $0^\circ \le x < 360^\circ$
Find the solutions of $$2 \cos^2 x β 2 \cos xβ 1 = 0$$ for all $0^\circ β€ x < 360^\circ$.
For $0^\circ \le x < 360^\circ$, I'm getting $x=111.5^\circ$ and $x=248.5^\circ$.
Is this correct? Thanks!
| Observe that $2 \cos^2 x - 2 \cos x - 1$ is irreducible and cannot be factored.
Apply the quadratic formula:
$$\cos x = \frac{-(-2)\pm\sqrt{(-2)^2-4(2)(-1)}}{2(2)}=\frac{2\pm\sqrt{12}}{4}=\frac{2\pm2\sqrt{3}}{4}=\frac{1\pm\sqrt{3}}{2}$$
But $\frac{1 + \sqrt{3}}{2}$ is not in between $-1$ and $1$, so we can discard it accordingly. Thus,
$$\cos x = \frac{-1-\sqrt{3}}{2}$$
and so $$x=\cos^{-1} \bigg(\frac{-1-\sqrt{3}}{2}\bigg)\approx \boxed{111.5,248.5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/768401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How prove this inequality $\frac{1}{n}\sum_{k=0}^{n-1}\binom{k}{a}\binom{k}{b}\le\frac{1}{a+b+1}\binom{n}{a}\binom{n}{b}$
let $a,b,n$ be positive integer numbers,and such $a,b\le n$, show that
$$\dfrac{1}{n}\sum_{k=0}^{n-1}\binom{k}{a}\binom{k}{b}\le\dfrac{1}{a+b+1}\binom{n}{a}\binom{n}{b}$$
this inequality maybe can use Integral inequality to solve it.because
$$\int_{0}^{1}x^{a+b}dx=\dfrac{1}{a+b+1},\int_{0}^{1}x^{n-1}dx=\dfrac{1}{n}$$
Thank you for you help.
| Since the inequality is symmetric in $a$ and $b$ we may (and will), suppose $a\leq b\leq n$.
Moreover, the inequality is trivially satisfied if $n=b$. So let us suppose that the inequality is proved for some $n\geq b$. It follows that
$$\eqalign{
\sum_{k=0}^n\binom{k}{a}\binom{k}{b}&=\sum_{k=0}^{n-1}\binom{k}{a}\binom{k}{b}+
\binom{n}{a}\binom{n}{b}
\cr
&\leq \frac{n}{a+b+1}\binom{n}{a}\binom{n}{b}+\binom{n}{a}\binom{n}{b}\cr
&\leq \frac{a+b+n+1}{a+b+1}\binom{n}{a}\binom{n}{b}\cr}
$$
Now the desired inequality for $n+1$ would follow if we have
$$
\frac{a+b+n+1}{a+b+1}\binom{n}{a}\binom{n}{b}\leq \frac{n+1}{a+b+1}\binom{n+1}{a}\binom{n+1}{b}
$$
This is equivalent to
$$
(n+1+a+b)(n+1-a)(n+1-b)\leq(n+1)^3
$$
or
$$
n+1 \geq \frac{a^2b+b^2a}{a^2+b^2+ab}=b-\frac{b^3}{a^2+ab+b^2}
$$
and this inequality is trivially satisfied since $n+1>b$.
This ends the proof by induction.$\qquad\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/768645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
set problem of integers Consider the following set $F=\{F^0, F^1, F^2, \ldots\}$. This set consists of positive integers which satisfy the following properties:
*
*$F^0= F^1=1$
AND
*$F^n= F^{n-1} + F^{n-2}$ for all positive integers $n\geq2$.
Prove that for all positive integers $n$, the elements of the set $F$ satisfy the following identity:
$$\begin{vmatrix}
1 & 1 \\
1 & 0
\end{vmatrix}^{n+1}
=
\begin{vmatrix}
F^{n+1} & F^n \\
F^n & F^{n-1}
\end{vmatrix}$$
where $\begin{vmatrix}\end{vmatrix}$ denotes the determinant.
| Hint:
$$\begin{pmatrix} F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}\begin{pmatrix} 1&1\\1&0
\end{pmatrix}=\begin{pmatrix} F_{n+1}+F_n&F_{n+1}\\F_{n}+F_{n-1}&F_n\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/769404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate the product ST, and infer from it the inverse of T. S=\begin{pmatrix}
1/2 & 1/2 & 0\\
1 & 0 & 0\\
-3/2 & 0 & 1/2
\end{pmatrix}
T=
\begin{pmatrix}
0 & 1 & 0\\
2 & -1 & 4\\
0 & 3 & 2
\end{pmatrix}
I have calculated ST to be =
\begin{pmatrix}
1 & 0 & 2\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}
But i'm now unsure of how to continue? This is obviously a significant matrix but I can't spot the answer.
Thanks.
| Think in terms of Gaussian elimination. It should be clear that
$$
\pmatrix{
1&0&-2\\
0&1&0\\
0&0&1
}
\pmatrix{
1&0&2\\
0&1&0\\
0&0&1
} = I
$$
That is, we have
$$
\pmatrix{
1&0&-2\\
0&1&0\\
0&0&1
}
\;(ST) =
\left(\pmatrix{
1&0&-2\\
0&1&0\\
0&0&1
}
\;S \right)\cdot T = I
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
} |
How prove this $100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2\ge 0$
let $a,b,c\ge 0$, and such $$a+b+c=6$$
show that
$$100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2\ge 0$$
My idea: since
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=36-2(ab+bc+ac)$$
$$a^2b^2+b^2c^2+a^2c^2=(ab+ca+bc)^2-2abc(a+b+c)=(ab+bc+ac)^2-12abc$$
so let $p=a+b+c=6, q=ab+bc+ac,r=abc$,so
$$\Longleftrightarrow 100+5[36-2q]-2[q^2-12r]-r^2\ge 0$$
$$\Longleftrightarrow r^2+2q^2-24r+10q\le 280$$
then I can't.Thank you
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition does not depend on $w^3$ and we need to prove that $f(w^3)\geq0$,
where $f$ is a concave function.
But the concave function gets a minimal value for an extremal value of $w^3$,
which happens for an equality case of two variables or when maybe $w^3=0$.
Thus, it's enough to prove our inequality in the following cases.
*
*$b=a$ and $c=6-2a$, where $0\leq a\leq 3$, which gives
$$(a-2)^2(a^2+5)(7+4a-2a^2)\geq0,$$
which is obvious for $0\leq a\leq 3$;
*$c=0$ and $b=6-a$, where $0\leq a\leq6$.
In this case we get
$$140-30a-31a^2+12a^3-a^4\geq0,$$
which is true for $0\leq a\leq6$.
Done!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
First order Cauchy problem I'm trying to solve the PDE
\begin{align*}
au_x + (bx+cu-1)u_y = d \\
u(x,0) = 0.
\end{align*}
So far, I got the solution for the case $cd-ab \neq 0$ as
\begin{align*}
u(x,y) = \frac{d}{cd-ab}\left( 1-bx-\sqrt{(1-bx)^2+2(cd+ab)y} \right)
\end{align*}
Now, I'm stuck on the case $ab = cd$. Can you point me the way to go?
| Solve the equation by the method of characteristics.
The characteristic ODE are as follows.
$\dfrac{d x}{ d t} =a$, $\dfrac{d u}{ d t} = d$, $\dfrac{d y}{ d t} =b x +c u -1$,
with initial conditions $x(0) = x_0$, $u(0) = 0 $, $y(0) = 0$ at $t=0$.
By solving the first two characteristic ODE, there are
$x = x_0 + a t $ and $u = d t$.
By substituting $x,u$ into the third characteristic ODE, there is
$\dfrac{d y}{ d t} = b x_0 -1 +(ab +cd) t $, with $y(0) = 0$.
The solution is $y = (b x_0 - 1) t + \dfrac{1}{2} (ab + cd) t^2$.
By substituting $x_0$ into $y$, there is
$y = (b(x - at) - 1) t + \dfrac{1}{2} (ab + cd) t^2 = (b x - 1) t + \dfrac{1}{2} (cd - ab) t^2 $, which is
$\dfrac{1}{2} (cd - ab) t^2 + (b x - 1) t - y =0$.
(1) If $cd \neq ab $, then $t = \dfrac{1-bx \pm \sqrt {(bx -1)^2 + 2 (cd - ab) y}}{cd - ab}$.
Notice that $y = 0$ at $t = 0$, $ t = \dfrac{1-bx - \sqrt {(1-bx)^2 + 2 (cd - ab) y}}{cd - ab}$.
Thus, the solution of the PDE is $u = \dfrac{d}{cd -ab} \Bigg(1-bx - \sqrt {(1-bx)^2 + 2 (cd - ab) y} \Bigg)$.
(2) If $cd = ab $, then $t = \dfrac{ y }{bx - 1}$.
Thus, the solution of the PDE is $u = \dfrac{ d y }{bx - 1}$.
Remark: Note that the solution is not defined on the whole plane and the regularity of the solution needs further discussion.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculation of $f_{\omega^3}(2)$ in the fast growing hierarchy How is the number
$$\large f_{\omega^3}(2)$$
in the fast growing hierarchy calculated ?
My only idea is to convert to
$$\large f_{\omega^2 2}(2)$$
but now I have no idea how to continue.
| Too large to calculate!
I assume you are referring to the definition on Wikipedia: http://en.wikipedia.org/wiki/Fast-growing_hierarchy
If so, then, as the article points out, $f_{\omega + 1}(64)$ is already larger than Graham's number. $f_{\omega^3}$ is much larger than this.
But perhaps you're asking what an algorithm could be if we have unbounded time and space? Well here's a start:
$f_{\omega^3}(2) = f_{\omega^2 \cdot 2}(2) = f_{\omega^2 + \omega^2}(2) = f_{\omega^2 + \omega \cdot 2}(2) = f_{\omega^2 + \omega + \omega}(2) = f_{\omega^2 + \omega + 2}(2) = f_{\omega^2 + \omega + 1}^2(2) = f_{\omega^2 + \omega + 1}(f_{\omega^2 + \omega + 1}(2)) = f_{\omega^2 + \omega + 1}(f_{\omega^2 + \omega}^2(2)) = f_{\omega^2 + \omega +1}(f_{\omega^2 + \omega}(f_{\omega^2 + \omega}(2))) = f_{\omega^2 + \omega +1}(f_{\omega^2 + \omega}(f_{\omega^2 + 2}(2))) = f_{\omega^2 + \omega +1}(f_{\omega^2 + \omega}(f_{\omega^2 + 1}(f_{\omega^2 + 1}(2)))) = f_{\omega^2 + \omega +1}(f_{\omega^2 + \omega}(f_{\omega^2 + 1}(f_{\omega^2}(f_{\omega^2}(2))))) = f_{\omega^2 + \omega +1}(f_{\omega^2 + \omega}(f_{\omega^2 + 1}(f_{\omega^2}(f_{\omega + \omega}(2))))) = f_{\omega^2 + \omega +1}(f_{\omega^2 + \omega}(f_{\omega^2 + 1}(f_{\omega^2}(f_{\omega + 2}(2))))) = f_{\omega^2 + \omega +1}(f_{\omega^2 + \omega}(f_{\omega^2 + 1}(f_{\omega^2}(f_{\omega + 1}(f_{\omega + 1}(2)))))) = f_{\omega^2 + \omega +1}(f_{\omega^2 + \omega}(f_{\omega^2 + 1}(f_{\omega^2}(f_{\omega + 1}(f_{\omega}(f_{\omega}(2))))))) = f_{\omega^2 + \omega +1}(f_{\omega^2 + \omega}(f_{\omega^2 + 1}(f_{\omega^2}(f_{\omega + 1}(f_{\omega}(8)))))) = f_{\omega^2 + \omega +1}(f_{\omega^2 + \omega}(f_{\omega^2 + 1}(f_{\omega^2}(f_{\omega + 1}(f_8(8)))))) = ...$
At this point I get bored of writing this out by hand, but you get the point, you just keep applying the rules. At some point the numbers will get too large to write down.
Why do you ask about calculating this number?
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $\sum_{k=1}^n \left(k \sum_{i=0}^{k-1} {n \choose i}\right)$ How do I calculate the following summation?
$$\sum_{k=1}^n \left[k \sum_{i=0}^{k-1} {n \choose i}\right]$$
| (Note: I have attempted to incorporate
OP's edit to the problem.)
$\begin{array}\\
\sum_{k=1}^n (k \sum_{i=0}^{k-1} {n \choose i})
&=\sum_{k=1}^n \sum_{i=0}^{k-1} (k {n \choose i})\\
&=\sum_{i=0}^{n-1} \sum_{k=i+1}^n (k {n \choose i})\\
&=\sum_{i=0}^{n-1} {n \choose i}\sum_{k=i+1}^n k \\
&=\sum_{i=0}^{n-1} {n \choose i}\left(\frac{n(n+1)}{2}-\frac{i(i+1)}{2}\right) \\
&=\frac{n(n+1)}{2}\sum_{i=0}^{n-1} {n \choose i}-\sum_{i=0}^{n-1} {n \choose i}\frac{i(i+1)}{2} \\
&= S-T\\
\end{array}
$
$S
=\frac{n(n+1)}{2}\sum_{i=0}^{n-1} {n \choose i}
=\frac{n(n+1)}{2}(\sum_{i=0}^{n} {n \choose i}-{n \choose n})
=\frac{n(n+1)}{2}(2^n-1)
$.
$T = \sum_{i=0}^{n-1} {n \choose i}\frac{i(i+1)}{2}$.
Consider $t_i = {n \choose i}\frac{i(i+1)}{2}$.
$t_0 = 0, t_1=n$.
For $i \ge 1$,
I would like to use the identity
${ n \choose m}{m \choose k}
={n \choose k}{n-k \choose m-k}
$.
$\begin{array}\\
t_i
&={n \choose i}\frac{i(i+1)}{2}\\
&={n \choose i}\left(\frac{i(i-1)}{2}+i\right)\\
&={n \choose i}({i \choose 2}+{i \choose 1})\\
&={n \choose i}{i \choose 2}+{n \choose i}{i \choose 1}\\
&={n \choose 2}{n-2 \choose i-2}+{n \choose 1}{n-1 \choose i-1}\\
\end{array}
$
$\begin{array}\\
T
&= t_0+t_1+\sum_{i=2}^{n-1} t_i\\
&= n+\sum_{i=2}^{n-1} \left( {n \choose 2}{n-2 \choose i-2}+{n \choose 1}{n-1 \choose i-1}\right)\\
&= n+\sum_{i=2}^{n-1} {n \choose 2}{n-2 \choose i-2}+\sum_{i=2}^{n-1}{n \choose 1}{n-1 \choose i-1}\\
&= n+ {n \choose 2}\sum_{i=2}^{n-1}{n-2 \choose i-2}+{n \choose 1}\sum_{i=2}^{n-1}{n-1 \choose i-1}\\
&= n+ {n \choose 2}\sum_{i=0}^{n-3}{n-2 \choose i}+{n \choose 1}\sum_{i=1}^{n-1}{n-1 \choose i}\\
&= n+ {n \choose 2}\left(\sum_{i=0}^{n-2}{n-2 \choose i}-1\right)+{n \choose 1}\sum_{i=0}^{n-1}{n-1 \choose i}-1\\
&= n+ {n \choose 2}\left(2^{n-2}-1\right)+{n \choose 1}(2^{n-1}-1)\\
&= n+ \frac{n(n+1)}{ 2}\left(2^{n-2}-1\right)+n(2^{n-1}-1)\\
&= \frac{n(n+1)}{ 2}\left(2^{n-2}-1\right)+n2^{n-1}\\
\end{array}
$
$\begin{array}\\
S-T
&=\frac{n(n+1)}{2}(2^n-1)-\left( \frac{n(n+1)}{ 2}\left(2^{n-2}-1\right)+n2^{n-1}\right)\\
&=2^{n-2}\left(2n(n+1)-\frac{n(n+1)}{ 2} -2n\right) -\frac{n(n+1)}{2}- \frac{n(n+1)}{ 2}\\
&=2^{n-2}\frac{3n(n+1)-4n}{ 2}-n(n+1)\\
&=2^{n-2}\frac{n(3n-1)}{ 2}-n(n+1)\\
\end{array}
$
As usual,
this was done off the top of my head,
so there is a good chance of error,
but it's late and I'm tired,
so I'll stop here.
All corrections will be appreciated.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Picard Approximation Consider the initial value problem
$$\frac{dy}{dx} = y^2 + 3x^2 - 1, \\
y(1) = 1$$
on D = {|x-1| <= 1, |y-1| <= 1}
Find the second approximation to the solution and estimate the error term
If someone could maybe walk me through the steps that would be most appreciated =)
| We are given:
$$\tag 1 \frac{dy}{dx} = y^2 + 3x^2 - 1, y(1) = 1, ~\mbox{on}~~ D = {|x-1| \le 1, |y-1| \le 1}$$
The Picard-LindelΓΆf iteration is given by:
$$\tag 2 \displaystyle y_0(x) = y_0, ~~y_{n+1}(x) = y_0 + \int^x_{x_0} f(s, y_n(s))ds$$
For $(1)$, we have: $f(s, y_n(s)) = y_n^2+3s^2-1$ and using $(2)$, yields:
*
*$\displaystyle y_0(1) = y_0 = 1$
*$\displaystyle y_1(x) = y_0 + \int^x_{x_0} f(s, y_0(s))ds = 1 + \int^x_{1} ((1)^2+3s^2-1) ds = x^3$
*$\displaystyle y_2(x) = y_0 + \int^x_{x_0} f(s, y_1(s))ds = 1 + \int^x_{1} ((s^3)^2+3s^2-1) ds = \dfrac{x^7}{7}+x^3-x +\dfrac{6}{7}$
If you need one more iteration, you would get:
$$y_3(x) = \frac{x^{15}}{735}+\frac{2 x^{11}}{77}-\frac{2 x^9}{63}+\frac{3 x^8}{98}+\frac{x^7}{7}-\frac{2 x^5}{5}+\frac{3 x^4}{7}+\frac{4 x^3}{3}-\frac{6 x^2}{7}-\frac{13 x}{49}+\frac{4099}{6930}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
calculus radical integration question Does anyone know how to calculate integral of $\sqrt{ 1-\cos (x)}$ ? I tried several methods resulting in $-2\sqrt2 \cos (x/2) + c$, but this is wrong in accordance with the text book, so i dont know how to proceed...
| \begin{align*}
\int \sqrt{ 1-\cos (x)} dx &= \int\sqrt 2 \left| \sin \left( \frac x 2 \right ) \right | dx\\
&= -2\sqrt 2 \cos \left( \frac x 2 \right ) \mathrm{sgn} \left( \sin \left( \frac x 2 \right )\right)\\
&= -2\sqrt 2 \cot \left( \frac x 2 \right ) \left| \sin \left( \frac x 2\right)\right|\\
&= -2 \cot \left(\frac x 2 \right )\sqrt{ 2\sin ^2 \left(\frac x 2 \right )}\\
&= -2 \cot \left(\frac x 2 \right ) \sqrt{1 - \cos x}
\end{align*}
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the inverse Laplace transform of $\frac{s^2-4s-4}{s^4+8s^2+16}$ $$F(s) = \frac{s^2-4s-4}{s^4+8s^2+16}$$
My work is as follows,
$$\frac{s^2-4s-4}{(s^2+4)^2}=\frac{s^2+4}{(s^2+4)^2}-\frac{8}{(s^2+4)^2}-\frac{4s}{(s^2+4)^2}$$
The inverse laplace of the first term is, $\frac{1}{2} \sin(2t)$
The second one has no direct transform, perhaps using the convolution theorem would do.
The third and final term is -$t\sin(2t)$. Is there a faster way to solve this instead of using the convolution theorem for the second term.
| Note that
$$
\mathcal{L}^{-1}\left[\frac{a}{s^2+a^2}\right]=\sin at
$$
and
$$
\mathcal{L}^{-1}\left[\frac{s^2-a^2}{(s^2+a^2)^2}\right]=t\cos at,
$$
then
$$
\frac{a}{s^2+a^2}-a\cdot\frac{s^2-a^2}{(s^2+a^2)^2}=\frac{2a^3}{(s^2+a^2)^2}.
$$
Hence
$$
\mathcal{L}^{-1}\left[\frac{2a^3}{(s^2+a^2)^2}\right]=\sin at-at\cos at
$$
and
$$
\mathcal{L}^{-1}\left[\frac{8}{(s^2+4)^2}\right]=\frac12(\sin 2t-2t\cos 2t).
$$
| {
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"url": "https://math.stackexchange.com/questions/784688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to integrate $\sqrt{1-\sin 2x}$? I want to solve the following integral without substitution:
$$\int{\sqrt{1-\sin2x}} \space dx$$
I have:
$$\int{\sqrt{1-\sin2x}} \space dx = \int{\sqrt{1-2\sin x\cos x}} \space dx = \int{\sqrt{\sin^2x + \cos^2x -2\sin x\cos x}} \space dx$$
but this can be written in two ways: $\int{\sqrt{(\sin x - \cos x)^2}} \space dx$ or $\int{\sqrt{(\cos x - \sin x)^2} \space dx}$
and it seems to be pretty far from what the real result is: http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427ecvsatkj0te&mail=1
Can I get any hints?
EDIT:
Thank you for your answers! So as you all showed, we have:
$\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\sin x - \cos x)^2}} \space dx$
or
$\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\cos x - \sin x)^2} \space dx} = \int{\sqrt{(-(\sin x - \cos x))^2} \space dx}$
combined:
$\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\pm(\sin x - \cos x))^2} \space dx}$
so we actually have:
$$\int{|\sin x - \cos x|} \space dx$$
I drew myself a trigonometric circle and if I concluded correctly, we have:
1. $\int{\sin x - \cos x} \space dx$ for $x \in [ \frac{\pi}{4} + 2k\pi , \frac{5\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$
2. $\int{\cos x - \sin x} \space dx$ for $x \in [ -\frac{3\pi}{4} + 2k\pi , \frac{\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$
... which means:
1. $x \in [ \frac{\pi}{4} + 2k\pi , \frac{5\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$ :
$$\int{\sqrt{1-\sin2x}} \space dx = \int{\sin x - \cos x} \space dx = -\cos x - \sin x + C$$
2. $x \in [ -\frac{3\pi}{4} + 2k\pi , \frac{\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$ :
$$\int{\sqrt{1-\sin2x}} \space dx = \int{\cos x - \sin x} \space dx = \sin x + \cos x + C$$
| The result in wolfram is given by :
$$\frac{\sqrt{1-\sin(2x)}(\cos x+\sin x)}{\cos x-\sin x}\tag{1}$$
Note your results from simplifying the integration. You want to check that $(1)$ is equal to either$$\int\sqrt{(\sin x-\cos x)^2}\mathrm dx=\int\sin x\mathrm dx-\int\cos x\mathrm dx=-\cos x-\sin x.\tag{2}$$
or
$$\int\sqrt{\cos x-\sin x)^2}\mathrm dx=\int\cos x\mathrm dx-\int\sin x\mathrm dx=\sin x+\cos x\tag{3}$$
Notice that $(2)$ and $(3)$ are equal to $(1)$. It is a trigonometric identity. I will show $(2)$ is equal to $(1)$ here. I will leave you to show that $(3)$ is equal to $(1)$.
Proof of $(1)=(2)$:
$$\begin{aligned}&\frac{\sqrt{1-\sin(2x)}(\cos x+\sin x)}{\cos x-\sin x}=-\cos x-\sin x\\&\iff\sqrt{1-\sin(2x)}(\cos x+\sin x)=-(\cos x+\sin x)(\cos x-\sin x)\\&\iff\sqrt{1-\sin(2x)}=-(\cos x-\sin x)\\&\iff \sqrt{1-\sin(2x)}=\sin x-\cos x\\&\iff 1-\sin(2x)=\sin^2 x-2\sin x\cos x+\cos^2 x\\&\iff 1-2\sin(2x)=1-2\sin x\cos x\\&\iff 1-\sin(2x)=1-\sin(2x)\end{aligned}$$
This shows that $$\int\sqrt{1-\sin(2x)}\mathrm dx=-\cos x-\sin x$$even though Wolfram gives you a result that makes you unsure of your own result! Now, just show for yourself that $(1)=(3)$ and you should be convinced that you came up with a much simpler solution than Wolfram.
| {
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"timestamp": "2023-03-29T00:00:00",
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Infinite Sum of products What is the infinite sum
$$S = {1 + \frac{1}{3} + \frac{1\cdot 3}{3\cdot 6} + \frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}+ ....}?$$
I attempted messing around with the $n$ th term in the series but didnt see any solution.
How should I proceed?
| HINT:
$n^{th}$ term is $$\frac{1\cdot3\cdot5\cdots(2n-1)}{3^n\cdot n!}$$
$$=\frac{2^n\cdot n!\cdot1\cdot3\cdot5\cdots(2n-1)}{6^n\cdot n!\cdot n!}$$
$$=\frac{2n!}{6^n\cdot n!\cdot n!}$$
$$=\frac{^{2n}C_n}{6^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/786162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$4$ variable system of equations Find all the solutions to this system of equations:
$$\begin{cases}
-2d^3+3a^2d+d+2c^3-3a^2c-c=0\\
2d^3-3db^2-d-2c^3+3b^2c+c=0\\
-6c^2b+b^3+b+6c^2a-a^3-a=0\\
6d^2b-b^3-b-6d^2a+a^3+a=0
\end{cases}$$
After factoring:
$$\begin{cases}
d(-2d^2+3a^2+1)+c(2c^2-3a^2-1)=0\\
d(2d^2-3b^2-1)-c(2c^2-3b^2-1)=0\\
b(-6c^2+b^2+1)+a(6c^2-a^2-1)=0\\
b(6d^2-b^2-1)-a(6d^2-a^2-1)=0
\end{cases}$$
How should I proceed next?
System of equation comes from following problem:
Find the line where line integral
$$\int_L (y^3-y)dx+2x^3dy$$
is the largest
From Green's theorem:
$$\begin{cases} F_x=6x^2 \\ Q_y=3y^2-1 \end{cases}$$
$$\underset{D\ \ }{\iint}(6x^2-3y^2-1)dxdy = \int_{a}^{b}(2x^3-3y^2x-x)\Bigg|_{c}^{d}dy=\int_{a}^{b}((2d^3-3y^2d-d)-(2c^3-3y^2c-c))dy=$$
$$=\left((2d^3y-y^3d-dy)-(2c^3y-y^3c-cy)\right)\Bigg|_{a}^{b}=$$
$$=\left((2d^3b-b^3d-db)-(2c^3b-b^3c-cb)\right)-\left((2d^3a-a^3d-ad)-(2c^3a-a^3c-ac)\right)=$$
$$=2d^3b-db^3-db-2c^3b+b^3c+cb-2d^3a+a^3d+ad+2c^3a-a^3c-ac$$
I need to find $t(a,b,c,d)$ partial derivatives:
$$t_a=-2d^3+3a^2d+d+2c^3-3a^2c-c$$
$$t_b=2d^3-3db^2-d-2c^3+3b^2c+c$$
$$t_c=-6c^2b+b^3+b+6c^2a-a^3-a$$
$$t_d=6d^2b-b^3-b-6d^2a+a^3+a$$
Now I equal all of them $0$, and find global maximum.
| Adding $1$st and $2$nd equations, one can get
$$
3d(a^2-b^2)-3c(a^2-b^2)=0,
$$
$$
(d-c)(a-b)(a+b)=0\tag{1};
$$
adding $3$rd and $4$th equation, one can get
$$
b(6d^2-6c^2)-a(6d^2-6c^2)=0,
$$
$$
(b-a)(d-c)(d+c)=0\tag{2}.
$$
Equations $(1), (2)$ can help here?
Other (similar) way:
write system in form:
$$\begin{cases}
2(d^3-c^3)-3a^2(d-c)-(d-c)=0\\
2(d^3-c^3)-3b^2(d-c)-(d-c)=0\\
6c^2(b-a)-(b^3-a^3)-(b-a)=0\\
6d^2(b-a)-(b^3-a^3)-(b-a)=0\\
\end{cases}$$
$$\begin{cases}
(d-c)(2d^2+2dc+2c^2-3a^2-1)=0\\
(d-c)(2d^2+2dc+2c^2-3b^2-1)=0\\
(b-a)(6c^2-b^2-ab-a^2-1)=0\\
(b-a)(6d^2-b^2-ab-a^2-1)=0\\
\end{cases}$$
$$\begin{cases}
(d-c)(2d^2+2dc+2c^2-3a^2-1)=0\\
(d-c)(b-a)(b+a)=0\\
(b-a)(6c^2-b^2-ab-a^2-1)=0\\
(b-a)(d-c)(d+c)=0\\
\end{cases}$$
Cases:
*
*$a=b$, $c=d$: then system has solution for any $a,c$;
*$a=b$, $c\ne d$: then remains $1$ equation: $2(d^2+dc+c^2)-3a^2-1=0$;
*$a\ne b$, $c=d$: then remains $1$ equation: $6c^2-(a^2+ab+b^2)-1=0$;
*$a\ne b$, $c\ne d$: then $b=-a$, $d=-c$, and
$$
\begin{cases}
2c^2-3a^2=1;\\
6c^2-a^2=1;\\
\end{cases}
$$
$\qquad$(linear system on $a^2, c^2$).
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a=\dfrac{3+\sqrt{5}}{2}.$ Show that $\lfloor a \lfloor an \rfloor \rfloor+n$ is divisible by $3$.
Let $a=\dfrac{3+\sqrt{5}}{2}.$
Show that for all $n\in\mathbb N$, $\lfloor a \lfloor an \rfloor \rfloor+n$ is divisible by $3$.
My teacher solve this problem with induction, I am just curious if we can do this exercise without it ?
| Let $\alpha=\dfrac{1+\sqrt{5}}{2}, \beta=\dfrac{3+\sqrt{5}}{2}$, then
$$\beta= \alpha^2= \alpha+1$$
Since
$$\alpha\lfloor\beta n\rfloor-\lfloor\beta n\rfloor=\alpha n-(\alpha-1)\{\alpha n\}=\lfloor\alpha n\rfloor+(2-\alpha)\{\alpha n\}$$
so $\lfloor\alpha n\rfloor\lt \alpha\lfloor\beta n\rfloor-\lfloor\beta n\rfloor\lt \lfloor\alpha n\rfloor+1$. we get that
$$\lfloor\alpha\lfloor\beta n\rfloor\rfloor=\lfloor\alpha n\rfloor+\lfloor\beta n\rfloor$$
then, we have
\begin{equation}
\begin{split}\lfloor\beta\lfloor\beta n\rfloor\rfloor+n &=\lfloor\alpha\lfloor\beta n\rfloor+\lfloor\beta n\rfloor \rfloor+n\\&=\lfloor\alpha\lfloor\beta n\rfloor\rfloor+\lfloor\beta n\rfloor +n\\
&=\lfloor\alpha n\rfloor+\lfloor\beta n\rfloor+\lfloor\beta n\rfloor +n\\
&=\lfloor\alpha n\rfloor+2\lfloor\beta n\rfloor +n\\
&=\lfloor\alpha n\rfloor+2\lfloor\alpha n +n\rfloor +n\\
&=3\lfloor\alpha n\rfloor +3n
\end{split}
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
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Positive integer solutions of $a^3 + b^3 = c$ Is there any fast way to solve for positive integer solutions of $$a^3 + b^3 = c$$ knowing $c$?
My current method is checking if $c - a^3$ is a perfect cube for a range of numbers for $a$, but this takes a lot of time for larger numbers. I know $c = (a+b)(a^2-ab+b^2)$, but I can't think of a way this would speed up the process. Factoring numbers can be done fairly quickly but then the values of $a$ and $b$ have to be chosen. This is for a previous question I had about Fibonacci numbers. Any relevant literature is appreciated.
| $c=(a+b)(a^2βab+b^2)$ should actually be quite helpful. After factoring $c$, you find all possibly ways of writing $c = xy$. Let $x = a + b$ or $b = x - a$, so
$$
a^2-ab+b^2 = a^2 - a(x-a) + (x-a)^2 = 3a^2 - 3ax + x^2 = y
$$
Solve this quadratic equation for a and check that a is an integer.
Obviously many factors need not be examined: If we keep the sum $x = a+b$ fixed, then $a^3 + b^3$ is between $x^3/4$ and $x^3$, so $x$ must be between $c^{1/3}$ and $(4c)^{1/3}$.
After reading previous posts, this seems to be a published result, cited as (Broughan, 2003). Had a look at the paper. Oh well, you can sure make this complicated.
For not very large c, say $c < 10^9$, brute force may be quicker: Let a = 0, b = $c^{1/3}$, rounded down. As long as $a^3 + b^3 < c$ increase a by 1. If $a^3 + b^3 = c$ note that you found a solution. Then decrease b by 1, repeat until $a > b$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inverse of Symmetric Matrix Plus Diagonal Matrix if Square Matrix's Inverse Is Known Let $A$ be an $n \times n$ symmetric matrix of rank $n$ with known inverse $A^{-1}$. Let $D$ be a diagonal matrix with the same dimensions and rank. What is the fastest way to compute $(A+D)^{-1}$? Assume all diagonal elements of $A+D$ are positive and that $A+D$ is invertible.
| Since $(A+D)^{-1} = (I+ A^{-1} D)^{-1} A^{-1}$, and $A^{-1}$ is known, we just need to find out how to calculate $B ^{-1}$ where $B$ is the (known) matrix $1+A^{-1} D$.
Following Greg Martin's observation, consider first the case where the only nonzero entry of $D$ is in the upper left corner. In this case, all columns of $A^{-1} D$ are zero except the first one so $B$ has the form
$$ B = \begin{pmatrix}
v_1 & 0 & 0 &\cdots & 0\\
v_2 & 1 & 0 & \cdots & 0 \\
v_3 & 0 & 1 & \cdots & 0 \\
\vdots & & & & \vdots \\
v_n & 0 & 0 & \cdots & 1 \\
\end{pmatrix} =
\begin{pmatrix}
v_1 & 0 & 0 &\cdots & 0\\
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & & & & \vdots \\
0 & 0 & 0 & \cdots & 1 \\
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 &\cdots & 0\\
v_2 & 1 & 0 & \cdots & 0 \\
v_3 & 0 & 1 & \cdots & 0 \\
\vdots & & & & \vdots \\
v_n & 0 & 0 & \cdots & 1 \\
\end{pmatrix}$$
Invertibility of $B$ is equivalent to the condition $v_1 \neq 0$. In this case both factors above are invertible. The inverse of the diagonal factor on the left is trivial to compute, and the inverse of the right factor happens to be
$$ \begin{pmatrix}
1 & 0 & 0 &\cdots & 0\\
-v_2 & 1 & 0 & \cdots & 0 \\
-v_3 & 0 & 1 & \cdots & 0 \\
\vdots & & & & \vdots \\
-v_n & 0 & 0 & \cdots & 1 \\
\end{pmatrix}$$
so the inverse of $B$ is easy to recover. Namely:
$$B^{-1} = \begin{pmatrix}
\frac{1}{v_1} & 0 & 0 &\cdots & 0\\
-\frac{v_2}{v_1} & 1 & 0 & \cdots & 0 \\
-\frac{v_3}{v_1} & 0 & 1 & \cdots & 0 \\
\vdots & & & & \vdots \\
-\frac{v_n}{v_1} & 0 & 0 & \cdots & 1 \\
\end{pmatrix}.$$
| {
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Proving $\prod_{i=1}^np_i+1$ is not a perfect square
Let $m=\displaystyle{\prod_{i=1}^np_n}$ be the product of the first $n$ primes $(n>1)$. prove that $m+1$ cannot be a perfect square.
I think that the opposite it correct: $m+1$ is not a complete square iff $(m+1)^{\frac {p-1}{2}}\equiv -1\pmod p$ but for each prime in the first $n$ primes $\displaystyle(m+1)^{\frac {p-1}{2}}=\sum_{k=0}^p\binom{n}{k}m^k$ and since $p\mid m$, $(m+1)^{\frac {p-1}{2}}\equiv 1\pmod p$ which means that allegedly it's a complete square.
Am I right?
| Every prime $p$ greater than $2$ is congruent to $1$ or $3 \pmod{4}$. Next, note that $3^2 \equiv 1 \pmod{4}$. Hence, $\prod_{i=2}^n p_i \equiv 1$ or $3 \pmod{4}$. Once we include the first prime, $2$, we get $\prod_{i=i}^n p_i \equiv 2 \pmod{4}$. Thus:
$$\prod_{i=i}^n p_i + 1 \equiv 3 \pmod{4}$$
This cannot be a perfect square since all perfect squares are equivalent to $0$ or $1 \pmod{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int \sqrt{x^2-3x+2}\ dx$ How to evaluate $$\int_3^{17} \sqrt{x^2-3x+2}\ dx \ ?$$ I tried Euler's substitution, that is $$\sqrt{x^2-3x+2}=x+t \Longleftrightarrow \frac{t^2-2}{-3-2t}+t=\frac{t^2+3t+2}{2t+3}\ ,$$ which I obtained from $$x^2+2tx+t^2=x^2-3x+2\Longleftrightarrow x=\frac{t^2-2}{-3-2t}$$ $$dx=\frac{2t(-3-2t)-(t^2-2)(-2)}{(-3-2t)^2}\ dt=-\frac{2t^2+6t+4}{(3+2t)^2} \ dt\ .$$ $x\in[3,17]$, so $t\in[\sqrt{2}-3,\sqrt{240}-17]$ (right?). Then we got $$\int_{\sqrt{2}-3}^{\sqrt{240}-17} \frac{t^2+3t+2}{2t+3}(-\frac{2t^2+6t+4}{(2t+3)^2})\ dt=-2\int_{\sqrt{2}-3}^{\sqrt{240}-17} \frac{(t^2+3t+2)^2}{(2t+3)^3}\ dt$$ and it's where I stuck...
EDIT
Ok I finished it my way, but I've heard there's a possibility to do it using $(x-\frac{3}{2})=\tan (t)$ substitution. How exactly?
| An Euler Substitution will work, remember that there are several such substitutions and you have the root of a quadratic which factors. Another approach in addition to the excellent idea above is to write the integral as
$$\int (x-1)\sqrt{\frac{x-2}{x-1}}$$
and let
$$y^2=\frac{x-2}{x-1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Arc Length polar curve $$r=a\sin^3\left(\frac{\theta}{3}\right) $$
I tried solving it using the equation for arc length with $dr/d\theta$ and $r^2$. Comes out messy and complicated.
| The portion of the arclength calculation in which we determine the infinitesimal arclength element $ \ ds \ $ just requires applying the Chain Rule carefully and using the Pythagorean Identity:
$$ \frac{dr}{d\theta} \ = \ \frac{d}{d\theta} \left[ \ a \ \sin^3 \left(\frac{\theta}{3}\right) \right] \ = \ a \ \cdot \ 3 \ \sin^2\left(\frac{\theta}{3}\right) \cdot \cos\left(\frac{\theta}{3}\right) \cdot \frac{1}{3} \ = \ a \ \sin^2\left(\frac{\theta}{3}\right) \cos\left(\frac{\theta}{3}\right) $$
$$ \Rightarrow \ \ ds \ = \ \sqrt{ \ r^2 + \left(\frac{dr}{d\theta}\right)^2} \ \ d\theta \ = \ \sqrt{ \ \left[ \ a \ \sin^3 \left(\frac{\theta}{3}\right) \right]^2 + \left[ \ a \ \sin^2\left(\frac{\theta}{3}\right) \cos\left(\frac{\theta}{3}\right) \right]^2} \ \ d\theta $$
$$ = \ \sqrt{ \ a^2 \ \sin^6 \left(\frac{\theta}{3}\right) + \ \ a^2 \ \sin^4\left(\frac{\theta}{3}\right) \cos^2\left(\frac{\theta}{3}\right)} \ \ d\theta $$
$$ = \ \sqrt{ \ a^2 \ \sin^4 \left(\frac{\theta}{3}\right) \left[ \ \sin^2 \left(\frac{\theta}{3}\right) + \ \cos^2\left(\frac{\theta}{3}\right) \right]} \ \ d\theta $$
$$ = \ \sqrt{ \ a^2 \ \sin^4 \left(\frac{\theta}{3}\right) } \ \ d\theta \ = \ a \ \sin^2 \left(\frac{\theta}{3}\right) \ d\theta \ \ . $$
What turns out to be the tricky part is in finding the limits of integration in order to cover this curve fully. Here is a graph of the curve described by this polar equation (for $ \ a \ = \ 1 ) \ $ :
A graph of $ \ r \ $ as a function of $ \ \theta \ $ shows that, while the period of the function is $ \ 6 \pi \ $ , the radius is negative in the interval $ \ ( \ 3 \pi \ , \ 6 \ \pi \ ) \ $ .
The curve resembles a limaçon , but the interval of negative radii retraces the curve already "swept out" by positive radii in $ \ ( \ 0 \ , \ 3 \pi \ ) \ $ . Thus, we will find the arclength correctly by integrating arclength over this interval, rather than only up to $ \ 2 \pi \ $ or all the way to $ \ 6 \pi \ $ .
Our arclength integral is therefore
$$ \int_0^{3 \pi} \ ds \ \ = \ \ \int_0^{3 \pi} a \ \sin^2 \left(\frac{\theta}{3}\right) \ \ d\theta \ \ = \ \ a \ \int_0^{3 \pi} \ \frac{1}{2} \left[ \ 1 \ - \ \cos \left(\frac{2\theta}{3}\right) \right] \ \ d\theta $$
$$ = \ \ \frac{a}{2} \left[ \ \theta \ - \ \frac{3}{2} \sin \left(\frac{2\theta}{3}\right) \right] \vert_0^{3 \pi} \ = \ \frac{a}{2} \left[ \ \left(3 \pi \ - \ \frac{3}{2} \sin \left[\frac{6 \pi}{3}\right] \ \right) \ - \ \left(0 \ - \ \frac{3}{2} \sin \left[\frac{0}{3}\right] \right) \right] $$
$$ = \ \frac{a}{2} \ \left(3 \pi \ - \ \frac{3}{2} \sin \ 2 \pi \ \ - \ 0 \ + \ 0 \right) \ = \ \frac{3 \pi}{2} a \ \ . $$
As a check, we can see from the graph above that the arclength of the curve will be somewhat larger than the circumference of a circle of diameter $ \ \frac{5}{4} a \ $ , which is $ \ \frac{5 \pi}{4} a \ $ (the small "self-crossing" loop of this curve does not add much to the total arclength).
| {
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Is this a theorem (is it correct?)? My instruction notes have specified a theorem of matrix transpose that be there two compatible matrices $A$ and $B$ in respect of their sums and products, then:
$(AB)^T =A^TB^T$
So I set on to verify if indeed this is correct:
Let $A = \begin{pmatrix}2 & 2 \\ 6 & 4\end{pmatrix} $
Let $B = \begin{pmatrix}1 & 0 \\ 7 & 3\end{pmatrix} $
$A^T=\begin{pmatrix}2 & 6 \\ 2 & 4\end{pmatrix} $
$B^T=\begin{pmatrix}1 & 7 \\ 0 & 3\end{pmatrix} $
$(AB) = \begin{pmatrix}2 & 6 \\ 2 & 4\end{pmatrix}\begin{pmatrix}1 & 7 \\ 0 & 3\end{pmatrix}$=$\begin{pmatrix}16 & 6 \\ 34 & 12\end{pmatrix} $
hence $(AB)^T=\begin{pmatrix}16 & 34 \\ 6 & 12\end{pmatrix}$
and....
$A^TB^T=\begin{pmatrix}2 & 6 \\ 2 & 4\end{pmatrix}\begin{pmatrix}1 & 7 \\ 0 & 3\end{pmatrix}=\begin{pmatrix}2 & 32 \\ 2 & 26\end{pmatrix}$
$\therefore (AB)^T\neq A^TB^T $
Is the theorem wrong or am having a rough time understanding?
| The assertion is false; what is true is that
$$
(AB)^T=B^TA^T
$$
which can be easily proved. Actually, the product $A^TB^T$ need not be defined when the product $AB$ is defined, so the equality doesn't make sense unless both matrices are square. However, the equality doesn't generally hold in this case.
Let's see what happens if you also have, for two square matrices, $(AB)^T=A^TB^T$:
\begin{gather}
A^TB^T=B^TA^T\\
(A^TB^T)^T=(B^TA^T)^T\\
(B^T)^T(A^T)^T=(A^T)^T(B^T)^T\\
BA=AB
\end{gather}
Conversely, it's obvious that from $AB=BA$ you can deduce $(AB)^T=A^TB^T$.
So the equality you read in your instructions is wrong.
| {
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"url": "https://math.stackexchange.com/questions/800241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Lagrange's Trigonometric Identity Lagrange's Trig identity is
$$
1+\cos\theta+\cos 2\theta +\cdots + \cos n \theta=\frac{1}{2}+\frac{\sin\frac{(2n+1)\theta}{2}}{2\sin \frac{\theta}{2}},\quad (0<\theta <2\pi).
$$
How can we prove this identity using series method and complex variables? I tried to use
$$
\sum_{n=0}^\infty z^n=\frac{1}{1-z}\quad |z|<1
$$
and writing left hand side as
$$
\sum_{n=0}^\infty (\cos \theta)^n=\frac{1}{2}+\frac{\sin\frac{(2n+1)\theta}{2}}{2\sin \frac{\theta}{2}}.
$$
Re-arranging this I get
$$
\sum_{n=0}^\infty (\cos \theta)^n-\frac{1}{2}=\frac{1}{2}+
\sum_{n=1}^\infty (\cos \theta)^n=\frac{\sin\frac{(2n+1)\theta}{2}}{2\sin \frac{\theta}{2}}.
$$
Now if I deal with the right hand side
$$
2\sin \frac{\theta}{2}=2\Im (e^{i\theta/2} ),\quad \sin\frac{(2n+1)\theta}{2}=\Im \big(e^{i(2n+1)\theta/2}\big).
$$
This is where I am stuck now. Thank you for reading
| If $S = 1+z+z^2+\cdots+z^n$, then $$S-zS=(1+z+z^2+\cdots+z^n)-(z+z^2+z^3+\cdots+z^{n+1})=1-z^{n+1}.$$
Therefore, $S=\frac{1-z^{n+1}}{1-z}$, with $z \not=1$. Equating both expressions of $S$, we have $$1+z+z^2+\cdots+z^n=\frac{1-z^{n+1}}{1-z}$$
Substitute $z=e^{i\theta}$, with $0 < \theta < 2\pi$, into the expression, and we get $$1+e^{i\theta}+e^{2i\theta}+\cdots+e^{ni\theta}=\frac{1-e^{(n+1)i\theta}}{1-e^{i\theta}}$$
The real component of the left side is $1+\cos\theta+\cos 2\theta + \cdots + \cos n\theta$. For the right side, the real component is (skipping a few steps on this one, but it is) $$\frac{1}{2}+\frac{\sin \frac{(2n+1)\theta}{2}}{2 \sin \frac{\theta}{2}}$$
If we equate the real components from both sides, we get $$1+\cos\theta+\cos 2\theta + \cdots + \cos n\theta=\frac{1}{2}+\frac{\sin \frac{(2n+1)\theta}{2}}{2 \sin \frac{\theta}{2}}$$
| {
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"source": "stackexchange",
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} |
Solving $df/dx = x\log x$ I have the following differential equation
$$\frac{df}{dx} = Kx\log x,$$
$K$ a constant. I'm wandering how one might solve for $f$.
| Integrating by parts we know that:
$$fg = \int f'g+\int fg'$$
Now take $f' = x,\; g=\log x$ (I use $e$ as the base for the logarithm). This means that $f= \frac{1}{2}x^2$ and $g'= \frac{1}{x}$ , so we get:
$$\int f'g = \int x \log x \;dx = \frac{1}{2}x^2\log x - \int \frac{1}{2}x^2\frac{1}{x}\;dx$$
$$= \frac{1}{2}x^2\log x - \frac{1}{2}\int x\;dx = \frac{1}{2}x^2\log x -\frac{1}{4}x^2 + C$$
So the answer is:
$$K\left( \frac{1}{2}x^2\log x -\frac{1}{4}x^2 \right) + C$$
| {
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} |
Show how to compute $2^{343}$ using the least multiplication. Show how to compute $2^{343}$ using the least multiplication.
| If you have $x$, you can calculate $x^7$ in four integer multiplications:
*
*$x \cdot x = x^2$
*$x^2 \cdot x^2 = x^4$
*$x^4 \cdot x^2 = x^6$
*$x^6 \cdot x = x^7$.
Since $343=7^3$, this means you can calculate $2^{343}$ in twelve multiplications: four to go from $2$ to $2^7$, four to go from $2^7$ to $(2^7)^7=2^{49}$, and four to go from $2^{49}$ to $(2^{49})^7=2^{343}$.
| {
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"source": "stackexchange",
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Double integral for $\int_{0}^{1} \int_{-1}^{0} \frac {xy}{x^2 + y^2 + 1}\ dy\ dx$ I'm trying to evaluate this
$$\int_{0}^{1} \int_{-1}^{0} \frac {xy}{x^2 + y^2 + 1}\ dy\ dx$$
tried substition
$$ u = {(x^2+y^2+1)}^{-1} \ \ du = \ln {(x^2+y^2+1)}$$
but du is not found in the given equation. I have the feeling that I should use arctan but I do not know how to apply that in two variables.
| $$\int_0^1\int_{-1}^0\frac{xy}{x^2+y^2+1}dydx = \int_0^1x\int_{-1}^0y(x^2+y^2+1)^{-1}dydx\\= \int_0^1\frac{x}{2}\int_{-1}^0(x^2+y^2+1)^{-1}(2y)dydx \\= \int_0^1\frac{x}{2}\log(x^2+y^2+1)_{y=-1}^{0} dx \\= \int_0^1\frac{x}{2}(\log(x^2+1)-\log(x^2+2))dx \\= \frac{1}{4}\int_0^1\log(x^2+1)(2x)dx-\frac{1}{4}\int_0^1\log(x^2+2)(2x)dx\\=\frac{1}{4}(-x^2-1+(x^2+1)\log(x^2+1))-\frac{1}{4}(-x^2-2+(x^2+2)\log(x^2+2))_{x=0}^1\\=\frac{1}{4}(4 \log(2)-3\log(3))$$
| {
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First order recurrence relation I have to solve this relation:
$$a_1 = k \\ a_n = \frac{10}{9} a_{n-1} + k + 1 - n$$
(k is constant)
How can I do it??
| Use the generating function $A(x) = \sum_{n \ge 1}a_nx^n$ to capture the sequence $\{a_n\}$.
We are given (after adjusting indices $n \rightarrow n + 1$)
$$a_{n+1} = \frac{10}{9}a_n + k - n \text{ for } n \ge 1$$
Multiply by $x^n$ throughout and sum for all values of $n$ for which the recurrence holds.
$\begin{eqnarray}&\sum_{n \ge 1}a_{n+1}x^n &= \frac{10}{9}\sum_{n \ge 1}a_{n}x^n + k\sum_{n \ge 1}x^n - \sum_{n \ge 1}nx^n\\\implies&\frac{A(x) - a_1x}{x} &=\frac{10}{9}A(x) + k\frac{x}{1-x} - \frac{x^2}{(1-x)^2} - \frac{x}{1-x}\\\implies&\frac{9 - 10x}{9x}A(x) &= k + (k-1)\frac{x}{1-x} - \frac{x^2}{(1-x)^2}\\\implies &A(x) &=\frac{9kx}{9-10x} + \frac{9(k-1)x^2}{(1-x)(9-10x)} - \frac{9x^3}{(1-x)^2(9-10x)}\end{eqnarray}$
Partial fraction decomposition gives
$$A(x) = \frac{81(k-9)}{9-10x} + \frac{9(8-k)}{1-x} +\frac{9}{(1-x)^2}$$
We thus have,
$$\begin{align}a_n &= [x^n]A(x)\\&=[x^n]\frac{81(k-9)}{9-10x} + [x^n]\frac{9(8-k)}{1-x} +[x^n]\frac{9}{(1-x)^2}\\&=9(k-9)\left(\frac{10}{9}\right)^n + 9(8-k) + 9(n+1)\end{align}$$
Note:
*
*The coefficient of $x^n$ in $\frac{c}{(1-\alpha x)^m} = c\binom{-m}{n}\alpha^n = c\binom{m + n - 1}{n}\alpha^n$
*The sum $\sum_{n \ge 1}nx^n = x\sum_{n \ge 1}nx^{n-1} = x\frac{d}{dx}\sum_{n \ge 1}x^n = x\frac{d}{dx}(\frac{x}{1-x})$
| {
"language": "en",
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"source": "stackexchange",
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Cant solve this differential equation using laplace transforms $$\left\{\begin{array}{ccc}
y''(t) &=& x(t) - 2y(t)\\
x''(t) &=& - 2x(t) + y(t)
\end{array}\right.$$
$y(0)=1,\;x(0)=1,\; y'(0) = \sqrt3,\; x'(0) = -\sqrt3$.
I am trying to solve this differential system using laplace transforms but the terms make it very hard to calculate it. I got
$$
x = \mathcal{L}^{-1} \left\{\frac{s^3 - \sqrt{3} s^2 + 3s - \sqrt{3}}{(s^2+2)^2 - 1} \right\}
$$
But that looks insanely difficult to find the laplace inverse. Is there anyway to make this simpler to do by hand?
| Taking the LT of both equations and initial values yields
$$(s^2+2) Y(s)-X(s) = s+\sqrt{3}$$
$$(s^2+2) X(s)-Y(s) = s-\sqrt{3}$$
Solution yields
$$X(s) = \frac{s}{s^2+1} - \frac{\sqrt{3}}{s^2+3} $$
$$Y(s) = \frac{s}{s^2+1} + \frac{\sqrt{3}}{s^2+3} $$
Inversion of these, individually, yields
$$x(t) = \cos{t} - \sin{\sqrt{3} t}$$
$$y(t) = \cos{t} + \sin{\sqrt{3} t}$$
Plugging this solution into the original equation yields agreement.
ADDENDUM
The algebra involved in the solution is a little tricky. To get $Y$, multiply both sides of the first equation by $s^2+2$ and add to the second equation to get
$$\left [ (s^2+2)^2-1\right ] Y(s) = (s^2+2) (s+\sqrt{3}) + s-\sqrt{3} = s(s^2+3) + \sqrt{3} (s^2+1)$$
Note that $(s^2+2)^2-1 = (s^2+1) (s^2+3)$. Dividing through yields the solution.
To get $X$, multiply the second equation through by $s^2+2$ and add to the first equation. The manipulations are virtually identical.
| {
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Using the substitution $u=x^3$, find the general solution of $xy''+y'+9x^5=0$. Using the substitution $u=x^3$, find the general solution of $xy''+y'+9x^5=0$.
I have no idea about above question? Could somebody suggest me a solution or resource for such problems?
| HINT:
$$u=x^3\implies\frac{du}{dx}=3x^2$$
Using Chain Rule, $$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=3x^2\cdot\frac{dy}{du}$$
$$\implies x\frac{dy}{dx}=3u\cdot\frac{dy}{du}$$
Differentiating wrt $x$ $$x\frac{d^2y}{dx^2}+\frac{dy}{dx}=3\left[\frac{du}{dx}\cdot\frac{dy}{du}+u\cdot\frac{d\left(\dfrac{dy}{du}\right)}{dx}\right]$$
$$=3\left[3x^2\cdot\frac{dy}{du}+u\cdot\frac{d\dfrac{dy}{du}}{du}\cdot\frac{du}{dx}\right]$$
$$\implies x\frac{d^2y}{dx^2}+\frac{dy}{dx}=9x^2\cdot\frac{dy}{du}+u\cdot\frac{d^2y}{du^2}\cdot3x^2$$
Then replace $x^3$ with $u$ and use this
| {
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A logarithmic integral $\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$ How to prove the following
$$\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx=\frac{\pi^2}{2}$$
I thought of separating the two integrals and use the beta or hypergeometric functions but I thought these are not best ideas to approach the problem.
Any other ideas ?
| Denote the integral as $I$. Define
$$ I(a)=\int_0^1 \frac{\log\left(\frac{1+ax}{1-x}\right)}{x\sqrt{1-x^2}} dx. $$
Then $I(-1)=0$, $I(1)=I$ and
\begin{eqnarray}
\frac{\partial I(a)}{\partial a}&=&\int_0^1 \frac{1}{(1+ax)\sqrt{1-x^2}}dx\\
&=&\int_0^{\frac{\pi}{2}} \frac{1}{1+a\sin t}dt \quad(x=\sin t).
\end{eqnarray}
From this post, we have
$$ \frac{\partial I(a)}{\partial a}=\frac{2}{\sqrt{1-a^2}}\arctan \left(\sqrt{\frac{1-a}{1+a}}\right) $$
and hence
\begin{eqnarray}
I&=&\int_{-1}^1\frac{2}{\sqrt{1-a^2}}\arctan \left(\sqrt{\frac{1-a}{1+a}}\right)da\quad(a=\sin t)\\
&=&2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arctan \left(\sqrt{\frac{1-\sin t}{1+\sin t }}\right)dt\\
&=&2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arctan \left(\sqrt{\frac{1+\cos(\frac{\pi}{2}+t)}{1-\cos(\frac{\pi}{2}+ t)}}\right)dt\\
&=&2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arctan \left(\sqrt{\frac{2\cos^2(\frac{\pi}{4}+\frac{t}{2})}{2\sin^2(\frac{\pi}{4}+\frac{t}{2})}}\right)dt\\
&=&2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arctan \tan(\frac{\pi}{4}+\frac{t}{2})dt\\
&=&2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\frac{\pi}{4}+\frac{t}{2})dt\\
&=&\frac{\pi^2}{2}.
\end{eqnarray}
| {
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Proving $\left(A-1+\frac1B\right)\left(B-1+\frac1C\right)\left(C-1+\frac1A\right)\leq1$ $A,B,C$ are positive reals with product 1. Prove that $$\left(A-1+\frac1B\right)\left(B-1+\frac1C\right)\left(C-1+\frac1A\right)\leq1$$
How can I prove this inequality. I just need a hint to get me started. Thanks
| Given that (w.l.o.g.) $0\lt A\le B\le C, A\cdot B\cdot C=1,$ we wish to show
$$\left(A-1+\frac1B\right)\left(B-1+\frac1C\right)\left(C-1+\frac1A\right)\leq1\tag 1$$
First, noting that $A-1+\dfrac 1B=AC+A-1$ and also $AC+A-1=(A-1)(C+1)+C,$ we transform $(1)$ as follows:
$$(AC+A-1)(AB+B-1)(BC+C-1)\\
=(A-1)(C+1)(B-1)(A+1)(C-1)(B+1)+C(AB+B-1)(BC+C-1)+A(A-1)(C+1)(BC+C-1)+B(A-1)(C+1)(A+1)(B-1)\\
=(A^2-1)(B^2-1)(C^2-1)+(1+BC-C)(BC+C-1)+(A-1)(C+1)(1+AC-A)+(1+BA-B-BC)(A+1)(B-1)$$
$$=(A^2-1)(B^2-1)(C^2-1)\\+B^2C^2-(1-C)^2+A^2C^2-(1-A)^2-C(1+AC-A)+\left(1-\frac 1A\right)(C+1)(A+1)(B-1)\tag 2$$
Can you take it forward from $(2)$?
| {
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Power Factoring Contest Question The question was as follows:
Compute the smallest positive integer $n$ such that $n^n$ has at least $1,000,000$ positive divisors.
I did some work, finding that if $n=2^a*3^b*5^c*7^d$ then the $n^n= 2^{na}*3^{nb}*5^{nc}*7^{nd}$ and that the number of positive divisors is $(an+1)(bn+1)(cn+1)(dn+1)$ I wasn't sure how to proceed after that point.
More Information: This was a team round, with 20 minutes to do it (and other problems) with no calculators allowed.
| Already $n=2\cdot 3\cdot 17= {102}$ gives us $(n+1)^3>10^6$ divisors.
Any number with at least four distinct prime divisors is $\ge 2\cdot 3\cdot 5\cdot 7>102$ and can thus be ignored: We need to check only up to three prime divisors.
If $n=p^kq^m$, we get $(kn+1)(nm+1)$ divisors (the case $m=0$ is not excluded). To make this $\ge 10^6$, one of the factors, say $kn+1$, needs to be $\ge 1000$. With $n<102$, we conclude $k>9$, but then $n\ge2^9>102$: We need only check numbers with at least, hence with exactly three prime divisors.
So $n=p^aq^br^c$ with $p,q,r$ distinct primes, $a,b,c\in\mathbb N$.
Then $a\ge3$ would imply $n\ge 2^3\cdot 3\cdot 5=120>102$. Hence alle exponents are $\le2$. If $a=b=2$, then $n\ge 2^23^25=180>102$, hence at most one exponent is $>1$, i.e. $n=pqr$ or $n=p^2qr$.
As in the first paragraph, $n=pqr$ gives us $(n+1)^3$ divisors, so we need $n\ge 99$, but $n=99, 100, 101$ are not of this form.
Remains the case $n=p^2qr$ with $(2n+1)(n+1)^2$ divisors. As $n<102$, we find $pqr\le 50$, so $pqr=2\cdot3\cdot 5=30$ or $pqr=2\cdot 3\cdot 7=42$ and that's it. Thus the $n$ we need to test are $60, 90, 84$. As $(2\cdot 60+1)(60+1)^2<10^6$, the minimal $n$ is indeed $$n=84.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove this inequality $\frac{a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}\le\cos{\frac{\pi}{n+1}}$
Let $a_{1},a_{2},\cdots,a_{n},n\ge 2$ be real numbers,show that
$$\dfrac{a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}\le\cos{\dfrac{\pi}{n+1}}$$
I think this result is interesting.
When $n=2$, clearly
$$\dfrac{a_{1}a_{2}}{a^2_{1}+a^2_{2}}\le\dfrac{1}{2}=\cos{\dfrac{\pi}{3}}=\dfrac{1}{2}$$
When $n=3$,
$$\dfrac{a_{1}a_{2}+a_{2}a_{3}}{a^2_{1}+a^2_{2}+a^2_{3}}\le\dfrac{\sqrt{2}}{2}.$$
This is true, because
$$a^2_{1}+\dfrac{1}{2}a^2_{2}\ge \sqrt{2}a_{1}a_{2},$$
$$\dfrac{1}{2}a^2_{2}+a^2_{3}\ge\sqrt{2}a_{2}a_{3}.$$
But for general $n$ I cannot prove it.
| Here is a derivation using the calculus of variations.
Given that
$$
\sum_{k=1}^na_k^2=1\tag{1}
$$
we want to find the maximum of
$$
\sum_{k=1}^na_ka_{k-1}\tag{2}
$$
where $a_0=a_{n+1}=0$.
We want to find $a_k$ so that $(2)$ is stationary, that is,
$$
\begin{align}
0
&=\sum_{k=1}^na_k\,\delta a_{k-1}+a_{k-1}\,\delta a_k\\
&=\sum_{k=1}^n(a_{k-1}+a_{k+1})\,\delta a_k\tag{3}
\end{align}
$$
for all $\delta a_k$ so that $(1)$ is constant:
$$
0=\sum_{k=1}^na_k\,\delta a_k\tag{4}
$$
For both the $a_{k-1}+a_{k+1}$ in $(3)$ and the $a_k$ in $(4)$ to be perpendicular to all the same $\delta a_k$, we must have some $\lambda$ so that
$$
a_{k+1}+a_{k-1}=\lambda a_k\tag{5}
$$
Solving $(5)$, so that $a_0=a_{n+1}=0$, using the standard methods for linear recurrences yields
$$
a_k=r\sin\left(\frac{\pi mk}{n+1}\right)\tag{6}
$$
for $m\in\mathbb{Z}$.
We can compute $r$ using $(1)$ and $(6)$:
$$
\begin{align}
\sum_{k=1}^na_k^2
&=\sum_{k=1}^nr^2\sin^2\left(\frac{\pi mk}{n+1}\right)\\
&=-\frac{r^2}{4}\sum_{k=1}^n\left[\exp\left(\frac{2\pi imk}{n+1}\right)+\exp\left(\frac{-2\pi imk}{n+1}\right)-2\right]\\
&=-\frac{r^2}{4}(-2n-2)\ \Big[m\not\equiv0\bmod{(n+1)}\Big]\\
&=r^2\frac{n+1}{2}\ \Big[m\not\equiv0\bmod{(n+1)}\Big]\tag{7}
\end{align}
$$
We cannot satisfy $(1)$ when $m\equiv0\bmod{(n+1)}$; therefore, we require $m\not\equiv0\bmod{(n+1)}$, in which case,
$$
r=\sqrt{\dfrac2{n+1}}\tag{8}
$$
Finally, we need to compute $(2)$ for each $m\not\equiv0\bmod{(n+1)}$:
$$
\begin{align}
&\frac2{n+1}\sum_{k=1}^n\sin\left(\frac{\pi mk\vphantom{()}}{n+1}\right)\sin\left(\frac{\pi m(k-1)}{n+1}\right)\\
&=\frac1{n+1}\sum_{k=1}^n\left[\cos\left(\frac{\pi\vphantom{()}m}{n+1}\right)-\cos\left(\frac{\pi m(2k-1)}{n+1}\right)\right]\\
&=\frac{n}{n+1}\cos\left(\frac{\pi m}{n+1}\right)+\frac1{n+1}\cos\left(\frac{\pi m}{n+1}\right)\\
&=\cos\left(\frac{\pi m}{n+1}\right)\tag{9}
\end{align}
$$
The largest that $(9)$ can be if $m\not\equiv0\bmod{(n+1)}$ is $\cos\left(\frac\pi{n+1}\right)$.
Therefore, by homogeneity,
$$
\frac{\displaystyle\sum_{k=1}^na_ka_{k-1}}{\displaystyle\sum_{k=1}^na_k^2}\le\cos\left(\frac\pi{n+1}\right)\tag{10}
$$
| {
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Prove that the gradient of a unit vector equals 2/magnitude of the vector Let $\vec r=(x,y,z)$
Firstly find $\vec \nabla (\frac 1 r)$ where r is the magnitude of $\vec r$.
I think I've done this correctly to get $-x(x^2+y^2+z^2)^{-\frac32} \hat i-y(x^2+y^2+z^2)^{-\frac32} \hat j-z(x^2+y^2+z^2)^{-\frac32} \hat k$
Secondly prove that $\vec \nabla. \frac{\vec r}{r}=\frac2r$
I've really got no idea for the second part.
| I'm unsure of the context to this problem for you, but it first came up in my E+M physics class, but I'll try to create a more rigorous proof than I was initially shown.
First we have
$$
\vec{r} = (x,y,z) \implies r = \|r\| = \sqrt{x^2+y^2+z^2} \\
\implies \frac{\vec{r}}{r} =
\left(\frac{x}{\sqrt{x^2+y^2+z^2}}, \frac{y}{\sqrt{x^2+y^2+z^2}}, \frac{z}{\sqrt{x^2+y^2+z^2}} \right)
$$
Now as you know in cartesian coordinates the divergence is the sum of the partial derivatives of the components with respect to the respective basis vector:
$$
\begin{eqnarray*}
\frac{\partial \frac{\vec{r}}{r}_x}{\partial x} & = & (x^2+y^2+z^2)^{-1 / 2}+x(-1 / 2)(x^2 + y^2 + z^2)^{-3/2}(2x) \\
& = & \frac{1}{\sqrt{x^2+y^2+z^2}} - \frac{x^2}{(x^2+y^2+z^2)^{3/2}} \\
& = & \frac{x^2 + y^2 + z^2}{(x^2+y^2+z^2)^{3/2}} - \frac{x^2}{(x^2+y^2+z^2)^{3/2}} \\
& = & \frac{y^2 + z^2}{r^3}
\end{eqnarray*}
$$
Now when we do the same thing above for $y$ (resp. $z$) we're going to get the same denominator but the numerator will consist of the sum of the components squared except the coordinate we're taking the derivative with respect to since the second term will always get said component squared and thus you'll get cancelation with the first term, so we get*:
$$
\frac{\partial \frac{\vec{r}}{r}_y}{\partial y} = \frac{x^2 + z^2}{r^3} \\
\frac{\partial \frac{\vec{r}}{r}_z}{\partial y} = \frac{x^2 + y^2}{r^3}
$$
thus we get
$$
\begin{eqnarray*}
\vec{\nabla} \cdot \frac{\vec{r}}{r} & = & \frac{y^2 + z^2}{r^3} + \frac{x^2+z^2}{r^3} + \frac{x^2 + y^2}{r^3} \\
& = & \frac{2 \left(x^2 + y^2 + z^2 \right)}{r^3} \\
& = & \frac{2 r^2}{r^3} \\
& = & \frac{2}{r}
\end{eqnarray*}
$$
Let me know if you want me to clarify (*). Note I can put this in some sort of summation notation if that would be easier to understand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/807853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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} |
Log trig integral with radical Show that:
$$\int_{0}^{\Large\frac{\pi}{2}}\frac{\log(\sin\theta)}{\sqrt{1+\sin^{2}\theta}}\ d\theta=-\frac{\Gamma^{2}\left(\dfrac14\right)\sqrt{\dfrac\pi2}}{16},$$ or some other equivalent form.
| Consider the integral
\begin{align}
I = \int_{0}^{\pi/2} \frac{\ln(\sin\theta)}{\sqrt{1+\sin^{2}\theta}} \ d\theta.
\end{align}
By making the substitution $t = \sin\theta$ the integral is reduced to
\begin{align}
I = \int_{0}^{1} \ln(t) \ (1-t^{4})^{-1/2} \ dt.
\end{align}
Making the substitution $u = t^{4}$ yields
\begin{align}
I &= \frac{1}{16} \int_{0}^{1} \ln(u) \ u^{-3/4} \ (1-u)^{-1/2} \ du \\
&= \frac{1}{16} \partial_{\mu} \left. \int_{0}^{1} u^{\mu-1} (1-u)^{-1/2}
\ du \right|_{\mu=1/4} \\
&= \frac{1}{16} \ B(1/4, 1/2) \left[ \psi(1/4) - \psi(3/4) \right].
\end{align}
Using
\begin{align}
\psi(1/4) &= - \gamma - \frac{\pi}{2} - 3 \ln(2) \\
\psi(3/4) &= - \gamma + \frac{\pi}{2} - 3 \ln(2) \\
\Gamma(3/4) &= \frac{\sqrt{2} \pi}{\Gamma(1/4)}
\end{align}
leads to
\begin{align}
I = - \frac{\Gamma^{2}\left( \frac{1}{4} \right)}{16} \sqrt{\frac{\pi}{2}}.
\end{align}
Hence
\begin{align}
\int_{0}^{\pi/2} \frac{\ln(\sin\theta)}{\sqrt{1+\sin^{2}\theta}} \ d\theta
= - \frac{\Gamma^{2}\left( \frac{1}{4} \right)}{16} \sqrt{\frac{\pi}{2}}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/808193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the difference to compute double integral? I see a double integral on web with a strange way to calculate. Please help me to make it clear.
Here is the integral:
$$\int_1^2\int_1^2(x+y)\ dx\ dy$$
As my way, I calculate it:
$$\int_1^2\int_1^2(x+y)\ dx\ dy= \int_1^2\left. {\left( {\frac{{{x^2}}}{2} + xy} \right)} \right|_{x = 1}^{x = 2}dy =\int_1^2 (\frac{3}{2}+y)\ dy $$.
But in the solution, they work is very strange:
\begin{array}{l}
\int\limits_1^2 {\int\limits_1^2 {\left( {x + y} \right)dxdy = \frac{1}{2}\int\limits_1^2 {\left. {{{\left( {x + y} \right)}^2}} \right|_{x = 1}^{x = 2}dy} } } \\
= \frac{1}{2}\int\limits_1^2 {\left[ {{{\left( {y + 2} \right)}^2} - {{\left( {y + 1} \right)}^2}} \right]dy} \\
= \frac{1}{6}\left[ {{{\left( {y + 2} \right)}^3} - {{\left( {y + 1} \right)}^3}} \right]_{y = 1}^{y = 2}
\end{array}.
Two way out two results, but their result is right.
Pls help me explain?
| The second one approach is using substitution $u=x+y\;\Rightarrow\;du=dx$, then the inner integral turns out to be
$$
\int_{1}^2 (x+y)\ dx = \int_{x=1}^2 u\ du = \left.\frac12u^2\right|_{x=1}^2 = \left.\frac12(x+y)^2\right|_{x=1}^2=\frac12\left[(y+2)^2-(y+1)^2\right].
$$
Similarly, the next integral is solved by using substitution $u=y+2\;\Rightarrow\;du=dy$ and $v=y+1\;\Rightarrow\;dv=dy$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/808536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Find values of the parameter a so that equation has equal roots. $x^2+2a\sqrt{a^2-3}x+4=0$
My final result was 2 and -0.5. Was it correct?
| To get a pair of equal roots, we need to have $2a\sqrt{a^2 - 3} = 4$.
$$\begin{align} 2a\sqrt{a^2-3}=4 &\iff \sqrt{a^4 - 3a^2}= 2 \\ \\ &\iff a^4 - 3a^2 = 4\\ \\ & \iff a^4 - 3a^2 - 4 = 0 \\ \\ & \iff (a^2 - 4)(a^2 +1) = 0\end{align} $$
$a^2 + 1 \geq 1$, so can never be zero. So we solve $$a^2 - 4 = (a+2)(a-2)=0$$ $$ \iff a = -2 \text{ or } a = 2$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Asymptotic expansions for the roots of $\epsilon^2x^4-\epsilon x^3-2x^2+2=0$ I'm trying to compute the asymptotic expansion for each of the four roots to the following equation, as $\epsilon \rightarrow 0$:
$\epsilon^2x^4-\epsilon x^3-2x^2+2=0$
I'd like my expansions to go up through terms of size $O(\epsilon^2)$.
IΒ΄ve made the change of variables $x=\delta y$, performed dominant balance and found out that the only two valid options are: (rescaled eq: $\epsilon^2\delta^4y^4-\epsilon \delta^3 y^3-2\delta^2 y^2+2=0$)
*
*$\delta^2 \sim 2 \Rightarrow \delta=O(1)$
*$\epsilon^2\delta^4 \sim \epsilon \delta^3 \sim \delta^2 \Rightarrow \delta=O(\epsilon^{-1})$
How do I proceed?
| Two of the roots will be near the roots $1$ and $-1$ of $-2x^2 + 2$, the other two will be near $\infty$. Solving numerically for $\epsilon = .01$,
those roots are approximately $-100$ and $200$.
So it looks like you want $x \sim c \epsilon^{-1}$. Indeed with $x = y/\epsilon$ the equation becomes $y^4 - y^3 - 2 y^2 + 2 \epsilon^2 = 0$.
EDIT: At $\epsilon = 0$ the second equation becomes $y^4 - y^3 - 2 y^2 = 0$, which has roots $y = -1$ and $y = 2$ as well as $y=0$ (which corresponds to $x=1$ and $x=-1$). For the solution near $y=-1$, for example, take $y = -1 + v$ and expand to get $ v^4 - 5 v^3 + 7 v^2 - 3 v + 2 \epsilon^2 = 0$ where $v$ should be small. To first order, $v = (2/3) \epsilon^2$. Take $v = (2/3) \epsilon^2 + w$
and the equation becomes
$$
{w}^{4}+ \left( \dfrac{8 \epsilon^2}{3}-5 \right) {w}^{3}+ \left(\dfrac{8 {
\epsilon}^{4}}{3}-10\,{\epsilon}^{2}+7 \right) {w}^{2}+ \left( {\frac {32
\,{\epsilon}^{6}}{27}}-{\frac {20\,{\epsilon}^{4}}{3}}+{\frac {28\,{
\epsilon}^{2}}{3}}-3 \right) w+{\frac {16\,{\epsilon}^{8}}{81}}-{
\frac {40\,{\epsilon}^{6}}{27}}+{\frac {28\,{\epsilon}^{4}}{9}} = 0
$$
from which $w = (28/27) \epsilon^4 + \ldots$. Thus this solution is
$$ x = -\epsilon^{-1} + \dfrac{2}{3} \epsilon + \dfrac{28}{27} \epsilon^3 + \ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Solving the Integral using $\ln|u|$ Can Someone help me solve this
$$
\int\frac{19\tan^{-1}x}{x^{2}}\,dx
$$
We have been told to use $\ln|u|$ and $C$.
Thanks!
| Using integration by parts we see:
$$ \int \frac{ \arctan(x) }{x^2} dx = -\frac{\arctan(x) }{x} + \int \frac{dx}{ (x^2+1) x} $$
Now if we use a partial fraction decomposition we obtain:
$$\int \frac{dx}{ (x^2+1) x} = \int \left ( \frac{1}{x} - \frac{x}{x^2 +1} \right ) dx $$
From here it should be clear that
$$\int \frac{ \arctan(x) }{x^2} dx = \ln (x) - \frac{1}{2} \ln (x^2 +1) - \frac{\arctan(x)}{x} +C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/813209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there always an alternative way to compute integral other than complex integral? Sometimes we have to compute integrals that are not easy to calculate so that we need to depend on the method of complex integrals like the residue method. But I became curious about possibility of alternative method of evaluation of integrals other than complex integrals.
For example, do we have an ordinary, 'real' method of integral evaluation method for calculating
$$\int_{0}^{\infty}x^{1-\alpha}\cos(\omega x)dx$$
or
$$\int_{-\infty}^{\infty}{\cos(ax)\over{b^2-x^2}}dx$$
?
In this question I mean 'real' method in the sense that one does not visit the complex plane to evaluate the integral.
| For $1 <a <2$,
$$ \begin{align} \int_{0}^{\infty} x^{1-a} \cos (wx)\ dx &= w^{a-2} \int_{0}^{\infty} u ^{1-a} \cos (u) \ du \\ &= \frac{w^{a-2}}{\Gamma(a-1)} \int_{0}^{\infty} \int_{0}^{\infty} \cos (u) \ t^{a-2} e^{-ut} \ dt \ du
\\ &= \frac{w^{a-2}}{\Gamma(a-1)} \int_{0}^{\infty}t^{a-2} \int_{0}^{\infty} \cos (u)e^{-tu} \ du \ dt \\ & = \frac{w^{a-2}}{\Gamma(a-1)} \int_{0}^{\infty} t^{a-2} \frac{t}{1+t^{2}} \ dt \\ &= \frac{w^{a-2}}{\Gamma(a-1)} \int_{0}^{\infty} \frac{t^{a-1}}{1+t^{2}} \ dt \\ &= \frac{w^{a-2}}{\Gamma(a-1)} \frac{1}{2} \int_{0}^{\infty} \frac{v^{\frac{a}{2}-1}}{1+v} \ dv \\ &= \frac{w^{a-2}}{\Gamma(a-1)} \frac{1}{2} B \left(\frac{a}{2}, 1- \frac{a}{2} \right) \\ &= \frac{w^{a-2}}{\Gamma(a-1)} \frac{\pi}{2} \csc \left(\frac{\pi a}{2} \right) \\ &= \frac{w^{a-2}}{\Gamma(a-1)} \frac{\pi}{2} \frac{2 \cos \left(\frac{\pi a}{2} \right)}{\sin (\pi a)} \\ &= w^{a-2} \frac{\Gamma(a) \Gamma(1-a) \cos \left(\frac{\pi a}{2} \right)}{\Gamma(a-1)} \\ &= w^{a-2} \ (a-1) \Gamma(1-a) \cos \left(\frac{\pi a}{2} \right) \\ &=- w^{a-2} \ \cos \left(\frac{\pi a}{2} \right) \Gamma(2-a) \end{align}$$
which is the answer given by Wolfram Alpha
If you want to be more rigorous, integrate by parts at the beginning and choose $1+ \sin u$ for the antiderivative of $\cos u$. Then when you switch the order of integration, it's easily justified by Tonelli's theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/814854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Compute and find 2009th decimal(2009th digit after the point), without automation, the following sum Compute and find 2009th decimal of (2009th digit after the point), without automation, the following sum
$$\frac{10}{11}+\frac{10^2}{1221}+\frac{10^3}{123321}+ \cdots +\frac{10^9}{123456789987654321}$$
| Let
$$a_n=\sum\limits_{b=1}^n\frac{10^b}{12\ldots bb\ldots 21}$$
where $b$ is a digit less than or equal to $9$.
We prove inductively that $a_n=1-\dfrac{1}{\underbrace{11\ldots11}_{n+1}}$.
Base case: $\dfrac{10}{11}=1-\dfrac{1}{11}$
Now assume
$$a_k=1-\dfrac{1}{\underbrace{11\ldots11}_{k+1}}$$
then
$$\begin{align}
a_{k+1}&=1-\dfrac{1}{\underbrace{11\ldots11}_{k+1}}+\frac{10^{k+1}}{12\ldots (k+1)(k+1)\ldots 21}
\\&=1-\dfrac{1}{\underbrace{11\ldots11}_{k+1}}+\frac{10^{k+1}}{\underbrace{(11\ldots11)}_{k+1}~\underbrace{(11\ldots11)}_{k+2}}
\\&=1-\frac{\overbrace{(11\ldots11)}^{k+2}-10^{k+1}}{\underbrace{(11\ldots11)}_{k+1}~\underbrace{(11\ldots11)}_{k+2}}
\\&=1-\frac{\overbrace{(11\ldots11)}^{k+1}}{\underbrace{(11\ldots11)}_{k+1}~\underbrace{(11\ldots11)}_{k+2}}
\\&=1-\frac{1}{\underbrace{(11\ldots11)}_{k+2}}
\end{align}$$
as desired.
Now we just need to find the $2009$th decimal of
$$a_9=1-\frac{1}{\underbrace{(11\ldots11)}_{10}}$$
but this is easy because this decimal is just $0.\overline{9999999990}$ $~~$ ($9$s everywhere except for every $10$th digit which is a $0$). Since $10\not|2009$, the digit we want is a $9$.
Note the easy decimal expansion comes from the fact that $\dfrac{1}{\underbrace{99\ldots99}_{n}}=0.\overline{\underbrace{00\ldots00}_{n-1}1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Why does $\frac{(x^2 + x-6)}{x-2} = \frac{(x+3)(x-2)}{x-2}$? I'm not the best at algebra and would be grateful if someone could explain how you can get from,
$$\frac{x^2 + x-6}{x-2}$$
to,
$$\frac{(x+3)(x-2)}{x-2}$$
| Try working backwards. Notice that the denominator does not change, so let's focus on the numerator:
$$(x+3)(x-2) = x\cdot x-2x+3x-2\cdot 3 \,\,\,\,\,\,\text{FOIL}.$$
If we look at like terms (meaning equal powers of $x$), we see that we have two pieces that have $x$ that we can simplify: $-2x+3x = (-2+3)x = x.$ As for the other bits, we have $x\cdot x$ which we can simplify to $x^2$ and also $2\cdot 3 = 6$. This gives us
$$(x+3)(x-2) = x^2+x-6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/817424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to find the following sum? $\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} $ I want to calculate the sum with complex analysis (residue)
$$
1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ...
$$ $$
1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = 1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ...
$$
I ask
$$
f\left( z \right) = - \frac{2}{{\left( {4z + 9} \right)\left( {4z + 7}\right)}}
$$
is to :
$$\sum\limits_{n = - \infty }^\infty {\frac{2}{{\left( {4n + 9} \right)\left( {4n + 7}\right)}}} = \left( {\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] + \mathop {\lim }\limits_{z \to - \frac{7}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}}\right] } \right)$$
I found:
\begin{array}{l}
\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{9}{4}} \right)\left( {4z + 7} \right)}}} \right] \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\frac{{\pi \cot \left( {\pi z} \right)}}{{4z + 7}}} \right] = \frac{1}{4}\left[ {\frac{{ - \pi }}{{ - 2}}} \right] = \frac{\pi }{8} \\
\mathop {\lim }\limits_{z \to - \frac{7}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{7}{4}} \right)\left( {4z + 9} \right)}}} \right] \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)}}} \right] = \frac{\pi }{8} \\
\end{array}
\begin{array}{l}
\sum\limits_{n = - \infty }^\infty {\frac{2}{{\left( {4n + 9} \right)4n + 7}}} = - \frac{\pi }{4} = - \left( {\frac{\pi }{8} + \frac{\pi }{8}} \right) \\
\Rightarrow s = 1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = 1 - \frac{\pi }{8} = \frac{{7 - \pi }}{8} \\
\end{array}
I have a question for the result
$$\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = - \frac{1}{5} \Rightarrow s = 1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = \frac{4}{5} \ne \frac{{7 - \pi }}{8}$$
thank you in advance
| Here is a method without complex analysis. I use the following two:
$$\int_0^1 x^{4n+8}\,dx=\frac{1}{4n+9}$$
$$\int_0^1 x^{4n+6}\,dx=\frac{1}{4n+7}$$
to get:
$$\sum_{n=0}^{\infty} \left(\frac{1}{4n+9}-\frac{1}{4n+7}\right)=\int_0^1 \sum_{n=0}^{\infty} \left(x^{4n+8}-x^{4n+6}\right)\,dx=\int_0^1 \frac{x^8-x^6}{1-x^4}\,dx$$
$$\Rightarrow \int_0^1 \frac{x^8-x^6}{1-x^4}\,dx=\int_0^1 \frac{-x^6}{1+x^2}\,dx=-\left(\int_0^1 \frac{1+x^6-1}{1+x^2}\,dx\right)$$
$$=-\int_0^1 \frac{1+x^6}{1+x^2}\,dx+\int_0^1 \frac{1}{1+x^2}\,dx$$
Write $1+x^6=(1+x^2)(1-x^2+x^4)$ to obtain:
$$-\int_0^1 (x^4-x^2+1)\,dx+\int_0^1 \frac{1}{1+x^2}\,dx$$
Both the integrals are easy to evaluate, hence the final answer is:
$$\boxed{\dfrac{\pi}{4}-\dfrac{13}{15}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/817911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Limits of trig functions How can I find the following problems using elementary trigonometry?
$$\lim_{x\to 0}\frac{1β\cos x}{x^2}.$$
$$\lim_{x\to0}\frac{\tan xβ\sin x}{x^3}.
$$
Have attempted trig identities, didn't help.
| HINT :
\begin{align}
\lim_{x\to 0}\frac{1β\cos x}{x^2}&=\lim_{x\to 0}\frac{1β\cos x}{x^2}\cdot\frac{1+\cos x}{1+\cos x}\\
&=\lim_{x\to 0}\frac{1β\cos^2 x}{x^2(1+\cos x)}\\
&=\lim_{x\to 0}\frac{\sin^2 x}{x^2(1+\cos x)}\\
&=\lim_{x\to 0}\frac{\sin x\cdot\sin x}{x\cdot x\cdot(1+\cos x)}\\
&=\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac{1}{1+\cos x}\\
&=1\cdot1\cdot\frac1{1+1}
\end{align}
and
\begin{align}
\lim_{x\to 0}\frac{\tan xβ\sin x}{x^3}&=\lim_{x\to 0}\frac{\tan xβ\sin x}{x^3}\cdot\frac{\cos x}{\cos x}\\
&=\lim_{x\to 0}\frac{\sin x(1β\cos x)}{x^3\cos x}\\
&=\lim_{x\to 0}\frac{\sin x(1β\cos x)}{x^3\cos x}\cdot\frac{1+\cos x}{1+\cos x}\\
&=\lim_{x\to 0}\frac{\sin^3 x}{x^3\cos x(1+\cos x)},\\
\end{align}
where $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/818000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$ Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$
My approach :
I used $\sin A +\sin B = 2\sin(A+B)/2\times\cos(A-B)/2 $
$\Rightarrow \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7} = 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(8\pi/7) $
$= 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(\pi + \pi/7) $
$= 2\sin(3\pi)/7\times\cos(\pi/7)-\sin(\pi/7) $
I am not getting any clue how to proceed further or whether it is correct or not. Please help thanks..
| You can evaluate this by using complex methods.
Let $\alpha=e^{2\pi i/7}$. Then
$$\sin\frac{2\pi}{7}=\frac{\alpha-\alpha^{-1}}{2i}\ ,\quad
\sin\frac{4\pi}{7}=\frac{\alpha^2-\alpha^{-2}}{2i}\ ,\quad
\sin\frac{8\pi}{7}=\frac{\alpha^4-\alpha^{-4}}{2i}\ .$$
Now let $S$ be the sum of these three numbers. Write $S$ in terms of powers of $\alpha$ and calculate $S^2$. Initially it's a mess, but you can use the relations
$$\alpha^7=1\quad\hbox{and}\quad \alpha^6+\alpha^5+\cdots+\alpha=-1$$
to show that it simplifies, amazingly, to
$$S^2=\frac{7}{4}\ .$$
It's not hard to see that $S$ is positive, so
$$S=\frac{\sqrt7}{2}\ .$$
Comment. You can also write the sum as
$$S=\frac{1}{2i}\sum_{k=1}^6 \Bigl(\frac{k}{7}\Bigr)\alpha^k\ ,$$
where $(\frac{k}{7})$ is a Legendre symbol, and this connects the sum with some very interesting and often difficult mathematics.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/818749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
} |
Finding the equation of the tangent I was working on the following question:
The curve $C$ has equation $y=(x+3)^2$ and the point $A$, with $x$ coordinate $-5$, and lies on $C$.
* $a)$ Find the equation of the tangent line to $C$ at $A$, giving your answer in the form $y= mx+c$
* $b)$ Another point $B$ also lies on $C$. The tangents to $C$ at $A$ and $B$ are parallel. Find the $x$ coordinate of $B$.
$\frac{d}{dx}(x+3)Β² = 2x+6 \text{ for } x = -5; 2x+6 = -4 = m$
I'm a bit lost :/
| The slope of the tangent line, at $A$ is equal to $y\prime(-5)$. Then you can just plug that back into the equation $y-y_1=m(x-x_1)$ and solve for $y$.
$$\frac{d}{dx}(x+3)^2 = 2(x+3) = 2x + 6 \implies y\prime(-5) = -10+6 = -4$$
$$y-y(-5) =y\prime(-5)(x+5)$$
$$y-4 = -4(x+5) \implies y-4=-4x-20$$
$$\boxed{y=-4x-16}$$
For part b, you have to find another point where the slope of the graph is $-4$, and therefore where $y\prime(x) = -4$
$$y\prime(x) = -4 \implies 2x+6 = -4 \implies 2x = -10 \implies x=-5$$
Therefore the only $x$ where the tangent has a slope of $-4$ is at $x=-5$. This means that in order to have a line tangent to $C$, and parallel to $A$, it must equal to $A$. This is why the word "another" is misleading, since there isn't a tangent line parallel to $A$ with different different $x$ and $y$ values where it is tangent to $C$.
Therefore, $B=A= (-5, 2)$ so the $x$ coordinate of the point is $-5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Answer to simple algebraic formula manipulation I have to show that
$2y+(x + 1) = 3 \cdot 2^{x+1}β (x + 1) β 2$
is equal to
$y=3β
2^xβxβ2$
I can get this far:
$2y+(x+1)=3β
2^{x+1}β(x+1)β2$
$2y+(x+1)=3β
2^{x+1}βxβ1β2$
$2y+(x+1)=3β
2^{x+1}βxβ3$
$2y+x+1=3β
2^{x+1}βxβ3$
$2y=3β
2^{x+1}βxβ3β1βx$
$2y=3β
2^{x+1}β2xβ4$
Now I should divide by 2, but the 3 on the right side throws me off. Some help would be appreciated!
| I think you missed one $x$ to be an exponent. Go ahead and divide by $2$ to get:
$$
y= 3\cdot \frac{2^{x+1}}2+x-2=3\cdot {2^{x}}+x-2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Factoring in the derivative of a rational function Given that
$$
f(x) = \frac{x}{1+x^2}
$$
I have to find
$$\frac{f(x) - f(a)}{x-a}$$
So some progressing shows that:
$$
\frac{\left(\frac{x}{1+x^2}\right) - \left(\frac{a}{1+a^2}\right)}{x-a} =
\frac{(x)(1+a^2)-(a)(1+x^2)}{(1+x^2)(1+a^2)}\cdot\frac{1}{x-a} =
\frac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)}
$$
Now, is it possible to factor $x+xa^2-a-ax^2$? I can't seem to find a way, as for simplifying the whole thing. Is there any rule I can use, and I'm unable to see?
| By the quotient rule for the difference $\ f'(x)\, :=\, \dfrac{f(x)-f(a)}{x-a}$
$$\quad \begin{eqnarray} (g/h)'(x) &=\,\ & \dfrac{\color{#c00}{g'(x)} h(a) - g(a)\color{#0a0}{h'(x)}}{h(a)h(x)}\\
\begin{array}{l}\ g(x)=x\qquad\Rightarrow\,\color{#c00}{g'(x) = 1}\\ h(x) = 1+x^2\,\Rightarrow\,\color{#0a0}{h'(x) = x+a}\\\end{array}\ \Bigg\}\!\!\!\!\!&=& \dfrac{\color{#c00}1\cdot (1+a^2)\overset{\phantom{I^I}}-a(\color{#0a0}{x+a})}{(1+a^2)(1+x^2)}\ =\ \dfrac{1-ax}{(1+a^2)(1+x^2)}\end{eqnarray}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Integrate : $\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}dx$
$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$
My approach :
Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then
$$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\tan^2x)}{x^2\tan x -x\tan^2x+x-\tan x}dx$$
But I am not getting any relation between numerator and denominator so that I will take any substitution and solve further please suggest whether it is correct and how to proceed in this. Thanks.
| $\displaystyle u=x\sin x+\cos x,v=x\cos x-\sin x \Rightarrow \displaystyle du=x\cos xdx,dv=-x\sin xdx$
$\begin{align}\displaystyle\int\frac{x^{2}dx}{(x\sin x+\cos x)(x\cos x-\sin x)}&=\int\left(\frac{x\cos x}{x\sin x+\cos x}-\frac{-x\sin x}{x\cos x-\sin x}\right)dx\\&=\displaystyle\int\frac{du}{u}-\int\frac{dv}{v}\\&=\ln\left| \frac{u}{v} \right|+c\end{align}$
$\displaystyle\int\frac{x^{2}dx}{(x\sin x+\cos x)(x\cos x-\sin x)}=\ln\left| \frac{x\sin x+\cos x}{x\cos x-\sin x} \right|+c $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/821862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 5,
"answer_id": 4
} |
Quick induction proof I am trying to prove $n^3<n!$ for all integers $n\geq 6.$ It would be trivial to do this by induction if $(n+1)^3<(n+1)n^3$ holds. I looked this up, and I found this is true for integers $n\geq 3.$ However, does anybody know of any more insightful method of proof for either statement, ie. not induction? Thank you.
| You can avoid induction if you allow division in the following way:
Let $n\geq 6$. Notice that
\begin{align*}
n!&=n(n-1)(n-2)(n-3)(n-4)(n-5)!\\
&\geq n(n-1)(n-2)3\cdot 2\cdot 1\tag{1}\\&=6n(n-1)(n-2).
\end{align*}
Now, we can make the following evaluation:
\begin{align*}
\dfrac{n^3}{n(n-1)(n-2)}=\dfrac{n}{n-1}\dfrac{n}{n-2}.\tag{2}
\end{align*}
It's easily proven that, for a given $k$, the function $f_k:n\mapsto\dfrac{n}{n-k}$ (defined for $n>k$) is decreasing. In particular, since $n\geq 6$, then
\begin{align*}
\dfrac{n}{n-1}=f_1(n)\leq f_1(6)=\dfrac{6}{5}\quad\text{and}\quad\dfrac{n}{n-2}=f_2(n)\leq f_2(6)=\dfrac{6}{4},\tag{3}
\end{align*}
so, using (2) and (3), we obtain
$$\dfrac{n^3}{n(n-1)(n-2)}\leq\dfrac{6}{5}\dfrac{6}{4}=\dfrac{36}{20}<6,\tag{4}$$
hence, by (4) and (1),
$$n^3\leq6n(n-1)(n-2)\leq n!$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/825506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Random Variable Problems? Can someone show me how to work this out? I can't get the answers in the boxes.
| Let's $Y$ be the number of spots on the dice.
Then:
$$Y = 1 \Rightarrow X = (1 - 4)^2 = 9$$
$$Y = 2 \Rightarrow X = (2 - 4)^2 = 4$$
$$Y = 3 \Rightarrow X = (3 - 4)^2 = 1$$
$$Y = 4 \Rightarrow X = (4 - 4)^2 = 0$$
$$Y = 5 \Rightarrow X = (5 - 4)^2 = 1$$
$$Y = 6 \Rightarrow X = (6 - 4)^2 = 4$$
You can write that:
$$P(X = 0) = \frac{1}{6}$$
$$P(X=1) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6}$$
$$P(X=4) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6}$$
$$P(X=9) = \frac{1}{6}$$
Also, $P(X = k) = 0$ for all $k \not \in \{0, 1, 4, 9\}$.
Now you have all the ingredients you need to finish the exercise yourself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/827065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $a+b$ is a perfect square $a, b, c$ are natural numbers such that $1/a + 1/b = 1/c$ and $gcd(a,b,c)=1$. Prove $a+b$ is a perfect square.
| $\dfrac{a+b}{ab} = \dfrac{1}{c} \to a+b = \dfrac{ab}{c}$. Thus:
$c|ab$. Write: $a = dp$, and $b = dq$ with $(p,q) = 1$. Thus: $\dfrac{ab}{c} = d^2\cdot \dfrac{pq}{c}$.
Claim: $c = pq$.
Proof: We have: $b-c = \dfrac{bc}{a}$. Thus: $a|bc \to dp|dqc \to p|qc$. Similarly:
$a - c = \dfrac{ac}{b}$. Thus: $b|ac \to dq|dpc \to q|pc$. Since: $(p,q) = 1$, we have:
$p|c$, and $q|c$. Thus: $pq|c$. So we can write: $c = kpq$. To finish the proof we show: $k = 1$. If $k > 1$, then let $m$ be a prime divisor of $k$, then: from $\dfrac{d^2pq}{c} = \dfrac{d^2pq}{kpq} = \dfrac{d^2}{k} \in \mathbb{N}$, we have: $k|d^2$. So: $m|d^2$ since $m|k$. But $m$ is a prime number, so $m|d$. So: $m|a$, $m|b$, and $m|c$, and $m > 1$. So: $(a,b,c) \geq m > 1$, contradiction. Thus: $k = 1$, and $a+b = d^2$ which is a perfect square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/828918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove infinite series $$
\frac{1}{x}+\frac{2}{x^2} + \frac{3}{x^3} + \frac{4}{x^4} + \cdots =\frac{x}{(x-1)^2}
$$
I can feel it. I can't prove it. I have tested it, and it seems to work. Domain-wise, I think it might be $x>1$, the question doesn't specify. Putting the LHS into Wolfram Alpha doesn't generate the RHS (it times out).
| Let's put this together cleanly:
For the sum to converge, $|x| > 1$ which means
$$\sum_{n=0}^\infty x^{-n} = \frac{x}{x-1}$$
Squaring this gives
$$\frac{x^2}{(x-1)^2} = \sum_{n=0}^\infty \sum_{m=0}^\infty x^{-(n+m)} $$
$$ = \sum_{r=0}^\infty (r+1)x^{-r}$$
(because each value of $(n+m)$ occurs $(n+m+1)$ times in the infinite sum (0 = 0+0, 1 = 1+0 or 0+1, 2=0+2 or 1+1 or 2+0, and so on).
Divide by $x$ gives:
$$ \frac{x}{(x-1)^2} = \sum_{r=0}^\infty (r+1)x^{-(r+1)} $$
$$ = \frac1{x} + \frac{2}{x^2} + \frac{3}{x^3} + ... $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/829168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Prove by induction that $(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}$ Prove by induction that $$(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}.$$
I got up to:
$n=1$ is true, and assuming $n=k$ prove for $n=k+1$.
Prove...
$$\frac{k(2k+1)(7k+1)+6(2(k+1))^2}{6} = \frac{(k+1)(2k+3)(7k+8)}{6}$$
I keep trying to expand to $6(2k+2)^2$ and factorising but I end up being short on one factor, e.g., I end up with $\frac{(k+1)(2k+3)(7k+6)}{6}$.
| Inductive step spelled out in detail:
\begin{align*}
&\hphantom{=}\frac{n(2n+1)(7n+1)}6+(2n+1)^2+(2n+2)^2-(n+1)^2\\
&=\frac{n(2n+1)(7n+1)}6+(2n+1)^2+3(n+1)^2\\
&=(2n+1)\left(\frac{n(7n+1)}6+2n+1\right)+3(n+1)^2\\
&=(2n+1)\frac{n(7n+1)+12n+6}6+3(n+1)^2\\
&=(2n+1)\frac{7n^2+13n+6 }6+3(n+1)^2\\
&=\frac{(2n+1)(n+1)(7n+6)}6+3(n+1)^2\\
&=(n+1)\left(\frac{(2n+1)(7n+6)}6+3(n+1)\right)\\
&=(n+1)\frac{(2n+1)(7n+6)+18(n+1)}6\\
&=(n+1)\frac{14n^2+19n+6+18n+18}6\\
&=(n+1)\frac{14n^2+37n+24}6\\
&=\frac{(n+1)(2n+3)(7n+8)}6
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/831521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
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