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Maclaurin expansion of $\arcsin x$ I'm trying to find the first five terms of the Maclaurin expansion of $\arcsin x$, possibly using the fact that
$$\arcsin x = \int_0^x \frac{dt}{(1-t^2)^{1/2}}.$$
I can only see that I can interchange differentiation and integration but not sure how to go about this. Thanks!
| Considering the binomial expansion
\begin{align}
\frac 1{\sqrt{1-u}}&=1+\frac12 u+\frac12\cdot\frac32\frac{u^2}{2!}+\frac12\cdot\frac32\cdot\frac52\frac{u^3}{3!}++\frac12\cdot\frac32\cdot\frac52\cdot\frac 72\frac{u^4}{4!}+\dotsm \\
&=1+\frac12 u+\frac{1\cdot3}{2^2\,2!}u^2 +\frac{1\cdot3\cdot 5}{2^3\,3!}u^3 + +\frac{1\cdot3\cdot 5\cdot 7}{2^4\,4!}u^4\dotsm ,
\end{align}
and substituting $u=x^2$, then integrating term by term, you obtain
\begin{align}
\arcsin x&= x+ \frac12 \frac{x^3}3+\frac{1\cdot3}{2^2\,2!}\frac{x^5}5 +\frac{1\cdot3\cdot 5}{2^3\,3!}\frac{x^7}7 + +\frac{1\cdot3\cdot 5\cdot 7}{2^4\,4!}\frac{x^9}9+\dotsm \\[1ex]
&=x+ \frac12 \frac{x^3}3+\frac{1\cdot3}{2\cdot 4}\frac{x^5}5 +\frac{1\cdot3\cdot 5}{2\cdot4\cdot 6}\frac{x^7}7 +\frac{1\cdot3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}\frac{x^9}9+\dotsm \\[1ex]
&=\sum_{n=1}^\infty\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1}.
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/197874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 8,
"answer_id": 3
} |
Completing the square with negative x coefficients I know how to complete the square with positive $x$ coefficients but how do you complete the square with negative $x$ coefficients?
For example:
\begin{align*}
f(x) & = x^2 + 6x + 11 \\
& = (x^2 + 6x) + 11 \\
& = (x^2 + 6x + \mathbf{9}) + 11 - \mathbf{9} \\
& = (x+3)^2 + 2.
\end{align*}
For positive you would add 9 inside the parenthesis and subtract 9 outside. However, I get the wrong answer when I do it with a negative coefficient. Do you do the same thing when there is a negative coefficient in front of $x^2$ or is it the other way around? (subtract inside parenthesis and add outside of parenthesis).
Here is an example of a negative coefficient:
$$f(x) = -3x^2 + 5x + 1.$$
I tried to solve this and entered in my answer online but it was wrong.
| $$(2ax+b)^2 = 4a^2x^2 + 4abx + b^2$$
So, multiplying by $4a$, or in this case $-4a$, will avoid fractions until the last step.
\begin{align}
f(x) &= -3x^2 + 5x + 1 \\
-12f(x) &= 36x^2 -60x - 12 \\
-12f(x) &= 36x^2 - 60x + 25 - 25 - 12 \\
-12f(x) &= (6x-5)^2 - 37 & \left(b = \frac{4ab}{4a} = \dfrac{-60}{12}=-5 \right)\\
f(x) &= -\dfrac{1}{12}(6x-5)^2 + \dfrac{37}{12}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Help me prove $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$ Please help me prove this Leibniz equation: $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$. Thanks!
| $\sqrt{1+\sqrt{-3}}+\sqrt{1-\sqrt{-3}}=\sqrt{6}$
Is an addition of complex conjugates of the form
$(a+b\sqrt{-1}) + (a-b\sqrt{-1})$
So then
$1+\sqrt{-3}=(a+b\sqrt{-1})^2$
$=a^2+2ab\sqrt{-1}-b^2$
If we equal the real and complex parts we get:
$1=a^2-b^2$ and $\sqrt{-3}=2ab\sqrt{-1}$
Solving for a in the complex equation gives:
$a=\frac{\sqrt{3}}{2b}$
Plugging in $a=\frac{\sqrt{3}}{2b}$ into $1=a^2-b^2$ and multiplying both sides by $b^2$ yields:
$b^4+b^2-\frac{3}{4}$ which is a quadratic for $b^2$
The quadratic equations then gives $b^2=\frac{-1\pm2}{2}$
Using only the positive root means $b=\frac{\sqrt{2}}{2}$
Plug in $b=\frac{\sqrt{2}}{2}$ into $1=a^2-b^2$ to get $a=\frac{\sqrt{6}}{2}$
Therefore
$\sqrt{1+\sqrt{-3}}+\sqrt{1-\sqrt{-3}}=(\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}i)+(\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2}i)=\frac{\sqrt{6}}{2}+\frac{\sqrt{6}}{2}=\sqrt{6}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Euclidean Algorithm in Polynomials $\Bbb Z_2[x]$ Express the greatest common divisor of the following pair of polynomials as a combination of polynomials:
$f(x) = x^3 + x^2 +x +1$ and $g(x) = x^4 + x^2 + 1$.
I've been trying to understand this, but still can't get how I should do it in $\mathbb{Z}_2[x]$.
| There’s really nothing different from solving such problems in $\Bbb Z$.
Using the Euclidean algorithm in its most straightforward form, not trying to be mechanically efficient:
$$\begin{align*}
x^4+x^2+1&=(x+1)(x^3+x^2+x+1)+x^2\\
x^3+x^2+x+1&=(x+1)x^2+(x+1)\\
x^2&=x(x+1)+x\\
x+1&=1\cdot x+\color{red}{1}
\end{align*}$$
The gcd is in red. Working upwards, and taking advantage of the fact that subtraction in $\Bbb Z_2[x]$ is addition, we have
$$\begin{align*}
1&=1\cdot(x+1)+1\cdot x\\
&=1\cdot(x+1)+1\cdot\left(x^2+x(x+1)\right)\\
&=(x+1)(x+1)+1\cdot x^2\\
&=(x+1)\left(f(x)+(x+1)x^2\right)+1\cdot x^2\\
&=(x+1)f(x)+\left((x+1)^2+1\right)x^2\\
&=(x+1)f(x)+x^2\cdot x^2\\
&=(x+1)f(x)+x^2\Big(g(x)+(x+1)f(x)\Big)\\
&=\left(x+1+x^2(x+1)\right)f(x)+x^2 g(x)\\
&=(x+1)\left(x^2+1\right)f(x)+x^2 g(x)\\
&=(x+1)^3 f(x)+x^2 g(x)\;.
\end{align*}$$
As Bill Dubuque points out, and as is illustrated for numerical problems in the Wikipedia article on the extended Euclidean algorithm, there are more efficient ways to carry out the computations, but this is perhaps the easiest way to see clearly exactly what is going on.
| {
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Find all $n>1$ such that $\frac{2^n+1}{n^2}$ is an integer.
Find all $n>1$ such that $\dfrac{2^n+1}{n^2}$ is an integer.
I know that $n$ must be odd, then I don't know how to carry on. Please help. Thank you.
| Let's consider $$\frac{2^n+1}{n^k}$$
If $p$ be the smallest prime that divide $n$
Let $$\operatorname{ord}_p2=d,d\mid(p-1,2n)\implies d\mid 2$$ as $p-1<$ all other primes, so it implies
$$ p\mid (2^2-1)\implies p=3.$$
Let $$3^r||n, 2^{2n}\equiv 1{\pmod {3^{kr}}}\implies \phi(3^{kr})|2n$$ as $2$ is a primitive root of $3^s$ for all $s\ge 1$ (as mentioned in Example 8.1 in the Naoki Sato's solution mentioned in Pantelis Damianou's answer).
This implies $$2\cdot 3^{kr-1}|2n \implies kr-1\le r$$
as $(3^{kr-1},\frac{n}{3^{kr}})=1$.
So, $r(k-1)\le 1$.
(1)If $k>2$, there will be no solution.
(2)If $k=2$ then $r=1;$ let $q>p=3$ be next smallest prime that divides $n$. Now $\operatorname{ord}_q2$ must divide $(q-1,2\cdot 3\cdot \frac{n}{3})$
Then $\operatorname{ord}_q2\mid 6$ as $q-1<$ all primes greater than $3\implies (q-1,\frac{n}{3})=1$
So, $q\mid (2^6-1)\implies q=7$, but $2^7+1=129$ is not divisible by $7$.
So, there is no prime$>3$ that satisfies the given condition $\implies n=3$ if $k=2$.
(3)If $k=1$, there is no restriction on $r>0$
Here $\operatorname{ord}_q2$ must divide $(q-1,2\cdot 3^r\cdot \frac{n}{3^r})=(q-1,2\cdot 3^r)$
So, $q-1=2^c3^d$ as $q<$ any other primes, which implies $ (q-1,2\cdot 3^r)=2\cdot 3^{\min(c,r)}$
Programmatically I have observed that $n=3^s$ keeps $\frac{2^n+1}{n}$ an integer, where $s$ is natural number which can be verified as follows:
As $2$ is a primitive root of $3^s$ for all $s\ge 1,$ so, if $n=3^s$, $\operatorname{ord}_{(3^s)}2=\phi(3^s)=2\cdot 3^{s-1}\implies 2^{\frac{\phi(n)}2}\equiv -1\pmod n$
Now, $\frac{\phi(n)}2=3^{s-1}\implies 2^{3^{s-1}}\equiv -1\pmod {3^s}$.
This implies $$(2^{3^{s-1}})^3\equiv -1\implies 2^{3^s}\equiv -1\pmod {3^s}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/203976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Derivative of $\frac{1}{\sqrt{x+5}}$
I'm trying to find the derivative of $\dfrac{1}{\sqrt{x+5}}$
using $\displaystyle \lim_{h\to 0} \frac {f(x+h)-f(x)}{h}$
So,
$$\begin{align*}
\lim_{h\to 0} \frac{\dfrac{1}{\sqrt{x+h+5}}-\dfrac{1}{\sqrt{x+5}}}{h} &= \frac{\dfrac{\sqrt{x+5}-\sqrt{x+h+5}}{(\sqrt{x+h+5})(\sqrt{x+5})}}{h}\\\\
&= \frac{\dfrac{x+5-x-h-5}{(\sqrt{x+h+5})(\sqrt{x+5})}}{\dfrac{h}{\sqrt{x+5}}+\dfrac{h}{\sqrt{x+h+5}}}
\end{align*}$$
I do not know if this is correct or not.. please help. I'm stuck.
| I would suggest, start it over..
We can introduce a variable, say $u:=x+5$. Then it's
$$\begin{align*} \lim_{h\to 0} \frac1h\cdot \left( \frac1{\sqrt{u+h}} -\frac1{\sqrt{u}} \right)
&= \lim_{h\to 0} \frac1h\cdot\left(\frac{\sqrt u-\sqrt{u+h}}{\sqrt{u(u+h)}} \right) = \\
&= \lim_{h\to 0} \frac1h\cdot\left(\frac{\sqrt u-\sqrt{u+h}}{\sqrt{u(u+h)}} \cdot \frac{\sqrt{u}+\sqrt{u+h}}{\sqrt{u}+\sqrt{u+h}} \right) = \\
&= \lim_{h\to 0} \frac1h\cdot\left(\frac{u-(u+h)}{\sqrt{u(u+h)}\left(\sqrt{u+h}+\sqrt u\right)}\right) = \\
&=\lim_{h\to 0}\frac{-1}{\sqrt{u(u+h)}\left(\sqrt{u+h}+\sqrt u\right)} =\\
&=\frac{-1}{u\cdot 2\sqrt u}
\end{align*}$$
Finally, rewrite back $u=x+5$.
| {
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How to solve the cubic equation $x^3-12x+16=0$ Please help me for solving this equation $x^3-12x+16=0$
| Can you see by inspection that 2 is a root? If so, we can write your equation as $x^{3}-12x+16=(x-2)(ax^2+bx+c)=ax^{3}+(b-2a)x^{2}+(c-2b)x-2c$ by the factor theorem.
Thus, $a=1$, $b-2a=0$, $c-2b=-12 $ and $-2c=16$
Which gives $a=1$, $b=2$ and $c=-8$.
Therefore $x^{3}-12x+16=(x-2)(x^2+2x-8)=(x-2)(x-2)(x+4)=(x-2)^{2}(x+4)$
So there is a repeated root at $x=2$ and a second at $x=-4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the distance between the line and plane Find the distance between the line $x-z=3, x+2y+4z=6$ and the plane $3x+2y+2z=5$.
What I have done so far,
Found the vector by crossing $1,0-1$ and $1,2,4$ from the first line
The line is parallel to plane $\langle 2,-5,2\rangle \cdot \langle 3,2,2 \rangle = 0$
I think my next step is to find a point on Line 1 which satisfies both equations and then insert those values into the plane $3(x)+2(y)+2(z)=5$ and use the formula, $$
\frac{(3(x)+2(y)+2(z)-5)}{(3^2+2^2+2^2)}$$ to find the distance.
The values that I calculuate do not match the posted answer of $7/\sqrt{17}$
| If $z=t,$ for the given line $x=z+3=t+3$ and
$x+2y+4z=6$ or, $t+3+2y+4(t)=6$ or, $2y=3-5t$
So, any point on the given line is $(t+3,t,\frac{3-5t}2)$
The distance of $(t+3,t,\frac{3-5t}2)$ from $3x+2y+2z=5$ is $$\left|\frac{3(t+3)+(3-5t)+2(t)-5}{\sqrt{3^2+2^2+2^2}}\right|=\frac 7{\sqrt {17}}$$
Had it not been a constant(which is due to the parallelism), we had to find $t$such that the distance is minimum.
| {
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Prove that for all non-negative integers $m,n$, $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ is an integer.
Prove that for all non-negative integers $m,n$, $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ is an integer.
I'm not familiar to factorial and I don't have much idea, can someone show me how to prove this? Thank you.
| Consider a prime $p$. The highest power of $p$ dividing the numerator is $$\alpha_{Nr} = \left(\left \lfloor \dfrac{2m}{p}\right \rfloor + \left \lfloor \dfrac{2m}{p^2}\right \rfloor + \left \lfloor \dfrac{2m}{p^3}\right \rfloor + \cdots \right) + \left(\left \lfloor \dfrac{2n}{p}\right \rfloor + \left \lfloor \dfrac{2n}{p^2}\right \rfloor + \left \lfloor \dfrac{2n}{p^3}\right \rfloor + \cdots \right)$$
The highest power $p$ dividing the denominator is $$\alpha_{Dr} = \left(\left \lfloor \dfrac{m}{p}\right \rfloor + \left \lfloor \dfrac{m}{p^2}\right \rfloor + \left \lfloor \dfrac{m}{p^3}\right \rfloor + \cdots \right) + \left(\left \lfloor \dfrac{n}{p}\right \rfloor + \left \lfloor \dfrac{n}{p^2}\right \rfloor + \left \lfloor \dfrac{n}{p^3}\right \rfloor + \cdots \right) + \left(\left \lfloor \dfrac{m+n}{p}\right \rfloor + \left \lfloor \dfrac{m+n}{p^2}\right \rfloor + \left \lfloor \dfrac{m+n}{p^3}\right \rfloor + \cdots \right)$$
You may want to look at this post that discusses the highest power of a prime dividing $N!$.
The goal now is to show that $\alpha_{Nr} > \alpha_{Dr}$. If we prove that $$\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$$ we are then done. Can you now prove this one?
We have $x = \lfloor x \rfloor + \{x \}$ and $y = \lfloor y \rfloor + \{y \}$.
We will consider 4 cases.
Case $1$: $\{x\} <1/2$, $\{y\} <1/2$. Then $\lfloor 2x \rfloor = 2 \lfloor x \rfloor$ and $\lfloor 2y \rfloor = 2 \lfloor y \rfloor$ and $\lfloor x + y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor$. Hence, we have $$\lfloor 2x \rfloor + \lfloor 2y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$$
Case $2$: $\{x\} <1/2$, $\{y\} >1/2$ or $\{x\} >1/2$, $\{y\} <1/2$. Then $$\lfloor 2x \rfloor + \lfloor 2y \rfloor = 2 \lfloor x \rfloor + 2 \lfloor y \rfloor + 1$$
Further, $$\lfloor x+y \rfloor \leq \lfloor x \rfloor + \lfloor y \rfloor + 1$$ Hence, we have $$\lfloor 2x \rfloor + \lfloor 2y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$$
Case $3$: $\{x\} >1/2$, $\{y\} >1/2$. Then $$\lfloor 2x \rfloor + \lfloor 2y \rfloor = 2 \lfloor x \rfloor + 2 \lfloor y \rfloor + 2$$
Further, $$\lfloor x+y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor + 1$$ Hence, we have $$\lfloor 2x \rfloor + \lfloor 2y \rfloor > \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is there a pattern for $2^n=x \pmod {13}$ for all integers n? I was doing a question yesterday on very elementary number theory, and I came across this pattern.
$2^1=2 \pmod {13}$
$2^2=4 \pmod {13}$
$2^3=8 \pmod {13}$
$2^4=3 \pmod {13}$
$2^5=6 \pmod {13}$
$2^6=12 \pmod {13}$
$2^7=11 \pmod {13}$
$2^8=9 \pmod {13}$
$2^9=5 \pmod {13}$
$2^{10}=10 \pmod {13}$
$2^{11}=7 \pmod {13}$
$2^{12}=1 \pmod {13}$
$2^{13}=2 \pmod {13}$
I noticed the remainder always is unique for every block multiples of powers 13. And it repeats!.
Also I tried this for base 3 and base 4 base 5. All showed similar patterns.
Why?
| Simpler answer: For all these modulo questions, it is enough to find some exponent that gives a modulo of 1. Consider
$2^n = 1 (mod 13)$ for some n. Then $2^{n + 1} = 2 (mod 13)$ which is same as $2^{0 + 1} = 2 (mod 13)$ and $2^{n + 2} = 4 (mod 13)$ which is $2^{1 + 1} = 4 (mod 13)$, in other words, they repeat. And also, notice that, if $2^n = k (mod 13)$ and $2^m = k (mod 13)$, then $2^{n - m} = 1 (mod 13)$ also. Since there are only 13 numbers below 13 that could be modulos, it is bound to repeat after some exponent n.
Salahuddin
http://maths-on-line.blogspot.in/
| {
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Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$ prove the following inequality:
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers.
Thanks :)
| Note that using the AM-GM inequality, we have
$$\frac{a^3}{a+b}=a^2-\frac{a^2b}{a+b}\geq a^2-\frac{a^2b}{2\sqrt{ab}}=a^2-\frac 12\sqrt{a^3b},$$
And so summing up cyclically, it is enough to check that
$$2(a^2+b^2+c^2)\geq\sqrt{a^3b}+\sqrt{b^3c}+\sqrt{c^3a}+ab+bc+ca,$$
Which is obviously true since $\sqrt{a^3b}\leq \dfrac{a^2+ab}2$ and $ab+bc+ca\leq a^2+b^2+c^2$ from the AM-GM inequality.
Equality holds iff $a=b=c.$
$\Box$
| {
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Solving system of recurrence relations Base Case:
$$
\left\{
\begin{array}{c}
T(1) = 1 \\ T(2) = 1 \\T(3) = 4\end{array}
\right.
$$
I have the system:
$$
\left\{
\begin{array}{c}
T(N) = G(N-1) + F(N-1) \\
G(N) = F(N-1) + G(N-1) \\
F(N) = 2H(N-1) + F(N-2) \\
H(N) = H(N-1) + F(N-1)\end{array}
\right.
$$
I seems $$T(N) = H(N) = G(N)$$ so we now have only two equations:
$$
\left\{
\begin{array}{c}
T(N) = T(N-1) + F(N-1) \\
F(N) = 2T(N-1) + F(N-2)\end{array}
\right.
$$
I figured
$T(N) = T(N-k) + \sum_{i=1}^k F(N-i) $
and
$F(N) = F(N-2) +2T(N-k) + 2\sum_{i=2}^k F(N-i) $
But after mixing these expressions in a similar way, I came unstuck.
I tried following another example but it didn't help.
I would like to find $T(10^{12})$. Probably by using matrix exponentiation.
| Note that
$$
\begin{pmatrix}
T_n \\
F_n \\
F_{n - 1}
\end{pmatrix} = \begin{pmatrix}
1 & 1 & 0 \\
2 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
T_{n-1}\\
F_{n-1}\\
F_{n-2}
\end{pmatrix}.
$$
So starting with $(T_1, F_1, F_0)$ you can find $(T_n, F_n, F_{n-1})$ quickly by matrix exponentiation. The latter can be done by the squaring method.
| {
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Find formula for ${-\frac{2\ln3}{3} , \frac{4\ln4}{9} ,\frac{6\ln5}{27},\frac{8\ln6}{81} ,....}$ The question is to find a general formula for the nth term, $a_n$, of the sequence:
$${-\frac{2\ln3}{3} , \frac{4\ln4}{9} ,\frac{6\ln5}{27}, \frac{8\ln6}{81} ,\frac{10\ln7}{243},...}$$
Here is what I got, but when I plugged in a few terms, it did not work out right.
I got the formula to be: $$a_n = (-1)^n \frac{2\ln(2+1)\cdot 2^{(n-1)}} {3^n}$$
Please help. Thanks
| Each term has a form like $\pm \frac{x\ln y}{z}$.
The sign alternates so it is $(-1)^n$.
$x$ is $2, 4, 6, 8, \dots$, so $x$ is $2n$.
$y$ is $3,4,5,\dots$ so $y$ is $n+2$.
$z$ is $3,9,27, 81, \dots$ so $z$ is $3^n$.
Combining all contributions, we get $a_n = (-1)^n \frac{2n \ln(n+2)}{3^n}$.
| {
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Finding an unknown angle Geometry: Auxiliary Lines
As shown in the figure:
| Let central point is $O$.
$\angle OBC = 180^\circ - (46^\circ+12^\circ) - (62^\circ+8^\circ) - X = 52^\circ-X$.
Using the sine formula, we have:
$$\frac{AO}{\sin 62^\circ} = \frac{CO}{\sin 12^\circ},$$
$$\frac{AO}{\sin X} = \frac{BO}{\sin 46^\circ},$$
$$\frac{CO}{\sin (52^\circ-X)} = \frac{BO}{\sin 8^\circ},$$
so,
$$\sin 12^\circ \cdot \sin 8^\circ \cdot \sin X = \sin 62^\circ \cdot \sin 46^\circ \cdot \sin (52^\circ-X).$$
(Here we can find that $X=50^\circ$. Uniqueness of solution is discussed in comments).
In fact we need to prove that
$$\sin 8^\circ \cdot \sin 12^\circ \cdot \sin 50^\circ \mathop{=}\limits^{???} \sin 2^\circ \cdot \sin 46^\circ \cdot \sin 62^\circ.$$
Using formulas
$$2\cdot \sin\alpha \cdot \sin\beta = \cos(\alpha-\beta) - \cos(\alpha+\beta),$$
$$2\cdot \sin\alpha \cdot \cos\beta = \sin(\alpha-\beta) + \sin(\alpha+\beta),$$
$$\sin(180^\circ-\varphi) = \sin(\varphi), \quad \cos(180^\circ-\varphi) = -\cos(\varphi),$$
we have
$$\sin 8^\circ \cdot (\cos 38^\circ - \cos 62^\circ) \mathop{=}\limits^{???} \sin 2^\circ \cdot (\cos 16^\circ + \cos 72^\circ),$$
$$-\sin 30^\circ + \sin 46^\circ + \sin 54^\circ - \sin 70^\circ \mathop{=}\limits^{???} -\sin 14^\circ + \sin 18^\circ - \sin 70^\circ + \sin 74^\circ,$$
$$-\sin 30^\circ + \sin 46^\circ + \sin 54^\circ + \sin 14^\circ - \sin 18^\circ - \sin 74^\circ \mathop{=}\limits^{???} 0,$$
$$(\sin 14^\circ + \sin 46^\circ - \sin 74^\circ) + ( - \sin 18^\circ + \sin 54^\circ -\frac{1}{2} ) \mathop{=}\limits^{???} 0,$$
$$(\sin 14^\circ + \sin 134^\circ + \sin 254^\circ) + \frac{1}{2}( - \sin 18^\circ + \sin 54^\circ +\sin 126^\circ + \sin 198^\circ + \sin270^\circ ) \mathop{=}\limits^{???} 0,$$
it is equality, because left side is equal to
$$\mathrm{Im} \left( p(\omega_3^0 + \omega_3^1 + \omega_3^2) + \frac{q}{2}( \omega_5^0 + \omega_5^1 +\omega_5^2 + \omega_5^3 + \omega_5^4) \right)=$$
$$\mathrm{Im} \left( p \cdot \frac{\omega_3^3-1}{\omega_3-1} + \frac{q}{2}\cdot \frac{\omega_5^5-1}{\omega_5-1} \right)=
\mathrm{Im} \left( p \cdot \frac{1-1}{\omega_3-1} + \frac{q}{2}\cdot \frac{1-1}{\omega_5-1} \right)=
\mathrm{Im} \left( p \cdot 0 + \frac{q}{2} \cdot 0 \right)= 0,$$
where
$p = \exp(i\pi 7/90)$, $q = \exp(-i\pi /10)$,
$w_{...}$ $-$ roots of unity:
$w_3 = \exp(i 2 \pi / 3) = \cos 2\pi/3 + i \sin 2\pi/3 = \cos 120^\circ + i \sin 120^\circ$,
$w_5 = \exp(i 2 \pi / 5) = \cos 2\pi/5 + i \sin 2\pi/5 = \cos 72^\circ + i \sin 72^\circ$.
| {
"language": "en",
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prove, that following formula is correct As in the statement, I got problems with:$$\binom{n}{0} +\binom{n}{2}+\binom{n}{4}+\cdots+\binom{n}{2[\frac{n}{2}]}=2^{n-1}$$ I started with Newton conjecture, trying to work with $(1+1)^n=\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}$ but actually it does not guide me anywhere. i would appreciate any hints, since it looks a bit strange to me
| The root of the solution lies in the fact, $\binom n {2[\frac{n}{2}]}$ is the last term of $(1+1)^n+(1-1)^n$ as proved below:
for $n=2m, [\frac n2]=[\frac{2m}2]=m, \binom n {2[\frac{n}{2}]}=\binom{2m}{2m}$
for $n=2m+1, [\frac n2]=[\frac{2m+1}2]=m ,\binom n {2[\frac{n}{2}]}=\binom{2m+1}{2m}$
Now,
$$2^{2m}=(1+1)^{2m}=\binom{2m}{0} +\binom{2m}1 +\binom{2m}2+\cdots+\binom{2m}{2m-1}+\binom{2m}{2m}$$
$$0=(1-1)^{2m}=\binom{2m}{0} -\binom{2m}1 +\binom{2m}2+\cdots-\binom{2m}{2m-1}+\binom{2m}{2m}$$
Adding we get, $2^{2m}=2\left(\binom{2m}{0}+\binom{2m}2+\cdots+\binom{2m}{2m-2}+\binom{2m}{2m} \right)$
$2^{2m-1}=\binom{2m}{0}+\binom{2m}2+\cdots+\binom{2m}{2m-2}+\binom{2m}{2m}--->(1)$
Replacing $2m$ with $n$ and $\binom{2m}{2m}$ with $\binom n {2[\frac{n}{2}]}$
we get,
$2^{n-1}=\binom n 0+\binom n2+\cdots+\binom n{n-2}+\binom n {2[\frac{n}{2}]}$
Similarly,
$$2^{2m+1}=(1+1)^{2m+1}=\binom{2m+1}{0} +\binom{2m+1}1 +\binom{2m+1}2+\cdots+\binom{2m+1}{2m}+\binom{2m+1}{2m+1}$$
$$0=(1-1)^{2m+1}=\binom{2m+1}{0} -\binom{2m+1}1 +\binom{2m+1}2-\cdots+\binom{2m+1}{2m}-\binom{2m+1}{2m+1}$$
Adding we get, $2^{2m+1}=2\left(\binom{2m+1}{0}+\binom{2m+1}2+\cdots+\binom{2m+1}{2m-2}+\binom{2m+1}{2m} \right)$
$2^{2m}=\binom{2m+1}{0}+\binom{2m+1}2+\cdots+\binom{2m+1}{2m-2}+\binom{2m+1}{2m}--->(2)$
Replacing $2m+1$ with $n$ and $\binom{2m+1}{2m}$ with $\binom n {2[\frac{n}{2}]}$
we get,
$2^{n-1}=\binom n 0+\binom n2+\cdots+\binom n{n-2}+\binom n {2[\frac{n}{2}]}$
| {
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Solving $\frac{x-1}{x+3}>\frac{x}{x-2}$ I'm having little trouble solving$$\frac{x-1}{x+3}>\frac{x}{x-2}$$
What steps should I take?
Need this to write the topological spaces of the set defined by this inequation.
| $$\frac{x-1}{x+3}>\frac{x}{x-2}$$
$$\frac{x-1}{x+3}-\frac{x}{x-2}>0$$
$$
\frac{(x-1)(x-2)}{(x+3)(x-2)}-\frac{x(x+3)}{(x-2)(x+3)}>0
$$
$$
\frac{(x-1)(x-2)-x(x+3)}{(x-2)(x+3)}>0
$$
$$
\frac{(x^2 -3x +2)-(x^2+3x)}{(x-2)(x+3)}>0
$$
$$
\frac{-6x +2}{(x-2)(x+3)}>0
$$
$$
\frac{x -(1/3)}{(x-2)(x+3)}<0
$$
This fraction undergoes sign changes at $-3$, $1/3$, and $2$. So ascertain in each of the four intervals whether it is positive or negative on that interval.
| {
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Use Lagrange multiplier to find absolute maximum and minimum Use Lagrange multiplier to find absolute maximum and minimum of $f(x,y) =x^2+xy+y^2, x^2+y^2 =8$.
What i've done so far..
$f_x = \lambda g_x \Rightarrow 2x+y =\lambda2x, \\f_y = \lambda g_y \Rightarrow x+2y = \lambda 2y,\\g(x,y) = x^2+y^2 -8 =0$
May I know how should i proceed from here?
| On a side note, this problem can be solved very nicely with substitution since it's equivalent to:
$$f(\theta) = 8 \left ( 1 + \frac{1}{2}\sin 2\theta \right )$$
where $ x = r \cos\theta$, $y = r \sin \theta$ and $r = 2\sqrt{2}$.
Since $f(\theta)$ is the same as maximizing $\sin 2 \theta$, $f(\theta)$ is maximum and minimum at $\theta = \pi n + \frac{\pi}{4}$ and $\theta = \pi n - \frac{\pi}{4}$ respectively.
So if $n = 1$, we have a maximum at $x = -2, y = -2$ and a minimum at $x = -2, y = 2$.
| {
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Find an equation of the plane that passes through the point $(1,2,3)$, and cuts off the smallest volume in the first octant. *help needed please* Find an equation of the plane that passes through the point $(1,2,3)$, and cuts off the smallest volume in the first octant.
This is what i've done so far....
Let $a,b,c$ be some points that the plane cuts the $x,y,z$ axes. --> $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where $a,b,c >0$.
I saw a solution for this question was to use Lagrange multiplier. The solution goes as follows...
The product $abc$ will be equal to $6$ times the volume of the tetrahedron $OABC$ (could someone explain to my why is this so?)
$f(a,b,c) = abc$ given the condition $(\frac1a + \frac2b + \frac3b -1)$
$f(a,b,c) = abc + \lambda (\frac1a + \frac2b + \frac3c -1)$
2nd query to the question...
$f_a = \lambda g_a \Rightarrow bc - \frac\lambda {a^2} ; a = \sqrt \frac \lambda {bc}
\\f_b = \lambda g_b \Rightarrow ac - \frac\lambda {b^2} ; b = \sqrt \frac {2\lambda}{ac}
\\f_c = \lambda g_c \Rightarrow ab - \frac\lambda {c^2} ; c = \sqrt \frac {3\lambda}{ab}$
using values of $a,b,c$ into $\frac1a+\frac1b+\frac1c = 1\Rightarrow \lambda =\frac{abc}{a+2b+3c}$.
May i know how should i proceed to solve the unknowns?
| BTW: you don't need Lagrange multipliers for this problem:
To find the minimum of $abc$ for $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}=1$ you can use $AM-GM$:
$$\frac{1}{3}=\frac{\frac{1}{a}+\frac{2}{b}+\frac{3}{c}}{3} \geq \sqrt[3]{\frac{6}{abc}}$$
Thus
$$abc \geq 3^3 \cdot 6 \,,$$
with equality if and only if
$$\frac{1}{a}=\frac{2}{b}=\frac{3}{c}$$
| {
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"source": "stackexchange",
"question_score": "3",
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Jordan canonical form of a matrix for distinct eigenvalues How can i find the Jordan canonical form of this matrix $$A=\begin{pmatrix}1 &0 &0 &0 \\ 1& 2& 0& 0\\ 1 &0& 2& 0\\ 1 &1& 0& 2\end{pmatrix}.$$
In my book there are examples but all the matrices in these examples have only one eigenvalue repeated n times(for $n\times n$ matrices) but the matrix $A$ has eigenvalues $1$ and $2$(multiple of $3$). What is the way of finding $A$'s jordan canonical form?
Thanks
| The properties listed here can help you.
In this case you have three possibilities for the Jordan Canonical Form.These are:
$$J_1=\begin{pmatrix}1 &0 &0 &0 \\ 0& 2& 0& 0\\ 0 &0& 2& 0\\ 0 &0& 0& 2\end{pmatrix} , \ J_2=\begin{pmatrix}1 &0 &0 &0 \\ 0& 2& 0& 0\\ 0 &0& 2& 1\\ 0 &0& 0& 2\end{pmatrix}, \ J_3=\begin{pmatrix}1 &0 &0 &0 \\ 0& 2& 1& 0\\ 0 &0& 2& 1\\ 0 &0& 0& 2 \end{pmatrix}.$$
If $(A-I)(A-2I)=0$ is $J_1$. If $(A-I)(A-2I)\neq 0$ and $(A-I)(A-2I)^2= 0$ is $J_2$. Else is $J_3.$
| {
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Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ Prove that
$$
\frac{1}{\sqrt{1-4t}} \left(\frac{1-\sqrt{1-4t}}{2t}\right)^k = \sum\limits_{n=0}^{\infty}\binom{2n+k}{n}t^n,
\quad
\forall k\in\mathbb{N}.
$$
I tried already by induction over $k$ but i have problems showing the statement holds for $k=0$ or $k=1$.
| This can be shown using a variant of Lagrange Inversion. Introduce
$$T(z) = w = \sqrt{1-4z}$$ so that
$$z = \frac{1}{4} (1-w^2)$$ and
$$dz = -\frac{1}{2} w \; dw.$$
Then we seek to compute
$$[z^n] \frac{1}{T(z)} \left(\frac{1-T(z)}{2z}\right)^k
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{1}{T(z)} \left(\frac{1-T(z)}{2z}\right)^k dz.$$
Using the substitution this becomes
$$- \frac{1}{2\pi i}
\int_{|w-1| = \epsilon} \frac{4^{n+1}}{(1-w^2)^{n+1}}
\frac{1}{w} \left(\frac{1-w}{1/2(1-w^2)}\right)^k
\frac{1}{2} w \; dw
\\ = - \frac{1}{2} \frac{1}{2\pi i}
\int_{|w-1| = \epsilon} \frac{4^{n+1}}{(1-w^2)^{n+1}}
\left(\frac{2}{1+w}\right)^k dw
\\ = - \frac{1}{2\pi i}
\int_{|w-1| = \epsilon} \frac{1}{(1-w)^{n+1}}
\frac{2^{2n+k+1}}{(1+w)^{n+k+1}} dw
\\ = (-1)^n \frac{1}{2\pi i}
\int_{|w-1| = \epsilon} \frac{1}{(w-1)^{n+1}}
\frac{2^{2n+k+1}}{(1+w)^{n+k+1}} dw.$$
Now expand the second fraction about $w=1$ to get
$$\frac{1}{(1+w)^{n+k+1}}
= \frac{1}{(2+(w-1))^{n+k+1}}
= \frac{1}{2^{n+k+1}} \frac{1}{(1+1/2(w-1))^{n+k+1}}
\\ = \frac{1}{2^{n+k+1}}
\sum_{q\ge 0} {q+n+k\choose q} (-1)^q \frac{(w-1)^q}{2^q}.$$
We need $q=n$ for the residue in the integral, getting the final answer
$$(-1)^n \times 2^{2n+k+1} \times
\frac{1}{2^{n+k+1}}
{2n+k\choose n} (-1)^n \frac{1}{2^n}
= {2n+k\choose n}.$$
This MSE link has another calculation that is quite similar.
| {
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Solve the recurrence relation:$ T(n) = \sqrt{n} T \left(\sqrt n \right) + n$ $$T(n) = \sqrt{n} T \left(\sqrt n \right) + n$$
Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable.
The answer is $Θ (n \log \log n)$.
Can anyone arrive at the solution.
| Yes, the Master Theorem can be applied. Here's how:
$$T(n) = \sqrt{n} T(\sqrt{n}) + n = \sqrt{n} T(\sqrt{n}) + \mathcal{O}(n)$$
Let $n = 2^k$, $\sqrt{n} = 2^{k/2}$, and $k = \log{n}$. Substituting in above equation, we get:
$$T(2^k) = 2^{k/2} T(2^{k/2}) + 2^k \tag{1}$$
Dividing (1) by $2^k$, we get:
$$\frac{T(2^k)}{2^k} = \frac{2^{k/2}T(2^{k/2})}{2^k} + 1\tag{2}$$
Simpliifying $\frac{2^{k/2}}{2^k} = \frac{1}{2^{k/2}}$ in $(2)$ gives us:
$$\frac{T(2^k)}{2^k} = \frac{T(2^{k/2})}{2^{k/2}} + 1\tag{3}$$
Let $y(k) = \frac{T(2^k)}{2^k}$, then $(3)$ gives us:
$$y(k) = y\left(\frac{k}{2}\right) + 1\tag{4}$$
Now, we apply the Master Theorem ($T(n) = aT\left(\frac{n}{b}\right) + n^d$), where $a=1, b=2$, and $d=0$, $a=b^d = 1$.
Because $d=0$, we have:
$$y(k) = k^d\log{k} = \log{k}\tag{5}$$
But, we also know that $T(2^k) = 2^ky(k)$, then
$$T(2^k) = 2^k\log{k} \implies T(n) = n\log{\log{n}}$$
since $n=2^k$ and $k = \log{n}$.
Finally: $T(n) = \Theta(n\log{\log{n}})$
| {
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integration by parts from Apostol I working through Apostol's calculus, and I need to prove integrating by parts that :
$\int (a^2 - x^2)^n \,dx = \frac{x (a^2 - x^2)^n}{2n + 1} + \frac{2 a^2 n}{2n+1} \int (a^2 - x^2)^{n-1} \,dx + C $
Now, using the integration by parts formula after first division the integral to parts I arrive at:
$\int (a^2 - x^2)^n \,dx = x(a^2 - x^2)^n + 2n \int x^2 (a^2 - x^2)^{n-1} \,dx $
I could substitute and solve the integral, but I need to do something else. I multiply the first expression on the right by $ \frac{2n + 1}{2n +1}$ which leads to:
$x(a^2 - x^2)^n + 2n \int (a^2 - x^2)^{n-1} \,dx = \frac{x (a^2 - x^2)^n}{2n + 1} + \frac{2nx (a^2 - x^2)^n}{2n + 1} + 2n\int x^2(a^2 - x^2)^{n-1} \,dx$
and I am somewhat close. If I try something else, I end up even closer:
$\int (a^2 - x^2)^n \,dx = \begin{pmatrix}
f(x)= a^2 - x^2 | f'(x)=-2x\\
g'(x)=(a^2-x^2)^{n-1} |g(x)=\int(a^2-x^2)^{n-1} \,dx
\end{pmatrix} = $
$(a^2-x^2)\int(a^2-x^2)^{n-1} \,dx + 2\int x \Big(\int(a^2-x^2)^{n-1} \,dx \Big)\,dx
=$
$(a^2-x^2)\int(a^2-x^2)^{n-1} \,dx +x^2\int(a^2-x^2)^{n-1} \,dx = a^2\int(a^2-x^2)^{n-1} \,dx $
But I think I made a mistake somewhere... could somebody help me out? I'm really stuck!
Thanks!
| OK. Another problem. Miserably stuck again :/. Integration by parts. Need to show that:
If $I_n(x)=\int_{0} ^{x}t^n(t^2+a^2)^{-\frac{1}{2}}dt$
Then: $nI_n(x) = x^{n-1}\sqrt{x^2+a^2}-(n-1)a^2I_{n-2(x)}$ if $x\geq2$
I can get to the point where:
$nI_{n}=x^{n+1}(x^2+a^2)^{-\frac{1}{2}}-a^2I_{n-2}+a^4\int t^{n-2}(t^2+a^2)^{-\frac{3}{2}}dt$
I get there by dividing it into parts and then using the trick user49685 suggested using.
Now, is this a good start or should I have taken another route? Because I can not find a way out :/
| {
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How many Arithmetic Progressions having $3$ terms can be made from integers $1$ to $n$? How many Arithmetic Progressions having $3$ terms can be made from integers $1$ to $n$? The numbers in the AP must be distinct.
For example if $n=6$ then number of AP's possible are $6$
*
*$1,2,3$
*$2,3,4$
*$3,4,5$
*$4,5,6$
*$2,4,6$
| Let the three term AP be $a,a+d,a+2d$ where $a,d$ are natural numbers.
So, $a\ge1$ and $a+2d\le n\implies 1\le a\le n-2d \implies d\le \frac{n-1}2$
If $d=1,a$ can assume $1,2,\cdots, n-2$ i.e., $n-2$ values.
If $d=2,a$ can assume $1,2,\cdots, n-4$ i.e., $n-4$ values.
If $n$ is even, $d_{max}=\frac{n-2}2$
for $d=\frac {n-2}2,1\le a\le 2$ i.e., $a$ has 2 values.
If $n$ is even, the number of AP is $(n-2)+(n-4)+\cdots+2=\frac{(n-2)}4(2+n-2)=\frac{n(n-2)}4$
If $n$ is odd, $d_{max}=\frac{n-1}2$
So, in that case the number of AP will be $1+3+\cdots+(n-4)+(n-2)=\frac{(n-1)^2}4$
| {
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How to solve $ \sum_{i=1}^{n-1} i^2 \equiv \;? \pmod n$ I solved expressions below . it is ok?
My attempt:
$$1)\;\;\;\;\; ?\cong\pmod n\sum_{i=1}^{n-1} i^2 \stackrel{?}= {(n-1)^3\over3}+{(n-1)^2\over2}+{(n-1)\over6}={n^3\over3}-{n^2\over6}+{n\over6}=n\left({n^2\over3}-{n\over2}+{1\over6}\right)
$$
(continue?)how to solve for all n?
$$2)\;\;\;\;\;?\cong\pmod n \sum_{i=1}^{n-1} i^3 = \left( {(n-1)n\over2} \right)^2$$
(continue?)how to solve for all n?
Thank you.
| It never hurts to gather some numerical data. For your first problem:
$$\begin{array}{cc|cc|cc}
n&\left(\sum_{k=1}^{n-1}k^2\right)\bmod n&n&\left(\sum_{k=1}^{n-1}k^2\right)\bmod n&n&\left(\sum_{k=1}^{n-1}k^2\right)\bmod n\\ \hline
2&1&8&4&14&7\\
3&2&9&6&15&10\\
4&2&10&5&16&8\\
5&0&11&0&17&0\\
6&1&12&2&18&3\\
7&0&13&0&19&0
\end{array}$$
Examination of that table should suggest that
$$\left(\sum_{k=1}^{n-1}k^2\right)\bmod n=\begin{cases}
\frac{n}2,&\text{if }n\equiv 2\!\!\!\pmod 6\\\\
\frac{2n}3,&\text{if }n\equiv 3\!\!\!\pmod 6\\\\
\frac{n}2,&\text{if }n\equiv 4\!\!\!\pmod 6\\\\
0,&\text{if }n\equiv 5\!\!\!\pmod 6\\\\
\frac{n}6,&\text{if }n\equiv 0\!\!\!\pmod 6\\\\
0,&\text{if }n\equiv 1\!\!\!\pmod 6\;;
\end{cases}\tag{1}$$
since you know that $$\sum_{k=1}^{n-1}k^2=n\left(\frac{n^2}3-\frac{n}2+\frac16\right)=\frac16n(n-1)(2n-1)\;,$$ it shouldn’t be too hard to prove $(1)$.
Added: For the second problem, did you try calculating
$$\left(\sum_{k=1}^{n-1}k^3\right)\bmod n=\frac{n^2(n-1)^2}4\bmod n$$
for $n=2,3,4,\dots$ as I did for the first problem? Doing so should let you guess right away that the value is $0$ for odd $n$, and discovering the mathematical reason for this isn’t hard. If $n$ is odd, then $\frac{(n-1)^2}4$ is an integer, and clearly $n^2\frac{(n-1)^2}4\bmod n=0$.
If $n$ is even, then
$$\frac{n^2(n-1)^2}4\equiv\left(\frac{n}2\right)^2(n-1)^2\equiv\left(\frac{n}2\right)(-1)^2\equiv\left(\frac{n}2\right)^2\pmod n\;.$$
Now make a table for $n=2,4,6,\dots$:
$$\begin{array}{rccc}
n:&2&4&6&8&10&12&14&16\\ \hline
\left(\frac{n}2\right)^2\bmod n:&1&0&3&0&5&0&7&0
\end{array}$$
The pattern is pretty obvious: it definitely appears that
$$\left(\frac{n}2\right)^2\bmod n=\begin{cases}
0,&\text{if }n\equiv 0\!\!\!\pmod 4\\\\
n/2,&\text{if }n\equiv 2\!\!\!\pmod 4\;,
\end{cases}$$
and you shouldn’t have too much trouble proving that this is actually the case.
| {
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"url": "https://math.stackexchange.com/questions/249082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving a system of linear equations to find two eigenvectors. So I have
\begin{align*}
x - 2y + z & = 0 \\
-2x + 4y - 2z & = 0 \\
x - 2y + z & = 0
\end{align*}
I know I need to find two eigenvectors for the eigenspace with eigenvalue 2 as I know the matrix is diagonalizable and I've already found the eigenvector for the eigenspace with eigenvalue 8...
I can't seem to solve this. Please help!
| $$ \begin{align*}
x - 2y + z & = 0 \\
-2x + 4y - 2z & = 0 \\
x - 2y + z & = 0
\end{align*} $$
$$ \implies \begin{align*}
x - 2y + z & = 0 \\
0 & = 0 \\
0 & = 0.
\end{align*}$$
So you have two free variables. Assuming $y=t, z=s,\, t,s\in \mathbb{R} $, then the solution is given by
$$ \begin{bmatrix}
x \\
y\\
z\end{bmatrix} = \begin{bmatrix}
2t-s \\
t\\
s\end{bmatrix} = \begin{bmatrix}
2t \\
t\\
0\end{bmatrix} + \begin{bmatrix}
-s \\
0\\
s\end{bmatrix} = t\begin{bmatrix}
2 \\
1\\
0\end{bmatrix} + s\begin{bmatrix}
-1 \\
0\\
1\end{bmatrix}.$$
Can you see the two eigenvectors now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Orbit of $\frac{1}{2}$ of the dynamical system defined by $f:[0,1] \to [0,\frac{3}{4}]$ where $f(x)=3x(1-x)$ converges. I view the orbit of $\frac{1}{2}$ of the dynamical system defined by $f:[0,1] \to [0,\frac{3}{4}]$ where $f(x)=3x(1-x)$ as the sequence $(x_{n})$. So
$$x_n = \frac{1}{2}, f\bigg(\frac{1}{2}\bigg), f\bigg(f\bigg(\frac{1}{2}\bigg)\bigg), \dots$$
$(x_n)$ appears to dance around the fixed point $\frac{2}{3}$ while slowly converging to it.
The odd-numbered iterates are increasing and the even-number iterates are decreasing. Let $(a_{k})$ be the sequence of odd terms of $(x_{n})$ where $k = 2n-1$ and $(b_{k})$ be the sequence of even terms of $(x_{n})$ where $k = 2n$.
Show $(a_k)$ is monotonic & bounded.
1. Increasing:
$a_{k}<a_{k+1} \forall \ k \in \ \mathbb{N}$. In terms of $(x_{n})$, show $x_{2n-2}<x_{2n+1} \forall \ n \in \ \mathbb{N}$.
$$x_{2n+1} - x_{2n-1}>0$$
$$(-27x_{2n-1}^4+54x_{2n-1}^3-36x_{2n-1}^2+9x_{2n-1})-(x_{2n-1}) > 0$$
$$-27x_{2n-1}^4+54x_{2n-1}^3-36x_{2n-1}^2+8x_{2n-1}> 0$$
$$x_{2n-1}(2-3x_{2n-1})^3 > 0$$
$$0<x_{2n-1}<\frac{2}{3}$$
$$0<a_{k}<\frac{2}{3}$$
Hence, the subsequence $(a_k)$ is strictly increasing and
$(a_k)$ is bounded below by the first point of the subsequence
$a_1=\frac{1}{2}$.
Question:
How do I show $a_{k}$ is bounded above by the fixed point $\frac{2}{3}$?
Maybe, use epsilon delta proof to show convergence, which implies $(x_n)$. Or by induction, assume $x_{2n-1}< \frac{2}{3}$ show $x_{2n+1}=f(f(x_{2n-1}))<\frac{2}{3}$
| Note that $f(\frac{1}{2})=\frac{3}{4}$(as commented by bonext), $f(\frac{2}{3})=\frac{2}{3}$(as you have noticed) and $f(\frac{3}{4})=\frac{9}{16}\in(\frac{1}{2},\frac{2}{3})$. Also note that $f$ is strictly decreasing on $[\frac{1}{2},1]$. Then we have
$$f((\frac{1}{2},\frac{2}{3}))\subset(\frac{2}{3},\frac{3}{4})\quad\mbox{and}\quad f((\frac{2}{3},\frac{3}{4}))\subset(\frac{1}{2},\frac{2}{3}).$$
Now by induction, it is easy to see $x_{2n-1}\in(\frac{1}{2},\frac{2}{3})$ and $x_{2n}\in(\frac{2}{3},\frac{3}{4})$ when $n\ge 2$.
| {
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Solving a quadratic equation via a tangent half-angle formula (Maybe I'll post my own answer here, but maybe others will make that redundant.)
This is a fun (?) trivia item that fell out of a bit of geometry I was thinking about.
One of the tangent half-angle formulas says
$$
\tan\frac\alpha2 = \frac{\sin\alpha}{1+\cos\alpha}.
$$
Consider the quadratic equation
$$
x^2 + 2bx - 1 = 0.
$$
Solving for $b$, we get
$$
b = \frac{1-x^2}{2x}.
$$
Let $\alpha=\arctan b$. Since $\tan\alpha=(1-x^2)/2x$, we have $\sin\alpha=(1-x^2)/(1+x^2)$ and $\cos\alpha=2x/(1-x^2)$, so
$$
\tan\frac\alpha2=\frac{1-x^2}{(1+x)^2}.
$$
Lo and behold, this fraction is not in lowest terms, and we have
$$
\tan\frac\alpha2= \frac{\sin\alpha}{1+\cos\alpha} = \frac{1-x}{1+x}.
$$
This implies
$$
(1+x)\tan\frac\alpha2= 1-x
$$
and so we seem to have reduced a second-degree equation to a first-degree equation by using the tangent half-angle formula. Solving for $x$, we get
$$
x=\frac{1-\tan\frac\alpha2}{1+\tan\frac\alpha2}.\tag{1}
$$
Since $\tan\alpha=b$, we have $\sin\alpha=b/\sqrt{1+b^2}$ and $\cos\alpha=1/\sqrt{1+b^2}$, so
$$
\tan\frac\alpha2=\frac{\sin\alpha}{1+\cos\alpha}=\frac{b}{1+\sqrt{1+b^2}}.
$$
Sustituting this back into $(1)$, we get
$$
x = \frac{1+\sqrt{b^2-1}-b}{1+\sqrt{b^2+1}+b} = -b+\sqrt{b^2+1}.
$$
This is one of the two solutions of the quadratic equation. What happened to the other one?
| When you took $\sqrt{1+b^2}$ you did not consider the negative square root.
| {
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Prove that every element of $a_{n+2013}=\frac{a_{n+1}a_{n+2}...a_{n+2012}+1}{a_n}$ is an integer
Given $\displaystyle a_1=a_2=\cdots=a_{2013}=1$ and $\displaystyle a_{n+2013}=\frac{a_{n+1}a_{n+2}\cdots a_{n+2012}+1}{a_n}$.
Prove that $a_{n+2013}\in\mathbb{N}$ for all $n\in\mathbb{N}$.
I tried to prove it with induction, but failed. Please help. Thank you.
| For these kind of questions, it is clear that $2013$ is arbitrary. Let us replace $2013$ by $k \geq 2$, so
$$a_1, a_2, \ldots , a_k=1, a_{n+k}=\frac{a_{n+1}a_{n+2} \ldots a_{n+k-1}+1}{a_n}, n \geq 1$$
It is clear that $a_{n+k}>0$.
We prove by induction on $n \geq 1$ that $a_{n+k} \in \mathbb{Z}$.
When $1 \leq n \leq k$, $a_n=1$, so clearly $a_{n+k}=\frac{a_{n+1}a_{n+2} \ldots a_{n+k-1}+1}{a_n}=(a_{n+1}a_{n+2} \ldots a_{n+k-1}+1) \in \mathbb{Z}$.
Suppose that the statement holds for $1 \leq n \leq m, m \geq k$. Then taking $\pmod{a_{m+1}}$,
\begin{align}
& a_{m+2}a_{m+3} \ldots a_{m+k}+1 \\
& \equiv -(a_{m+1}a_{m+1-k}-1)(a_{m+2}a_{m+3} \ldots a_{m+k})+1\\
& \equiv -(a_{m-(k-2)}a_{m-(k-3)} \ldots a_{m})(a_{m+2}a_{m+3} \ldots a_{m+k})+1 \\
& \equiv -\prod_{i=1}^{k-1}{(a_{m+1-k+i}a_{m+1+i})}+1 \\
& \equiv -\prod_{i=1}^{k-1}{(a_{(m+1-k+i)+1}a_{(m+1-k+i)+2} \ldots a_{m+1} \ldots a_{(m+1-k+i)+(k-1)}+1)}+1 \\
& \equiv 0 \pmod{a_{m+1}}
\end{align}
Thus $a_{(m+1)+k}=\frac{a_{m+2}a_{m+3} \ldots a_{m+k}+1}{a_{m+1}} \in \mathbb{Z}$, and we are done by induction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the point is the midpoint of a segment Let a circle (O) and a point A outside of the circle. AB, AC are tangents of (O) (B,C $\in$ (O)). BD is an diameter of (O). CK perpendicular to BD (K $\in$ BD). Let I is intersection of CK and AD. Prove that I is midpoint of KC
|
Without any loss of generality we can assume $B(a,0),O(0,0)$ so that $D(-a,0)$ the equation of the circle $(x-0)^2+(y-0)^2=(a-0)^2\implies x^2+y^2=a^2$
Using this, the equation of the tangent at any point $P(a\cos\theta,a\sin\theta)$ will be
$xa\cos \theta+ya\sin\theta=a^2\implies x\cos \theta+y\sin \theta=a$
So, $AB: x=a--->(1)$ (putting $\theta=0$)
Let $C$ be $(a\cos \beta,a\sin \beta)$ so $CA: x\cos\beta+y\sin\beta=a--->(2)$ (putting $\theta=\beta$)
Solving for $y$ we get, $y=\frac{a(1-\cos\beta)}{\sin \beta}$ so $A(a,\frac{a(1-\cos\beta)}{\sin \beta})$
$AD$ will be $$\frac{y-0}{x+a}=\frac{0-\frac{a(1-\cos\beta)}{\sin \beta})}{-a-a}=\frac{1-\cos\beta}{2\sin \beta}\implies 2y=(x+a)\frac{1-\cos\beta}{\sin \beta}$$
Clearly, the abscissa of $I$ will be $a\cos \beta$
So, $2y=(a\cos \beta+a)\frac{1-\cos\beta}{\sin \beta}=a\sin\beta\implies y=\frac{a\sin\beta}2$
Hence, $I(a\cos\beta,\frac{a\sin\beta}2)$ is the midpoint of $C(a\cos\beta,a\sin\beta)$ and $K(a\cos\beta,0)$
| {
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"question_score": "3",
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How to compute the determinant of a tridiagonal matrix with constant diagonals? How to show that the determinant of the following $(n\times n)$ matrix
$$\begin{pmatrix}
5 & 2 & 0 & 0 & 0 & \cdots & 0 \\
2 & 5 & 2 & 0 & 0 & \cdots & 0 \\
0 & 2 & 5 & 2 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots & \vdots & \vdots \\
0 & \cdots & \cdots & 0 & 2 & 5 & 2 \\
0 & \cdots & \cdots & \cdots & \cdots & 2 & 5
\end{pmatrix}$$
is equal to $\frac13(4^{n+1}-1)$?
More generally:
How does one compute the determinant of the following tridiagonal matrix (where the three diagonals are constant)?
$$M_n(a,b,c) = \begin{pmatrix}
a & b & 0 & 0 & 0 & \cdots & 0 \\
c & a & b & 0 & 0 & \cdots & 0 \\
0 & c & a & b & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots& \vdots & \vdots \\
0 & \cdots & \cdots & 0 & c & a & b \\
0 & \cdots & \cdots & \cdots & \cdots & c & a
\end{pmatrix}$$
Here $a,b,c$ can be taken to be real numbers, or complex numbers.
In other words, the matrix $M_n(a,b,c) = (m_{ij})_{1 \le i,j \le n}$ is such that
$$m_{ij} = \begin{cases}
a & i = j, \\
b & i = j - 1, \\
c & i = j + 1, \\
0 & \text{otherwise.}
\end{cases}$$
There does not seem to be an easy pattern to use induction: the matrix is not a diagonal block matrix of the type $M = \bigl(\begin{smallmatrix} A & C \\ 0 & B \end{smallmatrix}\bigr)$ (where we could use $\det(M) = \det(A) \det(B)$ for the induction step), and there are no lines or columns with only one nonzero entry, so Laplace expansion gets complicated quickly.
Is there a general pattern that one could use? Or is the answer only known on a case-by-case basis? It's possible to compute the determinant by hand for small $n$:
$$\begin{align}
\det(M_1(a,b,c)) & = \begin{vmatrix} a \end{vmatrix} = a \\
\det(M_2(a,b,c)) & = \begin{vmatrix} a & b \\ c & a \end{vmatrix} = a^2 - bc \\
\det(M_3(a,b,c)) & = \begin{vmatrix} a & b & 0 \\ c & a & b \\ 0 & c & a \end{vmatrix} = a^3 - 2abc
\end{align}$$
But there is no readily apparent pattern and the computation becomes very difficult when $n$ gets large.
| Your determinant is equal to
$$
2^n\det\begin{bmatrix}2x & 1 & 0 & 0 & 0 & \cdots & 0\\ 1 & 2x & 1 & 0 & 0 & \cdots & 0\\ 0 & 1 & 2x & 1 & 0 & \cdots & 0\\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots &\vdots\\ 0 &
\cdots & 0 & 1 & 2x & 1 & 0\\ 0 & \cdots & 0 & 0 & 1 & 2x & 1\\ 0 & \cdots & 0 & 0 & 0 & 1 & 2x\end{bmatrix}=:2^nD_n(x),
$$
with $x=5/4$. As in Calvin Lin's answer, $D_n(x)$ satisfies a recurrence, namely $D_n(x)=2xD_{n-1}(x)-D_{n-2}(x)$, which can be obtained by expanding $D_n(x)$ by minors on its first row and then expanding the $(n-1)$-by-$(n-1)$ determinant one gets in the second term by minors on its first column. This is the defining recurrence for the Chebyshev polynomials of the first and second kinds, which are denoted $T_n(x)$ and $U_n(x)$. Furthermore, $U_0(x)=1=D_0(x)$ and $U_1(x)=2x=D_1(x)$. So $D_n(x)=U_n(x)$.
The Chebyshev Polynomials are related to expansions of trigonometric or hyperbolic functions. In the case of polynomials of the second kind,
$$
\begin{aligned}
U_n(\cos t)&=\frac{\sin (n+1)t}{\sin t},\\
U_n(\cosh t)&=\frac{\sinh (n+1)t}{\sinh t}.
\end{aligned}
$$
Using the second of these, we let $\cosh t=\frac{5}{4}$ and find that $e^t=\frac{1}{2}$ or $2$. Choosing one of these solutions, say $e^t=2$, one can evaluate both $\sinh (n+1)t$ and $\sinh t$. The end result is the given formula.
This works for general $x,$ but as noted in Marc van Leeuwen's comment to user17762's answer, special care is required when $x=\pm1.$ Solving $\cosh t=x$ or $e^t+e^{-t}=2x$ we find $e^t=x\pm\sqrt{x^2-1},$ which yields
$$
\begin{aligned}
U_n(x)=U_n(\cosh t)&=\frac{(x+\sqrt{x^2-1})^{n+1}-(x-\sqrt{x^2-1})^{n+1}}{x+\sqrt{x^2-1}-(x-\sqrt{x^2-1})}\\
&=\frac{(x+\sqrt{x^2-1})^{n+1}-(x-\sqrt{x^2-1})^{n+1}}{2\sqrt{x^2-1}}.
\end{aligned}
$$
This clearly reduces to a polynomial in $x,$ but is more useful than the polynomial form for evaluation. When $x=\pm1,$ which is equivalent to $t=im\pi,$ l'Hôpital's rule is needed in the evaluation:
$$
\begin{aligned}
U_n(\pm1)=U_n(\cosh im\pi)&=\lim_{t\to im\pi}\frac{e^{(n+1)t}-e^{-(n+1)t}}{e^t-e^{-t}}\\
&=\lim_{t\to im\pi}\frac{(n+1)e^{(n+1)t}+(n+1)e^{-(n+1)t}}{e^t+e^{-t}}=(n+1)(\pm1)^n.
\end{aligned}
$$
Revised question: The method can be adapted to handle the more general form in the revised question of February 2015. If $bc=0$ the determinant is clearly $a^n.$ Otherwise, let $g$ be a solution to $g^2=bc$ and define $x:=\frac{a}{2g},$ $y:=\frac{b}{g}.$ Since $\frac{c}{g}=\frac{g}{b}=y^{-1},$ the more general determinant is equal to
$$
g^n\det\begin{bmatrix}2x & y & 0 & 0 & 0 & \cdots & 0\\ y^{-1} & 2x & y & 0 & 0 & \cdots & 0\\ 0 & y^{-1} & 2x & y & 0 & \cdots & 0\\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots &\vdots\\ 0 &
\cdots & 0 & y^{-1} & 2x & y & 0\\ 0 & \cdots & 0 & 0 & y^{-1} & 2x & y\\ 0 & \cdots & 0 & 0 & 0 & y^{-1} & 2x\end{bmatrix}=:g^nD_n(x,y).
$$
But $D_n(x,y)$ satisfies the same recurrence as $D_n(x)$ and the same initial conditions, and so $D_n(x,y)$ also equals $U_n(x).$ (It is independent of $y.$)
| {
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$ \frac{1}{3a^2+1}+\frac{1}{3b^2+1}+\frac{1}{3c^2+1}+\frac{1}{3d^2+1} \geq \frac{16}{7}$ Let $a,b,c,d >0$ and $a+b+c+d=2$. Prove this:
$$ \frac{1}{3a^2+1}+\frac{1}{3b^2+1}+\frac{1}{3c^2+1}+\frac{1}{3d^2+1} \geq \frac{16}{7}$$
| This is a brute force approach.
First, let's show it for $a,b,c,d\geq 0$, since it is true in that case, too, and this set is compact, so if there is a minimum, it is reached somwhere.
Let $f(x)=\frac{1}{3x^2+1}$. Then $f''(x)=\frac{6(9x^2-1)}{(3x^2+1)^3}$. So $f(x)$ is convex on $[1/3,2]$. In particular, if $a,b,c,d\geq 1/3$ then $$f(a)+f(b)+f(c)+f(d)\geq 4f\left(\frac{a+b+c+d}4\right)=4f\left(\frac 1 2\right) = \frac{16}7$$
So, to find a counter-example, we need some of $a,b,c,d$ to be in $[0,1/3)$. For now, assume $a\leq b\leq c\leq d$. By convexity of $f(x)$, we can assume that any values $\geq 1/3$ are equal.
It can be shown pretty directly that if $a,b,c<1/3$ and $d=2-(a+b+c)$ that:
$$f(a)+f(b)+f(c)+f(d)\geq f(1/3)+f(1/3)+f(1/3)+f(2) > 16/7$$
So, if $f(a)+f(b)+f(c)+f(d)$ takes any value smaller than $16/7$, it must be with:
$$a<1/3, c=d\geq 1/3$$
Now, if $a,b\in (0,1/3)$, then consider $f(a-\delta)+f(b+\delta)$ for small $\delta>0$. By the intermediate value theorem, $f(a-\delta) = f(a) - f'(a_0)\delta$ for some $a-\delta<a_0<a$ and $f(b+\delta) = f(b)+f'(b_0)\delta$ for some $b<b_0<b+\delta$. So $$f(a-\delta) + f(b+\delta) = f(a)+f(b) + \delta(f'(b_0)-f'(a_0))$$
Since $f''$ is negative on $(0,1/3)$, and $b_0>a_0$, then $f'(b_0)<f'(a_0)$, so we get:
$$f(a+\delta)+f(b-\delta) < f(a)+f(b)$$
So if there is a minimum reached with $0\leq a\leq b\leq 1/3$, then that minimum must be reached with either $a=0$ or $b=1/3$. But once $b\in[1/3,2]$, it is in the region of convexity, so we can get a miminum with $a\in[0,1/3)$ and $b=c=d$.
So we've reduced the cases to:
$$a=0\leq b < 1/3 <c=d=1-\frac{b}{2}$$
and
$$0\leq a < 1/3 < b=c=d=\frac{2-a}{3}$$
Note the minimum is actually reached when $a=0$ and $b=c=d=2/3$, so we have to take care with each of these cases. We essentially need to minimize the two formulas:
$$1+f(b) + 2f\left(1-\frac{b}{2}\right), 0\leq b<1/3$$
and
$$f(a) + 3f\left(\frac{2-a}{3}\right), 0\leq a< 1/3$$
The fact that the first is minimized when $b=1/3$ and the second when $a=0$ are not obvious to me, but that is what Wolfram Alpha says. We compute from there see we get at least $16/7$ in both cases.
Given that the region $a,b,c,d\geq 0$ and $a+b+c+d=2$ is a regular tetrahedron, and the minimum value is assumed at the center of mass and at the centers of the faces, it seems like you might be able to make a geometric argument, rather than this brute force approach. I'm just not seeing it.
| {
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Listing all the subfield and corresponding subgroup Here is an exercise in the book of Dummit Foote :
Find the Galois group of the splitting field of $(x^2-2)(x^2-3)(x^2-5)$ over $\mathbb{Q}$. Then list all the subgroups and the corresponding subfield
Here is my argument :
It is not hard to see that $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})=K$ is the splitting field for the above polynomial. Since the degree of the extension $K/\mathbb{Q}$ is 8, so the order of the Galois group is 8.
The automorphism in $Gal(K/\mathbb{Q})$ are :
\begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}
\begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}
\begin{cases}
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\sqrt{2}\mapsto -\sqrt{2}\\
\end{cases}
\begin{cases}
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}
\begin{cases}
\sqrt{5}\mapsto \sqrt{5}\\
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\end{cases}
\begin{cases}
\sqrt{5}\mapsto \sqrt{5}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{2}\mapsto -\sqrt{2}\\
\end{cases}
\begin{cases}
\sqrt{5}\mapsto -\sqrt{5}\\
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\end{cases}
\begin{cases}
\sqrt{5}\mapsto \sqrt{5}\\
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\end{cases}
From there, we see that except the identity map(the last map) are of order 2. But I do not know any group of order 8 having that property.
My first question is : what is the Galois group in this case ?
Here is the other argument :
Any element of $K$ can be represented in the form :
$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}+e\sqrt{5}+m\sqrt{10}+n\sqrt{15}+p\sqrt{30}$
Under the map : \begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}
this element turns to : $a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6}-e\sqrt{5}+m\sqrt{10}+n\sqrt{15}-p\sqrt{30}$.
This element is fixed under that map iff $b=c=e=p=0$ or it must has the form : $a+b\sqrt{6}+c\sqrt{10}+d\sqrt{15}$.
But from that I can not deduce the corresponding subfield with the subgroup generated by the above map. My second question is : What is that corresponding subfield ?
| For the first question, consider $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Note this relationship for "independent" automorphisms.
Alright, since I don't know how to do the larger tables in latex here (usually I use xymatrix!) lets (diagrammatically) look at the example at the beginning of 14.2 in Dummit and Foote. The Galois group of $\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$ gives the table
$$ \begin{array}{cccccccc}
& \{1\} & \\
\nearrow & \uparrow & \nwarrow \\
\{1,\tau\} & \{1,\sigma\tau\} & \{1,\sigma\} \\
\nwarrow & \uparrow & \nearrow \\
& \{1,\sigma,\tau,\sigma\tau\} \end{array}$$
In particular, when both $\sqrt{2}$ and $\sqrt{3}$ are permuted, the fixed field is $\mathbb{Q}(\sqrt{6})$.
Now, in our larger example, if $\sqrt{6}$ is fixed, the automorphism $\sigma_2\sigma_3$ is being applied (where $\sigma_n:\sqrt{n}\mapsto -\sqrt{n}$). Similarly, when $\sqrt{10}$ and $\sqrt{15}$ are fixed, the respective automrophisms are $\sigma_2\sigma_5$ and $\sigma_3\sigma_5$. Here's where a complete description of the Galois group is useful:
$$ \operatorname{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})/\mathbb{Q}) \simeq \{1,\sigma_2,\sigma_3,\sigma_5,\sigma_2\sigma_3,\sigma_2\sigma_5,\sigma_3\sigma_5,\sigma_2\sigma_3\sigma_5\} $$
The subgroup you're interested in is then $\{1,\sigma_2\sigma_3,\sigma_2\sigma_5,\sigma_3\sigma_5\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/270780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate value of expression $(\sin^6 x+\cos^6 x)/(\sin^4 x+\cos^4 x)$
Calculate the value of expresion:
$$
E(x)=\frac{\sin^6 x+\cos^6 x}{\sin^4 x+\cos^4 x}
$$
for $\tan(x) = 2$.
Here is the solution but I don't know why $\sin^6 x + \cos^6 x = ( \cos^6 x(\tan^6 x + 1) )$, see this:
Can you explain to me why they solved it like that?
| $$\tan x=2\implies \frac{\sin x}2=\frac{\cos x}1\implies \frac{\sin^2x}4=\frac{\cos^2x}1=\frac{\sin^2x+\cos^2x}{4+1}=\frac15$$
$$\frac{\sin^6 x+\cos^6 x}{\sin^4 x+\cos^4 x}=\frac{\left(\frac45\right)^3+\left(\frac15\right)^3}{\left(\frac45\right)^2+\left(\frac15\right)^2}=\frac{(4^3+1)5^2}{5^3(4^2+1)}=\frac{13}{17}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/271458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If x,y,z are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what If $x,y, z$ are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what?
$108$ , $216$ , $405$ , $1048$
| The problem can be restated as minimizing ${1 \over x^2} + {8 \over y^2} + {27 \over z^2}$ subject to $x + y + z = 1$. By Lagrange multipliers you seek an $(x,y,z)$ and $\lambda$ such that $-{2 \over x^3} = \lambda = -{16 \over y^3} = -{54 \over z^3}$. Dividing by $-2$ and taking cube roots leads to
$${1 \over x} = {2 \over y} = {3 \over z}$$
This can be rewritten as
$$y = 2x,\,\,\,\,\,\,\,\,\,\,\, z = {3 \over 2} y = 3x$$
Plugging this back into the constraint gives
$$x + 2x + 3x = 1$$
So $x = {1 \over 6}$, whereupon $y = {1 \over 3}$ and $z = {1 \over 2}$. In this case, ${1 \over x^2} + {8 \over y^2} + {27 \over z^2} = 36 + 72 + 27*4 = 216$. This has to be the absolute minimum since the function ${1 \over x^2} + {8 \over y^2} + {27 \over z^2}$ goes to infinity as one approaches the hyperplanes $x = 0$, $y = 0$, or $z = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/275153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Evaluating the series: $1 +(1/3)(1/4) +(1/5)(1/4^2)+(1/7)(1/4^3)+ \cdots$
Sum the series:
$$1 + \dfrac13 \cdot \dfrac14 + \dfrac15 \cdot \dfrac1{4^2} + \dfrac17 \cdot \dfrac1{4^3} + \cdots$$
How can I solve this? I am totally stuck on this problem.
| Using Infinite geometric series, $$1+x+x^2+\cdots=\frac1{1-x}$$ for $|x|<1$
Applying integration wrt to $x, \log(1-x)=-x-\frac{x^2}2-\frac{x^3}3+\cdots+c$ where $c$ is an arbitrary constant of indefinite integral.
Putting $x=0,\log(1)=c\implies c=0$
So, $$\log(1-x)=-x-\frac{x^2}2-\frac{x^3}3+\cdots$$
Putting $x=y$ we get, $$\log(1-y)=-y-\frac{y^2}2-\frac{y^3}3+\cdots$$
Putting $x=-y$ we get, $$\log(1+y)=y-\frac{y^2}2+\frac{y^3}3+\cdots$$
So, $$\log(1+y)-\log(1-y)=2(y+\frac{y^3}3+\frac{y^5}5+\cdots)$$ for $|y|<1$
Consequently, $$1 + \dfrac13 \cdot \dfrac14 + \dfrac15 \cdot \dfrac1{4^2} + \dfrac17 \cdot \dfrac1{4^3} + \cdots=2\left(\frac12+\frac{\left(\frac12\right)^3}3+\frac{\left(\frac12\right)^5}5+\cdots\right)$$
$$=\log(1+\frac12)-\log(1-\frac12)=\log\left(\frac{\frac32}{\frac12}\right)=\log 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/276753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum of the form $r+r^2+r^4+\dots+r^{2^k} = \sum_{i=1}^k r^{2^i}$ I am wondering if there exists any formula for the following power series :
$$S = r + r^2 + r^4 + r^8 + r^{16} + r^{32} + ...... + r^{2^k}$$
Is there any way to calculate the sum of above series (if $k$ is given) ?
| Consider the the sequence $a_n$ defined as:
If $n=1$ then $a_n=1$, else if $n>1$ then $a_n=-1$.
This is the sequence $a_n = 1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,...$
The Dirichlet inverse (call it $b_n$) of $a_n$ is the oeis sequence called
"Number of ordered factorizations of n" http://oeis.org/A074206 , starting: $b_n = 0, 1, 1, 1, 2, 1, 3, 1, 4, 2, 3, 1, 8,...$, which is also a Dirichlet series. Notice that the first term: $b_0 = 0$.
For reasons I can not explain clearly, $b_n$ has the ordinary generating function:
$$\sum\limits_ {n = 1}^{\infty} b_n r^n =
r + \sum\limits_ {a =
2}^{\infty} r^{a} + \sum\limits_ {a =
2}^{\infty}\sum\limits_ {b =
2}^{\infty} r^{ab} + \sum\limits_ {a =
2}^{\infty}\sum\limits_ {b = 2}^{\infty}\sum\limits_ {c =
2}^{\infty} r^{abc} + \sum\limits_ {a =
2}^{\infty}\sum\limits_ {b = 2}^{\infty}\sum\limits_ {c =
2}^{\infty}\sum\limits_ {d = 2}^{\infty} r^{abcd} + ... $$
Evaluate each sum partially to the summation index equal to $2$:
$$r + \sum\limits_ {a = 2}^{2} r^{a} + \sum\limits_ {a = 2}^{2}\sum\limits_ {b =
2}^{2} r^{ab} + \sum\limits_ {a = 2}^{2}\sum\limits_ {b =
2}^{2}\sum\limits_ {c =
2}^{2} r^{abc} + \sum\limits_ {a = 2}^{2}\sum\limits_ {b =
2}^{2}\sum\limits_ {c = 2}^{2}\sum\limits_ {d =
2}^{2} r^{abcd} + ... $$
Simplify:
$$r + r^{2} + r^{2*2} + r^{2*2*2} + r^{2*2*2*2} + r^{2*2*2*2*2} + ...... + r^{2^k}$$
multiply out:
$$r + r^{2} + r^{4} + r^{8} + r^{16} + r^{32} + ...... + r^{2^k}$$
call it $S$ and you are done.
The sum:
$$\sum\limits_ {n = 1}^{\infty} b_n r^n$$
above, probably does not have a closed form other than the multiple sums, and therefore $S$ probably does not have a closed form either. But it points out what the scope of generating functions is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/276892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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"answer_id": 4
} |
Definite integral of inverse cosine I want to calculate the definite integral $\Large{\int_0^{\frac \pi 2} \frac{1}{2 + \cos x} \; dx}$
I tried it with the properties of definite integrals but it was of no help.
| Firstly, rewrite integral as
$$\displaystyle \int\limits_{0}^{\frac{\pi}{2}}{\dfrac{dx}{2+\cos{x}}}=\int\limits_{0}^{\frac{\pi}{2}}{\dfrac{dx}{1+\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}+ \cos^2\dfrac{x}{2}-\sin^2\dfrac{x}{2}}}=\int\limits_{0}^{\frac{\pi}{2}}{\dfrac{dx}{1+2\cos^2\dfrac{x}{2}}}.$$
Substitute
\begin{gather}
t=\tan{\dfrac{x}{2}}, \\
x=2\arctan{t}, \\
dx=\dfrac{2 \, dt}{1+t^2}, \\
x\in \left[0,\, {\frac{\pi}{2}} \right] \Leftrightarrow t\in \left[0,\, {1} \right], \\
1+t^2=1+{\tan}^2{\dfrac{x}{2}}=\dfrac{1}{\cos^2\tfrac{x}{2}}, \\
\cos^2\dfrac{x}{2}=\dfrac{1}{1+t^2}.
\end{gather}
Now
\begin{gather}
\int\limits_{0}^{\frac{\pi}{2}}{\dfrac{dx}{1+2\cos^2\dfrac{x}{2}}}=2\int\limits_{0}^{1}{\dfrac{dt}{\left(1+\dfrac{2}{1+t^2}\right) \left(1+t^2 \right)}}=2\int\limits_{0}^{1}{\dfrac{dt}{3+t^2}}.
\end{gather}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/279135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
the value of :$ ((a-b)(b-c)(c-a))^2$ If the polynomial : $f(x)=x^3-3x+2$ have the roots :$a,b,c$
How to find the value of :$$ ((a-b)(b-c)(c-a))^2$$
| Notice that $f(x)=(x+2)(x-1)(x-1)$. This means the polynomial has three real roots which are $-2$, $1$ and $1$. Therefore you have that $((a-b)(b-c)(c-a))^2=(a-b)^2(b-c)^2(c-a)^2$. Notice that you have a cyclic product of squares. This means the product will be invariant of your choice of $a$, $b$ and $c$. You get $((a-b)(b-c)(c-a))^2=(-2-1)^2(1-1)^2(1-2)^2=0$. Finally, if you are wondering how to get the factoring of the polynomial, a good strategy is to guess a root and then if the root is $a\in\mathbb{R}$ divide the polynomial by $x-a$. The likelihood of guessing the root in little time will come with exposure to such polynomial ring exercises.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/281950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\frac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$ Let $a,b,c$ be positive numbers. Prove that $$\dfrac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$$
| You labeled this as homework, but I am not sure what methods you have learnt. (Are you in HSGS of Vietnam for example?) I would assume that you know the tangent line method, and will add more explanation if you need it.
Since LHS is homogeneous, we may assume that $a+b+c = 3$. So we need to prove that
$$\frac{(3-2a)^2}{(3-a)^2+a^2} + \frac{(3-2b)^2}{(3-b)^2+b^2} + \frac{(3-2c)^2}{(3-c)^2+c^2} \ge \frac{3}{5}$$
Note that
$$\frac{(3-2a)^2}{(3-a)^2+a^2} \ge \frac{1}{5} - \frac{18}{25}(a-1)$$
(This is equivalent to $(2a+1)(a-1)^2 \ge 0$) So
$$\frac{(3-2a)^2}{(3-a)^2+a^2} + \frac{(3-2b)^2}{(3-b)^2+b^2} + \frac{(3-2c)^2}{(3-c)^2+c^2} \ge \frac{3}{5} - \frac{18}{25}(a-1+b-1+c-1) = \frac{3}{5}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluate the limit $\lim_{n\to\infty}\left[n+{n^2}\log{\frac{n}{n+1}}\right]$ Evaluate
$$\lim_{n\to\infty}\left[n+{n^2}\log{\frac{n}{n+1}}\right]$$
I know that this limit is equal to $\frac1{2}$ but I don't know how to do it.
Thanks!
| Recall that $\log(1-x) = -x - \dfrac{x^2}2 + \mathcal{O}(x^3)$. Hence,
\begin{align}
\log \left(\dfrac{n}{n+1}\right) & = \log \left(1-\dfrac1{n+1}\right)\\
& = -\dfrac1{n+1} - \dfrac1{2(n+1)^2} + \mathcal{O} \left(\dfrac1{(n+1)^3}\right)\\
n^2\log \left(\dfrac{n}{n+1}\right) & = -\dfrac{n^2}{n+1} - \dfrac{n^2}{2(n+1)^2}+ \mathcal{O} \left(\dfrac1{(n+1)}\right)\\
n + n^2\log \left(\dfrac{n}{n+1}\right) & = n -\dfrac{n^2}{n+1} - \dfrac{n^2}{2(n+1)^2} + \mathcal{O} \left(\dfrac1{(n+1)}\right)\\
& = \dfrac{n^2+n-n^2}{n+1} - \dfrac{n^2}{2(n+1)^2} + \mathcal{O} \left(\dfrac1{(n+1)}\right)\\
& = \dfrac{n}{n+1} - \dfrac{n^2}{2(n+1)^2} + \mathcal{O} \left(\dfrac1{(n+1)}\right)\\
\end{align}
Hence, the limit is $\dfrac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/282972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How would I find the area of a triangle given three sides and using either the sine/cosine laws? Triangle ABC has sides $8.5m$ (a), $7.1$ (b), and $9$ (c). I have been asked to find the area of the triangle using trigonometry.
| If you must use Trigonometry, we can use this formula:
$$ K = \frac{1}{2} ab \sin C $$
In order to find the $ \sin C $, we can use the law of cosines:
$$ c^2 = a^2 + b^2 - 2ab \cos C $$
$$ 9^2 = 8.5^2 + 7.1^2 - 2 \cdot 8.5 \cdot 7.1 \cos C $$
$$ -41.66 = -120.7 \cos C $$
$$ 0.34515327257 = \cos C $$
$$ C \approx 69.80^{\circ} $$
Now, we have:
$$ K = \frac{1}{2} \cdot 8.5 \cdot 7.1 \sin 69.8 $$
$$ K \approx 30.175 \sin 69.8 $$
$$ \color{green}{K \approx 28.32} $$
If you only have a scientific calculator, you can also avoid the inverse cosine by using:
$$ \sin x = \sqrt{1 - \cos^2 x}$$
So:
$$ K = \frac{1}{2} \cdot 8.5 \cdot 7.1 \sqrt{1 - 0.34515327257^2} $$
$$ K = \frac{1}{2} \cdot 8.5 \cdot 7.1 \cdot 0.9385463325985667 $$
$$ K = 28.3206355862 $$
Applying Heron's Formula verifies this result. We have $s = 12.3$.
$$ \sqrt{12.3(12.3 - 8.5)(12.3 - 7.1)(12.3 - 9)} $$
$$ \sqrt{802.058} \approx 28.3206 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/283216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the order of a recurrence relation I have some homework that I'm working on where there is a whole section of problems I need to solve taking the following form:
"Assume that T(1) = 1, and find the order of function T(n)."
I have no idea what this means, really, and I'm having difficulty finding anything specific enough online. What I've found (I think) is that a function of order k is a function that can be solved using the k previous terms of the sequence.
However, I don't really know how to apply this. An example of what I'm working with is as follows:
$$T(n) = 3T(\frac{n}{2}) + n$$
If someone could walk me through exactly what I need to do to find the order of this function, that would be great; I'm really lost.
Thanks in advance!
| We want to find $T(n)$ given the recurrence $$T(n) = 3T(n/2) + n$$ Let us write $n= 2^m$ and call $g(m) = T(2^m)$. We then have
\begin{align}
g(m) & = 3 g(m-1) + 2^m = 2^m + 3 (2^{m-1} + 3g(m-2)) = 2^m + 3 \cdot 2^{m-1} + 3^2 g(m-2)\\
& = 2^m + 3 \cdot 2^{m-1} + 3^2 \cdot 2^{m-2} + 3^3 g(m-3)\\
& = 2^m + 3 \cdot 2^{m-1} + 3^2 \cdot 2^{m-2} + 3^3 \cdot 2^{m-3} + 3^4 g(m-4)
\end{align}
Hence, using induction, you can prove that
\begin{align}
g(m) & = 2^m + \left(\dfrac32\right) \cdot 2^m + \left(\dfrac32\right)^2 \cdot 2^m + \cdots + \left(\dfrac32\right)^{m-1} \cdot 2^m + 3^m g(0)\\
& = 2^m \dfrac{(3/2)^m-1}{3/2-1} + 3^m g(0) = 2 \cdot (3^m-2^m) + 3^m g(0)\\
& = (g(0) + 2) 3^m - 2^{m+1} = (g(0) + 2) 3^{\log_2(n)} - 2n\\
& = (g(0) + 2) n^{\log_2(3)} - 2n\\
\end{align}
Hence, $T(n)$ grows as $n^{\log_2(3)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/283793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$ I am trying to prove that
$$\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$$
I know how to deal with integrals involving cyclotomic polynomials and nested logarithms but I have no idea with this one.
| $$
\begin{align}
\int_0^1\frac{t+1}{t^2+1}\frac{t-1}{\log(t)}\,\mathrm{d}t
&=\int_0^1\frac{t+1}{t^2+1}\int_0^1t^x\,\mathrm{d}x\,\mathrm{d}t\\
&=\int_0^1\int_0^1\frac{t^{x+1}+t^x}{t^2+1}\,\mathrm{d}t\,\mathrm{d}x\\
&=\int_0^1\left(\frac1{x+1}+\frac1{x+2}-\frac1{x+3}-\frac1{x+4}+\dots\right)\,\mathrm{d}x\\
&=\left(\log\left(\frac21\right)+\log\left(\frac32\right)\right)
-\left(\log\left(\frac43\right)+\log\left(\frac54\right)\right)+\dots\\
&=\log\left(\frac31\right)-\log\left(\frac53\right)+\log\left(\frac75\right)-\log\left(\frac97\right)+\log\left(\frac{11}9\right)-\dots\\
&=\log\left(\frac31\cdot\frac35\cdot\frac75\cdot\frac79\cdot\frac{11}9\cdots\right)\\
&=\lim_{n\to\infty}\log\left(\frac{\Gamma\left(\frac54\right)^2}{\Gamma\left(\frac34\right)^2}
\frac{\Gamma\left(\frac{4n+3}4\right)^2}{\Gamma\left(\frac{4n+5}4\right)^2}(4n+3)\right)\\
&=2\log\left(2\frac{\Gamma\left(\frac54\right)}{\Gamma\left(\frac34\right)}\right)
\end{align}
$$
The last equality is due to Gautschi's inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
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Solving lyapunov equation, Matlab has different solution, why? I need to solve the lyapunov equation i.e. $A^TP + PA = -Q$. With $A = \begin{bmatrix} -2 & 1 \\ -1 & 0 \end{bmatrix}$ and $Q = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$.
Hence...
$$
\begin{bmatrix}
-2 & -1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
P_{11} & P_{12} \\
P_{12} & P_{22}
\end{bmatrix}
+
\begin{bmatrix}
P_{11} & P_{12} \\
P_{12} & P_{22}
\end{bmatrix}
\begin{bmatrix}
-2 & 1 \\
-1 & 0
\end{bmatrix} +
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} = 0
$$
$$
\begin{bmatrix}
-4P_{11} - 2P_{12} + 1 & P_{11} -2P_{12} - P_{22} \\
P_{11} - 2P_{12} - P_{22} & 2P_{12} + 1
\end{bmatrix} = 0
$$
$$
\begin{bmatrix}
-4 & -2 & 0 \\
1 & -2 & -1 \\
1 & -2 & -1 \\
0 & 2 & 0
\end{bmatrix}
\begin{bmatrix}
P_{11} \\
P_{12} \\
P_{22}
\end{bmatrix} =
\begin{bmatrix}
-1 \\
0 \\
0 \\
-1
\end{bmatrix}
\Rightarrow
\begin{bmatrix}
-4 & -2 & 0 \\
1 & -2 & -1 \\
0 & 2 & 0
\end{bmatrix}
\begin{bmatrix}
P_{11} \\
P_{12} \\
P_{22}
\end{bmatrix} =
\begin{bmatrix}
-1 \\
0 \\
-1
\end{bmatrix}$$
And such we get that $P = \begin{bmatrix} 1/2 & -1/2 \\ -1/2 & 3/2\end{bmatrix}$.
This is the same solution as given by my professor. I wanted to check however if I can also find the solution using Matlab. I entered the following:
A = [-2 1; -1 0];
Q = [1 0; 0 1];
P = lyap(A,Q)
This however tells me that $P = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 3/2\end{bmatrix}$.
What is going on here? Is Matlab correct or wrong? Or is my solution wrong? Or are we both correct?
| from MATLAB documents it actually solves $AP+PA'+Q=0$ so that the result is different than your equation $A'P+PA+Q=0$
to get the same answer in MATLAB use $A'$ instead of $A$ since $A'P+P(A')'+Q=0$ equals
$A'P+PA+Q=0$ which is equivalent to your equation. eventually the following
lyap(A',Q) matches your results
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Simple question - Proof How is $\frac{1}{2}ln(2x+2) = \frac{1}{2}ln(x+1) $ ?
As $\frac{1}{2}ln(2x+2)$ = $\frac{1}{2}ln(2(x+1))$, how does this become$ \frac{1}{2}ln(x+1)$?
Initial question was $ \int \frac{1}{2x+2} $
What I done was $ \int \frac{1}{u}du $ with $u=2x+2$ which lead to $\frac{1}{2}ln(2x+2) $ but WolframAlpha stated it was $\frac{1}{2}ln(x+1)$
| $\frac{1}{2}\ln(2(x+1))=\ln\sqrt{2}+\frac{1}{2}\ln(x+1)\neq \frac{1}{2}\ln(x+1)$
But as far as $$\int\frac{1}{(2x+2)}dx$$ is concerned , it comes out to be $$\frac{1}{2}\ln(2x+2)+c$$ or you can write it as $$\frac{1}{2}\int\frac{1}{(x+1)}dx$$ which comes out to be $$\frac{1}{2}\ln(x+1)+k$$
Here , the extra $\ln\sqrt{2}$ has been taken care of in constant terms of integration.so, as far as indefinite integration is concerned, it gives family of curves as its solution, and they both represent same family of curves and differ by a constant only.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/286032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding $B_{2\times 2}$ such that $A^{51}=B$ Assuming $$ A=\left(\begin{matrix}0 & \sqrt{3}\\\sqrt{3}\ &4\end{matrix}\right)$$ how to find B such that $$A^{51}=B$$
My attempts:
if find $P,D$ that $D$ be diagonal matrix $A=P^{-1} DP$ then $$A^{51}=P^{-1} D^{51}P$$ therefore $B=P^{-1} D^{51}P$ but how to find $D$ and $P$?
Thanks for any hints.
| Here's another way. The generating function of $A^n$ is $$G(t) = \sum_{n=0}^\infty t^n A^n = (I-tA)^{-1} =
\dfrac{1}{-1+4t+3t^2} \left( \begin {array}{cc} -1+4\,t&-t\sqrt {3}\\ -t
\sqrt {3}&-1\end {array} \right)
$$
Now using a partial fraction decomposition
$$
\eqalign{\frac{1}{-1+4t+3t^2} &={\frac {\sqrt {7}}{14(t+2/3-\sqrt {7}/3)}}-{\frac {\sqrt {7
}}{14(t+2/3+\sqrt {7}/3)}}\cr
&= -\frac{\sqrt{7}}{14} \sum_{n=0}^\infty \frac{t^n}{(-2/3 + \sqrt{7}/3)^{n+1}}
+\frac{\sqrt{7}}{14} \sum_{n=0}^\infty \frac{t^n}{(-2/3 - \sqrt{7}/3)^{n+1}}
}$$
Thus with $\alpha = -2/3 + \sqrt{7}/3$ and $\beta = -2/3 - \sqrt{7}/3$,
the coefficient of $t^{51}$ in $G(t)$ is
$$ \dfrac{\sqrt{7}}{14} \left(\frac{-1}{\alpha^{52}} + \frac{1}{\beta^{52}}\right) \pmatrix{-1 & 0\cr 0 & -1\cr} + \dfrac{\sqrt{7}}{14} \left(\frac{-1}{\alpha^{51}} + \frac{1}{\beta^{51}}\right) \pmatrix{4 & -\sqrt{3}\cr -\sqrt{3} & 0\cr}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/288269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Permutations of elements of a matrix I'm a bit confused about the second part of the question I'm working on. The question is as follows
Let A be the $4 \times 4$ matrix
$$A=\begin{bmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34} \\
a_{41} & a_{42} & a_{43} & a_{44}
\end{bmatrix}$$
(a) Write out all of the permutations $\sigma \in S_4$ with $\sigma(1)=4$ (there are six of them)
(b) Using the general formula for det($A$), write out all the terms corresponding to the permutations you found in part (a)
From my notes I have
To describe $\sigma$ in $S_n$ write $1,2,...,n$
$$\begin{pmatrix}
1 & 2 & 3 & \cdots & 5 \\
\sigma(1) & \sigma(2) & \sigma(3) & \cdots & \sigma(n)
\end{pmatrix}$$
So the six permutations I end up with are
$$\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 1 & 2 & 3
\end{pmatrix}
\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 2 & 1 & 3
\end{pmatrix}
\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 1 & 2
\end{pmatrix}
\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 2 & 3 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 1 & 3 & 2
\end{pmatrix}$$
Now, I am not sure about part (b). I have 6 different permutations of 4 numbers. What terms from the general formula for det($A$) correspond to the permutations I found?
| I believe the following is meant:
The general formula (complete expansion) of the determinant is:
$$\det A = \sum_{\textrm{permutations p}} \left( \textrm{sign p} \right) a_{1, p_1} a_{2, p_2} \textrm{...} a_{n, p_n}.$$
In the case of your first permutation, the term would thus be:
$$\left( -1 \right) a_{1, 4} a_{2, 1} a_{3,2} a_{4, 3},$$
with the others being similarly translated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/288391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Bell Numbers: How to put EGF $e^{e^x-1}$ into a series? I'm working on exponential generating functions, especially on the EGF for the Bell numbers $B_n$.
I found on the internet the EGF $f(x)=e^{e^x-1}$ for Bell numbers. Now I tried to use this EGF to compute $B_3$ (should be 15). I know that I have to put the EGF into a series and have a look at the coefficients. Using $e^{f(x)}=1+f(x)+\frac{f(x)^2}{2!}+\frac{f(x)^3}{3!}+\ldots$ I get
\begin{eqnarray*}
e^{e^x-1}&=&1+(e^x-1)+\frac{(e^x-1)^2}{2!}+\frac{(e^x-1)^3}{3!}\\
&=&1+e^x-1+\frac{e^{x^2}-2e^x+1}{2!}+\frac{e^{x^3}-3e^{x^2}+3e^x-1}{3!}\\
&=&e^x+\frac{e^{x^2}}{2!}-e^x+\frac{1}{2!}+\frac{e^{x^3}}{3!}-\frac{e^{x^2}}{2!}+\frac{e^{x}}{2!}-\frac{1}{3!}\\
&=&\frac{e^{x}}{2!}+\frac{e^{x^3}}{3!}-\frac{1}{2!}+\frac{1}{3!}\\
&=&\frac{1}{2!}e^x+\frac{1}{3!}e^{x^3}-\frac{1}{3}\\
&=&\frac{1}{2!}\left( 1+x^2+\frac{x^4}{2!}+\frac{x^8}{3!} \right)+\frac{1}{3!}\left( 1+x^3+\frac{x^6}{2!}+\frac{x^9}{3!}\right)\\
&=&\frac{1}{2}+\frac{x^2}{2}+\frac{x^4}{4}+\frac{x^8}{24}+\frac{1}{6}+\frac{x^3}{3!}+\frac{x^6}{12}+\frac{x^9}{36}\\
&=&\frac{1}{2}+\frac{x^2}{2}+\frac{x^3}{6}+\ldots
\end{eqnarray*}
I think I can stop here, because the coefficient in front of $\frac{x^3}{3!}$ is not $15$.
Perhaps someone can help me out and give a hint to find my mistake?
| It's probably easiest to expand the exponential in the exponent first, since that will lead to a finite number of terms to be evaluated:
$$\begin{align}e^{(e^x-1)} &= \exp\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right)\\
&=1+\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right) +\frac{1}{2}\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right)^2
+\frac{1}{6}\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right)^3+O(x^4)\\
&=1+\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right)+\frac{1}{2}\left(x^2+2(x)\left(\frac{x^2}{2}\right)+O(x^4)\right)+\frac{1}{6}\left(x^3+O(x^4)\right)\\
&=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^3}{6}+O(x^4)\\
&=1+x+x^2+\frac{5x^3}{6}+O(x^4)
\end{align}$$
From which the coefficients can be read off straightforwardly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/289097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Factorization problem Find $m + n$ if $m$ and $n$ are natural numbers such that: $$\frac {m+n} {m^2+mn+n^2} = \frac {4} {49}\;.$$
My reasoning:
Say: $$m+n = 4k$$ $$m^2+mn+n^2 = 49k$$
It follows:$$(m+n)^2 = (4k)^2 = 16k^2 \Rightarrow m^2+mn+n^2 + mn = 16k^2 \Rightarrow mn = 16k^2 - 49k$$
Since: $$mn\gt0 \Rightarrow 16k^2 - 49k\gt0 \Rightarrow k\gt3$$
Then no more progress.
| It is useful to write the stipulation of the problem in a few different ways:
$$\frac{49}{4}=\frac{(m+n)^2-mn}{m+n}=m+n-\frac{mn}{m+n},$$
$$\frac{49}{4}=\frac{(m+n)n+m^2}{m+n}=n+\frac{m^2}{m+n},$$
$$\frac{49}{4}=\frac{(m+n)m+n^2}{m+n}=m+\frac{n^2}{m+n}.$$
From the first, we glean that $m+n>12$ because $49/4=12+1/4$ and $m,n>0$. From the second and third, we see that $n\le 12$ and $m\le 12$.
Another thing to notice is that $mn/(m+n)$ must reduce to a fraction with a denominator of $4$. This means that
$$4mn=c(m+n)$$
for some $c$ relatively prime to $4$. This implies that $4$ divides $m+n$, so there are only three cases to consider: $m+n=16$, $m+n=20$, $m+n=24$. From the first, we have
$$12+\frac{1}{4}=16-\frac{m(16-m)}{16}$$
which easily gives the solution $\{m,n\}=\{6,10\}$. The other two cases result in quadratic equations with no real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/290166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Help with $\large \intop\frac{\sqrt{x^2-4}}{x} dx$ I have to $\large \intop\frac{\sqrt{x^2-4}}{x} dx$. Can you tell me what to substitute?
Should I substitute $x$ or $\sqrt{x^2-4}$? Would it be better If I'd substitute $x=\tan u$
or should I substitute $\sqrt{x^2-4}= \tan u$?
| You may try several substitutions, perhaps more elementary and maybe easier:
$$(1)\;\;\text{First substitution :}\;\;u=x^2-4\Longrightarrow du=2xdx\Longrightarrow dx=\frac{du}{2\sqrt{u+4}}\;\;,\;\;\text{so :}$$
$$\int\frac{\sqrt{x^2-4}}{x}dx=\int\frac{\sqrt u}{2(u+4)}du$$
$$(2)\;\;\text{Second substitution :}\;\;\;\;\;\;\;\;y^2=u\Longrightarrow 2ydy=du\;\;,\;\;\text{so :}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$$
$$\int\frac{\sqrt u}{2(u+4)}du=\int\frac{y^2}{y^2+4}dy=\int\left(1-\frac{4}{y^2+4}\right)dy=$$
$$=y+2\int\frac{\frac{1}{2}dy}{1+\left(\frac{y}{2}\right)^2}=y+\arctan\frac{y}{2}+C$$
And now go back to your original variable to get:
$$\int\frac{\sqrt{x^2-4}}{x}dx=\sqrt{x^2-4}-2\arctan\frac{\sqrt{x^2-4}}{2}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/290757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Differentiation question? How would I solve the following problem?
Where would the function $|2x-1|$ not be differentiable?
I am thinking it would not be differentiable at $x=1/2$ because there it would be zero.
| Let $f(x) = |2x-1|$. Then, if $x<\frac{1}{2}$, $f(x) = 1-2x$, if $x\geq \frac{1}{2}$, then $f(x) = 2x-1$.
Hence $\lim_{x \downarrow \frac{1}{2}}\frac{f(x) - f(\frac{1}{2})}{x-\frac{1}{2}} = \lim_{x \downarrow \frac{1}{2}}\frac{2x-1}{x-\frac{1}{2}} = +2$, but
$\lim_{x \uparrow \frac{1}{2}}\frac{f(x) - f(\frac{1}{2})}{x-\frac{1}{2}} = \lim_{x \uparrow \frac{1}{2}}\frac{1-2x}{x-\frac{1}{2}} = -2$.
So, the limit $x \to \frac{1}{2}$ does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/291634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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To get addition formula of $\tan (x)$ via analytic methods Assume that we only know $\tan (0)=0$ and also given the relation $\tan'(x)=1+\tan^2(x)$ about $\tan (x)$ and we do not know other $\tan (x)$ relations of trigonometry.
How can I get the additon formula $$ \tan (x+h)=\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)}$$ via using the differential equation ($\tan'(x)=1+\tan^2(x)$) and analytic methods ? (without using geometry)
Could you please provide me with an easy way to get addition formula of $\tan (x+h)$ ?
My attempt:
$$\tan'(x)=1+\tan^2(x)$$
$$\int \frac{d\tan(x)}{1+\tan^2(x)}=\int dx$$
$$\tan(x)- \frac{\tan^3(x)}{3}+ \frac{\tan^5(x)}{5}+....=x$$
$$\tan(h)- \frac{\tan^3(h)}{3}+ \frac{\tan^5(h)}{5}+....=h$$
$$+$$
$$\tan(x)+\tan(h)- (\frac{\tan^3(x)}{3}+\frac{\tan^3(h)}{3})+ (\frac{\tan^5(x)}{5}+\frac{\tan^5(h)}{5})+....=x+h$$
$$\tan(x+h)- \frac{\tan^3(x+h)}{3}+ \frac{\tan^5(x+h)}{5}+....=x+h$$
$$\tan(x+h)- \frac{\tan^3(x+h)}{3}+ \frac{\tan^5(x+h)}{5}+....=\tan(x)+\tan(h)- (\frac{\tan^3(x)}{3}+\frac{\tan^3(h)}{3})+ (\frac{\tan^5(x)}{5}+\frac{\tan^5(h)}{5})+....=x+h$$
Let's define that $\tan(x+h)=\tan(x)+\tan(h)+P(x,h)$
$$(\tan(x)+\tan(h)+P(x,h))- \frac{(\tan(x)+\tan(h)+P(x,h))^3}{3}+ \frac{(\tan(x)+\tan(h)+P(x,h))^5}{5}+....=\tan(x)+\tan(h)- (\frac{\tan^3(x)}{3}+\frac{\tan^3(h)}{3})+ (\frac{\tan^5(x)}{5}+\frac{\tan^5(h)}{5})+....$$
$$P(x,h)- \frac{(\tan(x)+\tan(h))^3+ 3 (\tan(x)+\tan(h))^2 P(x,h)+3 (\tan(x)+\tan(h)) P(x,h)^2+P(x,h)^3}{3}+ \frac{(\tan(x)+\tan(h)+P(x,h))^5}{5}+....=- (\frac{\tan^3(x)}{3}+\frac{\tan^3(h)}{3})+ (\frac{\tan^5(x)}{5}+\frac{\tan^5(h)}{5})+....$$
Let's define that
$$P(x,h)= \tan(x)\tan(h)(\tan(x)+\tan(h))+G(x,h) $$
thus we get
$\tan(x+h)=\tan(x)+\tan(h)+\tan(x)\tan(h)(\tan(x)+\tan(h))+G(x,h)$
I know we can find the solution in my method but so many calculations are needed in that method. It is very very long way.
And Finally I need to get that $$ \tan (x+h)=\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)}=(\tan(x)+\tan(h))(1-\tan(x)\tan(h))^{-1}=(\tan(x)+\tan(h))(1+\tan(x)\tan(h)+\tan^2(x)\tan^2(h)+......)$$
Note:I try to get addition formulas from a given differential equations and initial conditions.
I wish to find a method to get a closed form addition formulas for the problem as shown below.
$U(0)=0$ and also given the relation $U'(x)=1+U^n(x)$ where $n>2$
Thanks a lot for answers
| Here is my approach:
Let
$$
\arctan(x) := \int_0^x \frac{dt}{1+t^2}
$$
and $\tan(x)$ is defined as the inverse of $\arctan(x)$, so that $\tan(0) = 0$.
Consider the differential equation
$$
\frac{d x}{1+x^2} + \frac{dy}{1+y^2} = 0 \tag{DE},
$$
one solution is
$$
\arctan x + \arctan y = c,
$$
but the equation has also the solution
$$
\frac{x + y}{1 - x y} = C.
$$
Since the differential equation has but one distinct solution, the two solutions must be related to one another in a definite way. This relation is expressed by the equation
$$
C = f(c)
$$
Now, let
$$
x = \tan u, \quad y = \tan v,
$$
then
\begin{align}
u + v &= c \\ \\
\frac{\tan u + \tan v}{1 - \tan u \tan v} &= f(c) = f(u +v)
\end{align}
Let $v = 0$, then
$$
\tan u = f(u)
$$
and therefore
$$
\color{blue}{\frac{\tan u + \tan v}{1 - \tan u \tan v} = \tan(u + v).}
$$
Construction of the second solution
Let $x = \tan u$ and $y = \tan v$. By definition
$$
\frac{d x}{d u} = 1 + x^2 \quad \Longrightarrow \quad \frac{d^2 x}{d u^2} = 2x(1+x^2).
$$
Similarly
$$
\frac{d y}{d u} = -\frac{d y}{d v} = -(1+y^2), \mbox{ and } \frac{d^2 y}{d u^2} = \frac{d^2 y}{d v^2} = 2y(1+y^2)
$$
from which follows that
$$
x \frac{d^2 y}{d u^2} - y \frac{d^2 x}{d u^2} = 2xy(y^2 - x^2)
$$
and
$$
x^2\left(\frac{d y}{d u}\right)^2 - y^2 \left(\frac{d x}{d u}\right)^2 = (x^2 - y^2)(1 - x^2 y^2)
$$
Hence
$$
\frac{x \frac{d^2 y}{d u^2} - y \frac{d^2 x}{d u^2}}{x \frac{d y}{d u} - y \frac{d x}{d u}} = - \left(x \frac{d y}{d u} + y \frac{d x}{d u}\right) \frac{2 x y}{1-x^2y^2}
$$
This equation is immediately integrable; the solution is
$$
\log\left(x \frac{d y}{d u} - y \frac{d x}{d u}\right) = \mbox{const.} + \log(1 - x^2 y^2)
$$
or
$$
x \frac{d y}{d v} + y \frac{d x}{d u} = C(1- x^2 y^2).
$$
Using this information, we can see that
$$
\Phi(x,y) = \frac{x(1+y^2) + y(1+x^2)}{1 - x^2 y^2} = \frac{x + y}{1 - x y} = C
$$
is also a solution of (DE).
Other Examples
*
*$\color{green}{\sin(x)}$
Let $y' = \sqrt{1-y^2}$. If we consider the (DE)
$$
\frac{d x}{\sqrt{1-x^2}} + \frac{d y}{\sqrt{1-y^2}} = 0
$$
and define
$$
\arcsin(x) = \int_0^x \frac{d t}{\sqrt{1-t^2}}dt
$$
where $\sin(x)$ is defined as the inverse of $\arcsin(x)$, so that $\sin(0) = 0$, and $\cos(x)$ is defined as $\sqrt{1-\sin^2 (x)}$, with the condition that $\cos(0) = 1$.
One solution for the (DE) is
$$
\arcsin x + \arcsin y = c.
$$
Using the same method as with $\tan(x)$, we can build a second solution:
Let $x = \sin u$, $y = \sin v$, by definition
$$
\frac{d x}{d u} = \sqrt{1-x^2}, \qquad \frac{d^2 x}{d u^2} = -x
$$
Similarly
$$
\frac{d y}{d u} = -\frac{d y}{d u} = -\sqrt{1-y^2}, \qquad \frac{d^2 y}{d u^2} = -y
$$
from which follows
$$
x \frac{d^2 y}{d u^2} - y \frac{d^2 x}{d u^2} = 0
$$
Hence
$$
\frac{x \frac{d^2 y}{d u^2} - y \frac{d^2 x}{d u^2}}{x \frac{d y}{d u} - y \frac{d x}{d u}} = 0
$$
Integrating
$$
x \frac{d y}{d v} + y \frac{d x}{d u} = C
$$
and another solution of the (DE) is
$$
x\sqrt{1-y^2} + y \sqrt{1-x^2} = C
$$
Recapitulating:
The (DE) has two solutions
$$
\arcsin x + \arcsin y = c,
$$
$$
x\sqrt{1-y^2} + y \sqrt{1-x^2} = C
$$
As with $\tan(x)$, the (DE) has but one solution, and the two solutions must be related to one another in a definite way $f(c) =C$.
Let $x = \sin u$ and $y = \sin v$, then
$$
u + v = c
$$
$$
\sin u \cos v + \sin v \cos u = f(c) = f(u+v)
$$
Setting $v = 0$ implies $f(u) = \sin u$ so
$$
\color{blue}{\sin u \cos v + \sin v \cos u = \sin(u +v)}
$$
*
*$\color{green}{\mbox{sn}(x)}$
Let $y' = (1-y^2)^{\frac{1}{2}}(1-k^2y^2)^{\frac{1}{2}}$. The (DE)
$$
\frac{dx}{(1-x^2)^{\frac{1}{2}}(1-k^2x^2)^{\frac{1}{2}}} +\frac{dy}{(1-y^2)^{\frac{1}{2}}(1-k^2y^2)^{\frac{1}{2}}} = 0
$$
has the solutions (using the same technique)
$$
\mbox{argsn } x + \mbox{argsn } y = c
$$
$$
x \frac{d y}{d v} + y \frac{d x}{d u} = C(1-k^2 x^2 y^2)
$$
Let $x = \mbox{sn }u$, $y = \mbox{sn }v$, then
$$
\color{blue}{\mbox{sn}(u+v) = \frac{\mbox{sn }u \,\mbox{sn}'v + \mbox{sn }v \,\mbox{sn}'u}{1-k^2 \mbox{sn}^2u \,\mbox{sn}^2v}}
$$
*
*$\color{green}{\wp(x)}$
Let $y' = \sqrt{4x^3-g_2x-g_3}$. In this case, the (DE) is of the form
$$
\frac{dx}{\sqrt{4x^3-g_2x-g_3}} + \frac{dy}{\sqrt{4y^3-g_2y-g_3}}
$$
and we can derive the addition formula
$$
\color{blue}{\wp(u + v) = - \wp(u) - \wp(v) + \frac{1}{4} \left\{\frac{\wp'(u)-\wp'(v)}{\wp(u)-\wp(v)} \right\}^2}
$$
| {
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"url": "https://math.stackexchange.com/questions/292125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
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Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Let $s_n$ be a sequence defined as given below for $n \geq 1$. Then find out $\lim\limits_{n \to
\infty} s_n$.
\begin{align}
s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx
\end{align}
I have written a solution of my own, but I would like to know it is completely correct, and also I would like if people post more alternative solutions.
| We simplify the formulate for $s_n$ by integrating by parts.
\begin{align}
s_n &= \int\limits_0^1 \frac{nx^{n-1}}{1+x} d x \\
&= \left[
\frac{1}{1+x} \int nx^{n-1} d x
- \int \frac{1}{\left(1+x\right)^2} \left(\int nx^{n-1} d x\right) d x
\right]^1_0 \\
&= \left[\frac{1}{1+x} \int nx^{n-1} d x\right]^1_0
- \left[\int \frac{1}{\left(1+x\right)^2} \left(\int nx^{n-1} dx\right) d x\right]^1_0 \\
&= \left[\frac{x^n}{1+x}\right]^1_0
- \left[\int \frac{x^n}{\left(1+x\right)^2} d x\right]^1_0 \\
&= \frac{1}{2} - \int\limits_0^1 \frac{x^n}{\left(1+x\right)^2} d x \\
\end{align}
Now we estimate the remaining integral in the expression
\begin{align}
I(n) &= \int\limits_0^1 \frac{x^n}{\left(1+x \right)^2} d x \\
&\leq \int\limits_0^1 x^n d x \\
&= \frac{1}{n+1}
\end{align}
Hence, $I(n) \to 0$ as $n \to \infty$.
And so, the expression can be rewritten as
\begin{align}
\lim\limits_{n \to \infty} s_n = \frac{1}{2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/292251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 3
} |
Evaluate:: $ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 +\cdots + \frac 1n\right) $ How to evaluate the series:
$$ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 + \cdots + \frac 1n\right) $$
According to Mathematica, this converges to $ (\log 2)^2 $.
| Recall that, formally,
$$
\left(\sum_{n=1}^{\infty} a_n\right)\left(\sum_{n=1}^{\infty} b_n\right) = \sum_{n=1}^{\infty} c_{n+1},$$
where
$$
c_n = \sum_{k=1}^{n-1} a_k b_{n-k}.
$$
If the series $\sum c_{n+1}$ converges, then the above equality is actually true. You seem to know how to show this, so I'll just demonstrate the formal aspect of the problem.
Let $a_n = b_n = \frac{(-1)^{n}}{n}$. Then
$$
a_k b_{n-k} = \frac{(-1)^n}{k(n-k)} = \frac{(-1)^n}{n}\left(\frac{1}{k}+\frac{1}{n-k}\right),
$$
so that
$$
\begin{align*}
c_n &= \frac{(-1)^n}{n} \sum_{k=1}^{n-1} \left(\frac{1}{k}+\frac{1}{n-k}\right) \\
&= 2\frac{(-1)^n}{n} \sum_{k=1}^{n-1} \frac{1}{k}.
\end{align*}
$$
We therefore have
$$
2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1} \sum_{k=1}^{n} \frac{1}{k} = \left(\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)^2 = (-\log 2)^2 = (\log 2)^2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/292973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 6,
"answer_id": 1
} |
Prove that $1^3 + 2^3 + ... + n^3 = (1+ 2 + ... + n)^2$ This is what I've been able to do:
Base case: $n = 1$
$L.H.S: 1^3 = 1$
$R.H.S: (1)^2 = 1$
Therefore it's true for $n = 1$.
I.H.: Assume that, for some $k \in \Bbb N$, $1^3 + 2^3 + ... + k^3 = (1 + 2 +...+ k)^2$.
Want to show that $1^3 + 2^3 + ... + (k+1)^3 = (1 + 2 +...+ (k+1))^2$
$1^3 + 2^3 + ... + (k+1)^3$
$ = 1^3 + 2^3 + ... + k^3 + (k+1)^3$
$ = (1+2+...+k)^2 + (k+1)^3$ by I.H.
Annnnd I'm stuck. Not sure how to proceed from here on.
| Consider the case where $n = 1$. We have $1^3 = 1^2$. Now suppose $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for some $n \in \mathbb N$. Recall first that $\displaystyle (1 + 2 + 3 + \cdots + n) = \frac{n(n+1)}{2}$ so we know $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 = \bigg(\frac{n(n+1)}{2}\bigg)^2$. Now consider $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 + (n + 1)^3 = \bigg(\frac{n(n+1)}{2}\bigg)^2 + (n+1)^3 = \frac{n^2 (n+1)^2 + 4(n+1)^3}{4} = \bigg( \frac{(n+1)(n+2)}{2} \bigg)^2$. Hence, the statement holds for the $n + 1$ case. Thus by the principle of mathematical induction $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for each $n \in \mathbb N$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/294213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Rightmost digit of $ \left \lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor $ How could I find $$ 0 \leq a \leq 9 $$ such that
$$ \left \lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor \equiv a \mod 10 $$
?
|
This didn't work as a comment to Hagen von Eitzen's answer, so I post it here.
Writing the fraction as
$$
\small\begin{align}
&\frac{10^{20000}}{10^{100}}\frac1{1+3\cdot10^{-100}}\\
&=10^{19900}\left(1-3\cdot10^{-100}+\left(3\cdot10^{-100}\right)^2-\dots
\color{#00A000}{-\left(3\cdot10^{-100}\right)^{199}}\color{#C00000}{+\left(3\cdot10^{-100}\right)^{200}-\dots}\right)\\
&\equiv\color{#00A000}{-3^{199}}\color{#C00000}{+\epsilon}\pmod{10^{100}}
\end{align}
$$
where $0\le\epsilon\le\left(\frac{9}{10}\right)^{100}$
Thus,
$$
\begin{align}
\left\lfloor\frac{10^{20000}}{10^{100}+3}\right\rfloor
&\equiv-3^{199}\pmod{10^{100}}\\
&\equiv-\left(3^4\right)^{49}\,3^3\pmod{10}\\[6pt]
&\equiv-1^{49}\,7\pmod{10}\\[12pt]
&\equiv3\pmod{10}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/294400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Maximum Value of Trig Expression What is the general method for finding the maximum and minimum value of a trig expression without the use of a calculator. For example, given the expression :
$$\sin(3x) + 2 \cos(3x) \text{ where } - \infty < x < \infty$$
How would one go about finding the maximum and minimum values achieved in function such as these and others with more than two trig functions.
| $$f(x) = \sin{(3 x)} + 2 \cos{(3 x)}$$
$$f'(x) = 3 \cos{(3 x)} - 6 \sin{(3 x)} $$
Set $f'(x)$ equal to zero for maxima or minima.
$$f'(x) = 0 \implies 3 \cos{(3 x)} - 6 \sin{(3 x)} = 0 $$
or
$$\tan{(3 x)} = \frac{1}{2} \implies x = \frac{1}{3} \arctan{\left ( \frac{1}{2} \right )} + \frac{k \pi}{3}$$
where $k \in \mathbb{Z}$. Determine if max or min using $f''(x)$:
$$f''(x) = -9 \sin{(3 x)} - 18 \cos{(3 x)} \implies f''{\left [ \frac{1}{3} \arctan{\left ( \frac{1}{2} \right )} \right ]} = -\frac{9}{\sqrt{5}} - \frac{36}{\sqrt{5}}<0$$
so that this point is a maximum.
On the other hand,
$$f''{\left [ \frac{1}{3} \arctan{\left ( \frac{1}{2} \right )+ \pi } \right ]} = \frac{9}{\sqrt{5}} + \frac{36}{\sqrt{5}}>0$$
so this point is a minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/297669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Quadratic Diophantine Equations I note that the Diophantine equation, $x^2 + y^2 = z^2$, with $x, y, z \in \mathbb{N}$, has infinitely many solutions. Indeed, $(x, y, z) = (3,4,5)$ provides a solution, and for any $k \in \mathbb{N}$ : $(kx, ky, kz ) = (3k, 4k, 5k)$ provides a solution.
However, assuming $x, y, z \in \mathbb{N}$ with $x, y > 1$, is the same true for the Diophantine equations,
$x^2 +y^2 = z^2 + 1$,
$x^2 + y^2 = z^2 + 2$,
$x^2 + y^2 = z^2 + 3$
and more generally, for $x^2 + y^2 = z^2 + n$, for any $n \in \mathbb{N}$?
In particular, are there infinitely-many triples $(x, y, z) \in \mathbb{N}^3$ for which $x^2 + y^2 = z^2 + n$ is true for infinitely-many values of $n \in \mathbb{N}$?
| For the special case when the number of is a square and is given us.
That is, in the equation:
$X^2+Y^2=Z^2+q^2$
where the number of $q$ - given us.
Then the solutions of the equation can be written ospolzovavshis Pell: $p^2-2k(k-1)s^2=\pm{q}$
$k$ - given us can be anything.
$X=p^2+2(k-1)ps+2k(k-1)s^2$
$Y=p^2+2kps+2k(k-1)s^2$
$Z=p^2+2(2k-1)ps+2k(k-1)s^2$
If we use the solutions of Pell's equation: $p^2-2s^2=\pm1$
Solutions have the form:
$X=sp^2+2qps\pm(qp^2+2(p+q)s^2)$
$Y=2qps+2s^3\pm(qp^2+2(p+q)s^2)$
$Z=sp^2+4qps+2s^2\pm(qp^2+2(p+q)s^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/298053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Inequality problem $(a^2-b^2)(a^4-b^4)\le (a^3-b^3)^2$ This problem that I have been unable to solve is from the book "Introduction to Inequalities" by Beckenbach and Bellman, chapter 2, page 22, problem 4.
Problem 4. Show that $$(a^2-b^2)(a^4-b^4)\le (a^3-b^3)^2$$ and $$(a^2+b^2)(a^4+b^4)\ge (a^3+b^3)^2$$ for all $a,b$.
My try for first part of the problem, just so you know that I have done some work, is this:
We have $$(a^2-b^2)(a^4-b^4)\le (a^3-b^3)^2$$ by factoring out, we are left with $$(a-b)(a+b)(a-b)(a+b)(a^2+b^2)\le (a^3-b^3)^2,$$ $$(a-b)^2(a+b)^2(a^2+b^2)\le (a-b)^2(a^2+ab+b^2)^2,$$ $$a^4+2a^2b^2+2a^3b+2ab^3+b^4\le a^4+2a^3b+ab^3+2a^2b^2+2ab^3+b^4,$$we get $$ab^2\ge 0$$which is true but for only $a,b\ge 0$, so my solution is false, can anyone hint me to correct way of solving this?
| If $a = 0$ or $b = 0$, the two sides are equal, and hence the inequality holds. Now, let us assume $a \neq 0$ and $b \neq 0$:
$(a^2 - b^2)(a^4 - b^4) \leq (a^3 - b^3)^2$
iff $a^6 + b^6 - a^2b^2(a^2 + b^2) \leq a^6 + b^6 - 2a^3b^3$
iff $- a^2b^2(a^2 + b^2) \leq - 2a^3b^3$
iff $a^2b^2(a^2 + b^2) \geq 2a^3b^3$
iff $(a^2 + b^2) \geq 2ab\ \ \ $ ($a, b \neq 0$)
iff $a^2 + b^2 - 2ab \geq 0$
iff $(a - b)^2 \geq 0$
which is always true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/298670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How many solutions to prime = $a^3+b^3+c^3 - 3abc$ Let $a,b,c$ be integers.
Let $p$ be a given prime.
How to find the number of solutions to $p = a^3+b^3+c^3 - 3abc$ ?
Another question is ; let $w$ be a positive integer. Let $f(w)$ be the number of primes of type prime = $a^3+b^3+c^3 - 3abc$ below $w$. How does the function $f(w)$ behave ? How fast does it grow ? Are those primes of type $A$ $mod$ $B$ for some integers $A$ and $B$ ?
How to deal with this ?
Can this be solved without computing the class number ?
| Andre's quadratic form $a^2 + b^2 + c^2 - bc - ca - ab$ is only positive semidefinite. It is 0 if $a=b=c.$ Meanwhile,
$$a^2 + b^2 + c^2 - bc - ca - ab \; \geq \; \frac{3}{4} \; (b-c)^2,$$
$$a^2 + b^2 + c^2 - bc - ca - ab \; \geq \; \frac{3}{4} \; (c-a)^2,$$
$$a^2 + b^2 + c^2 - bc - ca - ab \; \geq \; \frac{3}{4} \; (a-b)^2.$$
So the quadratic form cannot be 1 unless all three letters are within 1 of each other. For example, we cannot have some negative and some positive. Indeed, all possible 1's are arrangements of $(n,n,n+1)$ or $(n,n+1,n+1).$ Taking $n$ as nonnegative, we see that we get the value of the cubic either $3n+1$ or $3n+2.$ So all primes not 3 are represented. Not sure about 3.
EEEEDDDIITTT: Now that I think of it, cubes are $0,1,-1 \pmod 9.$ So, in order to get $a^3 + b^3 + c^3 - 3 a b c \equiv 0 \pmod 3,$ our choices are (A) $a,b,c$ are all divisible by 3, or (B) one of them is divisible by 3, the other two cubes give cancelling $\pm 1\pmod 9.$ In either case, the cubic is divisible by 9. So the prime 3 is not represented.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/299717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Does $\sin^2 x - \cos^2 x = 1-2\cos^2 x$? I am finishing a proof. It seems like I can use $\cos^2 + \sin^2 = 1$ to figure this out, but I just can't see how it works. So I've got two questions.
Does $\sin^2 x - \cos^2 x = 1-2\cos^2 x$?
And if it does, then how?
| Observe that
$$
\begin{align*}
\sin^2(x)-\cos^2(x)&=\sin^2(x)+\bigl(\cos^2(x)-\cos^2(x)\bigr)-\cos^2(x)\\
&= (\sin^2(x)+\cos^2(x))-2\cos^2(x)\\
&= 1-2\cos^2(x).
\end{align*}
$$
More easily, just subtract $2\cos^2(x)$ from both sides of $\sin^2(x)+\cos^2(x)=1$ to get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/300900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Solving non-linear congruence $x^2+2x+2\equiv{0}\mod(5)$, $7x\equiv{3}\mod(11)$
My attempt:
$x^2+2x+2\equiv{0}\mod(5)$
$(x+1)^2\equiv-1\mod(5)$, we have $x+1\equiv-1\mod(5)$
since $5$ and $11$ are coprime. We have a solution in $\mathbb{Z}_{11}$
With $[3]$ represent $3$, $[13]$ works for $7x\equiv3\mod11$
so $[3]$ is the solution in $\mathbb{Z}_{55}$
General solution is $x=-2+k55$.
But the answer is wrong..
| As you said, $(x+1)² \equiv -1 \equiv 4 \pmod 5$, hence $(x+1)\equiv 2 or -2 \pmod 5$. Now, if $(x+1)\equiv -2 \equiv 3 \pmod 5$, then, since also $x\equiv 2 \pmod {11}$ is a solution to the last congruence, one solution is $x\equiv 2 \pmod {55}$. On the other hand, for $x\equiv 1 \pmod {5}$, we should solve for $x\equiv 2 \pmod {11}$. Using Chinese Reminder Theorem, we conclude that the other solution is given by $x\equiv 11-20\equiv -9 \pmod{55}$.
P.S. Here we might view CRT as a decomposition of rings. But in essence, it talks about finding integers $x_i$ such that $\Sigma _ia_ix_i \pmod b$ is a solution of the congruences $x\equiv a_i \pmod {b_i}$, where $b=\Pi_ib_i$.
Reference
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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contour integration of logarithm I must compute the following integral
$$\displaystyle\int_{0}^{+\infty}\frac{\log x}{1+x^3}dx$$
Can someone suggest me the right circuit in the complex plane over which to do the integration? I tried different paths, avoiding the origin, but unsuccessfully
| Let
$$I = \int_0^{\infty} \dfrac{\log(x)}{1+x^3} dx = \underbrace{\int_0^1 \dfrac{\log(x)}{1+x^3} dx}_J + \underbrace{\int_1^{\infty} \dfrac{\log(x)}{1+x^3} dx}_K$$
$$K = \int_1^{\infty} \dfrac{\log(x)}{1+x^3} dx = \int_1^0 \dfrac{\log(1/x)}{1+1/x^3} \left(-\dfrac{dx}{x^2}\right) = - \int_0^1 \dfrac{x \log(x)}{1+x^3} dx$$
Now recall that
$$\int_0^1 x^m \log(x) dx = - \dfrac1{(m+1)^2}$$
Hence,
$$J = \int_0^1 \dfrac{\log(x)}{1+x^3} dx = \int_0^1 \sum_{k=0}^{\infty}(-x^3)^k \log(x) dx = \sum_{k=0}^{\infty}(-1)^k \int_0^1 x^{3k} \log(x) dx = \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{(3k+1)^2}$$
$$K = -\int_0^1 \dfrac{x\log(x)}{1+x^3} dx = \sum_{k=0}^{\infty}(-1)^{k+1} \int_0^1 x^{3k+1} \log(x) dx = \sum_{k=0}^{\infty} \dfrac{(-1)^{k}}{(3k+2)^2}$$
Hence,
$$J = -\sum_{k=0}^{\infty} \dfrac1{(6k+1)^2} + \sum_{k=0}^{\infty} \dfrac1{(6k+4)^2} = f_4 - f_1$$
$$K = \sum_{k=0}^{\infty} \dfrac1{(6k+2)^2} - \sum_{k=0}^{\infty} \dfrac1{(6k+5)^2} = f_2 - f_5$$
where $$f_l = \sum_{k=0}^{\infty} \dfrac1{(6k+l)^2}$$
Note that $f_6 = \dfrac{\pi^2}{216}$, $f_3 = \dfrac19 \cdot \dfrac{\pi^2}8 = \dfrac{\pi^2}{72}$. Let $\zeta$ be the $6^{th}$ root of unity i.e. $\zeta = e^{\pi i /3}$.
We have
$$\text{Li}_2(\zeta) = \sum_{k=1}^{\infty} \dfrac{\zeta^k}{k^2} = \zeta f_1 + \zeta^2 f_2 - f_3 - \zeta f_4 - \zeta^2 f_5 + f_6 = - \dfrac{\pi^2}{108} - \zeta J + \zeta^2 K$$
$$\text{Li}_2(\zeta^5) = \sum_{k=1}^{\infty} \dfrac{\zeta^{5k}}{k^2} = \zeta^5 f_1 + \zeta^{10} f_2 + \zeta^{15} f_3 + \zeta^{20} f_4 + \zeta^{25} f_5 + f_6 = - \dfrac{\pi^2}{108} + \zeta^2 J - \zeta K$$
where $\text{Li}_s(x)$ is the polylogarithm function defined as
$$\text{Li}_s(x) = \sum_{k=0}^{\infty} \dfrac{x^k}{k^s}$$
Polylgarithm function satisfies a nice identity namely
$$\text{Li}_n(e^{2 \pi ix}) + (-1)^n \text{Li}_n(e^{-2 \pi ix}) = - \dfrac{(2\pi i)^n}{n!}B_n(x)$$ where $B_n(x)$ are Bernoulli polynomials. Take $n=2$ and $x = 1/6$ to get that
$$\text{Li}_2(\zeta) + \text{Li}_2(\zeta^5) = - \dfrac{(2\pi i)^2}{2!}B_2(1/6) = - \dfrac{(2\pi i)^2}{2!} \dfrac1{36} = \dfrac{\pi^2}{18}$$
Hence,
\begin{align}
\zeta^2(J+K) - \zeta(J+K) - \dfrac{\pi^2}{54} & = \dfrac{\pi^2}{18}\\
\overbrace{(\zeta^2-\zeta)}^{-1}(J+K) & = \dfrac{\pi^2}{18} + \dfrac{\pi^2}{54}\\
-(J+K) & = \dfrac{2\pi^2}{27}
\end{align}
Hence,
$$I = J+K = -\dfrac{2 \pi^2}{27}$$
| {
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"url": "https://math.stackexchange.com/questions/301573",
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"source": "stackexchange",
"question_score": "11",
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Pursuit curves solution For our math class we have to do some calculations with respect to pursuit curves. The chased object starts at point $(p,0)$. Chaser starts at $(0,0)$.(x,y) Speed of the chased object is $u$. Speed chaser = $v$.
We have that for the chaser
$$ \frac{dy}{dx}=\frac{ut-y}{p-x} $$
Then the length of the path is
$$ s = \int \sqrt{1+(\frac{dy}{dx})^2}=vt=\frac{vy}{u}-\frac{v(p-x)dy}{u\times dx}$$
the first derivative of both sides gives
$$ -\frac{u}{v}\sqrt{1+(\frac{dy}{dx})^2}=\frac{(p-x)d(\frac{dy}{dx})}{dx} $$
Now, we are asked to:
| You have some wrong signs in the last two equations.
Based on the equation of the derivative of the pursuit curve $y=f(x)$ described by the chaser object that you indicate
$$
\frac{dy}{dx}=\frac{ut-y}{p-x}\tag{1}
$$
I assume that the chased object moves along the straight line $x=p$, as I commented above. Assume further that it moves upwards. (See remark 2). Then
$$
t=\frac{y}{u}+\frac{p-x}{u}\frac{dy}{dx}.\tag{2}
$$
and from
$$
s=\int_{0}^{x}\sqrt{1+(f^{\prime }(\xi ))^{2}}d\xi =vt\tag{3}
$$
we conclude that
$$
t=\frac{1}{v}\int_{0}^{x}\sqrt{1+(f^{\prime }(\xi ))^{2}}d\xi .\tag{4}
$$
Equating the two equations for $t$ $(2)$ and $(4)$ we get
$$
\frac{1}{v}\int_{0}^{x}\sqrt{1+(f^{\prime }(\xi ))^{2}}d\xi =\frac{y}{u}+
\frac{p-x}{u}f(x).
$$
Differentiating both sides we obtain the equation (note that the LHS is
positive)
$$
\frac{1}{v}\sqrt{1+(f^{\prime }(x))^{2}}=\frac{p-x}{u}f^{\prime \prime }(x).\tag{5}
$$
If we let $w=\frac{dw}{dx}=f^{\prime }(x)$ this equation corresponds to the following one in $w$ and $w^{\prime }=\frac{dw}{dx}$
$$
\sqrt{1+w^{2}}=k(p-x)\frac{dw}{dx}\qquad w=f^{\prime }(x),k=\frac{v}{u},\tag{6}
$$
which can be rewritten as
$$
\frac{dw}{\sqrt{1+w^{2}}}=\frac{1}{k}\frac{dx}{p-x}\tag{7}
$$
by applying the method of separation of variables to $x$ and $w$. The integration is easy
$$
\begin{eqnarray*}
\int \frac{dw}{\sqrt{1+w^{2}}} &=&\int \frac{1}{k}\frac{dx}{p-x}+\log C \\
\text{arcsinh }w &=&-\frac{1}{k}\log \left( p-x\right) +\log C.
\end{eqnarray*}
$$
The initial condition $x=0,w=f^{\prime }(0)=0$ yields
$$
\begin{eqnarray*}
-\frac{1}{k}\log p+\log C &=&\text{arcsinh }0=0 \\
&\Rightarrow &\log C =\frac{1}{k}\log p ,
\end{eqnarray*}
$$
which means that
$$
\text{arcsinh }w=-\frac{1}{k}\log \left( p-x\right) +\frac{1}{k}\log p=-\frac{1}{k}\log \frac{p-x}{p}.\tag{8}
$$
Solving for $w$ we get
$$
\begin{eqnarray*}
\frac{dy}{dx} &=&w=\sinh \left( -\frac{1}{k}\log \frac{p-x}{p}\right) =\frac{1}{2}\left( e^{-\frac{1}{k}\log \frac{p-x}{p}}-e^{\frac{1}{k}\log \frac{p-x}{p}}\right) \\
&=&\frac{1}{2} \left( \frac{p-x}{p}\right) ^{-1/k}-\frac{1}{2}\left( \frac{p-x}{p}\right) ^{1/k}\tag{9}
\end{eqnarray*}
$$
To integrate this equation consider two cases.
*
*(a) $k=\frac{v}{u}>1$
$$\begin{eqnarray*}
y &=&\frac{1}{2}\int \left( \frac{p-x}{p}\right) ^{-1/k}-\left( \frac{p-x}{p}\right) ^{1/k} dx \\
&=&-\frac{1}{2}\frac{pk}{k-1}\left( \frac{p-x}{p}\right) ^{1-1/k}+\frac{1}{2}\frac{pk}{k+1}\left( \frac{p-x}{p}\right) ^{1+1/k}+C.
\end{eqnarray*}$$
The constant of integration $C$ is defined by the initial condition $x=0,y=0$
$$\begin{eqnarray*}
0 &=&-\frac{1}{2}\frac{pk}{k-1}+\frac{1}{2}\frac{pk}{k+1}+C \\
&\Rightarrow &C=\frac{pk}{k^{2}-1}.
\end{eqnarray*}$$
Hence
$$y=-\frac{1}{2}\frac{pk}{k-1}\left( \frac{p-x}{p}\right) ^{1-1/k}+\frac{1}{2}\frac{pk}{k+1}\left( \frac{p-x}{p}\right) ^{1+1/k}+\frac{pk}{k^{2}-1}$$
$$\tag{10}$$
The chaser overtakes the chased object at the point $(p,f(p))$, with $f(p)=
\frac{pk}{k^{2}-1}$.
*(b) $k=\frac{v}{u}=1$. We have
$$\frac{dy}{dx}=\frac{1}{2} \left( \frac{p-x}{p}\right) ^{-1}-\frac{1}{2}\left(
\frac{p-x}{p}\right)
$$
and
$$\begin{eqnarray*}
y &=&\frac{1}{2}\int \left( \frac{p-x}{p}\right) ^{-1}-\left( \frac{
p-x}{p}\right) dx \\
&=&-\frac{1}{2}p\ln \left( p-x\right) -\frac{1}{2}x+\frac{1}{4p}x^{2}+C.
\end{eqnarray*}$$
The same initial condition $x=0,y=0$ yields now
$$\begin{eqnarray*}
C &=&\frac{1}{2}p\ln \left( p\right) \\
&& \\
y &=&-\frac{1}{2}p\ln \left( \frac{p-x}{p}\right) -\frac{1}{2}x+\frac{1}{4p}x^{2}.\tag{11}
\end{eqnarray*}$$
The chaser never overtakes the chased object.
Example for (a): graph of $y=f(x)$ for $k=2,p=50$
Example for (b): graph of $y=f(x)$ for $k=1,p=50$
Remarks:
*
*This answer is similar to the answer of mine to the question Cat Dog problem using integration.
*It was inspired by Helmut Knaust's The Curve of Pursuit.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve equation $ \frac{1}{2} (\sqrt{x^2-16} + \sqrt{x^2-9}) = 1$? $$ \dfrac{1}{2} (\sqrt{x^2-16} + \sqrt{x^2-9}) = 1$$
How can I solve this equation in the easiest way?
| We have the equation
$$
\frac{1}{2}(\sqrt{x^2-16} + \sqrt{x^2-9}) = 1
$$
Let's multiply it by $\sqrt{x^2-16} - \sqrt{x^2-9}$ to get
$$
-\frac{7}{2}=\sqrt{x^2-16} - \sqrt{x^2-9}
$$
Hence
$$
2\sqrt{x^2-16} = (\sqrt{x^2-16} + \sqrt{x^2-9}) + (\sqrt{x^2-16} - \sqrt{x^2-9})=2-\frac{7}{2}<0
$$
This is imossible so there is no real solution for this equation
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Polynomial factorization over finite fields How can i factorize the polynomial $x^{12}-1$ as product of irreducibles polynomials over $\mathbb{F}_4$? Anyone can help me?
| Let $\alpha$ and $\beta = \alpha^2 = \alpha+1$ denote the two elements of $\mathbb F_4-\mathbb F_2$. Then, $\alpha$ and $\beta$ are roots of $x^2+x+1$ and
we have that
$$x^3-1 = (x-1)(x^2+x+1) = (x-1)(x-\alpha)(x-\beta).$$
In a field of characteristic $p$, $(a-b)^{p^i} = a^{p^i} - b^{p^i}$ and so we have
that
$$x^{12} - 1 = (x^3-1)^4 = (x-1)^4(x-\alpha)^4(x-\beta)^4$$
factors into $12$ linear factors that are three distinct polynomials
each occurring with multiplicity $4$.
Modified approach using cyclotomic polynomials
In a field of characteristic $p$, $(a-b)^{p^i} = a^{p^i} - b^{p^i}$ and so we have
that
$x^{12} - 1 = (x^3-1)^4$. But, $$x^3-1 = Q_1(x)Q_3(x) = (x-1)(x^2+x+1)$$
where $Q_1(x) = x-1$ and $Q_3(x) = x^2+x+1$ are cyclotomic polynomials.
Also, $\mathbb F_4$ is the splitting field of $x^3-1$ since the
multiplicative group $\mathbb F_4^{\times}$ is a cyclic group of order
$3$, and so $x^3-1 = (x-1)(x-\alpha)(x-\beta)$ where
$\alpha, \beta \in \mathbb F_4 - \mathbb F_2$. So,
$x^{12}-1 = (x^3-1)^4$ factors into $12$ linear factors over $\mathbb F_4$,
and these factors are three distinct polynomials
each occurring with multiplicity $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Is this simple proof about symmetric sums correct? I'm asked to prove
$$x \gt 0, y \gt 0, z \gt 0 \rightarrow$$
$$\left(\frac{x+y}{x+y+z}\right)^\frac{1}{2}+\left(\frac{x+z}{x+y+z}\right)^\frac{1}{2} + \left(\frac{y+z}{x+y+z}\right)^\frac{1}{2} \le 6^\frac{1}{2}$$
I rewrite the summands and say that it is sufficient to prove:
$$A^\frac{1}{2} + B^\frac{1}{2} + C^\frac{1}{2} \le 6^\frac{1}{2} $$
$$ A +B +C = 2$$
$$ 0 \lt A, B, C \le 1$$
Now I just square both sides to get:
$$A + (AB)^\frac{1}{2} + (AC)^\frac{1}{2} + B + (BC)^\frac{1}{2} + C \le 6$$
This seems simple:
$$2 + (AB)^\frac{1}{2} + (AC)^\frac{1}{2} + (BC)^\frac{1}{2} \le$$
$$ 2 + 1 + 1 + 1 \le 5 \le 6$$
So I ended up proving that the original bounds were too loose. That makes me worry that I messed up my proof somewhere.
| Another approach is to use Lagrange multipliers. With $A$, $B$, and $C$ as you chose, the minimum value of $f(A,B,C) = \sqrt{A} + \sqrt{B} + \sqrt{C}$ subject to the constraint $A+B+C=2$ must occur at a local minimum of $F(A,B,C,\lambda) = \sqrt{A} + \sqrt{B} + \sqrt{C} + \lambda (A+B+C-2)$.
This can only occur if ${\partial{F}\over\partial{A}}$, ${\partial{F}\over\partial{B}}$, and ${\partial{F}\over\partial{C}}$ are simultaneously zero. Calculating the derivatives, we get $${1\over{2\sqrt{A}}}+\lambda={1\over{2\sqrt{B}}}+\lambda={1\over{2\sqrt{C}}}+\lambda=0\textrm{,}$$ which implies that $A=B=C$. Using the fact that $A+B+C=2$, it follows that $A=B=C={2\over3}$. The minimum value of $\sqrt{A} + \sqrt{B} + \sqrt{C}$ is then $3\sqrt{2\over3}$, which equals $\sqrt6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Permutation question. $qpq^{-1}$, $q,p,r,s \in S_{8}$. Let $p,r,s,q \in S_{8}$ be the permutation given by the following products of cycles:
$$p=(1,4,3,8,2)(1,2)(1,5)$$
$$q=(1,2,3)(4,5,6,8)$$
$$r=(1,2,3,8,7,4,3)(5,6)$$
$$s=(1,3,4)(2,3,5,7)(1,8,4,6)$$
Compute $qpq^{-1}$ and $r^{-2}sr^{2}.$
thanks for your help.
I want to write the following permutations like :
$p=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\4&1&8&3&?&?&?&2\end{pmatrix}$
$q=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\2&3&1&5&6&8&7&4\end{pmatrix}$
$r=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 2&3&8&3&6&5&4&7\end{pmatrix}$
$s=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 3&?&4&1&?&?&?&?\end{pmatrix}$
can you help me plese to fill the $?$ mark. Is there another method to compute $qpq^{-1}$?
thanks:)
| There's actually a quicker way to do this. As an exercise, prove the following lemma:
Lemma. Suppose that $\alpha$ is an arbitrary permutation and $\beta$ is an $n$-cycle, written $\beta=(\beta_1\hspace{5pt} \beta_2 \hspace{5pt}\ldots \hspace{5pt}\beta_n)$. Then $$\alpha^{-1}\beta \alpha=(\alpha(\beta_1)\hspace{5pt}\alpha(\beta_2)\hspace{5pt}\ldots \hspace{5pt}\alpha(\beta_n)).$$ In other words, $\alpha^{-1}\beta\alpha$ is the $n$-cycle with $\alpha$ applied to its letters.
Since any permutation can be decomposed into the product of disjoint cycles, say $\gamma=a_1a_2\ldots a_k$ where $a_i$ are disjoint cycles, we can write $$\alpha^{-1}\gamma\alpha=\alpha^{-1}a_1a_2\ldots a_k\alpha=(\alpha^{-1}a_1\alpha)(\alpha^{-1}a_2\alpha)(\alpha^{-1}\ldots \alpha )(\alpha^{-1}a_k\alpha)$$
so by induction the lemma shows that $\alpha^{-1}\gamma\alpha$ is the permutation with the same cycle structure as $\gamma$ with $\alpha$ applied to the letters in each of its disjoint cycles.
Now, $p$ in disjoint cycles is just $p=(1,4,3,8,5)$ and $s$ in disjoint cycles is just $s=(1,5,7,2,3,6)(4,8)$.
So, we apply $q$ to the letters in $p$ and obtain $q^{-1}pq=(1,4,6,2,5)$, then $r^2$ to the letters in each disjoint cycle of $s$ and obtain $r^{-2}sr^2=(1,8,7,6,3,5)(2,4)$.
I'll leave it to you to put these back into permutation notation (which is easy, since the cycles are disjoint!)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Absolute value inequality - Please guide further
Prove that if the numbers $x$, $y$ are of one sign, then $\left|\frac{x+y}{2}-\sqrt{xy}\right|+\left|\frac{x+y}{2}+\sqrt{xy}\right|=|x|+|y|$.
Expanding the LHS,
$$\left|\frac{x+y}{2}-\sqrt{xy}\right|= \left|\frac{x+y -2\sqrt{xy}}{2}\right| = \frac{(\sqrt{x}-\sqrt{y})^2}{2}$$
and similarly
$$\left|\frac{x+y}{2}+\sqrt{xy}\right| = \left|\frac{ (\sqrt{x}+\sqrt{y})^2}{2}\right|$$
| Use AM-GM to conclude $\frac{x+y}{2} \ge \sqrt{xy}$ if $x,y > 0$ and $\frac{x+y}{2} \le -\sqrt{xy}$ if $x,y < 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Least value of $a$ for which at least one solution exists? What is the least value of $a$ for which
$$\frac{4}{\sin(x)}+\frac{1}{1-\sin(x)}=a$$
has atleast one solution in the interval $(0,\frac{\pi}{2})$?
I first calculate $f'(x)$ and put it equal to $0$ to find out the critical points.
This gives
$$\sin(x)=\frac{2}{3}$$
as $\cos(x)$ is not $0$ in $(0,\frac{\pi}{2})$. I calculate $f''(x)$ and at $\sin(x)=\frac{2}{3}$, I get a minima. Now to have at least one solution, putting $\sin(x)=\frac{2}{3}$ in the main equation, I get $f=9-a$, which should be greater than or equal to $0$. I then get the 'maximum' value of $a$ as $9$.
Where did I go wrong?
[Note the function is $f(x)=LHS-RHS$ of the main equation.]
| My Solution:: Using the Cauchy-Schwarz inequality:: $\displaystyle \frac{a^2}{x}+\frac{b^2}{y}\geq \frac{(a+b)^2}{x+y}$
and equality holds when $\displaystyle \frac{a}{x} = \frac{b}{y}.$
So here $\displaystyle \frac{2^2}{\sin x}+\frac{1^2}{1-\sin x}\geq \frac{(2+1)^2}{\sin x+1-\sin x}\Rightarrow a\geq 9$
and equality holds when $\displaystyle \frac{2}{\sin x} = \frac{1}{1-\sin x}\Rightarrow \sin x = \frac{2}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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integrating $\oint_C \frac{3z^3 + 2}{(z-1)(z^2 + 9)}dz$ on $|z|=4$ I am doing $\oint_C \dfrac{3z^3 + 2}{(z-1)(z^2 + 9)}dz$ on $|z|=4$ and I find that there are poles within the contour at $z = 1$ and at $z = 3i$, both simple poles. I find that the integral $I = 2\pi i \,\text{Res}(1) + 2 \pi i \,\text{Res}(3i)$, or $I = \pi i + 2 \pi i \,\,\,\text{Res}(3i)$, with only that second residue left to find. Is this correct, or should I include $\text{Res}(-3i)$ as well?
| There are 3 poles: $z=1$ and $z=\pm 3 i$.
$$\mathrm{Res}_{z=1} \frac{3 z^3+2}{(z-1)(z^2+9)} = \frac{5}{10} = \frac{1}{2}$$
$$\mathrm{Res}_{z=3 i} \frac{3 z^3+2}{(z-1)(z^2+9)} = \frac{2-i 81}{(-1+3 i)(6 i)}$$
$$\mathrm{Res}_{z=-3 i} \frac{3 z^3+2}{(z-1)(z^2+9)} = \frac{2+i 81}{(-1-3 i)(-6 i)}$$
The integral value is $i 2 \pi$ times the sum of these.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of a sequence with indeterminate form Let $\displaystyle u_n =\frac{n}{2}-\sum_{k=1}^n\frac{n^2}{(n+k)^2}$. The question is: Find the limit of the sequence $(u_n)$.
The problem is if we write $\displaystyle u_n=n\left(\frac{1}{2}-\frac{1}{n}\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}\right)$ and we use the fact that the limit of Riemann sum $\displaystyle \frac{1}{n}\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}$ is $\displaystyle \int_0^1 \frac{dx}{(1+x)^2}=\frac{1}{2}$ we find the indeterminate form $\infty\times 0$. How can we avoid this problem? Thanks for help.
| Here's a rigorous derivation.
$$
\frac{n}{2} - \sum_{k=1}^n \frac{n^2}{(n+k)^2} = n \int_0^1 \frac{dx}{(1+x)^2} - \sum_{k=1}^n \frac{1}{(1+\frac{k}{n})^2} \\
= \sum_{k=1}^n \left(\int_{\frac{k-1}{n}}^{\frac{k}{n}} \frac{ n\, dx}{(1+x)^2} - \frac{1}{(1+\frac{k}{n})^2}\right) \\
= \sum_{k=1}^n \left(\int_{-\frac{1}{n}}^0 \frac{ n\, dx}{(1+\frac{k}{n} + x)^2} - \frac{1}{(1+\frac{k}{n})^2}\right) \\
= \sum_{k=1}^n \left(n\int_{-\frac{1}{n}}^0 \left(\frac{1}{(1+\frac{k}{n})^2} - \frac{2x}{(1+\frac{k}{n})^3} + \frac{3 x^2_*(x,n,k)}{(1+\frac{k}{n} +x_*(x,n,k))^4}\right)dx - \frac{1}{(1+\frac{k}{n})^2}\right)\\
= \sum_{k=1}^n n \int_{-\frac{1}{n}}^0 \left(- \frac{2x}{(1+\frac{k}{n})^3} + \frac{3 x^2_*(x,n,k)}{(1+\frac{k}{n}+x_*(x,n,k))^4}\right)dx.
$$
In the last two lines, we've used Taylor's theorem for $(1+\frac{k}{n}+x)^{-2}$. By that theorem with the Lagrange form of the remainder, $x_*(x,n,k)$ is some point in $(-\frac{1}{n},0)$. Given this bound on $x_*$, it is easy to see the term involving $x_*$ is $O(n^{-1})$, so we can ignore it in the limit $n\to\infty$. Continuing, we evaluate the limit of the remaining term:
$$
\lim_{n\to\infty} \sum_{k=1}^n n \int_{-\frac{1}{n}}^0 \left(- \frac{2x}{(1+\frac{k}{n})^3} \right)dx = \lim_{n\to\infty} \sum_{k=1}^n n \left(\frac{1}{n^2(1+\frac{k}{n})^3} \right)dx \\
= \int_0^1 \frac{dx}{(1+x)^3} = \frac{3}{8}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
} |
If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then find the largest possible value of $|a-b|$ I came across the following problem that says:
If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then the largest possible value of $|a-b|$ is which of the following?
$(1)\sqrt 7$, $(2)\sqrt{7/2}$ , $(3)\sqrt {35}$ $(4)7$
My Attempt: We notice, $(a-b)^2=7-(4ab+b^2)$ and hence $|a-b|=\sqrt {7-(4ab+b^2)}$ and so $|a-b|$ will be maximum whenever $(4ab+b^2)$ will be minimum. But now I am not sure how to progress further hereon.Can someone point me in the right direction? Thanks in advance for your time.
| Instead of maximizing $|\,\cdot\,|$ you can equivalently maximize $|\,\cdot\,|^2$, which has the advantage of being smooth. Then you can set
\begin{align*}
f(a,b) &= a^2 - 2ab + b^2\\
g(a,b) &= a^2 + 2ab + 2b^2\\
c &= 7
\end{align*}
and solve the corresponding Lagrange problem
\begin{align*}
(d/da):&& a - b + L*(a + b) &= 0\\
(d/db):&& b - a + L*(2b + a) &= 0\\
(d/dL):&& a^2 + 2ab + 2b^2 - c &= 0
\end{align*}
If you add (d/da) and (d/db) and assume $L \ne 0$ (otherwise you obviously get the minimum), then this implies $2a+3b = 0$ and thus $b = -2/3a$. From (d/dL) it now follows that $a^2 = 63/5$ and also $b^2 = 28/5$. With $a=\sqrt{63/5}$ and $b=-\sqrt{28/5}$, we get $|a-b| = \sqrt{35} \approx 5.9$ (you can also change the sign on both $a$ and $b$ as long as you do it at the same time).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/310348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
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Finding the affine transformation that will change a given ellipse into the unit circle $x^2$ + $y^2$ =1. We are given that ellipse $E$ is given by $x^2+4y^2-2x+16y+1=0$ and we are asked to find $t_2 \in A(2)$ such that $t_2(E)$ is the unit circle.
| Transformation is - tranlsation + scaling
$$
x^2+4y^2-2x+16y+1=x^2-2x+1+4(y^2+4y+4)-16=(x-1)^2+4(y+2)^2-16=0 \\
\frac{(x-1)^2}{16}+\frac{(y+2)^2}4=1
$$
So if you change coordinates
$$
\xi = \frac{x-1}4 \\
\eta= \frac{y+2}2
$$
your ellipse will become $\xi^2+\eta^2=1$
Update
Transformation itself can be represented in vector form $\mathbf x' = A \mathbf x + \mathbf b$
$$
\left [ \begin{array}{c}
\xi \\ \eta
\end{array}\right ] = f \left ( \left [ \begin{array}{c}
x \\ y
\end{array}\right ]\right ) = \left [ \begin{array}{c}
\frac x4 - \frac 14 \\ \frac y2 + 1
\end{array}\right ] = \left [ \begin{array}{cc}
\frac 14 & 0 \\
0 & \frac 12
\end{array} \right ] \left [ \begin{array}{c}
x \\ y
\end{array}\right ] + \left [ \begin{array}{c}
-\frac 14 \\ 1
\end{array}\right ]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/310776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A simultaneous system of equations Solve for $a,b,c$:
\begin{align}
2ab+a+2b=24\\
2bc+b+c=52\\
2ac+2c+a=74\\
\end{align}
Solving them simultaneously is leading to very difficult situation. Plz help.
| Here is a partial answer: User the first equation and third equation (add and substract) to obtain
\begin{align}
c-b&=\frac{25}{a+1} \\
c+b&=\frac{49-a}{a+1}
\end{align}
Adding and subtracting should again give
\begin{align}
c=&\frac{1}{2}\frac{74-a}{a+1} \\
b=&\frac{1}{2}\frac{24-a}{a+1}
\end{align}
Substituting this into the second equation should give a equation in $a$ alone which looks to be quartic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/311051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} = \mathcal{O}(\log(n)) $. Prove that $ 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n} = \mathcal{O}(\log(n)) $, with induction.
I get the intuition behind this question. Clearly, the given function isn’t even growing at a linear rate, but what is the ‘proper’ proofy way to say that $ \displaystyle \sum_{k=1}^{n} \frac{1}{k} \leq \mathcal{O}(\log(n)) $? I was unable to find any useful identities to use for such a summation.
| I'm expanding the answer by xan:
Define $H_n=\displaystyle\sum_{1\le k\le n} {1\over k}$, let's prove by induction that $H_{2^n}\le n+1$. This is true for $n=0$ since $H_{2^0}=H_1=1\le 1$.
Now suppose $H_{2^n}\le n+1$. We have:
$$\begin{align}
H_{2^{n+1}} &= \sum_{1\le k\le 2^{n+1}} {1\over k} \\
H_{2^{n+1}} &= \sum_{1\le k\le 2^n} {1\over k} + \sum_{2^n+1\le k\le 2^{n+1}} {1\over k} \\
H_{2^{n+1}} &= H_{2^n} + \sum_{2^n+1\le k\le 2^{n+1}} {1\over k} \\
H_{2^{n+1}} &\le H_{2^n} + \sum_{2^n+1\le k\le 2^{n+1}} {1\over 2^n} \\
H_{2^{n+1}} &\le H_{2^n} + (2^n-1){1\over 2^n} \\
H_{2^{n+1}} &\le H_{2^n} + (1-{1\over 2^n}) \\
H_{2^{n+1}} &\le H_{2^n} + 1 \\
H_{2^{n+1}} &\le (n+1) + 1 \\
H_{2^{n+1}} &\le n+2 \\
\end{align}
$$
Now let's make $m=2^n$, then $n=\lg m$, and:
$$H_{2^n}=H_m\le\lg m+1=\mathcal{O}(\log m)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/313426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Factorization of cyclic polynomial
Factorize $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
Since this is a cyclic polynomial, factors are also cyclic
$$f(a) = a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$f(b) = b(b^2-c^2)+b(c^2-b^2)+c(b^2-b^2) = 0 \Rightarrow a-b$$
is a factor of the given expression. Therefore, other factors are $(b-c)$ and $(c-a)$. The given expression may have a coefficient a constant factor which is nonzero. Let it be $m$.
$$\therefore a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2) = m(a-b)(b-c)(c-a)$$
Please guide further on how to find this coefficient.
| $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$=ab^2-ac^2+bc^2-a^2b+a^2c-cb^2$$
Now the method to factor cyclic expressions is to arrange the expression with the highest powers of the first variable, i.e:
We take powers of $a$ $$a^2c-a^2b+ab^2-ac^2+bc^2-cb^2$$
$$=a^2(c-b)-a(c^2-b^2)+bc(c-b)$$
$$=(c-b)(a^2-ac-ab+bc)$$
Now we look for powers of $b$,:
$$=(c-b)(bc-ab-ac+a^2)$$
$$=(c-b)(b(c-a)-a(c-a))$$
$$=(c-b)(c-a)(b-a)$$
$$=(a-b)(b-c)(c-a)$$
Thus, $m$ is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/314105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Integral $\int\limits_0^\infty \prod\limits_{k=0}^\infty\frac{1+\frac{x^2}{(b+1+k)^2}}{1+\frac{x^2}{(a+k)^2}} \ dx$ Does anybody know how to prove this identity?
$$\int_0^\infty \prod_{k=0}^\infty\frac{1+\frac{x^2}{(b+1+k)^2}}{1+\frac{x^2}{(a+k)^2}} \ dx=\frac{\sqrt{\pi}}{2}\frac{\Gamma \left(a+\frac{1}{2}\right)\Gamma(b+1)\Gamma \left(b-a+\frac{1}{2}\right)}{\Gamma(a)\Gamma \left(b+\frac{1}{2}\right)\Gamma(b-a+1)}$$
I found this on Wikipedia.
| To conclude this we actually do the simplification in terms of the $\Gamma$ function. We have
$$\prod_{k=0}^{b-a-1}\frac{1}{2a+2k+1} =
\frac{1}{2^{b-a}} \frac{\Gamma(a + 1/2)}{\Gamma(b+1/2)}.$$
Furthermore $$\prod_{k=0}^{b-a} (a+k) = \frac{\Gamma(b+1)}{\Gamma(a)}.$$
Finally, $$ \frac{1}{2^{b-a}} {2(b-a) - 1 \choose b-a} =
\frac{1}{2^{b-a}} \frac{\Gamma(2b-2a)}{\Gamma(b-a+1)\Gamma(b-a)}.$$
Putting these all together, we have
$$I(a, b) = \frac{\pi}{4^{b-a}} \frac{\Gamma(a + 1/2)}{\Gamma(b+1/2)}
\frac{\Gamma(b+1)}{\Gamma(a)} \frac{\Gamma(2b-2a)}{\Gamma(b-a+1)\Gamma(b-a)}$$
Now by the duplication formula,
$$ \Gamma(2b-2a) = \frac{2^{2(b-a)-1}}{\sqrt\pi} \Gamma(b-a) \Gamma(b-a+1/2)$$
so that finally
$$I(a, b) = \frac{\pi}{4^{b-a}} \frac{\Gamma(a + 1/2)}{\Gamma(b+1/2)}
\frac{\Gamma(b+1)}{\Gamma(a)} \frac{\Gamma(b-a+1/2)}{\Gamma(b-a+1)}
\frac{2^{2(b-a)-1}}{\sqrt\pi} \\
= \frac{\sqrt\pi}{2} \frac{\Gamma(a + 1/2)}{\Gamma(b+1/2)}
\frac{\Gamma(b+1)}{\Gamma(a)} \frac{\Gamma(b-a+1/2)}{\Gamma(b-a+1)}.$$
We have verfied the formula for $a$ and $b$ positive integers with $b>a.$ It should now follow by a continuity argument that we can extend it to real $a,b$ with $b>a>1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/314856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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general solution of the differential equation $\frac{dy}{dx} + \frac{x+y+a}{x+y+b}=0$ I'm trying to find the general solution of the differential equation $\frac{dy}{dx} + \frac{x+y+a}{x+y+b}=0$ where a and b are constants.
I have tried puttting z=x+y thus, $\frac{dz}{dx} = 1 + \frac{dy}{dx}$. I subbed this into the equation to get $\frac{dz}{dx} = 1 -(\frac{z+a}{z+b})$ which I simplified to $\frac{b-a}{z+b}$ and I separated variables and integrated giving:
$$\frac{z^2}{2} +bz = (b-a)x +c$$
I then multiplied through by 2 and subbed z=x+y back however this did not give me the correct answer which is $(x+y+b)^2 = 2(b-a)(x+c)$. Does anyone know how to get to this answer?
| Set $x+y+a=u$ so, $1+y'=u'$ and then $~~x+y+b=u-a+b~~$ so $~~~y'=-\frac{u}{u+b-a}$ or $$u'-1=\frac{-u}{u+b-a}$$ or $$u'=\frac{b-a}{u+b-a}$$ or $$(u+b-a)du=(b-a)dx$$ or $$\frac{u^2}{2}+(b-a)u=(b-a)x+C$$ wherein $x+y+a=u$. Now, by substiting $u$ we have: $$\frac{(x+y)^2}{2}+\frac{a^2}2+(x+y)b=(b-a)x+C$$ which is $$\frac{(x+y)^2}{2}+\frac{b^2}2+(x+y)b=(b-a)x+C-\frac{a^2}2+\frac{b^2}2=(b-a)x+k$$ and $k=C-\frac{a^2}2+\frac{b^2}2$. Therefore we get $$\frac{(x+y+b)^2}{2}=(b-a)x+(b-a)c$$ wherein $k=(b-a)c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/315973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find of $\int e^x \cos(2x) dx$ I did the following.
Using the LIATE rule:
$$\begin{align*}
u &=& \cos(2x)\\
u\prime &=& -2 \sin(2x)\\
v &=& e^x\\
v\prime&=&e^x
\end{align*}$$
We get:
$$\int e^x \cos(2x)dx = e^x \cos(2x) +2 \int e^x \sin(2x)dx $$
Now we do the second part.
$$\begin{align*}
u &=& \sin(2x)\\
u\prime &=& 2 \cos(2x)\\
v &=& e^x\\
v\prime&=&e^x
\end{align*}$$
We get:
$$\int e^x \sin(2x)dx = e^x \sin(2x) -2 \int e^x \cos(2x)dx$$
Putting it together we get:
$\begin{align*}
\int e^x \cos(2x)dx &=& e^x \cos(2x) +2 \int e^x \sin(2x)dx \\
&=&e^x \cos(2x) + 2[e^x \sin(2x) -2 \int e^x \cos(2x)dx]\\
&=&e^x \cos(2x) + 2e^x \sin(2x) -4 \int e^x \cos(2x)dx
\end{align*}$
$\begin{align*}
\int e^x \cos(2x)dx &=&e^x \cos(2x) + 2e^x \sin(2x) -4 \int e^x \cos(2x)dx\\
5\int e^x \cos(2x)dx &=&e^x(\cos(2x) + 2 \sin(2x))\\
\int e^x \cos(2x)dx &=&\frac{e^x( \cos(2x) + 2 \sin(2x))}{5}
\end{align*}$
I am not sure if this is right, if it is right, is there a better way of doing this.
| There is no need to ask whether it is right. You can differentiate to check. A little problem: the arbitrary constant of integration is missing,
It is good to get accustomed to the approach you took, it is useful to master it. Here is an alternative. My guess is that the answer will look like $(A\cos 2x+B\sin 2x)e^x$. Differentiate. We get
$$(A\cos 2x+B\sin 2x)e^x+(-2A\sin 2x+2B\cos 2x)e^x.$$
This will match our integrand if $-2A+B=0$ and $2B+A=1$. Solve. We get $A=\frac{1}{5}$ and $B=\frac{2}{5}$. So our answer is $(A\cos 2x+B\sin 2x)e^x+C$, with $A$ and $B$ as just calculated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/316639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Change of basis (to a more structured one) in a dynamic system, wrong result? Given a time-invariant homogeneous dynamic system $x(k +1) = Ax(k)$, where the system matrix is:
$$
A = \begin{bmatrix}
1 & -1 \\
2 & 4 \\
\end{bmatrix}
$$
And an initial state vector $x(0) = \begin{bmatrix}1 \\ 2\end{bmatrix}$, we know that solution is $x(k) = A^kx(0)$:
$$
x(2) = A^2x(0) = \begin{bmatrix}-1 & -5 \\ 10 & 14\end{bmatrix}\begin{bmatrix}1 \\ 2\end{bmatrix} = \begin{bmatrix}-11 \\ 38\end{bmatrix}
$$
With the help of basis change (new basis would be a matrix $M$, coordinates vector $z$, $x = Mz$), we could write:
$$
x(k+1) = Ax(k) \\
Mz(k+1) = AMz(k) \\
z(k+1) = M^{-1}AMz(k)
$$
Because $A$ is diagonalizable $M^{-1}AM= \Lambda$, we get the solution (with respect of basis $M$):
$$z(k) = \Lambda^kz(0)$$
Once we know $z(k)$ we can compute $x(k) = Mz(k)$.
Computation of $M$,$\Lambda$ and $M^{-1}$
$\Lambda$ is the diagonal matrix with the eigenvalues $\lambda_1...\lambda_n$ while $M$ is the matrix whose columns are the eigenvectors $v_1...v_n$ of $A$:
$$
\lambda_1 = 2, v_1 = \begin{bmatrix}-1 \\ 1\end{bmatrix},
\lambda_2 = 3, v_2 = \begin{bmatrix}-\frac12 \\ 1\end{bmatrix} \\
M = \begin{bmatrix}-1 & -\frac12 \\ 1 & 1\end{bmatrix},
\Lambda = \begin{bmatrix}2 & 0 \\ 0 & 3\end{bmatrix},
M^{-1} = \begin{bmatrix}-2 & -1 \\ 2 & 2\end{bmatrix},
$$
Computation of $z(0)$
Solution of the linear system $Mz(0) = x(0)$, that is:
$$
\begin{bmatrix}-1 & -\frac12 \\ 1 & 1\end{bmatrix}z(0) = \begin{bmatrix}1 \\ 2\end{bmatrix}
$$
We get $z(0) = \begin{bmatrix}-4 \\ 6\end{bmatrix}$
Computation of $z(2)$ and $x(2)$
Easily:
$$
z(2) = \Lambda^2z(0) = \begin{bmatrix}4 & 0 \\ 0 & 9\end{bmatrix}\begin{bmatrix}-4 \\ 6\end{bmatrix} = \begin{bmatrix}-8 \\ 27\end{bmatrix}\\
x(2) = Mz(2) = \begin{bmatrix}-1 & -\frac12 \\ 1 & 1\end{bmatrix}\begin{bmatrix}-8 \\ 27\end{bmatrix} = \begin{bmatrix}-\frac{11}{2} \\ -19\end{bmatrix}
$$
However, result is wrong, it should be $\begin{bmatrix}-11 \\ 38\end{bmatrix}$. Any help?
| You've written:
$$z(2) = \begin{bmatrix}4 & 0 \\ 0 & 9\end{bmatrix}\begin{bmatrix}-4 \\ 6\end{bmatrix} = \begin{bmatrix}-8 \\ 27\end{bmatrix}$$
However, it should be:
$$z(2) = \begin{bmatrix}4 & 0 \\ 0 & 9\end{bmatrix}\begin{bmatrix}-4 \\ 6\end{bmatrix} = \begin{bmatrix}-16 \\ 54\end{bmatrix}$$
after which $x(2)$ works out.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $(e+x)^{e-x}>(e-x)^{e+x}$ I get stuck with proving that $$(e+x)^{e-x}>(e-x)^{e+x}$$ for $x \in (0, e)$. All I know, is that it is doable with Jensen inequality, and I started with defining $$f(x)=(e+x)^{e-x}$$ and further $$g(x)=\ln \cdot f(x)$$ and... nothing more come to my mind, I kindly ask for any help & hints. Thanks
| Using power series,
$$
\begin{align}
(e-x)\log(e+x)
&=e-x+(e-x)\log\left(1+\frac xe\right)\\
&=e-x+(e-x)\left(\frac xe-\frac12\frac{x^2}{e^2}+\frac13\frac{x^3}{e^3}-\dots\right)\\
&=e-x+x-\frac32\frac{x^2}{e}+\frac56\frac{x^3}{e^2}-\dots\\
&=e-\frac32\frac{x^2}{e}+\frac56\frac{x^3}{e^2}-\frac7{12}\frac{x^4}{e^3}+\dots\tag{1}
\end{align}
$$
and
$$
\begin{align}
(e+x)\log(e-x)
&=e+x+(e+x)\log\left(1-\frac xe\right)\\
&=e+x-(e+x)\left(\frac xe+\frac12\frac{x^2}{e^2}+\frac13\frac{x^3}{e^3}+\dots\right)\\
&=e+x-x-\frac32\frac{x^2}{e}-\frac56\frac{x^3}{e^2}-\dots\\
&=e-\frac32\frac{x^2}{e}-\frac56\frac{x^3}{e^2}-\frac7{12}\frac{x^4}{e^3}-\dots\tag{2}
\end{align}
$$
Therefore,
$$
\begin{align}
&(e-x)\log(e+x)-(e+x)\log(e-x)\\
&=2\left(\frac56\frac{x^3}{e^2}+\frac9{20}\frac{x^5}{e^4}+\dots+\frac{4n+1}{2n(2n+1)}\frac{x^{2n+1}}{e^{2n}}+\dots\right)\tag{3}
\end{align}
$$
Thus, for $0\lt x\le e$ (so that the power series converge), this shows that
$$
(e+x)^{e-x}\gt(e-x)^{e+x}\tag{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Rank of matrix products Suppose $A_1$, $A_2$ and $A_3$ are $3$ by $3$ matrices of rank $2$ such that their kernels are linearly independent. Is the following true?
Define: $V_1=A_2A_1$, $V_2=A_3A_2$ and $V_3=A_1A_3$. Then each $V_i$ is rank $1$ and $V_iV_j=0$ for $i\neq j$.
Thanks for your help!
Stan
| If I interpret the question correctly the assumption is wrong, take
\begin{align*}
A_1&= \begin{pmatrix} 0& 1 & 0\\ 0&0 &1\\ 0 & 0 & 0\\ \end{pmatrix} \qquad \operatorname{kernel} \begin{pmatrix} \alpha \\0 \\ 0 \end{pmatrix} \\
A_2&= \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 &0 \\ 0 & 0 &0 \end{pmatrix}
\qquad \operatorname{kernel} \begin{pmatrix} 0 \\ 0 \\ \beta \\ \end{pmatrix}\\
A_3&= \begin{pmatrix} 0 & 0& 1\\ 0& 0 & 0 \\ 1 & 0 & 0\end{pmatrix} \qquad \operatorname{kernel}\begin{pmatrix} 0 \\\gamma \\ 0\\ \end{pmatrix}
\end{align*}
Than $V_1 V_2= A_2 A_1^2 A_3$ which is
\begin{align*}
V_1V_2&= \begin{pmatrix} 1 & 0 &0\\ 1 & 1 &0 \\ 0 & 0 & 0 \\ \end{pmatrix} \cdot
\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0& 0 & 0\\ \end{pmatrix}\cdot
\begin{pmatrix} 0 & 0 & 1 \\0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix} \\
&= \begin{pmatrix} 1 & 0 & 0 \\1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}
\end{align*}
checked with mathematica, so we have shown that $V_i V_j=0$ for $i \neq j$ is wrong, we will show what else won't work with this example.
$$V_1=A_2 A_1 = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix} $$
So rank of $V_1$ is 2 not 1.
For $V_2$ we have
$$A_3 A_2 = \begin{pmatrix} 0 & 0 & 0 \\0 & 0 & 0\\ 1 & 0 & 0 \end{pmatrix}
$$
Here it works, and even for $V_3$ we have
$$A_1 A_3 = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\0 & 0 & 0 \\ \end{pmatrix} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
minima of $\frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}$ Can anyone help me in finding minima of $\frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}$ where $k$ and $k'$, are constants. I found the differential but it was too big to be equated.
$\frac{dy}{dx}=\frac{(1-k)\log(x^2-x)[\ln x^2(x-k')x+2(1-k')(x-1)]-(1-k,)\ln x^2[\ln(x^2-x)(x-1)(1-k+x)+(2x-1)(1-k)}{((1-k')(x-1)\log x^2)^2}$
| Wolfram gave an easier derivative.
$$
\dfrac{d}{dx}\left(\frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}\right) = \dfrac{((k-1) ((2 x-1) \log(x^2)-\log((x-1) x) (\log(x^2)+2 x-2)))}{((k'-1) (x-1)^2 \log^2(x^2))}
$$
To see this result visit this link.
Setting the numerator of the derivative equal to $0$,
$$
(k-1) ((2 x-1)\log(x^2)-\log((x-1) x) (\log(x^2)+2 x-2))=0
$$
We get the roots:
$$
x_1 \approx -2.5915\\
x_2 \approx -0.352004\\
x_3 \approx 2.94663
$$
Evaluating $x_1$:
$$
\left. \frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}\right| _{x_1} \approx\frac{(0.845206 k-0.845206)}{(k'-1)}
$$
Evaluating $x_2$:
$$
\left. \frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}\right| _{x_2} \approx \frac{(0.092577 k-0.092577)}{(k'-1)}
$$
Evaluating $x_3$:
$$
\left. \frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}\right| _{x_3} \approx\frac{(1.22336 k-1.22336)}{(k'-1)}
$$
So $x_2$ appears to be satisfactory.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How does $\frac{1}{2} \sqrt{4 + 4e^4} = \sqrt{1 + e^4}$ My understanding would lead me to believe that:
$$\frac{1}{2} \sqrt{4 + 4e^4} = \frac{1}{2}(2 + 2e^4) = 1 + e^4$$
But it actually equals: $\sqrt{1 + e^4}$
Can you explain why?
| Your mistake is to say that $\sqrt{4(a+b)}=2(a+b)$, which is not true. For instance $\sqrt{4(3^2+4^2)}=10\not=2(3^2+4^2)$
The correct equalities you need is
$$\sqrt{(c^2a+c^2b)^2}=\sqrt{c^2(a+b)}=\sqrt{c^2}\sqrt{a+b}=c\sqrt{a+b}$$
Now $\frac{1}{2}\sqrt{4 + 4e^4} = \frac{1}{2}\sqrt{4(1+e^4)} = \frac{1}{2}\sqrt{2^2(1+e^4)} = \frac{1}{2}\sqrt{2^2}\sqrt{1 + e^4} = \frac{1}{2}2\sqrt{1+e^4} = \sqrt{1+e^4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$ \forall n \in \Bbb N , 18\mid1^n+2^n+\ldots+9^n-3(1+6^n+8^n)$ How to prove:$ \forall n \in \Bbb N , 18\mid1^n+2^n+\ldots+9^n-3(1+6^n+8^n)$ ?
| $$\sum_{1\le a\le 9}a^n-3(1+6^n+8^n)$$
$$=\sum_{0\le a\le \frac{9-1}2}\{a^n+(9-a)^n\}-3\{1^n+8^n\}-3\cdot6\cdot6^{n-1}\text{ for }n\ge1$$
If $n$ is odd, $9\mid \{a^n+(9-a)^n\}$ and $9\mid \{1^n+8^n\}$
$$\implies 9\mid \{\sum_{1\le a\le 9}a^n-3(1+6^n+8^n)\}\text{ if } n \text{ is odd} $$
If $n$ is even, $8^n\equiv(-1)^n\pmod 9\equiv1$ and $(9-a)^n\equiv(-a)^n\pmod9\equiv a^n$
$$\text{So,}\sum_{0\le a\le \frac{9-1}2}\{a^n+(9-a)^n\}-3\{1^n+8^n\}-3\cdot6\cdot6^{n-1}$$
$$\equiv2(1+2^n+3^n+4^n)-6\pmod 9$$
$$\equiv2(2^n+4^n-2)\pmod 9 \text{ and }9\mid3^n\text{ for }n\ge2$$
Now, $2^n+4^n-2=(3-1)^n+(3+1)^n-2=(1-3)^n+(1+3)^n-2$ as $n$ is even
$(1-3)^n+(1+3)^n-2$
$=\{1-\binom n13+\binom n23^2-\cdots\}+\{1+\binom n13+\binom n23^2+\cdots\}-2$
$\equiv1-3n+1+3n-2\pmod 9\equiv0\pmod 9$
Thomas has already hinted that $\sum_{1\le a\le 9}a^n-3(1+6^n+8^n)$ is always even which can be derived as follows $\sum_{1\le a\le 9}a^n-3(1+6^n+8^n)\equiv1+3+5+7+9-3\pmod2\equiv0$
So, lcm$(2,9)\mid \{\sum_{1\le a\le 9}a^n-3(1+6^n+8^n)\}$ for all natural number.
If $0$ is also considered $$\sum_{1\le a\le 9}a^n-3(1+6^n+8^n)$$
becomes $$\sum_{1\le a\le 9}1-3(1+1+1)=9-9=0\text{ which is divisible by any nonzero number including }18$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Proof of Wolstenholme's theorem According to the theorem, if
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots+\frac{1}{p-1} =\frac{r}{q}$$
then we have to prove that $r\equiv0 \pmod{p^2}$.
(Given $p>3$, otherwise $1+\dfrac{1}{2}=\dfrac{3}{2}$, $3 \not\equiv 0 \pmod 9$.)
I guess there's a $(\bmod p)$ solution for this, but I don't really get how to start it.
| $ (1 + \frac 1 {p-1}) + (\frac 1 2 + \frac 1 {p-2}) + \ldots + (\frac 1 {(p-1)/2} + \frac 1 {(p+1)/2}) = \frac p {p-1} + \frac p {2(p-2)} + \ldots + \frac p {((p-1)(p+1)/4)}
\\ = p \sum_{k=1}^{(p-1)/2} \frac 1 {k(p-k)} = \frac p 2 \sum_{k=1}^{p-1} \frac 1 {k(p-k)} $.
Now, working on this second sum in the field $\Bbb Z / p \Bbb Z$, and assuming $p>3$ you get :
$\sum_{k=1}^{p-1} \frac 1 {k(p-k)} = - \sum_{k=1}^{p-1} \frac 1 {k^2} = - \sum_{k=1}^{p-1} k^2 = -(p-1)p(2p-1)/6 = 0$.
If you don't know the sum of the first consecutive squares formula, you can still get it with a bit more work :
Pick a number $a \neq 1$ such that $a$ is a square modulo $p$ (again assuming $p>3$).
Then, $\sum k^2 (1-a) = \sum k^2 - \sum (ak^2) = \sum k^2 - \sum k^2 = 0$. Since $1-a \neq 0$, $\sum k^2 = 0$.
(when $p = 3$, $1+ \frac 1 2 = \frac 3 2$ which is not a multiple of $3^2$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/325491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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} |
need help Jordan base Need help, how to find Jordan base for matrix:
$A=\begin{pmatrix}
-1&-1 &-2 &4 \\
1&-3 &1 &-2 \\
0&0&2&-8\\ 0&0 & 2&-6
\end{pmatrix}$
I found the Minimal polynomial: $(x+2)^3$
and normal Jordan is $Aj=\begin{pmatrix}
-2&0 &0 &0 \\
0&-2 &1 &0 \\
0&0&-2&1\\ 0&0 & 0&-2
\end{pmatrix}$
Then I have problem with find $v_1, v_2, v_3, v_4 $that:
$(A+2I)v_1=0 $ $ (A+2I)v_2=0 $ $ (A+2I)v_3=v_2$ $ (A+2I)v_4=v_3$
when i'm trying to find it take $v_1=(1,1,0,0) $ $ v_2=(0,0,2,1)$ and later i get contradiction
| Ok i have as jordan basis
$$\frac{1}{3} \cdot \begin{pmatrix}
2 \\ 2\\ -4 \\ -2 \end{pmatrix}; \qquad \begin{pmatrix} 6 \\ 6\\0 \\ 0\\ \end{pmatrix}; \qquad
\begin{pmatrix} 4 \\ -2 \\ -8 \\ -4 \end{pmatrix}; \qquad \begin{pmatrix} 0 \\ 0\\ 0\\ 1 \end{pmatrix} $$
Got them with that ugly algorithm.
At first we compute $(A-\lambda I)^k$ for $k=1,2,3$ which is $(A+2 I)^k$
For $k=1$ this is not so difficult we have
$$\begin{pmatrix}
1 & -1 &-2&4\\ 1&-1&1&-2\\ 0&0&4&-8 \\ 0 & 0 & 2 & -4 \end{pmatrix}$$
As next we calculate $(A+2 I)^2$
$$\begin{pmatrix} 0 & 0& -3 & 6 \\ 0 & 0 & -3 & 6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ \end{pmatrix}$$
We see that $$\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ \end{pmatrix}$$ is not in the kernel of $(A+2I)^2$ (this is great!). So one jordan chain is
$$\begin{pmatrix}
1 & -1 &-2&4\\ 1&-1&1&-2\\ 0&0&4&-8 \\ 0 & 0 & 2 & -4 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 0 \\1 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\-8 \\ -4 \end{pmatrix}$$
and
$$\begin{pmatrix} 0 & 0& -3 & 6 \\ 0 & 0 & -3 & 6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 0 \\1 \end{pmatrix}= \begin{pmatrix} 6 \\ 6 \\ 0 \\ 0\\ \end{pmatrix}$$
As we only got one other eigenvector $$\begin{pmatrix} 0 \\ 0 \\2 \\ 1\\ \end{pmatrix}$$ and the minimalpolynom is $(A+ \lambda I)^3$ this is the only chain we have to calculate (what a luck).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/326546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$x^2-y^2=n$ has solution in $\Bbb Z$ IFF : $n \not \equiv 2 \pmod 4$ suppose $n\in \Bbb Z $ then how to prove this statement:
$x^2-y^2=n$ has solution in $\Bbb Z$ IFF : $n \not \equiv 2 \pmod 4$
| tnx for Gerry Myerson.
it's equal with this statement:
$x^2-y^2=n$ iff $ n$ is odd or $4|n$
$(x-y)(x+y)=x^2-y^2=n=ab , \exists a,b \in \Bbb Z$
so we have ($x+y=n$,$x-y=1$ or $x-y=n$,$x+y=1$ ) that $n$ must be odd or $x=\frac{a+b}{2}, y=\frac{a-b}{2}$ that so two case exists :
$I)$both $a,b$ is even so $4|ab=n$
$II)$ both $a,b$ is odd so $n=ab $ is odd
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/328195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solution to a linear recurrence What is the general solution to the recurrence: $x(n + 2) = 6x(n + 1) - 9x(n)$ for $n \geq 0$;
with $x(0) = 0; x(1) = 1$?
Solution. The first few values of $x(n)$ are $0,1,6,27,...$ The auxiliary equation for the
recurrence is $r^2-6r+9$ which factors as $(r-3)^2$. Thus, we have repeated roots so that the general solution to the recurrence will have the form $x(n) = c_13^n + c_2n3^n$. Substituting the values $n = 0$ and $n = 1$ gives us $x(0) = 0 = c_1$ and $x(1) = 1 = 0 + c_2\cdot3$. Thus, $c_1=1/3$ and $x(n) = nc^{n-1}$.
What doesn't make sense to me is why they added the additional $n$ in $c_2n3^n$ to get $x(n) = c_13^n + c_2n3^n$. When I did it myself I only got $x(n) = c_13^n$. Also why is the answer $x(n) = nc^{n-1}$, when I got $x(n)=(1/3)n3^n$ or $x(n)=n3^{n-1}$ ?
| The best technique for solving recurrences I've seen is given in Wilf's "generatingfunctionology".
Define the ordinary generating function:
$$
X(z) = \sum_{n \ge 0} x_n z^n
$$
From the recurrence, by the properties of ordinary generating functions (see section 2.2 in the cited book):
$$
\begin{align*}
\frac{X(z) - x_0 - x_1 z}{z^2} &= 6 \frac{X(z) - x_0}{z} - 9 X(z) \\
X(z) &= \frac{1}{3} \cdot \frac{1}{(1 - 3 z)^2}
- \frac{1}{3} \cdot \frac{1}{1 - 3 z}
\end{align*}
$$
The solution is thus (see the table of series in section 2 of the book):
$$
\begin{align*}
x_n &= \frac{1}{3} \binom{-2}{n} (-3)^n - \frac{1}{3} \cdot 3^n \\
&= \left( \binom{n + 2 - 1}{2 - 1} - 1 \right) 3^{n - 1} \\
&= n 3^{n - 1}
\end{align*}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $(n - 1)2^{n+1} + 2 + (n+1)2^{n+1} = n(2^{n+2})+2$ I'm having a really hard time showing this equality is true, I've tried several ways of going about it and I just can't seem to make it work. Help!
$(n - 1)2^{n+1} + 2 + (n+1)2^{n+1} = n(2^{n+2})+2$
Thanks!
| $$\begin{align}\color{darkred}{\bf(n - 1)}\color{blue}{\bf 2^{n+1}} + 2 + \color{darkred}{\bf (n+1)}\color{blue}{\bf 2^{n+1}} & = \color{darkred}{\bf(n - 1 + n+1)}\color{blue}{\bf 2^{n+1}} + 2 \\ \\
& = 2n\cdot2^{n+1} +2 \\ \\
& = n\cdot 2\cdot 2^{n+1} + 2 \\ \\
& = n\cdot 2^{n+2} + 2 \\ \\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/333375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\sum_{k=1}^{n}(k^2 \cdot (k+1)!)$ We have to evaluate the following:
$$1^2 \cdot 2! + 2^2 \cdot 3! + \cdots + n^2 \cdot (n+1)! =\sum_{k=1}^{n} k^2 \cdot (k+1)!$$
Any hints ?
| Using k2=(k+3)(k+2)−5(k+2)+4 we write
fk=k2(k+1)!=(k+3)!−5(k+2)!+4(k+1)!=Gk+1−Gk
where Gk=(k+2)!−4(k+1)!, therefore
∑k=1nfk=∑k=1n(Gk+1−Gk)=Gn+1−G1
Since G1=−2, and Gn+1=(n+3)!−4(n+2)!=(n+2)!(n−1) we get
∑k=1nk2(k+1)!=(n−1)(n+2)!+2
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Series expansion $(1-\cos{x})^{-1}$ How do i get the series expansion $(1-\cos{x})^{-1} = \frac{2}{x^2}+\frac{1}{6}+\frac{x^2}{120}+o(x^4)$ ?
| $$
\begin{align}
\frac{1}{1 - \cos x} &= \left(\frac{x^2}{2} - \frac{x^4}{24} + \frac{x^6}{720} + o(x^6)\right)^{-1}\\
& = \frac{2}{x^2}\left(1-\frac{x^2}{12} + \frac{x^4}{360} + o(x^4)\right)^{-1}\\
& = \frac{2}{x^2}\left(1 + \frac{x^2}{12} - \frac{x^4}{360} + \frac{x^4}{144} + o(x^4)\right)\\
& = \frac{2}{x^2} + \frac{1}{6} + \frac{x^2}{120} + o(x^2)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/334685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum up to number $N$ using $1,2$ and $3$ So the question asked was finding out the number of ways(combinations), a given number $N$ can be formed using the sum of $1,2$ or $3$.
(eg)
For $n = 8$, the answer is $10$
The given solution for this is simply
$$\left\lfloor\frac{(N+3)^2}{12}\right\rfloor$$
But I don't understand how this is working ?
| One approach to this sort of problem is generating functions. If $f(n)$ is the answer for $n$, then we have the formula:
$$\sum_{n=0}^\infty f(n)x^n =\frac{1}{1-x}\frac{1}{1-x^2}\frac{1}{1-x^3} = \frac{1}{(1-x)^3(1+x)(1+x+x^2)}$$
This in turns lets us see that there must be $a,b,c,d,e,f$ such that:
$$f(n)=an^2 + bn + c + d(-1)^n + ew^n + f\bar w^n$$
Where $w=\frac{-1+\sqrt{-3}}{2}$ is a primitive cube root of unity. We can then compute the first $6$ values, $f(0),f(1),\dots, f(5)$ to solve for $a,b,c,d,e,f$. I suspect it will then becomes clear that $|d(-1)^n + ew^n + f\bar w^n|$ is always negative and less than $\frac{1}{2}$, so that you get $f(n)=\lfloor an^2+bn+c +\frac{1}{2}\rfloor$.
This is the math nerd brute force technique. There might be a more inductive approach.
Wolfram alpha helps us compute the partial fractions:
$$\frac{1}{(1-x)(1-x^2)(1-x^3)} = \\\frac{17}{72}\frac{1}{1-x} +\frac{1}{4}\frac{1}{(1-x)^2} + \frac{1}{6}\frac{1}{(1-x)^3}+\frac{1}{8}\frac{1}{1+x} +\frac{1}{9}\frac{2+x}{1+x+x^2}$$
You then see that $$\frac{2+x}{1+x+x^2} = \frac{2-x-x^2}{1-x^3} = \sum w_n x^n$$ has the property that $w_0=2, w_1=-1, w_2=-1, w_{k+3}=w_k$.
This gives us the terms:
$$f(n) = \frac{17}{72} + \frac{1}{4}(n+1) +\frac{1}{12}(n+1)(n+2) + \frac{1}{8}(-1)^n + \frac{1}{9}w_n$$
$\frac{1}{8}(-1)^n + w_n$ is periodic of degree $6$, and we can easily show that $$|\frac{1}{8}(-1)^n + w_n|\leq \frac{25}{72}<\frac{1}{2}$$
So this gives us that $f(n)$ is the nearest integer to:
$$\frac{17}{72} + \frac{1}{4}(n+1)+\frac{1}{12}(n+1)(n+2)=\frac{n^2+6n+9}{12}-\frac{7}{72}$$
This still doesn't get you quite what you want, but it gets you very close.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$
Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a + b^3)(4)} \ge 3\sqrt{a} + \sqrt{b^3}$$
$$\sqrt{(3a + b^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2}$$
Also I get:
$$\sqrt{(3b + c^3)} \ge \frac{3\sqrt{b} + \sqrt{c^3}}{2}$$
$$\sqrt{(3c + a^3)} \ge \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
If I add add 3 inequalities I get:
$$\sqrt{(3a + b^3)} + \sqrt{(3b + c^3)} + \sqrt{(3c + a^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2} + \frac{3\sqrt{b} + \sqrt{c^3}}{2} + \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
Now i need to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} + \frac{3\sqrt{b} + \sqrt{b^3}}{2} + \frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge 6 = 2(a+b+c)$$
It's enough now to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} \ge b+c = 3-a$$
$$\frac{3\sqrt{b} + \sqrt{b^3}}{2} \ge a+c = 3-b$$
$$\frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge b+a = 3-c$$
All three inequalities are of the form:
$$\frac{3\sqrt{x} + \sqrt{x^3}}{2} \ge 3-x$$
$$3\sqrt{x} + \sqrt{x^3} \ge 6-2x$$
$$(3\sqrt{x} + \sqrt{x^3})^2 \ge (6-2x)^2$$
$$9x + x^3 + 6x^2 \ge 36 - 24x + 4x^2$$
$$x^3 + 2x^2 + 33x - 36 \ge 0$$
$$(x-1)(x^2 + 3x + 33) \ge 0$$
Case 1:
$$(x-1) \ge 0 \ \ \ \ \text{ for any }\ x \geq 1$$
$$(x^2 + 3x + 33) \ge 0 \ \ \ \ \text{ for any x in R} $$
Case 2:
$$0 \ge (x-1) \ \ \ \ \text{ for any }\ 1 \geq x$$
$$0 \ge (x^2 + 3x + 33) \ \ \ \ \text{there are no solutions in R} $$
This proves that for $$x \geq 1$$ $$(x-1)(x^2 + 3x + 33) \ge 0$$ is true and so it is
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$$, but a, b, c can be every non-negative number. I proved it's true for $$a,b,c \geq 1$$, but i can't for $$a,b,c \geq 0$$
| Cauchy-Schwarz inequality is not the way to go here. Notice that at $a = b = c =1$ your original inequality becomes exact, whereas your relaxed inequality is no longer satisfied. Moreover, you have lost a factor of 2 there.
Your new edits are very interesting, but still lead to a dead-end: you can check that when $a=b=0$, $c=3$ your inequality labeled "Now I need to prove that:" fails.
Also note that in general the case $a,b,c\ge 1$ contains only one point $a = b = c = 1$ in it (because $a+b+c=3$).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Calculate:$y'$ for $y = x^{x^{x^{x^{x^{.^{.^{.^{\infty}}}}}}}}$ and $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....\infty}}}}$ (1) If $y = x^{x^{x^{x^{x^{.^{.^{.^{\infty}}}}}}}}$
(2) If $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....\infty}}}}$
then find $y'$ in both cases
(3)If $ y= \sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+....\infty}}}}\ $ then find$\ (2y-1)\frac{dy}{dx}$
| $y = x^{x^{x^{x^{{.^{.^{.^{}}}}}}}} = x^{y}$
Take the log of both sides.
Then $\ln y = y \ln x$
$ \implies\frac{1}{y} \frac{dy}{dx} = \ln x \frac{dy}{dx} + \frac{y}{x}$
$ \implies \frac{dy}{dx} = \frac{y^{2}}{x(1-y \ln x)}$ and then substitute for $y$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "17",
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Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| \begin{gather*}
|x|<1:\quad f(x)=\sum_{n=1}^{\infty} x^n=\frac{x}{1-x} \\
xf'(x)=\sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2}
\end{gather*}
Let $x=\frac{1}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Complex numbers and trig identities: $\cos(3\theta) + i \sin(3\theta)$ Using the equally rule $a + bi = c + di$ and trigonometric identities how do I make...
$$\cos^3(\theta) - 3\sin^2(\theta)\ \cos(\theta) + 3i\ \sin(\theta)\ \cos^2(\theta) - i\ \sin^3(\theta)=
\cos(3\theta) + i\ \sin(3\theta)$$
Apparently it's easy but I can't see what trig identities to substitute
PLEASE HELP!
| $(\cos x+i\sin x)^3=(\cos 3x+i\sin 3x)$(by De Moivre's theorem)
But $(\cos x+i\sin x)^3=\cos^3x+i^3\sin^3x+3\cos x i \sin x(\cos x+i\sin x)=\cos^3 x-i\sin^3x+3i\cos^2 x\sin x-3\cos x\sin^2x=\cos^3x+i^3\sin^3x+3\cos x i \sin x(\cos x+i\sin x)=\cos^3 x-i\sin^3x+3i(1-\sin^2x) \sin x-3\cos x(1-\cos^2 x)=4\cos ^3x-3\cos x+i(3\sin x-4\sin^3x)$
Now by equating we have, $\cos 3x=4\cos ^3x-3\cos x$ and $\sin 3x=3\sin x-4\sin^3x$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Use the $\varepsilon$ - $\delta$ definition to prove $\lim_{x\to\,-1}\frac{x}{2x+1}=1$ Use the $\varepsilon$ - $\delta$ definition of limit to prove that $\displaystyle\lim_{x\to\,-1}\frac{x}{2x+1}=1$.
My working:
$\left|\frac{x}{2x+1}-1\right|=\left|\frac{-x-1}{2x+1}\right|=\frac{1}{\left|2x+1\right|}\cdot \left|x+1\right|$
First restrict $x$ to $0<\left|x+1\right|<\frac{1}{4}$ $\Rightarrow$ initial choice of $\delta=\frac{1}{4}$
$\left| 2x+1 \right|$ = $\left|2(x+1)-1\right|$
$\le$ $\left|2(x+1)\right|+\left|-1\right|$ = $2\left|x+1\right|+1$ $> 1$
Thus if $\left|x+1\right|<\frac{1}{4}$ , then
$\left|\frac{x}{2x+1}-1\right|=\frac{1}{\left|2x+1\right|}.\left|x+1\right|$
$<1.\left|x+1\right|$.
Therefore, $\delta = \min\{\frac{1}{4},\varepsilon\}$
$0<\left|x+1\right|<\delta$ $\Rightarrow$ $\left|\frac{x}{2x+1}-1\right| < 1\cdot\left|x+1\right| < 1\cdot\varepsilon = \varepsilon$
Thus, the limit is 1.
| You shouldn't change direction of your inequalities in a chain--for example, $$|2(x+1)-1|\le 2|x+1|+1>1$$ doesn't allow you to conclude that $|2(x+1)-1|>1,$ as transitivity breaks down when you switch directions.
Instead, we can use triangle inequality (why?) to say $$|2(x+1)-1|\ge1-2|x+1|,$$ so whenever $|x+1|<\frac14,$ we will have $$|2(x+1)-1|\ge1-2|x+1|>\frac12,$$ and so $$\left|\frac{x}{2x+1}-1\right|=\frac1{|2(x+1)-1|}|x+1|<2|x+1|.$$ A good choice would then be $$\delta=\min\left\{\frac14,\frac{\epsilon}2\right\}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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generating functions / combinatorics Calculate number of solutions of the following equations:
$$ x_1 + x_2 + x_3 + x_4 = 15 $$
where $ 0 \le x_i < i + 4 $
I try to solve it using generating functions/enumerators :
$$ (1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^4+x^5+x^6)(1+x+x^2+x^3+x^4+x^5+x^6+x^7)$$
and take coefficient near $15$. But I do not know how to quickly calculate it. Maybe there exists any faster way?
| The expression you have is
$$
\frac{1-x^5}{1-x}\frac{1-x^6}{1-x}\frac{1-x^7}{1-x}\frac{1-x^8}{1-x}
$$
Then treat this as
$$
(1-x^5)(1-x^6)(1-x^7)(1-x^8)(1-x)^{-4}
$$
The $(1-x)^{-4}$ can be treated by taking derivatives for the geometric series $(1-x)^{-1}$, and you can easily compute the product of the first four terms.
What you need is only the terms $c_k x^k$ where $k\leq 15$, and find $15-k$-th coefficient in $(1-x)^{-4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/339230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
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