Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Closed Form of the Chebyshev Polynomials of the First Kind [Proof Request] $$T_n(x) = \frac{n}{2} \sum_{r=0}^{\lfloor \frac{n}{2} \rfloor} \frac{(-1)^r}{n-r} \binom{n-r}{r} (2x)^{n-2r}$$
Searching on the web yielded no results, and the result is given without proof on OEIS and Wolfram MathWorld
| Starting from the OGF
$$\sum_{n\ge 0} T_n(x) t^n
= \frac{1-tx}{1-2tx+t^2}$$
We extract coefficients $[x^q] [t^n]$ where $0\le q\le n$.
First do the coefficient on $q=0$ which is $[t^n] \frac{1}{1+t^2}.$
Continue with $q\ge 1$
$$[x^q] [t^n] \frac{1-tx}{1-2tx+t^2}
= [x^q] [t^n] \frac{1-tx}{1+t^2} \frac{1}{1-2tx/(1+t^2)}
\\ = [t^n] \frac{1}{1+t^2} \frac{2^q t^q}{(1+t^2)^q}
- [t^n] \frac{1}{1+t^2}
\frac{2^{q-1} t^{q}}{(1+t^2)^{q-1}}
\\ = [t^n] \frac{2^q t^q}{(1+t^2)^{q+1}}
- [t^n]
\frac{2^{q-1} t^{q}}{(1+t^2)^{q}}.$$
Now we see here that $q$ and $n$ must have the same parity. Supposing
that is true we obtain
$$2^q [t^{n-q}] \frac{1}{(1+t^2)^{q+1}}
- 2^{q-1} [t^{n-q}] \frac{1}{(1+t^2)^q}
\\ = 2^q (-1)^{(n-q)/2} {(n-q)/2+q\choose q}
- 2^{q-1} (-1)^{(n-q)/2} {(n-q)/2+q-1\choose q-1}
\\ = 2^q (-1)^{(n-q)/2} {(n+q)/2\choose q}
- 2^{q-1} (-1)^{(n-q)/2} \frac{q}{(n+q)/2} {(n+q)/2\choose q}
\\ = \frac{n}{n+q} 2^q (-1)^{(n-q)/2} {(n+q)/2\choose q}.$$
Note that this also gives the correct value for $q=0$ so we can merge
that case into our formula. Observe also that we get zero from the
binomial coefficient when $(n+q)/2\lt q$ or $n\lt q.$
Now when $q$ and $n$ have the same parity with $n$ and $q$ non-negative
then $n-q=2r$ where $0\le r\le \lfloor n/2 \rfloor.$ We have $q=n-2r$
and obtain summing over $r$
$$\sum_{r=0}^{\lfloor n/2 \rfloor} x^{n-2r}
\frac{n}{2n-2r} 2^{n-2r} (-1)^r {n-r\choose n-2r}
\\ = \frac{n}{2} \sum_{r=0}^{\lfloor n/2 \rfloor} (2x)^{n-2r}
\frac{(-1)^r}{n-r} {n-r\choose r}.$$
This is the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4625372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove all numbers in a sequence $a_{n+3}=15a_{n+2}-15a_{n+1}+a_n$ are perfect squares $a_{n+3}=15a_{n+2}-15a_{n+1}+a_n$, here $a_1=a_2=1,a_3=9$
Prove all numbers in a sequence are perfect squares.
My attempt is first to use the general formula of $a_n$.
It is $a_n=\frac{1}{6}\left((2+\sqrt3)^{2n-3}+(2-\sqrt3)^{2n-3}+2\right)$
But I still cannot prove every number in the sequence is a perfect square.
| Consider the sequence $(b_n)$ defined by
$$ b_{n+2} = 4b_{n+1} - b_n, \qquad b_1 = b_2 = 1. $$
It is clear that every term of $(b_n)$ is an integer. To prove the claim, it is thus sufficient to establish:
Claim. $a_n = b_n^2$ for all $n$.
Indeed, it is easy to check that the claim is true for $n = 1, 2, 3$. Now suppose the claim is true for $n$, $n+1$, $n+2$. Then
\begin{align*}
a_{n+3}
&= 15a_{n+2} - 15a_{n+1} + a_n \\
&= 15b_{n+2}^2 - 15b_{n+1}^2 + b_n^2 \\
&= 15b_{n+2}^2 - 15b_{n+1}^2 + (-b_{n+2} + 4b_{n+1})^2 \\
&= 16b_{n+2}^2 + b_{n+1}^2 - 8b_{n+1}b_{n+2} \\
&= (4b_{n+2} - b_{n+1})^2 \\
&= b_{n+3}^2,
\end{align*}
and so, the claim is true for $n+3$ as well. Therefore the claim follows by the principle of mathematical induciton.
Motivation. The choice of the sequence $(b_n)$ is motivated by the fact that OP's formula for $(a_n)$ simplifies to
$$ a_{n} = \left[ \frac{\sqrt{3}-1}{2\sqrt{3}}(2+\sqrt{3})^{n-1} + \frac{\sqrt{3}+1}{2\sqrt{3}}(2 - \sqrt{3})^{n-1} \right]^2. $$
Using this, I reverse-engineered the recurrence relation with the characteristic equation having $2\pm\sqrt{3}$ as roots, namely $x^2 - 4x + 1 = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4626791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $\lambda x^2+5y^2+\mu z^2-axy+2\lambda x-4\mu y-2az+d=0$ represents a sphere of radius $2.$ Find the value of $d$ If $\lambda x^2+5y^2+\mu z^2-axy+2\lambda x-4\mu y-2az+d=0$ represents a sphere of radius $2.$ Find the value of $d.$
My solution goes like this:
Since, $\lambda x^2+5y^2+\mu z^2-axy+2\lambda x-4\mu y-2az+d=0$ is an equation of a sphere, so, $\lambda=\mu=5$ and $a=0.$ Hence, $5x^2+5y^2+5z^2+10x-20y+d=0\implies x^2+y^2+z^2+2x-4y+\frac d5=0.$ Given, $\text{radius}(r)=2$ and hence $$\sqrt{1+4+0-\frac d5}=2\implies d=5.$$
Is the above solution correct? If not, where is it going wrong?
| You are indeed correct.
I would have added a couple steps between
$x^2+y^2+z^2+2x-4y+(d/5)=0$
and the formula you used to compute the radius. To wit, combine terms with like variables and complete the squares:
$(x^2+2x)+(y^2-4y)+z^2=-(d/5)$
$(x+1)^2+(y-2)^2+z^2=1+4-(d/5)$
making it easier to see where the formula for the radius comes from.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4627502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find Highest Common Factor of two polynomials $\ 2(x^4+9)-5x^2(x+1),\ 2x^3(2x-9)+81(x-1)$ Find the H.C.F. of $\ 2(x^4+9)-5x^2(x+1),\ 2x^3(2x-9)+81(x-1)$
I can't really see any pattern by inspection, so I carry out the multiplication with the brackets: $2x^4+18-5x^3-5x^2,\ 4x^4-18x^3+81x-81$
I tried to divide them into each other to find the common factor (Euler's method or something? I forgot). I did the obvious first step by multiplying $2x^4+18-5x^3-5x^2$ by $2$, then subtracting that from $4x^4-18x^3+81x-81$, but the result is $-8x^3+10x^2+81x-117$, so clearly continuing this way is almost impossible by hand. Now I don't know what to do. The answer is given as $2x^2-9x+9$, so I'm not understanding something here. Thanks for your help.
| You are thinking about The Euclidean algorithm applied to polynomials.
Certainly, the algorithm can be done by hand although it is a bit tedious:
$$\begin{align}
4x^4 - 18x^3 + 81x - 81 &= 2(2x^4 - 5x^3 - 5x^2 + 18) + (-8x^3 + 10x^2 + 81x - 117) \\
4(2x^4 - 5x^3 - 5x^2 + 18) &= (-x)(-8x^3 + 10x^2 + 81x - 117) + (-10x^3 + 61x^2 - 117x + 72) \\
5(-8x^3 + 10x^2 + 81x - 117) &= 4(-10x^3 + 61x^2 - 117x + 72) + 97(-2x^2 + 9x - 9) \\
97(-10x^3 + 61x^2 - 117x + 72) &= 5x(-2x^2 + 9x - 9) + 776(2x^2 - 9x + 9) \\
-2x^2 + 9x - 9 &= (-1)(2x^2 - 9x + 9) + 0.
\end{align}$$
All that remains is to check that $2x^2 - 9x + 9$ does in fact divide both polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $x^3+y^3$, given $x + y = 5$ and $xy = 1$
Given
$$x + y = 5 \qquad xy = 1$$
Find $x^3 + y^3$.
To solve this, I tried this:
$y = \frac{1} {x}$
$x + \frac{1}{x} = 5$
$x^2 - 5x + 1 = 0$
What formula needs to be used to find the value of $x$?
| Another method that requires some intuition and knowledge of cubics:
Observe that, $$\large{\displaystyle{(\color{red}{x + y})^3 = \color{purple}{x}^3 + 3x^2y + 3xy^2 + \color{purple}{y}^3} = \color{purple}{x}^3 + 3\color{green}{xy}\big(\color{red}{x+y}\big) + \color{purple}{y}^3}$$
Otherwise re-arranged, gives:
$$\large{\displaystyle{\color{purple}{x}^3 + \color{purple}{y}^3 =(\color{red}{x + y})^3 - 3\color{green}{xy}\big(\color{red}{x+y}\big)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Find the equation of the cylinder whose generators are parallel to the line $\frac x1=\frac {-y}{2}=\frac z3$ and whose guiding curve is the ellipse Find the equation of the cylinder whose generators are parallel to the line $\frac x1=\frac {-y}{2}=\frac z3$ and whose guiding curve is the ellipse $x^2+2y^2=1,z=3.$
My solution goes like this:
The generators are parallel to the line $\frac x1=\frac {-y}{2}=\frac z3.$ So, the equation of the generator is $\frac {x-\alpha}{1}=\frac {y-\beta}{-2}=\frac {z-\gamma}{3},$ where $(\alpha, \beta,\gamma)$ is a point on the cylinder. As we know, from the definition of a cylinder, it is a generator line which follows the trajectory of the guiding curve, just like a remote star. The guiding curve in this case is an ellipse. The cylinder extends infinitely in space and hence the cylinder also intersects in the $xy$ plane and in that plane the cylinder also follows the trajectory of the given ellipse. We can consider this situation alternatively, by the considering the guiding curve as $x^2+2y^2=1,z=0,$ because this will generate the same cylinder as the one in the case of $x^2+2y^2=1,z=3.$ So, in the $xy$ plane the point $(x,y,z)$ of the generater needs a closer examination. In the $xy$ plane, $z=0$ and hence $\frac {x-\alpha}{1}=\frac {y-\beta}{-2}=\frac {-\gamma}{3},$ due to which, $x=\alpha+\frac{-\gamma}{3}$ and $y=\beta+\frac{2\alpha}{3}.$ Thus, the point $(\alpha+\frac{-\gamma}{3},\beta+\frac{2\alpha}{3},0)$ on the generator satisfies the equation of $x^2+2y^2=1.$ Therefore, $(\alpha+\frac{-\gamma}{3})^2+2(\beta+\frac{2\alpha}{3})^2=1$ is the locus of $(\alpha, \beta,\gamma)$ which is the cylinder. Thus, the equation of the cylinder is $(\alpha+\frac{-\gamma}{3})^2+2(\beta+\frac{2\alpha}{3})^2=1,$ when written in terms of $x,y,z$ instead of $\alpha, \beta,\gamma$, i.e, $(x+\frac{-z}{3})^2+2(y+\frac{2x}{3})^2=1.$
Is the above solution correct? If not then where is it going wrong?
| Your solution is wrong at that precise point: "We can consider this situation alternatively, by the considering the guiding curve as $x^2+2y^2=1,z=0,$ because this will generate the same cylinder as the one in the case of $x^2+2y^2=1,z=3$" because the "generators" are not vertical.
There was a (small) second mistake when you solved $\frac {y-\beta}{-2}=\frac {-\gamma}{3}$: you wrote $y=\beta+\frac{2\alpha}{3}$ instead of $y=\beta+\frac{2\gamma}{3}.$
If you make the necessary corrections, your final equation will be
$$\left(x+\frac{-z}{3}+1\right)^2+2\left(y+\frac{2z}{3}-2\right)^2=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4629514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\int_{-\infty}^\infty\frac{x^2}{1+x^8}dx$ I wanted to practice using the Residue theorem to calculate integrals. I chose an integral of the form $$\int_{-\infty}^\infty\frac{x^n}{1+x^m}dx$$and then choose random numbers for $n$ and $m$, which happened to be $$\int_{-\infty}^\infty\frac{x^2}{1+x^8}dx$$I know there is a general formula for integrals of this form, which is $$\int_{-\infty}^\infty\frac{x^n}{1+x^m}dx=\frac{\pi}{m}\csc\left(\frac{\pi(n+1)}{m}\right)$$But let's forget about this right now. I used the Residue theorem and the semicircular contour to get that $$\int_{-\infty}^\infty\frac{x^2}{1+x^8}dx=\Re\left(\frac{\pi i}{4}\left(\frac{1}{e^{\frac{5\pi i}{8}}}+\frac{1}{e^{\frac{15\pi i}{8}}}+\frac{1}{e^{\frac{25\pi i}{8}}}+\frac{1}{e^{\frac{35\pi i}{8}}}\right)\right)$$
Is there any other ways to calculate this integral?
| Let
$$ I=\int_{-\infty}^\infty\frac{x^2}{1+x^8}dx=2\int_{0}^\infty\frac{x^2}{1+x^8}dx.$$
Under $x\to\frac1x$, one has
$$ I=\int_{-\infty}^\infty\frac{x^2}{1+x^8}dx=2\int_{0}^\infty\frac{x^4}{1+x^8}dx.\tag1$$
Clearly
$$ I=-2\int_{0}^\infty\frac{x^{4}}{1+x^{8}}d\bigg(\frac1x\bigg). \tag2$$
Adding (1) to (2) gives
\begin{eqnarray}
2I&=&2\int_{0}^\infty\frac{x^{4}}{1+x^{8}}d\bigg(x-\frac1x\bigg)\\
&=&2\int_{0}^\infty\frac{1}{x^4+x^{-4}}d\bigg(x-\frac1x\bigg)\\
&=&2\int_{0}^\infty\frac{1}{(x^2+x^{-2})^2-2}d\bigg(x-\frac1x\bigg)\\
&=&2\int_{0}^\infty\frac{1}{((x-x^{-1})^2+2)^2-2}d\bigg(x-\frac1x\bigg)\\
&=&2\int_{-\infty}^\infty\frac{1}{(x^2+2)^2-2}dx\\
&=&2\cdot\frac1{2\sqrt2}\int_{-\infty}^\infty\bigg(\frac{1}{x^2+2-\sqrt2}-\frac{1}{x^2+2+\sqrt2}\bigg)dx\\
&=&\frac1{\sqrt2}\bigg(\frac1{\sqrt{2-\sqrt2}}\arctan\bigg(\frac{x}{\sqrt{2-\sqrt2}}\bigg)-\frac1{\sqrt{2+\sqrt2}}\arctan\bigg(\frac{x}{\sqrt{2+\sqrt2}}\bigg)\bigg)\bigg|_{-\infty}^\infty\\
&=&\frac\pi{\sqrt2}\bigg(\frac1{\sqrt{2-\sqrt2}}-\frac1{\sqrt{2+\sqrt2}}\bigg)\\
&=&\frac\pi{2}(\sqrt{2+\sqrt2}-\sqrt{2-\sqrt2}).
\end{eqnarray}
So
$$ I= \frac\pi{4}(\sqrt{2+\sqrt2}-\sqrt{2-\sqrt2})=\frac\pi{2}\sqrt{1-\frac1{\sqrt2}}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4630398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Distributing 7 red balls and 2 blue balls into 3 containers such that in each container, there are 3 balls and at least 2 of them are red balls. How many ways are there to distribute 7 distinguishable red balls and 2 distinguishable blue balls into 3 indistinguishable containers such that in each container, there are 3 balls and at least 2 of them are red balls?
I got $\dfrac{\binom{7}{2}\cdot\binom{5}{2}\cdot\binom{3}{2}\cdot 3\cdot 2}{3!}=630$, because there are $\binom{7}{2}\cdot\binom{5}{2}\cdot\binom{3}{2}$ to distribute 6 red balls so each container has 2 balls. Next, there are $3$ containers for the last 7 ball and 2 ways to distribute the 2 blue balls. Lastly, divide by $3!$ because the containers are indistinguishable. However, this is the wrong answer. Can someone tell me what's wrong with my solution?
Thanks!
| For another view point :
As @trueblueanil said , the only possible distribution can be done when one boxes has $3$ red balls and the other two take $2$ red balls. The blue balls will go directly to complement them to $3$ balls
To make our calculation easier , think that the boxes are distinct in the beginning.
Step 1-) Determine which box will have $3$ red balls by $C(3,1)$.
Step 2-) Distribute red balls to $3$ distinct boxes such that one box has three ball , the others have two balls by $C(7,3)\times C(4,2) \times C(2,2)$
Step 3-) Distribute $2$ blue balls to those boxes which have $2$ red balls such that each box will take $1$ blue balls ,so we can do it by $C(2,1) \times C(1,1)$.
Now we found the number of distributing $7$ distinct red balls and $2$ distinct blue balls to $3$ distinct boxes such that one box has $3$ red balls and the other two boxes have $2$ red balls and $1$ blue ball. It is equal to $$C(3,1)\times C(7,3)\times C(4,2) \times C(2,2) \times C(2,1) \times C(1,1)=1260$$
However, we know that those three boxes were identical , so divide the result by $3!$ to convert distinct boxes into identical.
$$\frac{C(3,1)\times C(7,3)\times C(4,2) \times C(2,2) \times C(2,1) \times C(1,1)}{3!}=210$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proof by contraposition. Let $x$ be an integer. If $8$ does not divide $x^2-1$, then $x$ is even. Prove by contraposition: Let $x$ be an integer. If $8$ does not divide $x^2-1$, then $x$ is even.
Assume $x$ is odd.
Prove $8|x^2-1$
So, $x=2n+1$ for some integer $n$.
Then $x^2-1=4(n^2+n)$ for some $n^2+n$ integer by closure of Z.
By algebra, multiply through by $2$ to get $2(x^2-1)=8(n^2+n)$.
Then, $x^2-1=8(\frac{n^2+n}{2})$
I get stuck here, because I cannot say that the fraction is an integer because $\mathbb{Z}$ isn't closed for division. Help!
| $\pmod 8$ we have $$0^2\equiv 0\\1^2\equiv 1\\2^2\equiv 4\\3^2\equiv 1\\4^2\equiv 0\\5^2\equiv 1\\6^2\equiv 4\\7^2\equiv 1$$
So we have $0,2,4,$ or $6 \pmod8.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4632026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Solve $\sin^6x+\cos^2x+3\sin^2x\cos^2x=\sin x$ Solve $$\sin^6x+\cos^2x+3\sin^2x\cos^2x=\sin x$$
My try:
The equation is equivalent to $$\left(\sin^2x\right)^3+\dfrac{1+\cos 2x}{2}+3\dfrac{1-\cos 2x}{2}\dfrac{1+\cos2x}{2}=\sin x$$ which is $$\left(\dfrac{1-\cos2x}{2}\right)^3+\dfrac{1+\cos2x}{2}+3\dfrac{1-\cos^22x}{4}=\sin x \iff\\(1-\cos2x)^3+4(1+\cos2x)+6(1-\cos^22x)=8\sin x \iff\\-\cos^32x-3\cos^22x+\cos2x+7=8\sin x$$ which makes the impression this isn't the most straight-forward approach. Any hints would be appreciated.
| You have\begin{multline}\sin^6x+\cos^2x+3\sin^2x\cos^2x=\sin x\iff\\\iff\sin^6x+1-\sin^2x+3\sin^2x(1-\sin^2x)-\sin x=0.\end{multline}So, let$$f(s)=s^6+1-s^2+3s^2(1-s^2)-s=s^6-3 s^4+2 s^2-s+1.$$You want to solve the equation $f(s)=0$ when $s\in[-1,1]$. It is easy to see that $f(1)=0$. So, $f(s)$ is a multiple of $s-1$; in fact, $f(s)=(s-1)\left(s^5+s^4-2 s^3-2 s^2-1\right)$. Now, let $g(s)=s^5+s^4-2 s^3-2 s^2-1$; I will show that we will always have $g(s)\leqslant-1$ in $[-1,1]$. So, let$$h(s)=g(s)+1=s^5+s^4-2 s^3-2 s^2.$$Asserting that $(\forall s\in[-1,1]):g(s)\leqslant-1$ is equivalent to asserting that$$(\forall s\in[-1,1]):h(s)\leqslant0.\label{a}\tag1$$But $h(s)=s^2(s^2-2)(s+1)$ and therefore it is clear that \eqref{a} holds. So, since $f(s)=(s-1)g(s)$, $f(s)=0\iff s=1$ (assuming that $s\in[-1,1]$). And $\sin x=1\iff x=\frac\pi2+2k\pi$, for some $k\in\Bbb Z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4634405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving radical inequality by a substitution and $pqr$
(Zuyong) If $a,b,c>0$ and $abc=1$, prove that
\[a+b+c+6\ge\sqrt{a^2+8}+\sqrt{b^2+8}+\sqrt{c^2+8}.\]
I want to see if teomehai's substitution works, since it does deal with the square roots very well.
We have $a=\dfrac{2-x^2}x$, $b=\dfrac{2-y^2}y$, $c=\dfrac{2-z^2}z$, then $abc=1$ implies that $\prod\left(2-x^2\right)=xyz$, and need to prove that $x+y+z\le\dfrac62=3$.
Let $p=x+y+z$, $q=xy+yz+zx$, $r=xyz$. The condition turns into
$$r^2+r+4p^2+4pr-8-8q-2q^2=0.$$
By Schur inequality, $r\ge\dfrac{4pq-p^3}9$, plug in to get
\begin{align*}&16 {{p}^{2}}{{q}^{2}}-162 {{q}^{2}}-8 {{p}^{4}} q+144 {{p}^{2}} q+36 p q
\\[3pt]{}-{}&648 q+{{p}^{6}}-36 {{p}^{4}}-9 {{p}^{3}}+324 {{p}^{2}}-648\le0\vphantom\strut.\end{align*}
Since $p^6$ can blow up faster than anything else, $p$ might get the desired up bound although I don't know how.
| Remark: Here is a complicated proof using pqr method.
Let $2x = \sqrt{a^2 + 8} - a,
2y = \sqrt{b^2 + 8} - b, 2z = \sqrt{c^2 + 8} - c$
(correspondingly, $a=\frac{2-x^2}x$, $b=\frac{2-y^2}y$, $c=\frac{2-z^2}z$).
Let $p = x + y + z, q = xy + yz + zx, r = xyz$.
The condition $abc = 1$ is written as
$$r^2 + r + 4p^2 + 4pr - 8 - 8q - 2q^2 = 0. \tag{1}$$
Also, clearly, $x, y, z \le \sqrt 2$. We have $x + y + z \le 3\sqrt 2$ and
$$(\sqrt 2 - x)(\sqrt 2 - y) + (\sqrt 2 - y)(\sqrt 2 - z)
+ (\sqrt 2 - z)(\sqrt 2 - x) \ge 0$$
which are respectively written as
$$p \le 3\sqrt 2, \tag{2}$$
and
$$6 - 2\sqrt 2\, p + q \ge 0. \tag{3}$$
We need to prove that $x + y + z \le 3$.
We split into two cases:
Case 1: $p^2 \ge 4q$
Using $q \le p^2/4$ and (3), we have
$$6 - 2\sqrt 2\, p + p^2/4 \ge 0$$
or
$$\frac14(6\sqrt 2 - p)(2\sqrt 2 - p) \ge 0$$
which, when combined with (2), results in $p \le 2\sqrt 2 < 3$.
Case 2: $p^2 < 4q$
Using $r \ge \frac{(p^2 - q)(4q - p^2)}{6p} \ge 0$ (degree four Schur), from (1), we have
\begin{align*}
&16\,{q}^{4}-40\,{p}^{2}{q}^{3}+ \left( 33\,{p}^{4}-168\,{p}^{2}-24\,p
\right) {q}^{2}\\
&\qquad + \left( -10\,{p}^{6}+120\,{p}^{4}+30\,{p}^{3}-288\,{p
}^{2} \right) q+{p}^{8}-24\,{p}^{6}-6\,{p}^{5}+144\,{p}^{4}-288\,{p}^{
2}\\
&\le 0. \tag{4}
\end{align*}
Denote LHS by $f(p, q)$.
We can prove that
$$\left. \begin{array}{r}
6 - \frac{14}{5} p + q > 0\\
p^2 \ge 3q\\
p\le 9/2\\
p > 3
\end{array}
\right\} \Longrightarrow f(p, q) > 0. \tag{5}$$
(The proof is given at the end.)
Note that $6 - 2\sqrt 2 \, p + q \ge 0$ implies $6 - \frac{14}{5}p + q > 0$, and $p \le 3\sqrt 2$ implies $p \le 9/2$. (Note: We avoid irrational numbers.)
From (4) and (5), we get a contradiction. As a result, we have $p \le 3$.
We are done.
Proof of (5):
We have
$$3 < p \le \frac92
\quad \iff \quad p = 3 \cdot \frac{s}{1 + s} + \frac92 \cdot \frac{1}{1 + s}, \quad s \ge 0. $$
Then we have
$$\frac{14}{5}p - 6 < q \le \frac{p^2}{3} \quad \iff
\quad q = \left(\frac{14}{5}p - 6\right)\cdot \frac{t}{1+t}
+ \frac{p^2}{3}\cdot \frac{1}{1+t}, \quad t \ge 0.$$
We have
$$f(p, q) = \frac{81}{160000(1+s)^8(1+t)^4}[g(s,t) + 4882500] > 0$$
where $g(s,t)$ is a polynomial with non-negative coefficients.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4636545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Nested Radicals from Brazil Show that
$$
\frac{ \sqrt{2} }{ \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} + 1} - \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} - 1}} = \sqrt[8]{ 1 + 2 \sqrt{ \sqrt{5} -2 } }.
$$
What I've tried so far.
$$
\begin{align*}
E & = \frac{ \sqrt{2} }{ \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} + 1} - \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} - 1}} \\
& = \frac{1}{\sqrt{2}} \cdot \left[ \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} + 1} + \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} - 1} \right] \\
& = \frac{1}{\sqrt{2}} \cdot \left[ \sqrt{ a + 1 } + \sqrt{a -1} \right]
\end{align*}
$$
Squaring $E$,
$$
\begin{align*}
E^2 & = \frac{1}{2} \left[ a + 1 + a - 1 + 2\sqrt{a+1} \sqrt{a-1} \right] \\
& = \frac{1}{2} \left[ 2a + 2 \sqrt{a^2-1} \right] \\
& = a + \sqrt{a^2-1}
\end{align*}
$$
Then, isolating the radical
$$
\begin{align*}
E^2 - a & = \sqrt{a^2-1} \\
E^4 + a^2 - 2aE^2 & = a^2 - 1
\end{align*}
$$
We obtain $a$,
$$
a = \frac{1}{2} \left( E^2 + \frac{1}{E^2} \right).
$$
And then, nothing useful comes up.
| With your notations:
$E^2 = a +\sqrt{a^2-1}\\
E^8 =a^4+4a^3 \sqrt{a^2-1} +6a^2(a^2-1)+4a(a^2-1)\sqrt{a^2-1}+(a^2-1)^2\\
E^8 = 8a^4-8a^2+1+(8a^3-4a)\sqrt{a^2-1}$
We can write that:
$2\sqrt{\sqrt{5}-2} = \dfrac{2 \sqrt{\sqrt{5}-2} \sqrt{\dfrac{\sqrt{5}+2}{4}}} {\sqrt{\dfrac{\sqrt{5}+2}{4}}} =\dfrac{1}{a^2} $
So, we want to show that $\ E^8 = 1+\dfrac{1}{a^2}$
We want to show that $\ 8a^4-8a^2 + (8a^3-4a)\sqrt{a^2-1}=\dfrac{1}{a^2}$
We want to show that $\ (8a^5-4a^3)\sqrt{a^2-1} = 1-8a^6+8a^4$
Let $\ x=(8a^5-4a^3)\sqrt{a^2-1}\ $ and $\ y=1-8a^6+8a^4\ $. It's easy to check that $x$ and $y$ are positive.
$x^2-y^2 = (8a^5-4a^3)^2 (a^2-1) - (1-8a^6+8a^4)^2\\
x^2-y^2 = (64a^{10}-64a^8+16a^6)(a^2-1)-(64a^{12}-128a^{10}+64a^8-16a^6+16a^4+1)\\
x^2-y^2 = 16a^8-16a^4-1$
And, now, it's easy to check that $x=y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4636861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral from MIT Integration Bee 2023 Quarterfinals - $\lim_{n\to\infty} \frac{1}{n} \int_{0}^{n} \cos^2(\pi x^2/\sqrt{2}) \, \mathrm{d}x$ This question is from the MIT Integration Bee 2023 Quarterfinal #2. I'm doing these integrals to improve my quick math skills, and was wondering if there's a way to solve this within two minutes.
The goal is to show that $$\lim_{n\to\infty} \frac{1}{n} \int_{0}^{n} \cos^2\left(\frac{\pi x^2}{\sqrt{2}}\right) \,\textrm{d}x = \frac{1}{2} $$
I've tried L'Hospital's Rule then FTC1 but I don't think the first part would even apply here. I'm not sure how to approach such a problem, especially if it's to be solved within two minutes.
| Let $x=nt$ and then
\begin{eqnarray}
&&\frac{1}{n} \int_{0}^{n} \cos^2\left(\frac{\pi x^2}{\sqrt{2}}\right) \,\textrm{d}x \\
&=&\frac12 + \frac12\int_0^1\cos\left(\sqrt2\pi n^2t^2\right) \textrm{d}t\\
&=&\frac12 + \frac12\int_0^1\frac{1}{2\sqrt2\pi n^2t}\textrm{d}\sin\left(\sqrt2\pi n^2t^2\right)\\
&=&\frac12 + \frac{1}{4\sqrt2\pi n^2t}\sin\left(\sqrt2\pi n^2t^2\right)\bigg|_0^1+\frac{1}{4\sqrt2\pi n^2}\int_0^1\frac{\sin\left(\sqrt2\pi n^2t^2\right)}{t^2}\textrm{d}x.
\end{eqnarray}
Note
$$ \frac{1}{4\sqrt2\pi n^2t}\sin\left(\sqrt2\pi n^2t^2\right)\bigg|_0^1=\frac{1}{4\sqrt2\pi n^2}\sin\left(\sqrt2\pi n^2\right). $$
Since
$$\frac{1}{4\sqrt2\pi n^2}\bigg|\frac{\sin\left(\sqrt2\pi n^2t^2\right)}{t^2}\bigg|\le\frac14,$$
one has, by DCT,
$$ \lim_{n\to\infty}\frac{1}{4\sqrt2\pi n^2}\int_0^1\frac{\sin\left(\sqrt2\pi n^2t^2\right)}{t^2}\textrm{d}x=\int_0^1\lim_{n\to\infty}\frac{1}{4\sqrt2\pi n^2}\frac{\sin\left(\sqrt2\pi n^2t^2\right)}{t^2}\textrm{d}x=0.$$
So
$$\lim_{n\to\infty} \frac{1}{n} \int_{0}^{n} \cos^2\left(\frac{\pi x^2}{\sqrt{2}}\right) \,\textrm{d}x = \frac{1}{2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4638503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\binom{40}{0}\binom{20}{10}+\binom{41}{1}\binom{19}{10}+\cdots+\binom{50}{10}\binom{10}{10}$
Finding value of
$$\binom{40}{0}\binom{20}{10}+\binom{41}{1}\binom{19}{10}+\cdots+\binom{50}{10}\binom{10}{10}$$
Using $\displaystyle \binom{n}{r}=\binom{n}{n-r}$
$\displaystyle \binom{40}{0}\binom{20}{10}+\binom{41}{1}\binom{19}{9}+\cdots +\binom{50}{10}\binom{10}{0}$
I have tried it using combinational argument
Suppose we have $60$ students in a class in which $40$ boys and $20$ girls and we have to select $10$ students in which $r$ are boys and $n-r$ are girls
So total ways $\displaystyle \binom{60}{10}$
But my answer did not match with right answer
please have a look
| $$
\sum_{k=0}^n \binom{4n+k}{k}\binom{2n-k}{n}
= \sum_{k=0}^n \binom{4n+k}{4n}\binom{2n-k}{n}
= \binom{6n+1}{5n+1} = \binom{6n+1}{n}
$$
Now take $n=10$ to obtain $\binom{61}{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4639436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Maximize $(1-a)(1-c)+(1-b)(1-d)$ over $a^2+b^2=c^2+d^2=1$. Let $a,b,c,d$ be real numbers such that $a^2+b^2=c^2+d^2=1$. Find the maximum value of $(1-a)(1-c)+(1-b)(1-d)$.
I tried substituting $a=\sin x, b =\cos x, c = \sin y, d=\cos y$, then expanded $(1-a)(1-c)+(1-b)(1-d)$. However this just leads to an ugly expression, and I'm not sure how to proceed
| WLOG, Let $a=-\sin2x,c=-\sin2y, b=-\cos2x,d=-\cos2y$
$$S=(1-a)(1-c)+(1-b)(1-d)$$
$$=2+(\sin2x+\sin2y+\cos2x+\cos2y)+\cos2(x-y)$$
$$=1+2\cos(x-y)(\sin(x+y)+\cos(x+y))+2\cos^2(x-y)$$ will be maximum
$(1)$ if $\cos(x-y)$ is maximum and so is $\sin(x+y)+\cos(x+y)$
$\cos(x-y)$ will be maximum $(=1)$ which needs $x=2n\pi+y$
$\sin(x+y)+\cos(x+y)$ reduces to $\sin2y+\cos2y=\sqrt2\cos\left(2y-\dfrac\pi4\right)$
which is $\le\sqrt2$ the equality occurs if $2y=2m\pi+\dfrac\pi4$
So, $$S\le1+2\cdot1\cdot\sqrt2+2\cdot1$$
$(2)$ or if $\cos(x-y)$ is minimum and $\cos(x+y)+\sin(x+y)$ is minimum i.e., $-\sqrt2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_{0}^{\pi/2} \ln\left[ \tan\left ( \frac{\theta}{2}\right) \right ]^2 K\left ( \sin\theta \right )\text{d}\theta$ Let us define $K(x)$ as complete elliptic integral of the first kind, where $x$ is elliptic modulus. A possible closed-form is ($G$ denotes Catalan's constant.)
$$
\int_{0}^{\pi/2}
\ln\left[ \tan\left ( \frac{\theta}{2}\right) \right ]^2
K\left ( \sin\theta \right )\text{d}\theta
=\frac{\Gamma\left ( \frac14 \right )^4G }{8\pi}.
$$
It looks like a "product" of two solvable integrals (both are elementary):
$$\int_{0}^{\pi/2}
\ln\left[ \tan\left ( \frac{\theta}{2}\right) \right ]^2\text{d}\theta
=\frac{\pi^3 }{8}$$
and
$$
\int_{0}^{\pi/2}
K\left ( \sin\theta \right )\text{d}\theta
=\frac{\Gamma\left ( \frac14 \right )^4 }{16\pi}.
$$
Question:
How can we evaluate the integral? I try to utilize the Fourier series $K(\sin\theta)
=\pi\sum_{n\ge0} \frac{\left ( \frac12 \right )_n^2 }{(n!)^2}
\sin\left ( \left ( 4n+1 \right )\theta \right )$ to prove, but seems not to go well. I appreciate for your help.
An Interesting Observation:
We find
$$
\int_{0}^{\pi/2}
\ln\left[ \tan\left ( \frac{\theta}{2}\right) \right ]^4
K\left ( \sin\theta \right )\text{d}\theta
=\frac{3\,\Gamma\left ( \frac14 \right )^4}{4\pi}(G^2+\beta(4))
$$
where $\beta(.)$ is Dirichlet's $\beta$ function.
| We have
$$\int_0^{\pi/2} \log^{2n} (\tan \frac{x}{2}) K(\sin x) dx = \int_0^1 \frac{2 \log^{2n} t}{1+t^2} K(\frac{2t}{1+t^2}) dt$$
Their values can be extracted by differentiating $$\tag{*}\int_0^1 \frac{t^{4a} + t^{-4a}}{1+t^2} K(\frac{2t}{1+t^2}) dt = \frac{\pi}{8}\cot(\pi(\frac{1}{4}-a))\frac{\Gamma \left(a+\frac{1}{4}\right)^2}{\Gamma \left(a+\frac{3}{4}\right)^2} \quad -1/4<\Re(a)<1/4$$
I give two different proofs of $(*)$, I come up with the longer proof first.
First proof of $(*)$: using quadratic transformation $(1+t^2)K(t^2) = K(2t/(1+t^2))$, we have
$$
\begin{aligned}\int_0^1 \frac{t^{4a}}{1+t^2} K(\frac{2t}{1+t^2}) dt &= \frac{1}{2} \int_0^1 t^{2a-1/2} K(t) dt \\ &= \frac{1}{2} \frac{\pi}{1+4a}{_3F_2}(\frac{1}{2}, \frac{1}{2}, \frac{1}{4} + a; 1, \frac{5}{4} + a; 1)
\end{aligned}$$
here we noted that $K(t)$ is a $_2F_1$. Using third formulas here gives
$${_3F_2}(\frac{1}{2}, \frac{1}{2}, \frac{1}{4} + a; 1, \frac{5}{4} + a; 1) = \frac{4 a+1}{4 a-1} {_3F_2(\frac{1}{2}, \frac{1}{2}, \frac{1}{4} - a; 1, \frac{5}{4} - a; 1)} +\frac{\Gamma \left(\frac{1}{4}-a\right) \Gamma \left(a+\frac{5}{4}\right)}{\Gamma \left(\frac{3}{4}-a\right) \Gamma \left(a+\frac{3}{4}\right)}$$
we see the $a$ in $_3F_2$ becomes $-a$, giving $(*)$. Q.E.D.
Second proof of $(*)$: It's much longer. See edit history.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Maximizing volume of a cylinder with $2$ cones given surface area $A$ A cylinder with a cone on either side of it. The cones have the same radius as the cylinder.
Cylinder area without the sides:
$$A_{cyl}=2\pi r h_1$$
Cone area without the base:
$$A_{cone}=\pi r \sqrt{h_2^2+r^2}$$
Total area of body:
$$A_{tot}=2\pi r h_1 + 2\pi r \sqrt{h_2^2+r^2}$$
Volume:
$$V_{tot}=\pi r^2 h_1 +\frac{2\pi r^2 h_2}{3}=\pi r^2 (h_1 + \frac{2}{3}h_2)$$
I have managed to calculate the maximum volume given surface area for the cone and the cylinder separately but together there are three variables. How would I go about calculating $r,h_1,h_2$ to maximize the volume?
| As I said yesterday, we can eliminate one variable, $h_1=\frac{A}{2\pi r}-\sqrt{h_2^2+r^2}$ to get
$$V(h_2,r)=\frac{Ar}{2}-\pi r^2\sqrt{h_2^2+r^2}+\frac{2\pi r^2h_2}{3}.\tag1$$
The system of equations for critical points $\frac{\partial V}{\partial h_2}=\frac{\partial V}{\partial r}$ on the open region $r,h_2>0$ gives
$$\color{green}{-\frac{\pi r^2h_2}{\sqrt{h_2^2+r^2}}+\frac{2\pi r^2}{3}=0}\,\,\text{and}\,\,\color{blue}{\frac A2-2\pi r\sqrt{h_2^2+r^2}-\frac{\pi r^3}{\sqrt{h_2^2+r^2}}+\frac{4\pi rh_2}{3}=0}$$
From green $h_2=\frac{2}{\sqrt 5}r$ and from blue and this, $r=\sqrt{\frac{A}{2\sqrt 5\pi}}$ and thus $h_2=\sqrt{\frac{2A}{5\sqrt 5\pi}}$.
To show that at $(h_2,r)=(\sqrt{\frac{2A}{5\sqrt 5\pi}},\sqrt{\frac{A}{2\sqrt 5\pi}})$ we have a maximum we need to show that
$$\frac{\partial^2 V}{\partial h_2^2}\frac{\partial^2 V}{\partial r^2}-(\frac{\partial^2 V}{\partial h_2\partial r})^2>0\,\, \text{and} \,\,\frac{\partial^2 V}{\partial h_2^2}>0.$$
But, this is messy. Instead, when $r\rightarrow\infty$ and/or $h_2\rightarrow\infty$, from $(1)$, we see that $V\rightarrow -\infty$, so it is a local maximum $V_{\max}=\frac{A\sqrt A}{3\sqrt{2}\sqrt\pi\sqrt{\sqrt 5}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Using Cayley-Hamilton theorem to find minimal polynomial I need to find the minimal polynomial of $\beta = i + \sqrt[3]{2}$ over $\mathbb{Q}[x]$ using Cayley-Hamilton theorem. So far I've found $\{1, \sqrt[3]{2}, \sqrt[3]{4}, i, \sqrt[3]{2}i, \sqrt[3]{4}i\}$ is base of $\mathbb{Q}[\sqrt[3]{2},i]$ as a vector space over $\mathbb{Q}$. Also, the matrix representing the multiplication by $\beta$ using the base is
$$
A =
\begin{pmatrix}
0& 0& 2&-1& 0& 0\\
1& 0& 0& 0&-1& 0\\
0& 1& 0& 0& 0&-1\\
1& 0& 0& 0& 0& 2\\
0& 1& 0& 1& 0& 0\\
0& 0& 1& 0& 1& 0\\
\end{pmatrix}
$$
where the element $\lambda_{1} + \lambda_{2}\sqrt[3]{2} + \lambda_{3}\sqrt[3]{4} + \lambda_{4}i + \lambda_{5}\sqrt[3]{2}i + \lambda_{6}\sqrt[3]{4}i$ in $\mathbb{Q}[\sqrt[3]{2},i]$ is represented by the vector
$$
\begin{pmatrix}
\lambda_{1}\\
\lambda_{2}\\
\lambda_{3}\\
\lambda_{4}\\
\lambda_{5}\\
\lambda_{6}\\
\end{pmatrix}.
$$
I couldn´t figure out how to use the characteristic polynomial of $A$ ($p(x) = x^{6} + 3x^{4} - 4x^{3} + 3x^{2}$) to find a polynomial $f(x)$ in $\mathbb{Q}[x]$ such that $f(\beta) = 0$.
| The charactestic polynomial should be $p(x)=x^6 + 3 x^4 - 4 x^3 + 3 x^2 + 12 x + 5.$ You need to check if $p$ is irreducible. There are many standard algorithm to check if a polynomial is irreducible and you can use them. Otherwise you can use sagemath or any other tools to check as done here SageMath.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4651327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
For any prime $p > 3$, why is $p^2-1$ always divisible by $24$?
Let $p>3$ be a prime. Prove that $24 \mid p^2-1$.
I know this is very basic and old hat to many, but I love this question and I am interested in seeing whether there are any proofs beyond the two I already know.
| Consider modulo $24$.. We have $i\mod 24$ for $0\leq i\leq 23$
We have to neglect $a\mod 24$ if $a$ is a multiple of $2$ or a multiple of $3$.. We are left with
*
*$1\mod 24$
*$5\mod 24$
*$7\mod 24$
*$11\mod 24$
*$13\mod 24\equiv -11\mod 24$
*$17\mod 24\equiv -7\mod 24$
*$19\mod 24\equiv -19\mod 24$
*$23\mod 24\equiv -1\mod 24$
considering squares, we have
*
*$1^2\mod 24\equiv 1\mod 24$
*$5^2\mod 24\equiv 1\mod 24$
*$7^2\mod 24\equiv 1\mod 24$
*$11^2\mod 24\equiv 1\mod 24$
For same reason as above, squares of below congruence classes are equal to $1\mod 24$.
*
*$13\mod 24\equiv -11\mod 24$
*$17\mod 24\equiv -7\mod 24$
*$19\mod 24\equiv -19\mod 24$
*$23\mod 24\equiv -1\mod 24$
So, $p^2\equiv 1\mod 24$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "96",
"answer_count": 21,
"answer_id": 6
} |
Example of a function whose Fourier Series fails to converge at One point Can one think of an example of a continuous $2\pi$ periodic function whose Fourier series fails to converge on $\mathbb{R}$.
I referred this in the wikipedia page but no avail: It might be interesting to note that Jean-Pierre Kahane and Yitzhak Katznelson proved that for any given set E of measure zero, there exists a continuous function ƒ such that the Fourier series of ƒ fails to converge on any point of E.
| All, please see this example.
Let $G_{n}$ denote the grouping of this $2n$ numbers, $$\frac{1}{2n-1},\frac{1}{2n-3},...,\frac{1}{3},1,-1,-\frac{1}{3},\cdots,-\frac{1}{2n-1}$$
We take a strictly increasing sequence of positive integers ${\lambda_n}$ and consider the groups $G_{\lambda_1},G_{\lambda_2},\cdots,$. We multiply each number of the group $G_{\lambda_n}$ by $n^{-2}$ and obtain the sequence $$\frac{1}{1^{2}(2\lambda_{1}-1)}, \cdots,-\frac{1}{1^{2}(2\lambda_{1}-1)}, \frac{1}{2^{2}(2\lambda_{2}-1)},...,-\frac{1}{2^2(2\lambda_{2}-1)},....,$$
say $\alpha_{1},\alpha_{2},\cdots$. Our aim is to show that $$\sum\limits_{n=1}^{\infty} \alpha_{n} \cos{nx}$$
is the fourier series of a continuous function. We group the terms in the following way $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2\lambda_{1}+1}^{2\lambda_{1}+2\lambda_{2}} \alpha_{n}\cos{nx} + \sum\limits_{n=2\lambda_{1}+2\lambda_{2}+1}^{2\lambda_{1}+2\lambda_{2}+2\lambda_{3}} \alpha_{n} \cos{nx}\cdots$$
The last series can be written as $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2}^{\infty} \frac{\phi(\lambda_{n},2\lambda_{1}+2\lambda_{2} + \cdots + 2\lambda_{n-1},x)}{n^2}$$
where $$\phi(n,r,x)= \frac{\cos{(r+1)x}}{2n-1} + \frac{\cos{(r+2)x}}{2n-3} + \cdots + \frac{\cos{(r+n)x}}{1} - \frac{\cos{(r+n+1)x}}{1} - \cdots - \frac{\cos{(r+2n)x}}{2n-1}$$
Now one can show that there is a constant $M$ (independent of $n,r$ and $x$) such that $|\phi(n,r,x)|\leq M$. From this it follows that the grouped series $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2}^{\infty} \frac{\phi(\lambda_{n},2\lambda_{1}+2\lambda_{2} + \cdots + 2\lambda_{n-1},x)}{n^2}$$
converges absolutely on $\mathbb{R}$, say to $f(x)$, and $f$ is continuous on $\mathbb{R}$. It is also easy to check that $$f(x) \sim \sum\limits_{n=1}^{\infty} \alpha_{n} \cos{nx}$$
We shall finally show that ${\lambda_n}$ can be chose so that the above series diverges at zero, that is $S_{n} = \alpha_{1} + \alpha_{2} + \cdots + \alpha_{n}$ diverges to infinity.
Since $$S_{2\lambda_{1}+2\lambda_{2}+ \cdots + 2 \lambda_{n-1} + \lambda_{n}} = \frac{1}{n^2} \Bigl( \frac{1}{2\lambda_{n}-1} + \cdots + \frac{1}{3} + 1 \Bigr)$$ behaves as $\frac{\ln{\lambda_{n}}}{{2n^{2}}}$ as $n \to \infty$, it is enough to take $\lambda_{n}=n^{n^2}$. Then the fourier series does not converge to $f$ at $x=2k\pi, \ k\in \mathbb{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
What is the best way to solve an equation involving multiple absolute values? An absolute value expression such as $|ax-b|$ can be rewritten in two cases as $|ax-b|=\begin{cases}
ax-b & \text{ if } x\ge \frac{b}{a} \\
b-ax & \text{ if } x< \frac{b}{a}
\end{cases}$, so an equation with $n$ separate absolute value expressions can be split up into $2^n$ cases, but is there a better way?
For example, with $|2x-5|+|x-1|+|4x+3|=13$, is there a better way to handle all the possible combinations of $x\ge\frac{5}{2}$ versus $x<\frac{5}{2}$, $x\ge 1$ versus $x< 1$, and $x\ge-\frac{3}{4}$ versus $x<-\frac{3}{4}$?
| For an equation with $n$ absolute values, the $n$ places where each absolute value splits into 2 cases divide the number line into $n+1$ regions where, within each region, each absolute value can be replaced by either the expression inside the absolute value or its opposite. Each resulting equation can then be solved, restricting solutions to the corresponding region on the number line.
In the given example,
$\small{\begin{matrix}
\leftarrow & -\frac{3}{4} & \text{---} & 1 & \text{---} & \frac{5}{2} & \rightarrow
\\
\begin{matrix}-(2x-5)-(x-1)\\-(4x+3)=13\end{matrix}
& \begin{matrix}|\\|\end{matrix} &
\begin{matrix}-(2x-5)-(x-1)\\+(4x+3)=13\end{matrix}
& \begin{matrix}|\\|\end{matrix} &
\begin{matrix}-(2x-5)+(x-1)\\+(4x+3)=13\end{matrix}
& \begin{matrix}|\\|\end{matrix} &
\begin{matrix}(2x-5)+(x-1)\\+(4x+3)=13\end{matrix}
\\
-7x+3=13
& | &
x+9=13
& | &
3x+7=13
& | &
7x-3=13
\\
x=-\frac{10}{7}
& | &
x=4\notin[-\frac{3}{4},1]
& | &
x=2
& | &
x=\frac{16}{7}\notin[\frac{5}{2},\infty)
\end{matrix}}$
So, the solutions are $x=-\frac{10}{7}$ and $x=2$ (the values that were solutions to an equation for a particular region and were within that region).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 1
} |
If a prime $p$ satisfies given condition then $p^{4} \mid ap-b$ Question If $p > 3$ is a prime and $$ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{p} = \frac{a}{b}$$ then prove that $p^{4} \mid (ap-b)$.
There is an exercise in Herstein which states, if $p > 3$ is prime and if $$1 + \frac{1}{2} + \cdots + \frac{1}{p-1} = \frac{a}{b}$$ then $p^{2} \mid a$. (Page 116, Problem 2). But i am not sure whether this helps or not. Hints, suggestions would be of great help!
| From Wolstenholme's Theorem we may assume that, $1+\frac{1}{2}+.....+\frac{1}{p-1}=p^2\frac{x}{y}$.
Then $\frac{a}{b}=\frac{p^2x}{y}+\frac{1}{p}$.
This gives us, $\frac{ap-b}{b}=\frac{p^3x}{y}$.
Or, $y(ap-b)=bp^3x$.
Obviously, $p\not|y,p|b$. Let $b=pk$. We have $y(ap-b)=kp^4x$.
This equation gives $p^4|y(ap-b)$
And since $\gcd(p,y)=1,p^4|ap-b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Rationalising the denominator: $\frac{11}{3\sqrt{3}+7}$ For my homework, I have been asked to rationalise and simplify this surd;
$$\frac{11}{3\sqrt{3}+7}$$
Each time I do this I get the wrong answer. The method I am using is;
$$ \frac{11}{3\sqrt3+7} \times \frac{3\sqrt3-7}{3\sqrt3-7} $$
I ended up with $$\frac{33+11\sqrt3-77}{9+3+21+7\sqrt3-21-7\sqrt3}$$
This ends up no where near the right answer, even once it is simplified. Can someone tell me where I'm going wrong?
Many thanks!
| Apparently you made a minor operational mistake when interpreting $ 3\sqrt{3} + 7 $ as its conjugate, $ 3\sqrt{3} - 7 $. The proper way to rationalize the denominator containing a radical is to remove that radical by multiplying the numerator and denominator by the original radical's denominator's conjugate, $ 3\sqrt{3} - 7 $:
$$ \frac{11}{3\sqrt{3} + 7} \times \frac{3\sqrt{3} - 7}{3\sqrt{3} - 7} $$
Multiplying the above expression yields:
$$ \frac{33\sqrt{3} - 77}{9\cdot3 - 21\sqrt{3} + 21\sqrt{3} - 49} $$
Canceling the $ 21\sqrt{3} $ in the denominator and simplifying,
$$ \frac{33\sqrt{3} - 77}{9\cdot3 - 21\sqrt{3} + 21\sqrt{3} - 49} = \frac{33\sqrt{3} - 77}{27 - 49} = \frac{33\sqrt{3} - 77}{-22} = -\frac{33\sqrt{3} - 77}{22}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/5238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Find the sum to n terms of the series $\frac{1} {1\cdot2\cdot3\cdot4} + \frac{1} {2\cdot3\cdot4\cdot5} + \frac{1} {3\cdot4\cdot5\cdot6}\ldots $ Find the sum to n terms of the series $\frac{1} {1\cdot2\cdot3\cdot4} + \frac{1} {2\cdot3\cdot4\cdot5} + \frac{1} {3\cdot4\cdot5\cdot6}\ldots $
Please suggest an approach for this task.
| $\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} (\dfrac{k+3}{k(k+1)(k+2)(k+3)} - \dfrac{k}{k(k+1)(k+2)(k+3)})$
$ = \dfrac{1}{3}(\dfrac{1}{k(k+1)(k+2)} - \dfrac{1}{(k+1)(k+2)(k+3)})$
$\sum_{k=1}^{\infty}\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} \dfrac{1}{2*3} = \dfrac{1}{18}$
@moron Yes, you are right. For the first n terms:
$\sum_{k=1}^{n}\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} (\dfrac{1}{1*2*3} - \dfrac{1}{(n+1)(n+2)(n+3)})$
[edit] more details
$\sum_{k=1}^{n}\dfrac{1}{k(k+1)(k+2)(k+3)} = \sum_{k=1}^{n} \dfrac{1}{3} (\dfrac{1}{k(k+1)(k+2)} - \dfrac{1}{(k+1)(k+2)(k+3)})$
$= \dfrac{1}{3} [\sum_{k=1}^{n} \dfrac{1}{k(k+1)(k+2)} - \sum_{k=1}^{n} \dfrac{1}{(k+1)(k+2)(k+3)}]$
$= \dfrac{1}{3} [\sum_{k=0}^{n-1} \dfrac{1}{(k+1)(k+2)(k+3)} - \sum_{k=1}^{n} \dfrac{1}{(k+1)(k+2)(k+3)}]$
$= \dfrac{1}{3} (\dfrac{1}{1*2*3} - \dfrac{1}{(n+1)(n+2)(n+3)})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/5558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
For $n \in \mathbb{N}$ $\lfloor{\sqrt{n} + \sqrt{n+1}\rfloor} = \lfloor{\sqrt{4n+2}\rfloor}$ This is Exercise 3.20 from Apostol's book. Many of them seem tough and here is one of them which I am struggling with.
For $n \in \mathbb{N}$, prove that this identity is true: $$\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} = \Bigl\lfloor{\sqrt{4n+2}\Bigl\rfloor}$$
| Case 1: $n^2\le x\le(n+1)^2-2$ (for $n\ge 3$), then
$$ n\le \sqrt{x}\le\sqrt{(n+1)^2-2}<n+1, n<\sqrt{n^2+1}\le \sqrt{x+1}<\sqrt{(n+1)^2-1}=n+1$$
and
$$ 2n<\sqrt{4n^2+2}\le \sqrt{4x+2}<\sqrt{4(n+1)^2-2}<2n+1$$
So
$$ 2n<\sqrt{x}+\sqrt{x+1}<2n+2, \sqrt{4x+2}=2n $$
and
$$ 2n+1\le[\sqrt{x}+\sqrt{x+1}]\le2n+2, [\sqrt{4x+2}]=2n+1. $$
Now we show
$$ \sqrt{x}+\sqrt{x+1}<2n+1. $$
Note $\sqrt{x}+\sqrt{x+1}$ is increasing and hence
$$\sqrt{x}+\sqrt{x+1}\le\sqrt{(n+1)^2-2}+\sqrt{(n+1)^2-1}.$$
Note
\begin{eqnarray}
&&\sqrt{(n+1)^2-2}+\sqrt{(n+1)^2-1}-2n-1\\
&=&\frac{(n+1)^2-2-n^2}{\sqrt{(n+1)^2-2}+n^2}+\frac{(n+1)^2-1-n^2}{\sqrt{(n+1)^2-1}+n^2}-1\\
&<&\frac{4n-1}{\sqrt{(n+1)^2-2}+n^2}-1\\
&=&\frac{4n-1}{\sqrt{n^2+2n-1}+n^2}-1\\
&<&\frac{4n-1}{n+n^2}-1\\
&=&-\frac{n^2-3n+1}{n+n^2}\\
&<0
\end{eqnarray}
for $n\ge 3$.
Thus
$$2n<\sqrt{(n+1)^2-2}+\sqrt{(n+1)^2-1}<2n+1$$
or
$$[\sqrt{x}+\sqrt{x+1}=2n+1$$
Case 2: $x=(n+1)^2-1$.
\begin{eqnarray}
&&\sqrt{x}+\sqrt{x+1}\\
&=&\sqrt{(n+1)^2-1}+n+1\\
&<&2(n+1)
\end{eqnarray}
and
\begin{eqnarray}
&&\sqrt{x}+\sqrt{x+1}\\
&=&\sqrt{(n+1)^2-1}+n+1\\
&>&2n+1
\end{eqnarray}
and hence
$$ [\sqrt{x}+\sqrt{x+1}]=2(n+1). $$
On the other hand
$$ \sqrt{4x+2}=\sqrt{4(n+1)^2-2}<2(n+1) $$
and
$$ \sqrt{4x+2}=\sqrt{4(n+1)^2-2}>2n+1 $$
and hence
$$ [\sqrt{4x+2}]=2n+2 $$
In both cases,
$$ [\sqrt{x}+\sqrt{x+1}]=[\sqrt{4x+2}]. $$
Case 3: for $n=1,2$, it is easy to check the inequality is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/6087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 4
} |
The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
| We can use the function $f(x)=x^{2}$ with $-\pi \leq x\leq \pi $ and find
its expansion into a trigonometric Fourier series
$$\dfrac{a_{0}}{2}+\sum_{n=1}^{\infty }(a_{n}\cos nx+b_{n}\sin nx),$$
which is periodic and converges to $f(x)$ in $[-\pi, \pi] $.
Observing that $f(x)$ is even, it is enough to determine the coefficients
$$a_{n}=\dfrac{1}{\pi }\int_{-\pi }^{\pi }f(x)\cos nx\;dx\qquad n=0,1,2,3,...,$$
because
$$b_{n}=\dfrac{1}{\pi }\int_{-\pi }^{\pi }f(x)\sin nx\;dx=0\qquad n=1,2,3,... .$$
For $n=0$ we have
$$a_{0}=\dfrac{1}{\pi }\int_{-\pi }^{\pi }x^{2}dx=\dfrac{2}{\pi }\int_{0}^{\pi
}x^{2}dx=\dfrac{2\pi ^{2}}{3}.$$
And for $n=1,2,3,...$ we get
$$a_{n}=\dfrac{1}{\pi }\int_{-\pi }^{\pi }x^{2}\cos nx\;dx$$
$$=\dfrac{2}{\pi }\int_{0}^{\pi }x^{2}\cos nx\;dx=\dfrac{2}{\pi }\times \dfrac{
2\pi }{n^{2}}(-1)^{n}=(-1)^{n}\dfrac{4}{n^{2}},$$
because
$$\int x^2\cos nx\;dx=\dfrac{2x}{n^{2}}\cos nx+\left( \frac{x^{2}}{
n}-\dfrac{2}{n^{3}}\right) \sin nx.$$
Thus
$$f(x)=\dfrac{\pi ^{2}}{3}+\sum_{n=1}^{\infty }\left( (-1)^{n}\dfrac{4}{n^{2}}
\cos nx\right) .$$
Since $f(\pi )=\pi ^{2}$, we obtain
$$\pi ^{2}=\dfrac{\pi ^{2}}{3}+\sum_{n=1}^{\infty }\left( (-1)^{n}\dfrac{4}{
n^{2}}\cos \left( n\pi \right) \right) $$
$$\pi ^{2}=\dfrac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\left( (-1)^{n}(-1)^{n}
\dfrac{1}{n^{2}}\right) $$
$$\pi ^{2}=\dfrac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}.$$
Therefore
$$\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}=\dfrac{\pi ^{2}}{4}-\dfrac{\pi ^{2}}{12}=
\dfrac{\pi ^{2}}{6}$$
Second method (available on-line a few years ago) by Eric Rowland. From
$$\log (1-t)=-\sum_{n=1}^{\infty}\dfrac{t^n}{n}$$
and making the substitution $t=e^{ix}$ one gets the series expansion
$$w=\text{Log}(1-e^{ix})=-\sum_{n=1}^{\infty }\dfrac{e^{inx}}{n}=-\sum_{n=1}^{
\infty }\dfrac{1}{n}\cos nx-i\sum_{n=1}^{\infty }\dfrac{1}{n}\sin nx,$$
whose radius of convergence is $1$. Now if we take the imaginary part of both sides, the RHS becomes
$$\Im w=-\sum_{n=1}^{\infty }\dfrac{1}{n}\sin nx,$$
and the LHS
$$\Im w=\arg \left( 1-\cos x-i\sin x\right) =\arctan \dfrac{-\sin x}{
1-\cos x}.$$
Since
$$\arctan \dfrac{-\sin x}{1-\cos x}=-\arctan \dfrac{2\sin \dfrac{x}{2}\cdot \cos \dfrac{x}{2}}{2\sin ^{2}\dfrac{x}{2}}$$
$$=-\arctan \cot \dfrac{x}{2}=-\arctan \tan \left( \dfrac{\pi }{2}-\dfrac{x}{2}
\right) =\dfrac{x}{2}-\dfrac{\pi }{2},$$
the following expansion holds
$$\dfrac{\pi }{2}-\frac{x}{2}=\sum_{n=1}^{\infty }\dfrac{1}{n}\sin nx.\qquad
(\ast )$$
Integrating the identity $(\ast )$, we obtain
$$\dfrac{\pi }{2}x-\dfrac{x^{2}}{4}+C=-\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}\cos
nx.\qquad (\ast \ast )$$
Setting $x=0$, we get the relation between $C$ and $\zeta (2)$
$$C=-\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}=-\zeta (2).$$
And for $x=\pi $, since
$$\zeta (2)=2\sum_{n=1}^{\infty }\dfrac{(-1)^{n-1}}{n^{2}},$$
we deduce
$$\dfrac{\pi ^{2}}{4}+C=-\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}\cos n\pi
=\sum_{n=1}^{\infty }\dfrac{(-1)^{n-1}}{n^{2}}=\dfrac{1}{2}\zeta (2)=-\dfrac{1}{
2}C.$$
Solving for $C$
$$C=-\dfrac{\pi ^{2}}{6},$$
we thus prove
$$\zeta (2)=\dfrac{\pi ^{2}}{6}.$$
Note: this 2nd method can generate all the zeta values $\zeta (2n)$ by integrating repeatedly $(\ast\ast )$. This is the reason why I appreciate it. Unfortunately it does not work for $\zeta (2n+1)$.
Note also the $$C=-\dfrac{\pi ^{2}}{6}$$ can be obtained by integrating $(\ast\ast )$ and substitute $$x=0,x=\pi$$ respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/8337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "814",
"answer_count": 48,
"answer_id": 22
} |
Summing a given series I got this problem from one of my mates, and i rearranged them, and got it in a summable form. From here could anyone tell me as to how i can sum up this interesting series.
How does one sum the given series: $$S = 1 + \Bigl(1 + \frac{1}{3}\Bigr) \cdot \frac{1}{5} \cdot \frac{1}{3} + \Bigl(1 + \frac{1}{3} + \frac{1}{5}\Bigr)\cdot \frac{1}{5^{2}} \cdot \frac{1}{5} + \Bigl(1 + \frac{1}{3}+ \frac{1}{5} + \frac{1}{7}\Bigr) \cdot \frac{1}{5^{3}} \cdot \frac{1}{7} + \cdots + \text{ad inf}$$
| I presume this is $\sum_{n=0}^\infty a_n 5^{-n}$
where $(2n+1)a_n=\sum_{k=0}^n 1/(2k+1)$.
Let
$$f(x)=\sum_{n=0}^\infty a_n x^{2n+1}$$
so that $f(0)=0$ and
$$f'(x)=\sum_{n=0}^\infty x^{2n}\sum_{k=0}^n\frac{1}{2k+1}.$$
Therefore
$$f'(x)=\frac{1}{1-x^2}\frac{\tanh^{-1}x}{x}.$$
Your sum equals
$$\int_0^{1/5}\frac{\tanh^{-1}x}{x(1-x^2)}dx.$$
If we let $x=\tanh y$ then $dx=(1-x^2)dy$ and we get
$$\int_0^t y\coth y\ dy$$
where $t=\tanh^{-1}(1/5)$.
I'm getting a nasty feeling that $\int y\coth y\ dy$
is one of those integrals that can't be done elementarily. :-(
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/8791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to solve 3rd order nonhomogeneous recurrence relation with n in equation? This is a homework question so specifically I am looking for a direction (help). Not an absolute answer.
I understand how to solve 2nd order nonhomogeneous (and I think 3rd order is similiar) recurrence relations. Whats kinking me up with this one is a term with n.
h[n] = 4h[n- 1] - 4h[n-2] + 3n + 1
Here I am using square brackets to indicate subscripts.
Now I can reorganize this equation into:
0 = h[n] - 4h[n-1] + 4h[n-2] -3n-1
From which I can derive the characteristic equation:
0 = x^3 - 4x^2 + 4x - 3n -1
Normally at this point I would solve the roots and be on my way, but here I have that extra n. How do I deal with this?
| Since presumably the homework was long due, tackle this with generating functions. I'll use proper subscripts in what follows.
Define $H(z) = \sum_{n \ge 0} h_n z^n$. Write the recurrence as:
$$
h_{n + 2} = 4 h_{n + 1} - 4 h_n + 3 n + 7
$$
Multiply by $z^n$, sum over $n \ge 0$, and recognize:
\begin{align}
\sum_{n \ge 0} h_{n + 1} z^n &= \frac{H(z) - h_0}{z} \\
\sum_{n \ge 0} h_{n + 2} z^n &= \frac{H(z) - h_0 - h_1 z}{z^2} \\
\sum_{n \ge 0} z^n &= \frac{1}{1 - z} \\
\sum_{n \ge 0} n z^n
&= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \\
&= \frac{z}{(1 - z)^2}
\end{align}
to get:
$$
\frac{H(z) - h_0 - h_1 z}{z^2}
= 4 \frac{H(z) - h_0}{z} - 4 H(z) + \frac{3 z}{(1 - z)^2} + \frac{7}{1 - z}
$$
Solving for $H(z)$ gives the forbidding expression:
\begin{align}
H(z)
&= \frac{h_0 + (h_1 - 6 h_0) z - (2 h_1 - 9 h_0 - 7) z^2
+ (h_1 - 4 h_0 - 4) z^3}
{1 - 6 z + 13 z^2 - 12 z^3 + 4 z^4} \\
&= \frac{10}{1 - z}
+ \frac{3}{(1 - z)^2}
- \frac{h_1 - 4 h_0 + 36}{2 (1 - 2 z)}
+ \frac{h_1 - 2 h_0 + 10}{2 (1 - 2 z)^2}
\end{align}
Using the generalized binomial theorem and geometric series, this tells us the coefficients:
\begin{align}
h_n &= 10 + 3 \cdot (-1)^n \binom{-2}{n}
- \frac{h_1 - 4 h_0 + 36}{2} \cdot 2^n
+ \frac{h_1 - 2 h_0 + 10}{2} \cdot \binom{-2}{n} (-2)^n
\end{align}
As $(-1)^n \binom{-2}{n} = n + 1$, this reduces to:
$$
h_n = \frac{1}{2} \left(
(h_1 - 2 h_0 + 10) n + 2 h_0 - 26) 2^n + 6 n + 26
\right)
$$
The algebra help from maxima is gratefully acknowledged.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/9815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to prove this inequality without use of computers? With help from Maple, I got
$$\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2-(c-a)^2-(c-b)^2$$
equal to
$$\frac{(c(x^3+y^3+z^3)+(a-c)(x^2y+y^2z+z^2x)+(b-c)(x^2z+y^2x+z^2y)-3(a+b-c)xyz)^2}{(x-y)^2(y-z)^2(x-z)^2}$$ which of course is $\ge 0$.
But with no help from a computer algebra, how would one prove:$$\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2\ge (c-a)^2+(c-b)^2 ?$$
| It is not a solution, but a simplification.
Claim: it is enough to consider the case $c,z=1$.
Proof(sketch only because long):
Denote $f(a,b,c,x,y,z):=\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2 - (c-a)^2-(c-b)^2$.
Then $f(a,b,c,x,y,z)=f(a,b,c,\frac{x}{z},\frac{y}{z},1)$. So we have to investigate two cases, case 1) $z=1$, case 2) $z=0$.
Case 1) $z=1$.
Define the function $f1(a,b,c,x,y):=\left(\frac{ax+by+c}{x-y}\right)^2+\left(\frac{ay+b+cx}{y-1}\right)^2+\left(\frac{a+bx+cy}{1-x}\right)^2 - (c-a)^2-(c-b)^2$.
Then $c^2 f1(\frac{a}{c},\frac{b}{c},1,x,y)=f1(a,b,c,x,y)$ so we have two cases (i) $c=0$, (ii) $c=1$.
(i) $c=0$. Now we have two subcases, (a) $a=0$, (b) $a=1$.They are simply enough to handle by a human.
(ii) $c=1$. This is not so simple, but a human can handle this case also.
Case 2) $z=0$.We have two subcases, (i) $c=1$, (ii) $c=0$.
The last one is again handable by a human.
So we have the case $c,z=1$. It is more complicated, but more or less we can say a human is able to recognize the complete square expression.
Edit: a confusing misprint is corrected. (Originally there was Case 1) $z=0$ Define the $\ldots$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/12062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Period of a function What is the value of n $\in \mathbb{Z} $ for which the function $\displaystyle f(x) = \frac{\sin nx} { \sin \biggl( \frac{x}{n} \biggr) } \text { has } 4\pi $ as period?
Also could it be possible to solve this if we need $x\pi$ as period ?I am interested in learning the general approach for this particular type of the problem.
| You want $$\frac{\sin n(x + 4\pi)}{\sin \frac{x + 4\pi}{n}} = \frac{\sin nx}{\sin \frac{x}{n}}.$$
This is equivalent with
$$\sin \frac{x + 4\pi}{n} = \sin \frac{x}{n}.$$
Therefore $\frac{x}{n} = \frac{x + 4\pi}{n} + 2k\pi$ or $\frac{x}{n} = \pi - \frac{x + 4\pi}{n} + 2k\pi$ for some $k \in \mathbb{Z}$. In the first case $x = x + 4 \pi + 2k \pi n$ and thus $n = \frac{4}{2k}$. In the second case $x = \pi n - x - 4\pi + 2k\pi n$ and thus $n = \frac{2x + 4\pi}{2k\pi + \pi}$, which is impossible since this should hold for every $x \in \mathbb{R}$.
Thus $n = \pm 1$ or $\pm 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/12522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A gamma function inequality I would like to prove $$\frac{\Gamma(n+\frac{1}{2})}{\Gamma(n+1)} \le \frac{1}{\sqrt{n}}$$ for all natural $n \ge 1$. The inequality does seem to be true numerically, but the proof eludes me.
| For a completely elementary proof let
$$I_n= \int_0^{ \pi/2} \sin^n x \textrm{ d}x $$
then integration by parts gives
$$I_n = \frac{n-1}{n} I_{n-2} . \quad (1) $$
From which we get
$$I_{2n}= \frac{2n-1}{2n} \frac{2n-3}{2n-2} \cdots \frac{1}{2} \frac{\pi}{2} \quad (2) $$
and
$$I_{2n+1}= \frac{2n}{2n+1} \frac{2n-2}{2n-1} \cdots \frac{2}{3} . \quad (3) $$
Since $ \sin^{2n+1} x < \sin^{2n} x $ for $ x \in (0,\pi/2) $ we have
$$I_{2n+1}< I_{2n} . \quad (4) $$
Also, from $(1)$ we have $(2n+1)I_{2n+1} = 2nI_{2n-1} > 2nI_{2n}$ since $I_{2n-1} >I_{2n}.$
Therefore using this and $(4)$
$$ \frac{2n+1}{2n} > \frac{ I_{2n} }{ I_{2n+1} } > 1 . \quad (5) $$
But from $(2)$ and $(3)$
$$ \frac{ I_{2n} }{ I_{2n+1} } = \frac{2n+1}{2} \frac{ (2n!)^2 }{ 4^{2n} (n!)^4 } \pi .$$
Putting this in $(5)$ gives
$$ \frac{1}{ \sqrt{n} } > \frac{1}{4^n} { 2n \choose n } \sqrt{ \pi } > \frac{1}{ \sqrt{ n + 1/2} }.$$
i.e.
$$ \frac{1}{ \sqrt{n} } > \frac{ \Gamma(n + 1/2) }{ \Gamma(n+1) } > \frac{1}{ \sqrt{ n + 1/2} }.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/15207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 0
} |
what does $(n+\sin n)/\sqrt{n^2+1}$ converge to? I have the answer of $1$. But for the answer they split it up into $b_n = n/\sqrt{n^2+1}$ and $c_n =\sin n/\sqrt{n^2+1}$ which is fine.
Then for $b_n$ they divide evrything through by $n^2$ including everything inside the square root to give $1/\sqrt{1+1/n^2}$ and I thought you couldn't do this and checked with real numbers and they are not equivalent.
For $c_n$ they said $\sin n/\sqrt{n^2+1}\leq 1/\sqrt{n^2+1}$. Fine. But then they say this is $< 1/n$ by continuity of the square root? and this tends to $0$ so $\sin n/\sqrt{n^2+1}$ tends to $0$. I thought for this to be true $1/n$ would have to tend to $0$ slower then $\sin n/\sqrt{n^2+1}$?
Could you help please
| This is what was done to the square root: $$\sqrt{n^2+1}=\sqrt{n^2 (1+1/n^2)}=\sqrt{n^2}\sqrt{1+1/n^2}.$$
For positive $n$ we have $\sqrt{n^2}=n$, and this $n$ then cancels the $n$ in the numerator.
For the other estimate, clearly $n^2+1 > n^2$, and since the square root function is increasing this gives $\sqrt{n^2+1} > \sqrt{n^2} = n$. The numbers are positive, so taking reciprocals reverses the inequality:
$$\frac{1}{\sqrt{n^2+1}} < \frac{1}{n}.$$
Therefore
$$0 \le \left| \frac{\sin n}{\sqrt{n^2+1}} \right| < \frac{1}{n},$$
and because of squeezing the limit of $c_n$ must be zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/16886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Inverse of the sum of matrices I have two square matrices: $A$ and $B$. $A^{-1}$ is known and I want to calculate $(A+B)^{-1}$. Are there theorems that help with calculating the inverse of the sum of matrices? In general case $B^{-1}$ is not known, but if it is necessary then it can be assumed that $B^{-1}$ is also known.
| This I found accidentally.
Suppose given $A$, and $B$, where $A$ and $A+B$ are invertible. Now we want to know the expression of $(A+B)^{-1}$ without imposing the all inverse. Now we follow the intuition like this. Suppose that we can express $(A+B)^{-1} = A^{-1} + X$, next we will present simple straight forward method to compute $X$
\begin{equation}
(A+B)^{-1} = A^{-1} + X
\end{equation}
\begin{equation}
(A^{-1} + X) (A + B) = I
\end{equation}
\begin{equation}
A^{-1} A + X A + A^{-1} B + X B = I
\end{equation}
\begin{equation}
X(A + B) = - A^{-1} B
\end{equation}
\begin{equation}
X = - A^{-1} B ( A + B)^{-1}
\end{equation}
\begin{equation}
X = - A^{-1} B (A^{-1} + X)
\end{equation}
\begin{equation}
(I + A^{-1}B) X = - A^{-1} B A^{-1}
\end{equation}
\begin{equation}
X = - (I + A^{-1}B)^{-1} A^{-1} B A^{-1}
\end{equation}
This lemma is simplification of lemma presented by Ken Miller, 1981
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/17776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "205",
"answer_count": 13,
"answer_id": 4
} |
How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
Possible Duplicate:
Highest power of a prime $p$ dividing $N!$
How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
| The number of zeros at the end of $N!$ is given by $$\left \lfloor \frac{N}{5} \right \rfloor + \left \lfloor \frac{N}{5^2} \right \rfloor + \left \lfloor \frac{N}{5^3} \right \rfloor + \cdots$$ where $\left \lfloor \frac{x}{y} \right \rfloor$ is the greatest integer $\leq \frac{x}{y}$.
To make it clear, write $N!$ as a product of primes $N! = 2^{\alpha_2} 3^{\alpha_2} 5^{\alpha_5} 7^{\alpha_7} 11^{\alpha_{11}} \ldots$ where $\alpha_i \in \mathbb{N}$.
Note that $\alpha_5 < \alpha_2$, $\forall N$. (Why?)
The number of zeros at the end of $N!$ is the highest power of $10$ dividing $N!$
If $10^{\alpha}$ divides $N!$ and since $10 = 2 \times 5$, $2^{\alpha} | N!$ and $5^{\alpha} | N!$. Further since $\alpha_5 < \alpha_2$, the highest power of $10$ dividing $N!$ is the highest power of $5$ dividing $N!$ which is $\alpha_5$.
So you will find that for $N \leq 24$, the number of zeros will be less than or equal to 4. However when $N$ hits $25$ you will get 2 additional zeros courtesy $25$ since $25 \times 2^2 = 100$. Hence, there will be a jump when you go from $24$ to $25$.
EDIT:
Note that there will be
*
*A jump of $1$ zero going from $(N-1)!$ to $N!$ if $5 || N$
*A jump of $2$ zero going from $(N-1)!$ to $N!$ if $5^2 || N$
*A jump of $3$ zero going from $(N-1)!$ to $N!$ if $5^3 || N$ and in general
*A jump of $k$ zero going from $(N-1)!$ to $N!$ if $5^k || N$
where $a || b$ means $a$ divides $b$ and gcd($a,\frac{b}{a}$) = 1
EDIT
Largest power of a prime dividing $N!$
In general, the highest power of a prime $p$ dividing $N!$ is given by
$$s_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots$$
The first term appears since you want to count the number of terms less than $N$ and are multiples of $p$ and each of these contribute one $p$ to $N!$. But then when you have multiples of $p^2$ you are not multiplying just one $p$ but you are multiplying two of these primes $p$ to the product. So you now count the number of multiple of $p^2$ less than $N$ and add them. This is captured by the second term $\displaystyle \left \lfloor \frac{N}{p^2} \right \rfloor$. Repeat this to account for higher powers of $p$ less than $N$.
In case of the current example, the largest prime dividing $10$ is $5$. Hence, the largest power of $10$ dividing $N!$ is the same as the largest power of $5$ dividing $N!$.
Largest power of a prime dividing other related products
In general, if we want to find the highest power of a prime $p$ dividing numbers like $\displaystyle 1 \times 3 \times 5 \times \cdots (2N-1)$, $\displaystyle P(N,r)$, $\displaystyle \binom{N}{r}$, the key is to write them in terms of factorials.
For instance, $$\displaystyle 1 \times 3 \times 5 \times \cdots (2N-1) = \frac{(2N)!}{2^N N!}.$$ Hence, the largest power of a prime, $p>2$, dividing $\displaystyle 1 \times 3 \times 5 \times \cdots (2N-1)$ is given by $s_p((2N)!) - s_p(N!)$, where $s_p(N!)$ is defined above. If $p = 2$, then the answer is $s_p((2N)!) - s_p(N!) - N$.
Similarly, $$\displaystyle P(N,r) = \frac{N!}{(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle P(N,r)$ is given by $s_p((N)!) - s_p((N-r)!)$, where $s_p(N!)$ is defined above.
Similarly, $$\displaystyle C(N,r) = \binom{N}{r} = \frac{N!}{r!(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle C(N,r)$ is given by $s_p((N)!) - s_p(r!) - s_p((N-r)!)$, where $s_p(N!)$ is defined above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/17916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 3,
"answer_id": 1
} |
If $2^{n+1} - 1$ is composite, is $2^{n+2} - 1$ prime? In 1556, Tartaglia claimed that the sums
1 + 2 + 4
1 + 2 + 4 + 8
1 + 2 + 4 + 8 + 16
are alternative prime and composite. Show that his conjecture is false.
With a simple counter example, $1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255$, apparently it's false. However, I want to prove it in general case instead of using a specific counter example, but I got stuck :( !
I tried:
The sum $\sum_{i=0}^n 2^i$ is equal to $2^{n+1} - 1$. I assumed that $2^{n+1} - 1$ is prime, then we must show that $2^{n+1} - 1 + 2^{n+1} = 2^{n+2} - 1$ is not composite. Or we assume $2^{n+1}$ is composite and we must show that $2^{n+2} - 1$ is not prime.
But I have no clue how $2^{n+2} - 1$ relates to its previous prime/composite. Any hint?
| It is easy to see that $2^{2n}-1 \equiv 0 \mod 3$. This shows that for $n \geq 2$, $2^{2n}-1$ is never prime.
Also, since $2^3 \equiv 1 \mod 7$ it follows that $2^{3n} -1$ is always divisible by $7$.
Hence $2^{6n+2}-1, 2^{6n+3}-1, 2^{6n+4}-1$ are always composite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/19996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
If p is an odd prime, prove that $1^2 \times 3^2 \times 5^2 \cdots \times (p-2)^2 \equiv (-1)^{(p+1)/2}\pmod{p}$ I also have to prove this for $$2^2 \times 4^2 \times 6^2 \cdots \times (p-1)^2 \equiv (-1)^{(p+1)/2} \pmod{p}$$
I made some progress so far and got stuck. I said that since p is odd, $(p+1)/2$ is even. Then we can say that $(-1)^{\mathrm{even}}= 1$, so $(-1)^{(p+1)/2}\pmod{p}$ can be written as $1\pmod{p}$. I know that the product of odds is odd and the product of evens is even, but I cant prove that the left side of this equation in either case is congruent to $1 \pmod{p}$. Any help here would be greatly appreciated.
| $p=2m+1$,
we can write $1\equiv -(p-1)\pmod p$,
Now write $1^2 \times 3^2 \dots(p-2)^2$ as $1\times(-(p-1))\times 3\times(-(p-3)\dots(p-2)(-(P-(p-2)))$,
$\implies 1^2 \times 3^2 \dots(p-2)^2 = 1\times(-(p-1))\times3\times(-(p-3)\dots(p-2)(-(P-(p-2)))\\\equiv-1^m (p-1)! \equiv-1^{m+1}\pmod p$ wilson theorem,
$1^2 \times 3^2 \dots(p-2)^2\equiv-1^{(p+1)/2} \pmod p$ (as taken $p=2m+1$)
similar for another case
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/22399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Help solving $\int {\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)}dx}$ $\displaystyle\int {\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)}\,\mathrm{d}x}$
I used partial fractions, solved $A = 2, C = 3$.
$$\frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} +\frac{(Dx+E)}{(x^2+2)}$$
\begin{align*}
&8x^4+15x^3+16x^2+22x+4\\
&\quad = A(x+1)^2(x^2+2)+B(x)(x+1)(x^2+2)+C(x)(x^2+2)+(Dx+E)(x)(x+1)^2
\end{align*}
Substitute in $x=0$ to get $4=A(1)(2)$, so $A = 2$
$$6x^4+11x^3+10x^2+14x = B(x)(x+1)(x^2+2)+C(x)(x^2+2)+(Dx+E)(x)(x+1)^2$$
Substitute in $x=-1$ to get
$$6-11+10-14 = C(-1)(1+2)$$
so $-9=-3C$, thus $C=3$.
Leaving me what I have below:
Which brings me to where I am currently stuck.
$$6x^4 +8x^3 +10x^2+8x = B(x)(x+1)(x^2+2) + (Dx + E) (x) (x+1)^2$$
Is the next best move to use substitution to solve for $B$?
| Following your original method from the point where you pose your question, here are a couple of tricks you can use, which were not mentioned in previous answers. (1) Set $x=\varepsilon -1$ and neglect $\varepsilon^2$, to get $B=4$ immediately. (2) Set $x=\mathrm{i}\sqrt2$ and equate real and imaginary parts to find $D$ and$E$ (independently of $B$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/23484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 3
} |
Solving $N\equiv1\pmod2,N\equiv2\pmod3,\dots,N\equiv n-1\pmod n$ If I have:
\begin{align*}
N &\equiv 1 &\pmod{2}\\
N &\equiv 2 &\pmod{3}\\
N &\equiv 3 &\pmod{4}\\
&\vdots\\
N &\equiv n - 1 &\pmod{n}
\end{align*}
How could I solve for $N$? Is there any property relates to this problem?
Update
Base on Moron hint, we have:
\begin{align*}
N + 1 &\equiv 0 &\pmod{2}\\
N + 1 &\equiv 0 &\pmod{3}\\
N + 1 &\equiv 0 &\pmod{4}\\
\vdots\\
N + 1 &\equiv 0 &\pmod{n}
\end{align*}
Hence,
$$N + 1 \equiv 0 \pmod{\mathrm{lcm}(2\cdot 3\cdots n}$$
$$\therefore N + 1 = lcm(2.3.4...n) * k \text{ for some integers k } $$
$$\implies N = lcm(2.3.4...n) * k - 1$$
Does it look right?
Thanks,
Chan
| Hint: Consider the possible values for $N+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/24110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
integration by partial fraction I can not figure out how to integrate
$$\int\frac{3x^2+x+4}{x^3+x} \, dx.$$
I got as far as factoring the denominator which gives us $x(x^2+1)$.
So from there I got
$$\frac{3x^2+x+4}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$$
which would give us: $(3x^2 + x + 4) = A(x^2 + 1) + (Bx+C)x$
This is where Im stuck because I can not figure out the values for $B$ and $C$. Please help.
| Okay, first a bunch of nitpicks:
*
*You are not trying to integrate the integral, you are trying to integrate the function itself.
*The integral does not equal the partial fraction decomposition, only the rational function itself does.
*You are missing the $dx$ in all the integrals.
So. You are trying to integrate $\displaystyle \frac{3x^2+x+4}{x^3+x}$.
First, you make sure the numerator has smaller degree than the denominator, doing long division if necessary to get it into that form (done).
Then, you factor the denominator completely (done).
Then you set up the partial fraction decomposition problem (done):
$$\frac{3x^2+x+4}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}.$$
Then, you perform the operation on the right hand side to the expression you want (done):
$$\frac{3x^2+x+4}{x(x^2+1)} = \frac{A(x^2+1) + (Bx+C)x}{x(x^2+1)}.$$
Then, you know the numerators are equal (done):
$$3x^2 + x+4 = A(x^2+1)+ (Bx+c)x.$$
Finally, you figure out the values of $A$, $B$, and $C$. There are two strategies:
*
*Do the operations on the right and write it as a polynomial; since two polynomials are equal if and only if they have the same coefficients, the coefficient of $x^2$ on the right equals $3$, etc. This will set up a system of linear equations for $A$, $B$, and $C$, which you can solve:
$$3x^2 + x + 4 = Ax^2 + A + Bx^2 + Cx = (A+B)x^2 + Cx + A$$
so $A+B=3$, $C=1$, and $A=4$. From this, you get $A=4$, $B=-1$, and $C=1$, so
$$\frac{3x^2+x+4}{x(x+1)} = \frac{4}{x} + \frac{-x+1}{x^2+1}.$$
Now you just need to do the simpler integrals
$$\int\frac{3x^2+x+4}{x^3+x}\,dx = \int\frac{4}{x}\,dx - \int\frac{x}{x^2+1}\,dx + \int\frac{1}{x^2+1}\,dx.$$
*Plug in some values of $x$ to get information about $A$, $B$, and $C$. Specifically, pick values that make some of the terms equal to $0$ (the roots of the original polynomial), and start simplifying. For example, from
$$3x^2 + x + 4 = A(x^2+1) + (Bx+C)x,$$
plugging in $x=0$ you get $4 = A(1) + 0$, so $A=4$; now we have
$$3x^2 + x + 4 = 4x^2 + 4 + (Bx+C)x.$$
Moving all the known factors to the left, we have
$$-x^2 + x = (Bx+C)x$$
and factoring out $x$, we get
$x(-x+1) = (Bx+C)x$, from which you can cancel $x$ to get
$$-x+1 = Bx+C$$
which immediately gives $B=-1$ and $C=1$, as before. Now that we know $A$, $B$, and $C$, proceed as in 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/26106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Finding a Lower Bound of a Function Let
$$
f(x, y) = \frac{x + y}{2e}
$$
and
$$
g(x, y) = \frac{(x+y)+\sqrt{x^2 + y^2 + 6xy}}{2e^{\frac{\sqrt{x^2 + y^2 + 6xy}}{2(x + y)}}}
$$
I am looking for the largest $K$ so that,
$$ \forall x, y > 0, \frac{g(x, y)}{f(x, y)} > K$$
By numerical computation I found that $K$ must be around $3.3$. But the best thing I could prove is about 2.28 for $K$ using the following method,
$$
x^2 + y^2 + 2xy < x^2 + y^2 + 6xy < (\sqrt{3}x)^2 + (\sqrt{3}y)^2 + 2(\sqrt{3}x)(\sqrt{3}y)
$$
Which results in,
$$
(x+y) < \sqrt{x^2 + y^2 + 6xy} < \sqrt{3}(x+y)
$$
So,
$$
\frac{g(x, y)}{f(x, y)} > \frac{\frac{x+y}{e^{\sqrt{3}/2}}}{\frac{x+y}{2e}} > 2.28
$$
I'm wondered if anyone can propose a way to prove a larger K?
Note: This is the improvement of the network throughput based on the method I proposed (g(.)) over an old method (f(.)). So, this is important for me to show that how much better my proposed method works.
Thanks in advance :)
| $K=(1+\sqrt{2})\mathrm{e}^{1-\sqrt{2}/2}=3.23576963$.
In short, this is because $x^2+y^2+6xy\le2(x+y)^2$.
For a more detailed proof, write the ratio $g(x,y)/f(x,y)$ as $h(z)=\mathrm{e}^{1-z/2}(1+z)$ with $z^2=(x^2+y^2+6xy)/(x+y)^2$. Now, $z\mapsto h(z)$ is a decreasing function on $z\ge1$ (differentiate the logarithm) and, for every positive $x$ and $y$, $z^2-1=4xy/(x+y)^2\le1$ hence $1\le z^2\le 2$. This means that $h(z)$ is minimal when $z^2=2$, that is, when $x=y$. This yields the value of $K$ given above, if one replaces the strict inequality sign $>$ in your post by a $\ge$ sign.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/27468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Trig integral $\int ( \cos{x} + \sin{x}\cos{x}) \, dx $ Assume we have:
$$ \int (\cos{x} + \sin{x}\cos{x}) \, dx$$
Two ways to do it:
Use $$\sin{x}\cos{x} = \frac{ \sin{2x} }{2} $$
Then
$$ \int \left(\cos{x} + \frac{\sin{2x}}{2} \right) \, dx = \sin{x} - \frac{ \cos{2x} }{ 4 } + C. $$
The other way, just see that $ u = \sin(x), du = \cos(x)dx $,
$$ \int ( \cos{x} + \sin{x}\cos{x}) \, dx = \sin{x} + \frac{\sin^2{x}}{2} + C. $$
Now the part I don't see fully is, why aren't these results completely equal?
Taking the 2nd result,
\begin{align}
\sin{x} + \frac{\sin^2{x}}{2} &= \sin{x} + \frac{1}{2} \, \left( \frac{1 - \cos{2x}}{2} \right) \\
&= \sin{x} + \frac{1}{4} - \frac{\cos{2x}}{4}.
\end{align}
So you have to absorb $\frac{1}{4}$ into $C$ for them to be equal.
Shouldn't they be equal straight away?
| If you write
$ \int{ \cos{x} + \frac{\sin{2x}}{2} dx }=\sin{x} - \frac{ cos{2x} }{ 4 } + C_1 $
and
$ \int{ \cos{x} + \sin{x}\cos{x} dx }= \sin{x} + \frac{\sin^2{x}}{2} + C_2 $,
then you can see what's going on with the constants in your two cases($C_1$ and $C_2$). Being careful with choosing the notation is good for understanding.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/33187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
problem: Linear equation question the question is:
$$(y^2+xy^3)\mathrm dx + (5y^2-xy+y^3\sin(y))\mathrm dy = 0$$
can any body tell me how to solve this linear equation?? when I tried to solve this the expression of integrating factor becomes too much difficult, may be i calculated it wrong...
Any help will be appreciated.
Thanks!
| $(y^2+xy^3)~dx+(5y^2-xy+y^3\sin y)~dy=0$
$(xy^3+y^2)~dx=(xy-5y^2-y^3\sin y)~dy$
$\left(x+\dfrac{1}{y}\right)\dfrac{dx}{dy}=\dfrac{x}{y^2}-\dfrac{5}{y}-\sin y$
Let $u=x+\dfrac{1}{y}$ ,
Then $x=u-\dfrac{1}{y}$
$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{1}{y^2}$
$\therefore u\left(\dfrac{du}{dy}+\dfrac{1}{y^2}\right)=\dfrac{1}{y^2}\left(u-\dfrac{1}{y}\right)-\dfrac{5}{y}-\sin y$
$u\dfrac{du}{dy}+\dfrac{u}{y^2}=\dfrac{u}{y^2}-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y$
$u\dfrac{du}{dy}=-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y$
$u~du=\left(-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y\right)~dy$
$\int u~du=\int\left(-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y\right)~dy$
$\dfrac{u^2}{2}=\dfrac{1}{2y^2}-5\ln y+\cos y+c$
$\left(x+\dfrac{1}{y}\right)^2=\dfrac{1}{y^2}-10\ln y+2\cos y+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/34140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Simpler way to compute a definite integral without resorting to partial fractions? I found the method of partial fractions very laborious to solve this definite integral :
$$\int_0^\infty \frac{\sqrt[3]{x}}{1 + x^2}\,dx$$
Is there a simpler way to do this ?
| Let's generalize the problem. We will evaluate
$$
\int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx.
$$
Let $$y=\dfrac{1}{1+x^b}\quad\Rightarrow\quad x=\left(\dfrac{1-y}{y}\right)^{\large\frac1b}\quad\Rightarrow\quad dx=-\left(\dfrac{1-y}{y}\right)^{\large\frac1b-1}\ \dfrac{dy}{by^2}\ ,$$ then
\begin{align}
\int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx&=\int_0^1 y\left(\dfrac{1-y}{y}\right)^{\large\frac{a-1}b}\left(\dfrac{1-y}{y}\right)^{\large\frac1b-1}\ \dfrac{dy}{by^2}\\&=\frac1b\int_0^1y^{\large1-\frac{a}{b}-1}(1-y)^{\large\frac{a}{b}-1}\ dy,
\end{align}
where the last integral in RHS is Beta function.
$$
\text{B}(x,y)=\int_0^1t^{\ \large x-1}\ (1-t)^{\ \large y-1}\ dt=\frac{\Gamma(x)\cdot\Gamma(y)}{\Gamma(x+y)}.
$$
Hence
\begin{align}
\int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx&=\frac1b\int_0^1y^{\large1-\frac{a}{b}-1}(1-y)^{\large\frac{a}{b}-1}\ dy\\&=\frac1b\cdot\Gamma\left(1-\frac{a}{b}\right)\cdot\Gamma\left(\frac{a}{b}\right)\\&=\large{\color{blue}{\frac{\pi}{b\sin\left(\frac{a\pi}{b}\right)}}}.
\end{align}
The last part uses Euler's reflection formula for Gamma function provided $0<a<b$. Thus
$$
\large\int_0^\infty\dfrac{\sqrt[3]{x}}{1+x^2}\ dx=\int_0^\infty\dfrac{x^{\large\frac43-1}}{1+x^2}\ dx=\frac{\pi}{2\sin\left(\frac{2\pi}{3}\right)}=\color{blue}{\frac\pi3\sqrt{3}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/34351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 8,
"answer_id": 0
} |
Why must the constant of the divisor be negated in synthetic division? Why must the constant of the divisor be negated in synthetic division?
For example, if one was dividing a polynomial by $x+a$, $a$ must be changed to $-a$. Why is this?
| Synthetic division is based on the fact that evaluating a polynomial at $x=a$ is the same finding the remainder when dividing by $x-a$. (The reason why it's $x-a$ and not $x+a$ is that $x=a$ if and only if $x-a=0$, so there is a connection between $x-a$ and $x=a$).
So if you want to divide by $x+a = x-(-a)$, you evaluate at $-a$, which is what synthetic division is essentially doing.
To see what I mean when I say that synthetic division is computing the quotient by evaluating, consider the problem of dividing $x^4 - 2x^2 + 2x - 7$ by $x-3$. The way synthetic division works is by writing out the coefficients of the polynomial, and evaluating at $3$; then you add the result to the next coefficient:
$$\begin{align*}
&\begin{array}{r|rrrrr}
&1 & 0 & -2 & 2 & -7\\
3& & & & &\\
\hline
& 1
\end{array}\\
&\begin{array}{r|rrrrr}
&1 & 0 & -2 & 2 & -7\\
3& & 3 & & &\\
\hline
& 1& 3
\end{array}\\
&\vdots\\
&\begin{array}{r|rrrrr}
&1 & 0 & -2 & 2 & -7\\
3& & 3 & 9 &21& 69 &\\
\hline
& 1 & 3 & 7 & 23 & 62
\end{array}
\end{align*}$$
What we are really doing is computing the value of the polynomial $x^4 -2x^2 + 2x - 7$ at $x=3$ by looking at the polynomial as
$$ x\Biggl(x\Bigl(x\bigl(x+0\bigr) -2\Bigr)+2\Biggr)-7$$
so we are evaluating "inside out". First doing $3+0$; then multiplying by $3$ and subtracting $2$; then multiplying by $3$ and adding $2$; then multiplying by $3$ and subtracting $7$:
$$\begin{align*}
3+0 &\longrightarrow 3(3+0) - 2 \\
&\longrightarrow 3\Bigl(3\bigl(3+0\bigr)-2\Bigr) + 2\\
&\longrightarrow 3\Biggl(3\Bigl(3\bigl(3+0\bigr)-2\Bigr)+2\Biggr)-7
\end{align*}$$
which is the same as evaluating $p(x)$ at $x=3$, which is the same as finding the remainder when you divide $p(x)$ by $x-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/37015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving the odd Bernoulli numbers are zero suppose we define the Bernoulli numbers $b_n, n = 1, 2, 3, \ldots$ by the Faulhaber's fomula
$$\begin{eqnarray*}1 + 2^k + 3^k + \ldots n^k &=& \frac{1}{k+1}[n^{k+1} + b_1 c(k+1,2) n^k + b_2 c(k+1,3)n^{k-1} + \ldots ]\\&=& \frac{(n+b)^{k+1}-b^{k+1}}{k+1}\end{eqnarray*}$$
with the proviso we replace $b^k$ by $b_k$ in the binomial expansion and $c(k,l)$ is $k$ choose $l.$
my question is, how do you show all the odd Bernoulli numbers except $b_1$ is zero without invoking heavy analytical tools so that the reason can be explained to
a student who has only had one or two semesters of calculus in high school.
| thanks for all the responses. i will collect the results. this is too large to fit into a comment to quanta's response.
we will use $\sigma_k = 1^k + 2^k + \ldots n^k$
proving the odd bernoulli constants $b_k$ are zero for $= k = 3, 5, \ldots.$
let us fix the notation and collect some elementary results.
$\sigma_k = 1^k + 2^k + \ldots n^k, \sigma_1 = \frac{n(n+1)}{2}, \sigma_3 = \sigma_1^2.$
the idea of the proof is
to show that $n^2$ divides $\sigma_k$ for all $k \ge 3$ by induction. clearly true for $k = 3.$
we will show that $\sigma_k$ is a linear combination of $\sigma_{k-2}, \ldots, \sigma_3$ and
$\sigma_1^k$ for all $k \ge 3.$ that concludes the proof.
now i will establish the needed linear dependence of $\sigma_k$ which is
$$\sigma_1^k = \frac{1}{2^{k-1}}
\cases{[ {k \choose 1}\sigma_{2k-1} + {k \choose 3}\sigma_{2k-3} +\ldots +\sigma_k ]
& for $k$ odd, \cr
[ {k \choose 1}\sigma_{2k-1} + \ldots + {k \choose 3}\sigma_{2k-3} ]
& for $k$ even.}
$$
i will show all the intermediate steps for $k = 5.$
it is easy to see the steps for the $k = 5$ carry over to general case.
\
$\begin{array}
\sigma_1^5 = (1+2\ldots n)^5 = \frac{n^5(n+1)^5}{2^5} \\
~~~~ = \frac{1}{2^5}\{[n^5(n+1)^5 -(n-1)^5n^5] + [(n-1)^5n^5 -(n-2)^5(n-1)^5]
+\ldots +[2^53^5 - 1^52^5] + 1^5 2^5 \}\\
~~~~ = \frac{1}{2^5}\{ n^5[(n+1)^5 -(n-1)^5] + ((n-1)^5[n^5 -(n-2)^5]
+\ldots +2^5[3^5 - 1^5] + 1^5 2^5 \}\\
~~~~ = \frac{1}{2^5}\{ 2n^5[{5 \choose 1}n^4 + {5 \choose 3}n^2 + 1] +
2(n-1)^5[{5 \choose 1}(n-1)^4 + {5 \choose 3}(n-3)^2 + 1]+\\
~~~~~~ \ldots + 2^5[{5 \choose 1}2^4 + {5 \choose 3}2^2 + 1] +
1^5[{5 \choose 1}1^4 + {5 \choose 3}1^2 + 1]
\}\\
~~~~~~ = \frac{1}{2^4}\{ {5 \choose 1}[n^9 + (n-1)^9 + \ldots + 2^9 + 1^9] +
{5 \choose 3}[n^7 + (n-1)^7 + \ldots + 2^7 + 1^7] +
[n^5 + (n-1)^5 + \ldots + 2^5 + 1^5]
\}\\
~~~~~ = \frac{1}{2^4}[ {5 \choose 1}\sigma_9 +
{5 \choose 3}\sigma_7 +
\sigma_5 ].
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/37186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
How to calculate $\lim_{x \to 0}\left(\frac1{x} + \frac{\ln(1-x)}{x^2}\right)$? How to calculate the following limit?
$$\lim_{x \to 0}\left(\frac1{x} + \frac{\ln(1-x)}{x^2}\right)$$
| A not so elegant way is to represent $\log(1-x)$ as a power series for $|x| < 1$ i.e. $$\log(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \ldots$$
Plug this in to get $$\frac1{x} + \frac{\log(1-x)}{x^2} = \frac1{x} - \frac{x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots}{x^2} = -\frac1{2} - \frac{x}{3} - \frac{x^2}{4} - \cdots$$
Hence, the desired limit is $-\frac1{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/37451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Divisibility of 9 and $(n-1)^3 + n^3 + (n+1)^3$ Question: Show that for all natural numbers $n$ which greater than or equal to 1, then 9 divides $(n-1)^3+n^3+(n+1)^3$.
Hence, $(n-1)^3+n^3+(n+1)^3 = 3n^3+6n$, then $9c = 3n^3+6n$, then $c=(n^3+2n)/3$.
Therefore $c$ should be integers, but I don't know how to do it at next step ?
| HINT $\rm\quad\displaystyle (n+1)^3 + n^3 + (n-1)^3\ =\ 3\:(n+1)\:n\:(n-1) + 9\:n\ =\ 18\:{n+1 \choose 3} - 9\:n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/37805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 6
} |
Problem calculating an integral over a surface I've been trying to solve this for awhile and can't find a way.
Given $ S={(x,y,z) \in R^3 : z = x^2 - y^2 , x^2 + y^2 \leq 1 } $ and $\phi :R^3 \to R $ defined as $\phi (x,y,z)= (4z +8y^2 + 1)^{3/2}$, find $\iint_{S} \phi dS $.
I think I have to use $ \int_{0}^{2\pi} \int_{0}^{1} \phi (T(r,\theta)) \left \| T_{\theta} \times T_{r} \right \| dr d\theta $
So I first parametrized S with $x= r \cos \theta $, $y= r \sin \theta $ and $ z = r [2\cos^2 \theta - 1] $. So I had $ T(r,\theta)= (r\cos\theta , r\sin\theta, r[\cos^2\theta - 1]) $.
If I'm not mistaken I have to use $ \int_{0}^{2\pi} \int_{0}^{1} \phi (T(r,\theta) \left \| T_{\theta} \times T_{r} \right \| dr d\theta $.
Now I need to find $ || T_{r} \times T_{\theta} || = || (r\cos\theta[2\cos^2 \theta-1+2\sin\theta], r\sin\theta, -r) ||$
Now, as you can imagine that ends up being one messed up thing once you square the first term, and I haven't found anyway to simplify anything, which leaves me with a pretty tough integral to solve.
This is the integral I got, but Wolfram Alpha couldn't solve it and neither could I.
$ \int_{0}^{2\pi} \int_{0}^{1} (8 r^2 \sin^2(\theta)+4 r(2 \cos^2(\theta)-1)+1)^{3/2} \sqrt((r \sin(\theta))^2+(2 r \cos^3(\theta)+2 r \sin^2(\theta) \cos(\theta)-r \cos(\theta))^2+(-r \cos^2(\theta)-r \sin^2(\theta)|^2) dr d\theta $
I'm guessing I either used a wrong parametrization, or there's another way to solve this not using $ \iint_{S} \phi(T(r,\theta)) \left || T_{r} \times T_{\theta} \right || dS $.
Any help would be greatly appreciated.
EDIT: As Didier pointed out, $z = r^2 [2 \cos^2 \theta +1] $. But that doesn't make matters a whole lot simpler since now $ || T_{r} \times T_{\theta} || = || (-4r^2 \cos \theta \sin^2 \theta - 2 r^2 \cos\theta [2\cos^2 \theta +1], -4 r^{2} \cos^{2} \theta \sin \theta + 2r^2 \sin \theta [2\cos^2\theta +1], r) || $.
EDIT2: Another retarded mistake, fixed by Didier.
$ || T_{r} \times T_{\theta} || = (-4r^2 \cos\theta\sin^2\theta-2r^2 \cos\theta[2\cos^2 \theta -1], 4r^2 \cos^2 \theta \sin\theta-2r^2\sin\theta[2\cos^2\theta-1], r) $, still not totally right though.
I must have another error somewhere but just can't seem to find it.
| I believe you're still looking for something to find your mistake with. Now your things aren't looking that bad from here so I guess just giving you the solution will let you see where you're wrong in your numbers crunching.
Let $S = \{ T(r,\theta) = (r \cos \theta, r \sin \theta, r^2 (\cos^2 \theta - \sin^2 \theta)) \, | \, \theta \in [0,2\pi], r \in [0,1] \}$ be the chosen parametrization of $S$. Therefore
$\phi(T(r,\theta)) = (4(r^2(\cos^2 \theta - \sin^2 \theta)) + 8 (r \sin \theta)^2 + 1)^{3/2} $
$ = (4r^2 \cos^2 \theta - 4 r^2 \sin^2 \theta + 8 r^2 \sin^2 \theta + 1)^{3/2} = (4r^2 + 1)^{3/2}$
and
$
T_r \times T_{\theta} =
\begin{vmatrix}
\vec i & \vec j & \vec k \\
\cos \theta & \sin \theta & 2r(\cos^2 \theta - \sin^2 \theta) \\
- r \sin \theta & r \cos \theta & r^2(-4 \sin \theta \cos \theta)
\end{vmatrix}
$
$
= (-4r^2\sin^2 \theta \cos \theta - 2r^2 \cos^3 \theta + 2r^2 \sin^2 \theta \cos \theta, 4r^2 \sin \theta \cos^2 \theta - 2r^2 \sin \theta \cos^2 \theta + 2r^2 \sin^3 \theta, r)
$
$
=(-2r^2 \cos \theta, 2 r^2 \sin \theta,r) \quad
$
so that $\quad ||T_r \times T_{\theta}|| = \sqrt{4r^4 + r^2} = r \sqrt{4r^2 + 1}$.
Now it's easy because
$$
\int_0^{2\pi} \int_0^1 (4r^2+1)^{3/2} (r\sqrt{4r^2 +1}) \, dr d\theta = 2\pi \int_0^1 r(4r^2 + 1)^2 \, dr
$$
which gives a number I am not in the mood for computing right now, but I think you'll be fine. =)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/38796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the greatest integer that divides $p^4-1$ for every prime number $p$ greater than $5$? What is the greatest integer that divides $p^4-1$ for every prime number $p > 5$? This was on a practice math GRE so it's probably really easy.
| I suppose this is more of a brute force method. By FlT, you know $p^4\equiv 1\pmod{5}$, so $5$ is a factor. Also, factoring, you get
$$p^4-1=(p-1)(p+1)(p^2+1),$$
and since $p$ is odd, each factor is divisible by $2$. Moreover, one of the factors $p-1$ or $p+1$ is divisible by $4$, so $4\cdot2\cdot 2=16$ is another factor. Finally, of the three consecutive integers, $p-1$, $p$, and $p+1$, $3$ is a factor of one. Since $p$ is prime and larger than $5$, $3\nmid p$, so $3|p^4-1$. Hence $16\cdot 3\cdot 5=240$ is a factor of $p^4-1$ for any prime $p\gt 5$.
It still remains to see that $240$ is the greatest such divisor. This can be confirmed by looking at $7^4-1=2400$ and $11^4-1=14640$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/40429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
} |
rank of a matrix First of all I am sorry because I have asked similar kind of question a few days ago.But I still have problem with row reductions when there are letters in a matrix.The question is asking the value of 'a' when the rank of matrix is 1 , 2 , 3 and 4. I am not good at row reductions.In each row operations the matrix became more confusing.If someone help me with the row reductions , I will be very happy.
$$
\left(
\begin{array}{ccc}
1&1&2&0\\
2&a+1&3&a-1\\
-3&a-2&a-5&a+1\\
a+2&2&a+4&-2a
\end{array}
\right)
$$
| You just need to work the algebra, just like you do with regular algebra.
Subtract twice the first row from the second row to get
$$\left(\begin{array}{rrrr}
1 & 1 & 2 & 0\\
0 & a-1 & -1 & a-1\\
-3 & a-2 & a-5 & a+1\\
a+2 & 2 & a+4 & -2a
\end{array}\right).$$
This is just algebra (e.g., the (2,2) entry is $a-1$ because $(a+1)-2 = a-1$).
Add three times the first row to the second row. The (3,2) entry will be $(a-2) + 3(1) = a+1$; the (3,3) entry will be $(a-5)+3(2) = a+1$. Etc. You get:
$$\left(\begin{array}{rrrr}
1 & 1 & 2 & 0\\
0 & a-1 & -1 & a-1\\
0 & a+1 & a+1 & a+1\\
a+2 & 2 & a+4 & -2a
\end{array}\right).$$
Subtract $a+2$ times the first row from the fourth row; that means subtracting $1(a+2)$ from the first entry; $1(a+2)$ from the second entry; $2(a+2)$ from the third entry; and $0$ from the fourth entry. You get:
$$\left(\begin{array}{rrrr}
1 & 1 & 2 & 0\\
0 & a-1 & -1 & a-1\\
0 & a+1 & a+1 & a+1\\
0 & -a & -a & -2a
\end{array}\right).$$
Add the fourth row to the third row to simplify things:
$$\left(\begin{array}{rrrr}
1 & 1 & 2 & 0\\
0 & a-1 & -1 & a-1\\
0 & 1 & 1 & -a+1\\
0 & -a & -a & -2a
\end{array}\right).$$
Exchange second and third row:
$$\left(\begin{array}{rrrr}
1 & 1 & 2 & 0\\
0 & 1 & 1 & -a+1\\
0 & a-1 & -1 & a-1\\
0 & -a & -a & -2a
\end{array}\right).$$
Add $1-a$ times the second row to the third row:
$$\left(\begin{array}{rrrr}
1 & 1 & 2 & 0\\
0 & 1 & 1 & -a+1\\
0 & 0 & -a & -1+2a-a^2\\
0 & -a & -a & -2a
\end{array}\right).$$
We are just doing algebra, only in several columns at the same time. Keep going until you get a matrix for which the rank will be easy to figure out, depending on the values of $a$. I've gotten you almost all of the way there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/46786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$ \cos(\hat{A})BC+ A\cos(\hat{B})C+ AB\cos(\hat{C})=\frac {A^2 + B^2 + C^2}{2} $ What more can be said about the identity derived from law of cosines (motivation below)$$ \cos(\widehat{A})BC+ A\cos(\widehat{B})C+ AB\cos(\widehat{C})+=\frac {A^2 + B^2 + C^2}{2} \tag{IV}$$
RHS seems as if operator $\cos(\widehat{\phantom{X}})$ is being applied consecutively to terms of ABC, I tried to represent it in an analogous way to the Laplacian operator convention, but maybe there are more common ways of representing RHS using some operator and sigma notation ( please let me know if there is).
My question is : Are there any identities/structures relating or looking similar to IV, I apologize if this looks like a general fishing expedition question but I can not think of anything more that I can add to this post at this stage. Thank you
Motivation for IV,
Let $A,B,C$ be a triangle.
Let $\widehat{C} = \widehat{AB}$ stand for the Angle opposite to side C between the sides A and B, then the law of cosines for all three sides can be written as $$ A^2 + B^2 - 2 AB\cos(\widehat{C}) = C^2 \tag{I} $$
$$ A^2 + C^2 - 2 AC\cos(\widehat{B}) = B^2 \tag{II} $$
$$ B^2 + C^2 - 2 BC\cos(\widehat{A}) = A^2 \tag{III} $$
Adding $I ,II,III$ and juggling the terms we get :
$$ AB\cos(\widehat{C})+AC\cos(\widehat{B})+BC\cos(\widehat{A}) =\frac {A^2 + B^2 + C^2}{2} \tag{IV} $$
| REMARK: It seems now to me that OP is looking for generalizations of $\text{IV}$ type relations valid for quadrilaterals, pentagons, etc., and not other triangle trigonometric relations, as I exemplified below.
Notation: Consider a triangle with angles $A$, $B$, $C$ and opposite sides $a$, $b$, $c$.
It is known that there exists only three distinct relations between
the angles and the sides. For instance the system
$$\begin{eqnarray}
\frac{a}{\sin A} &=&\frac{b}{\sin B} \\
\frac{a}{\sin A} &=&\frac{c}{\sin C}\tag{1} \\
A+B+C &=&\pi;
\end{eqnarray}$$
or this equivalent (yours (I),(II),(III))
$$\begin{eqnarray}
a^{2} &=&b^{2}+c^{2}-2bc\cos A \\
b^{2} &=&c^{2}+a^{2}-2ac\cos B \tag{2}\\
c^{2} &=&a^{2}+b^{2}-2ab\cos C
\end{eqnarray}$$
are two of them. Another is
$$\begin{eqnarray}
\frac{\tan \frac{A+B}{2}}{\tan \frac{A-B}{2}} &=&\frac{a+b}{a-b}
\\
\frac{\tan \frac{B+C}{2}}{\tan \frac{B-C}{2}} &=&\frac{b+c}{b-c} \tag{3}\\
\frac{\tan \frac{C+A}{2}}{\tan \frac{C-A}{2}} &=&\frac{c+a}{c-a},
\end{eqnarray}$$
from which one can derive
$$\begin{equation}
\frac{\tan \frac{A+B}{2}}{\tan \frac{A-B}{2}}+\frac{\tan \frac{B+C}{2}}{\tan
\frac{B-C}{2}}+\frac{\tan \frac{C+A}{2}}{\tan \frac{C-A}{2}}=\frac{a+b}{a-b}+%
\frac{b+c}{b-c}+\frac{c+a}{c-a}.\tag{4}
\end{equation}$$
Also from
$$\begin{eqnarray}
\tan \frac{A-B}{2} &=&\frac{a-b}{a+b}\cot \frac{C}{2} \\
\tan \frac{B-C}{2} &=&\frac{b-c}{b+c}\cot \frac{A}{2}\tag{5} \\
\tan \frac{C-A}{2} &=&\frac{c-a}{c+a}\cot \frac{B}{2},
\end{eqnarray}$$
follows
$$\begin{eqnarray}
\tan \frac{A-B}{2} \tan \frac{C}{2}+\tan \frac{B-C}{2} \tan \frac{A}{2}+\tan \frac{C-A}{2}\tan \frac{B}{2} &=&\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}. \; (6)\end{eqnarray}$$
As for a reference I looked at my old trigonometry text book Compêndio de
Trigonometria (in Portuguese) by J. Jorge Calado. The system $\left(
3\right) $ is the law of tangents (Wikipedia) and $\left(
5\right) $ can be deduct from $\left( 3\right) $ by applying the relation $A+B+C=\pi $ to get
\begin{eqnarray}
\tan \frac{A+B}{2} &=&\tan \left( \frac{\pi }{2}-\frac{C}{2}\right) =\cot
\frac{C}{2}=\left( \tan \frac{C}{2}\right) ^{-1}\tag{7}
\end{eqnarray}
and similar to $\tan \frac{B+C}{2}$ and $\tan \frac{C+A}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/49145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Combined AM GM QM inequality I came across this interesting inequality, and was looking for interesting proofs. $x,y,z \geq 0$
$$ 2\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}+3\sqrt [3]{xyz}\leq 5\left(\frac{x+y+z}{3}\right) $$
Addendum.
In general, when is
$$ a\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}+b\sqrt [3]{xyz}\leq (a+b)\left(\frac{x+y+z}{3}\right) $$
true?
| Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that $f(w^3)\geq0$, where $f(w^3)=5u-3w-2\sqrt{3u^2-2v^2}$.
We see that $f$ is a decreasing function,
which says that it's enough to prove our inequality for a maximal value of $w^3$,
which happens for equality case of two variables.
Indeed, $x$, $y$ and $z$ are non-negative roots of the equation
$$(X-x)(X-y)(X-z)=0$$ or
$$X^3-3uX^2+3v^2X-w^3=0$$ or
$$w^3=X^3-3uX^2+3v^2X$$ and we see that $w^3$ gets a maximal value,
when a line $Y=w^3$ is a tangent line to the graph of $Y=X^3-3uX^2+3v^2X$,
which happens for equality case of two variables.
Since our inequality is homogeneous, we can assume $x=y=1$ and $z=t^3$.
Thus, it's enough to prove that:
$$\frac{5}{3}(t^3+2)-3t\geq2\sqrt{\frac{t^6+2}{3}}$$ or
$$(5t^3-9t+10)^2\geq12(t^6+2)$$ or
$$(t-1)^2(13t^4+26t^3-51t^2-28t+76)\geq0,$$
which is obvious.
Done!
By the same way we can get a best values of $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/53853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
partial fraction question
$ \frac{125x^{2}+x+3}{x^{2}(x-5)} =
> \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{(x-5)} | * x^{2}(x-5)$
$125x^{2}+x+3 = Ax(x-5) + B(x-5) + C (x^{2})$
$125x^{2}+x+3 = A x^{2} - 5Ax + Bx -5B +Cx^{2}$
$125x^{2}+x+3 = x^{2}(A+C) -x(A+B)-5B$
$3 = -5B \Rightarrow B = \frac{-3}{5}$
$-1 = A+B \Rightarrow A = -1 - B \Rightarrow A = \frac{-5}{5} - \frac{-3}{5} \Rightarrow A=\frac{-8}{5}$
$125 = A+C$
Where I did wrong in calculating of variable $A$, because correct answer is $A = \frac{-8}{25}$, but I get $A = \frac{-8}{5}$.
| The expression is of the form
$$\frac{Ax + b}{ax^2 + bx + c} + \frac{C}{x + d}$$ i.e Product of quadratic and Linear Factors.
Keeping the above form in mind,this is what i did.
$$\frac{125x^2 + x + 3}{x^2(x-5)} = \frac{Ax + B}{x^2} + \frac{C}{x-5}$$
Look for a common denominator in RHS
$$\frac{(Ax + B)(x - 5)}{x^2(x-5)} + \frac{c(x^2)}{(x-5)(x^2)}$$
Adding the above fraction
$$\frac{(Ax + B)(x - 5)+c(x^2)}{x^2(x - 5)}$$
Now equating the numerator from LHS with the numerter obtained above {RHS} as the denominator of both LHS and RHS contain the same expression,you obtain,
$$125x^2 + x + 3 = Ax^2 - 5Ax + Bx - 5B + cx^2$$
Let A = 5,find out the value of C,I'am pretty sure even B would get lost.Obtain the value of B and C by choosing a suitable value for x.
I'am having difficuilty in determining the value of B and C,Let me know if you could.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/62090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
given $y = a + bx + cx^2$ fits three given points, find and solve the matrix equation for the unknowns $a,b$, and $c$
Given $y = a + bx + cx^2$ fits three given points, find and solve the matrix equation for the unknowns $a$, $b$, and $c$.
the equation fits the points $(1,0), (-1, -4),$ and $(2, 11)$
I really don't know what I'm supposed to do here. I tried setting it up like a normal matrix like
$$\begin{pmatrix}
1&-1&2\\
0&4&11
\end{pmatrix}
(x\quad y)=??$$
but this doesn't make any sense because the original equation has multiple x's in it and I feel like im not taking care of that.
Can you help me get started with this? I just don't get it and I'm so frustrated.
Thanks!
| First note that $y = a + bx + cx^2$ can also be written as
$$(1 \quad x \quad x^2) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) = y$$
You are given three points, $(x_1,y_1) = (1,0)$, $(x_2,y_2) = (-1,-4)$ and $(x_3,y_3) = (2,11)$. So you can set up three of these equations:
$$\begin{align}
(1 \quad x_1 \quad x_1^2) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) &= y_1 \\
(1 \quad x_2 \quad x_2^2) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) &= y_2 \\
(1 \quad x_3 \quad x_3^2) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) &= y_3
\end{align}$$
Or, in one matrix form:
$$\left(\begin{array}{ccc} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \end{array}\right) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) = \left(\begin{array}{c} y_1 \\ y_2 \\ y_3
\end{array}\right)$$
You know the values of $x_1, x_2, x_3, y_1, y_2, y_3$, so filling them in you will be left with an equation of the form $Ax = b$, with $A$ and $b$ known and $x$ unknown.
Edit: For completeness, filling in those values we get
$$\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & 4 \end{array}\right) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) = \left(\begin{array}{c} 0 \\ -4 \\ 11
\end{array}\right)$$
You could do a Gaussian elimination, but this example has a simple solution which we can just calculate by hand. Subtracting the second row from the first we get
$$(0 \quad 2 \quad 0) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) = 4 \quad \Longrightarrow \quad \fbox{b = 2}$$
Subtracting the first from the last we get
$$(0 \quad 1 \quad 3) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) = 11 \quad \Longrightarrow \quad b + 3c = 11 \quad \Longrightarrow \quad \fbox{c = 3}$$
Then finally you could use the first row to get
$$(1 \quad 1 \quad 1) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) = 0 \quad \Longrightarrow \quad a + b + c = 0 \quad \Longrightarrow \quad \fbox{a = -5}$$
(For the enthusiast, note that the matrix $A$ is a Vandermonde matrix.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/70658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the exact value of $ \sum \frac{4n-3}{n(n^2-4)} $ I would like to find the exact value of the series
$$\begin{align*} \sum_{n=3}^{\infty} \frac{4n-3}{n(n^2-4)}
\end{align*}$$
which is certainly a telescoping series. Do you have any idea of telescopic cancelling?
| $$\begin{align*} \frac{4n-3}{n(n^2-4)}=\frac{3}{4n}-\frac{11}{8(n+2)}+\frac{5}{8(n-2)}
\end{align*}$$
$$\begin{align*} \frac{4n-3}{n(n^2-4)}=\frac{1}{8}(6(\frac{1}{n}-\frac{1}{n+2})+5(\frac{1}{n-2}-\frac{1}{n+2}))
\end{align*}$$
$$\begin{align*} \sum_{n=3}^{m}\frac{1}{n}-\frac{1}{n+2}=\frac{7}{12}-\frac{1}{m+1}-\frac{1}{m+2}\rightarrow7/12\end{align*}$$
$$\begin{align*} \sum_{n=3}^{m}\frac{1}{n-2}-\frac{1}{n+2}=\frac{25}{12}-\frac{1}{m-1}-\frac{1}{m}-\frac{1}{m+1}-\frac{1}{m+2}\rightarrow25/12 \end{align*}$$
$$\begin{align*} \sum_{n=3}^{\infty} \frac{4n-3}{n(n^2-4)}=\frac{1}{8}(6\times7/12+5\times25/12)=\frac{167}{96}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/75815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Tough Inequality I was doing some problems for Olympiad training and encountered this:
How would you prove that
$(a+b+c+d)-(a+c)(b+d)\geq 1$?
We are told that $0<a,b,c,d<1$ and the product $abcd=(1-a)(1-b)(1-c)(1-d)$.
Thanks!!
| let $x = a+c$, $y=b+d$, you have to prove
$x+y - xy \geq 1$
$(x-1)(y-1) \leq 0$
Suppose this is false, then $x-1$ and $y-1$ are both positive or both negative. But:
$x-1 > 0$ and $y-1 > 0$
means
$a+c > 1$ and $b+d > 1$
so
$a >1-c$ and $c > 1-a$ and $b> 1-d$ and $d > 1-b$
Multiplying them all you get contradiction with assumption $abcd=(1-a)(1-b)(1-c)(1-d)$. Same argument when they are both negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/77214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to prove the identity $\frac{1}{\sin(z)} = \cot(z) + \tan(\frac{z}{2})$? $$\frac{1}{\sin(z)} = \cot (z) + \tan (\tfrac{z}{2})$$
I did this:
First attempt: $$\displaystyle{\frac{1}{\sin (z)} = \frac{\cos (z)}{\sin (z)} + \frac{\sin (\frac{z}{2})}{ \cos (\frac{z}{2})} = \frac{\cos (z) }{\sin (z)} + \frac{2\sin(\frac{z}{4})\cos(\frac{z}{4})}{\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4})}} = $$
$$\frac{\cos (z)(\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4}))+2\sin z \sin(\frac{z}{4})\cos(\frac{z}{4})}{\sin (z)(\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4}))}$$
Stuck.
Second attempt:
$$\displaystyle{\frac{1}{\sin z} = \left(\frac{1}{2i}(e^{iz}-e^{-iz})\right)^{-1} = 2i\left(\frac{1}{e^{iz}-e^{-iz}}\right)}$$
Stuck.
Does anybody see a way to continue?
| Start out with
$$
\frac{1-\cos(z)}{\sin(z)}=\frac{2\sin^2(\tfrac{z}{2})}{2\sin(\tfrac{z}{2})\cos(\tfrac{z}{2})}=\tan(\tfrac{z}{2})\tag{1}
$$
and add $\cot(z)$ to both sides:
$$
\frac{1}{\sin(z)}=\cot(z)+\tan(\tfrac{z}{2})\tag{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/77642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Prove that if $a$ and $b$ are odd, coprime numbers, then $\gcd(2^a +1, 2^b +1) = 3$
Prove that if $a$ and $b$ are odd, coprime numbers, then $\gcd(2^a +1, 2^b +1) = 3$.
I was thinking among the lines of:
Since $a$ and $b$ are coprime numbers, $\gcd(a,b)=1$. Then there exist integers $x$ and $y$ such that, $ax+by=1$.
Then, $a=(1-by)/x$, $b=(1-ax)/y$
So if I write $2^a+1$ as:
$2^{(1-by)/x}+1$
Then can I say that the above expression is equivalent to 0 (mod 3)?
| Note that both $2^a+1$ and $2^b+1$ are odd, so any common divisor is odd.
Now, if $d$ divides $2^a+1$ and $2^b+1$, then it also divides $(2^a+1)-(2^b+1) = 2^a-2^b$, hence it divides $2^{\min(a,b)}(2^{|a-b|}- 1)$. But since $d$ is odd, then $d$ divides $2^{|a-b|}-1$. That is, if, say, $a\geq b$, then
$$d|2^{a}-1,2^b-1 \Longleftrightarrow d|2^a-1, 2^b-1, 2^{a-b}-1.$$
But if $d$ divides both $2^{a-b}-1$ and $2^b-1$, then it divides
$$2^b(2^{a-b}-1) + (2^b-1) = 2^a-1.$$
So:
$$d|2^a-1,2^b-1 \Longleftrightarrow d|2^b-1, 2^{a-b}-1.$$
We are assuming $a\geq b$; be careful with that assumption; more generally, we have:
$$\text{For }a,b\text{ odd, }\gcd(2^a-1,2^b-1) = \gcd(2^{\min(a,b)}-1, 2^{|a-b|}-1).$$
This looks very much like the gcd identity
$$\gcd(x,y) = \gcd(y,x-y),$$
which is very useful. Maybe you can make it work for you?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/78239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Solving simultaneous congruences Trying to figure out how to solve linear congruence by following through the sample solution to the following problem:
$x \equiv 3$ (mod $7$) $x \equiv 2$ (mod $5$) $x \equiv 1$
(mod $3$)
Let:
$n_1$ = 7
$n_2 = 5$
$n_3 = 3$
$N = n_1 \cdot n_2 \cdot n_3 = 105$
$m_1 = \frac{N}{n_1} = 15$
$m_2 = \frac{N}{n_2} = 21$
$m_3 = \frac{N}{n_3} = 35$
$gcd(m_1,n_1)$ = $gcd(15,7) = 1 = 15 \times 1 - 7 \times 2$ so $y_1 = 1$ and $x_1 = 15$
$gcd(m_2,n_2)$ = $gcd(21,5) = 1 = 21 \times 1 - 5 \times 4$ so $y_2 = 1$ and $x_2 = 21$
$gcd(m_3,n_3)$ = $gcd(35,3) = 1 = -35 \times 1 + 3 \times 12$ so $y_3 = -1$ and $x_3 = -35$
I understand up to this point, but the next line I don't get:
So $x = 15 \times 3 + 21 \times 2 - 35 \times 1 \equiv 52$ (mod $105$)
Where is the $\times 3$, $\times 2$, $\times 1$ from? Is it just because there are 3 terms, so it starts from 3 then 2 then 1? And where is the 52 coming from?
| The $3,2,1$ are from the right hand side of your congruences.
We know that $x=3$ is a solution to the first congruence, but this doesn't work as a solution to the next 2 congruences. So Chinese remaindering tells you to compute $(3\cdot 5)^{-1}=15^{-1}$ mod $7$. You find that this is $1$ (since $15(1)+(-2)7=1$). Thus $x=3\cdot 15\cdot 1$ ($=3 \cdot 15 \cdot 15^{-1} = 3 \cdot 1 =3$ (mod $7$) is still a solution of $x \equiv 3 \;\mathrm{mod}\; 7$ since $15\cdot 1 \equiv 1 \;\mathrm{mod}\;7$.
What is special about this solution? Well, $x=3\cdot 15\cdot 1$ is also congruent to 0 modulo $3$ and $5$ (that's why we were messing with $(3\cdot 5)^{-1}=15^{-1}$).
Next, $x=2$ solves $x\equiv 2\;\mathrm{mod}\;5$, but again does not solve the other two congruences. So you compute $(3\cdot 7)^{-1}=21^{-1}$ mod $5$. You find that this is $1$ (since $21(1)+5(-4)=1$). Thus $x=2\cdot 21\cdot 1$ is still a solution of $x \equiv 2\;\mathrm{mod}\; 5$ while it is also congruent to 0 modulo $3$ and $7$. So now we've found a solution to the second congruence which doesn't interfere with the first and last congruences.
Finally, $x=1$ solves the third congruence but not the first two. So you compute $(5 \cdot 7)^{-1}=35^{-1}$ mod $3$. You find that this is $-1$ (since $35(-1)+3(12)=1$). Thus $x=1\cdot 35 \cdot (-1)$ is still a solution of $x\equiv 1\;\mathrm{mod}\;3$ while it is congruent to $0$ modulo $5$ and $7$.
So now each congruence has a solution which doesn't interfere with the other congruences.
Thus adding the solutions together will solve all 3 at the same time.
Therefore, $x = 3\cdot 15\cdot 1 + 2\cdot 21\cdot 1 + 1\cdot 35 \cdot (-1) = 45+42-35=52$ is a solution to all 3 congruences. Since $105$ ($=3\cdot 5 \cdot 7$) is $0$ modulo $3$, $5$, and $7$, adding multiples of $105$ will still leave us with a solution to all $3$ congruences.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/79282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Is $ f_n=\frac{(x+1)^n-(x^n+1)}{x}$ irreducible over $\mathbf{Z}$ for arbitrary $n$?
In this document on page $3$ I found an interesting polynomial:
$$f_n=\frac{(x+1)^n-(x^n+1)}{x}.$$
Question is whether this polynomial is irreducible over $\mathbf{Q}$ for arbitrary $n \geq 1$.
In the document you may find a proof that polynomial is irreducible whenever $n=2p , p \in \mathbf{P}$ and below is my attempt to prove that polynomial isn't irreducible whenever $n$ is odd number greater than $3$. Assuming that my proof is correct my question is :
Is this polynomial irreducible over $\mathbf{Q}$ when $n=2k , k\in \mathbf{Z^+}$ \ $\mathbf{P} $ ?
Lemma 1: For odd $n$, $a^n + b^n = (a+b)(a^{n-1} - a^{n-2} b + \cdots + b^{n-1})$.
Binomial expansion: $(a+b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n-1} b + \cdots + \binom{n}{n} b^n$.
$$\begin{align*} f(x) &= \frac{(x+1)^n - x^n-1}{x} = \frac{(x+1)^n - (x^n+1)}{x}
\\ &= \frac{(x+1) \cdot (x+1)^{n-1} - (x+1)(x^{n-1}-x^{n-2}+ \cdots + 1)}{x}
\\ &= \frac{(x+1) \cdot (x+1)^{n-1} - (x+1)(x^{n-1}-x^{n-2}+ \cdots + 1)}{x}
\\ &= \frac{(x+1) \cdot \left[ \Bigl(\binom{n-1}{0}x^{n-1}+ \binom{n-1}{1} x^{n-2} + \cdots + 1 \Bigr) - (x^{n-1}-x^{n-2}+ \cdots + 1) \right]}{x}
\\ &= \frac{(x+1) \cdot \left[ \Bigl(\binom{n-1}{0}x^{n-1}+ \binom{n-1}{1} x^{n-2} + \cdots + \binom{n-1}{n-2} x \Bigr) - (x^{n-1}-x^{n-2}+ \cdots - x) \right]}{x}
\\ &= (x+1) \cdot \small{\left[ \left(\binom{n-1}{0}x^{n-2}+ \binom{n-1}{1} x^{n-3} + \cdots + \binom{n-1}{n-2} \right) - (x^{n-2}-x^{n-3}+ \cdots - 1) \right]} \end{align*}$$
So, when $n$ is an odd number greater than $1$, $f_n$ has factor $x+1$. Therefore, $f_n$ isn't irreducible whenever $n$ is an odd number greater than $3$.
| For even $n$ we bound the function from below to show it has no real roots and thus no rational roots.
First, note that for $x\ge 0$
$$f_n(x) = \sum_{k=0}^{n-2}{n\choose k+1} x^k
= n + \ldots
\ge n,$$
where the function expressed by $\ldots$ is clearly positive semidefinite.
(We take the sum above to be the definition of $f_n$ as implied by the problem statement, i.e, $f_n(0)=n$.)
Thus, the function has no real roots for $x\ge 0$.
One can show that
$$f_n(x) = \sum_{j=0}^{(n-2)/2}
\frac{n(n-j-2)!}{(j+1)!(n-2j-2)!}
(-x)^j (x+1)^{n-2j-2}.$$
For even $n$ and $x\in (-\infty,-1)\cup(-1,0)$ each term in the sum is explicitly positive definite.
In addition, $f_n(-1)=2$.
Therefore, for even $n$, $f_n$ has no real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/82295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "50",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of the integral $\int_0^\infty \left(\frac{\pi^2}{4}-x^2\right)^{-2}\cdot\frac{\pi^2}{4}\cos^2 x\,dx$ What are the steps to evaluate the following definite integral? (Answer provided)
$$\int_0^\infty {{\pi^2\over 4}\cos^2x\over\left({\pi^2\over4}-x^2 \right)^2} dx={\pi\over 4}$$?
| Contour integration is how I would approach this integral.
$$
\begin{align}
\int_0^\infty {{\pi^2\over 4}\cos^2x\over\left({\pi^2\over4}-x^2 \right)^2}\;\mathrm{d}x
&=\frac{1}{2}\int_{-\infty}^\infty {{\pi^2\over 4}\cos^2x\over\left({\pi^2\over4}-x^2 \right)^2} \;\mathrm{d}x\tag{1}\\
&=\frac{\pi^2}{2}\int_{-\infty}^\infty\frac{e^{2ix}+2+e^{-2ix}}{(\pi^2-4x^2)^2}\mathrm{d}x\tag{2}\\
&=\frac{\pi^2}{2}\int_{-\infty-i\epsilon}^{\infty-i\epsilon}\frac{e^{2ix}+2+e^{-2ix}}{(\pi^2-4x^2)^2}\mathrm{d}x\tag{3}\\
&=\frac{\pi^2}{2}\oint_{\gamma^+}\frac{e^{2ix}+1}{(\pi^2-4x^2)^2}\mathrm{d}x\\
&+\frac{\pi^2}{2}\oint_{\gamma^-}\frac{e^{-2ix}+1}{(\pi^2-4x^2)^2}\mathrm{d}x\tag{4}\\
&=\frac{\pi^2}{2}\oint_{\gamma^+}\frac{e^{2ix}+1}{(\pi^2-4x^2)^2}\mathrm{d}x\tag{5}\\
\end{align}
$$
*
*make the path a bit nicer. Since the integrand is even, let's duplicate the domain of integration and divide by two
*expand $\cos^2(x)=\dfrac{e^{2ix}+2+e^{-2ix}}{4}$
*move the path of integration. We cross no non-removable singularities and the connections near $+\infty$ and $-\infty$ vanish.
*break up the integral into two contours: $\gamma^+$ which passes from $-\infty-i\epsilon$ to $+\infty-i\epsilon$ and then circles back counter-clockwise around the upper half-plane, and $\gamma^-$ which passes from $-\infty-i\epsilon$ to $+\infty-i\epsilon$ and then circles back clockwise around the lower half-plane.
*$\gamma^-$ circles no singularities, so its integral is $0$.
Account for the residues of the singularities in $(5)$.
$$
\small\begin{array}{c}
\text{singularity}&&\text{integrand}&&\text{first order}&&\text{residue}\\
x=\frac{\pi}{2}&:&\frac{1-e^{2i(x-\pi/2)}}{(-2(x-\pi/2)(2\pi+2(x-\pi/2)))^2}&\to&\frac{-2i(x-\pi/2)}{16\pi^2(x-\pi/2)^2}&\to&\frac{1}{4\pi}\\
x=-\frac{\pi}{2}&:&\frac{1-e^{2i(x+\pi/2)}}{(2(x+\pi/2)(2\pi-2(x+\pi/2)))^2}&\to&\frac{-2i(x+\pi/2)}{16\pi^2(x+\pi/2)^2}&\to&\frac{1}{4\pi}
\end{array}
$$
Thus, the sum of the residues is $\frac{1}{2\pi}$. Including the $\frac{\pi^2}{2}$ yields an integral of $\frac{\pi}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/82671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
where are my calculations wrong? Expected value I am given a pdf $f(x) = |x-1|$ for $0 \leq x \leq 2,$ 0 otherwise, and asked to find expected value of $X^{2} + X.$ I simply integrated $(x^{2} + x)|x-1|$ from 0 to 2, checked with mathematica, and got $\frac{5}{2}.$ The answer in the back of the book, however, is $\frac{13}6.$ What am I doing wrong?
Thanks.
| Resolved in comments.
The textbook answer is wrong. The correct answer is what the OP got.
$$
\begin{align*}
\mathbf E [X^2 + X]
&= \int_{-\infty}^{+\infty} f(x) (x^2+x) ~dx
\\ &= \int_{0}^{2} |x-1| (x^2+x) ~dx
\\ &= \int_{0}^{1} (1-x) (x^2+x) ~dx + \int_{1}^{2} (x-1) (x^2+x) ~dx
\\ &= \int_{0}^{1} (x-x^3) ~dx + \int_{1}^{2} (x^3-x) ~dx
\\ &= \left. \Big(\frac{x^2}{2}-\frac{x^4}{4} \Big)\right|_{0}^1 + \left. \Big(\frac{x^4}{4}-\frac{x^2}{2} \Big) \right|_{1}^2
\\ &= \Big( \frac{1}{2}-\frac{1}{4} \Big) + \Big(\frac{2^4}{4}-\frac{2^2}{2} \Big) - \Big( \frac{1}{4}-\frac{1}{2} \Big)
\\ &= \frac52 .
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/83275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum\limits_{k=1}^n \frac{1}{k^2+3k+1}$ is bounded above by $\frac{13}{20}$ I want ask a question about a sum. The exercise is as follows:
Prove the following inequality for every $n \geq 1$:
$$\sum\limits_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{13}{20} .$$
| Since $\frac{1}{k^2+3k+1}$ is monotone decreasing for $k\geq 0$, we have
$$\begin{align*}\sum_{k=1}^n \frac{1}{k^2+3k+1} &\leq \frac{1}{5} + \frac{1}{11} + \int_2^\infty \frac{1}{k^2+3k+1} dk\\
&< \frac{1}{5} + \frac{1}{11} + \int_2^\infty \frac{1}{k^2+2k+1} dk\\
&= \frac{1}{5}+ \frac{1}{11} + \frac{-1}{k+1}\Big\vert_2^\infty\\
&= \frac{1}{5} + \frac{1}{11} + \frac{1}{3}\\
&< \frac{13}{20}.
\end{align*}
$$
EDIT: I didn't realize this was tagged homework; I now feel a little guilty giving such an explicit solution. Here are the steps I took to get at this answer, which might be useful for solving similar problems.
*
*I remembered that monotonic series can be bounded by integrals, by thinking of the series as a right Riemann sum. This suggests I try the bound
$$\sum_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{1}{5} + \int_1^{\infty} \frac{1}{k^2+3k+1} dk.$$
*That integral on the right looks mighty unpleasant; the denominator doesn't factor so the antiderivative will have logs and arctans galore. But I can bound the integral by the much nicer perfect square
$$\sum_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{1}{5} + \int_1^{\infty} \frac{1}{k^2+2k+1} dk = \frac{1}{5} + \frac{1}{2} = \frac{14}{20}.$$
*Ack! The bound is barely not tight enough. Pulling out more terms from the sum should improve it, so I try
$$\sum_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{1}{5} + \frac{1}{11} + \int_2^{\infty} \frac{1}{k^2+2k+1} dk,$$
which after working out the details turns out to work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/85316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Counting all possible combinations of a playoff tournament I have a question about a chess tournament, and I need to count all possible different tournaments. It's a playoff so that after the first round there are $2^n$ players, after the second round there are $2^{(n-1)}$ and so on....
So, how do I count it? Thanks!
| For a tournament starting with $2^n$ players, let $A_n$ be the number off all possible match results. That is, $A_n$ is the number of different player pools for the next round.
So, I am calling two tournaments different if there is a round where the player pools are different. I am ignoring the different ways for which a particular player pool in round $i$
can be obtained from the games played in the previous round.
Then $A_{n+1} = B_{n+1}\cdot A_n$, where $B_{n+1}$ is the number of different match results in the first round of a tournament with $2^{n+1}$ players.
Now,
$B_n= { 2^n\choose 2^{n-1}}$.
$A_1$=2
$A_2= B_2 \cdot A_1= { { 4\choose 2}}\cdot 2 $
$A_3= B_3\cdot A_2= {8\choose 4} \cdot { { 4\choose 2}}\cdot2 $
$A_4= B_4\cdot A_3={16\choose 8}\cdot {8\choose 4}\cdot { { 4\choose 2}}\cdot2 $
$\ \ \ \vdots$
$A_n={2^n\choose 2^{n-1}}\cdot{2^{n-1}\choose 2^{n-2}}\cdot\cdots\cdot{8\choose 4} \cdot { { 4\choose 2}}\cdot2 $.
Of course, the above can be simplified considerably:
Note that $2^n-2^{n-1}=2^{n-1}$.
So ${ 2^n\choose 2^{n-1}}={(2^n)!\over (2^{n-1})!(2^{n-1})!}$, and we have:
$$
\eqalign{
A_n&=
{(2^n)!\over (2^{n-1})! (2^{n-1})!}\cdot{(2^{n-1})!\over(2^{n-2})!(2^{n-2})!}\cdot\cdots\cdot{8!\over 4!\cdot4! } \cdot { { 4!\over 2!\cdot2!}}\cdot2 \cr
&=
{(2^n)!\over (2^{n-1})!\cdot 1}\cdot{1\over(2^{n-2})!\cdot1}\cdot\cdots\cdot{1\over 4!\cdot1 } \cdot { {1\over 2!\cdot1}}\cdot1 \cr
&={(2^n)!\over (2^{n-1})!(2^{n-2})!\cdots4!\cdot 2!}.
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/85988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Hard elementary combinatorics problem How does one compute (without brute force) the smallest integer $n$ such that
$\binom{2n}{1}(-3)^0 + \binom{2n}{3}(-3)^1 + \binom{2n}{5}(-3)^2 + \cdots + \binom{2n}{2n-1}(-3)^{(n-1)} = 0$?
| $(1+i \sqrt{3})^{2n} = \binom{2n}{0} (i \sqrt{3})^0 + \binom{2n}{1} (i \sqrt{3})^1 + \binom{2n}{2} (i \sqrt{3})^2 + \cdots + \binom{2n}{2n} (i \sqrt{3})^{2n}$
$(1-i \sqrt{3})^{2n} = \binom{2n}{0} (i \sqrt{3})^0 + \binom{2n}{1} (-i \sqrt{3})^1 + \binom{2n}{2} (i \sqrt{3})^2 + \cdots + \binom{2n}{2n} (i \sqrt{3})^{2n}$
Subtract both to get,
$$\frac{(1+i \sqrt{3})^{2n} - (1-i \sqrt{3})^{2n}}{2} = \binom{2n}{1} (i \sqrt{3})^1 + \binom{2n}{3} (i \sqrt{3})^3 + \cdots + \binom{2n}{2n-1} (i \sqrt{3})^{2n-1}$$
$$\frac{(1+i \sqrt{3})^{2n} - (1-i \sqrt{3})^{2n}}{2 i \sqrt{3}} = \binom{2n}{1} + \binom{2n}{3} (i \sqrt{3})^2 + \binom{2n}{5} (i \sqrt{3})^4 + \cdots + \binom{2n}{2n-1} (i \sqrt{3})^{2n-2}$$
$$\frac{(1+i \sqrt{3})^{2n} - (1-i \sqrt{3})^{2n}}{2 i \sqrt{3}} = \frac{2^{2n}}{2i\sqrt{3}} \left(\cos \left(\frac{2n \pi}{3} \right) + i \sin \left(\frac{2n \pi}{3} \right) - \cos \left(\frac{2n \pi}{3} \right) + i \sin \left(\frac{2n \pi}{3} \right)\right)$$
Hence, we get $$\binom{2n}{1} + \binom{2n}{3} (- 3)^1 + \binom{2n}{5} (- 3)^2 + \cdots + \binom{2n}{2n-1} (- 3)^{n-1} = \frac{4^{n}}{\sqrt{3}} \sin \left( \frac{2n \pi}{3} \right)$$
Hence, whenever $n = \frac{3 k}{2}$ where $k \in \mathbb{Z}$, $$\binom{2n}{1} + \binom{2n}{3} (- 3)^1 + \binom{2n}{5} (- 3)^2 + \cdots + \binom{2n}{2n-1} (- 3)^{n-1} = 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/87222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Any idea how to solve this equation? Any idea how to solve this equation?
$$x^2\log_{3}x^2-(2x^2+3)\log_{9}(2x+3)=3\log_{3}\frac{x}{2x+3}$$
| Note that $\log_9(x) = \log_3(x)/\log_3(9) = \frac{1}{2}\log_3(x)$.
We need $2x+3\gt 0$, and hence $x\gt 0$, so that all logarithms can be evaluated.
So we can simplify the equation using the basic properties of the logarithms
$$\begin{align*}
x^2\log_{3}x^2-(2x^2+3)\log_{9}(2x+3) &= 3\log_{3}\left(\frac{x}{2x+3}\right)\\
2x^2\log_3(x) - \frac{2x^2+3}{2}\log_3(2x+3) &= 3\log_3(x) - 3\log_3(2x+3)\\
(2x^2-3)\log_3(x) &= \left(\frac{2x^2+3}{2} - 3\right)\log_3(2x+3)\\
(2x^2-3)\log_3(x) &= \frac{2x^2-3}{2}\log_3(2x+3).
\end{align*}$$
From this, it is clear that any solutions to $2x^2-3=0$ will satisfy the equation. The solutions are $x=\sqrt{3/2}$ and $x=-\sqrt{3/2}$; but the latter cannot be used in the original equation. So one possibility is
$$x = \sqrt{\frac{3}{2}}.$$
If $2x^2-3\neq 0$, then cancelling we get:
$$\begin{align*}
(2x^2-3)\log_3(x) &= \frac{2x^2-3}{2}\log_3(2x+3)\\
\log_3(x) &= \frac{1}{2}\log_3(2x+3)\\
2\log_3(x) - \log_3(2x+3) &= 0\\
\log_3\left(\frac{x^2}{2x+3}\right) &= 0\\
\frac{x^2}{2x+3} &= 1\\
x^2 -2x - 3 &= 0\\
(x-3)(x+1) &=0.
\end{align*}$$
Again, $x=-1$ cannot be used. So the only other solution is $x=3$.
So the solutions are $x=3$ and $x=\sqrt{\frac{3}{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/87378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove a linear combination of tensor product can not be written as a single tensor product Prove that in $\mathbb{C}^{3}\otimes\mathbb{C}^{3}$, the state vector $$\mathbf{h}=\frac{1}{\sqrt{8}}=e_{1}\otimes e_{1}+e_{2}\otimes e_{2}+e_{1}\otimes e_{2}+e_{2}\otimes e_{1}+e_{1}\otimes e_{3}+e_{3}\otimes e_{1}+e_{2}\otimes e_{3}+e_{3}\otimes e_{2}$$
cannot be written in the form $h_{1}\otimes h_{2}.$
=========
My proof goes as follows:
Without loss of generality, let $e_{1}=\left(\begin{array}{c}
1\\
0\\
0\end{array}\right), e_{2}=\left(\begin{array}{c}
0\\
1\\
0\end{array}\right), e_{3}=\left(\begin{array}{c}
0\\
0\\
1\end{array}\right),$
then $e_{1}\otimes e_{1}=\left(\begin{array}{c}
1\\
0\\
0\\
0\\
0\\
0\\
0\\
0\\
0\end{array}\right), e_{2}\otimes e_{2}=\left(\begin{array}{c}
0\\
0\\
0\\
0\\
1\\
0\\
0\\
0\\
0\end{array}\right), e_{1}\otimes e_{2}=\left(\begin{array}{c}
0\\
1\\
0\\
0\\
0\\
0\\
0\\
0\\
0\end{array}\right),$
$e_{2}\otimes e_{1}=\left(\begin{array}{c}
0\\
0\\
0\\
1\\
0\\
0\\
0\\
0\\
0\end{array}\right), e_{1}\otimes e_{3}=\left(\begin{array}{c}
0\\
0\\
1\\
0\\
0\\
0\\
0\\
0\\
0\end{array}\right), e_{3}\otimes e_{1}=\left(\begin{array}{c}
0\\
0\\
0\\
0\\
0\\
0\\
1\\
0\\
0\end{array}\right), e_{2}\otimes e_{3}=\left(\begin{array}{c}
0\\
0\\
0\\
0\\
0\\
1\\
0\\
0\\
0\end{array}\right), $
$e_{3}\otimes e_{2}=\left(\begin{array}{c}
0\\
0\\
0\\
0\\
0\\
0\\
0\\
1\\
0\end{array}\right), $ and $\mathbf{h=\frac{1}{\sqrt{8}}}\left(\begin{array}{c}
1\\
1\\
1\\
1\\
1\\
1\\
1\\
1\\
0\end{array}\right)$
And eventually we will get some contradiction. But can this method be really applied "without loss of generality"? What is the way to do this without writing all these big vectors?
| write $h_1, h_2$ as linear combinations of $e_1, e_2, e_3$ and compute the tensor product in this representation. Compare the coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/87823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
General Matrix Question have had some problem solving the following question: (I've done part 1.):
We have two $n \times n$ matrices $A$ and $B$ and it says that $A=I-AB$.
*
*Prove that $A$ is regular and $AB=BA$. (done this one is pretty easy)
*Prove that if $B$ is symmetrical, so is $A$.
*Prove that $B^3=0$ if and only if $A=I-B+B^2$.
Thanks in advance...
| For part 2, factor $A$ and obtain $A(I+B) = I$. Check the transpose and notice the symmetric parts
For part 3, the necessity direction is shown by
$$
\begin{align}
AB &= I-A\\
AB^2 &= B-AB = B-I+A\\
AB^3 &= B^2 - B + AB = B^2 - B + I - A
\end{align}
$$
if $B^3=0$ then, $B^2 - B + I - A = 0$.
For sufficiency,
$$\begin{align}
A &= I-B+B^2\\
I - AB &= I - B + B^2\\
-AB &= -B + B^2\\
-AB^2 &= -B^2 + B^3\\
(I-A)B^2 &= B^3\\
AB^3 &= B^3
\end{align}
$$
from this, we see that $(I-A)B^3=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/88304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Problem based on Composite Functions. Let $f(x) = \sin^{2}{x} + \sin^{2}\left(x + \frac{\pi}{3}\right) + \cos{x}\cos\left(x + \frac{\pi}{3}\right)$
and $g(x) = \left\{\begin{array}{rcc} 2x,\quad0\le x<1 \\x + \frac{1}{4},\quad 1 \le x<2 \end{array}\right.$ , then find $g\{f(x)\}$.
I am really confused on this one.
| Note that $f(x)$ can be written as
$$f(x) = \sin^{2}{x} + \sin^{2}\left(x + \frac{\pi}{3}\right) + \cos{x}\cos\left(x + \frac{\pi}{3}\right)$$
$$=1-\cos^{2}{x} + 1-\cos^{2}\left(x + \frac{\pi}{3}\right) + \cos{x}\cos\left(x + \frac{\pi}{3}\right)$$
$$=2-\frac{3}{4}\cos^{2}{x}-\left[\frac{1}{2}\cos{x}-\cos(x+\frac{\pi}{3})\right]^2.$$
Since
$$\cos(x+\frac{\pi}{3})=\cos x\cos\frac{\pi}{3}-\sin x\sin\frac{\pi}{3}=\frac{1}{2}\cos{x}-\frac{\sqrt{3}}{2}\sin x,$$
we have
$$f(x) = 2-\frac{3}{4}\cos^{2}{x}-\Big(\frac{\sqrt{3}}{2}\sin x\Big)^2=2-\frac{3}{4}\cos^{2}{x}-\frac{3}{4}\sin^{2}{x}=2-\frac{3}{4}=\frac{5}{4}.$$
Therefore,
$$g(f(x))=g(\frac{5}{4})=\frac{5}{4}+\frac{1}{4}=\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/90404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Power of a matrix using Sylvester's Formula I have been thinking about this question and I'm really confused, I have gone through past solutions and I really understand those, but this, I don't understand. I'm to use Sylvester's formula to find $ A^{100}$ where the matrix is
$$A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \\ \end{pmatrix}$$
Sylvester's formula is given as $ P([A]) = \sum\limits_{k=1}^{n} P(\lambda _{k}) Z_{k}([A])$ and $$ Z_{k}([A])=\frac{\prod\limits_{r\neq k} (A - \lambda_{k}I)}{\prod\limits_{r\neq k} (\lambda_{r}- \lambda_{k}I)} , k= 1, \dots, 100$$ $ \lambda_{k}$ are the eigenvalues.
| The Sylvester formula, as you write, is that if $A$ is diagonalizable, and $f(t)$ is a polynomial (in fact, any analytic function), then
$$f(A) = \sum_{i=1}^n f(\lambda_i)A_i$$
where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$ (hence, $n$ is the size of the matrix), and $A_i$ is the Frobenius covariant of $A$,
$$A_i = \prod_{\stackrel{j=1}{j\neq i}}^n \frac{1}{\lambda_i-\lambda_j}(A-\lambda_jI).$$
Here, your matrix is $2\times 2$, so you will have
$$f(A) = f(\lambda_1)A_1 + f(\lambda_2)A_2,$$
where
$$A_1 = \frac{1}{\lambda_1-\lambda_2}(A-\lambda_2I),\qquad A_2 = \frac{1}{\lambda_2-\lambda_1}(A-\lambda_1I).$$
So the first step is to find the two eigenvalues; then compute the two Frobenius covariants; then apply the formula to $f(t)=t^{100}$.
You can compute the characteristic polynomial directly (it's $(2-t)^2 - 1 = t^2-4t+3$) and find the eigenvalues; or any of the standard tricks: the determinant of $A$ is $3$, the trace is $4$, so you want to numbers that add up to $4$ and whose product is $3$: $\lambda_1=1$ and $\lambda_2 = 3$ will do.
So then
$$\begin{align*}
A_1 &= \frac{1}{1-3}(A-3I) = -\frac{1}{2}\left(\begin{array}{rr}
-1&1\\
1& -1
\end{array}\right)\\
A_2 &= \frac{1}{3-1}(A-I) = \frac{1}{2}\left(\begin{array}{cc}
1&1\\
1& 1
\end{array}\right).
\end{align*}$$
So the Sylvester formula tells you that
$$A^{100} = f(A) = \left(-\frac{f(1)}{2}\left(\begin{array}{rr}-1&1\\1&-1\end{array}\right)\right) + \left(\frac{f(3)}{2}\left(\begin{array}{cc}1&1\\1&1
\end{array}\right)\right).$$
Alternatively, $A$ is diagonalizable, with eigenvectors $(1,-1)$ (corresponding to $\lambda_1=1$) and $(1,1)$ (corresponding to $\lambda_2=3$). That means that
$$
\left(\begin{array}{rr}
\frac{1}{2} & -\frac{1}{2}\\
\frac{1}{2} & \frac{1}{2}
\end{array}\right) A \left(\begin{array}{rr}
1 & 1\\
-1 & 1
\end{array}\right) = \left(\begin{array}{cc}1 & 0\\0 & 3
\end{array}\right).$$
Since $(PAP^{-1})^n = PA^nP^{-1}$, then
$$\begin{align*}
\left(\begin{array}{rr}
\frac{1}{2} & -\frac{1}{2}\\
\frac{1}{2} & \frac{1}{2}
\end{array}\right) A^{100} \left(\begin{array}{rr}
1 & 1\\
-1 & 1
\end{array}\right) &= \left(\left(\begin{array}{rr}
\frac{1}{2} & -\frac{1}{2}\\
\frac{1}{2} & \frac{1}{2}
\end{array}\right) A \left(\begin{array}{rr}
1 & 1\\
-1 & 1
\end{array}\right) = \left(\begin{array}{cc}1 & 0\\0 & 3
\end{array}\right)\right)^{100} \\
&= \left(\begin{array}{cc}
1& 0\\
0 & 3
\end{array}\right)^{100}.
\end{align*}$$
Computing the last matrix is easy, and then we just need to multiply by
$$\left(\begin{array}{rr}1 & 1\\-1 & 1
\end{array}\right)$$ on the left and $$\left(\begin{array}{rr}
\frac{1}{2} & -\frac{1}{2}\\
\frac{1}{2} & \frac{1}{2}\end{array}\right)$$
on the right to get the value of $A^{100}$. (But this does not use Sylvester's formula).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/91239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$? Is $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$?
If $\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\mid a,b,c,d\in\mathbf{Q}\}$ and $\mathbf{Q}(\sqrt{6})= \{a+b\sqrt{6} | a,b \in \mathbf{Q}\}$
Assume an element in $\mathbf{Q}(\sqrt{6})$ , then obviously it is also in $\mathbf{Q}(\sqrt{2},\sqrt{3})$.
Assume an element in $\mathbf{Q}(\sqrt{2},\sqrt{3})$ , because we can write: $\sqrt{6} = \sqrt{3} \cdot \sqrt{2} $ and $\sqrt{6} = \sqrt{2} \cdot \sqrt{3}$ the sum of $b\sqrt{3} + c\sqrt{2}$ is equal to a fraction of $\sqrt{6}$.
So $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$.
| Suppose $\sqrt{2}\in \mathbb{Q}(\sqrt{6})$. Then there are rational numbers $a,b\in\mathbb{Q}$ such that $a+b\sqrt{6}=\sqrt{2}$. If $a=0$ but $b\neq 0$, then $\sqrt{3}=1/b\in\mathbb{Q}$. That's impossible, as we know that $\sqrt{3}$ is irrational. Similarly, if $b=0$ and $a\neq 0$, then $\sqrt{2}=a$, and again we reach a contradiction. Otherwise, assume $ab\neq 0$. Thus,
$$2 = a^2+6b^2+2ab\sqrt{6}.$$
In particular,
$$\sqrt{6} = \frac{2-a^2-6b^2}{2ab},$$
so $\sqrt{6}\in \mathbb{Q}$. This is absurd, so we have reached a contradiction. Our original assumption $\sqrt{2}\in \mathbb{Q}(\sqrt{6})$ was false, and $\mathbb{Q}(\sqrt{2})\not\subseteq \mathbb{Q}(\sqrt{6})$. In particular, $\mathbb{Q}(\sqrt{2},\sqrt{3})\neq \mathbb{Q}(\sqrt{6})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/93459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
square that can be written as mean of as many pairs of squares as possible If $b^2=\displaystyle\frac{a_1^2+c_1^2}2$ and $b^2=\displaystyle\frac{a_2^2+c_2^2}2$... and $b^2=\displaystyle\frac{a_n^2+c_n^2}2$, what is the largest possible value of $n$, or can $n$ be arbitrarily large?
e.g for $n=7$, $$325^2=\frac{49^2+457^2}2=\frac{65^2+455^2}2=\frac{115^2+445^2}2=\frac{175^2+425^2}2$$
$$=\frac{221^2+403^2}2=\frac{235^2+395^2}2=\frac{287^2+359^2}2,$$
and
$$425^2=\frac{7^2+601^2}2=\frac{85^2+595^2}2=\frac{175^2+575^2}2=\frac{205^2+565^2}2$$
$$=\frac{289^2+527^2}2=\frac{329^2+503^2}2=\frac{355^2+485^2}2.$$
| You are asking for an integer ($2b^2)$ that can be written as a sum of two squares in as many ways as possible, one of which is a sum of two equal squares. By the result that I cited in this answer, there is no bound to this number of ways. Concretely, for instance $2\times25^n$ can be written as a sum of two squares in $4(2n+1)$ ways; these include $4$ ways of the form $(\pm5^n)^2+(\pm5^n)^2$ that you don't want to count and the remaining $8n$ possibilities come in symmetry classes of $8$ (by signs and order of terms), for a total of $n$ classes.
In terms of your question, the square $(5^n)^2$ can be written as the average of two squares in $n$ non-equivalent ways. To find these expressions, take the Gaussian integers $1+\mathbf i$ and $2n$ copies of $2+\mathbf i$, conjugate $i$ of the latter for $0\leq i<n$ and multiply everything together; the resulting $n$ Gaussian integers are all of norm-squared equal to $2\times25^n$, and their real and imaginary parts provide $n$ non-equivalent pairs of numbers, the averages of whose squares is $25^n=(5^n)^2$.
This is not the most economic way to get lots of expressions; it would be better to combine distinct prime numbers congruent to $1$ modulo $4$, rather than to take powers of one of them namely $5$. This explains your examples $325=5^2\times13$ and $425=5^2\times17$.
Added: The general formula for the number of solutions for writing $N^2$ as the average of a set of two squares of distinct positive numbers, in terms of the prime factorization of $N$, is as follows: only the primes congruent to $1$ modulo $4$ contribute; multiply together for every nonzero multiplicity $m$ of such a prime the numbers $2m+1$, subtract $1$ from the the (odd) product so obtained, and divide by $2$ (which accounts for the ignored order). So for $N=325=5^2\times13$ and $N=425=5^2\times17$ one gets $\frac{5\times3-1}2=7$ solutions, as indicated. Another value with many solutions is $N=5\times13\times17=1105$, namely $\frac{3\times3\times3-1}2=13$ solutions.
If one wants to disqualify solutions with a nontrivial common factor, as Jyrki Lahtonen suggests (I wouldn't know why), then each appropriate prime only contributes a factor $2$ independently of its multiplicity rather than $2m+1$ (but subtracting $1$ is omitted). This is beacuse mixing a Gaussian integer in a product with its complex conjugate introduces an integer factor, which will be common to the real and imaginary parts. In this variant one retains only $\frac{2\times2}2=2$ solutions for $N\in\{325,425\}$, and only $\frac{2\times2\times2}2=4$ solutions for $N=1105$ (namely $(73,1561)$, $(367,1519)$, $(809,1337)$, $(1057,1151)$). Even with this restriction the number will still be unbounded as $N$ acquires more and more useful prime factors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/94382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
What are the common solutions of $x^2+y=31$ and $y^2+x=41$? A friend asked me if I have a certain algorithm to solve $x^2+y = 31$ and $y^2+x=41$ simultanously. We found the solutions but we didn't find a way to solve both equations.
Any ideas?
| As Srivatsan commented we can eliminate one of the variables from the system
$$\left\{
\begin{array}{c}
x^{2}+y=31 \\
y^{2}+x=41.
\end{array}
\right. $$
It is equivalent to
$$\left\{
\begin{array}{c}
y=31-x^{2} \\
x^{4}-62x^{2}+x+920=0.
\end{array}
\right. $$
A rational solution of the quartic equation has to be a divisor of $
920=2^{3}\times 5\times 23$. If we check $x=5$, we conclude that it is a
root, which we call $x_{0}$. Now we can easily factor the quartic
$$
x^{4}-62x^{2}+x+920=\left( x-5\right) \left( x^{3}+5x^{2}-37x-184\right) $$
remaining to solve the cubic equation
$$
x^{3}+bx^{2}+cx+d=0
$$
with coefficients $b=5,c=-37,d=-184$. To get the correspondent depressed cubic equation
$$
t^{3}+pt+q=0
$$
we need to make the change of variables $x=t-b/3$ thus finding the coefficients $p=-136/3$ and $q=-3053/27$. Now we can apply the Cardano's method. Since
the discriminant $q^{2}+4p^{3}/27<0$ all roots are real. One of
the roots is
$$
\begin{eqnarray*}
t_{1} &=&\left( -\frac{q}{2}+\frac{1}{2}\sqrt{q^{2}+\frac{4p^{3}}{27}}
\right) ^{1/3} +\left( -\frac{q}{2}-\frac{1}{2}\sqrt{q^{2}+\frac{4p^{3}}{27}}\right) ^{1/3}
\\
&\approx &7.742\qquad\text{(numerical evaluation in SWP)},
\end{eqnarray*}$$
which corresponds to $x_{1}=t_{1}-5/3\approx 6.075$. The radicals are chosen
in such a way that their product is
$$\left( -\frac{q}{2}+\frac{1}{2}\sqrt{q^{2}+\frac{4p^{3}}{27}}\right)
^{1/3}\left( -\frac{q}{2}-\frac{1}{2}\sqrt{q^{2}+\frac{4p^{3}}{27}}\right)
^{1/3}=-\frac{p}{3}.$$
The remaining roots could also be found by factoring the cubic. Numerically we
have got $t_{2}\approx -4.487,t_{3}\approx -3.255$ and $x_{2}\approx
-6.154,x_{3}\approx -4.922$. The equation $y=31-x^{2}$ yields the values
$y_{0}=6, y_{1}\approx -5.910$, $y_{2}\approx -6.867$, $y_{3}\approx 6.776$ (Note: rounding error of $0.001$ in comparison with bgins's evaluation).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/94957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Showing $\left|\frac{a+b}{2}\right|^p+\left|\frac{a-b}{2}\right|^p\leq\frac{1}{2}|a|^p+\frac{1}{2}|b|^p$ For $a,b \in \mathbb R$, $p\geq2$ I try to show $$\left|\frac{a+b}{2}\right|^p+\left|\frac{a-b}{2}\right|^p\leq\frac{1}{2}|a|^p+\frac{1}{2}|b|^p.$$
Is this a popular inequality (At least I could not find it in the list of popular inequalities from wikipedia)? It seems to be related to convexity but I did not succeed to show it. A related inequality seems to be for $p \geq 1,a,b\geq0$
$$\left(\frac{a+b}{2}\right)^p\leq \frac{1}{2}a^p+\frac{1}{2}b^p,$$
which directly follows from the convexity of $x^p$ for positive numbers.
| We have for all $x_1,x_2\geq 0$: $x_1^p+x_2^p\leqslant (x_1^2+x_2^2)^{p/2}$. Indeed, it suffice to show it when $x_2=1$, otherwise apply it to $\frac{x_1}{x_2}$. $f(t):=(t^2+1)^{p/2}-t^p-1$ is non-negative, since its derivative is $p(t^2+1)^{p/2-1}t-pt^{p-1}\geqslant 0$ and $f(0)=0$. Applying it to $x_1=\frac{a+b}2$ and $x_2=\frac{a-b}2$, we get
\begin{align*}
\left|\frac{a+b}2\right|^p+\left|\frac{a-b}2\right|^p&\leqslant \left(\left|\frac{a+b}2\right|^2+\left|\frac{a-b}2\right|^2\right)^{p/2}\\\
&=\left(\frac {2a^2+2b^2}4\right)^{p/2}\\\
&=\left(\frac {a^2}2+\frac{b^2}2\right)^{p/2}\\\
&\leqslant \frac 12|a|^p+\frac 12|b|^p,
\end{align*}
since the map $t\mapsto |t|^{p/2}$ is convex ($p\geqslant 2$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/95307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 1,
"answer_id": 0
} |
Canonical form of a Matrix My understanding of canonical form is very limited, and so may require some help.
Suppose a quadratic of the form:
$$ x_1*x_2+x_1*x_3=Q.$$
How would one go about putting that into canonical form, i.e.
$$z_1^2+z_2^2+\cdots.$$
Many thanks.
| I change the notation to $a,b,c$ just for the typing convenience. I hope you don't mind.
You can do this the fun way: Complete the square as
$$
\begin{align}
2ab + 2bc &= 2Q \\
2b(a+c)&=2Q\\
2b(a+c)+(a+c)^2+b^2-(a+c)^2-b^2 &=2Q\\
(a+b+c)^2-(a+c)^2-b^2 &=2Q\\
\frac{1}{2}\left((a+b+c)^2-(a+c)^2-b^2\right) &=Q
\end{align}
$$
or you use the matrix notation
$$
\begin{align}
2ab + 2bc &= 2Q \\
\begin{pmatrix}a\\b\\c\end{pmatrix}^T\begin{pmatrix}
0 &1 &0\\1 &0 &1\\0 &1 &0
\end{pmatrix}
\begin{pmatrix}a\\b\\c\end{pmatrix}&=2Q\\
\begin{pmatrix}a\\b\\c\end{pmatrix}^T
\begin{pmatrix}
1 &1 &0\\1 &0 &1\\1 &1 &0
\end{pmatrix}
\begin{pmatrix}
1 &0 &0\\0 &-1 &0\\0 &0 &-1
\end{pmatrix}
\begin{pmatrix}
1 &1 &1\\1 &0 &1\\0 &1 &0
\end{pmatrix}
\begin{pmatrix}a\\b\\c\end{pmatrix} &=2Q\\
\begin{pmatrix}a+b+c\\a+c\\b\end{pmatrix}^T
\begin{pmatrix}
1 &0 &0\\0 &-1 &0\\0 &0 &-1
\end{pmatrix}\begin{pmatrix}a+b+c\\a+c\\b\end{pmatrix} &=2Q\\
\begin{pmatrix}a+b+c\\a+c\\b\end{pmatrix}^T
\begin{pmatrix}
0.5 &0 &0\\0 &-0.5 &0\\0 &0 &-0.5
\end{pmatrix}\begin{pmatrix}a+b+c\\a+c\\b\end{pmatrix} &=Q
\end{align}
$$
as this shows explicitly what the coordinate rotations (that J.M. mentions in the comments) are. You might want to grease those congruence transformation gears, it takes some affinity to write quickly the matrix product. But it is not that difficult anyway.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/96904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Odd numbers expressed as : $x^2-y^2$ How to prove following statement :
Conjecture:
An odd number $n$ , $(n>1)$ can be uniquely expressed as : $n= x^2-y^2$ ; $x,y \in \mathbb{Z}^{*}$
if and only if $n$ is a prime number .
If $x-y=m$ , where $m>1$ then $m \mid n$
Proof :
$n=x^2-y^2=(y+m)^2-y^2=y^2+2\cdot y\cdot m +m^2-y^2 \Rightarrow$
$\Rightarrow n=m\cdot (2\cdot y+m) \Rightarrow m \mid n$
Therefore , if $m \neq 1$ it follows that $n$ is a composite number , but how to prove that every odd
composite number ,other than $1$ , has representation : $x^2-y^2$ , where $x-y>1$ ?
| If $n$ is odd and factors nontrivially as $ab$ with $a\le b$, then $n$ is the sum of $a$ consecutive odd numbers of which $b$ is the center one. Since the odd numbers are the first differences of the sequence of perfect squares, this means that $n$ is $(x+a)^2-x^2$ for some integer $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/97893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Use induction to prove that $n^3 + (n+1)^3 + (n+2)^3 $ is divisible by $9$ Prove that for all integers $n\geq 0, n^3 + (n+1)^3 + (n+2)^3 $ is divisible by 9.
*
*If $ n=1, 1+8+27 = 36 = 9 * x $
*Suppose $ n = k, k^3 + (k+1)^3 + (k+2)^3 $ is divisible by 9.
*Find out $ n = k + 1, $ is divisible by 9.
Because $2$ is divisible by $9$. if $3$ is divisible by $9$, $3$ - $2$ will be divisible by $9$
and the result of $3$ - $2$ = $ 9(k^2 + 3k + 3) $ which is divisible by $9$. So, it is proved.
What I found out the answer so far is them. Are they right way?
| Hint $\rm\ \ \ \bbox[1px,border:3px solid #c60]{\bbox[8px,border:1px solid #c00]{\forall\, n\ge 0\!:\ 9\ |\ f_n\!\iff\! [\,9\ |\ f_0\,\ {\rm and}\ \rm\:\forall\, n\ge 0\!:\ 9\ |\ \color{#0a0}{f_{\,n+1}\!-f_n}\,]}}\ \ $ (typical telescopy)
because, $ $ if $\rm\ \ 9\ |\ \color{#0a0}{f_{n+1}\!-f_n}\ $ then $\rm\ 9\ |\ \color{#c00}{f_n} \Rightarrow\ 9\ |\ f_{n+1} = \color{#0a0}{f_{n+1}\! - f_n} + \color{#c00}{f_n}\ \,$ (inductive step $\,\rm P_n\Rightarrow P_{n+1}$)
i.e. $\rm\: mod\ 9\!:\ if\ \ \color{#0a0}{f_{n+1}\equiv f_n}\ $ then $\rm\ \color{#c00}{f_n\equiv c}\ \,\Longrightarrow\,\ \color{#0a0}{f_{n+1}\equiv \color{#0a0}{f}_{\color{#c00} n}}\color{#c00}{\equiv c}\ $ so induction yields
that $\rm\:f\:$ is a constant function, so $\rm\ f_n\equiv f_0\equiv\: 0^3+1^3+2^3\equiv 9\equiv 0$.
This holds in OP by $\rm\ f_{n+1}-f_n = (n\!+\!3)^3\!-n^3 = 9\ (n^2\! +\! 3n\! +\! 3)\equiv 0\pmod{\!9}$.
This is a special case of telescopic induction. For further examples see many prior posts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/98039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limit $\frac{0}{0}$ which tends to $\frac{\pi}{2}$ I'm trying to evaluate the following limit:
$$\lim_{x\rightarrow\pi/2}\frac{\cos(x)}{(1-\sin(x))^{2/3}}$$
The limit has the form $\frac{0}{0}$, I've tried using L'Hopital's rule but I can't resolve it.
Any idea?
| $\displaystyle \lim_{x\rightarrow\pi/2}\frac{\cos(x)}{(1-\sin(x))^{2/3}}=\displaystyle \lim_{x\rightarrow\pi/2} \frac{\cos^2{\frac{x}{2}}-\sin^2{\frac{x}{2}}}{(\sin^2{\frac{x}{2}}-2\sin{\frac{x}{2}}\cdot \cos {\frac{x}{2}}+\cos^2{\frac{x}{2}})^{2/3}}=$
$=\displaystyle \lim_{x\rightarrow\pi/2} \frac{(\cos {\frac{x}{2}}-\sin {\frac{x}{2}})(\cos {\frac{x}{2}}+\sin {\frac{x}{2}})}{(\cos {\frac{x}{2}}-\sin {\frac{x}{2}})^{4/3}}=\displaystyle \lim_{x\rightarrow\pi/2} \frac {\cos {\frac{x}{2}}+\sin {\frac{x}{2}}}{(\cos {\frac{x}{2}}-\sin {\frac{x}{2}})^{1/3}}=\frac{\sqrt{2}}{0}=\infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/99056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
How do I evaluate this limit: $\lim_{n\to+\infty}\sum_{k=1}^{n} \frac{1}{k(k+1)\cdots(k+m)}$? How do I evaluate this limit?
$$\lim_{n\to+\infty}\sum_{k=1}^{n} \frac{1}{k(k+1)\cdots(k+m)} \qquad (m=1,2,3,\cdots)$$
Thanks in advance.
| Here is an answer without beta functions. Note that the residue of
$$f(z) = \frac{1}{z(z+1)\cdots(z+m)}$$
at $z=-q$ is
$$\mathrm{Res}(f(z); z=-q)
=\frac{1}{(-q)(-q+1)\cdots(-q+(q-1))(-q+(q+1))\cdots(-q+m)}$$
which is
$$\frac{(-1)^q}{q!\times (m-q)!} = \frac{(-1)^q}{m!} {m\choose q}.$$
Hence by the residue method for partial fractions we get that
$$f(z) = \frac{1}{m!} \sum_{q=0}^m {m\choose q} (-1)^q \frac{1}{z+q}.$$
Apply this to the sum to obtain
$$\sum_{k=1}^\infty \frac{1}{k(k+1)\cdots(k+m)} =
\frac{1}{m!} \sum_{k=1}^\infty \sum_{q=0}^m {m\choose q} (-1)^q \frac{1}{k+q}.$$
Now ask what the coefficient is of the term $1/(k+q) = 1/p$ to get
$$\frac{1}{m!} \sum_{p=1}^\infty \frac{1}{p}\sum_{q=0}^{p-1} {m\choose q} (-1)^q.$$
Observe that when $p-1\ge m$ the inner sum becomes zero, so that we are left with
$$\frac{1}{m!} \sum_{p=1}^m \frac{1}{p}\sum_{q=0}^{p-1} {m\choose q} (-1)^q.$$
It therefore remains to evaluate the sum term.
We seek to show that
$$\frac{1}{m} =\sum_{p=1}^m \frac{1}{p}\sum_{q=0}^{p-1} {m\choose q} (-1)^q.$$
This is the moment to apply what H. Wilf terms the 'snake oil' method in his famous text on generating functions. (Consult this Mathworld Entry.) Introduce the univariate generating function
$$g(z) =
\sum_{m\ge 1} z^m \sum_{p=1}^m \frac{1}{p}\sum_{q=0}^{p-1} {m\choose q} (-1)^q
= \sum_{p=1}^\infty \frac{1}{p}
\sum_{m=p}^\infty z^m \sum_{q=0}^{p-1} {m\choose q} (-1)^q
\\ = \sum_{p=1}^\infty \frac{z^p}{p}
\sum_{m=0}^\infty z^m \sum_{q=0}^{p-1} {m+p\choose q} (-1)^q.$$
Now observe that
$$ \sum_{q=0}^{p-1} {m+p\choose q} (-1)^q = (-1)^{p-1} {m+p-1\choose p-1}$$
as can be seen by induction on $m$, which we now carry out.
For $m=0$ we get equality with
$$\sum_{q=0}^{p-1} {p\choose q} (-1)^q = -(-1)^p = (-1)^{p-1} {p-1\choose p-1}.$$
and for the induction step we have
$$ \sum_{q=0}^{p-1} {m+1+p\choose q} (-1)^q
= 1 + \sum_{q=1}^{p-1} {m+1+p\choose q} (-1)^q
\\= 1 + \sum_{q=1}^{p-1} {m+p\choose q} (-1)^q
+ \sum_{q=1}^{p-1} {m+p\choose q-1} (-1)^q
\\= (-1)^{p-1} {m+p-1\choose p-1}
- \sum_{q=0}^{p-2} {m+p\choose q} (-1)^q
\\ = (-1)^{p-1} {m+p-1\choose p-1}
- \sum_{q=0}^{p-1} {m+p\choose q} (-1)^q + (-1)^{p-1} {m+1+p-1\choose p-1}
\\ = (-1)^{p-1} {m+1+p-1\choose p-1}.$$
Returning to $g(z)$ we thus obtain
$$g(z) = \sum_{p=1}^\infty \frac{z^p}{p}
\sum_{m=0}^\infty z^m (-1)^{p-1} {m+p-1\choose p-1}
= - \sum_{p=1}^\infty \frac{z^p}{p} (-1)^p \frac{1}{(1-z)^p}$$
by the Newton binomial.
Finally recall that $$\log\frac{1}{1-z} = \sum_{q\ge 1} \frac{z^q}{q}$$
so that $g(z)$ simplifies to
$$ -\log\frac{1}{1+z/(1-z)} = -\log\frac{1-z}{1-z+z} = -\log(1-z)
= \log\frac{1}{1-z}.$$
Therefore $$[z^m] g(z) = \frac{1}{m}$$ and we are done.
There is a similar but not quite identical computation at this MSE link.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/103255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Series involving zeta functions Let's have the following zeta function binomial: $\sum\limits_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)^2=\pi^2/3-3$.
Does anyone know the limit of the following zeta function binomial $\sum\limits_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)^4$?
| Employ $(a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4=a^4+6(ab)^2-4ab(a^2+b^2)$ on the following:
$$\blacksquare \; = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right)^4=
\sum_{n=1}^\infty \color{Red}{\frac{1}{n^4}}-\frac{4}{\color{Blue}{n^3(n+1)}}+\frac{6}{\color{Green}{n^2(n+1)^2}}-\frac{4}{\color{Blue}{n(n+1)^3}}+\color{Red}{\frac{1}{(n+1)^4}} $$
$$=\color{Red}{2\zeta(4)-1}+6\color{Green}{\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)^2}-4\color{Blue}{\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)\left(\frac{1}{n^2}+\frac{1}{(n+1)^2}\right)} $$
$$\frac{\pi^4}{45}+2\pi^2-19-4\color{Blue}{\blacktriangle}. $$
Note we used $\color{Green}{\bullet}=\pi^2/3-3$ and $\color{Red}{2\zeta(4)}=\pi^4/45$ above. Now we evaluate $\color{Blue}{\blacktriangle}$:
$$\color{Blue}{\blacktriangle}=\sum_{n=1}^\infty \frac{1}{n}\left(\color{Purple}{\frac{1}{n^2}}+\color{DarkOrange}{\frac{1}{(n+1)^2}}\right)-\frac{1}{n+1}\left(\color{DarkOrange}{\frac{1}{n^2}}+\color{Purple}{\frac{1}{(n+1)^2}}\right) $$
$$=\color{Purple}{\zeta(3)-[\zeta(3)-1]}+\color{DarkOrange}{\sum_{n=1}^\infty \frac{1}{n(n+1)}\left(\frac{1}{n+1}-\frac{1}{n}\right)}$$
$$=1-(\pi^2/3-3)=4-\pi^2/3. $$
Putting it all together, we obtain the final answer of
$$\blacksquare = \frac{\pi^4}{45}+\frac{10\pi^2}{3}-35. $$
The two general tools in my answer to this and your previous zeta question are:
*
*$\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} $ to either break or put together terms, and
*$a^nb^m+a^mb^n=(ab)^k(a^l+b^l)$, where $k=\min\{n,m\}$ and $l+k=\max\{n,m\}$.
The first breaks down terms and the second is just for some (perhaps strained) efficiency. It only takes these formulas (plus $\zeta(2s)$'s) to compute $\sum_{n=1}^\infty (n^2+n)^{-k}$ for any $k$, and any $\zeta(2s+1)$'s that appear will always cancel each other out (up to a rational number).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/110541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_0^1 {\ln(1+x)\over x}\,dx$. How would one evaluate $\int_0^1 {\ln(1+x)\over x}\,dx$?
I'd like to do this without approximations. Not quite sure where to start. What really bothers me is that I came across this while reviewing my old intro to calculus book... but I'm fairly certain I've exhausted all the basic methods they teach in that text.
| $$ \int^1_0 \frac{ \log (1+x) }{x} dx = \int^1_0 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n-1}}{n} dx$$
$$ =\sum_{n=1}^{\infty} (-1)^{n-1} \int^1_0 \frac{x^{n-1} }{n} dx = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^2}. $$
Denote $\displaystyle S = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.$ Then $$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^2} = S - 2\left( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \cdots \right) = S - \frac{S}{2} = \frac{\pi^2}{12}.$$
Thus $$\int^1_0 \frac{ \log (1+x) }{x} dx = \frac{\pi^2}{12}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/114664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Find the minimum value of $3x + 4y$ The minimum value of $3x + 4y$ subject to the condition
$$x^2 y^3 = 6$$
and $x$ and $y$ are positive .
| Write
$$3x = \frac{3x}{2} + \frac{3x}{2}$$
$$4y = \frac{4y}{3} + \frac{4y}{3} + \frac{4y}{3} $$
and use $\text{AM} \ge \text{GM}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/115637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding the angle between u and v. Struggling with the following Suppose that $u$ and $v$ are non-parallel unit vectors, $a=u+sv$ and $b=u-sv$, where $s$ is a real number. If the angle $\theta$ between $u$ and $v$ is the same as the angle between $a$ and $b$, show that:
$$\cos^{2}\theta= \frac {1-2s^{2}+s^{4}}{4s^{2}}$$
Hence find $s$ such that $\theta= \pi/4$ and $0<s<1$.
Many thanks in advance.
| The cosine of the angle between two vectors $x$ and $y$ is
$$\frac{x\cdot y}{\Vert x\Vert\Vert y\Vert}.$$
Since $u$ and $v$ are unit vectors, the cosine of the angle $\theta$ between them is $u\cdot v$.
The angle between $u+sv$ and $u-sv$ is also $\theta$.
So we calculate the cosine of the angle between $u+sv$ and $u-sv$. First find the dot product $(u+sv)\cdot(u-sv)$. Calculation gives, since $u$ and $v$ have norm $1$, that the dot product is $1-s^2$.
We also need the norms of $u+sv$ and $u-sv$. Take the dot product of $u+sv$ with itself. We get $1+s^2+2s\cos\theta$, since $u\cdot v=\cos\theta$. The norm is the square root of that. A similar expression can be found for the norm of $u-sv$. We conclude that
$$\cos\theta=\frac{1-s^2}{ \sqrt{(1+s^2+2s\cos\theta)(1+s^2-2s\cos\theta) }}.$$
Square both sides. Some manipulation brings us to a quadratic equation for $\cos^2\theta$ that can be rewritten as
$$\left(4s^2\cos^2\theta-(1-s^2)^2\right)\left(\cos^2\theta-1\right)=0.$$
Since $\cos^2\theta\ne 1$, the desired result follows.
Finally, we find $s$ when $\theta=\pi/4$. There, the cosine is $1/\sqrt{2}$, and we arrive at the equation $s^4-4s^2+1=0$. By the Quadratic Formula, $s^2=2\pm\sqrt{3}$. Since $s$ is between $0$ and $1$, $s=\sqrt{2-\sqrt{3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/115941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\int e^{2\theta} \cdot \sin{3\theta} \ d\theta$ I am working on an integration by parts problem that, compared to the student solutions manual, my answer is pretty close. Could someone please point out where I went wrong?
Find $\int e^{2\theta} \cdot \sin{3\theta} \ d\theta$
$u_1 = \sin{3\theta}$
$du_1 = \frac{1}{3}\cos{3\theta} \ d\theta$
$v_1 = \frac{1}{2} e^{2\theta}$
$dv_1 = e^{2\theta} \ d\theta$
$\underbrace{\sin{3\theta}}_{u_1} \cdot \underbrace{\frac{1}{2} e^{2\theta}}_{v_1} - \int\underbrace{\frac{1}{2} e^{2\theta}}_{v_1} \cdot \underbrace{\frac{1}{3} \cos{3\theta} \ d\theta}_{du_1}$
$\frac{1}{2}\sin{3\theta} \cdot e^{2\theta} - \frac{1}{6} \int \cos{3\theta} \cdot e^{2\theta} \ d\theta$
Doing integration by parts again...
$u_2 = \cos{3\theta}$
$du_2 = -\frac{1}{3} \sin{3\theta} \ d\theta$
$v_2 = \frac{1}{2} e^{2\theta}$
$dv_2 = e^{2\theta} \ d\theta$
$\underbrace{\sin{3\theta}}_{u_1} \cdot \underbrace{\frac{1}{2} e^{2\theta}}_{v_1} - \frac{1}{6}\left(\underbrace{\cos{3\theta}}_{u_2} \cdot \underbrace{\frac{1}{2} e^{2\theta}}_{v_2} - \int \underbrace{\frac{1}{2} e^{2\theta}}_{v_2} \cdot \underbrace{-\frac{1}{3} \sin{3\theta} \ d\theta}_{du_2}\right)$
$\frac{1}{2}\sin{3\theta} \cdot e^{2\theta} - \frac{1}{6}\left(\frac{1}{2}\cos{3\theta} \cdot e^{2\theta} + \frac{1}{6} \int \sin{3\theta} \cdot e^{2\theta} \ d\theta\right)$
$\frac{1}{2}\sin{3\theta} \cdot e^{2\theta} - \frac{1}{12}\cos{3\theta} \cdot e^{2\theta} - \frac{1}{36} \int \sin{3\theta} \cdot e^{2\theta} \ d\theta$
$\frac{37}{36} \int \sin{3\theta} \cdot e^{2\theta} \ d\theta = \frac{1}{2}\sin{3\theta} \cdot e^{2\theta} - \frac{1}{12}\cos{3\theta} \cdot e^{2\theta}$
$\int \sin{3\theta} \cdot e^{2\theta} \ d\theta = \begin{equation} \boxed{\frac{18}{37}\sin{3\theta} \cdot e^{2\theta} - \frac{1}{37}\cos{3\theta} \cdot e^{2\theta}} \end{equation}$
However, the boxed answer is incorrect. The answer should read:
$\frac{1}{13} e^{2\theta} \left(2\sin{3\theta} - 3\cos{3\theta}\right)$
| Ah ha! I got it!
$du = 3 \cos{3\theta} \ d\theta$ not $\frac{1}{3} \cos{3\theta} \ d\theta$
For some reason I must have been thinking of integrating the dv rather than differentiating.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/116034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Convergence of Trig Functions This is not a homework problem, but I'm trying to show that $\sin\frac{1}{n} \rightarrow 0$.
By definition I would start as follows: $|\sin\frac{1}{n} - 0 | = |\sin\frac{1}{n}|$
Since $\sin\frac{1}{n}$ could be negative or positive, I could consider both cases and get an expression $n \geq N(\epsilon)$ involving $\sin^{-1}$. I was wondering if there is a better way of showing that $\sin\frac{1}{n}$ converges to 0?
| A common way to prove this is to first show that $|\sin x| \le |x|$.
For fun we use another approach. Note that if $0 < x < \pi/2$, then $0 < \sin x < 1$ and $0< \cos x< 1$.
Recall the familiar identity
$$\sin 2x =2\sin x\cos x.$$
It is more convenient to rewrite this as
$$\sin u=2\sin \frac{u}{2}\cos \frac{u}{2}.$$
If $0<u<\pi/2$, we can rewrite this as
$$\sin \frac{u}{2}=\frac{1}{2}\frac{\sin u}{\cos\frac{u}{2}}.$$
But if $u<\pi/2$, then $\cos\frac{u}{2}>\frac{1}{\sqrt{2}}$, and therefore
$$\sin \frac{u}{2}<\frac{1}{\sqrt{2}}\sin u.\qquad(\ast)$$
Let $u=1$. Since $\sin 1<1$, we find by using $(\ast)$ that
$$\sin \frac{1}{2}<\frac{1}{\sqrt{2}}.\qquad (1)$$
Let $u=\frac{1}{2}$. By using $(\ast)$ again, and $(1)$,
we find that
$$\sin \frac{1}{4}<\left(\frac{1}{\sqrt{2}}\right)^2.\qquad (2)$$
Let $u=\frac{1}{4}$. By using $(\ast)$ and $(2)$, we find that
$$\sin\frac{1}{8}<\left(\frac{1}{\sqrt{2}}\right)^3. \qquad (3)$$
Continue. In general we have
$$0<\sin\frac{1}{2^k}<\left(\frac{1}{\sqrt{2}}\right)^k.$$
Thus
$$\lim_{k\to\infty} \sin\frac{1}{2^k}=0.$$
For $0<x<\pi/2$, the sine function is an increasing function. It follows that
$$\lim_{n\to \infty} \sin\frac{1}{n}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/116579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of the integral $\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$ How can I evaluate the integral
$$\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$$
I tried manipulating the known integral
$$\int_0^1 \frac{\ln(1 - x)}{x}dx = -\frac{\pi^2}{6}$$
but couldn't do anything with it.
| You can use the integral you want to use, and the Dilogarithm function as mentioned in the comments.
Below we give a complete proof, including a derivation of the value of the integral you wanted to use.
The Dilogarithm function is defined as
$$\text{Li}_2(z) = -\int_{0}^{z} \frac{\log (1-x)}{x} \text{dx} = \sum_{n=1}^{\infty} \frac{z^n}{n^2}, \quad |z| \le 1$$
The integral which you want to use is $\displaystyle -\text{Li}_2(1)$.
Note that $\displaystyle \text{Li}_2(1) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \zeta(2) = \frac{\pi^2}{6}$. (For multiple proofs of that, see here: Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$)
In your integral(whose value you want), make the substitution $\displaystyle x = 2t -1$ and we get
$$\int_{\frac{1}{2}}^{1} \frac{\log (2(1-t))}{t} \text{dt} = \log^2 2 + \int_{\frac{1}{2}}^{1} \frac{\log (1-t)}{t} \text{dt} = \log^2 2 + \text{Li}_2 \left(\frac{1}{2} \right) - \text{Li}_2(1) $$
Now the Dilogarithm function also satisfies the identity
$$\text{Li}_2(x) + \text{Li}_2(1-x) = \frac{\pi^2}{6}-\log x \log (1-x), 0 \lt x \lt 1$$
This identity can easily be proven by just differentiating and using the value of $\displaystyle \text{Li}_2(1)$:
$$\text{Li}_2'(x) - \text{Li}_2'(1-x) = -\frac{\log (1-x)}{x} + \frac{\log x}{1-x} = (-\log x \log (1-x))'$$
and so
$$\text{Li}_2(x) + \text{Li}_2(1-x) = C -\log x \log (1-x), 0 \lt x \lt 1$$
Taking limits as $\displaystyle x \to 1$ gives us $\displaystyle C = \frac{\pi^2}{6}$.
Thus
$$\text{Li}_2(x) + \text{Li}_2(1-x) = \frac{\pi^2}{6}-\log x \log (1-x), 0 \lt x \lt 1$$
Setting $\displaystyle x = \frac{1}{2}$ gives us the value of $\displaystyle \text{Li}_2\left(\frac{1}{2}\right) = \frac{\pi^2}{12} - \frac{\log^2 2}{2}$
Thus your integral is
$$\log^2 2 + \text{Li}_2 \left(\frac{1}{2} \right) - \text{Li}_2(1) = \frac{\log^2 2}{2} - \frac{\pi^2}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/117246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 7,
"answer_id": 4
} |
Trying to show $1-\frac12 -\frac {1} {4}+\frac {1} {3}-\frac {1} {6}-\frac {1} {8}+\frac {1} {5}-\cdots =\frac {1} {2}\log 2$ Now I think the lhs can be rewritten as $$\sum _{n=1}^\infty \left( \dfrac {1} {2n-1}-\dfrac {1} {2\cdot 3^{n-1}}-\dfrac {1} {2^{n+1}}\right) =\dfrac {1} {2}\log 2$$
I guess one way to do this may be to start from RHS and try to bring up an expression which is same as the LHS.
so using log series expansion $$\frac {1} {2}\log 2 = \log \sqrt {2} = \sum _{n=1}^\infty \dfrac {\left( -1\right) ^{n+1}} {n}\left( \sqrt {2}-1\right) ^{n}$$
Now Binomial expansion Theorem seems to be making a calling here.
$$\sum _{n=1}^{\infty }\dfrac {\left( -1\right) ^{n+1}} {n}\left( \sqrt {2}-1\right) ^{n} = \sum _{n=1}^{\infty }\dfrac {\left( -1\right) ^{n+1}} {n}\sum _{k=0}^{k=n}C_{k}^{n} \sqrt {2}^{k}\left( -1\right) ^{n-k} $$
I am unsure how to proceed from here and even if i am going down the right path.
Any help would be much appreciated.
| Rewrite $1-\dfrac {1} {2} -\dfrac {1} {4}+\dfrac {1} {3}-\dfrac {1} {6}-\dfrac {1} {8}+\dfrac {1} {5}-\cdots$ as
$$\left(1-\dfrac {1} {2}\right) -\dfrac {1} {4}+\left(\dfrac {1} {3}-\dfrac {1} {6}\right)-\dfrac {1} {8}+\left(\dfrac {1} {5} - \dfrac{1}{10}\right) - \dfrac{1}{12} \ldots$$
or
$$\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \frac{1}{10} - \frac{1}{12} + \cdots = \frac{1}{2}\left(\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots\right) = \frac{1}{2} \log(2).$$
Edit: More generally (without numbers), we write the left hand side as $\displaystyle\sum_{k=1}^\infty \frac{1}{2k-1} - \frac{1}{4k-2} - \frac{1}{4k} $ which is equal to $\displaystyle\sum_{k=1}^\infty \frac{1}{4k-2}- \frac{1}{4k} $ (since $\dfrac{1}{2k-1} - \dfrac{1}{4k-2} = \dfrac{1}{2k-1} - \left(\dfrac{1}{2k-1}\right)\dfrac{1}{2} = \left(\dfrac{1}{2k-1}\right)\dfrac{1}{2}$).
This is equal to $\left(\dfrac{1}{2}\right)\displaystyle\sum_{k=1}^\infty \frac{1}{2k-1}- \frac{1}{2k} $ $= \dfrac{1}{2}\displaystyle\sum_{k=1}^\infty \dfrac{(-1)^{k+1}}{k}$, or $\dfrac{\log(2)}{2}$ by examination of the power series of $\log$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/118666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Radius of convergence of power series $\sum_{n=0}^{\infty} \frac{(2x-5)^n}{n^2}$ Consider the below example:
$$\sum_{n=0}^{\infty}\frac{(2x-5)^n}{n^2},\qquad c_n=\frac{2^n}{n^2},\qquad R=2^{-1}=\frac{1}{2}.$$
$$\begin{align*}
\lim_{n\to\infty}\frac{c_{n+1}}{c_n} &= \lim_{n\to\infty}\frac{2^{n+1}/(n+1)^2}{2^n/n^2} \\
&=\lim_{n\to\infty}\frac{2n^2}{(n+1)^2}\\
&= \lim_{n\to\infty}\frac{2}{(1+\frac{1}{n})^2}\\
&= 2.
\end{align*}$$
How come the expression
$$\sum_{n=0}^{\infty} \frac{(2x-5)^n}{n^2}$$ does not appear to be used? Is $c_n$ derieved from that? How?
UPDATE
In another example I have:
$$\sum_{n=1}^{\infty} \frac{n^2(x-3)^{n+1}}{5^n}, \qquad c_{n+1}=\frac{n^2}{5^n}$$
If I try to express the series in
$$\sum_{n=1}^{\infty} c_n (x-a)^n$$
I get
$$\frac{n^2 (x-3)}{5^n} \cdot (x-3)^n, \qquad c_n = \frac{n^2 (x-3)}{5^n}$$
Or can the $(x-3)$
be removed somehow?
| Hint:
Let $$a_{0}+a_{1}z+a_{2}z^{2}+a_{3}z^{3}+a_{4}z^{4}+\cdots$$ be a power-series, and consider the series $$a_{1}+a_{2}z^{1}+a_{3}z^{2}+a_{4}z^{3}+\cdots$$ which is obtained by differentiating the power series term by term. The derived series has the same circle of convergence as the original series. Similarly the series $$\sum _{n =0}^{\infty }\dfrac {a_{n}z^{n+1}} {n+1},$$ obtained by integrating the original power-series term by term, has the same circle of convergence as $\sum _{n=0}^{n=\infty }a_{n}z^{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/118735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Complex Numbers $x,y,z$ Find $x^{2007}+y^{2007}+z^{2007}$ Let $x,y,z$ be complex numbers such that
$$x+y+z = x^{5}+y^{5}+z^{5} = 0, \hspace{10pt}
x^3+y^3+z^3=2$$
Find all possible values of
$$x^{2007}+y^{2007}+z^{2007}$$
| Since $x+y+z=0$,
$$x+y = -z\tag{A}.$$
If we raise to power $3$ both sides
$$ x^3+3x^{2}y+3y^{2}x+y^{3} = -z^{3} \quad \Rightarrow \quad x^{3}+y^{3}+z^{3} = -3xy(x+y).$$
Since $x^{3}+y^{3}+z^{3} =3$,
$$ x^{3}+y^{3}+z^{3} = -3xy(x+y).$$
We can conclude therefore that
$$ xy(x+y) = -1 \tag{B}.$$
If we take fifth power to $(A)$,
$$ x^5+5x^{4}y+10x^{3}y^{2}+10x^{2}y^{3}+5xy^{4}+y^{5} = -z^{5} $$
$$ x^{5}+y^{5}+z^{5} = -5xy(x^{3}+2x^{2}y+2xy^{2}+y^{3}).$$
And since $x^{5}+y^{5}+z^{5}=0$
$$ -5xy(x^{3}+2x^{2}y+2xy^{2}+y^{3}) = 0. $$
Since $xy \neq 0$,
$$
\begin{align*}
x^{3}+2x^{2}y+2xy^{2}+y^{3} &= 0\\
x^{3}+y^{3}+2xy(x+y) &= 0.
\end{align*}
$$
From $(B)$,
$$ x^{3}+y^{3} = 2.$$ Also
$$x^{3}+y^{3}+z^{3} = 1 \quad \Rightarrow \quad z^{3} = 1.$$
By symmetry
$$x^{3}=y^{3}=z^{3} = 1.$$
Therefore
$$x^{2007}+y^{2007}+z^{2007} = 3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/118974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Help evaluating $\int \frac{dx}{(x^2 + a^2)^2}$ I have following integral and it should be simple, however whatever substitution I use and no matter how many times I integrate it by parts (or combine both) I never get the correct solution (or any alternative solution):
$$\int \frac{dx}{(x^2 + a^2)^2}$$
I'm looking for what is on the Wolfram|Alpha in the alternative solutions section:
$$\frac{\arctan(\frac{x}{a})}{2a^3} + \frac{x}{2a^2(a^2 + x^2)} $$
| After a linear substitution we may instead look at
$$\frac{1}{(1+x^2)^2}= \frac{(1+x^2)}{(1+x^2)^2}-\frac{x^2}{(1+x^2)^2}=\frac{1}{1+x^2}-\frac{x^2}{(1+x^2)^2}$$
where the first term is easy. For the second term we may try integration by parts
$$\int x\cdot\frac{x}{(1+x^2)^2}dx=\left[x\cdot\frac{-1}{2(1+x^2)}\right]+\frac{1}{2}\int\frac{1}{1+x^2}dx=-\frac{x}{2(1+x^2)}+\frac12\arctan x$$
Ending up with
$$\int\frac{dx}{(1+x^2)^2} = \frac12\arctan x +\frac{x}{2(1+x^2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Integration: How to Begin?
Possible Duplicate:
Help evaluating $\int \frac{dx}{(x^2 + a^2)^2}$
How to I begin this integration problem?
$\begin{align}\int_{0}^{1} \frac{dx}{{\left(x^2 + 1\right)}^{2}}\end{align}$
I'm not really sure how setup the triangle to do trigonometric substitution for this problem, since I am squaring the bottom, not taking the square root of it.
Could someone please demonstrate/explain how I could do this?
Thank you for your time.
| There is a "trick" here: begin noticing that $1=x^2+1-x^2$ so
$$\frac 1{(x^2+1)^2}=\frac{x^2+1-x^2}{(x^2+1)^2}=\frac 1{x^2+1}-\frac {x^2}{(x^2+1)^2}.$$
A primitive of the first term is well-known, for the second we integrate by parts:
$$\int\frac{x\cdot x}{(x^2+1)^2}dx=-\frac x2\frac 1{1+x^2}+\frac 12\int \frac 1{1+x^2}dx$$
so finally
$$\int \frac{dx}{(1+x^2)^2}=\frac 12\arctan x+\frac x2\frac 1{1+x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Let $a,b,c >0$ Prove the inequality $\displaystyle{\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{1}{a+b+c+1} \geq 1}$ Let $a,b,c >0$
Prove the inequality $\displaystyle{\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{1}{a+b+c+1} \geq 1}$
I dont even know where to begin.
Only interested in hints (not solution)
| We have $a+1\leq a+b+c+1$, $b+1\leq a+b+c+1$ and $c+1\leq a+b+c+1$ so $\frac 1{x+1}\geq \frac 1{1+a+b+c}$ where $x=a,b$ or $c$. Now multiply by $x$ to get $\frac x{x+1}\geq \frac x{1+a+b+c}$ and sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the first 4 elements of a given field The polynomial $x^4 + x +1$ is unsplittable under $\mathbb{Z}_2$ .
Given the following $K$:
$K = \mathbb{Z}_2[x] / \mathbb{Z}_2[x] (x^4 + x +1)= {{a+bx+ cx^2 + dx^3 : a,b,c,d \in \mathbb{Z}_2}} $
I'm requested to find $4$ elements of sub-field of order $4$ of $K$.
The given $K$ is a field of order $16$. If $x^4 + x +1$ is unsplittable then I can't use it, I think (!??).
If so, do I need to find some other polynomial of order 4 and try to work with it ?
Something like $t^4 - t$ maybe ?
Regards
EDIT:
Sorry for the delay , I've some problems regarding the topic of finite fields .
First , we work with the field of $Z_{2}$ , and we know that $t^4 = t$ , and also that :
$t^4 - t = t(t-1)(t^2 +t +1)$
As we can see , $t=0$ and $t=1$ are in the field , and now we're missing two more .
Now , assume that $t = a+bx+cx^2+dx^3$ , and place it in $t^2 +t +1$ , then :
$(a+bx+cx^2+dx^3 )^2+a+bx+cx^2+dx^3+1 = a^2+b^2 x^2+c^2 x^4+d^2 x^6+(a+bx+cx^2+dx^3)==Break=$
Since $x^4=-x-1=x+1 $ and $x^6=x^2⋅x^4=x^2(1+x)=x^2+x^3 $
$==Continues=a+bx^2+cx^4+dx^6+a+bx+cx^2+dx^3+1=a+bx^2+c(x+1)+d(x^2+x^3 )+a+bx+cx^2+dx^3+1=$
{After a lot of arithmetics} $= x(b+c+d)+x^2 (b+c)+(c+d+1)⋅1$
Finally , if b=c=1 and d=0 , we'd get that :
$b+c+d=1+1+0=0$
$b+c=1+1=0$
$c+d+1=1+0+1=2=0$
Therefore the other two elements of this lovely field are :
$1+x+x^2 ;x^2+x$
Since they are both closed for addition and multiplication :
Addition : $ (x^2+x+1)+(x^2+x)=2x^2+2x+1=0+0+1=1$
Multiplication : $(x^2+x+1)(x^2+x)=x^4+x^3+x^3+x^2+x^2+x=x^4+2x^3+2x^2+1=x^4+0+0+x={x^4=x+1}=x+1+x=2x+1=0+1=1$
Now my questions are :
*
*Is this correct ?
*The addition and multiplication of the two , doesn't one of them supposed to give "0" zero ?
*Why is it that $t^4 = t$ , always ?
Thanks !
| Further to Jyrki's comment that the multiplicative group of a finite field is cyclic: This means that you can write the $|K|-1=15$ non-zero elements as $a,a^2,\dotsc,a^{15}$, where $a$ is any of the $\phi(15)=8$ primitive elements of $K^*$. So an efficient strategy is to find the multiplicative order of some arbitrary element $a$ of $K$ other than $0$ and $1$; if it turns out to be $3$, you're done; if it turns out to be $5$, you know which $4$ elements to avoid on the second try; and if it turns out to be primitive, the multiplicative subgroup of order $3$ consists of $1=a^0,a^5$ and $a^{10}$. The easiest candidate to work with is $x$, and you can readily find that $x$, $x^2$, $x^3$, $x^4$, $x^5$ are all not $1$, so $x$ must be primitive (since otherwise its order would have to divide $15$ and thus be at most $5$). Thus the multiplicative subgroup of order $3$ consists of $1=x^0$, $x^5$ and $x^{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the derivative respect to $x$? Question
$\sqrt{2x+1}$
$=\sqrt{2x+1}$
$=\sqrt{2x} + \sqrt{1}$
$=\dfrac{1}{2x^{1/2}}$
however the right answer is $\dfrac{1}{\sqrt{2x+1}}$
Can you please help me out?
this chapter name is (differentiationg rational power
$x^{p/q}$)
| $\sqrt{2x+1} \neq \sqrt{2x}+\sqrt{1}$: Witness $3 = \sqrt{9} = \sqrt{2 \cdot 4 + 1} \neq \sqrt{2 \cdot 4} + \sqrt{1} = 2 \sqrt{2} + 1$.
The right way to solve this is to apply the chain rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/127584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A tricky math problem about the difference of 2 squares Jose is given two 2-digit numbers AB and CD where (A, B, C,
and D represent unique digits) and is told to find the
difference between the squares of these numbers.
However, Jose has dyslexia and reverses the digits of each
number before finding the difference between the squares.
Miraculously, Jose gets the correct answer nonetheless.
Find the answer Jose obtained.
Hint: AB is a number that can be expressed as 10A + B.
| Even if $A,B,C,D$ must be distinct and non-zero, I seem to have four sets of solutions (or perhaps two pairs of sets) to $A^2-B^2=C^2-D^2$, such as
$$8^2-7^2 = 15 =4^2-1^2 \text{ so } 87^2-41^2 = 5888 = 78^2-14^2$$
$$8^2-4^2 = 48 = 7^2-1^2 \text{ so }84^2-71^2 = 2015 = 48^2-17^2$$
$$9^2-7^2 = 32 =6^2-2^2 \text{ so }97^2-62^2 = 5565 = 79^2-26^2$$
$$9^2-6^2 = 45 =7^2-2^2 \text{ so } 96^2-72^2 = 4032 = 69^2-27^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/128160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving that: $\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$ Let $a$ and $b$ be positive reals. Show that
$$\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$$
| You can use the following inequality:
$$ \sqrt{xy} \le \frac{x+y}{2} \le \sqrt[x+y]{x^x y^y}$$
The first inequality is straightforward, and the second one can be gotten by
$$ \frac{2}{x+y} = \frac{ x \times 1/x + y \times 1/y}{x+y} \ge \sqrt[x+y]{\frac{1}{x^x y^y}}$$
using the weighted $\text{AM} \ge \text{GM}$.
Setting $x = a^{1/n}$, $y = b^{1/n}$ and taking the $n^{th}$ powers gives us that the limit is $\sqrt{ab}$, by the squeeze theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/130497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 5
} |
How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?. How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.
Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator
| $$\cos(\pi/5) - \cos(2\pi/5)= {\rm Re}(e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}})=\frac{1}{2}( e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}+ \overline{e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}})=\frac{1}{2}( e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}+ e^{-i\frac{\pi}{5}}-e^{-i\frac{2\pi}{5}})=\frac{e^{\frac{-2i\pi}{5}}}{2}( -e^{i\frac{4\pi}{5}}+e^{i\frac{3\pi}{5}}+ e^{i\frac{\pi}{5}}-1)$$
To simplify the computations, let $\omega=e^{\frac{i \pi}{5}}$. Note that $\omega^5=-1$.
Then
$$\cos(\pi/5) - \cos(2\pi/5)= \frac{-1}{2\omega^2}(\omega^4-\omega^3-\omega+1)= \frac{-1}{2\omega^2}(\omega^4-\omega^3+\omega^2-\omega+1-\omega^2)$$
$$\cos(\pi/5) - \cos(2\pi/5)= \frac{-1}{2\omega^2}(\frac{\omega^5+1}{\omega+1} -\omega^2)=\frac{-1}{2\omega^2}(0 -\omega^2)=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/130817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 18,
"answer_id": 7
} |
How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but I don't know how.
| According to link#1, the product of 5 consecutive integers divisible by 5! and the product of 3 consecutive integers divisible by 3!.
Now $n(n-1)(n+1)(n-2)(n+2) = n^5-5n^3+4n $
$n^5-n= n^5-5n^3+4n + 5(n^3-n)$
=$n(n-1)(n+1)(n-2)(n+2) -5n(n-1)(n+1)$
So, the first part is divisible by 5! and the 2nd part is by 5(3!)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 21,
"answer_id": 12
} |
Direct Proof that $1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n$
Prove that for all integers $n$, $n \geq 1$,
$$1 + 3 + 5 + \cdots + (2n - 1) = n\cdot n$$
How would I go about proving this?
| By inspection note that $1 = 1 \cdot 1$, that $1 + 3 = 2 \cdot 2$, and that $1 + 3 + 5 = 3 \cdot 3$.
Next, witness that the difference between any two successive perfect squares $(n - 1)^2$ and $n^2$ is a particular odd integer:
$n^2 - (n - 1)^2 \implies n^2 - (n^2 - 2n + 1) \implies 2n - 1$
Note that this is the very formula which you are summing to generate your series.
So for instance the difference between $3 \cdot 3$ and $2 \cdot 2$ is $2 \cdot 3 - 1 = 5$, corresponding to the $5$ in $3\cdot3 = 1 + 3 + 5$.
If the sum of the series generated so far is a square (shown by inspection), and the next term is from the series generated by $2n - 1$, then after this term is added, the sum must be the next square because the each term obeys the formula for the difference between two successive squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/136237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 9,
"answer_id": 2
} |
Showing that $\cos\left(\frac{\pi}{5}\right)=\frac{1}{2}\phi?$ What is the usual way of proving things like
$$\cos\left(\frac{\pi}{5}\right)=\frac{1}{2}\phi?$$
I know that there is an identity which claims the above, but how was it derived? Are other identities used in the process? I am interested in knowing this.
For example, I could follow through as such:
$$\cos\left(\frac{\pi}{5}\right)=\cos\left(-\frac{\pi}{5}\right)=-\cos\left(\pi-\frac{\pi}{5}\right)=-\cos\left(\frac{4\pi}{5}\right)=1-2\cos^2\left(\frac{2\pi}{5}\right),$$
which gets me somewhat close to the golden ratio, but I must admit that I am stuck here.
Do you guys have any ideas?
| Here's a standard way of doing this with complex numbers:
Let $z=e^{\pi i/5}$, which is a primitive 10th root of unity. So it's a root of the 10th cyclotomic polynomial, which is $z^4-z^3+z^2-z+1$. In other words,
$$e^{4\pi i/5}-e^{3\pi i/5}+e^{2\pi i/5}-e^{\pi i/5}+1=0.$$
So the real part of this is also equal to zero. But the real part of this is
$$\cos\left(\frac{4\pi}5\right)-\cos\left(\frac{3\pi}5\right)+\cos\left(\frac{2\pi}5\right)-\cos\left(\frac{\pi}5\right)+1=2\cos\left(\frac{2\pi}5\right)-2\cos\left(\frac{\pi}5\right)+1.$$
Let $x=\cos\left(\frac{\pi}5\right)$. Then this can be written as
\begin{align*}2(2x^2-1)-2x+1 &= 0\\
4x^2-2x-1 &= 0,\end{align*}
whose positive root is $\frac{1+\sqrt 5}4=\frac{\phi}2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/136978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.