Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to solve the limit of $\ln(x \sin x)$ as $x$ approaches $0$? I have a limit:
$$k = \lim_{x\to0+} \ln(x \sin x)$$
How do I find this? Since $\ln(x)$ is continuous I tried:
$$k = \ln( \lim_{x\to0+} (x \sin x))$$
$$k = \ln(0)$$
Which to my understanding is undefined, but the answer is $-\infty$ somehow. Where did I go wrong?
| You could use Taylor series for approximating the value of $\log (x \sin (x))$ and then get the limit (and how it is approached).
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$
$$x\sin(x)=x^2-\frac{x^4}{6}+\frac{x^6}{120}+O\left(x^8\right)=x^2\left(1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right) \right)$$
$$\log (x \sin (x))=\log(x^2)+\log\left(1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right) \right)$$
$$\log (x \sin (x))=\log(x^2)-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right)$$
Just for curiosity, make $x=\frac \pi 6$. The exact value is
$$\log \left(\frac{\pi }{12}\right)\approx -1.34018$$ while the approximation would give
$$2\log\left(\frac \pi 6\right)-\frac{\pi ^2}{216}-\frac{\pi ^4}{233280}\approx -1.34017$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a_{n+1}=\sqrt{a_n+n}$ with $a_1=1$ then find $\lim_{n \to \infty} (a_n -\sqrt{n})$
Let $a_{n+1}=\sqrt{a_n+n}$ with $a_1=1$ then find
$\displaystyle \lim_{n \to \infty} (a_n -\sqrt{n}).$
My Try:
$$\lim_n( \sqrt{a_{n}+n}-\sqrt{n}) \times \dfrac{( \sqrt{a_{n}+n}+\sqrt{n}) }{( \sqrt{a_{n}+n}+\sqrt{n})}=\lim_n \dfrac{1}{( \sqrt{\dfrac{1}{a_{n}}+\dfrac{n}{a_{n}^2}}+\sqrt{\dfrac{n}{a_{n}^2}})}.$$
Now what do I do?
| We first show by induction that for any $n\geq 1$,
$$\sqrt{n}\leq a_n\leq \sqrt{n+2\sqrt{n}}\tag{1}.$$
The inequalities hold for $n=1$. As regards the inductive step we have that for $n\geq 1$,
$$a_{n+1}=\sqrt{a_n+n}\geq \sqrt{\sqrt{n}+n}\geq \sqrt{1+n},$$
and
$$a_{n+1}=\sqrt{a_n+n}\leq \sqrt{\sqrt{n+2\sqrt{n}}+n}
\leq \sqrt{(\sqrt{n}+1)+n}\leq \sqrt{2\sqrt{n+1}+n+1}.$$
Hence, by (1) and by the Squeeze Theorem,
$$1\leq \frac{a_n}{\sqrt{n}}\leq \sqrt{1+\frac{2}{\sqrt{n}}}\to 1\implies
\lim_{n\to \infty}\frac{a_n}{\sqrt{n}}=1$$
Finally, as $n$ goes to infinity,
$$
\begin{align}
a_{n+1} -\sqrt{n+1}&=\sqrt{a_n+n} -\sqrt{n+1}=\frac{a_n-1}{\sqrt{a_n+n} +\sqrt{n+1}}\\&=\frac{\frac{a_n}{\sqrt{n}}-\frac{1}{\sqrt{n}}}{\sqrt{\frac{a_n}{n}+1} +\sqrt{1+\frac{1}{n}}}\to \frac{1}{2}.
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Express $2\tan^{-1}x$ in terms of $\sin^{-1}$ and $\cos^{-1}$ $$
2\tan^{-1}x=\begin{cases}\tan^{-1}\frac{2x}{1-x^2} &\mbox{ if $|x|<1$}\\
\pi+\tan^{-1}\frac{2x}{1-x^2} &\mbox{ if $|x|>1$}
\end{cases}
$$
Similarly,
$$
2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2} &\mbox{ if $|x|\leq1$}\\
\text{____________} &\mbox{ if $|x|>1$}
\end{cases}
$$
and
$$
2\tan^{-1}x=\begin{cases}\cos^{-1}\frac{1-x^2}{1+x^2} &\mbox{ if $x\geq0$}\\
\text{____________} &\mbox{ if $x<0$}
\end{cases}
$$
How do I derive the missing cases ?
My Attempt:
Take $\tan^{-1}x=A\implies x=\tan A$ where $\frac{-\pi}{2}<A<\frac{\pi}{2}\implies -\pi<2A<\pi$.
Case 1 :
$$
\cos2A=\frac{1-\tan^2A}{1+\tan^2A}=\frac{1-x^2}{1+x^2}=\cos\Big(\cos^{-1}\frac{1-x^2}{1+x^2}\Big)\\
\implies \cos^{-1}\frac{1-x^2}{1+x^2}=2n\pi\pm2A=2n\pi\pm2\tan^{-1}x
$$
As $0\leq\cos^{-1}\leq\pi$,
If $0\leq2A\leq\pi$
$$
\cos^{-1}\frac{1-x^2}{1+x^2}=2A=2\tan^{-1}x
$$
If $-\pi<2A<0$
$$
\cos^{-1}\frac{1-x^2}{1+x^2}= \color{red}{\text{?}}
$$
Case 2 :
$$
\sin2A=\frac{2\tan A}{1+\tan^2A}=\frac{2x}{1+x^2}=\sin\Big(\sin^{-1}\frac{2x}{1+x^2}\Big)\\
\implies \sin^{-1}\frac{2x}{1+x^2}=n\pi+(-1)^n.2A=n\pi+(-1)^n.2\tan^{-1}x
$$
As $\frac{-\pi}{2}\leq\sin^{-1}\leq\frac{\pi}{2}$,
If $\frac{\pi}{2}\leq2A\leq{\pi}$
$$
\sin^{-1}\frac{2x}{1+x^2}= \color{red}{\text{?}}
$$
what value should $\cos^{-1}\frac{1-x^2}{1+x^2}$ and $\sin^{-1}\frac{2x}{1+x^2}$ take and how do I proceed further ?
My Understanding:
For,
$$
\tan2A=\tan\big(\tan^{-1}\frac{2x}{1-x^2}\big)\implies\tan^{-1}\frac{2x}{1-x^2}=n\pi+2A
$$
If $\frac{\pi}{2}<2A<{\pi}$ I would take
$$
\tan^{-1}\frac{2x}{1-x^2}=-\pi+2A=-\pi+2\tan^{-1}x
$$
| hint
If $A=\tan (\frac {x}{2}) $ then
$$\cos (x)=\frac {1-A^2}{1+A^2} $$
and
$$\sin (x)=\frac {2A}{1+A^2} $$
| {
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"source": "stackexchange",
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} |
Computing $\lim_{x \to 1}\frac{x^\frac{1}{5}-1}{x^\frac{1}{6} -1}$ I cannot figure out how to
get around the zero numerator and denominator in order to compute the limit below:
$$\lim_{x \to 1}\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}$$
I tried:
$$ \lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{1}{6} - 1) (x^\frac{1}{6} + 1) } $$
$$\lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{2}{6} - 1) } $$
| Hint:
As lcm$(5,6)=30,$ choose $x^{1/30}=y\implies x^{1/5}=y^6, x^{1/6}=?$
| {
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How do I find the quotient of the division of $x^{1993}+x^2+1$ by $x^2+x+1$? I have an exercise where I need to find the quotient of the division of polynomial $f=x^{1993}+x^2+1$ by $g=x^2+x+1$. I know to do it the long way but I think there is a trick to do it fast. By the way, I know the remainder is $0$.
| Since $(x-1)(x^2+x+1)=x^3-1$ you have $x^3\equiv 1 \bmod x^2+x+1$ whence $x^{1993}\equiv x$ and then it is trivial.
To expand: $f=X^{1993}+X^2+1=(X^{1993}-X)+(X^2+X+1)$
The first of these factors is $$X(X^{1992}-1)=X(X^3-1)(X^{1989}+X^{1986}+\dots +X^3+1)=$$$$=(X^2+X+1)(X^2-X)(X^{1989}+X^{1986}+\dots +X^3+1)$$ and dividing through by $(X^2+X+1)$ and multiplying through by $(X^2-X)$ gives a sum of pairs of terms of the form $X^{3n+2}-X^{3n+1}$ starting $X^{1991}-X^{1990}+X^{1988}-X^{1987}+\dots$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is every $N$th Fibonacci number where $N$ is divisible by $5$ itself divisible by $5$ There seems to be a pattern of sorts in the Fibonacci sequence:
The $5$th, $10$th, $15$th & $20$th values are:
$$5, 55, 610, 6765$$
Does this pattern continue ad infinitum? I've tried a few more using Binet's Formula and it seems to hold. So:
Is every $N$th Fibonacci number where $N$ is divisible by $5$ itself divisible by $5$?
Bonus Q: Are there any other patterns?
| Yes, if $5$ divides $n$ then $5$ divides $F_n$. Actually also the other way round is true, so $5$ divides $n$ if and only if $5$ divides $F_n$.
More generally, the following congruence holds
$$F_n\equiv n3^{n-1}\pmod{5}.$$
This congruence implies our claim and more than that. For example, it follows that the Fibonacci sequence modulo $5$ is periodic with period length $5\cdot 4=20$ where $5$ is the period of $n$ and $4$ the period of $3^{n-1}$ (recall that by Fermat's little theorem $3^{k(5-1)}\equiv 1 \pmod{5}$). The list of values $F_n\pmod{5}$ is
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
n \pmod{20}& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9\\
F_n\pmod{5} &\mathbf{0}& 1& 1& 2& 3& \mathbf{0}& 3& 3& 1& 4\\
\hline
n \pmod{20}&10& 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18&19\\
F_n\pmod{5} &\mathbf{0}& 4& 4& 3& 2& \mathbf{0}& 2& 2& 4& 1
\\\hline
\end{array}$$
Proof of the above congruence $F_n\equiv n3^{n-1}\pmod{5}$. By the Binet's formula and the binomial theorem we have that
$$\begin{align}F_n&=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right)\\
&=\frac{1}{2^{n}\sqrt{5}}\sum_{k=0}^{n}\binom{n}{k}\left((\sqrt{5})^k-(-\sqrt{5})^k\right)\\
&=\frac{1}{2^{n}\sqrt{5}}\sum_{j=0}^{\lfloor (n-1)/2\rfloor}\binom{n}{2j+1}\left((\sqrt{5})^{2j+1}-(-\sqrt{5})^{2j+1}\right)\\
&=\frac{1}{2^{n}\sqrt{5}}\sum_{j=0}^{\lfloor (n-1)/2\rfloor}\binom{n}{2j+1}(2\sqrt{5}\cdot 5^j)\\
&=\frac{1}{2^{n-1}}\sum_{j=0}^{\lfloor (n-1)/2\rfloor}\binom{n}{2j+1}5^j
\equiv 3^{n-1}\sum_{j=0}^{0}\binom{n}{2j+1}5^j\equiv n3^{n-1}\pmod{5}.
\end{align}$$
| {
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"source": "stackexchange",
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Chinese Remainder Theorem/Simultaneous congruences Find an integer N , 0 ≤ N < 105 such that
N ≡ 2 mod 3,
N ≡ 1 mod 5, and
N ≡ 4 mod 7.
What I have done:
So I have been able to start by splitting up the summation of $x$ into 3 sections:
$ x = 5*7 + 3*7 + 3*5 $
with the first multiplication corresponding with the mod 3 equation, the second multiplication corresponding with the mod 5 equation and the third multiplication corresponding with the mod 7 equation.
Therefore:
$x = 35 + 21 +15$
However, I know that this isn't complete. But I am not exactly sure on how to proceed.
Any help?
| $N \equiv 2 \mod 3$ so $N = 2 + 3a$.
$N \equiv 1 \mod 5$ so $N =1 + 5b$
So $2+3a = 1 + 5b$ so $5b - 3a = 1$. One solution is $b = 2$ and $a = 3$
So $N \equiv 2 + 3*3 = 1 + 2*5 = 11 \mod 3*5$.
So $N = 11 + 15c$.
$N \equiv 4 \mod 7$ so $N = 4 + 7d$.
So $11 + 15c = 4 + 7d$ so $15c - 7d = -7$. $c =0$ and $d = 1$ is a solution.
So $N \equiv 11 = 4+7 \equiv 11 \mod 3*5*7 = 105$.
And, indeed, $11 \equiv 2 \mod 3$ and $11 \equiv 1 \mod 5$ and $11\equiv 4 \mod 7$.
Trying to follow your partition reasoning.
$5*7 = 35 \equiv 2 \mod 3$ so that satisfies.
$3*7 = 21 \equiv 1 \mod 5$ so that satisifies.
$3*5 = 15 \equiv 1 \not \equiv 4 \mod 7$ so that does not satisfy.
But $4*3*5 \equiv 4 \mod 7$ so that does.
So $5*7 + 3*7 + 60 = 116$ solves all three. But $116 > 105$. But any $k \equiv 116 \mod 105$ so do so $116 - 105 = 11$ will do.
.... or ... when we hae $3*5 \equiv 1 \mod 7$ and we could have figured $4 \equiv -3 \mod 7$ so $-3*3*5 \equiv 4 \mod 7$.
So $N = 5*7 + 3*7 - 3*3*5 = 11$. (by taking a negative we know we won't get a number too large).
Actually, I had never done the "partitioning" before.
It works well. I like it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving linear system after Gaussian elimination If I had an initial $2 \times 3$ matrix $A$ and a $2 \times 1$ constant vector $b$, such that I have $A$ and $b$ of $Ax=b$:
$$A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$$
$$b = \begin{pmatrix} 7 \\ 8 \end{pmatrix}$$
Such that in augmented form it appears as such;
$$\left(\begin{array}{ccc|c} 1 & 2 & 3 & 7 \\ 4 & 5 & 6 & 8 \end{array}\right)$$
And valid operations were performed such that it was in $r.e.f.$ (the values here are just examples):
$$\left(\begin{array}{ccc|c} 1 & -3 & 2 & 11 \\ 0 & 1 & -9 & 2 \end{array}\right)$$
How does one solve this system, if there is not a row that only contain:
$$\left(\begin{array}{ccc|c} 0 & 0 & 1 & N \end{array}\right)$$
...so that you could back solve the entire system?
| Lets take a step back from the linear algebra to more basic algebra.
$x -3y +2z = 11\\
0x +1y -9z = 2$
Since you have two equations and 3 unknowns there will be a "line of solutions."
You can set $z = 0$ and find a pair of $x,y$ that will solve the equation.
i.e.
$x - 3y = 11\\
0x +1y = 2$
$(17,2,0)$ is a solution.
Then set $z = 1$
$x - 3y +2= 11\\
0x +1y -9= 2$
$(42, 11, 1)$ is also a solution
the full line of solutions is $(17,2,0) + (25,9,1) t$
It is worth noting that:
$\begin{bmatrix} 1&-3&2\\0&1&-9\end{bmatrix}\begin{bmatrix} 17\\2\\0\end{bmatrix} = \begin{bmatrix} 11\\2\end{bmatrix}$
and
$\begin{bmatrix} 1&-3&2\\0&1&-9\end{bmatrix}\begin{bmatrix} 29\\9\\1\end{bmatrix} = \bf 0$
| {
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"source": "stackexchange",
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Solve $n$th order determinant Solve $n$th order determinant:
$$
\begin{pmatrix}
1 & 2 & 3 & \cdots & n-2 & n-1 &n\\
2 & 3 & 4 & \cdots & n-1& n& 1\\
3 & 4 & 5 & \cdots & n &1 &2\\
\vdots& \vdots& \vdots& \ddots& \vdots& \vdots&\vdots\\
n-2 &n-1&n &\cdots &n-5 &n-4 &n-3\\
n-1 &n&1 &\cdots &n-4 &n-3 &n-2\\
n &1&2 &\cdots &n-3 &n-2 &n-1\\
\end{pmatrix}
$$
I tried adding $2$nd to $n$th row to the $1$st row, then the first row is:
$$
\begin{pmatrix}
\frac{n(n+1)}{2} &\frac{n(n+1)}{2} \cdots \frac{n(n+1)}{2}\\
\end{pmatrix}
$$
then I took out $\frac{n(n+1)}{2}$ from the first row to get
$$
\begin{pmatrix}
1 & 1 & \cdots& 1\\
\end{pmatrix}
$$
and then tried to work with that but I don't think this is the correct approach because anything I tried doesn't look nice.
| Let us denote by $A$ the matrix corresponding to the determinant we have to compute.
Our objective is to show that :
$$\tag{*}\det(A)=(-1)^{n-1}n^{n-2}\dfrac{n(n + 1)}{2}$$
As $C:=A^T$ is circulant, we can use the following result : (https://en.wikipedia.org/wiki/Circulant_matrix), (http://www.math.columbia.edu/~ums/pdf/cir-not5.pdf):
A $n \times n$ circulant matrix $C$ is diagonalizable under the form
$$C=F_n^{−1}DF_n$$
where $F_n$ is the $n$-th order matrix of DFT (Discrete Fourier Transform) with general term $\omega^{k\ell}$ where $\omega:=e^{-\tfrac{2 i \pi}{n}}.$
Moreover if we denote by $\lambda_k$ the entries $D_{kk}$ of diagonal matrix $D$, with $V=(\lambda_1, \lambda_2, ... \lambda_n)^T$, $V$ is the DFT of the first column $(1,2,...n)^T$ of $C$ , i.e,
$$\begin{pmatrix}\lambda_1\\ \lambda_2\\ \vdots \\ \vdots \\ \lambda_n\end{pmatrix}=\underbrace{\begin{pmatrix}\omega^{0 \times 0}&&\cdots&&\omega^{(n-1)0}\\ &&&& \\ &&\omega^{k \ell}&&\\ &&&&\\ \omega^{0(n-1)}&&\cdots&&\omega^{(n-1)(n-1)}\end{pmatrix}}_{\text{DFT matrix} \ \ F_n}\begin{pmatrix}1\\2\\ \vdots \\ \vdots \\ n\end{pmatrix}$$
Thus:
$$\tag{1}\det(C)=\det(D)=\prod_{k=1}^n \lambda_k=\prod_{k=0}^{n-1} p(\omega^k),$$
where polynomial $p$ is defined by
$$p(x):=1+2x+3x^2+4x^3+...+nx^{n-1}=q'(x)$$
$$\text{with} \ \ q(x):=1+x+x^2+\cdots+x^n=\dfrac{1-x^{n+1}}{1-x}$$
As an immediate consequence :
$$\tag{2}\text{if} \ x \neq 1, \ p(x)=q(x)'=\dfrac{nx^{n+1}-(n+1)x^n+1}{(1-x)^2}.$$
In (1), let us treat separately the case $k=0$:
$$p(\omega^0)=p(1)=1+2+3+\cdots+n=\dfrac{n(n+1)}{2}$$
Concerning the general case, using (2):
$$\tag{3}\text{for} \ k=1,2,... (n-1): \ \
p(\omega^k)= \dfrac{n(\omega^k-1)}{(\omega^k-1)^2}=n\dfrac{1}{\omega^k-1}$$
(because any $n$th root of unity at power $n$ is equal to $1$.)
As a consequence
$$\prod_{k=1}^{n-1} p(\omega^k)=n^{n-1}\prod_{k=1}^{n-1}\dfrac{1}{\omega^k-1}$$
In view of objective (*), it remains to prove that
$$\tag{4}\prod_{k=1}^{n-1}(\omega^k-1)=(-1)^{n-1}n.$$
This will be done by considering $r$ defined by
$$\tag{5}r(x):=1-x^n=(1-x)s(x) \ \text{with} \ \ s(x):=1+x+x^2+\cdots +x^{n-1}$$
But the roots of $s$ are the $\omega^k$ for $k=1\cdots (n-1)$ ; thus, the factorized form of $s$ is:
$$\tag{6}s(x)=\prod_{k=1}^{n-1}(x-\omega^k)$$
Setting $x=1$ in (6) and (5) resp. gives (4).
Remark: This proof is different from the direct proofs using lines and columns manipulations. Its interest lies in the facts that
*
*it is based on mathematics (circulant matrices, Fourier Transform, $n$th roots of unity) that are beyond "tricks of the trade".
*it explains in particular why we have factors $\dfrac{n(n+1)}{2}$ and $n^{n-2}$ in the result.
| {
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"question_score": "2",
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$9^x=5$, solve for $5\cdot27^{-x-1}$ Question:
Given $9^x=5$
Compute the value of $5\cdot27^{-x-1}$
My attempt:
$9^x=(3^x)^2=5\Rightarrow3^x=\sqrt{5}$
$5\cdot27^{-x-1}=3^{2x}\cdot(3^{3-x}\div3^{3})=3^{2x}\cdot3^{-x}=3^{x}=\sqrt{5}$
My answer is incorrect and I wonder why.
| You don't need to do that at all. $$5\cdot 27^{-x-1}=\dfrac{5}{27^{x+1}}=\dfrac{5}{27}\dfrac{1}{27^x}=\dfrac{5}{27}\dfrac{1}{(9^x)^{1.5}}=\dfrac{5}{27}\dfrac{1}{5\sqrt 5}=\dfrac{1}{27\sqrt 5}$$
| {
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Roots to an equation, $\frac{3-2x}{x-1}-\sqrt[4]{x^3}$
Find the number of roots to $\frac{3-2x}{x-1}-\sqrt[4]{x^3}=0$.
My method:
$\frac{3-2x}{x-1}-\sqrt[4]{x}$, we multiply $\sqrt[4]{x}$ and $x-1$ to both sides, getting $3\sqrt[4]{x}-2x\sqrt[4]{x}-x(x-1)=0$. We let $\sqrt[4]{x}=k$. Then we have $3k-2xk-x^2+x=0$. Finding the discriminant ($b^2-4ac$), getting $4k^2+8k+1$, evidently greater than $0$, which means there are 2 roots.
Plotting this on Desmos gives 1 root only. How am I wrong?
| Since $k$ is a function of $x$, the coefficients of your quadratric equation are functions of $x$.
The quadratic formula is useless in that context, since that would be solving for $x$ in terms of $x$.
Here's how I would approach the problem . . .
Let $f(x)={\displaystyle{{\small{\frac{3-2x}{x-1}}}-\sqrt[4]{x^3}}}$.
You want to count the number of real zeros of $f$.
Rather than trying to solve the equation $f(x) = 0$, instead, analyze the behavior of $f$.
The domain of $f$ is $[0,1)\cup(1,\infty)$.
On the interval $[0,1)$, we have
$$\frac{3-2x}{x-1}<0$$
$$-\sqrt[4]{x^3} \le 0$$
hence $f(x) < 0$.
Thus, $f$ has no real zeros in the interval $[0,1)$.
On the interval $(1,\infty)$, the functions
$$\frac{3-2x}{x-1}$$
$$-\sqrt[4]{x^3}$$
are both strictly decreasing, hence $f$ is strictly decreasing.
It follows that $f$ has at most one real root.
Letting $x$ approach $1$ from the right, we get
\begin{align*}
\lim_{x\to 1^{+}} f(x)
&=\;\lim_{x\to 1^{+}} \frac{3-2x}{x-1}-\sqrt[4]{x^3}\\[4pt]
&=\;\lim_{x\to 1^{+}} \frac{3-2x}{x-1}-\lim_{x\to 1^{+}}\sqrt[4]{x^3}\\[4pt]
&=\;\infty-1\\[4pt]
&=\;\infty\\[4pt]
\end{align*}
Letting $x$ approach $\infty$, we get
\begin{align*}
\lim_{x\to \infty} f(x)
&=\;\lim_{x\to \infty} \frac{3-2x}{x-1}-\sqrt[4]{x^3}\\[4pt]
&=\;\lim_{x\to \infty} \frac{3-2x}{x-1}-\lim_{x\to \infty}\sqrt[4]{x^3}\\[4pt]
&=\;-2-\infty\\[4pt]
&=\;-\infty\\[4pt]
\end{align*}
Since $f$ is continuous on $(1,\infty)$, it follows that $f$ has at least one real root.
Therefore $f$ has exactly one real root.
| {
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"url": "https://math.stackexchange.com/questions/2629685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find values for x such that A is not invertible. For $A$ not to be invertible, i.e. $\det(A) = 0$
$A =$ $$
\begin{bmatrix}
3 & 1 & 7-x \\
3 & 2-x & 4 \\
4 & 2 & 8 \\
\end{bmatrix}
$$
After some row operations (step 1: R1 - R2 => R1 then R2 - R3 => R2 and lastly R1+R2 => R1) I got:
$A =$ $$
\begin{bmatrix}
-1 & -1 & 7-x \\
-1 & -x & -4 \\
4 & 2 & 8 \\
\end{bmatrix}
$$
I computed the determinant.
Then checked when it equals $0$ and for values of $x$ I got $\dfrac {1}{4} (19\pm \sqrt{305})$.
Could you please confirm with me whether I am on the right track? Thanks.
| We have
$$A =
\begin{bmatrix}
3 & 1 & 7-x \\
3 & 2-x & 4 \\
4 & 2 & 8 \\
\end{bmatrix}
$$
Step 1: R1-R2 -> R1
$$\text{det}A= \left\vert
\begin{array}{ccc}
0 & -1+x & 3-x \\
3 & 2-x & 4 \\
4 & 2 & 8 \\
\end{array} \right\vert
$$
Step 2: R2-R3 -> R2
$$ \text{det}A= \left\vert\begin{array}{ccc}
0 & -1+x & 3-x \\
-1 & -x & -4 \\
4 & 2 & 8 \\
\end{array}\right\vert
$$
Step 3: R3 + $4$ R2 -> R3
$$\text{det}A=
\left\vert \begin{array}{ccc}
0 & -1+x & 3-x \\
-1 & -x & -4 \\
0 & 2-4x & -8 \\
\end{array}\right\vert
$$
Then, using minors of the first column, we obtain
$$ \text{det}A = (-1) (-1)^{2+1} \left\vert \begin{array}{cc} -1+x &3-x\\ 2-4x &-8 \end{array} \right\vert\\
= 8 - 8x - 6 +2x+12x-4x^2\\
= -4x^2+6x +2\\
=-2 (2x^2 -3x -1)$$
Then, $2x^2 - 3x-1 = 2\left(x-\dfrac{3+\sqrt{17}}{4}\right)\left(x-\dfrac{3-\sqrt{17}}{4}\right)$,
if I haven't made a mistake.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Little help with a residue integration with branch point
$$\int_0^{+\infty} \frac{dx}{\sqrt{x}(4x^2+1)(x+1)}$$
Now I transformed it into
$$\int_0^{+\infty} \frac{z\ dz}{z(4z^2+1)(z+1)}$$
To obtain the form
$$\frac{z^{\alpha}F(z)}{G(z)}$$
In such a wa I can apply residue calculus to get the answer as
$$\frac{2\pi i}{1 - e^{2\pi i \alpha}}\text{Res}[f, z]$$
Where a sum of residues is understood.
The poles are at $z = 0, -1, \pm \frac{1}{2}i$
The result shall be $\frac{2\pi }{5}$ but now the question is: how to deal with the poles at $\pm \frac{1}{2}i$? I would have a cumbersome square root of an imaginary unit.
Is there some "nicer" or smart way to deal with the integral?
Thank you!
| Substitute $x=u^2$ therefore$$\int_0^{+\infty} \frac{dx}{\sqrt{x}(4x^2+1)(x+1)}=\int_{0}^{\infty} \frac{2du}{(4u^4+1)(u^2+1)}=\int_{-\infty}^{\infty} \frac{du}{(4u^4+1)(u^2+1)}$$using the same famous contour for complex integration we have:$$\int_{C} \frac{dz}{(4z^4+1)(z^2+1)}=\int_{-R}^{R} \frac{du}{(4u^4+1)(u^2+1)}+\int_{0}^{\pi} \frac{iRe^{i\theta}d\theta}{(4R^4e^{4i\theta}+1)(R^2e^{2i\theta}+1)}$$also $f(z)=\dfrac{1}{(4z^4+1)(z^2+1)}$ has 6 roots among those 3 contained by the contour and all are of order 1. Those 3 are:$$r_1=i\\r_2=\dfrac{1}{\sqrt2}e^{i\dfrac{\pi}{4}}\\r_3=\dfrac{1}{\sqrt2}e^{i\dfrac{3\pi}{4}}$$if we denote the residue of $f(z)$ in these points by $R_1$, $R_2$ and $R_3$ we have:$$R_1=R_2+R_3=\dfrac{1}{10i}$$therefore$$R_1+R_2+R_3=\dfrac{1}{5i}$$and the integral would be$$\int_{C} \frac{dz}{(4z^4+1)(z^2+1)}=2\pi i(R_1+R_2+R_3)=\frac{2\pi}{5}$$which leads to $$I=\int_{-\infty}^{\infty} \frac{du}{(4u^4+1)(u^2+1)}=\frac{2\pi}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
How to calculate $\int \frac{dx}{(a^2 + x^2)^2}$? I'm trying to use a trig substitution but I'm stuck. Here's what I did so far:
$$\int \frac{dx}{(a^2 + x^2)^2}$$
Let $x = a\sin \theta, dx = a\cos \theta d\theta$
$$\int \frac{a cos\theta d\theta}{(a^2 + a^2 sin^2 \theta)^2} = \int \frac{a\cos \theta d\theta}{(a^2(1+sin^2\theta))^2} $$
$$\int \frac{a\cos \theta d\theta}{(a^2 \cos^2\theta)^2} =\int \frac{d\theta}{a^3cos^3\theta} $$
I don't know what to do anymore
| For any $A>0$ we have
$$ \int\frac{dx}{A+x^2}=C+\frac{1}{\sqrt{A}}\arctan\frac{x}{\sqrt{A}}$$
hence by differentiating both sides with respect to $A$ we get
$$ \int \frac{dx}{(x^2+A)^2} = D+\frac{x}{2A(x^2+A)}+\frac{1}{2A\sqrt{A}}\arctan\frac{x}{\sqrt{A}} $$
and now we just have to evaluate both sides at $A=a^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Proof $n \geq 3\hspace{0.5cm} \sum\limits_{k=1}^{n-2}{n-k\choose 2}={n\choose 3}$ How do you prove this expression trough mathematical induction ? It just ends up as a messy string of expressions when I try to solve it.
$n \geq 3\hspace{0.5cm} \sum\limits_{k=1}^{n-2}{n-k\choose 2}={n\choose 3}$
| To prove:
$$\sum_{k=1}^{n-2} \binom{n-k}{2} = \binom{n}{3}$$
Base case ($n = 3$):
$$ \sum_{k=1}^{1} \binom{3-k}{2} = \binom{2}{2} = 1 = \binom{3}{3}
$$
Induction hypothesis:
For fixed $n$ we may assume: $\sum_{k=1}^{n-2} \binom{n-k}{2} = \binom{n}{3}$
Induction step ($n \to n+1$):
\begin{align}
\sum_{k=1}^{n-1} \binom{n+1-k}{2} &= \binom{2}{2} + \sum_{k=1}^{n-2} \binom{n+1-k}{2} \\
&= 1 + \sum_{k=1}^{n-2} \left(\binom{n-k}{2} + n-k\right) \\
&= 1 + n(n-2) - \frac{(n-2)(n-1)}{2} +\sum_{k=1}^{n-2} \binom{n-k}{2} \\
&\overset{IH}{=} \frac{n(n-1)}{2} + \binom{n}{3} \\
&= \frac{n!}{2(n-2)!} + \binom{n}{3} \\
&= \binom{n}{2} + \binom{n}{3} \\
&= \binom{n+1}{3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Expressing complex numbers in exponential form I do not know why I am having so many issues with complex numbers. I am basically trying to teach myself, but keep doubting myself and getting very frustrated. I've no way of checking my answers so I'm just lost. So I want to write the following in polar form $re^{i\theta}$ (although I thought that was called exponential form?) and $-\pi < \theta \leq \pi$.
My first question is expressing $(\cos(\frac{2\pi}{9}) + i\sin(\frac{2\pi}{9}))^3$. Using De Moivre, this is $$\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3})$$ so it is $e^{i(\frac{2\pi}{3})}$ yes?
My next one is to express $\frac{2+2i}{-\sqrt{3} +i}$. So my attempt is to multiply by the complex conjugate and we get that this is $\frac{-\sqrt{3} +1}{2} + i\frac{-1 - \sqrt{3}}{2}$. Find $r$ which is the modulus which is $\sqrt2$ and $\theta$ is just something I cannot seem to find. I know if $z = a+ib$ then $\theta = \tan^{-1}(\frac{b}{a})$ but I'm stuck even with the $(\frac{b}{a})$ and am confusing myself of what quadrant to look in.
Lastly, I have no idea how to express the next one which is $\frac{4i}{3e^{(4+i)}}$.
Any help and direction at all is appreciated.
| You have $\frac {2+2i}{-\sqrt 3 + i}$ rather than rationalizing the denominator, and converting, convert numerator and denominator.
$2+2i = 2\sqrt2e^{\frac {\pi}{4}i}\\
-\sqrt 3 + i = 2e^{\frac {5\pi}{6}i}\\
\frac {2+2i}{-\sqrt 3 + i} = \sqrt 2\frac {e{\frac {\pi}{4}i}}{e{\frac {5\pi}{6}i}} = \sqrt2e^{(\frac {\pi}{4}i-\frac{5\pi}{6}i)}= \sqrt2e^{-\frac {7\pi}{12}i}$
Since you did get as far as:
$z = \frac {-\sqrt 3 + 1}{2} + \frac {-\sqrt 3 - 1}{2}i$
After doing it enough times in trig class I came to memorize $\cos \frac \pi{12} = \frac {\sqrt 6 + \sqrt 2}{4}, \sin \frac \pi{12} = \frac {\sqrt 6 - \sqrt 2}{4}$
Which might have helped.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Consider $T$ = $9 \times 99 \times 999 \times 9999 \times \cdots \times \underbrace{999....9}_{2015 \:nines}$ Consider $T$ = $9 \times 99 \times 999 \times 9999 \times
\cdots \times \underbrace{999....9}_{2015
\:nines}$
Find the last 3 digits of $T$
Advise: I wrote it down wrong the first time, it should be a product of "2015" numbers, i apologize about that, i realized my fault when i was travelling and i couldn't repare it in my cellphone.
My try
I know the last digit, i found it easily, but the struggle is with the others. I tried this:
$9 \times 99 \times 999 \times \cdots \times \underbrace{999 \ldots 9}_{2015
\:nines}$ $= 9 \times 9(11) \times 9(111) \times \cdots \times 9(\underbrace{111 \ldots 1}_{2015
\:ones})$
So $T$ = $9(1+11+111+ \cdots +\underbrace{111 \ldots 1}_{2015
\:ones})$
But from here i found nothing, any hints?
| Well, following from where you ended up:
$T = 9(1 \times 11 \times 111 \times ...)$
Notice that except for the first and second terms, the other 2013 terms are congruent to 111 (mod 1000).
So $T \equiv 9(1 \times 11\times 111^{2013}) \pmod {1000}$
We will use the property that $\varphi (n^m) = n^{m-1} \varphi (n)$
So $\varphi (10^3) = 10^2 \varphi (10) = 100 \times 4 = 400 $
Since $2013 = 400 \times 5 + 13$,
$9(1 \times 11\times 111^{2013} \equiv 9(11 \times 111^{13}) \equiv 9(11 \times 111^{10} \times 111^3) \equiv 9(11 \times (111^2 )^5 \times 1367631) \equiv 9(11 \times 12321^5 \times 631) \equiv 9(11 \times 321^5 \times 631) \equiv 9(11 \times 3408200705601 \times 631) \equiv 9(11 \times 601 \times 631) \equiv 9(4171541) \equiv 9(541) \equiv 4869 \equiv 869 \pmod {1000}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Prove that if $a$ and $b$ satisfy $2a+4b=1$, then $a^2+b^2\ge \frac {1}{20}$ Prove that if $a$ and $b$ satisfy $2a+4b=1$, then $a^2+b^2\ge \frac {1}{20}$
The only idea I have is that I could apply Cauchy-Schwartz but i don't see how, any hints?
| An alternative approach is to use AM-QM inequality as
$$\left(\frac{2a + b + b + b +b }{5}\right)^2 \le \frac{4a^2+b^2+b^2+b^2+b^2}{5} \implies a^2+b^2 \ge \frac{1}{20}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Find the coefficients in Euclid's algorithm $a(x)=3x^4-4x^3-11x^2+4x+9$
$b(x)=3x^3+5x^2+x-1$
I find the greatest common divisor:
$$
\frac {3x^4-4x^3-11x^2+4x+9}{3x^3+5x^2+x-1}=x-3 \;\ \;\ mod= 3x^2+8x+6
$$
$$
\frac {3x^3+5x^2+x-1}{3x^2+8x+6}=x-1 \;\ \;\ mod= 3x+5
$$
$$
\frac {3x^2+8x+6}{3x+5}=x+1 \;\ \;\ mod= 1
$$
Greatest common divisor $= 1$
I am looking for coefficients using the extended Euclidean algorithm, table:
$3x^4-4x^3-11x^2+4x+9 \;\ \;\ \;\ \;\ 1 \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ 0$
$3x^3+5x^2+x-1 \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ 0 \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ 1$
$3x^2+8x+6 \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ 1 \;\ \;\ \;\ \;\ -x+3$
$3x+5 \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ -x+1 \;\ \;\ \;\ \;\ \;\ \;\ ?$
$1 \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ $?$ \;\ \;\ \;\ \;\ \;\ \;\ \;\ $?$ $
Checking:
$ $a(x)$(3x^4-4x^3-11x^2+4x+9)+$b(x)$(3x^3+5x^2+x-1)=1$
| In general, $a = bq+r$, so $1 \cdot a + (-q) \cdot b = r$. To continue the calculations, we introduce the notation $(\alpha,\beta)$ to denote $\alpha \cdot a + \beta \cdot b$, so that it's convenient to calculate the linear combinations of $a$ and $b$. Therefore, $r$ becomes $(1,-q)$.
In the Extended Euclidean Algorithm, since $r_{i+1} = r_{i-1} - q_ir_i$, we have
$$\begin{aligned}
(\alpha_{i+1},\beta_{i+1}) &= (\alpha_{i-1},\beta_{i-1}) - q_i(\alpha_i,\beta_i) \\
&= (\alpha_{i-1}-q_i\alpha_i,\beta_{i-1}-q_i\beta_i).
\end{aligned}$$
The follow table illustrates the calculations for $\alpha_{i+1}$ and $\beta_{i+1}$ from the last two rows.
\begin{array}{|r|r|r|r|}\hline
\text{index} & \text{quotient } q_{i-1} & \hfill \text{remainder } r_i \hfill & \alpha_i & \beta_i \\ \hline
i-1 & q_{i-2} & r_{i-1} & \color{red}{\alpha_{i-1}} & \color{red}{\beta_{i-1}} \\ \hline
i & q_{i-1} & r_i & \color{blue}{\alpha_i} & \color{blue}{\beta_i} \\ \hline
i+1 & \color{blue}{q_i} & r_{i+1} & \color{red}{\alpha_{i-1}}-\color{blue}{q_i\alpha_i} & \color{red}{\beta_{i-1}}-\color{blue}{q_i\beta_i} \\ \hline
\end{array}
From your (correct) calculations for the Euclidean Algorithm, the first few rows of the following table are fixed.
\begin{array}{|r|r|r|r|}\hline
\text{index} & \text{quotient } q_{i-1} & \hfill \text{remainder } r_i \hfill & \alpha_i & \beta_i \\ \hline
1 & & 3x^4-4x^3-11x^2+4x+9 & 1 & 0 \\ \hline
2 & & 3x^3+5x^2+x-1 & 0 & 1 \\ \hline
3 & x-3 & 3x^2+8x+6 & 1 & -x+3 \\ \hline
4 & x-1 & 3x + 5 & -x+1 & x^2-4x+4\\ \hline
5 & x+1 & 1 & x^2 & -x^3+3x^2-x-1\\ \hline
\end{array}
Checking:
$$\begin{aligned}
& x^2 (3x^4-4x^3-11x^2+4x+9) + (-x^3+3x^2-x-1)(3x^3+5x^2+x-1) \\
&= 3x^6-4x^5-11x^4+4x^3+9x^2 + (-3x^6+4x^5+11x^4-4x^3-9x^2+1) \\
&= 1
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Solving without squaring If $\sec \theta - \csc \theta = \dfrac{4}{3}$, then prove that $\theta= \dfrac{1}{2}\arcsin {\dfrac{3}{4}}$
I have tried really hard to solve this without squaring both sides of the equation but it seems next to impossible.
The closest thing I reached is $\sin \theta = \dfrac{3(1- 2\sin^2(\theta/2))}{8\sin^2(\theta/2)- 1}$
I also obtained $\sin 2\theta = \dfrac{-3}{2}\dfrac{(1-\tan\theta/2)^2}{1+\tan^2(\theta/2)}$
| $$\implies3(\sin\theta-\cos\theta)=4\sin\theta\cos\theta=2\sin2\theta$$
$$\iff3\sqrt2\sin\left(\theta-\dfrac\pi4\right)=2\cos2\left(\theta-\dfrac\pi4\right)=2\left(1-2\sin^2\left(\theta-\dfrac\pi4\right)\right)$$
$$\iff4\sin^2\left(\theta-\dfrac\pi4\right)+3\sqrt2\sin\left(\theta-\dfrac\pi4\right)-2=0$$
$$\sin\left(\theta-\dfrac\pi4\right)=\dfrac{-3\sqrt2\pm5\sqrt2}8=-\sqrt2\text{ or } \dfrac1{2\sqrt2}$$
As $\sin\left(\theta-\dfrac\pi4\right)\ge-1,$
$$\sin\left(\theta-\dfrac\pi4\right)=\dfrac1{2\sqrt2}$$
$$\implies\sin2\theta=\cos2\left(\theta-\dfrac\pi4\right)=1-2\sin^2\left(\theta-\dfrac\pi4\right)=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2640729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$ This is a question from Math Olympiad.
If $\{x,y,z\}\subset\Bbb{R}^+$ and if $$x^2 + xy + y^2 = 3 \\ y^2 + yz + z^2 = 1 \\ x^2 + xz + z^2 = 4$$ find the value of $xy+yz+zx$.
I basically do not know how to approach this question. Please let me know how to approach this question, and if you attach full explanation, I will appreciate it. Thanks.
| We can obtain $yz+zx+xy=2$ simply by finding the values of $x$, $y$, and $z$. We are given $$y^2+yz+z^2=1,\qquad(1)$$$$z^2+zx+x^2=4,\qquad(2)$$$$x^2+xy+y^2=3,\,\qquad(3)$$with $x,y,z>0$. Subtracting eqn $3$ from eqn $2$, and noticing the factor $z-y$ on the LHS, gives $$(z-y)(x+y+z)=1.$$Similarly, from eqn $2$ minus eqn $1$, we get$$(x-y)(x+y+z)=3.$$The ratio of the latter two equations yields $x-y=3(z-y)$, or$$x=3z-2y.$$ Now substituting for $x$ from the above equation into eqn $3$ yields$$y^2-3yz+3z^2=1.$$By subtracting this from eqn $1$, we find $2yz-z^2=0$, or$$z=2y$$(since $z\neq0$). Substitution for $z$ into eqn $1$ now gives $7y^2=1$. It follows that $y=1/\surd7$, because $y>0$, and then $z=2/\surd7$ and $x=4/\surd7$, giving$$yz+zx+xy=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 6,
"answer_id": 0
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Solve in $\mathbb{Z}$ the equation $x^4 + 1 = 2y^2$. Find all pairs of intergers $(x,y)$ such that $x^4 + 1 = 2y^2$.
I'm thinking of Gaussian integers, since the LHS can be factored in $\mathbb{C}$. But I don't know how to continue here.
| Certainly, $x$ must be odd. Note that $(x^2+1)^2 = x^4+2x^2+1 = 2y^2+2x^2$ and $(x^2-1)^2 = x^4-2x^2+1 = 2y^2-2x^2.$ Multiply these two equations together to get
$$(x^4-1)^2 = 4(y^4-x^4).$$
Since $x^4\equiv 1 \pmod{4}$, we can write
$$\left(\frac{x^4-1}{2}\right)^2 = y^4-x^4.$$
This implies a solution to the better-known Diophantine equation
$$X^4-Y^4 =Z^2.$$
This equation can be proved to have only trivial solutions by infinite descent, so we must have $x^4-1 = 0$. Hence $x=\pm 1$ which forces $y=\pm 1$.
For the descent (switching to lower case):
We have $z^2 + (y^2)^2 = (x^2)^2.$ By the usual construction of solutions to the Pythagorean equation, either $y^2=2mn$ or $y^2=m^2-n^2.$ If $y^2 = 2mn,$ then $m = u^2,$ and
$n = 2v^2.$ Then $z^2 = m^2 - n^2 = u^4 - v^4,$ and $u^4 = m^2 < m^2 + n^2 <
x^2 < x^4,$ so we have a smaller positive solution. If $y^2 = m^2 - n^2,$
$z = 2mn,$
and $x^2 = m^2 + n^2,$ then $x^2y^2 = m^4 - n^4$ which is a smaller solution, because $m^2 < x^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Derivative of natural log with chain rule. Is there a better way? I'm a bit stuck on taking these two derivatives:
$$h(x) = \ln(x + \sqrt{x^2-1})$$
\begin{align}
h'(x) &= \frac{1}{x + \sqrt{x^2 - 1}} \cdot \frac{d }{dx} (x + \sqrt{x^2 -1} )\\
&= \frac{1}{x + \sqrt{x^2 - 1}} \biggl(1 + \frac{2x}{2\sqrt{x^2 - 1}}\biggr)
\end{align}
I'm a bit stuck on how to simplify from here? Was there a simpler way somewhere?
Also this one is giving me problems. I am thinking of changing the log forms first?:
$$\ln \frac{(2y+1)^5}{\sqrt{y^2 + 1}}= \ln(2y+1)^5 - \ln \sqrt{y^2 +1}=5 \ln(2y+1) - \frac{1}{2}\,\ln (y^2 +1)$$
so:
$$G'y = \frac{5}{2y+1} \cdot 2 - \frac{2y}{2(y^2 + 1)} = \frac{10}{2y+1} - \frac{y}{2 (y^2 +1)}$$
Is that right?
| \begin{align*}
f(y)&=\ln\dfrac{(2y+1)^{5}}{\sqrt{y^{2}+1}}\\
&=5\ln(2y+1)-\dfrac{1}{2}\ln(y^{2}+1),
\end{align*}
so
\begin{align*}
f'(y)=\dfrac{10}{2y+1}-\dfrac{y}{y^{2}+1}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
On the convergence of the series $ \sum_n \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} $ I've got some problem with this series
$$
\sum_{n=1}^{\infty} \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)}
$$
I know that the series diverges by a comparison test, but I've got a lot of trouble to prove it. I've tried some algebraic tricks like
$$
a - b = \frac{a^{3} - b^{3}}{a^{2} + ab + b^{2}},
$$
where $ a = \sqrt[3]{(n^{3}+n)} $ and $ b = \sqrt[3]{(n^3-n)} $.
It gave me
$$
\sum_{n=1}^{\infty} \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} = \sum_{n=1}^{\infty} \frac{2n}{\sqrt[3]{(n^{3}+n)^{2}} + \sqrt[3]{n^{6} - n^{2}} + \sqrt[3]{(n^3-n)^{2}}}
$$
but for me it looks like a blind valley. I would appreciate every help, thank you.
| Let
$$a_n=\sqrt[3]{n^3+n}-\sqrt[3]{n^3-n}=\sqrt[3]n\left(\sqrt[3]{n^2+1}-\sqrt[3]{n^2-1}\right)$$
Defining $f(x)=\sqrt[3]x$ we can apply Mean Value Theorem in $[n^2-1,n^2+1]$.
$$f'(x)=\frac1{3\sqrt[3]{x^2}}$$
For some $\xi\in[n^2-1,n^2+1]$, we have
$$a_n=\frac23\sqrt[3]{\frac n{\xi^2}}\ge\frac23\sqrt[3]{\frac n{(n^2+1)^2}}\ge\frac23\sqrt[3]{\frac n{4n^4}}\ge\frac1{3n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Limit of the sequence $\frac{1}{n^2+1}+\frac{2}{n^2+2}+...+\frac{n}{n^2+n}$ As in the title, we have to calculate the limit $$\frac{1}{n^2+1}+\frac{2}{n^2+2}+...+\frac{n}{n^2+n}$$ when $n\rightarrow+\infty$. It is easy to see that $$\lim_{n\to\infty} \frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}=0$$ and $$\lim_{n\to\infty} \frac{n}{n^2+1}+\frac{n}{n^2+2}+...+\frac{n}{n^2+n}=1$$
I don't know how to calculate the limit. I am pretty sure that we somehow apply the sandwich rule, but I am afraid that there is something similary with $$\lim_{n\to\infty} \frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}=\frac{\pi^2}{6}$$
| Another idea is to identify the sum
$$s = \sum _{k=1}^n \frac{k}{k+n^2}$$
in the limit $n\to \infty$ as a Riemann integral.
In fact we can write
$$s = \frac{1}{n} \sum _{k=1}^n \frac{k}{n \left(\frac{k}{n^2}+1\right)}$$
For $n\to\infty$ we can neglect $\frac{k}{n^2}$ against unity:
$$s = \frac{1}{n} \sum _{k=1}^n \frac{k}{n} $$
which asymptotically goes to
$$s \to \int_0^1 x \, dx = 1/2$$
An obvious generalization of the sum gives immediately
$$s_3 = \sum _{k=1}^n \frac{k^2}{k^2+n^3}\to \int_0^1 x^2 \, dx = 1/3 $$
and more generally
$$s_m = \sum _{k=1}^n \frac{k^{m-1}}{k^{m-1}+n^m}\to \int_0^1 x^{m-1} \, dx = 1/m $$
The latter relation holds even for any real $m \gt 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Series convergence or divergence . $\sum_{n=1}^\infty \left(\frac{n^4}{n^5+7}\right)$ $$\sum_{n=1}^\infty \left(\frac{n^4}{n^5+7}\right)$$
I try this series by the comparison test with $a_n\le b_n,$ $a_n=\frac{n^4}{n^5+7}$ and $b_n=\frac{n^4}{n^5}=\frac{1}{n}$
then $b_n$ diverges, dose the series diverges ?
| As an alternative to the limit comparison test note that by binomial series
$$\frac{n^4}{n^5+7}=\frac{1}{n+\frac7{n^4}}=\frac1n\left(1+\frac7{n^5}\right)^{-1}\ge \frac1n\left(1-\frac7{n^5}\right)=\frac1n-\frac7{n^6}$$
thus $\sum_{n=1}^{\infty} \left(\frac{n^4}{n^5+7}\right)$ diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
What can $\log_e(x) + \log_e(1+x) =0$ be written as? The equation $\log_e(x) + \log_e(1+x) =0$ can be written as:
a) $x^2+x-e=0$
b) $x^2+x-1=0$
c) $x^2+x+1=0$
d) $x^2+xe-e=0$
I tried differentiating both sides, then it becomes $\frac{1}{x}+\frac{1}{1+x}=0$, but I dont get any of the answers.
| Recall that $$\log_e x + \log_e(x+1) = \log_e\Big( x(x+1)\Big) = \log_e(x^2+x)$$ and that $$ \text{if } \log_e(x^2+x) = 0 \text{ then } x^2+x = e^0. $$
| {
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"url": "https://math.stackexchange.com/questions/2650628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question about $\lim_{n\rightarrow\infty} (\sqrt{n} - \sqrt{n - n^c})$ and $\lim_{n\rightarrow\infty} (\sqrt[3]{n + n^c} - \sqrt[3]{n})$ I have following exercises:
1) Find $\lim_{n\rightarrow\infty} (\sqrt{n} - \sqrt{n - n^c})$, $0\leq c\leq 1$
hint: make a case distinction for $c < \frac{1}{2}, c = \frac{1}{2}, c > \frac{1}{2}$
2) Find $\lim_{n\rightarrow\infty} (\sqrt[3]{n + n^c} - \sqrt[3]{n})$, $0\leq c\leq 1$
hint: make a case distinction for $c < \frac{2}{3}, c = \frac{2}{3}, c > \frac{2}{3}$
My question: without the hints, how could I derive the values $\frac{1}{2}$ and $\frac{2}{3}$ on my own?
| I like to use series for these. I will do the second one.
Cases $c=0,1$ are trivial, so assume $0<c<1$.
$$
\sqrt[3]{n+n^c} = (n+n^c)^{1/3} = n^{1/3}(1+n^{c-1})^{1/3}
$$
But $n^{c-1} \to 0$, so (Taylor series or Newton's binomial series)
$$
(1+n^{c-1})^{1/3} = 1+\frac{1}{3} n^{c-1}-\frac{1}{9} n^{2c-2}+\dots
\\
\sqrt[3]{n+n^c}
= n^{1/3}+\frac{1}{3} n^{c-2/3}-\frac{1}{9} n^{2c-5/3}+\dots
\\
\sqrt[3]{n+n^c} - \sqrt[3]{n} = \frac{1}{3} n^{c-2/3}-\frac{1}{9} n^{2c-5/3}+\dots
$$
Now we can see where the $2/3$ comes from. If $c>2/3$, then the leading term has a positive power on $n$, so we get limit $\infty$. If $c<2/3$ then all terms have negative exponent, so we get limit $0$. And if $c=2/3$, the leading term is $\frac{1}{3}$ and all remaining terms have negatvie exponent, so we get limit $\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2651783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove $15\mid 4^{2n}-1$ without induction
Prove without induction that $\forall n \in\mathbb Z$, $15\mid4^{2n}-1$.
$4^{2n} = (4^2)^n = 16^n$.
If $n=1,$ then $(4^2)^n-1=15$, and for $n=2$, it is $255$, which is divisible by $15$.
Using the congruence arithmetic, any $n$ in $4^{2n}-1$ can be expressed as a product of prime factors. Also, there is only one even prime, and any odd prime will be having for any positive integer $k$, an addition of $1+ k.$(even terms).
Say, for $n=5$, with $5=2^2+1$.
And it already proved that any even power of $n$ leads $4^{2n}-1$ to be a multiple of $15.$
edit Based on the selected answer, have observations, that are possibly implied in that answer, but not explicitly stated.
$4^{2n} -1 = (2^n -1)(2^n+1)(2^{2n}+1) = (2^n -1)(2^n+1)(4^n+1)$
Need consider both even and odd cases separately, for values w.r.t. two modulo $3,5$.
(A) $n$ is even :
(i) check w.r.t. modulo $3$
(a) $ 2^n-1 :: 2\equiv -1 \pmod 3 \implies 2^n \equiv 1 \pmod 3 \implies (2^n-1) \equiv 0 \pmod 3$
(b) $ 2^n+1 :: 2^n +1 \equiv 2 \pmod 3$
(c) $ 4^n+1 :: 4 \equiv 1 \pmod 3 \implies 4^n \equiv 1 \pmod 3$
Finding the product of the three terms w.r.t. modulo $3$, is :
$(\equiv 0 \pmod 3)(\equiv 2 \pmod 3)(\equiv 1 \pmod 3)$
As the first term of the product is a factor of $3$, so divisible by $3$.
(ii) check w.r.t. modulo $5$
(a) $ 2^n-1$ :: Take $n = 2n', 2^2 \equiv -1 \pmod 5\implies$$ 4^{n'} \equiv 1 \pmod 5 $ $\implies (2^n-1) \equiv 0 \pmod 5$
(b) $ 2^n+1$ :: Take $n = 2{n'}, 4^{n'} +1 \equiv 2 \pmod 5$
(c) $ 4^n+1 :: 4 \equiv -1 \pmod 5 \implies 4^n \equiv 1 \pmod 5$
Finding the product of the three terms w.r.t. modulo $5$, is :
$(\equiv 0 \pmod 5)(\equiv 2 \pmod 5)(\equiv 1 \pmod 5)$
As the first term of the product is a factor of $5$, so divisible by $5$.
(B) $n$ is odd :
(i) check w.r.t. modulo $3$
(a) $ 2^n-1 :: 2\equiv -1 \pmod 3 \implies 2^n \equiv -1 \pmod 3 \implies (2^n-1) \equiv -2 \pmod 3$
(b) $ 2^n+1 :: 2^n +1 \equiv 0 \pmod 3$
(c) $ 4^n+1 :: 4 \equiv 1 \pmod 3 \implies 4^n \equiv 1 \pmod 3$
Finding the product of the three terms w.r.t. modulo $3$, is :
$(\equiv -2 \pmod 3)(\equiv 0 \pmod 3)(\equiv 1 \pmod 3)$
As middle term of the product is a factor of $3$, so divisible by $3$.
(ii) check w.r.t. modulo $5$
(a) $ 2^n-1$ :: $2 \equiv 2 \pmod 5\implies 2^n \equiv ........$
(b) $ 2^n+1$ :: $2 \equiv 2 \pmod 5\implies 2^n \equiv ........$
(c) $ 4^n+1 :: 4 \equiv -1 \pmod 5 \implies 4^n +1\equiv 0 \pmod 5$
Finding the product of the three terms w.r.t. modulo $5$, is :
$(....)(...)(\equiv 0 \pmod 5)$
As the last term of the product is a factor of $5$, so divisible by $5$.
Don't know how to handle the last case's , sub cases (a), (b) for checking w.r.t. modulo $5$.
| $4 \equiv 1 \mod 3$. So $4^k - 1\equiv 1-1 \equiv 0 \mod 5$ for all $k$. So all $4^k-1$ and all $4^{2n} -1$ are divisible by $3$.
$4 \equiv -1 \mod 5$. So $4^{2n} -1 \equiv (-1^n)^2 - 1 \equiv 1 - 1\equiv 0 \mod 5$ for all even $2n$ so all $4^{2n} -1$ are divisible by $5$.
So $4^{2n} -1$ is divisible by $3*5=15$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Factoring High School Level Olympiad Problem Factor $x^2 - 3xy + 2y^2 + x -8y - 6$
Attempt at a solution:
I have factored these and don't know how to continue...
$x^2-3xy +2y^2 = (x - y) (x-2y)$
$x^2 + x -6 = (x + 3) (x - 2)$
$2y^2 - 8y + 6 = 2 (y - 3)(y - 1)$
| Since the polynomial is of degree $2$, we can use the
well established "tool-set" for the study of Quadrics
or Conic Sections.
So
$$
\eqalign{
& Q(x,y) = x^{\,2} - 3xy + 2y^{\,2} + x - 8y - 6 = \cr
& = \left( {x,y,1} \right)^T \left( {\matrix{
1 & { - 3/2} & {1/2} \cr
{ - 3/2} & 2 & { - 4} \cr
{1/2} & { - 4} & 6 \cr
} } \right)\left( {\matrix{
x \cr
y \cr
1 \cr
} } \right) \cr}
$$
But the determinant of the matrix defining the Conic
$$
{\bf A}_{\,Q} = \left( {\matrix{
1 & { - 3/2} & {1/2} \cr
{ - 3/2} & 2 & { - 4} \cr
{1/2} & { - 4} & 6 \cr
} } \right)
$$
is not null, which means that the conic is not degenerate
and thus $Q(x,y)$ cannot be factored, not even in the complex field.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve $3\sin^2 x - \cos^2 x - 2 =0$
Find all the angles between $0$ and $360^\circ$ that satisfy $$3\sin^2 x - \cos^2 x - 2 =0$$
My attempt -
$3\sin^2 x - (1-\sin^2x) - 2 =0$
$ 3 \sin^2 x + \sin^2 x = 3 $
$4\sin^2 x = 3 $
$ \sin x= \frac{\sqrt{3}}{2} $
I found that $x= 60,120 $
Why is the answer for this $60,120,240,300$ ? How do I find 240 and 300?
| Remember that when you take the square root of both sides you get $\pm$ that number. This means that $\sin(x) = \pm \frac{\sqrt{3}}{2}$. How does that change the solution?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
General formula for the $n^{\rm th}$ Smarandache number The $n^{\rm th}$ Smarandache number is defined as the concatenation of the first $n$ positive integers. The first few Smarandache numbers are thus the following: $1$, $12$, $123$, $\ldots$
For $n<9$, we can denote the $n^{\rm th}$ Smarandache number as:
$$
\sum_{k=1}^{n}10^{n-k}k = \frac{1}{81}(10^{n+1}-9n-10).
$$
Is there a general formula for any positive integer $n$?
| let $S_a$ be the $a^{th}$ smarandache number and $S_0=0$ $$\\$$
let $j\in\Bbb{N}\,|\,j$ is the smallest number that satisfies the inequality $10^j>n$.
$$S_n=S_{(10^{j-1}-1)}\cdot10^{j\cdot(n+1-10^{j-1})}+\frac{10^{j\cdot(n+2-10^{j-1})}-(10^j-1)(n+1-10^{j-1})-10^j}{(10^j-1)^2}+\frac{(10^{j-1}-1)(10^{j\cdot(n+1-10^j)}-1)}{10^j-1}$$
This formula is completed in $\lceil \log_{10}(n+1)\rceil$ steps.$$\\$$
For example $S_{101}$ is completed in three steps as shown below:
$$Step\,\,1:S_{101}=S_{99}\cdot10^{6}+\frac{10^9-(10^3-1)\cdot2-10^3}{999^2}+\frac{(10^2-1)(10^6-1)}{10^3-1}$$
$$Step\,\,2:S_{99}=S_{9}\cdot10^{180}+\frac{10^{182}-(10^2-1)\cdot90-10^2}{99^2}+\frac{(10^1-1)(10^{180}-1)}{10^2-1}$$
When n=9, $S_0$ is in the first term of the general expression so the first term equals zero. The last term of the general formula also equals zero when n=9. so,
$$Step\,\,3:S_{9}=\frac{10^{10}-81-10}{9^2}$$
Note: There are a few super-scripts that are smaller than the others and difficult to read. These super-scripts are $j$ or $j-1$.
| {
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"url": "https://math.stackexchange.com/questions/2658604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Number of solution set such that $a \geq b \geq c \geq d \geq 0$; $a, b, c, d \in \mathbb{Z}$ and $a + b + c + d = 10$ Given $(a, b, c, d)$ is a set of integers and $a \geq b \geq c \geq d \geq 0$. Find the number of solution sets for $a + b + c + d = 10$.
(This is problem from a competition, the answer key says its $23$, wherein my answer is $455$)
Solution
$x_1 + x_2 +1 + x_3 + 2 + x_4 + 3 = 10$
$x_1 + x_2 + x_3 + x_4 = 16$
$_{15}C_3$ = 455
| The solutions are:
$$\begin{align}(d,c,b,a)=
&(0,0,0,10) \cdots (0,0,5,5) \Rightarrow 6 \\
&(0,1,1,8) \cdots (0,1,4,5) \Rightarrow 4 \\
&(0,2,2,6), (0,2,4,4) \Rightarrow 3 \\
&(0,3,3,4) \Rightarrow 1 \\
&(1,1,1,7) \cdots (1,1,4,4) \Rightarrow 4 \\
&(1,2,2,5), (1,2,3,4) \Rightarrow 2 \\
&(1,3,3,3) \Rightarrow 1 \\
&(2,2,2,4),(2,2,3,3) \Rightarrow 2 \end{align}$$
Hence, there are $23$ solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to go from $u^{2/3}-5u^{1/3}+7=0$ to $u^2-20u+343=0$ I have absolutely no clue how to do this - any hints would be appreciated.
So far, I have tried multiplying each term by a power of u, but this leads nowhere...
| Cube both sides of $u^{2/3}=5u^{1/3}-7$ to get
$$
\begin{align}
u^2
&=125u\overbrace{-525u^{2/3}+735u^{1/3}}^{-105u}-343\\
&=20u-343
\end{align}
$$
where $-105u=-105\overbrace{\left(5u^{1/3}-7\right)}^{u^{2/3}}u^{1/3}$
Therefore,
$$
u^2-20u+343=0
$$
In General
Suppose we had $u^{2/3}=au^{1/3}+b$, then
$$
\begin{align}
u
&=au^{2/3}+bu^{1/3}\\
&=a\left(au^{1/3}+b\right)+bu^{1/3}\\
&=\color{#C00}{\left(a^2+b\right)u^{1/3}+ab}\\
u^2
&=\left(a^2+b\right)^2u^{2/3}+2ab\left(a^2+b\right)u^{1/3}+(ab)^2\\
&=\left(a^2+b\right)^2\left(au^{1/3}+b\right)+2ab\left(a^2+b\right)u^{1/3}+(ab)^2\\
&=\color{#C00}{\left(a^2+b\right)\left(a^3+3ab\right)u^{1/3}+b\left(a^4+3a^2b+b^2\right)}
\end{align}
$$
Then, we can combine the two red formulas to get
$$
u^2=\left(a^3+3ab\right)u+b^3
$$
Complex Approach
Let $v=u^{1/3}$ and $\omega=e^{2\pi i/3}$. Then note that
$$
\left(v^2-av-b\right)\left(v^2\omega^2-av\omega-b\right)\left(v^2\omega-av\omega^2-b\right)=v^6-\left(a^3+3ab\right)v^3-b^3
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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How are these steps in this integral done? $$\int\frac{du}{u^{2}-2}=\frac{1}{2\sqrt{2}}\int\frac{du}{u-\sqrt{2}}-\int\frac{du}{u+\sqrt{2}}=\frac{1}{2\sqrt{2}}\ln\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|$$
How to do the middle step?
| $$\int\frac{du}{u^{2}-2}=\frac{1}{2\sqrt{2}} \left(\int\frac{du}{u-\sqrt{2}}-\int\frac{du}{u+\sqrt{2}}\right )=\frac{1}{2\sqrt{2}}\ln\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|$$
Because of this...
$$\frac 1 {a^2-b^2}=\frac 1 {2b} \left (\frac 1 {a-b}-\frac 1 {a+b} \right)$$
$$\int \frac {da} {a^2-b^2}=\frac 1 {2b} \left (\int \frac 1 {a-b}da-\int \frac 1 {a+b}da \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int \sqrt{1+x^2}dx$ I was trying to do this integral
$$\int \sqrt{1+x^2}dx$$
I saw this question and its' use of hyperbolic functions. I did it with binomial differential method since the given integral is in a form of $\int x^m(a+bx^n)^p\,dx$ and I spent a lot of time on it so I would like to see if it can be done this way and where did I go wrong.
$$\int(1+x^2)^\frac{1}{2}dx$$
$m=0$, $n=2$, $p=\frac{1}{2}$
Because $\frac{m+1}{n}+p \in \mathbb Z$ I used substitution $x^{-2}+1=t^2$.
From there I got:
$$-\frac{dx}{x^3}=t\,dt$$
$$x=\frac{1}{\sqrt {t^2-1}}$$
$$t=\frac{\sqrt{1+x^2}}{x}$$
I expanded the original with $x^4$:
$$\int \frac{x^4\sqrt{1+x^2}dx}{x^4}=\int \left(\frac{1}{\sqrt{t^2-1}}\right)^4t(-tdt)=\int\frac{-t^2dt}{(t^2-1)^2}$$
Now I used partial integration:
$u=t$,
$du=dt$,
$dv=\frac{-tdt}{(t^2-1)^2}$,
$v=\frac{1}{2(t^2-1)}$
Then
$$\begin{align}
\int\frac{-t^2dt}{(t^2-1)^2}
&=\frac{t}{2(t^2-1)}-\frac{1}{2}\int \frac{dt}{t^2-1}= \\
&=\frac{t}{2(t^2-1)}-\frac{1}{2}\frac{1}{2}\ln\frac{t-1}{t+1}= \\
&=\frac{\frac{\sqrt{1+x^2}}{x}}{2\left(\left(\frac{\sqrt{1+x^2}}{x}\right)^2-1\right)}-\frac{1}{4}\ln\frac{\frac{\sqrt{1+x^2}}{x}-1}{\frac{\sqrt{1+x^2}}{x}+1}= \\
&=\frac{x\sqrt{1+x^2}}{2}-\frac{1}{4}\ln\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}+x}=\\
&=\frac{x\sqrt{1+x^2}}{2}-\frac{1}{4}\ln\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}+x}\cdot\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}-x}=\\
&=\frac{x\sqrt{1+x^2}}{2}-\frac{1}{4}\ln(\sqrt{1+x^2}-x)^2=\\
&=\frac{x\sqrt{1+x^2}}{2}-\frac{1}{2}\ln(\sqrt{1+x^2}-x)+C
\end{align}$$
The solution is
$$\frac{x\sqrt{1+x^2}}{2}+\frac{1}{2}\ln(x+\sqrt{1+x^2})+C$$
My solution looks very similar, so where did I go wrong?
| Let $ x = \sinh \theta$, then set $$I=\int \sqrt{1+x^2}dx$$
Whence
\begin{align}
I &= \int \sqrt{1+x^2}dx \\
&= \int \sqrt{1+\sinh^2 \theta}\cosh \theta d \theta \\
&= \int \cosh^2 \theta d \theta \\
&= \frac{1}{2}\int 1 + \cosh 2 \theta d \theta \\
&= \frac{1}{2}\theta+\frac{1}{4}\sinh 2 \theta + c\\
&= \frac{1}{2}\sinh^{-1}x+\frac{1}{2}x\sqrt{1+x^2}+c\\
&=\frac{1}{2}\ln(x + \sqrt{1+x^2})+\frac{1}{2}x\sqrt{1+x^2}+c
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2660140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Does $1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{4}+\frac{1}{4^2}-\dots$ converge?
Does
$$1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{4}+\frac{1}{4^2}-\dots$$
converge?
What I have tried:
The series diverges since $$1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{4}+\frac{1}{4^2}-\dots=(1-\frac{1}{2})+(\frac{1}{2^2}-\frac{1}{3})+(\frac{1}{3^2}-\frac{1}{4})+\dots=\sum(\frac{1}{k^2}-\frac{1}{k+1})$$ and it can be shown that $$\sum \frac{1}{k+1}$$ diverges.
However, I am not sure at all if this is correct or not. Am I allowed to re-group the terms like that? I think there would be a more elegant way to do this ...
| You can prove directly that the series diverges by doing, for $k\geq3$,
$$
\frac1 {k^2}-\frac1 {k+1}=\frac {k+1-k^2}{(k+1)k^2}<\frac {-k^2}{2 (k+1)k^2}=-\frac12\,\frac1 {k+1}
$$
and the series diverges by comparison.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show the polynomial $p(x,y) =y^3+xy^2+(x^2+x)y+(4x^3+2x)$ is irreducible in $\mathbb{Z}[x, y]$ I've thought about $p(x,y)$ being in $\mathbb{Z}[x][y]$ and tried to use Eisenstein's criterion, but I haven't had any luck. The best I've got is, we can suppose $p(x,y)$ is reducible and write it as
$$p(x,y)=(y+a)(y^2+by+c) = y^3 + (a+b)y^2 + (ab+c)y + ac$$
for some $a,b,c \in \mathbb{Z}[x]$.
Comparing coefficients we have
\begin{align}
&a+b=x \\
&ab+c = x^2+x \\
&ac = 4x^3+2x = 2x(2x^2+1)
\end{align}
From the last equation we have $a = 2x$ and $c=2x^2+1$ or alternatively $a = 2x^2+1$ and $c=2x$.
If $a=2x$, $c=2x^2+1$ then from the first equation, $2x+b=x$ so $b=-x$. From the second equation, $-2x^2+x-1=x^2+x$, but this doesn't hold for all $x$...
If $a=2x^2+1$, $c=2x$ then from the first equation, $2x^2+1+b=x$ so $b=-2x^2+x-1$. From the second equation, $(2x^2+1)(-2x^2+x-1) = x^2+x$, but again the l.h.s. and r.h.s. don't match since without even computing we see the l.h.s. is degree $4$.
Is there a better way to prove this?
| Take $x=3$ and use Eisenstein for $p=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solve: $\sqrt{x-\frac1x}-\sqrt{1-\frac1x}=1-\frac1x$ for $ x\neq0$ $$\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}{=1-\frac{1}{x}}, x\neq0$$
So my idea is that I moved the second sqrt to the RHS and left the first on the LHS. I then squared everything and ended up with $\frac{x^2-1}{x}=2-\frac{3}{x}+\frac{1}{x^2}$. I don't know what to do next? Can somebody give a hint?
| squaring two times and foctorizing we get
$$-\frac{(x-1)^2 \left(x^2-x-1\right)^2}{x^4}=0$$
if we do so
we get
$$x-\frac{1}{x}=(1-\frac{1}{x})^2+1-\frac{1}{x}+2(1-\frac{1}{x})\sqrt{1-\frac{1}{x}}$$
the square root is missing !!
and you must square again!!!!!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2666187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What is the minimal polynomial of A?
Let $f$ be an endomorphism of $\mathbb{R}^3$ with its matrix $A$ in the canonical basis $\mathcal{B}$ as
$$ A = \begin{pmatrix} 3 & 1 & 1 \\ -1 & 1 & - 1 \\ 0 & 0 & 1 \end{pmatrix}.$$
What is the minimal polynomial of $f$?
The characteristic polynomial of $f$ is:
$$P_{f}(X) = (1 - X)(X - 2)^2.$$
The minimal polynomial $m_{f}$ is the polynomial with the least degree that divides $P_{f}$, has the eigenvalues of $f$ as roots and $m_{f}(A) = 0$.
In this case, we have $m_{f}(X)= (1 - X)(2 - X)$ but $m_f(f) \neq 0 $.
How can I find the minimal polynomial of $A$ and what is the fastest method to determine it?
| The minimal polynomial and characteristic polynomial agree, which is equivalent to each eigenvalue occurring in exactly one Jordan block.
$$
R =
\left(
\begin{array}{rrr}
1 & 1 & 0 \\
-1 & -1 & 1 \\
-1 & 0 & 0
\end{array}
\right)
$$
$$
R^{-1} =
\left(
\begin{array}{rrr}
0 & 0 & -1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}
\right)
$$
and
$$ R^{-1} A R = J. $$
$$
J =
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{array}
\right)
$$
The direction that is actually useful is $R J R^{-1} = A.$ Useful for finding $e^A$ or $A^{100}$ or any $f(A)$ with $f$ single-variable analytic.
$$
\left(
\begin{array}{ccc}
1 & 1 & 0 \\
-1 & -1 & 1 \\
-1 & 0 & 0
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{array}
\right)
\left(
\begin{array}{rrr}
0 & 0 & -1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}
\right) =
\left(
\begin{array}{rrr}
3 & 1 & 1 \\
-1 & 1 & -1 \\
0 & 0 & 1
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2667415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solving this ODE near x=0 I could not get the correct solution on this ODE:
$$(1+x^2)y''-xy'-8y=0$$ around $x=0$.
I tried to solve this using MacLaurin expansion on $y$ and forming recurrence relations between the coefficients.
Can anyone teach me how to solve this by steps? Thank you.
| Using the ansatz
\begin{align}
y = \sum^\infty_{k=0} a_k x^k
\end{align}
then it follows
\begin{align}
y' = \sum^\infty_{k=1} k a_k x^{k-1} \ \ \text{ and } \ \ \ y''= \sum^\infty_{k=2} k(k-1) a_k x^{k-2}.
\end{align}
Plugging the power series back into the differential equation yields
\begin{align}
&(1+x^2)\sum^\infty_{k=2}k(k-1)a_k x^{k-2}-x\sum^\infty_{k=1}k a_k x^{k-1} -8 \sum^\infty_{k=0}a_k x^k \\
&= \sum^\infty_{k=0}(k+2)(k+1) a_{k+2}x^k +\sum^\infty_{k=2}k(k-1)a_k x^k-\sum^\infty_{k=1}ka_k x^k-8\sum^\infty_{k=0}a_k x^k\\
&= 2a_2+6a_3x+\sum^\infty_{k=2}\{(k+2)(k+1)a_{k+2}+k(k-1)a_k\}x^k -a_1 x -\sum^\infty_{k=2}ka_k x^k -8a_0-8a_1x\\
&-8\sum^\infty_{k=2}a_k x^k\\
&= (2a_2-8a_0)+(6a_3-9a_1)x+\sum^\infty_{k=2}\{(k+2)(k+1)a_{k+2}+(k^2-2k-8)a_k\}x^k=0.
\end{align}
So, it follows
\begin{align}
a_2 = 4a_0, \ a_3 = \frac{3}{2}a_1, \ a_{k+2} = -\frac{k-4}{k+1}a_k \ \text{ for } k\geq 2.
\end{align}
If $k$ is even then it follows
\begin{align}
a_{2(\ell + 1)} = -\frac{2\ell -4}{2\ell+1}a_{2\ell}= \frac{2\ell-4}{2\ell+1}\frac{2(\ell-1)-4}{2(\ell-1)+1}a_{2(\ell-1)}= (-1)^\ell\prod^{\ell}_{j=1}\frac{2j-4}{2j+1}a_2 = 4\left(\prod^{\ell}_{j=1}\frac{-2j+4}{2j+1}\right)a_0
\end{align}
for $\ell \geq 1$ and likewise when $k$ is odd we have
\begin{align}
a_{(2\ell+1)+2} = \frac{2\ell+1 -4}{2\ell+1+1}a_{2\ell+1} = \frac{3}{2}\left(\prod^{\ell}_{j=1}\frac{-2j+3}{2j+2}\right)a_1.
\end{align}
Finally, we see that
\begin{align}
y =&\ a_0+a_1x+a_2 x^2+a_3x^3 +\sum^\infty_{k=1} a_{2k+2}x^{2k+2}+\sum^\infty_{k=1} a_{2k+3}x^{2k+3}\\
=&\ a_0\left(1+4x^2+4\sum^\infty_{k=1}\left(\prod^{k}_{j=1}\frac{-2j+4}{2j+1}\right)x^{2k+2} \right)+a_1\left(x+\frac{3}{2}x^3+\frac{3}{2}\sum^\infty_{k=1}\left(\prod^{k}_{j=1}\frac{-2j+3}{2j+2}\right)x^{2k+3} \right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2669960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $\sin8\theta-\sin10\theta=\cot9\theta(\cos10\theta-\cos8\theta)$ $$\sin8\theta-\sin10\theta=\cot9\theta(\cos10\theta-\cos8\theta)$$
So far I've done
$$2\cos\left(\dfrac {18}2\right)\theta\sin\left(\dfrac {8-10}2\right)\theta$$
I am stuck, can anyone help?
Am I supposed to multiply them by $2$ to get rid of the $2$'s and then divide all by $\sin$?
| From the angle addition identities
$$\begin{align*}
\sin (a \pm b) &= \sin a \cos b \pm \sin b \cos a \\
\cos (a \pm b) &= \cos a \cos b \mp \sin a \sin b
\end{align*}$$
we find
$$\begin{align*}
\sin (a+b) - \sin(a-b) &= 2\sin b \cos a \\
\cos (a+b) - \cos(a-b) &= -2 \sin a \sin b.
\end{align*}$$
Then choosing $a = 9\theta$, $b = \theta$, we get
$$\begin{align*}
\sin 10\theta - \sin 8\theta &= 2 \sin \theta \cos 9\theta \\
\cos 10\theta - \cos 8\theta &= -2 \sin 9\theta \sin \theta.
\end{align*}$$
Consequently, $$\frac{\sin 8\theta - \sin 10\theta}{\cos 10\theta - \cos 8\theta} = \frac{-2\sin \theta \cos 9\theta}{-2 \sin 9\theta \sin \theta} = \cot 9\theta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2673102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to express $\theta$ in terms of $x$ where $3\sin(3\theta+x)=\frac{2.5}{\sin\theta}$? I tried to solve it by using compound angle formulas but in the end I could not leave $\theta$ alone.
It goes like this:
\begin{align}
& \frac{2.5}{3}=\sin(3\theta+x)\sin\theta \\[10pt]
& \sin(3\theta+x)=\sin(2\theta+\theta)\cos(x)+\sin(x)\cos(2\theta+\theta) \\[10pt]
= {} & [(\sin(2\theta)\cos(\theta)+\sin(\theta)\cos(2\theta))]\cos(x)+[(\cos(2\theta)\cos(\theta)-\sin(2\theta)\sin(\theta))]\sin(x)
\end{align}
Then I did a couple more steps but I couldn't solve it this way, is there any other way to solve it algebraically?
| your equation is equivalent to $$4\,\cos \left( x \right) \left( \sin \left( \theta \right) \right) ^
{2} \left( \cos \left( \theta \right) \right) ^{2}-\cos \left( x
\right) \left( \sin \left( \theta \right) \right) ^{2}+4\,\sin
\left( \theta \right) \sin \left( x \right) \left( \cos \left(
\theta \right) \right) ^{3}-3\,\sin \left( \theta \right) \sin
\left( x \right) \cos \left( \theta \right)
=2.5$$
you can Substitute
$$\sin(\theta)=\frac{2\tan(\frac{\theta}{2})}{1+\tan(\frac{\theta}{2})^2}$$
and $$\cos(\theta)=\frac{1-\tan(\frac{\theta}{2})^2}{1+\tan(\frac{\theta}{2})^2}$$
and then you can Substitute $$\tan(\frac{\theta}{2})=t$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2674973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\begin{vmatrix} xa&yb&zc\\ yc&za&xb\\ zb&xc&ya\\ \end{vmatrix}=xyz\begin{vmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{vmatrix}$ if $x+y+z=0$
If $x+y+z=0$, then prove that
$$
\begin{vmatrix}
xa&yb&zc\\
yc&za&xb\\
zb&xc&ya\\
\end{vmatrix}=xyz\begin{vmatrix}
a&b&c\\
c&a&b\\
b&c&a\\
\end{vmatrix}
$$
I can do it by Sarrus' law but how can I prove it by matrix operations without actually expanding the determinant ?
My Attempt
$$
\begin{vmatrix}
xa&yb&zc\\
yc&za&xb\\
zb&xc&ya\\
\end{vmatrix}=xyz\begin{vmatrix}
a&\frac{yb}{x}&\frac{zc}{x}\\
c&\frac{za}{y}&\frac{xb}{y}\\
b&\frac{xc}{z}&\frac{ya}{z}\\
\end{vmatrix}=xyz\begin{vmatrix}
a&-b-\frac{zb}{x}&-c-\frac{yc}{x}\\
c&-a-\frac{xa}{y}&-b-\frac{zb}{y}\\
b&-c-\frac{yc}{z}&-a-\frac{xa}{z}\\
\end{vmatrix}=xyz\begin{vmatrix}
a&b+\frac{zb}{x}&c+\frac{yc}{x}\\
c&a+\frac{xa}{y}&b+\frac{zb}{y}\\
b&c+\frac{yc}{z}&a+\frac{xa}{z}\\
\end{vmatrix}=xyz\begin{vmatrix}
a&b&c+\frac{yc}{x}\\
c&a&b+\frac{zb}{y}\\
b&c&a+\frac{xa}{z}\\
\end{vmatrix}+xyz\begin{vmatrix}
a&\frac{zb}{x}&c+\frac{yc}{x}\\
c&\frac{xa}{y}&b+\frac{zb}{y}\\
b&\frac{yc}{z}&a+\frac{xa}{z}\\
\end{vmatrix}\\
=xyz\bigg(\begin{vmatrix}
a&b&c\\
c&a&b\\
b&c&a\\
\end{vmatrix}+\begin{vmatrix}
a&b&\frac{yc}{x}\\
c&a&\frac{zb}{y}\\
b&c&\frac{xa}{z}\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&c\\
c&\frac{xa}{y}&b\\
b&\frac{yc}{z}&a\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&\frac{yc}{x}\\
c&\frac{xa}{y}&\frac{zb}{y}\\
b&\frac{yc}{z}&\frac{xa}{z}\\
\end{vmatrix}\bigg)\\
$$
I need to prove that the sum of last three terms is zero.
$$
\begin{vmatrix}
a&b&\frac{yc}{x}\\
c&a&\frac{zb}{y}\\
b&c&\frac{xa}{z}\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&c\\
c&\frac{xa}{y}&b\\
b&\frac{yc}{z}&a\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&\frac{yc}{x}\\
c&\frac{xa}{y}&\frac{zb}{y}\\
b&\frac{yc}{z}&\frac{xa}{z}\\
\end{vmatrix}=\begin{vmatrix}
a&b&\frac{yc}{x}\\
c&a&\frac{zb}{y}\\
b&c&\frac{xa}{z}\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&\frac{-zc}{x}\\
c&\frac{xa}{y}&\frac{-xb}{y}\\
b&\frac{yc}{z}&\frac{-ya}{z}\\
\end{vmatrix}\\
$$
Solution by expansion
$$
\Delta=\begin{matrix}
xa&yb&zc&xa&yb\\
yc&za&xb&yc&za\\
zb&xc&ya&zb&xc\\
\end{matrix}=xyz(a^3+b^3+c^3)-abc(x^3+y^3+z^3)
$$
We have $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0$ as $x+y+z=0$. Thus, $x^3+y^3+z^3=3xyz$
$$
\Delta=xyz(a^3+b^3+c^3)-abc(3xyz)=xyz(a^3+b^3+c^3-3abc)\\
=xyz\bigg[ a\big(a^2-bc\big)-b\big(ac-b^2\big)+c\big(c^2-ab\big) \bigg]\\
=xyz.\bigg[a\begin{vmatrix}
a&b\\
c&a\\
\end{vmatrix}-b\begin{vmatrix}
c&b\\
b&a\\
\end{vmatrix}+c\begin{vmatrix}
c&a\\
b&c\\
\end{vmatrix}\bigg]
=xyz\begin{vmatrix}
a&b&c\\
c&a&b\\
b&c&a\\
\end{vmatrix}
$$
| Possible direction, though not a proof.
Do note that curiously, the identity is only true for $n=3$, in higher dimensions it doesn't seem to work. Note also that the matrices do not actually have to be positive semi-definite matrices.
This result may be related to Oppenheim's inequality for Hadamard products, namely if $A,X$:
$$A=\begin{pmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{pmatrix},\quad X=\begin{pmatrix} x&y&z\\ y&z&x\\ z&x&y\\ \end{pmatrix}$$
Are two positive semi definite matrices, then:
$$\det(A\circ X)\ge (\det A)\prod_i x_{ii}$$
With equality if $A\circ X$ is singular. In your case, you have a circulant and anti-circulant matrices $A,X$ and the above theorem states that:
$$\det(A\circ X)\ge xyz\det(A)$$
$$\begin{vmatrix} xa&yb&zc\\ yc&za&xb\\ zb&xc&ya\\ \end{vmatrix} \ge xyz\begin{vmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{vmatrix}$$
You are looking for equality and we of course don't know that either $A$ or $X$ is actually PSD. Moreover we haven't used the condition $x+y+z=0$. Nevertheless, the overall form direction may be one worth pursuing.
Further Reading: The equality cases for the inequalities of Oppenheim and Schur for positive semi-definite matrices, Czechoslovak Mathematical Journal, Vol. 59 (2009), No. 1, 197–206
| {
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"url": "https://math.stackexchange.com/questions/2676190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Dealing with negative roots in code I have an equation in the following form that needs to be solved for $x$, with $x > 0$ and constants $a,b,c,d \geq 0$.
$\left(\frac{x - a}{b}\right)^2 + \left(\frac{x - c}{d}\right)^2 = 1$
From the following paper, which i am trying to implement:
https://grail.cs.washington.edu/projects/crowd-flows/78-treuille.pdf
Wolfram Alpha gives the following solution:
$x = \frac{\sqrt{b^2 d^2 (-a^2 + 2 a c + b^2 - c^2 + d^2} + a d^2 + b^2 c}{b^2 + d^2}$
http://www.wolframalpha.com/input/?i=((x+-+a)+%2F+b)+%5E+2+%2B+((x+-+c)+%2F+d)+%5E+2++%3D+1+solve+for+x
However this can result in the root of a negative number, and resulting in a complex number, which is not what the paper seems to want. I am unsure on how to solve this (in code) to get a real value for x.
So my question is, how do i solve this equation as desired by the paper?
PS. Limiting to real x still has the same problem
http://www.wolframalpha.com/input/?i=((x+-+a)+%2F+b)+%5E+2+%2B+((x+-+c)+%2F+d)+%5E+2++%3D+1+solve+for+real+x
| It's $$d^2(x^2-2ax+a^2)+b^2(x^2-2cx+c^2)=b^2d^2$$ or
$$(b^2+d^2)x^2-2(d^2a+b^2c)x+a^2d^2+b^2c^2-b^2d^2=0,$$ which gives
$$x=\frac{d^2a+b^2c\pm\sqrt{(d^2a+b^2c)^2-(b^2+d^2)(a^2d^2+b^2c^2-b^2d^2)}}{b^2+d^2}$$ or
$$x=\frac{d^2a+b^2c\pm\sqrt{d^4a^2+2b^2d^2ac+b^4c^2-b^2a^2d^2-b^4c^2+b^4d^2-d^4a^2-b^2c^2d^2+d^4b^2}}{b^2+d^2}$$ or
$$x=\frac{d^2a+b^2c\pm bd\sqrt{2ac-a^2-c^2+b^2+d^2}}{b^2+d^2}.$$
We need $b^2+d^2\neq0$ and $2ac-a^2-c^2+b^2+d^2\geq0$.
Thus, for the biggest root we obtain:
$$x=\frac{d^2a+b^2c+bd\sqrt{2ac-a^2-c^2+b^2+d^2}}{b^2+d^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Verifying Newton's Binomial Series for $(1+x)^{−2}$
Verify the consistency of Newton’s Binomial series for the function
$(1+x)^{−2}$ in two ways:
$(a)$ by multiplying the usual geometric series for $(1 +x)^{−1}$with
itself
$(b)$ by differentiation of the series for $(1 +x)^{−1}$
$(a)$ I have that Newton's Binomial series says that
$$(1+x)^p=1+px+\frac{p(p−1)}{2!}x^2+\frac{p(p−1)(p−2)}{3!}x^3+···$$
I have that the geometric series for $(1 +x)^{−1}$ is given by
$$\begin{align*}
\frac{1}{1+x}
&=\sum_{k=0}^{\infty}(-x)^k\\\\
&=-x^0+(-x)^1+(-x)^2+(-x)^3+...\\\\
&=1-x+x^2-x^3+...
\end{align*}$$
Thus
$$\begin{align*}
\left(\frac{1}{1+x}\right)^2
&=(1-x+x^2-x^3+...)(1-x+x^2-x^3+...)\\\\
&=(1-x+x^2-x^3+...)\\\\
& +(-x+x^2-x^3+...)\\\\
& +(x^2-x^3+...)\\\\
& +(-x^3+...)\\\\
&=1-2x+3x^2-4x^3+...
\end{align*}$$
Checking that this equals Newton's Binomial series
$$(1+x)^p=1+px+\frac{p(p−1)}{2!}x^2+\frac{p(p−1)(p−2)}{3!}x^3+···$$
where $p=-2$
$$\begin{align*}
(1+x)^{-2}
&=1-2x+\frac{-2(-3)}{2!}x^2+\frac{-2(-3)(-4)}{3!}x^3+···\\\\
&=1-2x+3x^2-4x^3+... \checkmark
\end{align*}$$
Is this a valid proof?
$(b)$
I have that $$\int_0^x\frac{dt}{1+t}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$$
but I'm not sure how to use this fact.
| Yes, first idea is right. More formally,
$$
\left(\sum_{i=0}^\infty (-1)^i x^i \right)
\left(\sum_{j=0}^\infty (-1)^j x^j \right)
= \sum_{n=0}^\infty \sum_{i=0}^n (-1)^i x^i (-1)^{n-i}x^{n-i}
= \sum_{n=0}^\infty \sum_{i=0}^n (-1)^n x^n
= \sum_{n=0}^\infty (n+1) (-1)^n x^n.
$$
On the second one, note that
$$
\frac{d}{dx} (1+x)^{-1} = -(1+x)^{-2},
$$
thus
$$
(1+x)^{-2}
= -\frac{d}{dx} (1+x)^{-1}
= -\frac{d}{dx} \sum_{i=0}^\infty (-1)^i x^i
= -\sum_{i=0}^\infty (-1)^i \frac{d}{dx} x^i
= \sum_{i=1}^\infty (-1)^{i-1}ix^{i-1}
= \sum_{i=0}^\infty (-1)^i (i+1)x^i
$$
| {
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"source": "stackexchange",
"question_score": "1",
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Range of $p$ in cubic equation
If $2x^3+px^2+qx+4=0$ has $3$ real roots, then what is the range of $p$? Here $p,q>0$.
My try: Let $\alpha,\beta,\gamma$ be the roots of $2x^3+px^2+qx+4=0$.$$
\begin{cases}
\alpha+\beta+\gamma=-\dfrac{p}{2}\\
\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{q}{2}\\
\alpha\beta\gamma=-2
\end{cases}
$$
Help me how to solve after that point.
| Since $p, q > 0$, the real roots to the equation $2x^3 + px^2 + qx + 4 = 0$ can only be negative. Replacing $x$ by $-x$ and dividing by $x$, the equation is equivalent to $2x^2 - px + q - \dfrac{4}{x} = 0$, where $x > 0$.
Define $f(x) = 2x^2 - px + q - \dfrac{4}{x}\ (x > 0)$, then$$
f'(x) =4x - p + \frac{4}{x^2},\ f''(x) = 4 - \frac{8}{x^3}.
$$
Thus the minimum of $f'(x)$ is $f'(\sqrt[3]{2}) = 6\sqrt[3]{2} - p$.
Case 1: $0 < p \leqslant 6\sqrt[3]{2}$. In this case, because $f'$ is positive on $(0, \sqrt[3]{2})$ and $(\sqrt[3]{2}, +∞)$, respectively, then $f$ is strictly increasing on $(0, \sqrt[3]{2})$ and $(\sqrt[3]{2}, +∞)$, and thus strictly increasing on $(0, +∞)$. Therefore, $f$ has no more than one real root.
Case 2: $p > 6\sqrt[3]{2}$. Now take $q = \sqrt[3]{2} p$, then $f(\sqrt[3]{2}) = 0$. Because $f'(\sqrt[3]{2}) < 0$, then there exist $x_1 < \sqrt[3]{2} < x_2$ such that $f(x_1) > f(\sqrt[3]{2}) = 0 > f(x_2)$. Note that$$
\lim_{x → 0^+} f(x) = -∞,\ \lim_{x →+∞} f(x) = +∞,
$$
then there exist $0 < x_3 < x_1$ and $x_4 > x_2$ such that $f(x_3) < 0 < f(x_4)$. Therefore, $f$ has roots on $(x_3, x_1)$ and $(x_2, x_4)$, respectively. Therefore, $f$ has three real roots.
Hence the range of $p$ is $(6\sqrt[3]{2}, +∞)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a series expansion for this integral, for large $\lambda > 0$ Given the integral $$Z(\lambda ) = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} dx e^{- \frac{x^2}{2!} - \frac{\lambda}{4!} x^4} $$ in quantum field theory, I need to find a series expansion for $Z(\lambda)$ when $\lambda >> 1$, of the form $$ Z_N ( \lambda) = \sum_{n=0}^N d_n \lambda^{-(2n+1)/4}. $$
I was thinking of writing $e^{- \frac{\lambda}{4!} x^4} \approx 1 - \frac{ \lambda}{4!} x^4$, because only the first two terms will dominate when $\lambda$ is large. But this doesn't seem to give me the right expansion when I plug it in the integral and do the Gaussian integration.
Anyone has any ideas/advice? Thanks in advance.
| Using the substitution $y=\frac{\lambda}{24}x^4$, we find that
\begin{align*}
&\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp\left( -\frac{x^2}{2} - \frac{\lambda}{24}x^4 \right) \, dx \\
&\hspace{4em} = \frac{1}{2\sqrt{\pi}} \int_{0}^{\infty} \frac{1}{y^{3/4}} \left(\frac{6}{\lambda}\right)^{1/4} \exp\left( -\sqrt{\frac{6}{\lambda}y} - y \right) \, dx \\
&\hspace{4em} = \frac{1}{2\sqrt{\pi}} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \left(\frac{6}{\lambda}\right)^{\frac{2n+1}{4}} \int_{0}^{\infty} y^{\frac{n}{2}-\frac{3}{4}} e^{-y} \, dx \\
&\hspace{4em} = \frac{1}{2\sqrt{\pi}} \sum_{n=0}^{\infty} (-1)^n \frac{\Gamma\left(\frac{n}{2}+\frac{1}{4}\right)}{n!} \left(\frac{6}{\lambda}\right)^{\frac{2n+1}{4}}.
\end{align*}
So you may let
$$ d_n = \frac{(-1)^n}{2\sqrt{\pi}} \cdot \frac{\Gamma\left(\frac{n}{2}+\frac{1}{4}\right)}{n!} \cdot 6^{\frac{2n+1}{4}}. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Evaluate $\lim_{x\to 2} \frac{\frac{1}{x} - \frac{1}{2}}{x-2}$ Evaluate:
$$\lim_{x\to 2} \frac{\frac{1}{x} - \frac{1}{2}}{x-2}$$
My attempt:
$$\lim _{ x\rightarrow 2 }{ \frac { \frac 1 x -\frac 1 2 }{ x-2 } } =\lim_{ x\rightarrow 2 }{ \frac { \frac { 2-x }{ 2x } }{ x-2 } } =-\lim_{ x\rightarrow 2 }{ \frac 1 {2x} =-\frac 1 4} $$
Is this correct?
| Yes. Your answer is correct. You can also do it by using L'Hospitals rule.
$$\begin{align}\lim_{x \to 2}\dfrac{\dfrac{1}{x}-\dfrac{1}{2}}{x-2}&=\lim_{x\to 2}\left(-\dfrac{1}{x^2}\right) \\ &=-\dfrac{1}{4}\end{align}\tag*{}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To find a remainder, when divided by $101$
There is a large number given in an exam. It is said to find a remainder when it is divided by $101$. The number is
$$1+11^{{(1+11+11^2+11^3+...+11^{100}~~)}^{(11^{101}~~+11^{102}~~+\cdots+11^{1000}~~)}}$$
We know $11$ and $101$ are itself prime then $(101,11)=1$ and $\phi(101)=100$
Now,
$$1+11+11^2+11^3+....+11^{100}\equiv r\mod 100 \\ \Rightarrow1+11+21+21*11+21^2+21^2*11+.....+(21^2)^{50}\equiv r \mod 100 \\ \Rightarrow (1+11+21+31+41+51+61+71+81-9)*10 \equiv r\mod 100 \\ \Rightarrow 0\equiv r\mod 100 \\ \therefore 0^{{(11^{101}+11^{102}+.....+11^{1000}~~)}}\equiv0 \mod 100 \\ \therefore 1+11^0\equiv1+1\equiv2 \mod 101$$
Is it correct? Any help is appreciated.
| Let's show $1+11+11^2+11^3+....+11^{100} \equiv 1 \mod 100$.
Easy to see that $11^{101}-1$ is divisible by $10$.
Not trivial, but not that hard to work that $11^{10} \equiv 1 \mod 100$.
With above two results we have
$$10(1+11+11^2+11^3+....+11^{100}) = 11^{101}-1 \equiv 11-1 \equiv 10 \mod 100$$
This yields $$1+11+11^2+11^3+....+11^{100} \equiv 1 \mod 100$$
$$1+11^{{(1+11+11^2+11^3+...+11^{100})}^{(11^{101}+11^{102}+.....+11^{1000})}} \\ \equiv 1 + 11^{1\mod 100} \\ \equiv 1 + 11 \\ \equiv 12 \mod 101$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2682105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Estimating error when calculating $\pi^2$ with $8 + \frac{8}{3^2} + \frac{8}{5^2} + \frac{8}{7^2} + \cdots$ While answering a CodeReview question (Approximating constant $\pi^2$ to within error), I noticed that when calculating the sum
$$8 + \dfrac{8}{3^2} + \dfrac{8}{5^2} + \dfrac{8}{7^2} + \dfrac{8}{9^2} + \cdots $$
and stopping at the smallest term larger than $\varepsilon$, the difference to $\pi^2$ seems very close to $\sqrt{2\varepsilon}$.
With $$n = \left\lfloor\frac{\sqrt{\frac{8}{\varepsilon}} - 1}{2} \right\rfloor,$$
it looks like:
$$\sum_{i=0}^n \frac{8}{(2i+1)^2} \approx \pi^2 - \sqrt{2\varepsilon}.$$
For example, with $\varepsilon = 10^{-10}$:
$n = 141421$
$\sum = 9.86959025 \dots$
$\pi^2 - \sum \approx 1.4142170*10^{-5}$
$(\pi^2 - \sum)^2 \approx 2.0000099*10^{-10} \approx 2\varepsilon$
I don't think it's a coincidence, but I don't know where to begin to link
$$\sum_{i=n+1}^\infty \frac{8}{(2i+1)^2}$$ to $\sqrt{2\varepsilon}$.
Wolfram Alpha expresses this sum in terms of a polygamma function but I don't know anything about it.
Is the approximation correct? Is there any simple way to prove it?
| Recall from calculus the integral test:
$$\displaystyle\sum a(n)\text{ converges if and only if }\int a(x)\text{ converges}$$ where the sequence $a(n)=a_n>0$ and $a_n$ is decreasing, and the function $a(x)$ is continuous.
From the proof of this statement, we can deduce the inequality: if $\sum_{n=1}^\infty a_n=L$, then $$\left|L-\sum_{i=0}^n a_n\right|\leq\int_{n}^\infty a(x)\,dx,$$(see here for help with providing visual intuition as to why the inequality is true) where the notation $a(x)$ is used to represent $a_n$ with the $n$ replaced by $x$. Thus, performing the necessary calculations, we find
\begin{align*}
\int_{n}^\infty\frac{8}{(2x+1)^2}\,dx&=\frac{4}{2n+1}\\
&\leq\frac{4}{\sqrt{\frac{8}\varepsilon}-1}\\
&=\frac{1}{\sqrt{\frac{1}{2\varepsilon}}-\frac14}\\
&=\frac{\sqrt{2\varepsilon}}{1-\frac{\sqrt{2\varepsilon}}{4}}\\
&\approx \sqrt{2\varepsilon},\end{align*}
where I've taken $$n=\left\lfloor\frac{\sqrt{\frac{8}{\varepsilon}}-1}{2}\right\rfloor.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2682525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Finding $a$ for which definite Integral is minimum
If $\displaystyle f(a)=\int^{\infty}_{0}\frac{x^a}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}$is minimum.Then real value of $a$ is
Try: $$f(a)=\int^{\infty}_{0}\frac{x^{a-3}}{2(x^6+x^{-6})+4(x^2+x^{-2})+3(x+x^{-1})+5}dx$$
given $x>0$. So using A.M$\geq$ G.M,
$x^k+x^{-k}\geq 2$ for $k=1,2,6$
Could some help me to solve it, Thanks
| $$f'(a)=\int_{0}^{+\infty}\frac{x^a\log(x)}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}\,dx $$
equals
$$ f'(a)=\int_{0}^{+\infty}\frac{x^{a-3}\log(x)}{2(x^3+x^{-3})+4(x^2+x^{-2})+3(x+x^{-1})+5}\,dx $$
or
$$ f'(a)=\int_{0}^{1}\frac{(x^{a-3}-x^{1-a})\log(x)}{2(x^3+x^{-3})+4(x^2+x^{-2})+3(x+x^{-1})+5}\,dx $$
so $f'(a)$ clearly equals zero when $a-3=1-a$, i.e. for $a=2$.
Due to the substitution $x\mapsto\frac{1}{x}$ we have $f(2-u)=f(2+u)$ and now it is not difficult to show that the minimum of $f$ is indeed $f(2)\approx 0.06822$, since $f$ is (log)convex.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $I=\int(x+3)\sqrt\frac{x+2}{x-2}\mathrm dx$
Compute $$I=\int(x+3)\sqrt\frac{x+2}{x-2}\mathrm dx$$
The way I approach this problem was to:
*
*Set $u=\sqrt{x-2}$ and arrive at $$I=2\int\frac{u^2(u^2+1)}{\sqrt{u^2-4}}\mathrm du$$
*Set $u=2\sec t\implies\mathrm du=2\sec t\tan t\mathrm dt$ to get $I=8\int\sec^3(t)(4\sec^2(t)+1)\mathrm dt$
Now this integral involves $\sec^5t$ and $\sec^3t$ which does not make me very happy. I think there should be a simpler method. Can anyone show me simpler steps? (the first few steps/substitutions would suffice)
| When you end up with $\int\sec\theta\,d\theta$ it often means that you would have been better off making a $\cosh\theta$ substitution instead. Here we are going to use $x=2\cosh\theta$ so $\sqrt{x^2-4}=2\sinh\theta$ and $dx=2\sinh\theta\,d\theta$
$$\begin{align}\int(x+3)\sqrt{\frac{x+2}{x-2}}dx&=\int\frac{(x+3)(x+2)}{\sqrt{x^2-4}}dx\\
&=\int\frac{(2\cosh\theta+3)(2\cosh\theta+2)}{2\sinh\theta}2\sinh\theta\,d\theta\\
&=\int\left(4\cosh^2\theta+10\cosh\theta+6\right)d\theta\\
&=\int(2\cosh2\theta+10\cosh\theta+8)d\theta\\
&=\sinh2\theta+10\sinh\theta+8\theta+C_1\\
&=2\sinh\theta\cosh\theta+10\sinh\theta+8\theta+C_1\\
&=2\frac{\sqrt{x^4-4}}2\frac x2+10\frac{\sqrt{x^2-4}}2+8\cosh^{-1}\left(\frac x2\right)+C_1\\
&=\left(\frac12x+5\right)\sqrt{x^2-4}+8\ln\left(\frac x2+\sqrt{\frac{x^2}4-1}\right)+C_1\\
&=\left(\frac12x+5\right)\sqrt{x^2-4}+8\ln\left(x+\sqrt{x^2-4}\right)+C\end{align}$$
Where $C=C_1-\ln2$. Differentiation verifies this result.
Given that the OP doesn't know about the hyperbolic functions:
$$\begin{align}\cosh x&=\frac{e^x+e^{-x}}2\\
\sinh x&=\frac{e^x-e^{-x}}2\end{align}$$
So
$$\begin{align}\cosh^2x-\sinh^2x&=\frac{e^{2x}+2+e^{-2x}}4-\frac{e^{2x}-2+e^{-2x}}4=1\\
\cosh^2x+\sinh^2x&=\frac{e^{2x}+2+e^{-2x}}4+\frac{e^{2x}-2+e^{-2x}}4=\frac{e^{2x}+e^{-2x}}2=\cosh2x\end{align}$$
On addition of the last $2$ formulas we have
$$2\cosh^2x=\cosh2x+1$$
And finally there is
$$\begin{align}\frac d{dx}\cosh x&=\frac d{dx}\frac{e^x+e^{-x}}2=\frac{e^x-e^{-x}}2=\sinh x\\
\frac d{dx}\sinh x&=\frac d{dx}\frac{e^x-e^{-x}}2=\frac{e^x+e^{-x}}2=\cosh x\end{align}$$
Oh, there is one more thing: if $y=\cosh^{-1}x$ then
$$x=\cosh y=\frac{e^y+e^{-y}}2$$
So
$$e^{2y}-2xe^y+1=0$$
Solving for $y$,
$$y=\cosh^{-1}x=\ln\left(x+\sqrt{x^2-1}\right)$$
Now you know!
| {
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Mathematical Induction prove that $n^3+5n$ is divisible by $6$ Sorry, I know this will be a duplicate on the site but the other solution I found confusing and the method look completely different to what I was taught.
Prove that $n^3 + 5n$ is divisible by $6$ by using induction
The question is
Prove by mathematical Induction $(n^3+5n)$ is divisible by $6$
Here is what I have done
Assume $n=k$ is true
$\sum_{1}^{k} k =\dfrac{(k^3+5k)}{6}$
Now assume $n=k+1$ is true
$\sum_{1}^{k+1} k+1 =\dfrac{(k+1)^3+5(k+1)}{6}$
Then now
$\dfrac{(k+1)^3+5(k+1)}{6}=\sum_{1}^{k} k + (k+1)$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+5k)}{6} + (k+1)$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+5k)}{6} + \dfrac{(6k+6)}{6}$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+11k+6)}{6}$
But the other side doesnt equate (LHS)
$\dfrac{(k^3+3k^2+3k+1)+5k+5}{6}$
$\dfrac{(k^3+3k^2+8k+6)}{6}\ne\dfrac{(k^3+11k+6)}{6}$
I hope my method was clear enough so you can see where I went wrong. It would be much more use to me if you solved it as I learn better from looking at solutions and then applying them to other questions.
| When you start, you assume the given statement is true for $n=k$. The given statement is that $n^3 + 5n$ is divisible by $6$. Then you have to assume $k^3+5k$ is divisible by $6$, i.e., $$6N = k^3+5k$$ for some integer $N$.
There's nothing in the statement about the sum from $1$ to $n$, so the $\sum_1^k k$ term doesn't belong.
From there, you now ask whether $(k+1)^3 + 5(k+1)$ is divisible by $6$, i.e.,
$$ 6M = (k+1)^3 + 5(k+1)$$ for some integer $M$ (with no relation to $N$, yet). To demonstrate such an integer $M$ exists we expand the right side:
$$ 6M = k^3 + 3k^2 + 8k + 6$$
$$6M = (k^3 + 5k) + (3k^2 + 3k + 6)$$
$$6M = 6N + (3k^2 + 3k + 6)$$
$$M = N + \frac{3k^2+3k+6}{6} = N + \frac{k^2+k+2}{2} = N+1+\frac{k^2+k}{2}$$
So the problem now reduces to showing that $k^2+k$ is divisible by $2$. But $k^2$ and $k$ have the same parity, so their sum is always even; thus we're done.
| {
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Recurrence series with matrix for $a_n=a_{n-1}-2a_{n-2}$
$a_0=0$, $a_1=1$, $a_n=a_{n-1} - 2a_{n-2}$ for $n\ge2$. Find a closed form for $a_n$.
So if I compute $a_2=1-0=1$, $a_3=1-2=-1$:
$$\begin{bmatrix} a_n\\a_{n-1} \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n-1} \\ a_{n-2}\end{bmatrix}$$
I think this matrix is true for $n\ge2$. But the idea is to find a matrix that satisfy $n\ge1$?
I can see that $\begin{bmatrix} 1 \\0\end{bmatrix}$ is base case? And if they ask to find general term of $a_n$, I should find expression that satisfy the recurrence or the base case?
I suppose that $\begin{bmatrix} a_n\\a_{n-1} \end{bmatrix} $ = $x_n$
and $\begin{bmatrix} a_{n-1} \\ a_{n-2}\end{bmatrix}=x_{n-1}$
$$x_{n}=Ax_{n-1}$$
$$x_{n-1}=A^{-1}x_n$$
clearly do not satisfy the equation, and what is the purpose of finding a matrix for this problem?
The right pattern is $x_n=A^{n-1}x_1$, I was confused why it has always to be $x_1$ and how can I find it? thanks
|
So if I compute $a_2=1-0=1$, $a_3=1-2=-1$:
$$\begin{bmatrix} a_n\\a_{n-1} \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n-1} \\ a_{n-2}\end{bmatrix}$$
Good start. Now continue that....
$$\begin{align}
\begin{bmatrix} a_{n} \\ a_{n - 1} \end{bmatrix} &=
\begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n - 1} \\ a_{n - 2} \end{bmatrix}
\\ &= \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n - 2} \\ a_{n - 3} \end{bmatrix}
\\ &= \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n - 3} \\ a_{n - 4} \end{bmatrix}
\\ &= \dots
\end{align}$$
In summary $$\begin{bmatrix} a_{n + 1} \\ a_{n} \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} a_1 \\ a_0 \end{bmatrix}$$
I think this matrix is true for $n\ge2$. But the idea is to find a matrix that satisfy $n\ge1$?
Well lets look at $n=1$:
$$\begin{bmatrix} a_1 \\a_{0} \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{0} \\ a_{-1}\end{bmatrix}$$
So the $n = 1$ case is the equation $a_1 = a_{0} - 2a_{-1}$. Since we have no information about $a_{-1}$, this is neither provable nor useful. So $n \ge 2$ is fine.
what is the purpose of finding a matrix for this problem?
Because it let's you reach $$\begin{bmatrix} a_{n + 1} \\ a_{n} \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} a_1 \\ a_0 \end{bmatrix}$$ after which you can use all kinds of nice matrix properties to say more things about the sequence. For example, you could diagonalize $\begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix}$ to get
$$\begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ \frac{2}{1 - \sqrt{-7}} & \frac{2}{1 + \sqrt{-7}} \end{bmatrix} \begin{bmatrix} \frac{1 - \sqrt{-7}}2 & 0 \\ 0 & \frac{1 + \sqrt{-7}}2 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ \frac{2}{1 - \sqrt{-7}} & \frac{2}{1 + \sqrt{-7}} \end{bmatrix}^{-1}$$
so
$$\begin{align}
\begin{bmatrix} a_{n + 1} \\ a_{n} \end{bmatrix} &= \left( \begin{bmatrix} 1 & 1 \\ \frac{2}{1 - \sqrt{-7}} & \frac{2}{1 + \sqrt{-7}} \end{bmatrix} \begin{bmatrix} \frac{1 - \sqrt{-7}}2 & 0 \\ 0 & \frac{1 + \sqrt{-7}}2 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ \frac{2}{1 - \sqrt{-7}} & \frac{2}{1 + \sqrt{-7}} \end{bmatrix}^{-1}\right)^n \begin{bmatrix} a_1 \\ a_0 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 1 \\ \frac{2}{1 - \sqrt{-7}} & \frac{2}{1 + \sqrt{-7}} \end{bmatrix} \begin{bmatrix} \frac{1 - \sqrt{-7}}2 & 0 \\ 0 & \frac{1 + \sqrt{-7}}2 \end{bmatrix}^n \begin{bmatrix} 1 & 1 \\ \frac{2}{1 - \sqrt{-7}} & \frac{2}{1 + \sqrt{-7}} \end{bmatrix}^{-1} \begin{bmatrix} a_1 \\ a_0 \end{bmatrix}
\end{align}$$
Ugh, this sequence doesn't simplify into anything pretty because of all the complex numbers. But it is a straightforward way of deriving a formula of the form $a_{n} = C_1 \phi^n + C_2 {\varphi}~^n$
Also matrices keep the problem only involving integers instead of complex numbers, so computation is so much nicer.
The right pattern is $x_n=A^{n-1}x_1$, I was confused why it has always to be $x_1$ and how can I find it? thanks
It is a bit cleaner to do $x_{n+1} = A^n x_1$. You use $x_1$ as the base case because $a_1$ and $a_0$ is what you were given in the statement of the problem. You can use any convenient $x$ that you can calculate as the base case, but why not just use the one given to you?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Monic polynomials whose roots are their remaining coefficients I was bored in class one day and wondered to myself if there were any quadratics $x^2+ax+b$ such that $a$ and $b$ are the zeros. I found two: $x^2+x-2,$ and $x^2 -{1\over2}x -{1\over2}$. The comments suggested $x^2+0x+0$, though this seems trivial. I wonder if this applies to other degree polynomials. Clearly it never works for a linear, except for $x+0=0$, as if $x+a=0$, $x=-a$, not $a$. What about cubics, quadratics, or even higher powers? In general, $x^n$ works. What about non-trivial solutions?
| Cubics:
\begin{align}
x^3+x^2-2\,x=0\quad &\Rightarrow 1,-2,\phantom{-}0
,\\
x^3+x^2-x-1=0\quad &\Rightarrow 1,-1,-1
.
\end{align}
Similarly,
\begin{align}
x^4+x^3-2x^2=0\quad &\Rightarrow 1,-2,0,0
,\\
x^4+x^3-x^2-x=0\quad &\Rightarrow 1,-1,-1,0
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2689332",
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"source": "stackexchange",
"question_score": "39",
"answer_count": 3,
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} |
What did I do wrong with this integral? PROBLEM
I have to compute this indefinite integral:
$$\int(2-x)^2\ln(4x)\,\,dx$$
MY ATTEMPT
So I did integration by parts:
$$\begin{matrix}
u=(2-x)^2 & dv=\ln(4x)\,\,dx\\
du=(2x-4)\,\,dx & v=\int\ln(4x)\\
\end{matrix}$$
$$\int\ln(4x)\,\,dx=\;\;?$$
$$\begin{matrix}
a=\ln(4x) & db=dx\\
da={1\over x}\,\,dx & b=x\\
\end{matrix}$$
$$\int\ln(4x)\,\,dx=x\ln(4x)-\int x\,\cdot\frac1x\,\,dx$$
$$\int\ln(4x)\,\,dx=x\ln(4x)-x$$
So because of that:
$$\int(2-x)^2\ln(4x)\,\,dx=(2-x)^2\Big(x\ln(4x)-x\Big)-\int(2x-4)\Big(x\ln(4x)-x\Big)\,\,dx$$
Now integrating by parts one more time:
$$\begin{matrix}
u=2x-4 & dv=\Big(x\ln(4x)-x\Big)\,\,dx\\
du=2\,\,dx & v=\int\Big(x\ln(4x)-x\Big)\,\,dx\\
\end{matrix}$$
$$\int\Big(x\ln(4x)-x\Big)\,\,dx=\int x\ln(4x)\,\,dx\;-\int x\,\,dx$$
$$\begin{matrix}
a=\ln(4x) & db=x\,\,dx\\
da={1\over x}\,\,dx & b=\frac{x^2}{2}\\
\end{matrix}$$
$$\int x\ln(4x)\,\,dx=\frac{x^2}{2}\ln(4x)\;-\int\frac{x^2}{2}\,\cdot\frac1x\,\,dx$$
$$\int x\ln(4x)\,\,dx=\frac{x^2}{2}\ln(4x)\;-\int\frac x2\,\,dx$$
$$\int x\ln(4x)\,\,dx=\frac{x^2}{2}\ln(4x)\;-\frac{x^2}{4}$$
$$\int\Big(x\ln(4x)-x\Big)\,\,dx=\frac{x^2}{2}\ln(4x)\;-\frac{x^2}{4}-\frac{x^2}{2}$$
Finally, after all this mess, plugging in in the equation of the original integral:
$$\int(2-x)^2\ln(4x)\,\,dx=(2-x)^2\Big(x\ln(4x)-x\Big)-\frac{x^2}{2}\ln(4x)\;+\frac{x^2}{4}+\frac{x^2}{2}+c$$
Now, I cleaned up this mess a little to give a nicer answer:
$$(2-x)^2\Big(x\ln(4x)-x\Big)-\frac{x^2}{2}\ln(4x)\;+\frac{x^2}{4}+\frac{x^2}{2}+c$$
$$x(x^2-4x+4)\Big(\ln(4x)-1\Big)-\frac{x^2}{2}\ln(4x)\;+\frac{x^2}{4}+\frac{x^2}{2}+c$$
$$(x^3-4x^2+4x)\ln(4x)-x^2+4x-4-\frac{x^2}{2}\ln(4x)\;+\frac34 x^2+c$$
$$(x^3-4x^2+4x)\ln(4x)-\frac{x^2}{2}\ln(4x)-x^2+4x-4\;+\frac34 x^2+c$$
$$\boxed{\Big(x^3-\frac{9}{2}x^2+4x\Big)\ln(4x)-\frac{1}{4}x^2+4x-4+c}$$
ANSWER FROM WOLFRAMALPHA
WolframAlpha gives me a different answer:
$$\frac{1}{9}x\Big(-x^2+3(x^2-6x+12)\ln(4x)+9x-36\Big)+c$$
Even after rearranging the WA's answer, I get similar answer but the coefficients and other details don't match. I won't bother rearranging the WolframAlpha's answer here, but trust me, after putting it into similar form to my answer, it doesn't work.
So what did I do wrong?
| Up to here
$$\int(2-x)^2\ln(4x)\,\,dx=(2-x)^2\Big(x\ln(4x)-x\Big)-\int(2x-4)\Big(x\ln(4x)-x\Big)\,\,dx$$
it seems to be correct but this
$$\int(2-x)^2\ln(4x)\,\,dx=(2-x)^2\Big(x\ln(4x)-x\Big)-\frac{x^2}{2}\ln(4x)\;+\frac{x^2}{4}+\frac{x^2}{2}+c$$
seems not correct since you didn't considered the term
$$A=(2x-4)\left(\frac{x^2}{2}\ln(4x)\;-\frac{x^2}{4}-\frac{x^2}{2}\right)$$
and also the calculation of this part $\left(\frac{x^2}{2}\ln(4x)\;-\frac{x^2}{4}-\frac{x^2}{2}\right)$ seems uncorrect indeed the integral should be
$$\int(2x-4)\Big(x\ln(4x)-x\Big)\,\,dx=-\frac89x^3+3x^2+\frac23(x^3-3x^2)\ln (4x)$$
Let try to fix form here, since
$$\int\Big(x\ln(4x)-x\Big)\,\,dx=\frac{x^2}{2}\ln(4x)\;-\frac{x^2}{4}-\frac{x^2}{2}=\frac{x^2}{2}\ln(4x)\;-\frac{3x^2}{4}$$
we have that
$$\int(2x-4)\Big(x\ln(4x)-x\Big)\,\,dx=\\
=(2x-4)\left(\frac{x^2}{2}\ln(4x)\;-\frac{3x^2}{4}\right)-2\int\Big(\frac{x^2}{2}\ln(4x)\;-\frac{3x^2}{4}\Big)\,\,dx=\\
=(2x-4)\left(\frac{x^2}{2}\ln(4x)\;-\frac{3x^2}{4}\right)+\frac{x^3}{2}-\int x^2\ln(4x)\,\,dx=\\
=(x^3-2x^2)\ln(4x)\;-(x^3-3x^2)-\frac13x^3\log (4x)+\frac19x^3=\\
-\frac89x^3+3x^2+\frac23(x^3-3x^2)\ln (4x)$$
| {
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"url": "https://math.stackexchange.com/questions/2693023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Proving that $3^{n+2}$ does not divide $2^{3^n} +1$ for any positive integer $n$ Proving that $3^{n+2}$ does not divide $2^{3^n} +1$ for any positive integer $n$
I am thinking of applying inductionon $n$ but am not being successful.
| Suppose $\left.3^{n+1}\,\middle|\,2^{3^n}+1\right.$ and $\left.3^{n+2}\not\,\middle|\,\,2^{3^n}+1\right.$; that is $2^{3^n}+1$ has exactly $n+1$ factors of $3$. Then
$$
\begin{align}
\left(2^{3^n}+1\right)^3
&=2^{3^{n+1}}+3\cdot2^{2\cdot3^n}+3\cdot2^{3^n}+1\\
&=2^{3^{n+1}}+3\cdot2^{2\cdot3^n}\left(2^{2\cdot3^n}+1\right)+1
\end{align}
$$
Therefore,
$$
2^{3^{n+1}}+1=\overbrace{\left(2^{3^n}+1\right)^3}^{\substack{\text{exactly }3n+3\\\text{ factors of }3}}-\overbrace{3\cdot2^{2\cdot3^n}\left(2^{2\cdot3^n}+1\right)\vphantom{\left(2^{3^n}\right)^3}}^{\substack{\text{exactly }n+2\\\text{ factors of }3}}
$$
has exactly $n+2$ factors of $3$. Since this holds for $n=0$, for all $n\ge0$,
$$
\left.3^{n+1}\,\middle|\,2^{3^n}+1\right.\quad\text{and}\quad\bbox[5px,border:2px solid #C0A000]{\left.3^{n+2}\not\,\middle|\,\,2^{3^n}+1\right.}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2693812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find and simplify $a^3+c^3-b^3c\pmod{19}$ when $a \equiv 2\pmod {19},b \equiv 4\pmod {19}, c \equiv 5\pmod {19}$ Suppose $a \equiv 2\pmod {19},b \equiv 4\pmod {19}, c \equiv 5\pmod {19}$. Find and simplify $a^3+c^3-b^3c\pmod{19}$
My attempt
$a^3+c^3-b^3c\pmod{19}=2^3+5^3-(4^3*5)\pmod{19}=8+125-320\pmod{19}=8+125-16=117$
Am I on the right track?
| Your idea is correct, but the notation is a bit redundant. In one line, you can have $$\begin{align}a^3+c^3-b^3c&\equiv2^3+5^3-4^3(5)\equiv8+125-320\equiv-187\equiv3\pmod{19}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2694047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Exercise A12.5 of Hubbards' Vector Calculus, Linear Algebra, and Differential Forms I have a question about the solution to part (a) of exercise A12.5 of Hubbards' Vector Calculus, Linear Algebra, and Differential Forms.
Here is the exercise:
Let $f $ be the function $f
\begin{pmatrix}
x \\
y \\
\end{pmatrix}=sgn(y) \sqrt{ \frac {-x+\sqrt{x^2+y^2}}{2}}
$ where $sgn(y)$ is the sign of y (i.e., $+1$ when $y>0$ , $0$ when $y=0$ ,$-1$ when $y<0$ )
a. show that $f$ is continuously differentiable on the complement of the half line $y=0, x\le 0$.
I've included part of the solution on the student solution manual to this exercise below:
To show that $f$ is continuously differentiable on the locus where $y=0,x>0$:
In a neighborhood of a point $\begin{pmatrix}
x_0 \\
y_0 \\
\end{pmatrix}$satisfying $x_0>0 $ , $y_0=0$ , we can write
$-x+\sqrt{x^2+y^2}=-x+x\sqrt{1+\frac{y^2}{x^2}}= -x+x(1+\frac{1}{2}\frac{y^2}{x^2}+o(\frac{y^2}{x^2}))$
$ =\frac{y^2}{2x}+o(\frac{y^2}{x})$,
and since $sgn(y)y^2$ is of class $C^1$, the function is of class $C^1$ on the half axis $y=0, x> 0$.
I don't understand the bold part of the solution. Could you explain why the fact that $sgn(y)y^2$ is of class $C^1$ means that the function $f$ is continuously differentiable on the half-axis $y=0, x>0$ ?
(The book uses the following definition for little $o$ notation:
a function $f$ is in $o(h)$ if $\lim \limits_{x \to 0}\frac{f(x)}{h(x)}=0.$
)
| Definitely strange. In fact, $f_y(x_0,y_0)$ does not exists for $x_0 > 0$, $y_0 = 0$:
First, ignore the irrelevant 2 and consider
$$f(x,y) = \hbox{sgn}(y)\sqrt{\sqrt{x^2+y^2} - x}.$$
$$
\lim_{y\to 0}\frac{\sqrt{\sqrt{x_0^2+y^2} - x_0} - 0}{y - 0} = \lim_{y\to 0}\sqrt{\frac1{2x_0} + \frac{o(y^2/x_0)}{y^2}} = \frac1{\sqrt{2x_0}}
,$$
so the lateral limits of
$$
\lim_{y\to 0}\frac{\hbox{sgn}(y)\sqrt{\sqrt{x_0^2+y^2} - x_0} - 0}{y - 0}
$$
are different.
Using
$$
\sqrt{x_0^2+y^2} - x_0 =
\frac{(\sqrt{x_0^2+y^2} - x_0)(\sqrt{x_0^2+y^2} + x_0)}{\sqrt{x_0^2+y^2} + x_0} =
\frac{y^2}{\sqrt{x_0^2+y^2} + x_0}
$$
gives the same result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Restriction of an ideal sheaf to a hypersurface This is from Example 3.82 of Janos Kollar's Lectures on resolution of singularities. Let $I = (x^3+xy+y^2z^4) \subset k[x,y,z]$ (for some field $k$) be an ideal sheaf of $\mathbb{A}^3_k$ and $H = (x+u_1xy^3+u_2y^2+u_3y^2z^2 = 0)$ be a hypersurface in $\mathbb{A}^3_k$, where $u_1,u_2,u_3$ are units. I would like to compute $I|_{H}$.
The book says that by the equation of $H$, we can substitute $x$ by $$x = -y^2(u_2+u_3z^2)(1+u_1y^3)^{-1}$$ which then tells us that $I|_H \subset (y^3,y^2z^4)$.
I'm confused by why we can take the inverse of $1+u_1y^3$ and how this substitution leads to the result. Thanks!
| First, note that $V(1+u_1y^3)$ does not contain $H$ - if we have a point $p=(x,y,z)$ so that $1+u_1y^3=0$, then $p\in H$ iff $u_2y^2+u_3y^2z^2=0$, and since $y\neq 0$, $u_2+u_3z^2=0$. So the intersection of $V(1+u_1y^3)$ with $H$ is at least codimension one in $H$. So the relation $x=-y^2(u_2+u_3z^2)(1+u_1y^3)^{-1}$ is valid on a dense open set $U$ of $H$.
On this dense open set, we can make the substitution $x=-y^2(u_2+u_3z^2)(1+u_1y^3)^{-1}$ and see that on $U$, $$x^3+xy+y^2z^4 = -y^6(u_2+u_3z^2)^3(1+u_1y^3)^{-3} -y^3(u_2+u_3z^2)(1+u_1y^3)^{-1} + y^2z^4$$ or after factoring, $$x^3+xy+y^2z^4 = -y^3(y^3(u_2+u_3z^2)^3(1+u_1y^3)^{-3} -(u_2+u_3z^2)(1+u_1y^3)^{-1}) + y^2z^4$$
and so on $U$, we have $I|_H\subset (y^3,y^2z^4)$.
Now we need to reason about the set $H\setminus U$. What we've demonstrated is that $V(y^3,y^2z^4)\cap U \subset V(I|_H)\cap U$ - if $V(y^3,y^2z^4)\cap (H\setminus U)$ is empty, then we're done. This is in fact the case - on closed points, this says that $y=0$, which contradicts $1+u_1y^3=0$.
| {
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"source": "stackexchange",
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Are the only solutions to $a^n + b^n =(a+b \forall n\geq 1$, 0, 1? Suppose $a^n + b^n = a+b$ for all $n \in \mathbb{N}$.
Is it true that the only solutions are:
$$a=1, b=0$$
$$a=0, b=1$$
$$a=b=0$$
?
If so, how do you prove it?
| Let $A:=\{(1,0),(0,1),(0,0),(1,1)\}$
*
*It is clear that if $(a,b) \in A$, then $a^n+b^n=a+b$.
*If $(a,b)=(1,-1)$ or $=(-1,1)$, then $a^2+b^2=2 \ne 0=a+b$.
*Let $a^n+b^n=a+b$ for all $n$. If $|a|>1$ or $|b|>1,$ then the sequence $(a^n+b^n)$ is unbounded, a cotradiction , since this sequence is constant.
Hence, by 1., 2. and 3. , we have only to investigate the case $|a| < 1$ and $|b| < 1$.
From $a^n+b^n=a+b$ for all $n$ we get with $ n \to \infty$ that $ a=-b$.
Hence $0=a+b=a^2+a^2=2a^2$, therefore $a=b=0$.
Conclusion: $a^n+b^n=a+b \iff (a,b) \in A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2699800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Condition that maybe implies convexity We have $f:R\to R, f(x+a)-f(x) \leq f(y+a)-f(y), \forall x,y,a \in R, x\leq y, a\ge 0$
We have to prove that the function satisfies the following:
$f((1-\frac{1}{2^k})x +\frac{1}{2^k}y) \leq (1-\frac{1}{2^k})f(x)+\frac{1}{2^k}f(y), \forall x,y \in R, \forall k$ positive integer. That inequality is reminiscent of the inequality for convex functions. Is there a connection? How could I approach the problem?
| Let $y\geq x$ and choose $a=y-x \geq 0$, than
$$ f(y)-f(x) \leq f(2y)-f(y) \quad \iff \quad 2 f(y) \leq f(2y-x)+f(x). $$
Now I prove the statement by induction on $k$.
*
*$k=1$, the inequality above gives
$$ f(y) \leq \frac{1}{2} f(2y-x) + \frac{1}{2} f(x). $$
Choose $y = \frac{1}{2} a + \frac{1}{2}b$, we get
$$ f\left( \frac{1}{2} a + \frac{1}{2}b \right) \leq \frac{1}{2} f(a+b-x) + \frac{1}{2} f(x). $$
Choosing $x = \min\{a,b\}\leq \frac{1}{2}a + \frac{1}{2}b$, we get the sought inequality
$$ f\left( \frac{1}{2} a + \frac{1}{2}b \right) \leq \frac{1}{2} f(a) + \frac{1}{2} f(b). $$
Assume, w.l.g, that $x=a$.
*$k \to k+1$, assuming it holds true for $k$, we know that
$$ f\left( \left( 1-\frac{1}{2^k} \right)a + \frac{1}{2^k}b \right) \leq \left( 1-\frac{1}{2^k} \right) f(a) + \frac{1}{2^k} f(b). $$
Now set $y=\left( 1-\frac{1}{2^{k+1}} \right)a + \frac{1}{2^{k+1}}b $ and $x=a$ in the first inequality and use the inductive step (IS)
\begin{align} f\left( \left( 1-\frac{1}{2^{k+1}} \right)a + \frac{1}{2^{k+1}}b \right) & \leq \frac{1}{2} f\left( 2a - \frac{1}{2^k}a + \frac{1}{2^k}b - a \right) + \frac{1}{2} f(a) \\
& = \frac{1}{2} f\left( a\left(1 - \frac{1}{2^k}\right) + \frac{1}{2^k}b \right) + \frac{1}{2} f(a) \\
& \overset{\text{IS}}{\leq} \frac{1}{2} \left( \left( 1-\frac{1}{2^k} \right) f(a) + \frac{1}{2^k} f(b) \right)+ \frac{1}{2} f(a) \\
& = \left( 1-\frac{1}{2^{k+1}} \right) f(a) + \frac{1}{2^{k+1}} f(b)
\end{align}
| {
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Prove $O(2)$ is non-abelian? $$O(2) = \{Q\in \mathbb{F}^{2\times 2} | Q^TQ= QQ^T=I\}$$
What is the most elegant way to prove that $O(2)$ is non-Abelian?
Here is my thinking: I know that $O(2)$ can be generated by reflections and moreover two reflections result in a rotation. Rotations commute with each other, but reflections do not
$$
ROT(\theta/2)=
\begin{bmatrix}
\cos\theta & \sin\theta \\ \sin\theta & -\cos\theta
\end{bmatrix}
$$
It suffices to show that $$ROT(\phi/2)ROT(\theta/2)\neq ROT(\theta/2)ROT(\phi/2)$$ for some $\phi,\theta\in (0,2\pi)$.
$$
\begin{bmatrix}
\cos\phi & \sin\phi \\ \sin\phi & -\cos\phi
\end{bmatrix}
\begin{bmatrix}
\cos\theta & \sin\theta \\ \sin\theta & -\cos\theta
\end{bmatrix}
=\begin{bmatrix}
\cos\phi\cos\theta + \sin\phi\sin\theta
&
\cos\phi\sin\theta-\sin\phi\cos\theta
\\
\sin\phi\cos\theta-\cos\phi\sin\theta
&
\sin\phi\sin\theta+\cos\phi\cos\theta
\end{bmatrix}
$$
$$
\begin{bmatrix}
\cos\theta & \sin\theta \\ \sin\theta & -\cos\theta
\end{bmatrix}
\begin{bmatrix}
\cos\phi & \sin\phi \\ \sin\phi & -\cos\phi
\end{bmatrix}
=\begin{bmatrix}
\cos\phi\cos\theta + \sin\phi\sin\theta
&
\sin\phi\cos\theta-\cos\phi\sin\theta
\\
\cos\phi\sin\theta-\sin\phi\cos\theta
&
\sin\phi\sin\theta+\cos\phi\cos\theta
\end{bmatrix}
$$
We see that indeed the two matrices do not commute since their top-right and bottom-left elements switch signs.
My question: Is there a purely algebraic proof that does not need geometric consideration? If not, what is the most common proof of this result?
| Take for example
$$\begin{pmatrix}1&0\\0&\!-1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}0&1\\\!-1&0\end{pmatrix}\neq\begin{pmatrix}0&\!-1\\1&0\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&\!-1\end{pmatrix}$$
...over any field with characteristic$\,\neq2\;$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2703205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How does $n^4 + 6n^3 + 11n^2 + 6n + 1 = (n^2 + 3n + 1)^2?$ C++ student here, not quite familiar with these type of expressions. Can someone explain how does this work? I'm familiar with $(a+b)^2$ etc. mathematics but this seems to be like $(a+b+c)^2$ and having searched online, the opened form for this formula doesn't look much alike. Any help will be appreciated. Thanks!
| This uses a simple rule called distributivity. This rule says that for all numbers $a,b,c$ we have:
$$a(b+c)=ab+ac$$
and:
$$(a+b)c=ac+bc$$
Now, let's say we have some numbers $a$ $b$ and $c$ and we want to compute $(a+b+c)^2$. Using our rule, we obtain:
\begin{align*}
(a+b+c)^2 &= (a+b+c)(a+b+c)\\
&=a(a+b+c)+b(a+b+c)+c(a+b+c)\\
&=a^2+ab+ac+ba+b^2+bc+ca+cb+c^2\\
&= a^2+b^2+c^2+2ab+2bc+2ca.
\end{align*}
If we put $n^2$ for $a$, $3n$ for $b$ and $1$ for $c$ to obtain:
$$(n^2+3n+1)^2=(n^2)^2+(3n)^2+1^2+2n^2\cdot 3n+2n^2\cdot 1+2\cdot 3n\cdot 1$$
Simplifying this gives:
\begin{align*}
(n^2+3n+1)^2&=(n^2)^2+(3n)^2+1^2+2n^2\cdot 3n+2n^2\cdot 1+2\cdot 3n\cdot 1\\
&= n^4+9n^2+1+6n^3+2n^2+6n\\
&=n^4+6n^3+11n^2+6n+1
\end{align*}
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Simplex Method Solution So I and 2 colleagues are arguing over a solution to an exam question and would like some clarification. It's an MCQ
Q: Consider the following tableau for a maximisation LP Problem:
\begin{array}{r|rrrr|r}
& x_1 & x_2 & x_3 & x_4 & \text{b} \\ \hline
x_3 & 1 & 1 & 1 & 0 & 15\\
x_4 & 1 & 2 & 0 & 1 & 20\\ \hline
z & -3 & 0 & 0 & -6 & -75
\end{array}
Which of the following tableau will be reached after performing one step of the Simplex Method?
a)\begin{array}{r|rrrr|r}
& x_1 & x_2 & x_3 & x_4 & \text{b} \\ \hline
x_1 & 1 & 1 & 1 & 0 & 15\\
x_4 & 0 & 1 & -1 & 1 & 5\\ \hline
z & 0 & 3 & 3 & 0 & -30
\end{array}
b)\begin{array}{r|rrrr|r}
& x_1 & x_2 & x_3 & x_4 & \text{b} \\ \hline
x_3 & 1 & 1 & 1 & 0 & 15\\
x_4 & 1 & 2 & 0 & 1 & 20\\ \hline
z & 3 & 12 & 0 & 0 & 45
\end{array}
c)\begin{array}{r|rrrr|r}
& x_1 & x_2 & x_3 & x_4 & \text{b} \\ \hline
x_3 & 0 & -1 & 1 & -1 & -5\\
x_1 & 1 & 2 & 0 & 1 & 20\\ \hline
z & 0 & 6 & 0 & -3 & -15
\end{array}
d)\begin{array}{r|rrrr|r}
& x_1 & x_2 & x_3 & x_4 & \text{b} \\ \hline
x_2 & -3 & 1 & 0 & 1 & 6\\
x_3 & -1 & 0 & 1 & -2 & 2\\ \hline
z & -2 & 0 & 0 & 1 & 20
\end{array}
e) We cannot perform an update on this tableau.
My one colleague says the answer is b) As your Pivot column is $x_4$ as it has the largest negative element. Most texts state that the pivot column is determined by the largest negative value.
The other colleague says it's e) as $x_2$ is nonbasic yet has a coefficient of 0 in the z row.
Now, I think there is a mistake in the actual question, since $x_4$ is in the basis, but does not have a 0 coefficient in the z row... which means this tableau is already wrong and should actually be something like
\begin{array}{r|rrrr|r}
& x_1 & x_2 & x_3 & x_4 & \text{b} \\ \hline
x_3 & 1 & 1 & 1 & 0 & 15\\
x_4 & 1 & 2 & 0 & 1 & 20\\ \hline
z & -3 & -6 & 0 & 0 & -75
\end{array}
So which is it? Is the question wrong from the start like I suggested.. or is someone actually right.. or are we all wrong and the solution is one of the other options?
| The first colleague's answer is right, but the argument is not. We have to observe the coefficients from a tableau with the $z$-row fixed. The $-6$ tells you nothing about the LP's optimality since it's in the $x_4$-column representing the current basis.
Since $x_4$ is in the basis, you have to clear the entry $-6$ at the $z$-row. To do so, multiply the $x_4$-row by $6$ and add the result to the $z$-row. This gives the tableau in (b).
\begin{array}{r|rrrr|r}
& x_1 & x_2 & x_3 & x_4 & \text{b} \\ \hline
z_0 & -3 & 0 & 0 & -6 & -75\\
+\; 6x_4 & 6 & 12 & 0 & 6 & 120\\ \hline
z_1 & 3 & 12 & 0 & 0 & 45
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Do I need to substitute in the expressions before calculating this partial derivative? Find $\frac{\partial f}{\partial x}$ when:
$f = 3xy^2z^3$, $y = 3x^2 + 2$, $z = \sqrt{x-1}$
I would need to replace $y$ and $z$ in $f$ before calculating the partial of $f$, correct?
(The alternative would be find the partial of $y$ and $z$, square and cube them respectively and multiply all that by 3, which I think is incorrect, but I'm not sure)
| We are free to either replace $y=y(x)$ and $z=z(x)$ or keep them in the calculation. If we decide to keep them in, which might also be convenient, we have to be aware that both $y$ and $z$ are functions of $x$ and have to treat them accordingly.
We consider
\begin{align*}
f(x,y,z)&=3xy^2z^3\\
y&=y(x)=3x^2+2\\
z&=z(x)=\sqrt{x-1}
\end{align*}
Differentiating $y=y(x)$ and $z=z(x)$ we obtain
\begin{align*}
y^\prime(x)=6x\qquad\qquad z^\prime(x)=\frac{1}{2\sqrt{x-1}}
\end{align*}
One way to do the partial differentiation is applying the product rule as following
\begin{align*}
\color{blue}{\frac{\partial}{\partial x}f(x,y,z)}&=((3xy^2)z^3)^{\prime}\\
&=(3xy^2)^\prime z^3+3xy^2(z^3)^\prime\\
&=(3y^2+3x\cdot 2yy^\prime)z^3+3xy^2\cdot 3z^2z^\prime\\
&\,\,\color{blue}{=3y^2z^3+6xyy^\prime z^3+9xy^2z^2z^\prime}\tag{1}\\
&=3(3x^2+2)^2(x-1)^{\frac{3}{2}}+6x(3x^2+2)6x(x-1)^{\frac{3}{2}}\\
&\qquad+9x(3x^2+2)^2\frac{1}{2}(x-1)^{\frac{1}{2}}\\
&\,\,\color{blue}{=\frac{3}{2}(x-1)^{\frac{1}{2}}(3x^2+2)(39x^3-30x^2+10x-4)}
\end{align*}
Similarly we obtain
\begin{align*}
\frac{\partial}{\partial x}f(x,y,z)&=(3x(y^2z^3))^{\prime}\\
&=(3x)^\prime y^2z^3+3x(y^2z^3)^\prime\\
&=3y^2z^3+3x(2yy^\prime z^3+3y^2z^2z^\prime)\\
&=3y^2z^3+6xyy^\prime z^3+9xy^2z^2z^\prime\\
\end{align*}
giving the same result as in (1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Why $\frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} dt0$? Why does the following inequality hold for $x > 0$?
$$
Q(x)<e^{-x^2/2}, \tag{1}
$$
where
$$
Q(x) = \frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} dt. \tag{2}
$$
The following is my attempt: Define
$$
\begin{align}
f(x)
& \triangleq e^{-x^2/2} - Q(x) \\
& = e^{-x^2/2} - \frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} dt \\
& = e^{-x^2/2} - \left( 1 - \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-t^2/2} dt \right) \\
& = e^{-x^2/2} - 1 + \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-t^2/2} dt.
\tag{3}
\end{align}
$$
Then
$$
\begin{align}
f'(x)
& = -\frac{2x}{2} e^{-x^2/2} + \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \\
& = \left(-x+\frac{1}{\sqrt{2\pi}} \right) e^{-x^2/2}. \tag{4}
\end{align}
$$
When $0<x<\frac{1}{\sqrt{2\pi}}$, $f'(x)>0 \Rightarrow f(x)$ is a monotonically increasing function.
When $x=\frac{1}{\sqrt{2\pi}}$, $f'(x)=0$.
When $x>\frac{1}{\sqrt{2\pi}}$, $f'(x)<0 \Rightarrow f(x)$ is a monotonically decreasing function.
So $f(x)$ has a local maximum at $x = \frac{1}{\sqrt{2\pi}}$.
$$
\lim_{x \rightarrow \infty} f(x) = 0 - 0 =0.
$$
$$
f(0) = e^{-0^2/2} - Q(0) = 1 - \frac{1}{2} = \frac{1}{2}.
$$
Then I don't know how to continue. Any comments and answers are welcome. Thanks in advance.
| $$e^{-x^2/2}=\int^{\infty}_xe^{-t^2/2}t\,dt$$
Therefore
$$Q(x)-e^{-x^2/2}=\frac{1}{\sqrt{2\pi}}\int^{\infty}_xe^{-t^2/2}\,dt-\int^{\infty}_xe^{-t^2/2}t\,dt=\int^{\infty}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt$$
If $x\geqslant1/\sqrt{2\pi}$ then
$$Q(x)-e^{-x^2/2}=\int^{\infty}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt<0$$
otherwise split the integral into two parts
$$\int^{\infty}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt=\int^{1/\sqrt{2\pi}}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt+\int^{\infty}_{1/\sqrt{2\pi}}\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt$$
The first is positive and the second part negative. You can estimate each part
$$\int^{1/\sqrt{2\pi}}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt\leqslant\int^{1/\sqrt{2\pi}}_0\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt<0.08$$
and
$$\Big|\int^{\infty}_{1/\sqrt{2\pi}}\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt\Big|>0.5$$
The negative part outweights the positive part. So whenever $x>0$ the inequality holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Find the rational number of a, b, c, solving $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{a}+ \sqrt[3]{b}+\sqrt[3]{c}$ I try as following
let
\begin{eqnarray}
x= \sqrt[3]{a} \\
y= \sqrt[3]{b} \\
z= \sqrt[3]{c} \\
x+y+z = \sqrt[3]{\sqrt[3]{2}-1 }\\
\end{eqnarray}
We know that
\begin{equation}
x^3+y^3+z^3 = (x+y+z)^3-3(x+y)(x+z)(y+z)
\end{equation}
that turns out to be
\begin{equation}
(x+y+z-1)(x+y+z)(x+y+z-1)=3(x+y)(x+z)(y+z)
\end{equation}
plug in $x+y+z$,
we will get
\begin{equation}
\sqrt[3]{2}-1 - \sqrt[3]{\sqrt[3]{2}-1 } = 3(x+y)(x+z)(y+z)
\end{equation}
From now I stuck to solve for the rational numbers of $a,b,c$. Could anyone show me the way how to continue to solve it?
| An observation:
\begin{align*}
(\sqrt[3]{4}-\sqrt[3]{2}+1)^3&=\frac{(\sqrt[3]{8}+1)^3}{(\sqrt[3]{2}+1)^3}\\
&=\frac{27}{2+3\sqrt[3]{4}+3\sqrt[3]{2}+1}\\
&=\frac{9}{\sqrt[3]{4}+\sqrt[3]{2}+1}\\
&=\frac{9(\sqrt[3]{2}-1)}{\sqrt[3]{8}-1}\\
&=9(\sqrt[3]{2}-1)\\
\left(\sqrt[3]{\frac{4}{9}}+\sqrt[3]{\frac{-2}{9}}+\sqrt[3]{\frac{1}{9}}\right)^3&=\sqrt[3]{2}-1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Help me understand how to Find f(n) I have this question:
*
*$f(0) = 0$
*$f(1) = 1$
*$f(n) = f(n-1) + f(n-2)$
help me understand and find $f(n)$.
| Do you know Fibonacci rabbits?
That's a famous recurrence relation that we can solve by trial solution in the form $f(n)=x^n$ that is
$$x^{n}=x^{n-1}+x^{n-2}\implies x^2-x-1=0 \implies x_{1,2}=\frac{1\pm\sqrt 5}{2}$$
then
$$f(n)=c_1x_1^n+c_2x_2^n=c_1\left(\frac{1+\sqrt 5}{2}\right)^n+c_2\left(\frac{1-\sqrt 5}{2}\right)^n$$
and from the initial conditions $f(0)=0$ and $f(1)=1$ we obtain
*
*$c_1+c_2=0\implies c_1=-c_2=c$
*$c\left(\frac{1+\sqrt 5}{2}-\frac{1-\sqrt 5}{2}\right)=1\implies c=\frac1{\sqrt 5}$
and finally
$$f(n)=\frac1{\sqrt 5}\left(\frac{1+\sqrt 5}{2}\right)^n-\frac1{\sqrt 5}\left(\frac{1-\sqrt 5}{2}\right)^n$$
and indicating with $\phi=\frac{1+\sqrt 5}{2}$ the golden ratio we have that $\frac{1-\sqrt 5}{2}=-\frac1{\phi}$ and thus
$$f(n)=\frac1{\sqrt 5}\left(\phi\right)^n-\frac1{\sqrt 5}\left(-\frac1{\phi}\right)^n=\frac{\phi^{n}-(-\phi)^{-n}}{\sqrt 5}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I show that $h_n=\begin{cases}n!2^{\frac{n}{2}-1} , \text {if $n$ is even} \\ 0, \text{if $n$ is odd} \end{cases}$
We have $n $ distinct cards. We want to split these cards into non-empty subsets so that each subset contains even number of elements. Then we want to order the cards in each subset. Finally we want to order these subsets into a line. Let $h_n$ be the number of ways to build this structure. $h_n$?
Now
$$
a_j= \begin{cases}
j!, &\text{if $j>0$ is even}\\
0, &\text{otherwise}
\end{cases},$$ and $b_k=k!, b_0=1$.
So, $$A(x)=\sum_{n\geq 1}(2n)!\frac{x^{2n}}{(2n)!}=\sum_{n\geq 1}x^{2n}=\frac{x^2}{1-x^2}$$
and
$$B(x)=\sum_{k\geq 0}(k)!\frac{x^{k}}{(k)!}=\sum_{k\geq 0}x^k=\frac{1}{1-x}$$
Then $h_0=1$ and
$$
\begin{split}
\sum_{n\geq 0}h_n\frac{x^n}{n!} &= H(x)\\
&= B(A(x))\\
&=\frac{1}{1-\frac{x^2}{1-x^2}}=\frac{1}{1-2x^2}-\frac{x^2}{1-2x^2}\\
&=\sum_{n\geq 0}2^nx^{2n}-\sum_{n\geq 0}2^nx^{2n+2}\\
&=1+\sum_{n\geq 1}2^nx^{2n}-\sum_{n\geq 1}2^{n-1}x^{2n}\\
&=1+\sum_{n\geq 1}(2^n-2^{n-1})x^{2n}\\
&=1+\sum_{n\geq 1}2^{n-1}x^{2n}\\
&=1+\sum_{n\geq 1}{2^{n-1}n!}\frac{x^{2n} }{n!}
\end{split}
$$
Now how do I show that $$h_n=\begin{cases}n!2^{n/2-1}, & \text{if $n$ is even} \\ 0, & \text{if $n$ is odd} \end{cases}$$
| Note
$$
\sum h_n \frac{x^n}{n!}
= 1 + \sum_{n \ge 1} 2^{n-1}x^{2n}
= 1 + \sum_{n \ge 1} 2^{n-1} (2n)! \frac{x^{2n}}{(2n)!}
$$
and using $k=2n$ to make sure $k$ is even, we get
$$
\sum_{k \text{ even}} 2^{k/2-1} k! \frac{x^k}{k!}
$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2711847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$
First,I tried to evaluate like this:
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)d\left(\frac{\sin x}{1+\cos x}\right)$$
$$=\int_{0}^{\frac{\pi}{2}}x^2d\log\left(\frac{\sin x}{1+\cos x}\right)=x^2\log\left(\frac{\sin x}{1+\cos x}\right)|_{0}^{\frac{\pi}{2}}-2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{\sin x}{1+\cos x}\right)dx$$
$$=0+2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{1+\cos x}{\sin x}\right)dx=2\int_{0}^{\frac{\pi}{2}}x\log\left(1+\cos x\right)dx-2\int_{0}^{\frac{\pi}{2}}x\log\left(\sin x\right)dx$$
$$=2\int_{0}^{\frac{\pi}{2}}x\log\cot \left(\frac{x}{2}\right)dx=8\int_{0}^{\frac{\pi}{4}}x\log\cot xdx$$
but I can't proceed next step,help me,thanks.
| $$\mathcal{J}=\int_{0}^{\pi/2}\frac{x^2}{\sin x}\,dx = \int_{0}^{1}\frac{\arcsin^2(x)}{x\sqrt{1-x^2}}\,dx=\sum_{n\geq 1}\frac{2^{2n-1}}{n^2\binom{2n}{n}}\int_{0}^{1}\frac{x^{2n-1}}{\sqrt{1-x^2}}\,dx \tag{1}$$
by the Maclaurin series of $\arcsin^2(x)$. Euler's Beta function then leads to
$$ \mathcal{J}=\sum_{n\geq 1}\frac{16^n}{4n^3 \binom{2n}{n}^2}=\phantom{}_4 F_3\left(1,1,1,1;\tfrac{3}{2},\tfrac{3}{2},2;1\right)\tag{2} $$
where the RHS is a manageable hypergeometric function (similar objects are evaluated both here and here) and as already shown by Claude Leibovici, $\mathcal{J}=4\int_{0}^{1}\frac{\arctan^2(u)}{u}\,du $ is simply given by a combination of a dilogarithm and a trilogarithm. Indeed
$$ \int_{0}^{\pi/2}\int_{0}^{\theta}\frac{u}{\sin u}\,du\,d\theta =-\pi G+\frac{7}{2}\zeta(3)\tag{3}$$
leading to $\mathcal{J}=2\pi G-\frac{7}{2}\zeta(3)$, has already been a key lemma in this historical thread.
An alternative way for proving this identity is just to write $\frac{x}{\sin x}$ and $|x|$ as Fourier cosine series.
The Shafer-Fink inequality leads to
$$ \int_{0}^{\pi/2}\frac{x^2}{\sin x}\,dx = 4 \int_{0}^{1}\frac{\arctan^2(u)}{u}\,du \approx \frac{6}{7}(3\sqrt{2}-5)+9\log\left(\frac{2\sqrt{2}+1}{3}\right)\approx 1.54.\tag{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 9,
"answer_id": 1
} |
Calculate $\int\limits_{0}^{1}{f\left( x \right)dx}$ when $f$ satisfies three given conditions.
Let $f$ be a function with $f'$ is continuous on $[0,1]$ such that $$f\left( 1 \right)=0,\quad\int\limits_{0}^{1}{{{\left[ f'(x) \right]}^{2}}dx=7},\quad\int\limits_{0}^{1}{{{x}^{2}}f\left( x \right)dx=\frac{1}{3}.}$$
Calculate $\displaystyle\int\limits_{0}^{1}{f\left( x \right)dx}$.
| Hint. Note that by Cauchy-Schwarz inequality
$$
\begin{align*}\frac{1}{3}=\int_0^1 x^2 f(x)dx&=\left[\frac{x^3}{3} f(x)\right]_0^1-\int_0^1\frac{x^3}{3}\, f'(x)dx=\frac{1}{3}\int_0^1 (-x^3) f'(x)dx\\
&\leq \frac{1}{3}\left(\int_{0}^{1}x^6 dx\right)^{1/2} \left(\int_{0}^{1} (f'(x))^2dx\right)^{1/2}=\frac{1}{3}\cdot\frac{1}{\sqrt{7}}\cdot \sqrt{7}= \frac{1}{3}.\end{align*}$$
Now recall that in the Cauchy-Schwarz inequality, the equality holds if and only if one function is a scalar multiple of the other.
P.S. "Without" the Cauchy-Schwarz inequality: from the first line $\int_0^1 x^3 f'(x)dx=-1$ and therefore
$$\int_0^1(7x^3+ f'(x))^2 dx=7^2\int_0^1 x^6 dx+2\cdot 7\int_0^1 x^3 f'(x) dx +\int_0^1(f'(x))^2 dx=0$$
which implies that the non-negative continuous function $(7x^3+ f'(x))^2$ is zero, that is $f'(x)=-7x^3$. Hence
$$f(x)=f(1)+\int_1^x f'(t)dt=0-\frac{7}{4}[t^4]_1^x=\frac{7(1-x^4)}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differentiate $f(x)=\frac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}$, $-2
Differentiate $f(x)=\frac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}$, $-2<x<2$
My Attempt
$$
\begin{align}
f'(x)&=\frac{\sqrt{2x+7}\frac{d}{dx}\cos^{-1}\big(\frac{x}{2}\big)-\cos^{-1}\big(\frac{x}{2}\big)\frac{d}{dx}\sqrt{2x+7}}{(\sqrt{2x+7})^2}\\
&=\frac{\sqrt{2x+7}\frac{\frac{-1}{2}}{\sqrt{1-\frac{x^2}{4}}}-\cos^{-1}\big(\frac{x}{2}\big)\frac{2}{2\sqrt{2x+7}}}{(\sqrt{2x+7})^2}\\
&=\frac{\frac{-\sqrt{2x+7}.\sqrt{4}}{2\sqrt{4-x^2}}-\frac{\cos^{-1}\big(\frac{x}{2}\big)}{\sqrt{2x+7}}}{2x+7}\\
&=\frac{\pm2}{2.\sqrt{2x+7}\sqrt{4-x^2}}-\frac{\cos^{-1}\big(\frac{x}{2}\big)}{(2x+7)^{3/2}}\\
&\color{blue}{=\frac{\pm1}{\sqrt{2x+7}\sqrt{4-x^2}}-\frac{\cos^{-1}\big(\frac{x}{2}\big)}{(2x+7)^{3/2}}\\}
\end{align}
$$
Result by Mathematica
D[(ArcCos[x/2]/Sqrt[2 x + 7]),x]
$$
f'(x)={\frac{-1}{\sqrt{2-x}\sqrt{2+x}\sqrt{7+2x}}-\frac{\arccos\big(\frac{x}{2}\big)}{(7+2x)^{3/2}}}\\
\color{blue}{=\frac{-1}{\sqrt{2x+7}\sqrt{4-x^2}}-\frac{\cos^{-1}\big(\frac{x}{2}\big)}{(2x+7)^{3/2}}\\}
$$
Mathematica seems to use only the principal root, ie. $\sqrt{4}=+2$ but in that case can I say the given solution is incomplete ?
Or is there any other domain or range considerations which eliminate the other case ?
| $\sqrt{4}=|2|=+2$ in line $3-5$ of your solution
Edit (After OP edited the question):
The derivativeof $\arccos(x)$ is found to be $\frac{-1}{\sqrt{1-x^2}}$, where we must take the positive sign so that the derivative always has negative sign (Otherwise why even bother with the minus sign too?). So, $\frac{d}{dx}\arccos(x/2)= \frac{-1/2}{\sqrt{1-x^2/4}}$.As you can see, the $\sqrt{4}$ causing all the trouble for you originates from here, and as I said we must only take the positive square root, because the derivative requires it.
| {
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Approximation of the number of partitions of n, denoted p(n) I managed to prove that $p(n)\ge \max_{1\le k\le n}{{n-1\choose k-1}\over k!}$,that wasn't hard. Now I need to use this result to prove that there exists a constant c>0 for which $p(n)\ge e^{c\sqrt n}$ for any $n\in \mathbb{N}$. I tried to use Stirling's approximation here, but didn't get the result I want. Any hints?
| First find for which $k$ the expression
$$a(n,k) = \frac{1}{k!}\binom{n-1}{k-1} = \frac{(n-1)!}{k!(k-1)!(n-k)!}$$
reaches its maximum. Since $a(n,k) > 0$ for $1 \leqslant k \leqslant n$ and
$$\frac{a(n,k+1)}{a(n,k)} = \frac{n-k}{(k+1)k}$$
we see that
$$a(n,k+1) < a(n,k) \iff n - k < (k+1)k \iff n+1 < (k+1)^2,$$
so the maximum is attained for $k = \lfloor \sqrt{n+1}\rfloor$.
Now recall Stirling's approximation
$$\log m! = \bigl(m + \tfrac{1}{2}\bigr)\log m - m + \log \sqrt{2\pi} + O\biggl(\frac{1}{m}\biggr)$$
and the equivalent
$$\log (m-1)! = \bigl(m - \tfrac{1}{2}\bigr)\log m - m + \log \sqrt{2\pi} + O\biggl(\frac{1}{m}\biggr)\,.$$
Plugging these into
$$\log p(n) \geqslant \log (n-1)! - \log (n-k)! - \log (k-1)! - \log k!$$
yields
\begin{align}
\log p(n)
&\geqslant \bigl(n - \tfrac{1}{2}\bigr) \log n - n + \log \sqrt{2\pi} + O\biggl(\frac{1}{n}\biggr) \\
&\qquad - \bigl(n - k + \tfrac{1}{2}\bigr)\log (n-k) + n - k - \log \sqrt{2\pi} + O\biggl(\frac{1}{n-k}\biggr) \\
&\qquad - \bigl(k - \tfrac{1}{2}\bigr)\log k + k - \log \sqrt{2\pi} + O\biggl(\frac{1}{k}\biggr) \\
&\qquad - \bigl(k + \tfrac{1}{2}\bigr)\log k + k - \log \sqrt{2\pi} + O\biggl(\frac{1}{k}\biggr) \\
&= (k-1)\log n - 2k\log k - \bigl(n - k + \tfrac{1}{2}\bigr)\log \biggl(1 - \frac{k}{n}\biggr) + k - \log (2\pi) + O\biggl(\frac{1}{\sqrt{n}}\biggr)
\end{align}
for $k = \lfloor \sqrt{n+1}\rfloor = \sqrt{n} + O(1)$. With $k^2 = n + O(\sqrt{n})$ and a low-order Taylor expansion of the logarithms this leads to
$$\log p(n) \geqslant 2\sqrt{n} + O(\log n)$$
and hence
$$\liminf_{n \to \infty} \frac{\log p(n)}{\sqrt{n}} \geqslant 2\,,$$
from which
$$a := \inf \: \biggl\{ \frac{\log p(n)}{\sqrt{n}} : n \in \mathbb{N}\setminus \{0\}\biggr\} > 0$$
follows. Thus $p(n) \geqslant e^{c\sqrt{n}}$ holds e.g. for $c = a > 0$.
| {
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infinitude of primes that are $11 \bmod 12$ Suppose there are finitely many primes $\{p_1, \ldots, p_k\}$ which are $11 \pmod {12}$ and consider $p = (p_1 \cdots p_k)^2 + 10$. Then $p_i \nmid p$ for any $i \leq k$, and $p \equiv 1 + 10 \equiv 11 \bmod 12$. Then $p$ is either a prime itself, contradiction, or $p$ has all prime divisors of the form $12n + 1, 12n + 5, 12n + 7, 12n + 11$. Not all prime factor are of the forms $12n + 1, 5, 7$ because then $p \equiv 1, 5, 7 \mod 12$. So it must have a prime factor $12n + 11$, but $p_i \nmid p$ and so we have a contradiction.
Is this proof valid?
| I see what you are trying to do. Here is the bad news. Generalizing a similar proof of that of Euclid's for the infinitude of primes isn't going to work. If you have a number $n=11\pmod {12}$, the first thing to realize is there exists a prime $p | n$ that is either congruent to $5$, $7$, or $11 \pmod {12}$. If there is no prime $p = 11 \pmod {12}$ dividing $n$, then there are two primes $q$ and $q_2$ such that $q = 5 \pmod {12}$ and $q_2 = 7 \pmod {12}$. The good news is that we can restrict integers $n$ of certain forms to only have prime factors congruent to $1$ or $11 \pmod {12}$ (excluding $2$ or $3$). Let $p$ be a prime and consider whether $x^2 = 3 \pmod p$ is solvable or not solvable. By the law of quadratic reciprocity, $p = 1$ or $11 \mod {12}$. Therefore, each prime $p$ dividing $x^2-3$ is either $1$ or $11 \pmod {12}$. If $x = 12k$ is a multiple of $12$, then $(12k)^2-3$ = $144k^2-3$ which is divsible by $3$. Removing this factor of $3$, we have $48k^2-1$, which is congruent to $11 \pmod {12}$. By the previous conditions, all primes $p$ dividing $48k^2-1$ are congruent to $1$ or $11 \pmod {12}$. It is simple to show that not all primes dividing $48k^2-1$ are of the form $1 \pmod {12}$ because if this were true it would imply the $48k^2-1$ is congruent to $1 \pmod {12}$, and $48k^2$ is congruent to $2 \pmod {12}$, but $48$ is a multiple of $12$, and so $48k^2$, a contradiction, therefore there is at least one prime $p$ dividing $48k^2-1$ congruent to $11 \pmod {12}$, which is relatively prime to $k$ obviously. At this point you'd see how this is similar to Euclid's Proof of infinitely many primes.
| {
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A polynomial $p(x)$ with real coefficients is of degree five. Find $p(x)$ in factorised form with real coefficients. A polynomial $p(x)$ with real coefficients is of degree five. The equation $p(x)=0$ has a complex root $2+i$. The graph $y=p(x)$ has the x-axis as a tangent at $(2,0)$ and intersects the cordinate axes at $(-1,0)$ and $(0,4)$. Find $p(x)$ in factorised form with real coefficients.
Firts I found the roots:
$r_1=2+i$, $r_2=2-i$, $r_3=-1$, $r_4=2$ and $r_5=a$, I was not able to find it ( for some reason the anwser considered $r_5=2$, but I was not able to prove it)
$p(x)=B(x+1)(x-2)(x-2+i)(x-2-i)(x-a)$
$p(x)=B(x+1)(x-2)(x^2-4x+5)(x-a)$
$p(0)=4$
then:
$4=B(1)(-2)(+5)(-a)$
$B=\frac{2}{5a}$
Observation: if we consider the value of $a=2$ then the answer is right. However I cannot show that $a=2$
| As the excellent hint in the comment stated, an $x$-axis tangent means a repeated root. That means that $(x-2)$ is a repeated factor of $p(x)$, i.e that $(x-1)^m, m \geq 2$ is a factor of $p(x)$.
The other obvious factors are $(x+1)$ and $(x-(2+i))(x-(2-i)) = (x^2 - 4x + 5)$, which, when multiplied together, gives you a cubic.
That means that the only remaining factor to achieve a fifth degree polynomial is of degree two, and therefore it must be $(x-2)^2$.
So $p(x) = k(x-2)^2(x+1)(x^2-4x+5)$
Apply the last piece of information, that $p(0) = 4$, which gives $k(4)(1)(5) = 4$ or $k = \frac 15$, so $p(x) = \frac 15(x-2)^2(x+1)(x^2-4x+5)$
| {
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How does one compute integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt[3]{5})$
Integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt[3]{5})$
I was told that it is $\mathbb{Z}[\sqrt[3]{5}]$. I do not find this clear. Is there an easy way to see this?
| Let $D$ be a cube-free integer, not $\pm 1$. Consider an element
$$x=a + b\sqrt[3]{D}+c \sqrt[3]{D^2}$$ of $\mathbb{Q}(\sqrt[3]{D})$. The Galois transforms of $x$ inside $\mathbb{Q}(\sqrt[3]{D}, \omega)$ are
$$x_0= a + b\sqrt[3]{D}+c \sqrt[3]{D^2}=x\\
x_1=a + b\sqrt[3]{D}\omega+c \sqrt[3]{D^2}\omega^2\\
x_2=a + b\sqrt[3]{D}\omega^2+c \sqrt[3]{D^2}\omega$$
Assume $x$ is an (algebraic) integer. Then $x_1$, $x_2$ are also integer. Summing the above equality we get $x_0+ x_1 + x_2=3a$, an integer.
Also, $x_0+ \omega^2 x_1 + \omega x_2 = 3 b \sqrt[3]{D}$ and $x_0+ \omega x_1 + \omega^2 x_2 = 3 c \sqrt[3]{D^2}$ are algebraic integers.
Raising to cube and using that $D$ is cube-free, we conclude $3b$, $3c$ are also integers.
The characteristic polynomial of $x$, $(X-x_0)(X-x_1)(X-x_2)$ equals
$$X^3 -3 a X^2 + (3a^2 - 3D b c )X -(a^3+Db^3 + D^2c^3-3D a b c)$$
Now $x$ is an integer if and only if the coefficients of the above polynomial are in $\mathbb{Z}$.
Assume $a$ is an integer.
If only one of the $b$, $c$ is not an integer then $a^3+Db^3 + D^2c^3-3D a b c$ will have at least $3$ in the denominator, contradiction.
If both $b$, $c$ are non-integers then $D$ must be divisible by $3$.
But then again we get denominators in the same expression, contradiction. Therefore, $b$, $c$ are also integers.
Assume $a\not \in \mathbb{Z}$. From $3a^2 - 3D b c$ we conclude that the denominator of $Dbc$ is $9$.
We get a contradiction right away if $D$ is divisible by $3$, since that would imply the denominator of $bc$ is $27$, not possible.
Otherwise, $a$, $b$, $c$ have denominator $3$. Replacing $x$ with $ x -(p + q \sqrt[3]{D}+ r\sqrt[3]{D^2}$ we only have to look
for possible solutions with $a,b,c= \pm \frac{1}{3}$. Therefore, we have to look for $A$, $B$,$C$ in $\{-1,1\}$ so that
$$3\mid (1 - D\cdot B C ) \\
27 \mid (A^3 + D B^3 + D^2 C^3 - 3 D A B C)$$
In the case $D = 5$ we do not get any solutions.
| {
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How to factorize $2x^2-9x+9$ by completing the square? I know that $x^2-bx+c=(x-k)^2=x^2-2kx+k^2$ if it is a complete square. If not we create one by adding and subtracting $\left(\frac{b}{2}\right)^2$
I tried $$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x^2-\frac{9}{2}x+\left(\frac{9/2}{2}\right)^2-\left(\frac{9/2}{2}\right)^2+\frac{9}{2}\right)$$
What am I missing and what is the best way to factorize when we have coefficients?(Is this the right term, ($a x^2$)?)
| Notice that $$x^2-\frac{9}{2}x+\left(\frac{9}{4}\right)^2-\left(\frac{9}{4}\right)^2+\frac{9}{2}=\left(x-\frac{9}{4}\right)^2-\left(\frac{3}{4}\right)^2=\left(x-\frac{9}{4}-\frac{3}{4}\right)\left(x-\frac{9}{4}+\frac{3}{4}\right)$$
| {
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Roots of composite polynomial function For a polynomial function a(x), is there a generalized solution for the roots of:
$$a(a_2(a_3(...(a_m(x))...)))$$
As an example, if:
$$a(x)=x^2-5x+2, m=2$$
How would I find the roots of
$$a(a(x))=(x^2-5x+2)^2-5(x^2-5x+2)+2$$
| Expand the polynomial and solve it the usual way..
In this case,
\begin{align}
(x^2 - 5x + 2)^2 - 5(x^2 - 5x + 2) + 2 &= x^4 - 10 x^3 + 24 x^2 + 5 x - 4
\end{align}
(by Wolfram Alpha)
Then carry out the usual root-finding methods for non-linear equations. For example, using Newton's method,
\begin{align}
f(x) &= x^4 - 10 x^3 + 24 x^2 + 5 x - 4\\
f'(x) &= 4x^3 - 30x^2 + 48x + 5\\
x_{i + 1} &= x_i - \frac{f(x_i)}{f'(x_i)}
\end{align}
With $x_0 = 4$ we get,
\begin{array}{c | c}
i & x_i\\
\hline
0 & 4 \\
1 & 4.5926 \\
2 & 4.6617 \\
3 & 4.6653 \\
4 & 4.6653 \\
\end{array}
Alternatively, you can also get already existing software like Wolfram Alpha to also solve your polynomial for you. The bottom line, however, is that there aren't really any shortcuts. You just need to expand the messy polynomial and do the root-finding. Thankfully, we have computers to take care of such a process.
EDIT:
I just realized that there is indeed an alernative..
You have $a_i(x) = x^2 - 5x + 2$. Then for $a_1(a_2(x)) = 0$,
\begin{align}
a_2^2 - 5a_2 + 2 &= 0\\
\Rightarrow a_2 &= \frac{5 \pm \sqrt{17}}{2}\\
\Rightarrow x^2 - 5x + 2 &= \frac{5 \pm \sqrt{17}}{2}\\
\Rightarrow x^2 - 5x + \frac{-1 \pm \sqrt{17}}{2} &= 0\\
\Rightarrow x &= \frac{5 \pm \sqrt{25 - 2\left(-1 \pm \sqrt{17} \right)}}{2}\\
\Rightarrow x &= \frac{5 \pm \sqrt{27 \pm 2 \sqrt{17}}}{2}
\end{align}
You can then repeat this again if with $a(x)$ instead of $x$ if you had $a_1(a_2(a_3(x)))$. That is, instead of $x$ on the LHS you'd have $x^2 - 5x + 2$ and you could use the quadratic formula once more.
| {
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Prove that all positive integers $n$, $(1-{\sqrt 5})^n$ can be written in the form $a-b{\sqrt 5}$ where $a$ and $b$ are positive integers Prove, by induction, that all positive integers $n$, $(1-{\sqrt 5})^n$ can be written in the form $a-b{\sqrt 5}$ where $a$ and $b$ are positive integers.
I understand these idea of proof by induction and this was a different type of question that I'm used too and wasn't sure on how to approach it as I'm not entirely confident with proving things with induction yet.
Any help would be appreciated.
| 1-Base Step: $$For\ n=1,\ {{\left( 1-\sqrt{5} \right)}^{1}}=a-b\sqrt{5},\ with\ a=1\,and\ b=1$$
2-Inductive step: Assume that ${{\left( 1-\sqrt{5} \right)}^{n}}=a-b\sqrt{5},\ Consider\ {{\left( 1-\sqrt{5} \right)}^{n+1}}$
$\begin{align}
& {{\left( 1-\sqrt{5} \right)}^{n+1}}=\left( 1-\sqrt{5} \right){{\left( 1-\sqrt{5} \right)}^{n}} \\
& \quad \quad \quad \quad \ \ =\left( 1-\sqrt{5} \right)\left( a-\sqrt{5}b \right) \\
& \quad \quad \quad \quad \ \ =a-a\sqrt{5}+5b-b\sqrt{5} \\
& \quad \quad \quad \quad \ \ =\left( a+5b \right)-\left( a+b \right)\sqrt{5} \\
\end{align}$
So the inductive case holds. Now by induction we see that the assumption is true.
| {
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Math Olympiad exercise $\frac{x}{\sqrt{3x+2y+z}}+\frac{y}{\sqrt{3y+2z+x}}+\frac{z}{\sqrt{3z+2x+y}}\leq\sqrt{\frac{x+y+z}{2}}$ I would like to prove the inequality
$$\frac{x}{\sqrt{3x+2y+z}}+\frac{y}{\sqrt{3y+2z+x}}+\frac{z}{\sqrt{3z+2x+y}}\leq\sqrt{\frac{x+y+z}{2}},$$
where $x,y,z>0$. What I did so far is the following:
The inequality is homogeneous, thus we can assume $x+y+z=1$. When we square both sides of the inequality and use Cauchy-Schwartz on the left-hand side, it suffices to show:
$$\frac{x}{3x+2y+z}+\frac{y}{3y+2z+x}+\frac{z}{3z+2x+y}\leq \frac{1}{2}.$$ But I couldn't make any real progress from there. For example, we can write the left-hand side, using $x+y+z=1$, as
$$\frac{x}{2+x-z}+\frac{y}{2+y-x}+\frac{z}{2+z-y}.$$
This seems somehow simpler, but how can I proceed?
| Now, we can use C-S again.
We need to prove that
$$\sum_{cyc}\left(\frac{x}{3x+2y+z}-\frac{1}{3}\right)\leq\frac{1}{2}-1$$ or
$$\sum_{cyc}\frac{2y+z}{3x+2y+z}\geq\frac{3}{2}$$ and
$$\sum_{cyc}\frac{2y+z}{3x+2y+z}=\sum_{cyc}\frac{(2y+z)^2}{(2y+z)(3x+2y+z)}\geq$$
$$\geq\frac{9(x+y+z)^2}{\sum\limits_{cyc}(2y+z)(3x+2y+z)}=\frac{9(x+y+z)^2}{\sum\limits_{cyc}(5x^2+13xy)}\geq$$
$$\geq\frac{9(x+y+z)^2}{\sum\limits_{cyc}(6x^2+12xy)}=\frac{3}{2}.$$
| {
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Prove $u$ is a unit if and only if $N(u) = 1$? The norm $N:\mathbb{Z}[\sqrt[3]{2}] \rightarrow \mathbb{N}$ defined by $N(a+b\sqrt[3]{2} + c\sqrt[3]{4}) = |a^3 + 2b^3 + 4c^3 - 6abc|$ is multiplicative. (Already proven). Show that $\alpha \in \mathbb{Z}[\sqrt[3]{2}]$ is a unit if and only if $N(\alpha) = 1$.
So far proved it one direction, but am struggling with the other.
Proof $\rightarrow$
Let $\alpha$ be a unit, so by defintion $\exists \beta$ such that $\alpha\beta = 1$. We can then use the norm, knowing that $N(\alpha\beta) = |(1)^3 + 2(0)^3 + 4(0)^3 - 6(1)(0)(0)| = 1$. Since the norm is multiplicative, $N(\alpha\beta) = N(\alpha)N(\beta) = 1$, and since the results of the norms have to be positive integers, we know $N(\alpha)=1$.
Proof $\leftarrow$ Let $N(\alpha)$ = 1. Then for $\alpha = a+b\sqrt[3]{2} + c\sqrt[3]{4}$, $|a^3 + 2b^3 + 4c^3 - 6abc|=1$
Any help on where to go?
| Hint: if you embed $\mathbb{Z}[\sqrt[3]{2}]$ into $\mathbb{Z}[\sqrt[3]{2}, \omega]$ where $\omega := e^{2 \pi i / 3}$, then
$$N(a + b \sqrt[3]{2} + c \sqrt[3]{4}) = \left| (a + b \sqrt[3]{2} + c \sqrt[3]{4}) (a + b \omega \sqrt[3]{2} + c \omega^2 \sqrt[3]{4}) (a + b \omega^2 \sqrt[3]{2} + c \omega \sqrt[3]{4}) \right|.$$
(This expression is heavily inspired by the definition of norm in Galois theory, of which your definition is close to being a special case.)
Now, multiply the last two factors on the right hand side, and you should get an expression for $\pm$ of the inverse of $a + b \sqrt[3]{2} + c \sqrt[3]{4}$ which lies in $\mathbb{Z}[\sqrt[3]{2}]$. (With the sign being determined by the sign of $a^3 + 2 b^3 + 4 c^3 - 6abc$.)
| {
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Limit of $\lim\limits_{x\to 1}\left(\frac{\sqrt{x^2+2x+5-8\sqrt{x}}}{\log(x)}\right)$
Given the limit:
$$
\lim\limits_{x\to 1}\left(\frac{\sqrt{x^2+2x+5-8\sqrt{x}}}{\log(x)}\right) = \alpha
$$
Find the value of $\alpha$
Could not get my head around on how to simplify the nominator.
Anyone feeling like this should be easy?
Thanks in advance.
| Let se $x=1+y$ with $y\to0$ then
$$\lim\limits_{x\to 1}\left(\frac{\sqrt{x^2+2x+5-8\sqrt{x}}}{\log(x)}\right) = \lim\limits_{y\to 0}\left(\frac{\sqrt{y^2+4y+8-8\sqrt{1+y}}}{\log(1+y)}\right)$$
and note that
*
*$\sqrt{1+y}=1+\frac12 y-\frac18y^2 +o(y^2)$
*$\log(1+y)=y+o(y)$
then
$$\frac{\sqrt{y^2+4y+8-8\sqrt{1+y}}}{\log(1+y)}=\frac{\sqrt{y^2+4y+8-8-4y+y^2+o(y^2)}}{y+o(y)}=\frac{\sqrt{2y^2+o(y^2)}}{y+o(y)}=\frac{|y|\sqrt{2+o(1)}}{y+o(y)}=\frac{|y|}{y}\frac{\sqrt{2+o(1)}}{{1+o(1)}}$$
thus the limit doesn't exist.
| {
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Why did I get extra solutions to $6\cos \theta + 7\sin \theta = 4 $?
Q. Find the values of $ \theta $ between $ 0^{\circ}$ and $360^{\circ}$ which satisfy the equation
$$ 6\cos \theta + 7\sin \theta = 4 $$
I solved this question differently to how I normally solve these questions, just to experiment and I ended up with two extra solutions. Please explain why this was so.
Workings:
Let $ \sin \theta = \frac{x}{1} $ then $ \cos \theta = \frac{\sqrt{1-x^2}}{1} $
$$6\times \sqrt{1-x^2} = 4-7x $$
$$ (6\times \sqrt{1-x^2})^2 = (4-7x)^2 $$
$$36 - 36x^2 = 16-56x + 49x^2 $$
$$85x^2 - 56x - 20 = 0 $$
$$ x = \frac{28\pm6\sqrt{69}}{85}$$
So $$ \sin \theta = \frac{28\pm6\sqrt{69}}{85} $$
$$ \theta = 66.3^{\circ}, 113.7^{\circ}, 194.9^{\circ}, 345.1^{\circ} $$
| There are two mistakes.
(1) $\cos\theta$ can be negative. It can be $-\sqrt{1-x^2}$.
(2) Squaring an equation may lead to extra answers. But this mistake is neutralized by the first one.
The correct version should be:
Let $\displaystyle \sin \theta = \frac{x}{1} $ then $\displaystyle \cos \theta = \pm\frac{\sqrt{1-x^2}}{1} $
$$6\times \pm\sqrt{1-x^2} = 4-7x $$
$$ (6\times \pm\sqrt{1-x^2})^2 = (4-7x)^2 $$
$$36 - 36x^2 = 16-56x + 49x^2 $$
$$85x^2 - 56x - 20 = 0 $$
$$ x = \frac{28\pm6\sqrt{69}}{85}$$
When $\displaystyle x=\frac{28+6\sqrt{69}}{85}$, $x>0$ and $4-7x<0$. So we should take negative square root $-\sqrt{1-x^2}$ on L.H.S. Now $\sin\theta>0$ and $\cos\theta<0$. $\theta $ is in Quadrant II.
$$\theta=113.7^\circ$$
When $\displaystyle x=\frac{28-6\sqrt{69}}{85}$, $x<0$ and $4-7x>0$. So we should take positive square root $\sqrt{1-x^2}$ on L.H.S. Now $\sin\theta<0$ and $\cos\theta>0$. $\theta $ is in Quadrant IV.
$$\theta=345.1^\circ$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve a linear system when there isn't an exact solution? I've been trying to solve this linear system using Gaussian elimination, but I can't seem to finish it.
$ x +0y - z + w = 0 $
$ 0x + 2y + 2z + 2w = 2 $
$ x + y + 0z + 2w = 1 $
$$ \left[
\begin{array}{cccc|c}
1 & 0 & -1 & 1 & 0 \\
0 & 2 & 2 & 2 & 2 \\
1 & 1 & 0 & 2 & 1
\end{array}
\right] \Rightarrow \left[
\begin{array}{cccc|c}
1 & 0 & -1 & 1 & 2 \\
0 & 1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 & 1
\end{array}
\right] $$
$$ \\ -R_1\rightarrow R_3 \\ \frac{1}{2} R_2 \\ $$
And what then? I couldn't come up with a conventional solution where I just start with finding the rightmost variable and all other variables show themselves easily then. $ -R_2 \rightarrow R_3 $ doesn't end up in anything sensible. Could you please give me a hint on how to move on from here?
p.s. First time using the formatting, pardon my mistakes.
| From
$\left[
\begin{array}{cccc|c}
1 & 0 & -1 & 1 & 0 \\
0 & 1 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 0
\end{array}
\right]$
add the free variables $z=s$, $w=t$ to get
$\left[
\begin{array}{cccc|c}
1 & 0 & -1 & 1 & 0 \\
0 & 1 & 1 & 1 & 1 \\
0 & 0 & 1 & 0 & s\\
0 & 0 & 0 & 1 & t
\end{array}
\right]\to
\left[\begin{array}{cccc|c}
1 & 0 & 0 & 1 & s \\
0 & 1 & 0 & 1 & 1-s \\
0 & 0 & 1 & 0 & s\\
0 & 0 & 0 & 1 & t
\end{array}
\right]\to
\left[\begin{array}{cccc|c}
1 & 0 & 0 & 0 & s-t \\
0 & 1 & 0 & 0 & 1-s-t \\
0 & 0 & 1 & 0 & s\\
0 & 0 & 0 & 1 & t
\end{array}
\right]$
obtaining
$$\left(
\begin{array}{c}
x\\y\\z\\w
\end{array}
\right)=
\left(
\begin{array}{c}
0\\1\\0\\0
\end{array}
\right)+
\left(\begin{array}{r}
1\\-1\\1\\0
\end{array}
\right)s+
\left(\begin{array}{r}
-1\\-1\\0\\1
\end{array}
\right)t$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Boundaries of a doughnut-shaped surface to simplify Stokes theorem problem This question is taken from a former exam for my calculus class.
Let $D$ be the surface given by $(\sqrt{x^2+y^2}-3)^2+z^2=1$ where $z\geq0$. The vectorfield $F$ is equal to $F(x,y,z) = [a, b, c]$.
a, b, c were some arbitrary combinations of variables where the curl of $F$ ended up having nasty $x$ and $y$ components, but with a simple $z$ component equal to $-(3+x)$
The question was to calcuate $\iint_D (curl F)*\hat N dS$ where $\hat N$ is the unitnormal for $S$ with positive $k$-component.
No idea about where to begin with the surface equation $(\sqrt{x^2+y^2}-3)^2+z^2=1$, I looked at the solution where they simply say let $D'$ be the surface $2 \leq \sqrt{x^2+y^2} \leq 4$ where $z=0$. Then $D$ and $D'$ share the same boundary (the two circles): $x^2+y^2=4$ and $x^2+y^2=16$. Positive $k$ component simplifies the integrand to the $z$ component only: $-(3+x)$. $x$ being an odd function, we are left with
$$-3\iint_D dxdy$$ which evaluates to $-3*\pi(4^2-2^2) = -36\pi$
There was a simple sketch of the doughnut shape the surface $D$ forms in the exam paper.
Working with $D$, setting $z=0$, I ended up with
$x^2+y^2-6\sqrt{x^2+y^2}+9=1$
$x^2+y^2-6\sqrt{x^2+y^2}=-8$
My question is; how am I supposed to make this realization?
Specifically, how do I go from this expression $x^2+y^2-6\sqrt{x^2+y^2}=-8$, to realizing that the boundaries can be expressed as two circles with $r=4$ and $r=16$, respectively, and thus simplifying the problem significantly?
Thank you in advance!
| The isosurfaces of a smooth function at a regular value are manifolds without boundary, so the isosurface
$$\left(\sqrt{x^2+y^2}-3\right)^2 + z^2 = 1$$
would have no boundary. A boundary only arises because this surface is being intersected with the upper half-space $z\geq 0$.
So, to find the boundary, we need to find where the surface is being cut by the plane $z=0$. This leads to
$$\left(\sqrt{x^2+y^2}-3\right)^2 = 1.$$
There is no need to square things out; just take the square root of both sides to get
$$\|(x,y)\| = 3\pm 1$$
which of course is a pair of circles on the $z=0$ plane with center at the origin.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving similarity for a triangle inside another triangle given ratio between sides.
In $\triangle ABC$, $A_1, B_1, C_1$ are points on $BC,CA,AB$ such that $$\frac{BA_1}{A_1C} = \frac{CB_1}{B_1A} = \frac{AC_1}{C_1B} = \delta$$ If $A_2,B_2,C_2$ are points on $B_1C_1, C_1A_1, A_1B_1$ such that $$\frac{B_1A_2}{A_2C_1} = \frac{C_1B_2}{B_2A_1} = \frac{A_1C_2}{C_2B_1} = \frac{1}{\delta}$$ then prove that $\triangle ABC$ is similar to $\triangle A_2B_2C_2$ and find the ratio of similitude.
After making a diagram it is pretty much clear that $A_2C_2 || AC, C_2B_2 || BC$ and $B_2A_2 || BA$. Proving this would be sufficient to prove the similarity.
| Working with vectors we have
$$B-A_1 = \delta(A_1-C)$$ so
$$A_1 = \frac{B+\delta C}{1+\delta}\\
B_1=\frac{C+\delta A}{1+\delta}\\
C_1= \frac{A+\delta B}{1+\delta}
$$
Similarly
$$A_2=\frac{\delta B_1 + C_1}{1+\delta}=\frac{\delta\frac{C+\delta A}{1+\delta}+\frac{A+\delta B}{1+\delta}}{1+\delta}=\frac{(\delta^2+1)A + \delta B+\delta C}{(1+\delta)^2}\\
B_2=\frac{(\delta^2+1)B + \delta C+\delta A}{(1+\delta)^2}\\
C_2=\frac{(\delta^2+1)C + \delta A+\delta B}{(1+\delta)^2}$$
Therefore
$$A_2B_2=\frac{\delta^2+1-\delta}{(1+\delta)^2} AB$$
and the others, so the triangles $ABC$, $A_2B_2C_2$ are similar.
We can rewrite the above equalities as
$$A_2= \frac{\delta^2-\delta+1}{(\delta+1)^2}A+\frac{3 \delta}{(\delta+1)^2}\frac{A+B+C}{3}$$
Therefore, the triangle $A_2B_2C_2$ is obtained from $ABC$ with a homothety with center the center of mass of $ABC$ and constant $\frac{\delta^2-\delta+1}{(\delta+1)^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Formalizing an approach to the sum $\sum_{k=0}^{\infty}ka^k$
Let a be $a$ be a real number such that $0<a<1$
find
$\begin{align}
\sum_{k=0}^{\infty}ka^k
\end{align}$
I know that
$a^0+a^1+...+a^\infty = \frac{1}{1-a}$
By stretching it as follows
$\begin{align}
\sum_{k=0}^{\infty}ka^k= \\a^1+a^2+a^3...+a^\infty \\+a^2+a^3+a^4...+a^\infty
\\+a^3+a^4+a^5...+a^\infty
\\ .\\.\\.\\+a^\infty \\= \sum_{k=1}^{\infty}\frac{a^k}{1-a}
\\ = \frac{a}{(1-a)^2}\end{align}$
Writing it a little better,
$\begin{align}
\sum_{k=0}^{\infty}ka^k=
\end{align}$
$\begin{gather}
a^1\rightarrow a^\infty+\\a^2\rightarrow a^\infty+\\ \downarrow \\a^\infty \end{gather}\\$$\begin{align}=\sum_{k=1}^\infty\frac{a^k}{1-a}= \frac{a}{(1-a)^2}\end{align}$
How to formalize this stuff?
| First of all, there is no last term, $a^{\infty}$. The proper way to write these sums would be $a_1+a_2+\cdots$.
Your argument is basically this: we have an infinite triangle of numbers that looks like
$$
\begin{array}{ccccc}
a^1 & a^2 & a^3 & \dots\\
& a^2 & a^3 & \dots\\
& & a^3 & \dots\\
& & & \ddots\\
\end{array}
$$
The given sum, $\sum ka^k$, is what happens when you add up all the columns, then add up all of those sums. What you have done is instead first add up the rows, then add up all of those sums. The fact that both summation methods agree is obvious for a finite array of numbers, but requires justification for infinite ones. It turns out that is always OK to do when all the terms are nonnegative, which they are in your case. This is known as Tonelli's theorem.
Here is another method, which is different and less intuitive than your method, but is easier to justify. By definition, $\sum_{k=1}^\infty ka^k=\lim_{n\to\infty} \sum_{k=1}^n ka^k$. We can compute all of these finite sums exactly using your trick:
$$
\begin{align}
\sum_{k=1}^n ka^k
&=(a^1+\dots+a^n)+(a^2+\dots+a^n)+\dots+(a^n)\\
&=\frac{a^1-a^{n+1}}{1-a}+\frac{a^2-a^{n+1}}{1-a}+\dots+\frac{a^n-a^{n+1}}{1-a}\\
&=\frac{a+\dots+a^n-na^{n+1}}{1-a}\\
&=\frac{\frac{a-a^{n+1}}{1-a}-na^{n+1}}{1-a}\end{align}$$
Letting $n\to\infty$, the $a^{n+1}$ and $ na^{n+1}$ both disappear, leaving the answer you expected.
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$
Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$
Proof. So i'm going to prove this by induction. The first case when $n=1$ is trivial since:
$$9+16=25,$$
implying $ 5 \mid 25$.
Now we need to show is divisible when $n=k+1$. We will use this later on, but $$9^k+4^{k+1}=5k.$$
The
n,
$$\begin{align}9^{k+1}+4^{k+2}
&= 9 \cdot 9^k+4 \cdot 4^{k+1} \\
&= 9 \cdot 9^k+(9-5) \cdot 4^{k+1} \\
&=9 \cdot 9^k+9 \cdot 4^{k+1} -5 \cdot 4^{k+1} \\
&= 9(9^k + 4^{k+1})-5 \cdot 4^{k+1} \\
&=\underbrace{9(5k)-5 \cdot 4^{k+1}}_{\text{This is where the $5k$ comes in}} \\
&= 5(9k-4^{k+1}),\end{align}$$
thus, the original expression a multiple of $5$.
Is my induction correct?
Edit: I see several answers that took a different approach, all is welcome it really helps me see it in a different way. Thank You!
| By the binomial theorem, $9^n+4^{n+1}=(5+4)^n+4^{n+1}=5a+4^n+4^{n+1}=5a+5\cdot4^n$.
| {
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"source": "stackexchange",
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"answer_count": 5,
"answer_id": 1
} |
Why doesn't the matrix $\begin{pmatrix} 10 & -20 & -20 \\ -20 & 40 & 40 \\ -20 & 40 & 40 \end{pmatrix}$ have an orthogonal set of eigenvectors?
Why doesn't the matrix
$$\begin{pmatrix}
10 & -20 & -20 \\
-20 & 40 & 40 \\
-20 & 40 & 40
\end{pmatrix}$$ have an orthogonal set of eigenvectors?
I have used a matrix calculator to diagonalize the matrix but I see that the eigenvectors:
$$\begin{pmatrix}
2 \\
1 \\
0
\end{pmatrix},
\begin{pmatrix}
2 \\
0 \\
1
\end{pmatrix},
\begin{pmatrix}
-1/2 \\
1 \\
1
\end{pmatrix}$$
are not orthogonal. According the spectral theorem, this matrix should have an orthogonal basis of eigenvectors. What is wrong?
| You need to do Gram-Schmidt on the first two eigenvectors to get orthogonal vectors. $$ \begin{pmatrix}
2 \\
1 \\
0
\end{pmatrix},
\begin{pmatrix}
2 \\
0 \\
1
\end{pmatrix},
\begin{pmatrix}
-1/2 \\
1 \\
1
\end{pmatrix}$$
They are eigenvectors associated with the same eigenvalue $\lambda =0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $\lim_{x\to 0} \frac {1-(\cos 2x)^3(\cos 5x)^5(\cos 7x)^7(\sec 4x)^9(\sec 6x) ^{11}}{x^2}$
Evaluate $$\lim_{x\to 0} \frac {1-(\cos 2x)^3(\cos 5x)^5(\cos 7x)^7(\sec 4x)^9(\sec 6x) ^{11}}{x^2}$$
Now I can apply L'Hospital rule twice but believe it or not it would seriously be a very tedious task to do. I also tried writing $\cos 2x=\frac {e^{i2x}+e^{-i2x}}{2}$ and so on but couldn't continue due to large powers. Moreover I don't see any standard limits popping out. The only sequence I could notice was in the powers of the trigonometric functions which follow the series $3,5,7,9,11$
Any hints would be appreciated
| Hint: Try a Taylor series expansion around $0$, and notice you can throw away many terms.
$$\begin{align}&\lim_{x \to 0} \frac {1-(\cos 2x)^3(\cos 5x)^5(\cos 7x)^7(\sec 4x)^9(\sec 6x) ^{11}}{x^2} \\
&=
\lim_{x \to 0} \frac {1-(1-2x^2)^3(1-\frac{25}{2}x^2)^5(1-\frac{49}{2}x^2)^7(1+8x^2)^9(1+18x^2) ^{11}}{x^2} \\
&=\lim_{x \to 0} \frac {1-(1-6x^2)(1-\frac{125}{2}x^2)(1-\frac{343}{2}x^2)(1+72x^2)(1+198x^2)}{x^2} \\
&=\lim_{x \to 0} \frac {1-\left(1 + x^2\left(-6-\frac{125}{2}-\frac{343}{2}+72+198\right)\right)}{x^2} \\
&=-\left(-6-\frac{125}{2}-\frac{343}{2}+72+198\right) \\
&= -30
\end{align}$$
I used the facts that $\sec x = 1 + \frac{1}{2}x^2 + \mathcal{O}(x^4)$, $\cos x = 1 - \frac{1}{2}x^2 + \mathcal{O}(x^4)$ and $(1+t)^n = 1+tn + \mathcal{O}(t^2)$.
| {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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An expression for sin(40) I am trying to find an expression for $\sin(40)$ in the following way:
Let
$\theta$ = 40
3$\theta$=120
$\sin(3\theta)$=sin(120)
$-4\sin^3(\theta)+3\sin(\theta)=\frac {\sqrt 3} 2 $
Put $\sin(\theta)=x$
$-4x^3+3x=\frac {\sqrt 3}2$
$x^3-\frac 34x=\frac {\sqrt 3}{-8}$
We have to solve this cubic equation. I will brought it in the the form $x^3+q=px$. Now I apply the Cardan Formula.The original formula shows roots for $x^3+px=q$. I am putting $-p$ for $p$ and $-q$ for $q$ to get the formula:
$x=\sqrt[3] {\frac {-q}2+ \sqrt{{\frac{q^2}4}-{\frac{p^3}{27}}}}+\sqrt[3] {\frac {-q}2- \sqrt{{\frac{q^2}4}-{\frac{p^3}{27}}}}$
The quantity under square root reduces to $\frac {-1}{256}$.Upon Simplyfying I found the that
$x=\sqrt[3] \frac {-\sqrt 3 +\sqrt{-1}} {16}+\sqrt[3] \frac {-\sqrt 3 -\sqrt{-1}} {16}$
$x=\frac 12{\sqrt[3] \frac {-\sqrt 3 +\sqrt{-1}} {2}+\sqrt[3] \frac {-\sqrt 3 -\sqrt{-1}} {2}}$
This is one complex root of the cubic. Now I simplified this cube roots:
$-\sqrt 3 +i=\text{(polar form)}\:2,150$
$\sqrt[3] {-\sqrt 3 +i}=2^{1/3},50=2^{1/3}(\cos(50)+i \sin(50))$
$\sqrt[3] {-\sqrt 3 -i}=2^{1/3},10=2^{1/3}(\cos(10)+i \sin(10))$
Am I right here? Otherwise How should I continue to find an expression for sin(40).
Please help.
| You'll get a cleaner answer using this identity:
$$\sin x=\frac{1}{2} i e^{-i x}-\frac{1}{2} i e^{i x}$$
Thus for $40^\circ=\frac{2\pi}9$, you get:
$$\begin{align}\sin 40^\circ&=\frac{1}{2} i e^{-\frac{1}{9} (2 i \pi )}-\frac{1}{2} i e^{\frac{2 i \pi }{9}}\\
&=-\frac{1}{2} (-1)^{5/18} \left((-1)^{4/9}-1\right)
\end{align}$$
From $f(x)=-4x^3+3x-\frac{\sqrt3}{2}=0$, we get the roots using Cardano's for $P(x)=ax^3+bx^2+cx+d$ formula using:
$$x_1=-\frac{b}{3 a}+S+T\\
x_2=-\frac{b}{3 a}+\frac{1}{2} \left(i \sqrt{3}\right) (S-T)-\frac{S+T}{2}\\
x_3=-\frac{b}{3 a}-\frac{1}{2} \left(i \sqrt{3}\right) (S-T)-\frac{S+T}{2}$$
Where:
$$S=\sqrt[3]{\sqrt{Q^3+R^2}+R}\\
T=\sqrt[3]{R-\sqrt{Q^3+R^2}}$$
Where:
$$Q=\frac{3 a c-b^2}{9 a^2}\\
R=\frac{-27 a^2 d+9 a b c-2 b^3}{54 a^3}$$
For $f(x)$, we get:
$$Q=-\frac{1}{4}\\
R=-\frac{\sqrt{3}}{16}$$
For $S$ and $T$, we get:
$$S=\sqrt[3]{-\frac{\sqrt{3}}{16}+\frac{i}{16}}\\
T=\sqrt[3]{-\frac{\sqrt{3}}{16}-\frac{i}{16}}$$
And thus we get roots $x_1,x_2,x_3$ as:
$$x_1=\frac{\sqrt[3]{-\sqrt{3}-i}+\sqrt[3]{-\sqrt{3}+i}}{2 \sqrt[3]{2}}\\
x_2=\frac{\left(-1-i \sqrt{3}\right) \sqrt[3]{-\sqrt{3}-i}+i \sqrt[3]{-\sqrt{3}+i} \left(\sqrt{3}+i\right)}{4 \sqrt[3]{2}}\\
x_3=\frac{i \left(\left(-\sqrt{3}+i\right)^{4/3}-\left(-\sqrt{3}-i\right)^{4/3}\right)}{4 \sqrt[3]{2}}$$
Evaluating these numerically, we get that $x_1$ is the only correct answer. Take not too that all these roots are real because $D<0$
| {
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"answer_count": 1,
"answer_id": 0
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We throw two dices distinc. What is the probability that there will be at least a 6 given that the dice show different results? We throw two dices distinc.
What is the probability that there will be at least a 6 given that the dice show different results?
My work
Let
$A=$We have a six,
$B=$Different result
Then $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{1}{6}$
This because: $P(B)=\frac{30}{36}$ and $P(A\cap B)=\frac{5}{36}$
Is good this?
| $A\cap B =$ one 6 and one non-6 = $\{(1,6), (2, 6), (3,6), (4,6), (5,6), (6,1), (6,2),(6,3), (6,4), (6,5)\}$.
So $|A\cap B| = 10, P(A\cap B)=\frac{10}{36}, P(A|B) = \frac{10}{30} = \frac{1}{3}$.
(Incidentally, $|A| = 11$, not $12$, because $A = (A \cap B) \cup \{(6,6)\}$.)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What's the Jordan form of $J^2$? $$J^2=\begin{pmatrix}0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\\ 0&0&0&0&0\end{pmatrix}$$
What's the Jordan form of $J^2$?
I know that it has two blocks and $2$ independent eigenvectors.
So it could be
$$\begin{pmatrix}0&1&0&0&0\\ 0&0&0&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{pmatrix}
\text{ or }
\begin{pmatrix}0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{pmatrix}$$
But why can't the Jordan chain have the length of $4$ and $1$?
| My first answer would have been any of the three existing answers. But here is another take. The Jordan form is unique up to permutation of the blocks. So if we achieve a Jordan form via similarities, it will the the Jordan form.
One particular case of similarity is to conjugate with permutation matrices. The net effect of conjugating with a transposition is to exchange both the row and the column. So
$$
\begin{bmatrix}0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\\ 0&0&0&0&0\end{bmatrix}
\xrightarrow{3\leftrightarrow4}\begin{bmatrix}0&0&0&1&0\\ 0&0&1&0&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{bmatrix}
\xrightarrow{1\leftrightarrow2}\begin{bmatrix}0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{bmatrix}
\xrightarrow{2\leftrightarrow3}\begin{bmatrix}0&1&0&0&0\\ 0&0&0&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{bmatrix}.
$$
At this stage, we have the Jordan form.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating the limit of a product Question: Let $a_1=1$ and $a_n=n(a_{n-1} +1)$ for $n=2,3, \ldots$
Define $P_n= (1+\frac{1}{a_1})(1+\frac{1}{a_2}) \cdots (1+\frac{1}{a_n}) ; \;\;\; n=1,2, \ldots$
Find $\lim_{n \to \infty} P_n$.
[I have outlined my (tedious) approach below, I'm merely curious whether there is a more elegant way to arrive at the result]
My Approach: I was able to arrive at the fact that $P_n=\frac{a_{n+1}}{(n+1)!}$
Then I expanded $a_{n+1}$ as follows
$$ a_{n+1}=(n+1)[a_n+1]$$
$$ = (n+1)[n(a_{n-1}+1)+1]$$
$$ = (n+1)(n)a_{n-1} + (n+1)(n)+(n+1)$$
$$ = (n+1)(n)[(n-1)\{a_{n-2}+1\}] + (n+1)(n)+(n+1)$$
$$ = (n+1)(n)(n-1)a_{n-2} + (n+1)(n)(n-1) + (n+1)(n)+(n+1)$$
Proceeding similarly, we arrive at the expansion
$$ a_{n+1} = (n+1)!a_1 + (n+1)! + \frac{(n+1)!}{2!} + \frac{(n+1)!}{3!} + \cdots + \frac{(n+1)!}{n!}$$
and thus $$ P_n= \frac{a_{n+1}}{(n+1)!}= 1+ \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{n!}$$
$$ \lim_{n\to\infty} P_n = 1+ \frac{1}{1!} + \frac{1}{2!} + \cdots = e$$
| Note that $$\frac{a_{n+1}}{(n+1)!}=\frac{a_n}{n!}+\frac{1}{n!}$$ this gives your final equation.
| {
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Integral $\int \frac {dx}{\sin^2 x + \tan^2x}$
I am trying to find the following integral:
$$\int \frac {dx}{\sin^2 x + \tan^2x}$$
I have tried the common thing to do when encountering rational functions that contains rational functions and converting everything in terms of $\tan \frac{x}{2}$ then substituting it.
$$\tan\frac{x}{2}=t\Rightarrow \sin x=\frac{2t}{1+t^2},\ \cos x=\frac{1-t^2}{1+t^2},\ dx=\frac{2}{1+t^2}dt$$
$$\Rightarrow \int \frac{dx}{\sin^2 x + \frac{\sin^2 x}{\cos^2 x}}=\int \frac{\frac{2}{1+t^2}}{\left(\frac{2t}{1+t^2}\right)^2+\left(\frac{2t}{1-t^2}\right)^2}dt$$
However I am stuck. Is there perhaps an easier way to approach it?
| Let $t=\tan(x)$, then $dx=dt/(1+t^2)$ and $\sin^2(x)=t^2/(1+t^2)$. Hence
$$\int \frac {dx}{\sin^2 x + \tan^2x}=\int \frac {dt}{(t^2+ (1+t^2)t^2)}=\int \frac {dt}{t^2(2+t^2)}=\frac{1}{2}\int \frac {dt}{t^2}-\frac{1}{2}\int \frac {dt}{2+t^2}.$$
Can you take it from here?
P.S. By letting $t=\tan(x/2)$, then $dx=2dt/(1+t^2)$, $\sin(x)=2t/(1+t^2)$ and $\tan(x)=2t/(1-t^2)$. Then we obtain the integral of a more complicated rational function.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Similar matrices for matrix with first column zero This is exercise in Artin's Algebra, chapter 4, exercise 3b:
Let $A=\begin{pmatrix}a&b\\c&d\\ \end{pmatrix}$ be a real matrix. Which matrices with $c=0$ are similar to a matrix in which the "$a$" entry is zero?
I tried: Suppose $\begin{pmatrix}p&q\\ r&s\end{pmatrix}$ is any invertible matrix. Then a similar matrix is $$\frac1{ps-qr}\begin{pmatrix}s&-q\\ \:-r&p\end{pmatrix}\begin{pmatrix}0&b\\ 0&d\end{pmatrix}\begin{pmatrix}p&q\\ r&s\end{pmatrix}=\frac1{ps-qr}\begin{pmatrix}r\left(bs-dq\right)&s\left(bs-dq\right)\\ r\left(dp-br\right)&s\left(dp-br\right)\end{pmatrix}.$$
That's it? Is only this required for exercise? or can I do better?
| Let the matrix be
$$
A =
\begin{pmatrix}
a & b\\
0 & d
\end{pmatrix}
$$
Claim 1: If $b\neq 0$ then we can find the similar desired matrix.
If $b\neq 0$, then
$$
\begin{pmatrix}
1 & 0\\
\frac{a}{b} & 1
\end{pmatrix}
\begin{pmatrix}
a & b\\
0 & d
\end{pmatrix}
\begin{pmatrix}
1 & 0\\
-\frac{a}{b} & 1
\end{pmatrix} =
\begin{pmatrix}
0 & b\\
\frac{-ad}{b} & a+d
\end{pmatrix}
$$
This proves claim 1
Claim 2: If $a\neq d$ then we can find the similar desired matrix.
The characteristic polynomial is $p(t) = t^2 - (a+d)t + ad$ whose discriminant is
$$
\Delta = (a+d)^2 - 4ad = (a-d)^2 \geq 0
$$ Now, if $a\neq d$ then $\Delta > 0$ and $p(t)$ will have two distinct real roots. In other words, $A$ will have two distinct real eigenvalues. So it has real independent eigenvectors. This means that $P^{-1}A P= \Lambda$ where $P$ is an invertible real matrix whose columns are the eigenvectors and $\Lambda$ is a diagonal matrix whose diagonal elements are the eigenvalues. Thus, any matrix $B$ that has the same trace and determinant as $A$ will have the same characterstic polynomial and the same eigenvalues so $B$ will be similar to $\Lambda$ and $A, B$ will be similar. Now let
$$
B =
\begin{pmatrix}
0 & x\\
y & z
\end{pmatrix}
$$
Then by choosing its elements such that $z = a+d$ and $ad = -xy$ we ensure that $A, B$ will have the same trace/determinant and by the argument above, they'll be similar. This proves claim 2.
The only case left is when $a = d$ and $b = 0$. Then $A = aI$. So for any invertible $P$ we have
$$
P^{-1}A P= P^{-1}a I P = aI PP^{-1} = aI
$$
which is obviously not similar unless $a = 0$ in which case we are left with the zero matrix.
To sum it all up, any matrix that is not of the form $aI, a\neq 0$ will do the job.
| {
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Is there a simple proof that there are no "off-by-1" solutions to Fermat's Last Theorem? $x^p + y^p = z^p$ has no integral solutions for $p>2$ due to the work of Andrew Wiles and others. My questions is: consider cases where $p>2$ and $y=x+1$ and $z=x+2$. We might call these "off-by-1" solutions. Is there any simple proof that there are no such off-by-1 solutions? (Admittedly, simple is somewhat subjective.)
| It is easy (and nice!) to discard $p$ odd (look at the J.G.'s answer) because of $(-1)^p\equiv 1\mod{(x+1)}$ impossible for $x\gt1$ and because the equation $1+2^p=3^p$ can be easily discarded so we consider just $p=4n$ and $p=4n+2$.
It is well known the periodicity modulo $4$ of classes modulo $10$. We have modulo $10$ the following table
$$\begin{array}{|c|c|}\hline x & 1 & 2 & 3 & 4&5&6&7&8&9&0 \\\hline x^2 & 1 & 4 &9&6&5&6&9&4&1&0\\\hline x^4 &1 &6 &1&6&5&6&1&6&1&0\\\hline\end{array}$$ It follows
► for $p\equiv 2\pmod4$.
$$\begin{cases}1^2+2^2=3^2\Rightarrow5\equiv9\pmod{10}\\2^2+3^2=4^2\Rightarrow3\equiv6\pmod{10}\\3^2+4^2=5^2\Rightarrow5\equiv5\pmod{10}\\4^2+5^2=6^2\Rightarrow1\equiv6\pmod{10}\\5^2+6^2=7^2\Rightarrow1\equiv9\pmod{10}\\6^2+7^2=8^2\Rightarrow5\equiv4\pmod{10}\\7^2+8^2=9^2\Rightarrow3\equiv1\pmod{10}\\8^2+9^2=0^2\Rightarrow5\equiv0\pmod{10}\\0^2+1^2=2^2\Rightarrow1\equiv4\pmod{10}\end{cases}$$
(This can be written directly from the table but has been made explicit for beginners).
We see that WLG we can proved just the impossibility of
$$(10m+3)^{4n+2}+(10m+4)^{4n+2}=(10m+5)^{4n+2}.....\space\space (*)$$
►for $p\equiv 0\pmod4$.
Similarly we should try only the equation
$$(10m+5)^{4n}+(10m+6)^{4n}=(10m+7)^{4n}.....\space\space (**)$$
So we have reduced the problem to just proving $(*)$ and $(**)$. I stop here until a new attempt and maybe someone can prove beforehand the impossibility of these two equations for which the Pythagorean triples could be perhaps a good way.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Taylor series $\sqrt{1-2x+x^2+o(x^3)}$ I need to solve this asymptotic expansion for $x\to0$
$$\sqrt{1-2x+x^2+o(x^3)}$$
This is the expansion to use:
$$\sqrt{1+t} = 1+ \frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}+o(x^3)$$
let:
$$t = -2x+x^2+o(x^3)$$
My question is, should I stop to the first order or keep going?
How can I decide when to stop?
I have tried first and second order and I get:
$$\sqrt{1+t}= 1+ \frac{-2x+x^2+o(x^3)}{2}+o(-2x+x^2+o(x^3) = 1-x+\frac{1}{2}x^3+o(x^3)$$
The second grade:
$$\sqrt{1+t}= 1+ \frac{-2x+x^2+o(x^3)}{2} - \frac{\big(-2x+x^2+o(x^3)\big)^2}{8} +o\big(-2x+x^2+o(x^3)\big)^2 = 1-x+4x^2+\frac{1}{2}x^3-4x^3-\frac{1}{8}x^4+x^4+o(x^4)$$
| Since we have an $o(x^3)$ term we can use the following
$$\sqrt{1+t} = 1+ \frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}+o(t^3)$$
that is
$$\sqrt{1-2x+x^2+o(x^3) }=1+ \frac{-2x+x^2+o(x^3)}{2}-\frac{(-2x+x^2+o(x^3))^2}{8}+\frac{(-2x+x^2+o(x^3))^3}{16}+o((-2x+x^2+o(x^3))^3)$$
keeping only the term with order less than $o(x^3)$, that is
$$1-x+\frac12x^2 -\frac12x^2+\frac12x^3 -\frac12x^3+o(x^3) =1-x+o(x^3)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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.