Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find limit of $\frac{n(nx_n-\frac{1}{3})}{\ln n}$ knowing that $x_{n+1}=x_n-3x_n^2$ Let $x_n$ a sequence with $\frac{1}{3}>x_1>0$ and $x_{n+1}=x_n-3x_n^2$. Find the limit
$$\lim_{n\to \infty} \frac{n(nx_n-\frac{1}{3})}{\ln n}.$$
I proved that $x_n$ is convergent to $0$ and that $\lim\limits_{n\to \infty} nx_n= \frac{1}{3}$ with Stolz-Cesaro theorem.
$$\lim_{n\to \infty} \frac{n}{\frac{1}{x_n}}=\lim_{n\to \infty} \frac{1}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}=\lim_{n\to \infty} \frac{x_{n+1}x_n}{x_n-x_{n+1}}=\lim_{n\to \infty}\frac{x_n(x_n-3x_n^2)}{3x_n^2}=\frac{1}{3}$$
I also know that $\lim\limits_{n\to \infty}\frac{n}{\ln n}=\infty$. The limit is $\infty \cdot 0$ and I am stuck.
Can I get a clue or a hint please?
| Firstly,
$$\lim_{n\to \infty} \frac{n\left(nx_n-\frac{1}{3}\right)}{\ln n}=\lim_{n\to \infty} nx_n \cdot \lim_{n\to \infty} \frac{n-\frac{1}{3x_n}}{\ln n}=\frac{1}{3} \lim_{n\to \infty}\frac{n-\frac{1}{3x_n}}{\ln n}$$
Now, from Cesaro-Stolz:
$$
\begin{aligned}
\lim_{n\to \infty}\frac{n-\frac{1}{3x_n}}{\ln n} &= \lim_{n\to \infty} \frac{n+1-\frac{1}{3x_{n+1}}-n+\frac{1}{3x_n}}{\ln(n+1)-\ln n}\\
&= \lim_{n\to \infty} \frac{1+\frac{1}{3x_n}-\frac{1}{3x_{n+1}}}{\ln\left(\frac{n+1}{n}\right)}\\
&=\lim_{n\to \infty} \frac{n\left(1+\frac{x_{n+1}-x_n}{3x_nx_{n+1}}\right)}{\ln\left(\frac{n+1}{n}\right)^n}\\
&= \lim_{n\to \infty} n\left(1-\frac{x_n}{x_{n+1}}\right)\\
&= \lim_{n\to \infty} \frac{n(x_{n+1}-x_n)}{x_{n+1}}\\
&= \lim_{n\to \infty} \frac{n(-3x_n^2)}{x_n-3x_n^2}\\
&= -\lim_{n\to \infty} nx_n \cdot \lim_{n\to \infty} \frac{3}{1-3x_n}\\
&=-1
\end{aligned}
$$
Thus, the final result is:
$$\lim_{n\to \infty} \frac{n\left(nx_n-\frac{1}{3}\right)}{\ln n}=-\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How do I solve this limit without l'Hopital? I tried the substitution $t=x-(\pi/3)$ but it doesn't help at all. I have also tried using $\sin(\pi/3)=\sqrt{3}/2$ but couldn't do anything useful then. I tried to factor the denominator and numerator, but it didn't help either. I want a solution without l'Hopital's rule.
$$\lim_{x\to \pi/3} \left[\dfrac{\sin^2(x) - \sin^2\left(\dfrac{\pi}{3}\right)}{x^2 -\left(\dfrac{\pi}{3}\right)^2}\right]$$
| $$ \lim_{x \to \frac{\pi}{3}}\frac{\sin^2 x-\sin^2 \frac{\pi}{3}}{x^2-(\frac{\pi}{3})^2} =\\ \lim_{x \to \frac{\pi}{3}}\frac{(\sin x-\sin \frac{\pi}{3})(\sin x+\sin \frac{\pi}{3})}{(x-\frac{\pi}{3})(x+\frac{\pi}{3})}=\\
\lim_{x \to \frac{\pi}{3}}\frac{2\cos(\frac{x}{2}+\frac{\pi}{6})\sin(\frac{x}{2}-\frac{\pi}{6})(\sin x+\sin \frac{\pi}{3})}{(x-\frac{\pi}{3})(x+\frac{\pi}{3})}=\\\lim_{y \to 0}\frac{2\cos(y+\frac{\pi}{3})\sin(y)(\sin (2y+\frac{\pi}{3})+\sin \frac{\pi}{3})}{2y(2y+\frac{2\pi}{3})}=\\\frac{2\cos \frac{\pi}{3}\sin \frac{\pi}{3}}{\frac{2\pi}{3}}=\frac{3\sqrt 3}{4\pi}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\sum _{n=0}^{\infty} \frac{1}{(n+1) (n+2)} \left(\frac{1}{\lfloor n \phi \rfloor +2}+\frac{1}{\lfloor n \phi ^{-1} \rfloor +2}\right)$ How to prove: $$\sum _{n=0}^{\infty} \frac{1}{(n+1) (n+2)} \left(\frac{1}{\lfloor n \phi \rfloor +2}+\frac{1}{\lfloor n \phi ^{-1} \rfloor +2}\right)=\frac{3}{4}$$
Here $\phi=\frac{1+\sqrt 5}{2}$ and $\lfloor \cdot \rfloor$ the floor function. I suspect this is related to number theory (continued fractions) which I'm not familiar with. Any help will be appreciated.
Update: Here is a related problem, solved by similar techniques.
| The term for $n=0$ gives us $\frac 1 2$.
For the others, use $\frac 1{(n+1)(n+2)} = \frac 1 {n+1} - \frac 1{n+2}$ break the sum into:
$$ \sum_{n=1}^\infty \frac 1{(n+1)(\lfloor n\phi\rfloor + 2)} - \sum_{n'=1}^\infty \frac 1 {(n'+2)(\lfloor n'\phi\rfloor + 2)} + \sum_{m'=1}^\infty \frac 1{(m'+1)(\lfloor m'\phi^{-1}\rfloor + 2)} - \sum_{m=1}^\infty \frac 1 {(m+2)(\lfloor m \phi^{-1}\rfloor + 2)}.$$
Note that each sum is absolutely convergent so there are no issues here. Next, use the following result.
Claim. For any integers $n, n' \ge 1$, we have
$$ m = \lfloor n\phi\rfloor \implies \lfloor m\phi^{-1}\rfloor = n-1, \qquad m' = \lfloor n'\phi\rfloor + 1 \implies \lfloor m'\phi^{-1}\rfloor = n'.$$
Proof. Since $n\phi$ is not an integer, $m = \lfloor n\phi\rfloor $ satisfies $n\phi - 1 < m < n\phi$. This gives $n - \phi^{-1} < m \phi^{-1} < n$ and thus $\lfloor m\phi^{-1} \rfloor = n-1$. The other case is similar.
Back to the problem. We see that every term in the first sum occurs in the fourth. Specifically, if $m = \lfloor n \phi\rfloor$ then $(m+2)(\lfloor m\phi^{-1}\rfloor + 2) = (\lfloor n\phi\rfloor + 2)(n + 1)$. Likewise, if $m' = \lfloor n'\phi\rfloor + 1$ then $(m'+1)(\lfloor m'\phi^{-1}\rfloor + 2) = (\lfloor n'\phi\rfloor + 2)(n' + 2)$. This results in a whole lot of cancellations, and the surviving terms are:
$$- \sum_{m\in A} \frac 1 {(m+2)(\lfloor m\phi^{-1}\rfloor + 2)} + \sum_{m' \in B} \frac 1 {(m'+1)(\lfloor m'\phi^{-1}\rfloor + 2)},
$$
where $A$ (resp. $B$) is the set of positive integers not expressible as $\lfloor n\phi\rfloor$ (resp. $\lfloor n'\phi\rfloor + 1$). Note that $B = \{1\} \cup \{m+1 : m\in A\}$. The case $1\in B$ gives us $\frac 1 4$. For the remaining terms, we claim that for all $m\in A$, we have $\lfloor (m+1)\phi^{-1}\rfloor = \lfloor m\phi^{-1}\rfloor$ which completes the proof since the two sums cancel each other out.
Claim. If positive integer $m$ is not expressible as $\lfloor n\phi\rfloor$, then $\lfloor (m+1)\phi^{-1}\rfloor = \lfloor m\phi^{-1}\rfloor$.
Proof. Since $1 < \phi < 2$, we have an integer $n$ such that
$$ \lfloor n \phi\rfloor = m-1, \qquad \lfloor (n+1)\phi \rfloor = m+1.$$
This gives the inequalities $m-1 < n\phi < m$ and $m+1 < (n + 1) \phi < m+2$ and thus
$n < m \phi^{-1} < n + 1 - \phi^{-1}$. Since $n+\phi^{-1} < (m+1)\phi^{-1} < n + 1$ both floors are $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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$\int_0^\pi\left|\frac{\sin {nx}}{x}\right|dx\ge \frac{2}{\pi}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)$
Question: Show that $$\int_0^\pi\left|\frac{\sin {nx}}{x}\right|dx\ge \frac{2}{\pi}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right).$$
My approach: We know that $$1+\frac{1}{2}+\cdots+\frac{1}{n}=\int_0^1\left(1+x+x^2+\cdots+x^{n-1}\right)dx=\int_0^1\frac{x^n-1}{x-1}dx.$$
Therefore we have $$\int_0^\pi\left|\frac{\sin {nx}}{x}\right|dx\ge \frac{2}{\pi}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right) \\ \Leftrightarrow \int_0^\pi\left|\frac{\sin {nx}}{x}\right|dx\ge \frac{2}{\pi}\int_0^1\frac{x^n-1}{x-1}dx.$$
Now substituting $x=\pi t$, we have $$\frac{2}{\pi}\int_0^1\frac{x^n-1}{x-1}dx=\frac{2}{\pi^{n+1}}\int_0^\pi \frac{t^n-\pi^n}{t-\pi}dt.$$
Thus we have $$\int_0^\pi\left|\frac{\sin {nx}}{x}\right|dx\ge \frac{2}{\pi}\int_0^1\frac{x^n-1}{x-1}dx \\ \Leftrightarrow \int_0^\pi\left|\frac{\sin {nx}}{x}\right|dx\ge \frac{2}{\pi^{n+1}}\int_0^\pi \frac{x^n-\pi^n}{x-\pi}dx\hspace{0.5 cm}...(1)$$
Therefore, if we show that $(1)$ is true, then we are done. Can someone provide me a hint?
| Let $t_k=\frac{k\pi}{n}$ for $k=0,1,2,\ldots,n$. We have
$$\int_{t_{k-1}}^{t_k}\left|\frac{\sin nx}{x}\right|dx=\int_0^\pi\frac{\sin u}{(k-1)\pi+u}du,$$
where $u=nx-(k-1)\pi$. For $0\le u \le \pi$, $(k-1)\pi \le (k-1)\pi+u\le k\pi$, so
$$\int_{t_{k-1}}^{t_k}\left|\frac{\sin nx}{x}\right|dx\ge \int_0^\pi\frac{\sin u}{k\pi}du=\frac{1}{k\pi}\big(-\cos u\big)\Big|_0^\pi=\frac{2}{k\pi}.$$
Hence
$$\int_0^\pi\left|\frac{\sin nx}{x}\right|dx=\sum_{k=1}^n\int_{t_{k-1}}^{t_k}\left|\frac{\sin nx}{x}\right|dx\ge \sum_{k=1}^n\frac{2}{k\pi},$$
which is what we want. In fact, we have
$$\frac{2}{\pi}\sum_{k=1}^n\frac1k\le \int_0^\pi\left|\frac{\sin nx}{x}\right|dx \le \operatorname{Si}(\pi)+\frac{2}{\pi}\sum_{k=1}^{n-1}\frac1k,$$
where $\operatorname{Si}$ is the sine integral ($\operatorname{Si}(\pi)\approx1.8519$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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LU factorization of a singular matrix I am trying to find the LU factorisation of the following matrix: $$A=\begin{pmatrix} 1 & 0 & 3 \\ 2 & 2 & 2 \\ 3 & 6 & -3 \end{pmatrix}.$$ Note that $A$ is singular.
I proceed with Gaussian elimination:
$$A=\begin{pmatrix} 1 & 0 & 3 \\ 2 & 2 & 2 \\ 3 & 6 & -3 \end{pmatrix}\implies \begin{pmatrix} 1 & 0 & 3 \\ 0 & 2 & -4 \\ 0 & 6 & -12 \end{pmatrix}\implies \begin{pmatrix} 1 & 0 & 3 \\ 0 & 2 & -4 \\ 0 & 0 & 0 \end{pmatrix}.$$ Gaussian elimination has not provided an upper triangular matrix, $U$, so how can an LU factorisation be aquired?
| $$A=\begin{pmatrix} 1 & 0 & 3 \\ 2 & 2 & 2 \\ 3 & 6 & -3 \end{pmatrix}=\begin{pmatrix}
a& 0 & 0\\
b& c &0 \\
d& e & f
\end{pmatrix}\times \begin{pmatrix}
1&0 &3 \\
0 & 2 &-4 \\
0 & 0 & 0
\end{pmatrix}$$
multiply it and equate to original matrix and find $a,b,c,d,e,f$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If the value of integral in the image below is π then what is the value of y? I could not simplify
$$
\int_0^1 \sqrt{-1 + \sqrt{\frac{1+y}{x} - y}}\ dx
$$
I tried integration it in an online integrator but trust me the result is seriously daunting to be back traced to $\pi$ as a value, thereby determining $y$. So I am looking for a rather clever trick to get through this one.
| Define
$$f \colon [-1,\infty) \to [0,\infty), \, f(y) = \int \limits_0^1 \sqrt{\sqrt{\frac{1+y}{x} - y}-1} \, \mathrm{d} x \, .$$
Obviously, $f(-1) = 0$. For $y > -1$ we have
\begin{align}
f(y) &= \int \limits_0^1 \sqrt{\sqrt{\frac{1+y}{x} - y}-1} \, \mathrm{d} x \stackrel{\frac{1+y}{x} - y = \frac{1}{t^2}}{=} (1+y) \int \limits_0^1 \frac{2t}{(1+y t^2)^2} \sqrt{\frac{1-t}{t}} \, \mathrm{d} t \\
&\!\stackrel{\text{IBP}}{=} (1+y) \int \limits_0^1 \frac{t^2}{1+y t^2} \frac{1}{2\sqrt{t^3(1-t)}} \, \mathrm{d} t = \frac{1+y}{2} \int \limits_0^1 \frac{\sqrt{\frac{t}{1-t}}}{1+yt^2} \, \mathrm{d} t \\
&\!\!\!\!\stackrel{t = \frac{u}{1+u}}{=} \frac{1+y}{2} \int \limits_0^\infty \frac{\sqrt{u}}{1 + 2u + (1+y) u^2} \, \mathrm{d} u \stackrel{u = \frac{v^2}{\sqrt{1+y}}}{=} (1+y)^{1/4} \int \limits_0^\infty \frac{v^2}{1 + \frac{2}{\sqrt{1+y}} v^2 + v^4} \, \mathrm{d} v \, .
\end{align}
The remaining integral can be computed using the residue theorem or the Cauchy-Schlömilch substitution discussed here. The second method is faster and yields
\begin{align}
f(y) &= \frac{(1+y)^{1/4}}{2} \int \limits_{-\infty}^\infty \frac{\mathrm{d} v}{\left(v - \frac{1}{v}\right)^2 + 2\left(1 + \frac{1}{\sqrt{1+y}}\right)} \stackrel{\text{CS}}{=} \frac{(1+y)^{1/4}}{2} \int \limits_{-\infty}^\infty \frac{\mathrm{d} w}{w^2 + 2\left(1 + \frac{1}{\sqrt{1+y}}\right)} \\
&= \frac{(1+y)^{1/4}}{2} \frac{\pi}{\sqrt{2\left(1 + \frac{1}{\sqrt{1+y}}\right)}} = \frac{\pi}{2} \sqrt{\frac{1+y}{2(1+\sqrt{1+y})}} \, .
\end{align}
As can be seen from the original definition, $f$ is strictly increasing from $0$ to $\infty$, so the inverse function $f^{-1} \colon [0,\infty) \to [-1,\infty)$ exists. For $x \geq 0$ we have
$$ x = \frac{\pi}{2} \sqrt{\frac{1+f^{-1}(x)}{2(1+\sqrt{1+f^{-1}(x)})}} \, ,$$
which can be rewritten as a simple quadratic equation for $\sqrt{1+f^{-1}(x)}$. Its solution is
$$ f^{-1} (x) = \left(\frac{2 x}{\pi}\right)^2 \left[\frac{2x}{\pi} + \sqrt{2 + \left(\frac{2x}{\pi}\right)^2}\right]^2 - 1 \stackrel{x > 0}{=} \left(\frac{2 x}{\pi}\right)^4 \left[1 + \sqrt{1 + 2 \left(\frac{\pi}{2x}\right)^2}\right]^2 - 1 \, ,$$
so
$$ f^{-1}(\pi) = 16 \left(1 + \sqrt{\frac{3}{2}}\right)^2 - 1 = 39 + 16 \sqrt{6} \, . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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how to show $B(n,m)=\frac{(n-1)!(m-1)!}{(n+m-1)!}$?
I have done the following transformation:
$(1-x)^{m-1}=u$ $\Rightarrow -(m-1)(1-x)^{m-2}dx=du$
$x^{n-1}dx=dv$ $\Rightarrow \frac{x^n}{n}=v$
Hence $$\int^1_0x^{n-1}(1-x)^{m-1}dx= (1-x)^{m-1}\frac{x^n}{n}\bigg |^1_0+ \int\frac{x^n}{n}(m-1)(1-x)^{m-2}dx=(1-1)^{m-1}\frac{1}{n}-1^{m-1}\frac{0}{n} + \frac{m-1}{n}\int^1_0x^n(1-x)^{m-2}dx= \frac{m-1}{n}\int^1_0x^n(1-x)^{m-2}dx=\frac{m-1}{n}B(n+1,m-1)$$
But i'm not exactly sure how to show $B(n,m)=\frac{(n-1)!(m-1)!}{(n+m-1)!}$?
| Beginning with the result you have shown, repeated application yields the following chain of equalities:
\begin{align*}
B(n,m) &= \frac{m-1}{n}\cdot B(n+1,m-1) \\
&= \frac{m-1}{n}\cdot \frac{m-2}{n+1}\cdot B(n+2,m-2) \\
&= \cdots \\
&= \frac{m-1}{n}\cdot \frac{m-2}{n+1} \cdot \cdots \cdot \frac{1}{n+m-2} \cdot B(n+m-1,1) \\
&= \frac{(m-1)!(n-1)!}{(n+m-2)!}\cdot B(n+m-1,1)
\end{align*}
It remains to calculate $B(n+m-1,1)$: $$B(n+m-1,1) = \int_0^1x^{n+m-2} \, \mathrm{d}x = \left[ \frac{x^{n+m-1}}{n+m-1} \right]_0^1 = \frac1{n+m-1}.$$
Hence, the result follows, namely: $$B(n,m) = \frac{(m-1)!(n-1)!}{(n+m-1)!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Complex number proof involving angles I need to show that (3 + i)^3 = 18 + 26i and use this to show that the angle AOC = 3AOB, where O, A, B, C are points in the plane given by O = (0, 0), A = (1, 0), B = (3, 1) and C = (18, 26).
This is what I have done:
Expand (3 + i)^3
(3 + i)^3 = (3 + i)(3 + i)(3 + i)
= 9 + 3i + 3i + i^2 * (3 + i)
= 9 + 6i - 1 * (3 + i)
= (8 + 6i) * (3 + i)
= 24 + 8i + 18i + 6i^2
= 24 + 26i - 6
= 18 + 26i
Find the magnitude of OB = 3 + i and OC = 18 + 26i to determine ->OB and ->OC
|OB| = sqrt(3^2 + 1^2) = sqrt(10)
|OC| = sqrt(18^2 + 26^2) = 10 * sqrt(10)
Therefore,
->OB = 3 + i = sqrt(10) * cis(AOB)
->OC = 18 + 26i = 10 * sqrt(10) * cis(AOC)
Cube ->OB and show to show AOC = 3AOB
->OB^3 = (3 + i)^3 = (sqrt(10))^3 * cis^3(AOB)
= 18 + 26i = 10 * sqrt(10) * cis^3(AOB)
We found that (3 + i)^3 = 18 + 26i and by cubing ->OB we find the magnitude is the same as ->OC. Since this is the case, this shows that the angle AOC = 3AOB.
Would this be the correct way to solve this question?
| Use $\sin 3\theta = 3\sin\theta - 4\sin^3 \theta$ (Mathworld [16])
$\sin\angle AOB=\frac1{\sqrt{10}}$ and $\sin\angle AOC=\frac{26}{10\sqrt{10}}$
so we have:
$RHS=\frac3{\sqrt{10}}-\frac4{10\sqrt{10}}=\frac{26}{10\sqrt{10}}=LHS$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3595535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate: $\sum_{k=0}^{n}(-4)^k\frac{{n+k \choose 2k}}{ak+b}$ From this Question:
How does one find the closed form for $(1)?$
$$\sum_{k=0}^{n}(-4)^k\frac{{n+k \choose 2k}}{ak+b}=F(a,b)\tag1$$
$$\sum_{k=0}^{n}(-4)^k\frac{{n+k \choose n-k}}{ak+b}=F(a,b)\tag2$$
This one is from the Question:
$F(0,1)=(-1)^n(2n+1)$
We got these two:
$F(1,1)=\frac{(-1)^n}{n}$
$F(2,1)=\frac{(-1)^n}{2n+1}$
| You can find a closed form as follow :
$$ F(a,b)=\left(\sum_{k=0}^n \int_0^1(-4t)^{ak+b-1}\binom{n+k}{2k}dt \right) $$
Finite sum
$$ F(a,b)=\left( \int_0^1\sum_{k=0}^n(-4t)^{ak+b-1}\binom{n+k}{n-k}dt \right) $$
Then using the coefficient extraction function $[X^i]$ for a given $i$.
\begin{align*}
\sum_{k=0}^n(-4t)^{ak+b-1}\binom{n+k}{n-k}
& = \sum_{k=0}^n(-4t)^{ak+b-1}[X^{n-k}](1+X)^{n+k} \\
& = \sum_{k=0}^n(-4t)^{ak+b-1}[X^n]z^k(1+X)^{n+k} \\
& = [X^n](1+X)^n(-t)^{a+b-1}\sum_{k=0}^n(-4^at^a)^{k}X^k(1+X)^{k} \\
& = [X^n](1+X)^n(-t)^{a+b-1}\dfrac{1-(-4^at^aX(1+X))^{n+1}}{1+4^at^aX(1+X)} \\
&= [X^n](1+X)^n(-t)^{a+b-1}\dfrac{1}{1+4^at^aX(1+X)} \\
&=[X^n](-t)^{a+b-1}\dfrac{(1+X)^{n-1}}{1+4^at^aX} \\
& = [X^n](-t)^{a+b-1}(1+X)^{n-1}\sum_{i=0}^\infty(-1)^i(4^at^aX)^i \\
& = (-t)^{a+b-1}\sum_{i=0}^\infty(-1)^i[X^{n-i}](1+X)^{n-1}(4^at^a)\\
& = (-t)^{a+b-1}\sum_{i=0}^\infty(-1)^i\binom{n-1}{i}(4^at^a)^i\\
& = (-t)^{a+b-1}(1-4^at^a)^{n-1}\\
\end{align*}
From here the exercise can be ended.
Credits
This answer is inspired from the work in the answer here : sum of binomial series with alternate terms.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the sum $\sum_{n=1}^{\infty}\Big((2n-1)\Big(\sum_{k=n}^{\infty}\frac{1}{k^2}\Big)-2\Big) $ Calculate the sum $$\sum_{n=1}^{\infty}\Big((2n-1)\Big(\sum_{k=n}^{\infty}\frac{1}{k^2}\Big)-2\Big) $$
The only thing I can think of is that $\sum_{n=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}.$ However, I can't go any further.
| For $ n\in\mathbb{N} $, define the sequence $ \left(u_{n,k}\right)_{k} $ as follows : $ u_{n,k}=\left\lbrace\begin{aligned}\frac{2n-1}{k^{2}\left(4k^{2}-1\right)},\ \ \ \textrm{If }k\geq n\\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \textrm{If }k<n\end{aligned}\right. $
Since $ \sum\limits_{n=1}^{+\infty}{u_{n,k}}=\frac{1}{k^{2}\left(4k^{2}-1\right)}\sum\limits_{n=1}^{k}{\left(2n-1\right)}=\frac{1}{4k^{2}-1} $, and $ \sum\limits_{k=1}^{+\infty}{\frac{1}{4k^{2}-1}}=\frac{1}{2}\sum\limits_{k=1}^{+\infty}{\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)}=\frac{1}{2} $, the family $ \left(u_{n,k}\right)_{\left(n,k\right)\in\left(\mathbb{N}^{*}\right)^{2}} $ is summable, and Fubini's theorem allows us to write the following : \begin{aligned} \sum_{n=1}^{+\infty}{\sum_{k=1}^{+\infty}{u_{n,k}}}&=\sum_{k=1}^{+\infty}{\sum_{n=1}^{+\infty}}{u_{n,k}}\\ &=\sum_{k=1}^{+\infty}{\frac{1}{4k^{2}-1}}\\ \sum_{n=1}^{+\infty}{\sum_{k=1}^{+\infty}{u_{n,k}}}&=\frac{1}{2} \end{aligned}
But since \begin{aligned} \sum_{n=1}^{+\infty}{\sum_{k=1}^{+\infty}{u_{n,k}}}&=\sum_{n=1}^{+\infty}{\left(2n-1\right)\sum_{k=n}^{+\infty}{\frac{1}{k^{2}\left(4k^{2}-1\right)}}}\\&=\sum_{n=1}^{+\infty}{\left(2n-1\right)\sum_{k=n}^{+\infty}{\left(\frac{4}{4k^{2}-1}-\frac{1}{k^{2}}\right)}}\\ &=\sum_{n=1}^{+\infty}{\left[\left(2n-1\right)\left(2\sum_{k=n}^{+\infty}{\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)}\right)-\left(2n-1\right)\sum_{k=n}^{+\infty}{\frac{1}{k^{2}}}\right]}\\ \sum_{n=1}^{+\infty}{\sum_{k=1}^{+\infty}{u_{n,k}}}&=\sum_{n=1}^{+\infty}{\left(2-\left(2n-1\right)\sum_{k=n}^{+\infty}{\frac{1}{k^{2}}}\right)} \end{aligned}
We get, $$ \sum_{n=1}^{+\infty}{\left(\left(2n-1\right)\left(\sum_{k=n}^{+\infty}{\frac{1}{k^{2}}}\right)-2\right)}=-\frac{1}{2} $$
| {
"language": "en",
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} |
Proving that $\lim_{(x;y)\rightarrow (0;0)} \frac{xy(x-y)}{x^4+y^4} = 0$ using the squeeze theorem My attempt:
$$0 \le \frac{|xy(x-y)|}{|x^4+y^4|} = \frac{|x^2y-xy^2|}{|x^4+y^4|} \le \frac{|x^2y|+|xy^2|}{x^4+y^4} \le \frac{|x|^2|y|+|x||y|^2}{x^4+y^4} = \frac{|x|^2|y|+|x||y|^2}{(x^2+y^2)^2-(\sqrt{2}xy)^2} = ?$$
What do I do next?
| This limit is not $0$ (I bet it doesnt even exist).
If we move towards $0$ on the line $2x=y$ we have
$$\frac{xy(x-y)}{x^4+y^4}=\frac{2y^3}{17y^4}=\frac{2}{17y}$$
and as $y\to0$ this is $\pm\infty$ (depending on whether $y>0$ or $y<0$).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
} |
limit of the sequence $a_n=n\left[\sin\left(\frac{1+n^3}{n^2}\right)-\sin n\right]$ I started by writting
$$
\begin{eqnarray*}
a_n &=& n \left[\sin\left(\frac{1+n^3}{n^2}\right)-\sin\, n\right]=\frac{\sin\left(\displaystyle\frac{1}{n^2}+n\right)-\sin\, n}{\displaystyle\frac{1}{n}} \\
&=&\frac{2\sin\left(\displaystyle\frac{1}{n^2}\right)\cos\left(\displaystyle\frac{1}{n^2}+2n\right)}{\displaystyle\frac{1}{n}}
\end{eqnarray*}
$$
after putting 2 in the denominator of the denominator, I got stuck.
Solution's manual of the book follows my reasoning but writting the result after using the trigonometric identity was different and I did not understood it. Did I do something wrong?
| The basic formula is $sin(x+y)-sin(x-y)=2cos(x)sin(y)$. Here $x+y=n+\frac{1}{n^2}$ and $x-y=n$, resulting in $x=n+\frac{1}{2n^2}$ and $y=\frac{1}{2n^2}$. So $a_n=2ncos(n+\frac{1}{2n^2})sin(\frac{1}{2n^2})$ which is the expression in the book.
Your error was in getting expressions for $x$ and $y$.
| {
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Prove that $\left(A+B\right)^{2}\nmid A^{2n+1}+B^{2n+1}$ Let it be A and B two coprime positive integers. I know how to prove by induction that $A+B\mid A^{2n+1}+B^{2n+1}$, but I am having a bit trouble proving that $\left(A+B\right)^{2}\nmid A^{2n+1}+B^{2n+1}$.
For the case $n=1$ I have managed to prove it in the following way:
$$A^{2n+1}+B^{2n+1}=A^{3}+B^{3}$$
$$A^{3}+B^{3}=A^{2}(A+B)+B\left(B^{2}-A^{2}\right)$$
$$A^{3}+B^{3}=A^{2}(A+B)+B\left(A+B\right)\left(B-A\right)$$
$$A^{3}+B^{3}=\left(A+B\right)\left(A^{2}+B^{2}-AB\right)$$
As $\left(A+B\right)^{2}=A^{2}+B^{2}+2AB$, then we have that $A^{2}+B^{2}-AB=\left(A+B\right)^{2}-3AB$.
Thus, substituting, we get that
$$A^{3}+B^{3}=\left(A+B\right)\left(\left(A+B\right)^{2}-3AB\right)$$
$\left(A+B\right)^{2}\mid A^{3}+B^{3}$ only if $\left(A+B\right)\mid\left(A+B\right)^{2}-3AB$. As $\left(A+B\right)\mid\left(A+B\right)^{2}$, then it follows that $\left(A+B\right)\mid\left(A+B\right)^{2}-3AB$ only if $\left(A+B\right)\mid3AB$.
As A and B are coprime, it follows that $A+B\nmid3AB$. Therefore,
$$\left(A+B\right)^{2}\nmid A^{3}+B^{3}$$
The problem is that it seems not easy (or maybe possible) to apply induction on this method.
My questions are:
*
*Is the above proof correct?
*How could it be proved (if possible) that $\left(A+B\right)^{2}\nmid A^{2n+1}+B^{2n+1}$?
Thanks in advance!
| The statement is false. We show how to find counterexamples.
Since we know that $$A^{2n+1}+B^{2n+1}=(A+B)(A^{2n}-A^{2n-1}B+\dots-AB^{2n-1}+B^{2n}),$$ it is equivalent to finding $A,B$ such that $$A+B\mid A^{2n}-A^{2n-1}B+\dots-AB^{2n-1}+B^{2n},$$ which is the same as $$A^{2n}-A^{2n-1}B+\dots-AB^{2n-1}+B^{2n}\equiv 0\pmod{A+B}.$$
However, we have from $A\equiv -B\pmod{A+B}$ that
\begin{align*}
A^{2n}-A^{2n-1}B+\dots-AB^{2n-1}+B^{2n}&\equiv (-B)^{2n}-(-B)^{2n-1}B+\dots-(-B)B^{2n-1}+B^{2n}\pmod{A+B}\\
&\equiv (2n+1)\cdot B^{2n}\pmod{A+B}.
\end{align*}
As such, we see that counterexamples can be constructed by selecting the value of $n$ such that $A+B\mid 2n+1$. For example, $(A,B,n)=(3,4,3)$ is a counterexample because $$(3+4)^2=49\mid 18571=3^7+4^7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3606907",
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How to prove $\frac{v}{\pi} \int_0^1 \frac{1}{x^2 \sqrt{1-x^2}} e^{\frac{-v^2}{2x^2}} dx =\frac{1}{\sqrt{2\pi}} e^{-\frac{v^2}{2}}$ If $v>0$ why
$$\frac{v}{\pi} \int_0^1 \frac{1}{x^2 \sqrt{1-x^2}} e^{\frac{-v^2}{2x^2}} dx
=\frac{1}{\sqrt{2\pi}} e^{-\frac{v^2}{2}}$$.
I saw it in
I tried to solve it by $t=\frac{1}{x^2}-1$
$$\frac{v}{\pi} \int_0^1 \frac{1}{x^2 \sqrt{1-x^2}} e^{\frac{-v^2}{2x^2}} dx
=\frac{v}{\pi}\int_0^{\infty} \frac{1}{\frac{1}{t+1} \sqrt{1-\frac{1}{t+1}}} e^{\frac{-v^2}{2}(t+1)} \, \frac{1}{2}(t+1)^{\frac{-3}{2}}dt $$
$$=\frac{v e^{\frac{-v^2}{2}} }{2\pi}\int_0^{\infty}\frac{1}{\sqrt{t}} e^{\frac{-v^2}{2}t} \, dt $$
Thanks in advance for any help you are able to provide.
| First substitute $y=\frac{v^2}{2x^2}$. Then $x^2=\frac{v^2}{2y}$ and $\mathrm dx=\color{green}{-\frac{v}{2\sqrt 2} y^{-\frac32}}\,\mathrm dy$. Thus $$\int_0^1 \frac{1}{\color{blue}{x^2} \color{orange}{\sqrt{1-x^2}}} e^{\frac{-v^2}{2x^2}} \,\mathrm dx=\color{green}-\frac{\color{blue}2\color{green}v}{\color{green}{2\sqrt 2} \color{blue}{v^2}}\int_\infty^{\frac{v^2}2} \frac{\color{blue}y\exp(-y)}{\color{green}{\sqrt{y^3}}\color{orange}{\sqrt{1-\frac{v^2}{2y}}}}\,\mathrm dy=\frac{1}{v\sqrt 2}\int_{\frac{v^2}2}^\infty \frac{\exp(-y)}{\sqrt{y-\frac{v^2}2}}\,\mathrm dy.$$
Now use $z=y-\frac{v^2}2$ and this turns into $$\frac{1}{v\sqrt 2}\int_0^\infty \frac{\exp\left(-z-\frac{v^2}2\right)}{\sqrt{z}}\,\mathrm dz=\frac{\exp\left(-\frac{v^2}2\right)}{v\sqrt 2} \color{violet}{\int_0^\infty \frac{\exp(-z)}{\sqrt z}\,\mathrm dz}=\frac{\exp\left(-\frac{v^2}2\right)}{v\sqrt 2} \color{violet}{\Gamma\left(\frac12\right)}=\frac{\color{violet}{\sqrt \pi}\exp\left(-\frac{v^2}2\right)}{v\sqrt 2}.$$
Hence $$\frac{v}{\pi}\int_0^1 \frac{1}{{x^2} {\sqrt{1-x^2}}} e^{\frac{-v^2}{2x^2}} \,\mathrm dx=\frac{v}{\pi}\frac{{\sqrt \pi}\exp\left(-\frac{v^2}2\right)}{v\sqrt 2}=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{v^2}2\right).$$ QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3607629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how to compute $\lim_{n \to \infty}\sqrt{n}\int_{0}^{1}(1-x^2)^n$? I need to compute $$\lim_{n \to \infty}\sqrt{n}\int_{0}^{1}(1-x^2)^n dx.$$
I proved that
for $n\ge1$,
$$\int_{0}^{1}(1-x^2)^ndx={(2n)!!\over (2n+1)!!},$$
but I don't know how to continue from here.
I also need to calculate $\int_{0}^{1}(1-x^2)^ndx$ for $n=50$ with a $1$% accuracy. I thought about using Taylor series but also failed.
| Redoing what has been done
many many many times before.
$\begin{array}\\
I_n
&=\int_0^1 (1-x^2)^n dx\\
I_0
&=\int_0^1 dx\\
&= 1\\
I_1
&=\int_0^1 (1-x^2) dx\\
&=1-\dfrac13\\
&=\dfrac23\\
I_n
&=\int_0^1 (1-x^2)^n dx\\
&=x(1-x^2)^n|_0^1+\int_0^1 2x^2n(1-x^2)^{n-1} dx\\
&\qquad\text{integrating by parts}\\
&\qquad f = (1-x^2)^n, f' = -2xn(1-x^2)^{n-1}, g' = 1, g = x\\
&=2n\int_0^1 x^2(1-x^2)^{n-1} dx\\
&=2n\int_0^1 (x^2-1+1)(1-x^2)^{n-1} dx\\
&=2n\int_0^1 (1-(1-x^2))(1-x^2)^{n-1} dx\\
&=2n\int_0^1 (1-x^2)^{n-1} dx-2n\int_0^1 (1-x^2)^{n} dx\\
&=2nI_{n-1}-2nI_n\\
\text{so}\\
I_n
&=\dfrac{2n}{2n+1}I_{n-1}\\
\dfrac{I_n}{I_{n-1}}
&=\dfrac{2n}{2n+1}\\
I_n
&=\dfrac{I_n}{I_{0}}\\
&=\prod_{k=1}^n\dfrac{I_k}{I_{k-1}}\\
&=\prod_{k=1}^n\dfrac{2k}{2k+1}\\
&=\dfrac{\prod_{k=1}^n(2k)}{\prod_{k=1}^n(2k+1)}\\
&=\dfrac{\prod_{k=1}^n(2k)\prod_{k=1}^n(2k)}{\prod_{k=1}^n(2k)\prod_{k=1}^n(2k+1)}\\
&=\dfrac{4^nn!^2}{(2n+1)!}\\
&=\dfrac{4^nn!^2}{(2n)!(2n+1)}\\
&\approx\dfrac{4^n(\sqrt{2\pi n}(n/e)^n)^2}{\sqrt{2\pi 2n}(2n/e)^{2n}(2n+1)}
\qquad\text{Stirling strikes twice}\\
&=\dfrac{4^n((2\pi n)(n^{2n}/e^{2n})}{2\sqrt{\pi n}4^nn^{2n}e^{2n}(2n+1)}\\
&=\dfrac{2\pi n}{2\sqrt{\pi n}(2n+1)}\\
&=\dfrac{\sqrt{\pi n}}{(2n+1)}\\
&=\dfrac{\sqrt{\pi n}}{2n(1+1/(2n))}\\
&=\dfrac{\sqrt{\pi n}}{2n}\dfrac1{1+1/(2n)}\\
&=\dfrac{\sqrt{\pi }}{2\sqrt{n}}\dfrac1{1+1/(2n)}\\
&=\dfrac{\sqrt{\pi }}{2\sqrt{n}}(1-\dfrac1{2n}+O(\dfrac1{n^2}))\\
\end{array}
$
so
$\sqrt{n}I_n
=\dfrac{\sqrt{\pi }}{2}(1-\dfrac1{2n}+O(\dfrac1{n^2}))
\to\dfrac{\sqrt{\pi }}{2}
$.
| {
"language": "en",
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"answer_count": 4,
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Laurent expansion of $f(z)=\frac{1}{1+z^2}+\frac{1}{3-z}$ in the region $|z|>3$. Laurent expansion of $f(z)=\frac{1}{1+z^2}+\frac{1}{3-z}$ in the region $|z|>3$.
According to the solution the expansion is
$$\sum_{n=0}^{\infty}(-1)^nz^{-2n}+\sum_{n=0}^{\infty}\frac{3^n}{z^{n+1}}.$$
But I don't understand why. My work is as follows:
\begin{align*}
f(z)&=\frac{1}{z^2}\frac{1}{1-(-z^{-2})}-\frac{1}{z}\frac{1}{1-\frac{3}{z}}\\
&=\frac{1}{z^2}\sum_{n=0}^{\infty}(-1)^nz^{-2n}-\frac{1}{z}\sum_{n=0}^{\infty}\left(\frac{3}{z}\right)^n\\
&=\sum_{n=0}^{\infty}(-1)^nz^{-2n-2}-\sum_{n=0}^{\infty}\frac{3^n}{z^{n+1}}
\end{align*}
But in the correct answer, what happens to the $\frac{1}{z^2}$ and the negative from the $-\frac{1}{z}$ in the second term?
| I think the given answer, the first summatoy BEGINS in $n=1$.
$$\frac{1}{z^2}\frac{1}{1-(-z^2)} = - \frac{-z^{-2}}{1-(-z^2)}= -\sum_{n=1}^\infty (-z^2)^n=\sum_{n=1}^\infty (-1)^{n+1} z^{2n}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Players and tournaments 16 players take part in a tennis tournament.
The order of the matches is chosen at random.
If there is a player better than the other one then the better one wins find the :
a)probability that all the four best place reach the semifinals
b)probability that sixth best reaches the semifinals
My attempt:
I did realise that we need to select 8 players out of 12 less better players so (12C8*8C2*6C2*4C2*2*2*2) in the numerator /16C2*8C2 in the denominator.
But I can't understand why this comes out to be >1. This means I am doing a mistake
Please help me out.
| I am assuming the way the tournament works is that there is a first round with $8$ games, then the $8$ winners of the first round play a second round of $4$ games, then the winners of the second round enter the semi-final. In order for the top $4$ players to all play in the semi-final, no two of them can play against each other in either of the first two rounds.
For the sake of exposition, let's say the top four are the players numbered $1,2,3$ and $4$. So what is the probability that player $1$ does not play another of the top four in his first game? There are $15$ other players, of which $3$ are also in the top $4$. So the probability that player $1$ does not play another of the top $4$ is $12/15$.
Assuming player $1$ does not play another player in the top $4$, what is the probability that player $2$ does not play another in the top $4$? Excluding player $1$ and his opponent, there are $13$ players left, of which player $2$ must avoid $2$. So the probability that player $2$ does not play another in the top $4$, given that player $1$ doesn't, is $11/13$.
Proceeding in this way, we see that the probability that all top $4$ players make it through the first round is
$$\frac{12}{15} \cdot \frac{11}{13} \cdot \frac{10}{11} \cdot \frac{9}{9}$$
and the probability that all top $4$ players make it through the second round, given that they made it through the first round, is
$$\frac{4}{7} \cdot \frac{3}{5} \cdot \frac{2}{3} \cdot \frac{1}{1}$$
So the probability that the top $4$ all make it to the semi-final is
$$\frac{12}{15} \cdot \frac{11}{13} \cdot \frac{10}{11} \cdot \frac{9}{9} \cdot \frac{4}{7} \cdot \frac{3}{5} \cdot \frac{2}{3} \cdot \frac{1}{1}$$
* EDIT *
The second part of the problem asks for the probability that player $6$ survives to the semi-finals. Let's say his portion of the tournament bracket looks like this:
6 A B C
| | | |
+-6-+ +-?-+
| |
+---6---+
Player $6$ survives to the semi-final exactly when all the other three players in his bracket of four are not in the top $6$, i.e, are in the set $\{7, 8, 9, \dots ,16\}$. There are $\binom{15}{3}$ ways to select the other three players in the bracket. Of these, $\binom{10}{3}$ consist entirely of players not in the top $6$. So the probability that player $6$ survives to the semi-final is
$$\frac{\binom{10}{3}}{\binom{15}{3}} = \frac{10 \cdot 9 \cdot 8}{15 \cdot 14 \cdot 13}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $abcd \le 3+ ab+ac+ad+bc+bd+cd$ when $(a^2+1)(b^2+1)(c^2+1)(d^2+1)=16$ Mathematica proves the inequality (the code on demand) $$abcd \le 3+ ab+ac+ad+bc+bd+cd$$ under the condition $$(a^2+1)(b^2+1)(c^2+1)(d^2+1)=16$$
The equality is attained at
$$\begin{align}
a &=\frac{\frac{927}{512}-\frac{73 \left(1196032-\sqrt{578562030719}\right)}{83867680}}{-\frac{609 \left(1196032-\sqrt{578562030719}\right)}{1341882880}+\frac{\sqrt{578562030719}-1196032}{2620865}+\frac{73}{32}} \\[4pt]
b &= -\frac{87}{32} \\[4pt]
c &= \frac{7}{16} \\[4pt]
d &= \frac{1196032-\sqrt{578562030719}}{2620865}
\end{align}$$
The question arises: How to prove it by hand?
| Since $$(x^2+y^2)(z^2+t^2)=(xz+yt)^2+(xt-yz)^2,$$ we obtain:$$16=(a^2+1)(b^2+1)(c^2+1)(d^2+1)=((ab-1)^2+(a+b)^2)((cd-1)^2+(c+d)^2)=$$
$$=((ab-1)(cd-1)-(a+b)(c+d))^2+((ab-1)(c+d)+(cd-1)(a+b))^2=$$
$$=(1+abcd-ab-ac-ad-bc-bd-cd)^2+$$
$$+(abc+abd+acd+bcd-a-b-c-d)^2\geq$$
$$\geq(abcd+1-ab-ac-ad-bc-bd-cd)^2,$$ which gives
$$-4\leq abcd+1-ab-ac-ad-bc-bd-cd\leq4$$ and
$$3+ab+ac+ad+bc+bd+cd\geq abcd.$$
The equality occurs for $$abc+abd+acd+bcd=a+b+c+d$$ and $$\prod_{cyc}(a^2+1)=16.$$
| {
"language": "en",
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How do I find this matrix? a) Let $\mathbf{P}$ be the $2\times 2$ matrix that projects vectors onto $\mathbf{u} = \begin{pmatrix}2\\ -1 \end{pmatrix}$. That is,
$\mathbf{P} {v} = \operatorname{proj}_{u} ({v}) = \text{Projection of $\mathbf{v}$ onto $\mathbf{u}$}.$
Using the geometric meaning of the matrix and the picture, find
$\mathbf{P} \begin{pmatrix}2 \\ -1 \end{pmatrix} \text{ and } \mathbf{P} \begin{pmatrix}1 \\ 2 \end{pmatrix} .$
The answer to part a)
\begin{align*}
\mathbf{P} \begin{pmatrix}2 \\ -1 \end{pmatrix} = \boxed{\begin{pmatrix} 2 \\- 1\end{pmatrix}}, \mathbf{P} \begin{pmatrix}1 \\ 2 \end{pmatrix} = \boxed{\begin{pmatrix} 0 \\ 0\end{pmatrix}}.
\end{align*}
b) Now, let $\mathbf{P}$ be the $2\times 2$ matrix that projects vectors onto $\mathbf{u} = \begin{pmatrix}2\\ -1 \end{pmatrix}$. That is,
$\mathbf{P} {v} = \operatorname{proj}_{{u}} ({v}) = \text{Projection of $\mathbf{v}$ onto $\mathbf{u}$}.$Use your answers from part (a) to figure out $\mathbf{P}$.
I set $P = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and I have $\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}.$ How should I continue?
| You should set $(x,y)^T=(2,-1)^T$ as you did for your answer in part (a). This will not give you the values you seek, but will produce simultaneous equations for them. By repeating the same process, but for the other values obtained at the end of part (a), you will arrive at the values of $a,b,c,d$ and therefore $P$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3618121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Four roots of the polynomial $x^4+px^3+qx^2+rx+1$ Let $a, b, c, d$ be four roots of the polynomial $x^4+px^3+qx^2+rx+1$ prove that $(a^4+1)(b^4+1)(c^4+1)(d^4+1)=(p^2+r^2)^2+q^4-4pq^2r$
Please provide hint.
| $$\prod_{cyc}(a^4+1)=((a^2b^2-1)^2+(a^2+b^2)^2)((c^2d^2-1)^2+(c^2+d^2)^2)=$$
$$=((a^2b^2-1)(c^2+d^2)+(c^2d^2-1)(a^2+b^2))^2+$$
$$+((a^2b^2-1)(c^2d^2-1)-(a^2+b^2)(c^2+d^2))^2=$$
$$=\left(\sum_{cyc}a^2b^2c^2-\sum_{cyc}a^2\right)^2+\left(a^2b^2c^2d^2-\frac{1}{6}\sum_{sym}a^2b^2\right)^2.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3619135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of ways to color in a $2\times5$ grid
How many ways are there to color in each unit square of a $2\times5$ grid red or blue so that no $2\times2$ square is allowed to be the same color?
So I thought about doing an inclusion-exclusion method to approach this problem, but my main issue with this approach was that it was too hard to bash out all the cases involved with such an approach (that is, considering the cases where there exist $2\times2$ squares which are the same color). Any thoughts on a better approach I could take?
| Let $a_n$ denote the number of good colorings of a $2\times n$ grid. You want to compute $a_5$. The initial conditions are $a_0 = 1$ and $a_1 = 4$. To derive a recurrence relation for $a_n$, consider the four possibilities for the two cells in the leftmost column. You will find that $a_n = 3 a_{n-1} + 2 a_{n-2}$, yielding:
\begin{align}
a_2 &= 3 a_1 + 2 a_0 = 3\cdot4 +2\cdot 1 = 14 \\
a_3 &= 3 a_2 + 2 a_1 = 3\cdot14 +2\cdot 4 = 50 \\
a_4 &= 3 a_3 + 2 a_2 = 3\cdot50 +2\cdot 14 = 178 \\
a_5 &= 3 a_4 + 2 a_3 = 3\cdot178 +2\cdot 50 = \color{red}{634}
\end{align}
Alternatively, if you just care about $n=5$, the inclusion-exclusion approach yields
\begin{align}
a_5 &= 2^{10} - 2(5-1)2^6
+ (2(5-2)2^4 + 2^2\cdot 3\cdot 2^2)
- (2(5-3)2^2 + 2^3\cdot 2^0)
+ 2(5-4)2^0 \\
&= 1024 - 512 + (96 + 48) - (16 + 8) + 2 \\
&= \color{red}{634}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3619775",
"timestamp": "2023-03-29T00:00:00",
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Langley's Adventitious Angles${}$ I've been running in circles and couldn't give a rigorous mathematical proof that the angle is x = 20°.
Any idea? This is my try:
I got the answer $x=20^\circ$ using a computer program: https://www.geogebra.org/classic/qt79hpec
| This is a variant of the original Langley's puzzle, which has a straightforward trigonometric solution. Apply the sine rule to the triangles ADE, ADB and BDE
$$\frac{\sin x}{\sin 10}\cdot \frac{\sin 20}{\sin (30+x)}\cdot \frac{\sin 80}{\sin 60}
=\frac{DA}{DE}\cdot \frac{DE}{DB}\cdot \frac{DB}{DA} = 1$$
which simplifies to
$$2\cos^210\sin x = \sin60\sin(30+x) =\frac{\sqrt3}4\cos x + \frac{3}4\sin x$$
Solve for $\tan x$,
$$\begin{align}
\tan x & = \frac{\sqrt3}{1+4\cos 20} = \frac{\sqrt3\sin 20}{(\sin 20 +\sin 40 )+ \sin40} \\
& = \frac{\sqrt3\sin 20}{2\sin30\cos10 +\sin 40} = \frac{\sqrt3\sin 20}{\sin 80 +\sin 40} = \frac{\sqrt3\sin 20}{\sqrt3\cos 20} =\tan 20 \\
\end{align}$$
Thus, $x = 20$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Showing that $\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}$ via Fourier Transform If $a,b>0$, how can I prove this using Fourier Series
$$\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}.$$
I tried to split the product and calculate the integral using Parceval's Theorem, but $\frac{x}{(x^2+a^2)}$ and $\frac{x^2}{(x^2+a^2)}$ aren't in $L^1(\mathbb{R})$.
Any hints hold be appreciated.
| I know what is following is not what is asked but it's a simpler way.
$a\neq b$
\begin{align}J&=\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx\\
&=2\int_{0}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx\\
&=\frac{2b^2}{b^2-a^2}\int_0^\infty \frac{1}{x^2+b^2}\,dx-\frac{2a^2}{b^2-a^2}\int_0^\infty \frac{1}{x^2+a^2}\,dx\\
&=\frac{2b^2}{b(b^2-a^2)}\left[\arctan\left(\frac{x}{b}\right)\right]_0^\infty -\frac{2a^2}{a(b^2-a^2)}\left[\arctan\left(\frac{x}{a}\right)\right]_0^\infty\\
&=\frac{\pi b}{b^2-a^2}-\frac{\pi a}{b^2-a^2}\\
&=\boxed{\frac{\pi }{a+b}}
\end{align}
By continuity, the formula is also true for $a=b$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is: If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is:
$A)$ $\le3$ , $B)$ $\ge3$, $(C)$ $\ge2$, $(D)$ $\le2$
My attempt is as follows:-
$$\dfrac{1}{2}\left(\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}\right)$$
Let $y=\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}$
$$A.M\ge H.M$$
$$\dfrac{\dfrac{s-a}{a}+\dfrac{s-b}{b}+\dfrac{s-c}{c}}{3}\ge \dfrac{3}{y}$$
$$\dfrac{\dfrac{s\cdot (ab+bc+ca)}{abc}-3}{3}\ge\dfrac{3}{y}$$
$$y\ge\dfrac{9}{\dfrac{(a+b+c)(ab+bc+ca)}{2abc}-3}\tag{1}$$
Let $z=\dfrac{(a+b+c)(ab+bc+ca)}{abc}$
Equation $(1)$ will give us $y_{min}$, so for that we need to find maximum value of $z$
But unfortunately I was able to find minimum value of $z$ in the following way
$$\dfrac{a+b+c}{3}\ge \dfrac{3abc}{ab+bc+ca}$$
$$\dfrac{(a+b+c)(ab+bc+ca)}{abc}\ge 9$$
But nevertheless, I tried plugging this minimum value of $z$ into equation $(1)$ and I got $y\ge 6$ and as the original expression was $\dfrac{y}{2}$, so $\dfrac{y}{2}\ge 3$ and surprisingly this answer is correct. What am I missing here?
| Note that $f(x)=\frac x{s-x}$ is a convex function. Then, Jensen’s inequality leads to,
$$\begin{align}
& \dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}\\
=& \frac12 \left(\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}\right) \\
\ge &\frac12\cdot3\cdot\frac{ \frac{a+b+c}3}{s -\frac{a+b+c}3}=\frac32\cdot\frac{\frac{2s}3}{s-\frac{2s}3}=3
\end{align} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Find complex roots of quartic function $(3z + 1)(4z + 1)(6z + 1)(12z + 1) = 2$ I found a math problem involving complex number
Find all complex number z such that
$$(3z + 1)(4z + 1)(6z + 1)(12z + 1) = 2$$
The complex number form is z = a + bi
If I multiply all the factor to make it to the standard form of a polynomial, It would be too long to solve. Not even substituting z = a + bi to the polynomial yet.
I don't think that's a good way to solve
I want to know if there's a simpler way to solve this problem, can anyone shows me a hint?
| Let $z=\frac{x}{12}.$
Thus, $$(3z + 1)(4z + 1)(6z + 1)(12z + 1)-2=$$
$$=\left(\frac{x}{4}+1\right)\left(\frac{x}{3}+1\right)\left(\frac{x}{2}+1\right)(x+1)-2=$$
$$=\frac{1}{24}((x+4)(x+3)(x+2)(x+1)-48)=$$
$$=\frac{1}{24}((x^2+5x+4)(x^2+5x+6)-48)=$$
$$=\frac{1}{24}(x^2+5x+12)(x^2+5x-2)=$$
$$=(12z^2+5z+1)(72z^2+30z-1).$$
Can you end it now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show value of $\frac{\tan(x+A)}{\tan(x-A)}$ can not lie between $\tan^2(\pi /4 -A)$ and $\tan^2(\pi /4 +A)$ Let $$y=\frac{\tan(x+A)}{\tan(x-A)}.$$
Show that $y$ can't lie between $\tan^2(\pi /4 -A)$ and $\tan^2(\pi /4 +A)$.
I tried by contradiction. Assumed $$\tan^2\left(\frac{\pi}{4} -A\right)<y$$
and the counterpart's expression. Got a quadratic inequality in $\tan x$ from the former.
Solved and obtained a range of $\tan x$. (A large expression). Do I have to put the range of $\tan x$ in $y$, or is there any other method?
| Use the short hands $t=\tan x$, $a=\tan A$ to rewrite $y=\frac{\tan(x+A)}{\tan(x-A)}$ as a quadratic equation in $t^2$ as you attempted,
$$(1+y)at^2+(1+a)(1-y)t+a(1+y)=0$$
Considering that the range over which $y$ can't lie corresponds to no real solutions for $t$, which requires that the discriminant is negative, i.e.
$$(1+a^2)^2(1-y)^2-4a^2(1+y)^2<0$$
Reexpress the inequalities as
$$ -\frac{2|a|}{1+a^2} < \frac{1-y}{1+y} < \frac{2|a|}{1+a^2} $$
Rearrange to get
$$ \left( \frac{1-|a|}{1+|a|} \right)^2 < y < \left( \frac{1+|a|}{1-|a|} \right)^2 \tag 1$$
Now, examine two cases of $A$:
Case 1) A in the 1st and 3rd quadrants. Then, $|a| = \tan A$ and the inequalities becomes
$$ \left( \frac{1-\tan A}{1+\tan A} \right)^2 < y < \left( \frac{1+\tan A}{1-\tan A} \right)^2$$
which is
$$ \tan^2 (\pi /4-A)< y < \tan^2 (\pi /4+A)$$
Case 2) A in the 2nd and 4th quadrants. Then, $|a| = -\tan A$ and the inequalities becomes
$$ \tan^2 (\pi /4+A)< y < \tan^2 (\pi /4-A)$$
Thus, for any value of $A$, $y$ can't lie between $\tan^2(\pi /4 -A)$ and $\tan^2(\pi /4 +A)$.
| {
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"url": "https://math.stackexchange.com/questions/3629741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$3^n$ does not divide $4^n+5$ for $n\geq 2$ Question as in the title : does anyone know how to prove that $3^n$ does not divide $4^n+5$ for $n\geq 2$ or find a counterexample ?
My thoughts : (1) I have checked that this is true for $n\leq 1000$.
(2) I asked a similar question recently, and it was successfully solved with a method that uses a "lifting exponent lemma" which ultimately reduces to the identity $x^k-y^k=(x-y)(x^{k-1}+x^{k-2}y+\ldots+y^{k-1})$. Since $4^n+5$ cannot be so factored, this does not seem to apply here.
(3) For $r\geq 0$, denote by $q_r$ the smallest positive integer such that $4^{q_r}+5$ is divisible by $3^r$. It is easy to see that the order of $4$ modulo $3^r$ is exactly $3^{r-1}$, and hence $3^r$ divides $4^n+5$ iff $n\equiv q_r \ \pmod{3^{r-1}}$. It follows that $q_{r+1}\equiv q_r \ \pmod{3^{r-1}}$ and so we have a decomposition in base three, $q_r=\sum_{j=0}^{r-1}\varepsilon_j 3^j$ (where $\varepsilon_0=q_0$ and $\varepsilon_k=\frac{q_k-q_{k-1}}{3^{k-1}}\in\lbrace 0,1,2\rbrace$ for $k\geq 1$). The first terms of the $\varepsilon$ sequence are
$$
\varepsilon_0=1,\varepsilon_1=2, 2, 1, 1, 0, 0, 2, 1, 0, 0, 0, 1, 1, 2, 0, 0, 1, 2
$$
No pattern seems to emerge at this point.
| This follows from an effective abc conjecture.
If $4^n+5=3^nm$ then the quality of this $(a,b,c)$-triple is
\begin{align*}
q(4^n,5,3^nm)&=\frac{\log(3^nm)}{\log(\mathrm{rad}(4^n\cdot5\cdot3^nm))}\geq\frac{\log(4^n+5)}{\log(30m)}=\frac{\log(4^n+5)}{\log(30)+\log(4^n+5)-\log(3^n)}
\end{align*}
which is larger than 2 for $n\geq9$, larger than 3 for $n\geq20$, and larger than 4 for $n\geq58$.
Conjecturally, there are no such $(a,b,c)$-triples.
Below is an unrelated attempt to figure out what's going on algebraic-number-theoretically.
In the ring of integers $\mathbb{Z}[\sqrt{-5}]$, we have the factorization of ideals
$$(4^n+5)=(2^n+\sqrt{-5})(2^n-\sqrt{-5}).$$
Let $I=(2^n+\sqrt{-5})$ and let $I^\prime=(2^n-\sqrt{-5})$.
Note that $(2\sqrt{-5})\subseteq I+I^\prime$.
Then $I+I^\prime$ divides both $(2\sqrt{-5})$ and $(4^n+5)$.
However, $(2\sqrt{-5})$ has norm $20$ and $(4^n+5)$ has norm $(4^n+5)^2$ (which is coprime to $20$.
Thus, $I+I^\prime=1$ which shows that $I$ and $I^\prime$ are coprime.
Now suppose that $4^n+5$ is divisible by $3^n$.
We have the factorization of ideals
$$(3^n)=(3,1+\sqrt{-5})^n(3,1-\sqrt{-5})^n$$
where $\mathfrak p=(3,1+\sqrt{-5})$ and $\mathfrak q=(3,1-\sqrt{-5})$ are conjugate prime ideals of $\mathbb{Z}[\sqrt{-5}]$.
Since $I$ and $I^\prime$ are coprime, exactly one of the two possibilities holds:
*
*$\mathfrak p^n$ divides $I$ and $\mathfrak q^n$ divides $I^\prime$
*$\mathfrak q^n$ divides $I$ and $\mathfrak p^n$ divides $I^\prime$
The first case occurs when $n$ is even ($\mathfrak p$ contains both $2^n+\sqrt{-5}$ and $1+\sqrt{-5}$ so $\mathfrak p$ contains $2^n-1$ so $3\bigm|2^n-1$ so $n$ is even).
The second case occurs when $n$ is odd ($\mathfrak p$ contains both $2^n-\sqrt{-5}$ and $1-\sqrt{-5}$ so $\mathfrak p$ contains $2^n+1$ so $3\bigm|2^n+1$ so $n$ is odd).
| {
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"url": "https://math.stackexchange.com/questions/3631619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Prove that sum of matrices equals zero
The matrix $A$ has size $3 \times 3$ and we know that for any column
vector $v\in \mathbb{R}^{3}$ the vectors $Av$ and $v$ are orthogonal. Prove that
$A^{T} + A = 0$, where $A^{T}$ is the transposed matrix $A$.
So if
$$A = \begin{pmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{pmatrix}$$
and
$$v = \begin{pmatrix}
x\\
y\\
z
\end{pmatrix}$$
orthogonality of $Av$ and $v$ brought me to the equation
$x \cdot y \cdot (a_{12}+a_{21}) + x \cdot z \cdot (a_{13}+a_{31}) + y \cdot z \cdot (a_{23}+a_{32}) + a_{11} \cdot x^{2} + a_{22} \cdot y^{22} + a_{33} \cdot z^{2} = 0$
and the matrix $A^{T} + A$ equals
$$A^{T} + A = \begin{pmatrix}
2a_{11} & a_{12} + a_{21} & a_{13} + a_{31}\\
a_{12} + a_{21} & 2a_{22} & a_{23} + a_{32}\\
a_{13} + a_{31} & a_{23} + a_{32} & 2a_{33}
\end{pmatrix}$$
However I don't see how then prove that $A^{T} + A = 0$.
| It suffices to show $\langle (A^\intercal + A) v, v \rangle = 0$ for all $v$. This is true since
\begin{align} \langle (A^\intercal + A) v, v \rangle &= \langle A^\intercal v, v \rangle + \langle Av, v \rangle \\ &= \langle v, Av \rangle + \langle Av, v \rangle \\ &= 2 \cdot \langle Av, v \rangle \\ &= 2 \cdot 0 \\ &= 0.\end{align}
Remark: Note that this holds for all real square matrices $A$.
| {
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If $\sec x + \csc x =p$ has four distinct solutions between $(0,2\pi)$, then which if the following is incorrect?
a) $p^2-8>0$
b) $p=\sqrt 2$
c) $p=-\sqrt 2$
d) $p=0$
My attempt
$$\frac{\sec x +\csc x}{2} \ge \sqrt {\sec x \csc x}$$
$$\frac{\sin x +\cos x}{\sin x \cos x }\ge 2\sqrt {\frac{1}{\sin x \cos x}}$$
$$\sin x +\cos x \ge 2\sqrt {\sin x \cos x}$$
$$(\sqrt {\sin x}-\sqrt {\cos x})^2\ge 0$$
$$\sin x \ge \cos x$$
I realise that some of that squaring might have removed or added some roots, but I don’t know what else to do
From this result, the interval for $x$ is $[\frac{\pi}{4}, \pi]\cup [\frac{5\pi}{4}, \frac{3\pi}{2}]$
Don’t know what do next. Can I get some insight?
Another attempt
$$\sin x +\cos x =p \sin x \cos x$$
$$1+2\sin x\cos x =\frac 14 p^2 4\sin^2x \cos ^2x$$
$$1+\sin 2x =\frac 14 p\sin^2 2x$$
$$p\sin^22x-4\sin 2x -4=0$$
| Squaring an equation may introduce unwanted roots, so it is generally avoided. Rewrite the equation, instead, as
$$p(x) = \frac{\sin x +\cos x}{\frac12\sin 2x} =\frac{2\sqrt2 \cos(x-\frac\pi4)}{2\cos^2(x-\frac \pi4)-1}$$
So, $p(x)$ is a function of $\cos(x-\frac\pi4)$ and symmetric around $\frac\pi4$ and $\frac{5\pi}4$, which are also the two local extrema over $(0,2\pi)$, i.e.
$$p_{min}(\frac\pi4) = 2\sqrt2,\>\>\>\>\>p_{max}(\frac{5\pi}4) = -2\sqrt2$$
Thus, for any $p$ satisfying $p^2>(\pm 2\sqrt2)^2=8$, it meets the curve $p=\sec x+ \csc x$ four times, i.e. four distinctive roots in the shaded area of the graph below,
| {
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If $a+b+c=0$, then $a^3+b^3+c^3$ is ... $0$? $1$? $a^3b^3c^3$? $3abc$? Many mistakes in this post. ( See comments below). I let it as it is, as an example of what shouldn't be done.
If $a+b+c=0$, then $a^3+b^3+c^3 = \ldots $
A. $\;0\quad$ B. $\;1\quad$ C. $\;a^3b^3c^3\quad$ D. $\;3abc$
Source: 4/12/2020, Competitive Exams Reasoning Sample Paper 3- Translation in Hindi, Kannada, Malayalam, Marathi, Punjabi, Sindhi, Sindhi, Tamil, Telgu - Examrace. Downloaded from examrace.com
I can only see the pitfall consisting in inferring that all 3 numbers must be equal to 0.
What I can conclude from the premise is that one of the 3 numbers is the additive inverse of the sum of the 2 others.
Admitting it is number $c$, we get
$$a+b+c = 0= (a+b) + \left( - (a+b) \right) \tag{1}$$
In that case
$$c^3 = [- (a+b)]^3 = - (a+b) (a+b)(a+b) = - ( a^3 +2a^2b+2ab^2+b^3) \tag{2}$$
So
$$\begin{align}
a^3+b^3+c^3 &= a^3+b^3 - ( a^3 +2a^2b+2ab^2+b^3) \tag{3} \\
&= a^3+b^3 - a^3 - 2a^2b- 2ab^2- b^3 \tag{4}\\
&= 2a^2b - 2ab^2 \tag{5} \\
&=2 ( a^2b - b^2a) \tag{6} \\
&= 2 ( a) (ab-b^2) \tag{7} \\
&= 2 ( a) (b) (a-b) \tag{8} \\
&= 2 ( a) (b) (- c) \quad\text{[ Since $c = -(a+b) = b - a = - (a-b) $]} \tag{9} \\
&= - 2 ( a) (b) (c) \tag{10}
\end{align}$$
However, this isn't one of the possible answers.
What did I miss? Was I wrong in supposing that I could take any number $a$, $b$, or $c$ to play the role of additive inverse of the sum of the two others?
| Normally $$a^3+b^3+c^3 = (a+b+c)^3-3(a+b)(a+c)(b+c)$$
If $a+b+c = 0$
Then $a^3+b^3+c^3 = -3(a+b)(a+c)(b+c)$
Setting $c = -a-b$
$$a^3+b^3+c^3 = -3ab(b+a)$$
$$a^3+b^3+c^3 = 3abc$$
Recheck you expansion of $(a+b)^3$ in the second line of your work, you've replace $3$ with $2$
| {
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Prove by induction $u_{n+2}+u_n=4u_{n+1}$ for all $n\in \mathbb{N}$, where $u_n=(2+\sqrt{3})^n+(2-\sqrt{3})^n$. Prove by induction $u_{n+2}+u_n=4u_{n+1}$ for all $n\in \mathbb{N}$, where $u_n=(2+\sqrt{3})^n+(2-\sqrt{3})^n$.
For $n=1$,
My efforts:
LHS= $u_3+u_1=(2+\sqrt{3})^3+(2-\sqrt{3})^3+ (2+\sqrt{3})^1+(2-\sqrt{3})^1=56$.
RHS= $4u_2=4\left[(2+\sqrt{3})^2+(2-\sqrt{3})^2 \right]=56$.
Let us assume that the result is true for $n=k$ for some $k\in \mathbb{N}$.
Then $u_{k+2}+u_k=4u_{k+1}$.
Now, $u_{k+3}+u_{k+1}=4u_{k+1}$
How can I show the result for $n=k+1$?
| To make it a bit easier to deal with, the relation you're asked to prove is true for all $n \in \mathbb{N}$ can be rewritten as
$$u_{n+2} = 4u_{n+1} - u_{n} \tag{1}\label{eq1A}$$
You're given that
$$u_n = (2 + \sqrt{3})^n + (2 - \sqrt{3})^n \tag{2}\label{eq2A}$$
for all $n \in \mathbb{N}$. Using \eqref{eq2A} to define $u_1$ and $u_2$, you've done the base case by proving that \eqref{eq1A} is true for $u_3$. Next, use strong induction to assume \eqref{eq1A} is true for all $n \le k$ for some $k \ge 1$. Now, you need to use the values in \eqref{eq2A} of $u_j$ where $j \le k + 2$ to show \eqref{eq1A} is also true for $n = k + 1$. To do this, since the RHS of \eqref{eq1A} can be expressed using those terms, simplify its value as shown below to get
$$\begin{equation}\begin{aligned}
4u_{k+2} - u_{k+1} & = 4\left((2 + \sqrt{3})^{k+2} + (2 - \sqrt{3})^{k+2}\right) - \left((2 + \sqrt{3})^{k+1} + (2 - \sqrt{3})^{k+1}\right) \\
& = 4(2 + \sqrt{3})^{k+2} - (2 + \sqrt{3})^{k+1} + 4(2 - \sqrt{3})^{k+2} - (2 - \sqrt{3})^{k+1} \\
& = (4(2 + \sqrt{3}) - 1)(2 + \sqrt{3})^{k+1} + (4(2 - \sqrt{3}) - 1)(2 - \sqrt{3})^{k+1} \\
& = (7 + 4\sqrt{3})(2 + \sqrt{3})^{k+1} + (7 - 4\sqrt{3})(2 - \sqrt{3})^{k+1} \\
& = (2 + \sqrt{3})^2(2 + \sqrt{3})^{k+1} + (2 - \sqrt{3})^2(2 - \sqrt{3})^{k+1} \\
& = (2 + \sqrt{3})^{k+3} + (2 - \sqrt{3})^{k+3} \\
& = u_{k+3}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This shows the RHS is equal to the LHS of \eqref{eq1A}, so it's also true for $n = k + 1$. Thus, by induction, \eqref{eq1A} is true for all $n \in \mathbb{N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3638022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this optimization problem with equality constraints
Here's what I have tried:
I multiplied the second equation by $\sqrt[3]{abc}$ and get
$$\sqrt[3]{a^5c}+\sqrt[3]{b^5a}+\sqrt[3]{c^5b}=0$$
By manipulating the first expression, I obtain
$$
{1\over a^4b^4c^4}(a^5c+b^5a+c^5b)^2
$$
For simplicity, I attempt to let $x^3=a^5c,y^3=b^5a,z^3=c^5b$, and, as a result, the problem becomes the following:
Maximize $\displaystyle{\left(x^3+y^3+z^3\over xyz\right)^2}$ subjected to $x+y+z=0$
I know that this problem is likely to be solved by AM-GM inequality, so I wonder if anybody could provide me some help.
| Because $x+y+z=0$, $z=-x-y$, and the objective function $f(x,y,z)$ becomes
$$
f(x,y,z)\triangleq\left(x^3+y^3+z^3\over xyz\right)^2=\left[x^3+y^3-(x+y)^3\over xy(-x-y)\right]^2
$$
According to binomial theorem, we have $(x+y)^3=x^3+3x^2y+3xy^2+y^3$, so we can further simplify $f(x,y,z)$ into
$$
\begin{aligned}
f(x,y,z)
&=\left[-3x^2y-3xy^2\over xy(-x-y)\right]^2 \\
&=\left[3x^2y+3xy^2\over xy(x+y)\right]^2 \\
&=\left[3x+3y\over x+y\right]^2=3^2=9
\end{aligned}
$$
As a result, $f(x,y,z)$ remains constant in the domain of $\{x,y,z\in\mathbb{R}^3|x+y+z=0\}$, so its maximum value is also 9.
| {
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"url": "https://math.stackexchange.com/questions/3638556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the supremum and infimum of a set of natural numbers
Let $A = \left\{ 2(-1)^{n+1} + (-1)^{ \frac{n(n+1)}{2} } \left( 2 +
\dfrac{3}{n} \right) : n \in \mathbb{N} \right\}.$ Our goal is find $\sup A$ and $\inf
A$.
Attempt
At first glance, it looks like a very formidable set, and I can't see any obvious way but to separate in cases whether $n$ is odd or even.
If $n=2k$, then $n+1$ is off and $\dfrac{n(n+1)}{2} = k(2k+1)$ and thus our set takes the form
$$ A = \{ -2 + (-1)^k (2 + 3/2k) : k \in \mathbb{N} \} $$
We observe that for large values of $k$, the value $3/(2k)$ is negligible and so we have $-2 + (-1)^k 2$. In other words, we can have either $-4 $ or $0$ and so we claim that
$$ \sup A = 0 \; \; \text{and} \; \; \inf A = -4 $$
I am getting stuck on trying to actually prove these claims rigorously. Can I get some advice in how to do so?
Update:
Maybe itd be easy if we write $A_1 = \{ 2 (-1)^{n+1} \}$ and $A_2 = \left\{ (-1)^{ \frac{n(n+1)}{2} } \left( 2 +
\dfrac{3}{n} \right) \right\} $
and use $\sup(A_1 + A_2) = \sup A_1 + \sup A_2 $ may help
| Let
$$F(n) = 2(-1)^{n+1} + (-1)^\frac{n(n+1)}{2}\left(2+\frac{3}{n}\right)$$
For $n \ge 3$ we have $| F(n)| \le 5$, so $-5 \le F(n) \le 5$ when $n \ge 3$.
Since $F(1) = -3$, $F(2)=-5.5$ and $F(3)=5$, we have $\sup A = 5$ and $\inf A = -5.5$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Geometry Problem on $\triangle ABC$ and Angle Chasing
$\triangle ABC$ is an isosceles triangle with $AB=BC$ and $\angle ABD=60^{\circ}$, $\angle DBC=20^{\circ}$ and $\angle DCB=10^{\circ}$. Find $\angle BDA$.
My approach: Let $\angle BDA=x$. Let $AB=BC=p$. Applying sine law in $\triangle ADB$, $\dfrac{p}{\sin x}=\dfrac{BD}{\sin (60+x)}$. Applying sine law in $\triangle BDC$, $\dfrac{p}{\sin150^{\circ}}=\dfrac{BD}{\sin 10^{\circ}}$. Using the two equations, we get $\dfrac{1}{2\sin 10^\circ}=\dfrac{\sin x}{\sin (60^\circ +x)} \implies 2\sin 10^\circ=\dfrac{\sqrt{3}}{2}\cot x + \dfrac{1}{2} \\ \implies x = \text{arccot} \left(\dfrac{4\sin 10^\circ-1}{\sqrt{3}}\right)$.
Now I am stuck. I know that the answer is $100^\circ$ but no matter how hard I try I cannot seem to simplify it any further. Please help. If anybody has a better solution (involving simple Euclidean Geometry), I would be grateful if you provide it too.
Edit: I am extremely sorry. The original problem was when $AB=BC$. Sorry for the inconvenience caused. I have rectified my mistake. Also, I have changed the answer to $100 ^\circ$.
|
$\angle ABC=\angle ABD+\angle DBC=80^\circ$.
\begin{align*}
AB&=BC\\
\implies \angle CAB&=\angle BCA=(180^\circ-\angle ABC)/2=50^\circ.
\end{align*}
Erect an equilateral triangle $ACE$ on base $AC$.
Then $\triangle$s $ABE, CBE$ are congruent in opposite sense because $AB=CB$, $AE=CE$ and $BE$ is common. Thus $$\angle AEB=\angle BEC=30^\circ.$$
$$\angle CDB=180^\circ-\angle DBC-\angle BCD=150^\circ.$$ Thus quadrilateral $BDCE$ is cyclic because its angles $D$ and $E$ are supplementary. Thus
$$\angle DEC=\angle DBC=20^\circ.$$
\begin{align*}
\angle ECB&=\angle ECA-\angle BCA=10^\circ\\
\implies \angle ECD&=\angle ECB+\angle BCD=20^\circ=\angle DEC.
\end{align*}
Thus triangle $CED$ is isosceles on base $CE$, so $CD=DE$. Thus $\triangle$s $ACD, AED$ are congruent in opposite sense because $AC=AE$, $CD=ED$ and $AD$ is common. Thus
\begin{align*}
\angle CAD&=\angle DAE=30^\circ\\
\angle BAE&=\angle CAE-\angle CAB=10^\circ\\
\implies \angle DAB&=\angle DAE-\angle BAE=20^\circ\\
\implies \angle BDA&=180^\circ-\angle DAB-\angle ABD=100^\circ.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluating $ 4\sum^{30}_{n=1} n\;T(n)$, where $T(n) = \cos^2(30^\circ -n) - \cos(30^\circ -n)\cos(30^\circ +n) +\cos^2(30^\circ +n)$
For $n$ measured in degrees, let
$$T(n) = \cos^2(30^\circ -n) - \cos(30^\circ -n)\cos(30^\circ +n) +\cos^2(30^\circ +n)$$
Evaluate $$ 4\sum^{30}_{n=1} n \cdot T(n)$$
I have tried to use double-angle identities but got stuck with the coefficient $n$. I am new to trig, so I probably miss some advanced concepts.
| Since
$$\cos^2(30^\circ-n)-\cos(30^\circ-n)\cos(30^\circ+n)+\cos^2(30^\circ+n)=\cos(30^\circ-n)[\cos(30^\circ-n)-\cos(30^\circ+n)] +\cos^2(30^\circ+n)=(\frac{\sqrt{3}}{2}\cos n +\frac{1}{2}\sin n )2\sin30°\sin n +(\frac{\sqrt{3}}{2}\cos n -\frac{1}{2}\sin n )^2=
\frac{3}{4}$$
$$ 4\sum^{30}_{n=1} n \cdot T(n)=4\frac{3}{4}(1+2+\cdots+30)=1395 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the indicial equation for $r_1$ and $r_2$ The given expression is:
$$x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
First we need to convert the given into its indicial equation:
$$x^2\sum^\infty_{n=0}(n+r)(n+r-1)C_nx^{n+r-2}+x\sum^\infty_{n=0}(n+r)C_nx^{n+r-1}+x^2\sum^\infty_{n=0}C_nx^{n+r}-\frac{1}{4}\sum^\infty_{n=0}C_nx^{n+r}=0$$
Now factoring through the $x$ coefficients
$$\sum^\infty_{n=0}(n+r)(n+r-1)C_nx^{n+r}+\sum^\infty_{n=0}(n+r)C_nx^{n+r}+\sum^\infty_{n=0}C_nx^{n+r+2}-\frac{1}{4}\sum^\infty_{n=0}C_nx^{n+r}=0$$
Now bringing first, second and fourth series into a single series
$$\sum^\infty_{n=0}\left[(n+r)(n+r-1)+(n+r)-\frac{1}{4}\right]C_nx^{n+r}+\sum^\infty_{n=0}C_nx^{n+r+2}=0$$
Now to evaluate the expression with $n=0$ and $n=1$; And just as a personal preference, I like to factor out the $x^r$ during this step:
$$x^r\left[C_0\left(r^2-r+r+x^2-\frac{1}{4} \right) x^0\right]+x^r\left[(1+r)(r)C_1x+(1+r)C_1x+C_1x^3-\frac{1}{4}C_1x \right]+\sum^\infty_{n=0}...=0$$
Now from here I calculated my $r$ values and this is where I got confused. For the first bracket group of the expression above I got $r=\frac{1}{2}$ but when I tried to calculate $r$ for the second bracket group I got $r=-\frac{1}{2}$ and $r=-\frac{3}{2}$. This is where I know I messed up somewhere as I am only supposed to get two values of $r$. I re-did the problem from the start and got the same answers, where did I go wrong?
| $$\sum^\infty_{n=0}\left[(n+r)(n+r-1)+(n+r)-\frac{1}{4}\right]C_nx^{n+r}+\sum^\infty_{n=0}C_nx^{n+r+2}=0$$
$$\sum^\infty_{n=0}\left[(n+r)^2-\frac{1}{4}\right]C_nx^{n+r}+\sum^\infty_{n=2}C_{n-2}x^{n+r}=0$$
Changing the indices:
$$ \color {blue} {C_0x^r(r^2-\frac 14)}+C_1x^{r+1}\left(r^2+2r+\frac 34 \right)+\sum^\infty_{n=2}(\left[(n+r)^2-\frac{1}{4}\right]C_n+C_{n-2})x^{n+r}=0$$
Your indicial polynomial is the coefficient of the lower power of $x$:
$$P(r)=r^2-\frac 14$$
The indicial equation is:
$$\left(r^2-\frac 14 \right)=0$$
$$\implies S_r=\left \{ -\frac 12, \frac 12 \right \}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does $\sum_{i,j}^\infty \frac{1}{i(i^2+j^2)}$ converge? According to Mathematica the series $\sum_{i,j=1}^\infty \frac{1}{i(i^2+j^2)}$ converges. Is there a simple argument to obtain the convergence? Maybe a dominating convergent series?
| By the integral test
$$\sum_{i=1}^\infty \frac{1}{i^3+ij^2}$$
is convergent and
$$\sum_{i=1}^\infty \frac{1}{i^3+ij^2} \leq \frac{1}{1+j^2}+ \int_{x=1}^\infty \frac{1}{x^3+xj^2} dx = \frac{1}{1+j^2}+ \int_{x=1}^\infty \frac{x}{x^4+x^2j^2} dx \\
=\frac{1}{1+j^2}+ \frac{1}{2}\int_{u=1}^\infty \frac{1}{u(u+j^2)} du \\ =\frac{1}{1+j^2}+ \frac{1}{2}\lim_R \to \infty \int_1^R \frac{1}{j^2} \left( \frac{1}{u}-\frac{1}{u+j^2} \right) du \\=\frac{1}{1+j^2}+ \frac{1}{2j^2} \lim_R\left( \ln(u)-\ln(u+j^2) \right)_1^\infty=\frac{1}{1+j^2}+ \frac{\ln(1+j^2)}{2j^2}$$
Next,
$$\sum_{j=1}^\infty \frac{\ln(1+j^2)}{2j^2} < \infty$$
by (Limit) comparison test (compare to $\sum \frac{1}{j^{\frac{3}{2}}}$)
and $\sum_{j=1}^\infty \frac{1}{1+j^2}$ is convergent.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $x^4y^{\prime\prime} = (y-xy^\prime)^3, y(1) = y^\prime(1) = 1$ $\lambda^4x^4\lambda^{n-2}y^{\prime\prime} = (\lambda^ny-\lambda x\lambda^{n-1}y^\prime)^3 \Rightarrow \lambda^{n+2}x^4y^{\prime\prime} = (\lambda^n(y-xy^\prime))^3$.
$\lambda^{n+2} = \lambda^{3n} \Rightarrow n+2 = 3n \Rightarrow n = 1$.
Let $x = e^t, y = ue^{nt} = ue^t$.
Now, $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{u^\prime e^t + ue^t}{e^t} = u^\prime + u$. And, $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}}{\frac{dx}{dt}}(\frac{dy}{dx}) = \frac{\frac{d}{dt}(u^\prime + u)}{e^t} = e^{-t}(\frac{d^2u}{dt^2} + \frac{du}{dt})$.
Thus,
$e^{4t}(e^{-t}(\frac{du^2}{dt^2} + \frac{du}{dt})) = (ue^{t}-e^t(u+\frac{du}{dt}))^3 \Rightarrow e^{3t}(\frac{d^2u}{t^2} + \frac{du}{dt}) = e^{3t}(u-(u+\frac{du}{dt}))^3 \Rightarrow (\frac{d^2u}{dt^2} + \frac{du}{dt}) = (\frac{du}{dt})^3$.
Let $p = \frac{du}{dt}, p^\prime = \frac{d^2u}{dt^2} = \frac{dp}{du}\frac{du}{dt} = p\frac{dp}{du}$.
Thus,
$(p + p\frac{dp}{du}) = p^3 \Rightarrow 1 + \frac{dp}{du} = p^2 \Rightarrow \frac{dp}{du} = p^2-1 \Rightarrow u + c_1 = \int\frac{dp}{p^2-1} = \int\frac{dp}{(p-1)(p+1)} = \int(\frac{1}{2(p-1)} - \frac{1}{2(p+1)})dp = \frac{1}{2}\ln(p-1)-\frac{1}{2}\ln(p+1) = \ln(\frac{\sqrt{p-1}}{\sqrt{p+1}}) \Rightarrow c_1e^u = \frac{\sqrt{p-1}}{\sqrt{p+1}} \Rightarrow c_1e^{2u} = \frac{p-1}{p+1} \Rightarrow p-1 = c_1e^{2u}p + c_1e^{2u} \Rightarrow p = \frac{c_1e^{2u}+1}{1-c_1e^{2u}} = \frac{du}{dt}$.
But then I think that I have to use the initial conditions to get $c_1$, but I don know how to do that.
| Taking $y = u x$, the equation becomes
$$ u'' = - \dfrac{u' ((u')^2 x^2 + 2)}{x} $$
with $u(1) = 1$, $u'(1) = 0$.
If $u' = v$, this is
$$ v' = - \dfrac{v (v^2 x^2 + 2)}{x},\ v(1) = 0$$
Now we could go on to solve this differential equation in general, but for this specific initial condition it's obvious that the solution is $v = 0$. Then $u$ must be a constant, namely $u=1$ from the initial condition, so $y=x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $ \frac1{\sqrt{(z+a)(z+b)}}=\frac1\pi \int_a^b \frac1{\sqrt{(t-a)(b-t)}}\frac1{t+z}dt $ Reading papers online i found the following definition
$$ \frac1{\sqrt{(z+a)(z+b)}}=\frac1\pi \int_a^b \frac1{\sqrt{(t-a)(b-t)}}\frac1{t+z}dt $$
for $$ b>a \,\, and \,\,\,z \in \Bbb C |(-\infty,-a] $$
Question : How can this be proven? would the proof for this above definition be similar for another expression of type:
$$\frac{f(x)}{\sqrt{(z+a)(z+b)}}$$
Thank you kindly for your help and time.
| Let $t = \frac{a+b}{2}+\frac{b-a}{2}\sin\theta$. Then
$$\pi I(z) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d\theta}{\frac{a+b}{2}+\frac{b-a}{2}\sin\theta+z} = \int_0^\pi \frac{d\theta}{\frac{a+b}{2}+\frac{b-a}{2}\cos\theta+z}$$
Then denoting $\frac{a+b}{2}+z \equiv k_1$ and $\frac{b-a}{2} \equiv k_2$ we get that
$$\pi I(z) = \int_0^\pi \frac{d\theta}{(k_1+k_2)\cos^2\left(\frac{\theta}{2}\right)+(k_1-k_2)\sin^2\left(\frac{\theta}{2}\right)}$$
$$= \frac{2}{k_1-k_2}\int_0^{\pi}\frac{\frac{1}{2}\sec^2\left(\frac{\theta}{2}\right) \:d\theta}{\frac{k_1+k_2}{k_1-k_2}+\tan^2\left(\frac{\theta}{2}\right)} = \frac{2}{\sqrt{k_1^2-k_2^2}}\tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\sqrt{\frac{k_1+k_2}{k_1-k_2}}\right)\Biggr|_0^\pi = \frac{\pi}{\sqrt{k_1^2-k_2^2}}$$
Then plugging in for $k_1$ and $k_2$ gives us that
$$I(z) = \frac{1}{\sqrt{(k_1-k_2)(k_1+k_2)}} = \frac{1}{\sqrt{(z+a)(z+b)}}$$
| {
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"url": "https://math.stackexchange.com/questions/3654645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculating matrix determinant This is a very easy determinant to calculate, but I get two different results when I calculate it in two different ways.
\begin{equation}
A = \begin{bmatrix} 1 & 2 & 3 \\
0 & 1 & 2 \\
2 & 2 & 4
\end{bmatrix}
\end{equation}
When I used Laplace expansion right away I got:
\begin{equation}
\det(A) = 1 \cdot \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} - 0 \cdot \begin{bmatrix} 2 & 3 \\ 2 & 4 \end{bmatrix} + 2 \cdot \begin{bmatrix} 2 & 3 \\ 1 & 2\end{bmatrix} = 1 \cdot (4 - 4) + 2 \cdot (4 - 3) = 2
\end{equation}
But when I rearrange the rows in the matrix and then try to calculate the determinant:
\begin{equation}
A = \begin{bmatrix} 1 & 2 & 3 \\
0 & 1 & 2 \\
2 & 2 & 4 \\
\end{bmatrix} \overset{r_1 \leftarrow 2 \cdot r_1 - r_3}{\longrightarrow} \begin{bmatrix} 0 & 2 & 2 \\
0 & 1 & 2 \\
2 & 2 & 4 \\
\end{bmatrix}\\
\det(A) = 0 \cdot \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} - 0 \cdot \begin{bmatrix} 2 & 2 \\ 2 & 4 \end{bmatrix} + 2 \cdot \begin{bmatrix} 2 & 2 \\ 1 & 2\end{bmatrix} = 2 \cdot (4 - 2) = 4
\end{equation}
I have probably made a simple mistake, but I can't figure out where and I really want to get the basics down, before I move to harder examples.
| $r_1 \leftarrow 2r_1 -r_3$ is not allowed.
If you want to do row addition it is:
$r_1 \leftarrow r_1+\lambda r_2$ or $r_1 \leftarrow r_1+\lambda r_3$
| {
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"url": "https://math.stackexchange.com/questions/3656372",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove and give geometric meaning of $|\sqrt{a^2+b^2} - \sqrt{a^2+c^2} | < |b-c| $
Carefully, prove that
$$ | \sqrt{a^2+b^2} - \sqrt{a^2+c^2} | \leq |b-c| $$
and give a geometric interpretation.
pf
We observe that $| \sqrt{a^2+b^2} - \sqrt{a^2+c^2} | = \dfrac{ |b^2 - c^2 | }{\sqrt{a^2+b^2} + \sqrt{b^2+c^2} } \leq \dfrac{ |b^2-c^2| }{\sqrt{b^2} + \sqrt{c^2} } = \dfrac{ |b+c | }{|b| + |c| } \cdot |b-c| $
Since by the triangle ineqality $|b+c| \leq |b| + |c| \implies \dfrac{ |b+c| }{|b| + |c| } \leq 1$, then the required result is proved!
Now, as for the hardest part, I dont see a geometric interpretation of this inequality. Can someone enlighten me?
| Let $B(a,b)$ and $C(a,c)$ be the two vecrors. Then,
$$| |B|-|C||<| B-C| = |BC|$$
which translates To
$$| \sqrt{a^2+b^2} - \sqrt{a^2+c^2} | \leq |b-c| $$
Geometrically, the above inequality means the length difference of two triangle sides is less than the length of the third side.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there formula that solves quartic equation $ax^4+bx+c=0$ In general form, a quartic equation is $ax^4+bx^3+cx^2+dx+e=0$. I was thinking of the quartic equation of the form
$$ax^4+bx+c=0$$
which resembles a depressed cubic equation.
Does anyone know a formula that solves this type of depressed quartic equation, simlilar to the formula
$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
for the quadratic equation $ax^2+bx+c=0$, or
$$\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$ for the cubic equation $x^3+px +q=0$.
| The quartic equation of the form
$$ax^4+bx+c=0$$
has the following formulas for its four roots
$$x= -s \pm \sqrt{ \frac b{4as}-s^2},\>\>\> s \pm \sqrt{ -\frac b{4as}-s^2} $$
where $s^2$ satisfies the cubic equation $s^6 -\frac c{4a} s^2-\left(\frac b{8a}\right)^2=0$. Take the example of
$$x^4+2x+\frac12=0$$
for which $s^6 -\frac 18 s^2-\frac1{16}=0$ or $s^2=\frac1{2}$.
Plug $s$ into the formulas above to obtain
$$x=\frac{-1\pm\sqrt{\sqrt2-1}}{\sqrt2},\>\>\>
\frac{1\pm i\sqrt{\sqrt2+1}}{\sqrt2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Homework Problem, Power Series Limit I am looking to find the solution for:
$$\lim_{x\rightarrow 0} {5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$$
A hint was provided:
transform: ${5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$ to $5y{5^{y-1} - 1 \over y-1}, y = {1\over cos^s(x)} \rightarrow 1$
The transformation is straigt forward:
$tan^2(x) + 1 = {1\over cos^2(x)} = y \text{ and } \\1-cos^2(x) = ycos^2(x) - cos^2(x) = cos^2(x)(y-1)$ combined we have:
$$5y{(5^{y-1} - 1) \over y-1}$$
as $y \rightarrow 1$ $5y{(5^{y-1} - 1) \over y-1}$ is not defined. Since both, nominator and denominator are $0$ I tried L'Hopital but ended at:
$5 5^{y-1} + 5y(y-1)5^{y-2}-5$ with $\lim_{y \rightarrow 1} = 5 + 0 - 5 = 0$ and $(y-1)' = 1$
Here I have to stop with no solution. I have also tried to use the quotient rule to differentiate the expression which did not get me anywhere.
| I’m not sure you’re doing the differentiation right. We have
$$\lim_{y\to 1} \ 5y \ \frac{5^{y-1} -1}{y-1} $$
$$=5(1)\cdot \ \lim_{y\to 1} \frac{(5^{y-1} -1)’}{(y-1)’}$$
$$= 5\cdot\lim_{y\to 1} \frac{5^{y-1} \cdot \ln 5}{1}$$
$$=5\ln 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Maclaurin series expansion for $\dfrac{1}{2t}\left(1+t-\dfrac{(1-t)^2}{2 \sqrt {t}}\log \left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)\right)$
Show that the expansion of
$$\dfrac{1}{2t}\left(1+t-\dfrac{(1-t)^2}{2 \sqrt{t}} \log \left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)\right)$$
as a Maclaurin series in power of $t$ to obtain
$$\dfrac{4}{3}-4\sum_{n=1}^{\infty}\dfrac{t^n}{(4n^2-1)(2n+3)}$$
I am not sure how to do this. Below are some steps I have managed to get which gets more complicated and am not sure how to continue. Help is appreciated.
Expand into:
$$\dfrac{1}{2t}+\dfrac{1}{2} - \dfrac{(1-t)^2}{4t\sqrt{t}}\log\left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)$$
$$=\dfrac{1}{2t}+\dfrac{1}{2}- \dfrac{\log\left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)}{4t\sqrt{t}}(-t+1)^2$$
Multiply by congugate:
$$=\dfrac{1}{2t}+\dfrac{1}{2}-\dfrac{\log\left(\dfrac{(1+\sqrt{t})^2}{-t+1}\right)}{4t\sqrt{t}}(-t+1)^2$$
| We have $\frac12\log\frac{1+z}{1-z}=\sum_{n=0}^\infty\frac{z^{2n+1}}{2n+1}$, hence $\frac{1}{2\sqrt{t}}\log\frac{1+\sqrt{t}}{1-\sqrt{t}}=\sum_{n=0}^{\infty}\frac{t^n}{2n+1}$ and $$\frac{(1-t)^2}{2\sqrt{t}}\log\frac{1+\sqrt{t}}{1-\sqrt{t}}=(1-2t+t^2)\sum_{n=0}^{\infty}\frac{t^n}{2n+1}\\=1+\left(\frac13-2\right)t+\sum_{n=2}^{\infty}\left(\frac{1}{2n+1}-\frac{2}{2n-1}+\frac{1}{2n-3}\right)t^n\\=1-\frac53t+\sum_{n=2}^{\infty}\frac{8t^n}{(2n+1)(2n-1)(2n-3)}=1-\frac53t+8\sum_{n=\color{blue}{1}}^{\infty}\frac{t^{n\color{blue}{+1}}}{(4n^2-1)(2n+3)}.$$ Now substitute and get the expected result.
| {
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"url": "https://math.stackexchange.com/questions/3662808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\prod_{k=1}^{\infty}\frac{{5^\frac {1}{2}}^k+{3^\frac{1}{2}}^k}{2}$ While doing questions on series and products, I tried hard to solve this infinite product, but no methods has worked so far.Well, this is from Johns Hopkins math tournament
Question:- Evaluate $\prod_{k=1}^{\infty}\frac{{5^\frac {1}{2}}^k+{3^\frac{1}{2}}^k}{2}$
Can anybody help me out!
| Let $\alpha = 5, \beta = 3$. Recall the identity
$$x + y = \frac{x^2 - y^2}{x-y}$$
The product you have is a telescoping one.
$$\prod_{k=1}^N \frac{\alpha^{2^{-k}} + \beta^{2^{-k}}}{2}
= \frac{1}{2^N} \prod_{k=1}^N
\frac{\alpha^{2^{1-k}} - \beta^{2^{1-k}}}{\alpha^{2^{-k}} - \beta^{2^{-k}}}
= \frac{\alpha - \beta}{\frac{\alpha^{2^{-N}} - \beta^{2^{-N}}}{2^{-N}}}
\tag{*1}
$$
Since $2^{-N} \to 0$ as $N \to \infty$, we can use the fact
$$\lim_{x\to0} \frac{u^x - 1}{x} = \left.\frac{du^x}{dx}\right|_{x=0} = \log u\quad\text{ for }\quad u = \alpha,\beta
$$
to compute the limit of the denominator in $(*1)$.
$$\lim_{N\to\infty}
\frac{\alpha^{2^{-N}} - \beta^{2^{-N}}}{2^{-N}}
= \lim_{x\to 0}
\frac{\alpha^x - \beta^x}{x}
= \lim_{x\to 0}
\left[\frac{\alpha^x - 1}{x} - \frac{\beta^x - 1}{x}\right]
= \log\frac{\alpha}{\beta}$$
As a result,
$$\prod_{k=1}^\infty \frac{5^{2^{-k}} + 3^{2^{-k}}}{2}
= \frac{5 - 3}{\log\frac53} = \frac{2}{\log\frac{5}{3}} \approx 3.91523037794...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Binomial theorem and multinomial coefficient I have a question to which I could not find an answer in the forum.
I am trying to solve a bracelet problem. Actually I already solved it but with some kind of a cheating (using wolframaplha to expand the Dihedral group formula). The problem that I try to solve is: how many different bracelets we can create from 12 beads (3 red, 4 green 5 blue).
The Dihedral group is:
$$ D_n = \frac{f_1^{12} + f_2^6 + 2f_3^4 + 2f_4^3 + 2f_6^2 + 4f_{12}^1 + 6f_1^2f_2^5 + 6f_2^6}{24} $$
and if we replace in the formula for the three colors we got:
$$ D_n = \frac{(x + y + z)^{12} + (x^2 + y^2 + z^2)^6 + 2(x^3 + y^3 + z^3)^4 + 2(x^4 + y^4+z^4)^3 + 2(x^6 + y^6 + z^6)^2 + 4(x^{12} + y^{12} + z^{12}) + 6(x + y + z)^2(x^2 + y^2 + z^2)^5 + 6(x^2 + y^2 + z^2)^6}{24}$$
From here by using wolframalpha to expand each part I found that only:
$(x + y + z)^{12}$
and
$6(x + y + z)^2(x^2 + y^2 + z^2)^5$
can make $x^3y^4z^5$
From there I also get the coeficent before them and found the answer 1170 bracelets from 3 red, 4 green and 5 blue beads.
However, I have two questions because I do not want to cheat with wolframalpha:
1. How I can found which parts of the formula could expand to $x^3y^4z^5$
2. How I can use multinomial coefficient for calculating them without expanding. I know that by using it I do not need to expand the formula. I need for example the formula for something hard:
$$(x^2 + y^2 + z^2)^5$$
The problem that I have is solved, I just need an explanation how can I approach the hard part without using a tool.
Thanks for the answers.
| For the first question, note that for example in $(x^2+y^2+z^2)^6$, each term will be of the form $kx^{2a}y^{2b}z^{2c}$ for some integers $k,a,b,c$. So it's not possible to have an exponent of $3$ or $5$. A similar consideration eliminates every term but the two you have mentioned.
For multinomial coefficients in the example you give, $$\left(x^2+y^2+z^2\right)^5=\sum_{a+b+c=5}\binom{5}{a,b,c}x^{2a}y^{2b}z^{2c},\tag1$$ where the sum is over all ordered triples $(a,b,c)$ of nonnegative integers such that $a+b+c=5$.
In general, we have the multinomial formula: $$\left(x_1+x_2+\cdots+x_k\right)^n=\sum_{n_1+\cdots+n_k=n}\binom{n}{n_1,\dots,n_k}x_1^{n_1}\cdots x_k^{n_k}\tag2$$
If you have an expression with powers of the variables, just substitute them in the formula. That's how to get $(1)$ from $(2)$.
EDIT
So to finish the necklace problem, from the first term we get $$\frac1{24}\binom{12}{3,4,5}=1155$$. Consider the second term, $$6(x + y + z)^2\left(x^2 + y^2 + z^2\right)^5$$ Since the second factor will give only even powers, we need odd powers of $x$ and $z$ from the first term. Since the exponent is $2$, these powers can only be $1$. From the second term then we need $x^{2a}y^{2b}z^{2c}=x^2y^4z^4$ so $a=1,b=2,c=2$. That gives $$\frac6{24}\binom{2}{1,0,1}\binom{5}{1,2,2}=15$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Combinations of elements from multiple sets with constraints Given four sets of numbers:
$$A=\{1,2,3,4\}$$
$$B=\{1,2,3,4\}$$
$$C=\{1,2,3,5\}$$
$$D=\{1,2,6\}$$
I need to find number of all possible combinations of numbers from these sets in following format:
$$(a,b,c,d)$$
where $a\in A, b \in B, c \in C, d \in D$ and where each number can be present only once per combination (e.g. combination $(1,2,3,4)$ is valid, but combination $(1,1,2,3)$ is not (number 1 repeats) and neither is combination $(2,3,5,2)$ (number 2 repeats)).
My idea is to use inclusion-exclusion principle. First I calculate number of all possible combinations:
$$\vert{A}\vert \cdot \vert{B}\vert \cdot \vert{C}\vert \cdot \vert{D}\vert = 192$$
And then I need to remove all combinations where elements are repeating 2 or more times. These sets are:
$$\vert{A\cap B}\vert \cdot \vert{C}\vert \cdot \vert{D}\vert = 48 $$
$$\vert{A\cap C}\vert \cdot \vert{B}\vert \cdot \vert{D}\vert = 36 $$
$$\vert{A\cap D}\vert \cdot \vert{B}\vert \cdot \vert{C}\vert = 32 $$
$$\vert{B\cap C}\vert \cdot \vert{A}\vert \cdot \vert{D}\vert = 36 $$
$$\vert{B\cap D}\vert \cdot \vert{A}\vert \cdot \vert{C}\vert = 32 $$
$$\vert{C\cap D}\vert \cdot \vert{A}\vert \cdot \vert{B}\vert = 32 $$
The problem here is that I have duplicates during removal, since sum of all combinations above is $216$ (some combinations that I remove - I remove multiple times).
My questions is - how to find intersections between all these sets, in order to get correct number for removal (in this case it should be $142$ and not $216$) - so correct answer in the end should be $192-142=50$.
I guess I need to find following intersections but not sure how to calculate all of these:
$$\vert(\vert{A\cap B}\vert \cdot \vert{C}\vert \cdot \vert{D}\vert) \cap (\vert{A\cap B}\vert \cdot \vert{C}\vert \cdot \vert{D}\vert)\vert = 9 \tag{1}\label{1}$$
$$\vert(\vert{A\cap B}\vert \cdot \vert{C}\vert \cdot \vert{D}\vert) \cap (\vert{A\cap C}\vert \cdot \vert{B}\vert \cdot \vert{D}\vert)\vert = 8$$
$$\vert(\vert{A\cap B}\vert \cdot \vert{C}\vert \cdot \vert{D}\vert) \cap (\vert{A\cap D}\vert \cdot \vert{B}\vert \cdot \vert{C}\vert)\vert = 9$$
$$\vert(\vert{A\cap B}\vert \cdot \vert{C}\vert \cdot \vert{D}\vert) \cap (\vert{B\cap C}\vert \cdot \vert{A}\vert \cdot \vert{D}\vert)\vert = 8$$
$$\vert(\vert{A\cap B}\vert \cdot \vert{C}\vert \cdot \vert{D}\vert) \cap (\vert{B\cap D}\vert \cdot \vert{A}\vert \cdot \vert{C}\vert)\vert = 8$$
$$\vert(\vert{A\cap B}\vert \cdot \vert{C}\vert \cdot \vert{D}\vert) \cap (\vert{C\cap D}\vert \cdot \vert{A}\vert \cdot \vert{B}\vert)\vert = 8$$
$$...$$
$$\vert(\vert{A\cap D}\vert \cdot \vert{B}\vert \cdot \vert{C}\vert) \cap (\vert{B\cap C}\vert \cdot \vert{A}\vert \cdot \vert{D}\vert)\vert = 6 \tag{2}\label{2}$$
$$...$$
$\eqref{1}$ duplicates are $(1,1,1,1),(1,1,1,2),(1,1,1,6),(2,2,2,1),(2,2,2,2),(2,2,2,6),(3,3,3,1),(3,3,3,2),(3,3,3,6)$
$\eqref{2}$ duplicates are $(1,1,1,1),(1,2,2,1),(1,3,3,1),(2,1,1,2),(2,2,2,2),(2,3,3,2)$ -> what is the rule here if I have more than 4 sets like here?
and then I need all 3 intersections:
$$\vert((\vert{A\cap B}\vert \cdot \vert{C}\vert \cdot \vert{D}\vert) \cap (\vert{A\cap C}\vert \cdot \vert{B}\vert \cdot \vert{D}\vert) \cap (\vert{A\cap D}\vert \cdot \vert{B}\vert \cdot \vert{C}\vert))\vert = 2$$
$$...$$
and then 4,5 intersections (not shown here) and finally I need remaining (6) intersection of all:
$$\vert((\vert{A\cap B}\vert \cdot \vert{C}\vert \cdot \vert{D}\vert) \cap (\vert{A\cap C}\vert \cdot \vert{B}\vert \cdot \vert{D}\vert) \cap (\vert{A\cap D}\vert \cdot \vert{B}\vert \cdot \vert{C}\vert) \cap (\vert{B\cap C}\vert \cdot \vert{A}\vert \cdot \vert{D}\vert) \cap (\vert{B\cap D}\vert \cdot \vert{A}\vert \cdot \vert{C}\vert) \cap (\vert{C\cap D}\vert \cdot \vert{A}\vert \cdot \vert{B}\vert))\vert = 2$$
In the end I get result: $192-(216-119+65-30+12-2)=50$
I tried to generalize this to more than 4 sets, but cannot find strict rule to calculate these intersections that I wrote above. Any help would be appreciated.
UPDATE
Based on answer below, I found a formula that makes computation much easier per partition:
$$ Part(p) = \prod_{n=1}^{g} (-1)^{l_n} \cdot l_n! \cdot s_n $$
Here, $p$ is one of partitions - e.g. $\{\{A,B\},\{C\},\{D\}\}$; $g$ is number of groups in that partition (in this case 3 - $\{A,B\}$ ,$\{C\}$ and $\{D\}$); $l_n$ is number of elements in a group minus 1 - in this case it would be $2-1=1$, $1-1=0$ and $1-1=0$ per group; $s_n$ is number of elements in that group - in this case $\vert\{A,B\}\vert=4$, $\vert\{C\}\vert=4$ and finally $\vert\{D\}\vert=3$.
You can see all the formula for this example in the uploaded picture below:
| We calculate the wanted number with the help of rook polynomials. We consider a $(6\times4)$-rectangular board, where the columns represent the sets $A,B,C,D$ and the rows the elements of the sets. A valid quadruple $(a,b,c,d)\in A\times B\times C\times D$ can be represented by placing four non-attacking rooks on the board, with forbidden positions marked as red squares. A valid arrangement is given showing $(a,b,c,d)=(2,4,1,6)$.
Since the number of forbidden squares is smaller than the number of valid squares, we calculate the rook polynomial for the forbidden squares and subtract the corresponding number of non-attacking rook arrangements from the non-attacking rook arrangements of the complete rectangle.
The rook polynomial $r(x)$ we are looking for is
\begin{align*}
r_0+r_1x+r_2x^2+r_3x^3+r_4x^4\tag{1}
\end{align*}
with $r_j \ (0\leq j\leq 4)$ denoting the number of placing exactly $j$ non-attacking rooks on the forbidden red squares.
*
*$r_0$: There is one arrangement placing no rooks at all.
*$r_1$: We have $9$ red squares where we can place a rook, giving $r_1=9$.
*$r_2$: We count all valid arrangements placing two non-attacking rooks on the forbidden squares. We start with placing a rook on $(A,6)$ which admits $5$ arrangements by placing the other rook on $(B,5),(C,4),(D,5),(D,4),(D,3)$. We continue systematically and obtain:
\begin{align*}
&(A,6)\to 5&(A,5)\to 5\\
&(B,6)\to 4&(B,5)\to 4\\
&(C,6)\to 3&(C,4)\to 2\\
\end{align*}
from which $r_2=5+5+4+4+3+2=23$ follows.
*$r_3$: We count all valid arrangements placing two non-attacking rooks on the forbidden squares. We start with placing two rooks on $(A,6),(B,5)$ which admits $3$ arrangements by placing the third rook on $(C,4),(D,4),(D,3)$. We continue systematically and obtain
\begin{align*}
&(A,6),(B,5)\to 3&(A,5),(B,6)\to 3\\
&(A,6),(C,4)\to 2&(A,5),(C,6)\to 2\\
&(A,5),(C,4)\to 1&(B,6),(C,4)\to 2\\
&(B,5),(C,6)\to 2&(B,5),(C,4)\to 1
\end{align*}
from which $r_3=3+3+2+2+1+2+2+1=16$ follows.
*$r_4$: We have two valid arrangements with four rooks, namely $(A,6),(B,5),(C,4),(D,3)$ and $(A,5),(B,6),(C,4),(D,3)$ giving $r_4=2$.
The rook polynomial of the forbidden squares is
\begin{align*}
r(x)=1+9x+23x^2+16x^3+2x^4
\end{align*}
Now its time to harvest. The number of all arrangements of four non-attacking rooks on the $(6\times 4)$ board is
\begin{align*}
6\cdot5\cdot4\cdot3=\frac{6!}{2}\tag{2}
\end{align*}
We now use the coefficients of the rook polynomial $r(x)$ and apply the inclusion-exclusion principle. We subtract from (2) the number of arrangements with one rook on the forbidden squares, then adding the number of arrangements with two rooks on the forbidden squares. We continue this way and obtain
\begin{align*}
&\frac{6!}{2}-9\cdot\frac{5!}{2}+23\cdot\frac{4!}{2}-16\cdot\frac{3}{2}+2\frac{2!}{2}\\
&\qquad=360-540+276-48+2\\
&\qquad\,\,\color{blue}{=50}
\end{align*}
in accordance with the other answers.
| {
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"url": "https://math.stackexchange.com/questions/3666963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Evaluate logarithmic integral $\int_{-\infty}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx$ I need help to evaluate the integral with the residue theorem:
$$\int_{-\infty}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx$$
where a,b>0 real numbers.
I think I could consider the contour integral where C is the half circle in the first two quadrant.
But I'm not sure how to continue.
Could someone help me?
| Let $I(a)=2\int_{0}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx$ and evaluate
$$I’(a) =\int_{0}^{\infty}\frac{4adx}{(x^2+a^2)(x^2+b^2)}=
\frac{2\pi}{b(a+b)}$$
Then,
\begin{align}
\int_{-\infty}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx
& = I(a)=I(0)+ \int_{0}^{a}I’(t)dt\\
&=2 \int_{0}^{\infty} \frac{\ln{x^2}}{x^2+b^2}\,dx
+ \frac{2\pi}b \int_{0}^{a}\frac1{t+b}dt\\
&= \frac{2\pi }b \ln b + \frac{2\pi }b \ln \frac{a+b}b= \frac{2\pi }b \ln (a+b)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3671716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Wirtinger's inequality variation
If $f \in C^1[0,1]$ with $f'(0) = f(1) = 0$, then$$\|f\|_2\leq\frac2\pi\|f'\|_2.$$
Elaboration:
Assume the Sturm-Liouville operator $A: D \longrightarrow L^2(0,1)$ where the domain is
$$
D = \{f \in C^1[0,1]: f'' \in L^2(0,1), f'(0) = f(1) =0\}
$$
and
$$
Af(x) = f''(x)-\lambda f(x), \, \lambda \in \mathbb{R}, \, x \in [0,1]
$$
The eigenfuctions of $A$: $\phi_n(x) = \sqrt{2} \cos\left( \frac{(2n-1)\pi}{2}\right), \, n =1,2,\dots$ is an orthonormal basis of $L^2(0,1)$.
Then for an $f \in C_1[0,1]$ with $f'(0) = f(1) = 0$ we have:
$$
f(x) = \sum_{n=1}^\infty b_n \sqrt{2} \cos\left( \frac{(2n-1)\pi}{2}x\right), \, n =1,2,\dots
$$
Now it'd be very nice if
$$
f'(x) = \sum_{n=1}^\infty a_n \sqrt{2} \sin\left( \frac{(2n-1)\pi}{2}x\right), \, n =1,2,\dots \tag{$*$}
$$
so that, by integrating both sides
$$
\int_1^x f'(s)\,\mathrm ds = \sum_{n=1}^\infty a_n \sqrt{2} \int_1^x\sin\left( \frac{(2n-1)\pi}{2}s\right)\,\mathrm ds\\
f(x) = \sum_{n=1}^\infty \frac{-2a_n}{\pi(2n-1)}\sqrt{2}\cos\left( \frac{(2n-1)\pi}{2}x\right)
$$
and thus by using Parseval theorem:
$$
\|f\|_2^2 = \sum_{n=1}^\infty \frac{4a^2_n}{\pi^2(2n-1)^2} \leq \frac{4}{\pi^2}\sum_{n=1}^\infty a_n^2 = \frac{4}{\pi^2}\|f'\|^2_2
$$
and therefore:
$$
\|f\|_2^2 \leq \frac{2}{\pi} \|f'\|_2^2
$$
Is equation $(*)$ (or some variation of it) true and why?
In other words, can the Fourier series expansion of $f$ be term by term differentiated and why?
| $\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$Note that after changing the domain $D$ to$$
D_1 = \{f \in C^1([0, 1]) \mid f'' \in L^2([0, 1]),\ f'(1) = f(0) = 0\},
$$
the Sturm-Liouville theorem implies that $\{ψ_n(x) \mid n \in \mathbb{N}_+\}$ is also an orthonormal basis of $C^1([0, 1])$, where $ψ_n(x) = \sqrt{2} \sin\left( \dfrac{1}{2} (2n - 1)π x \right)$, thus there exist a sequence of constants $\{a_n\}$ such that$$
f'(x) = \sum_{n = 1}^∞ a_n \sqrt{2} \sin\left( \frac{1}{2} (2n - 1)π x \right)
$$
if $f \in C^2([0, 1])$. But for any $f \in C^1([0, 1])$, there exist a sequence of functions $\{f_n\} \subseteq C^2([0, 1])$ that $f_n'$ uniformly converges to $f'$ and $\lim\limits_{n → ∞} f_n(0) = f(0)$, so this suffices for the proof of the inequality.
Actually, there is an identity:
Proposition: If $f \in C^1([0, 1])$ satisfies $f'(0) = f(1) = 0$, then$$
\frac{4}{π^2} \int_0^1 (f'(x))^2 \,\d x - \int_0^1 (f(x))^2 \,\d x = \int_0^1 \left( \frac{2}{π} f'(x) + f(x) \tan\left( \frac{π}{2} x \right) \right)^2 \,\d x.
$$
Proof: For $0 < δ < 1$,\begin{gather*}
\int_0^{1 - δ} \left( \frac{2}{π} f'(x) + f(x) \tan\left( \frac{π}{2} x \right) \right)^2 \,\d x\\
{\small= \frac{4}{π^2} \int_0^{1 - δ} (f'(x))^2 \,\d x + \frac{4}{π} \int_0^{1 - δ} f(x) f'(x) \tan\left( \frac{π}{2} x \right) \,\d x + \int_0^{1 - δ} (f(x))^2 \tan^2\left( \frac{π}{2} x \right) \,\d x,}\tag{1}
\end{gather*}
and\begin{align*}
&\peq \int_0^{1 - δ} f(x) f'(x) \tan\left( \frac{π}{2} x \right) \,\d x = \frac{1}{2} \int_0^{1 - δ} \tan\left( \frac{π}{2} x \right) \,\d((f(x))^2)\\
&= \frac{1}{2} \left. (f(x))^2 \tan\left( \frac{π}{2} x \right) \right|_0^{1 - δ} - \frac{π}{4} \int_0^{1 - δ} (f(x))^2 \sec^2\left( \frac{π}{2} x \right) \,\d x\\
&= (f(1 - δ))^2 \tan\left( \frac{π}{2} (1 - δ) \right) - \frac{π}{4} \int_0^{1 - δ} (f(x))^2 \sec^2\left( \frac{π}{2} x \right) \,\d x.
\end{align*}
Since $\tan^2 α - \sec^2 α = -1$, then\begin{gather*}
\small(1) = \frac{4}{π^2} \int_0^{1 - δ} (f'(x))^2 \,\d x - \int_0^{1 - δ} (f(x))^2 \,\d x + \frac{4}{π} (f(1 - δ))^2 \tan\left( \frac{π}{2} (1 - δ) \right).\tag{2}
\end{gather*}
Note that as $δ → 0^+$,$$
f(1 - δ) = \int_{1 - δ}^1 f'(x) \,\d x \sim f'(1) δ,\quad \tan\left( \frac{π}{2} (1 - δ) \right) = \cot\left( \frac{π}{2} δ \right) \sim \frac{2}{πδ},
$$
thus making $δ → 0^+$ in (2) yields$$
\int_0^1 \left( \frac{2}{π} f'(x) + f(x) \tan\left( \frac{π}{2} x \right) \right)^2 \,\d x = \frac{4}{π^2} \int_0^1 (f'(x))^2 \,\d x - \int_0^1 (f(x))^2 \,\d x.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using a solution to show no others are possible in positive odd integer If I have an equation say $$3(1+x+x^2)(1+y+y^2)(1+z+z^2)+1=4x^2y^2z^2 \quad (1)$$ and I know a non negative integer solution $x=4, y=64, z=262144$ then no odd positive integer solutions can possibly exist. I know that one way to show it for this example would be to compute but I am looking for a proof that does not rely on computation like using wolfram as I want to generalize this. I will happily award the bounty to anyone who can prove it.
What I have tried is minimal. I suppose that $p,q,r \in \mathbb N$ and they satisfy equation $(1)$. I think if we let $p$ be the smallest integer of the solution $p<4$ is impossible so assume that $p\ge 5$ but this might lead to a contradiction since the coefficient of $4p^2q^2r^2 \quad $ is $4<5$ and that might be impossible?
| The given equation is $3(1+x+x^2)(1+y+y^2)(1+z+z^2)=4x^2y^2z^2-1$.
Let $x \leq y \leq z$. Then $x=1$ has no solution as in this case the LHS is at least $9y^2z^2>4y^2z^2$.
Thus, let $x \geq 3$.
Also note that $3$ divides the LHS so $3$ does not divide $xyz$; hence $x \geq 5$.
Divide the original equation by $x^2y^2z^2$ on both sides and upper-bound each sum by an infinite geometric series to get:
$3\dfrac{xyz}{(x-1)(y-1)(z-1)}>4-\dfrac{1}{x^2y^2z^2}>4-\dfrac{1}{xyz(x-1)(y-1)(z-1)}$.
Thus, $3xyz\geq 4(x-1)(y-1)(z-1)$, so that
$$\left(1-\dfrac{1}{x}\right)\left(1-\dfrac{1}{y}\right)\left(1-\dfrac{1}{z}\right)\leq\dfrac{3}{4}.$$
From the above, the minimum of $x,y,z$ must be at most $9$, otherwise the LHS is at least $(10/11)^3>3/4$.
This means that $x \in \{5,7\}$.
If $x=5$, then $y \leq 29$ and if $x=7$, then $y \leq 15$, both upper bounds obtained from the previous inequality.
When $x=5$, considering the cases modulo 5, we must have $y \equiv 0/2/4$ (mod 5).
Thus, we are finally left with the following pairs of $(x,y)$ to check and eliminate:
$(5,5),(5,7),(5,17),(5,19),(5,25),(5,29),(7,7),(7,13),(7,15)$.
| {
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Find min,max value of $P=2x-y$ Let $x,y$ such that $\sqrt{\left(x+2\right)^2+y^2}+\sqrt{\left(x-2\right)^2+y^2}=6$. Find minimize, maximize value of $$P=2x-y$$
We have :
$$y=2x-P$$ so from condition:
$$\sqrt{\left(x+2\right)^2+\left(2x-P\right)^2}+\sqrt{\left(x-2\right)^2+\left(2x-P\right)^2}=6$$
i solved if $x=\frac {18} {\sqrt {41}}$ then $P=\sqrt {41}$ and $x=-\frac {18} {\sqrt {41}}\rightarrow P=-\sqrt {41}$ but i do not know how to prove it is max,min value
| The condition it's the equation of the ellipse with focuses $(2,0)$ and $(-2,0)$ and $a=3,$
which gives $b^2=5$ and an equation of the ellipse:
$$\frac{x^2}{9}+\frac{y^2}{5}=1.$$
Also, easy to get this equation by squaring twice.
Now, let $2x-y=k$.
Thus, $$y=2x-k$$ and the equation $$\frac{x^2}{9}+\frac{(2x-k)^2}{5}=1$$ or
$$41x^2-36kx+9k^2-45=0$$ has real roots.
Id est, for the discriminant of this equation we obtain:
$$324k^2-41(9k^2-45)\geq0$$ or
$$k^2\leq41$$ or
$$-\sqrt{41}\leq k\leq\sqrt{41}.$$
For $k=-\sqrt{41}$ the line $y=2x+\sqrt{41}$ is a tangent line to the ellipse and the equality occurs in the touching point, which says that $-\sqrt{41}$ is a minimal value.
$\sqrt{41}$ is a maximal value because $y=2x-\sqrt{41}$ is a tangent to the ellipse.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of straight lines can be drawn
How many different straight lines can be drawn using $9$ points on the triangle of the figure below?
My try: Considering points other than $A$, $B$ and $C$ we got $2\times3 + 2\times1 + 3\times1 = 11$. For those three points, number of lines are $1 \times1 + 2 \times 1 + 3\times1 = 6$. So the total number is $6 + 11 + 3 = 20$ counting sides as possible lines but the solution gives $24$. What are the other lines which I'm not considering here? Also I think there is a combinatorial solution but I didn't find that.
| We have $9$ points and a straight line is determined by two distinct points.
We can use the binomial coefficients.
In how many ways can we choose pairs of points?
$$\binom{9}{2}=\frac{9\cdot 8}{2}=36$$
But, $A,D,E, B$ are collinear as well as $A,F,G,H,C$ and $B,I,C$.
So, we have $\binom{4}{2}+\binom{5}{2}+\binom{3}{2}$ non-distinct lines.
$$\binom{4}{2}+\binom{5}{2}+\binom{3}{2}=\frac{4\cdot 3}{2}+\frac{5\cdot 4}{2}+\frac{3\cdot 2}{2}=19$$
Now, we have to take $AB, BC$ and $AC$ into account, so the final result is:
$$36-19+3=20$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the integral of $\sqrt{\tan x}$ I just need to know why is my method wrong:
Let $I=\int\sqrt{\tan x} dx$, let $\tan(x)=t^2$ then, \begin{align*}&\sec^2(x)dx=2tdt\\
\implies&(1+\tan^2(x))dx=2tdt\\
\implies &dx=2tdt/(1+t^2)\end{align*}
So \begin{align*}I &= \int t\cdot \frac{2t}{1+t^2} dt\\
&= \int\frac{2t^2}{1+t^2}dt\end{align*} which can be solved easily. Is this method correct?
| Let's flesh out @QuantumApple's hint. With $\tan x=t^2$,$$\begin{align}\int\sqrt{\tan x}dx&=\int\frac{2t^2dt}{1+t^4}\\&=\int\frac{1}{2}\left(\tfrac{1}{1-\sqrt{2}t+t^{2}}+\tfrac{1}{1+\sqrt{2}t+t^{2}}-\tfrac{1}{\sqrt{2}}\left(\tfrac{\sqrt{2}-2t}{1-\sqrt{2}t+t^{2}}+\tfrac{\sqrt{2}+2t}{1+\sqrt{2}t+t^{2}}\right)\right)dt\\&=\frac{1}{\sqrt{2}}\arctan(\sqrt{2\tan x}-1)+\frac{1}{\sqrt{2}}\arctan(\sqrt{2\tan x}+1)\\&+\frac{1}{2\sqrt{2}}\ln\left|\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}\right|+C.\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Express $ \operatorname{gcd}\left(5^{m}+7^{m}, 5^{n}+7^{n}\right) $ in terms of $m$ and $n$
Let $m$ be a positive integer with $\operatorname{gcd}(m, n)=1 .$ Express
$
\operatorname{gcd}\left(5^{m}+7^{m}, 5^{n}+7^{n}\right)
$
in terms of $m$ and $n$
My work -
let $d=\operatorname{gcd}(5^m +7^m,5^n +7^n)$ then
$5^{2m} \equiv 7^{2m}$ mod(d)
$5^{2n} \equiv 7^{2n}$ mod(d)
and obviously $gcd(5,d)=gcd(7,d)=1$ so,
$5^{gcd(2m,2n)} \equiv 7^{gcd(2m,2n)}$ (mod d)
$5^2 \equiv 7^2$ (mod d)
$d= 1,2,3,4,6,8,12,24$
now i find values of d how to express this in terms of $m$ and $n$ ???
| For $k$ odd we have that
$$5^k+7^k\equiv 0\mod 3$$
and
$$5^k+7^k\equiv 5+7\equiv 4\mod 8$$
For $k$ even we have that
$$5^k+7^k\equiv 2\mod 3$$
and
$$5^k+7^k\equiv 1+1\equiv 2\mod 8$$
From here we have that if $m,n$ are both odd(i.e. if $m+n$ is even), then $\gcd(5^m+7^m,5^n+7^n)=12$.
Otherwise, if between $m$ and $n$ there is an odd and an even number(i.e. if $n+m$ is odd), then $\gcd(5^m+7^m,5^n+7^n)=2$.
So, for $\gcd(m,n)=1$, you could express
$$\gcd(5^m+7^m,5^n+7^n)=2\cdot
3^{(m+n+1)\%2}\cdot 2^{(m+n+1)\%2}$$
Where $a\%b$ denotes the remainder that we obtain when dividing $a÷b$.
Ps. This is clearly not a favorable answer since it doesn't seem generalizable.
| {
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Find the point where L intersects the curve The curve $x^2 - 2y^2 = 1$ includes the point $(1, 0)$. Let $L$ be the line through $(1, 0)$ having slope $m$. Find the other point where $L$ intersects the curve.
The line can be written as $y=m(x-1)$. Substituting this value of y in the original curve gives us:
$x^2(1-2m^2) + 4m^2x + 2m^2 - 1 = 0$.
I think we can divide this equation by $x-1$ so we get another solution but I couldn't do that. Can someone help me?
| Note the equation should actually be $$x^2(1-2m^2) +4m^2x \color{red} - 2m^2 -1=0$$ Performing long division by $(x-1)$ yields $$(x-1) \left( (1-2m^2)x +(2m^2+1) \right) =0 \\ \implies x=1, \frac{2m^2+1}{2m^2-1}$$ and the required point is $$\left ( \frac{2m^2+1}{2m^2-1} , \frac{2m}{2m^2-1} \right)$$
| {
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Can this function be defined in a way to make it continuous at $x=0$? We have $$f=\frac{x}{\vert x-1 \vert - \vert x +1 \vert}$$
If we want to "define" this function to be continuous at $x=0$, it's limit at $0$ must equal $f(0)$. So we should find this limit and assign it to be equal to $f(0)$, then the function is continuous at $0$. Since we are looking at the function when $x\to 0$, $x\neq 0$. Lets divide both sides by $x$.
$$f=\frac{x}{\vert x-1 \vert - \vert x +1 \vert}=\frac{1}{\frac{\vert x-1 \vert}{x}-\frac{\vert x+1\vert}{x}}=\frac{1}{\vert 1-\frac{1}{x}\vert - \vert 1+ \frac{1}{x}\vert }$$
We can use $\lim \phi(x)^{-1}=\frac{1}{\lim \phi(x)}$ here ( the limit $\neq$ 0, by hypothesis ). The inverse of the limit of $\phi(x)=\vert 1 - \frac{1}{x} \vert-\vert 1+\frac{1}{x}\vert$, when $x\to 0$. If $x<1$, we have that $$\frac{1}{x}>1\implies0>1-\frac{1}{x}\implies \Bigg\vert 1-\frac{1}{x}\Bigg\vert=-\Big(1-\frac{1}{x}\Big)$$
Now if $x>0$, we have that $$\Bigg\vert 1 - \frac{1}{x} \Bigg\vert-\Bigg\vert 1+\frac{1}{x}\Bigg\vert=-2$$
and if $x<0$, then $$\Bigg\vert 1 - \frac{1}{x} \Bigg\vert-\Bigg\vert 1+\frac{1}{x}\Bigg\vert=1-\frac{1}{x}-1-\frac{1}{x}=\frac{(-2)}{x}$$
The limit of $f$ when $x\to 0$ appears to be $\frac{-1}{2}$. Could anyone tell me what errors I made in the limit finding process?
| There is an error at the end of your calculations. In the case $-1 < x < 0$ we have
$$
\left\vert 1 - \frac{1}{x} \right\vert-\left\vert 1+\frac{1}{x}\right\vert=
\left(1-\frac{1}{x}\right)+\left(1+\frac{1}{x}\right)=2
$$
because the argument of the first absolute value is positive, and the argument of the second absolute value is negative.
A slightly simpler approach:
Since you are interested in the limit at $x=0$ it suffices to consider $x = (-1, 1)$, $x\ne 0$. For these arguments is $x-1 <0$ and $x+1> 0$, and therefore
$$
f(x)=\frac{x}{\vert x-1 \vert - \vert x +1 \vert} = \frac{x}{-(x-1) - ( x +1)} = \frac{x}{-2x} =-\frac 12
$$
so that
$$
\lim_{x \to 0 }f(x) = -\frac 12 \, .
$$
| {
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Let chord of contact be drawn from every point on the circle $x^2+y^2=100$ to the ellipse [CONT..]
Let chord of contact be drawn from every point on the circle $x^2+y^2=100$ to the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ such that all lines touch a standard ellipse. Find $e$ for the ellipse
Let the point $(h,k)$ lie on the given circle
The chord of the contact drawn to the given ellipse is
$$\frac{hx}{4}+\frac{ky}{9}-1=0$$
This line is coincident with the the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$$y=mx\pm \sqrt{a^2m^2+b^2}$$
Then comparing the two equations
$$m=\frac{-9h}{4k}$$
And $$\frac{81}{k^2}=a^2m^2+b^2$$
$$\frac{81}{k^2}=\frac{81a^2h^2}{16k^2}+b^2$$
$$(81)(16)=81a^2h^2+16k^2b^2$$
How do I proceed from here? Simply substituting $h^2=100-k^2$ doesn’t give any details for $a$ and $b$
| What you are looking for is the curve that the chord of contact envelops. You can find (one half of) the solution ellipse by letting $k=\sqrt{100-h^2}$ and substituting into the chord of contact $\frac{hx}{4}+\frac{ky}{9}-1=0.$ The curve the family of lines $\frac{hx}{4}+\frac{\sqrt{100-h^2}y}{9}-1=0$ envelops can be found by the method in the wikipedia link above. In maxima CAS
eq1:h*x/4+sqrt(100-h^2)*y/9-1;
solve([diff(eq1,h),eq1],[x,y]);
[[x = h/25,y = (9*sqrt(100-h^2))/100]]
The answer is the parametrized curve
$$(x(h),y(h))=(\frac{h}{25}, \frac{9\sqrt{100-h^2}}{100}).$$
Implicitizing by putting $h=25x$ into $y^2=\frac{81(100-h^2)}{100^2}$ you get
$2025x^2+400y^2-324=0,$ when you factor out a $5^2.$ This is $(x/(2/5))^2 +(y/(9/10))^2=1 $ in the standard form.
Now to answer the question an ellipse with semi-axes $\frac9{10},\frac25$ has eccentricity $\sqrt{1-b^2/a^2}=\sqrt{1-\frac{4/25}{81/100}}=\sqrt{65}/9.$
| {
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Closed form of $\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx$
Can a closed form solution for the following integral be found:
$$\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx\,?$$
I have tried all the standard tricks such as integration by parts, various substitutions, and parametric differentiation (Feynman's trick), but all to no avail.
An attempt is letting
$$f(t):=\int_0^\infty\,\arctan^2\left(\frac{2tx}{1+x^2}\right)\,\text{d}x\,.$$
Therefore,
$$f'(t)=\int_0^\infty\,\frac{8x^2(x^2+1)}{\big(x^4+2(2t^2+1)x^2+1\big)^2}\,\left(1+x^2-4tx\arctan\left(\frac{2tx}{1+x^2}\right)^{\vphantom{a^2}}\right)\,\text{d}x\,.$$
This doesn't seem to go anywhere. Help!
| Here is another solution with a generalization:
Let $r=\sinh\alpha$ and $s=\sinh\beta$. Then
$$\begin{aligned}
&\int_{0}^{\infty} \arctan\left(\frac{2rx}{1+x^2}\right)\arctan\left(\frac{2sx}{1+x^2}\right) \, \mathrm{d}x\\
&= \pi \left( \alpha \sinh\beta+\beta\sinh\alpha+(\cosh\alpha+\cosh\beta)\log\left(\frac{e^{\alpha}+e^{\beta}}{1+e^{\alpha+\beta}}\right) \right)
\end{aligned} \tag{*}$$
Proof. Let $J = J(\alpha,\beta)$ denote the right-hand side of $\text{(*)}$. Then
$$
J(0, \beta) = 0, \qquad
J_{\alpha}(\alpha, 0) = 0, \qquad
J_{\alpha\beta} = \pi \left( \frac{1+\cosh\alpha\cosh\beta}{\cosh\alpha + \cosh\beta} \right).
$$
Now let $I = I(\alpha, \beta)$ denote the left-hand side of $\text{(*)}$. Then by the substitution $x=\tan(\theta/2)$, we get
$$ I = \frac{1}{2}\int_{0}^{\pi} \frac{\arctan(\sinh\alpha\sin\theta)\arctan(\sinh\beta\sin\theta)}{1+\cos\theta} \, \mathrm{d}\theta. $$
From this, we easily check that $I$ also satisfies
$$ I(0, \beta) = 0, \qquad I_{\alpha}(\alpha, 0) = 0. $$
Moreover,
\begin{align*}
\require{cancel}
I_{\alpha\beta}
&= \frac{1}{2}\int_{0}^{\pi} \frac{\cosh\alpha\cosh\beta(1-\cos\theta)}{(1+\sinh^2\alpha\sin^2\theta)(1+\sinh^2\beta\sin^2\theta)} \, \mathrm{d}\theta \\
&= \frac{1}{2}\int_{0}^{\pi} \frac{\cosh\alpha\cosh\beta}{(1+\sinh^2\alpha\sin^2\theta)(1+\sinh^2\beta\sin^2\theta)} \, \mathrm{d}\theta \\
&\quad - \cancelto{0}{\frac{1}{2}\int_{0}^{\pi} \frac{\cosh\alpha\cosh\beta}{(1+\sinh^2\alpha\sin^2\theta)(1+\sinh^2\beta\sin^2\theta)} \, \mathrm{d}\sin\theta}\\
&= \frac{1}{2}\int_{0}^{\infty} \frac{\cosh\alpha\cosh\beta (1 + t^2)}{(t^2 + \cosh^2\alpha)(t^2 + \cosh^2\beta)} \, \mathrm{d}t \tag{$t=\cot\theta$}
\end{align*}
It is not hard to check that the last integral is equal to $J_{\alpha\beta}$. Therefore we get $I = J$.
| {
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Find angles of triangle $DEC$ $ABC$ is an equaliterial triangle. $AC∥MN$ and $M$ and $N$ lies at $AB$ and $BC$ respectively. $D$ is centroid of $MBN$, $E$ is a midpoint of $AN$. Then, find the angles of the triangle $DEC$.
First of all I am sure that it can be solved by rotation. Unfortunately, I haven't managed to process much. Let $AC⊥BF$ and $F∈AC$. $B$,$F$,$D$ are collinear and $AF$=$FC$. Since $AE$=$EN$. $EF∥NC$ and $NC$=$2EF$.
I am suspecting angles of $DEC$ are 30°,60°,90°. So I am now trying to prove $CDEF$ is circumcircle.
| Another way to solve it is by analytical geometry. Considering the figure of the preceding answer, setting the vertices $B(0,0)$, $C(1,0)$, and $A(1/2,\sqrt{3}/2)$, and placing the point $N$ in $(k,0)$ (with $0 \leq k \leq 1$), we directy get that:
*
*the coordinates of $D$ are $\Big(k/2,k/(2 \sqrt{3})\Big)$;
*the coordinates of $E$, midpoint between $A$ and $N$, are $\Big((2k+1)/4,\sqrt{3}/4)\Big)$.
Now we can find the equation of the $DE$ line by solving the system
$\left\{
\begin{array}{ll}
\frac{k}{2 \sqrt{3}}=a \frac{ k}{2} + b\\
\frac{\sqrt{3}}{4}= a \frac{2k+1}{4} +b
\end{array}
\right.
$
whose solutions lead to the line $$y=\frac{3-2k}{\sqrt{3}}+\frac{ k(k-1)}{\sqrt{3}}$$
Similarly, we can find the equation of the $CE$ line by solving the system
$\left\{
\begin{array}{ll}
0=a+ b\\
\frac{\sqrt{3}}{4}= a \frac{2k+1}{4} +b
\end{array}
\right.
$
whose solutions lead to the line $$y=-\frac{\sqrt{3}}{3-2k}+\frac{ \sqrt{3}}{3-2k}$$
Since the slopes are negative reciprocals, the two lines are perpendicular and $\angle{DEC}=90°$. Now we can note that
$$\overline{DE}=\sqrt{\Big((2k+1)/4-k/2\Big)^2 +\Big(\sqrt{3}/4- k/(2 \sqrt{3})\Big) ^2}\\=\frac{\sqrt{k^2-3k+3}}{2 \sqrt{3}}$$
and
$$\overline{CE}=\sqrt{\Big((2k+1)/4-1\Big)^2 +\Big(\sqrt{3}/4\Big) ^2}\\=\frac{\sqrt{k^2-3k+3}}{2}$$
Since $\overline{CE}= \sqrt{3}\cdot \overline{DE } $, we directly get $\angle{DCE}=30°$ and $\angle{CDE}=60°$.
| {
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Why does simplifying $\arcsin(2x\sqrt{1-x^2})$ two ways give different results?
Find derivative of
$y=\arcsin(2x \sqrt{1-x^2}) $ in domain $\frac{-1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$
If you put $x=\sin\theta$ then
$$ y= \arcsin(2sin(\theta) \sqrt{1- sin^2 ( \theta)})$$
$$y= \arcsin( sin2\theta)$$
$$y= 2\theta$$
$$y = 2 \arcsin(x)$$
But, if you put $x=\cos\theta$ then , again,
$$y=2\theta$$
But, resubstituting
$$ y= 2 \arccos(x)$$
But derivatives of both are different.
Now where's the mistake? Is it something related to the original domain I took?
| There is nothing wrong with your algebra. It’s just that you need to take extra care when dealing with the domain. Note that $\arcsin(\sin \theta=\theta$ only when $-\frac{\pi}{2} \le \theta\le \frac{\pi}{2}$.
Substituting $x=\sin\theta$ gives $$y= \arcsin(2\sin\theta|\cos\theta|)=\arcsin|\sin 2\theta| $$ Now, $$-\frac{1}{\sqrt 2} \lt \sin\theta \lt \frac{1}{\sqrt 2} \\ \implies -\frac{\pi}{4} \le \theta\le \frac{\pi}{4} \\\implies -\frac{\pi}{2} \lt2\theta \lt \frac{\pi}{2}$$ and we can safely say that $$y=2\theta=2\arcsin x$$
But when $x=\cos \theta$, we have $$ -\frac{1}{\sqrt 2} \lt \cos\theta \lt \frac{1}{\sqrt 2} \\ \implies \frac{\pi}{4} \lt \theta \lt \frac{3\pi}{4} \\ \implies \frac{\pi}{2} \lt 2\theta \lt \frac{3\pi}{2}$$ This time, we need to shift by $\pi$ so that $$-\frac{\pi}{2}\lt 2\theta-\pi\lt\frac{\pi}{2}$$So, $$ y =\arcsin(\sin 2\theta) =\arcsin\left(-\sin(2\theta-\pi)\right) = \pi-2\arccos x \\ = 2\left(\frac{\pi}{2}-\arccos x\right) \\ = 2\arcsin x$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a, b, c, d \in R^+$ such that $a + b + c + d = 1$. Prove that $\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq \frac{1}{8}$ Let $a, b, c, d \in R^+$ such that $a + b + c + d = 1$. Prove that,
$$\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq \frac{1}{8}$$
Well from their sum we do get that $\frac{1}{4} \geq \sqrt[4]{abcd}$
$$\Rightarrow \frac{1}{4^8} \geq a^2b^2c^2d^2$$
and applying AM-GM on LHS of the given inequality we get ,
$$\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq 4 \cdot \sqrt[4]{\frac{a^3b^3c^3d^3}{(a+b)(b+c)(c+d)(d+a)}}$$
and $(a+b)(b+c)(c+d)(d+a) \geq 16 \cdot abcd$ or
$$\frac{a^3b^3c^3d^3}{(a+b)(b+c)(c+d)(d+a)} \leq 16 \cdot a^2b^2c^2d^2 \leq 4^2 \cdot \frac{1}{4^8}= \frac{1}{4^6}$$
$$\Rightarrow \frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq 4 \cdot\frac{1}{8}=\frac{1}{2} > \frac{1}{8} \blacksquare.$$
Is this proof correct ? Did i miss any details? My doubt really stems from the fact that i didnt get $\frac{1}{8}$ directly but $\frac{1}{2}$, which makes my resultant inequality strict instead of being $\geq$ and it makes me wonder whether my proof is right. Thanks.
EDIT: Well guys i haven't read Titu's Lemma or Holder's inequality just yet though both of them do seem very powerful. I guess i'll just come to this question later when m done with those topics. Thanks for your help. Also I was just wondering whether it is possible to do it purely using AM-GM or maybe WAM-WGM ? Thanks again.
| No. Your proof is not correct.
After your first step we need to prove a wrong for $a\rightarrow0^+$ inequality.
I think it's better to use Holder: $$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(b+c)}=\frac{1}{8}.$$
| {
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Evaluate: $\int \frac{t^5-t^3}{t^{10}-1}\, dt. $ None of my attempts (such as letting $x=t^5$ or splitting up the integrand into two parts etc.) succeeded. Basically I couldn't find a useful thing on manipulating the given integrand. Please give me some hints.. Thanks in advance.
| And I'll finish the thing. $I=\int\frac{t^5-t^3}{t^{10}-1}\,\mathrm{d}t$
Let $u=t^2,du=2t\,dt$, by J.G.'s answer
$$I=\frac12\int\frac{u\,\mathrm{d}u}{u^4+u^3+u^2+u+1}$$
Using $$u^4+u^3+u^2+u+1=\left(u^2+\frac{1+\sqrt{5}}{2}u+1\right)
\left(u^2+\frac{1-\sqrt{5}}{2}u+1\right)$$ both parenthesis have no real roots, so we will have to boil that down to $\int\frac{ax+b}{x^2+1}\mathrm{d}x$ for some $x$ after partial decomposition. So complete the squares:
$$\left(u^2+\frac{1+\sqrt{5}}{2}u+1\right)
\left(u^2+\frac{1-\sqrt{5}}{2}u+1\right)=\\
\left(\left(\frac{4u+1+\sqrt{5}}{4}\right)^2+\frac{5-\sqrt{5}}{8}\right)
\left(\left(\frac{-4u-1+\sqrt{5}}{4}\right)^2+\frac{5+\sqrt{5}}{8}\right)$$
As we see $4u+1$ shows up in both cases, let $4u+1=x$, $u=\frac{x-1}{4}$, $\mathrm{d}u=\frac{1}{4}\mathrm{d}x$ and
$$I=8\int\frac{x-1}{x^4 + 10 x^2 + 40 x + 205}\mathrm{d}x$$
$$=8\int\frac{x-1}{\left((x - \sqrt{5})^2 + 2\sqrt{5} + 10\right)
\left((x + \sqrt{5})^2 - 2\sqrt{5} + 10\right)}\mathrm{d}x$$
and the partial fraction decomposition is (details are here and here)$$
\frac{x-1}{x^4 + 10 x^2 + 40 x + 205}=
\frac{1}{4\sqrt{5}}\left(\frac{1}{(x - \sqrt{5})^2 + 2\sqrt{5} + 10}-\frac{1}{(x + \sqrt{5})^2 - 2\sqrt{5} + 10}\right)$$
$$I=\frac{2\sqrt{5}}{5}\left(
\frac{1}{\sqrt{2\sqrt{5} + 10}}\int\frac{\mathrm{d}\left(\frac{x-\sqrt{5}}{\sqrt{2\sqrt{5} + 10}}\right)}{\left(\frac{x-\sqrt{5}}{\sqrt{2\sqrt{5} + 10}}\right)^2+1}-
\frac{1}{\sqrt{-2\sqrt{5} + 10}}\int\frac{\mathrm{d}\left(\frac{x+\sqrt{5}}{\sqrt{-2\sqrt{5} + 10}}\right)}{\left(\frac{x+\sqrt{5}}{\sqrt{-2\sqrt{5} + 10}}\right)^2+1}
\right)$$
$$=\frac{2\sqrt{5}}{5}\left(
\frac{1}{\sqrt{2\sqrt{5} + 10}}\arctan\left(\frac{x-\sqrt{5}}{\sqrt{2\sqrt{5} + 10}}\right)-
\frac{1}{\sqrt{-2\sqrt{5} + 10}}\arctan\left(\frac{x+\sqrt{5}}{\sqrt{-2\sqrt{5} + 10}}\right)
\right)+C$$
where $x=4t^2+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all functions $f : \mathbb R \to \mathbb R$ such that: $f\left(x^3\right)+f\left(y^3\right) = (x + y)f\left(x^2\right)+f\left(y^2\right)-f(xy)$ Here is the question:
Find all functions $f : \mathbb R \to \mathbb R$ such that:
$$f\left(x^3\right)+f\left(y^3\right) = (x + y)f\left(x^2\right)+f\left(y^2\right)-f(xy)$$
What I tried:
Note that the function $f\left(x^3\right)+f\left(y^3\right)$ is symmetric. From here we get
$$(x + y)f\left(x^2\right) + f\left(y^2\right) - f(xy) = f(x + y)f\left(y^2\right) + f\left(x^2\right) - f(xy)$$
which gives $(x + y - 1)f\left(y^2\right) = (x + y - 1)f\left(x^2\right)$.
From here I cannot proceed further. I could have cancelled $(x + y - 1)$ but I haven't proved that $(x + y - 1) \neq 0$ and neither I don't know how to proceed with $\left(y^2\right) = f\left(x^2\right)$ even if we can cancel $x + y - 1$. Any hints or suggestions will be greatly appreciated.
| Here's how to get $f(x^2) = f(y^2)$.
If $x+y-1 \neq 0$, we must have $f(x^2) = f(y^2)$. F any $x^2$ and $y^2$, pick a $z$ such that $x+z-1$ and $y+z-1$ are nonzero, which is always possible by taking $z \neq 1-x, 1-y$. Then $f(x^2) = f(z^2) = f(y^2)$. So $f(x^2) = f(y^2)$ for all reals $x, y$, and hence $f$ is constant on nonnegative reals.
| {
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Compute a matrix $P$ such that $P$ satisfies $3I+P+P^2=\left(\begin{smallmatrix}3&0&0\\3&6&0\\0&0&6\end{smallmatrix}\right)$ Question: Compute a matrix $P$ such that the matrix $P$ satisfies $3I + P + P^2 = \begin{pmatrix} 3 & 0 & 0\\ 3 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix}$
Attempt: First, I simplify the equation and get $P + P^2 = P(I + P) = \begin{pmatrix} 0 & 0 & 0 \\3 & 3 & 0\\ 0& 0 & 3\end{pmatrix}$. And then I don't know how to solve further. I think this question somehow is related to the characteristic polynomial, but I am not sure. I get stuck on this question for hours, could someone please help me out, thanks!
| This is a bit of a brute-force diagonalisation approach - I do feel as though there should be a simpler method, but this at least gives you an answer.
We can reduce this to solving an equation of the form $X^2=C$ by completing the square, as we would do in the real case. Note that (dropping the now-implicit identity matrix for scalars)
$$\left(P + \frac{1}{2}\right)^2 = P^2 + P + \frac{1}{4} \implies P^2 + P = \left(P + \frac{1}{2}\right)^2 - \frac{1}{4}$$
Writing $Q$ for the RHS of your original equation, we have
$$P^2 + P + 3 = Q \iff \left(P + \frac{1}{2}\right)^2 = Q - \frac{11}{4}$$
We now seek a matrix $X$ such that $X^2 = Q'$ where $Q' = Q - 11/4$. One way to do this is to diagonalise $Q'$ as
$$Q' = V^{-1} D V$$
where $D$ is diagonal, since then we find that taking
$$X = V^{-1} \sqrt{D} V \implies X^2 = Q'$$
and $\sqrt{D}$ is easy to calculate since we just take the square root of the diagonal entries. Indeed, such a decomposition is possible, with
$$V = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}, \quad D = \frac{1}{4} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 13 & 0 \\ 0 & 0 & 13 \end{pmatrix}$$
Computing $X$, as indicated above, one can show that
$$
X = \frac{1}{2} \begin{pmatrix} 1 & 0 & 0 \\ \sqrt{13}-1 & \sqrt{13} & 0 \\ 0 & 0 & \sqrt{13} \end{pmatrix}$$
and hence
$$P = \frac{1}{2} \begin{pmatrix} 0 & 0 & 0 \\ \sqrt{13}-1 & \sqrt{13}-1 & 0 \\ 0 & 0 & \sqrt{13}-1 \end{pmatrix} = \frac{\sqrt{13}-1}{2} \begin{pmatrix} 0 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}
$$
| {
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Can any cyclic polynomial in $a, b, c$ be expressed in terms of $a^2b+b^2c+c^2a$, $a+b+c$, $ab+bc+ca$ and $abc$? Problem. Let $f(a,b,c)$ be a cyclic polynomial in $a, b, c$.
Can $f$ always be expressed as $g(a+b+c, ab+bc+ca, abc, a^2b+b^2c + c^2a)$ for some polynomial $g(p, q, r, Q)$?
Motivation: If we want to prove $g(a+b+c, ab+bc+ca, abc, a^2b+b^2c + c^2a)\ge 0$, and $g(p, q, r, Q)$ is non-increasing with $Q$,
by using the known inequality $a^2b + b^2c + c^2a \le \frac{4}{27}(a+b+c)^3 - abc$(see How to prove this inequality? $a^{2}+b^{2}+c^{2}\leq 3$),
sometimes, we may prove $g(a+b+c, ab+bc+ca, abc, \frac{4}{27}(a+b+c)^3 - abc)\ge 0$.
As a result, we deal with a symmetric inequality rather than the original cyclic inequality.
Is it known? Is it easily to prove?
Let us see some examples. Denote $p = a+b+c, q = ab+bc+ca, r = abc$ and $Q = a^2b+b^2c+c^2a$.
1) $ab^2 + bc^2 + ca^2 = (a+b+c)(a^2+b^2+c^2) - a^3-b^3-c^3 - (a^2b+b^2c+c^2a)$
2) From the known identity $a^3b+b^3c+c^3a +(ab+bc+ca)^2 = (a+b+c)(a^2b+b^2c+c^2a+abc)$, we have
$a^3b+b^3c+c^3a = p(Q + r) - q^2$.
3) $a^3b^2+b^3c^2+c^3a^2 = Qq - r(p^2-2q) - rq$.
4) $(a^2+b)(b^2+c)(c^2+a) = \cdots$
Any comments and solutions are welcome.
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Thus, $$\sum_{cyc}a^2b=\frac{1}{2}\sum_{cyc}2a^2b=\frac{1}{2}\sum_{cyc}(a^2b+a^2c-(a^2c-a^2b))=$$
$$=\frac{1}{2}(9uv^2-3w^3)+\frac{1}{2}(a-b)(a-c)(b-c).$$
Now, for $a\geq b\geq c$ we obtain:
$$\sum_{cyc}a^2b=\frac{1}{2}(9uv^2-3w^3)+\frac{1}{2}\sqrt{27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)}$$ and for $a\leq b\leq c$ we obtain:
$$\sum_{cyc}a^2b=\frac{1}{2}(9uv^2-3w^3)-\frac{1}{2}\sqrt{27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)}.$$
By the similar way we can write any cyclic polynomial as function of elementary symmetric polynomials.
Because any Schur's polynomial we can write as $\prod\limits_{cyc}(a-b)P(a,b,c),$ where $P$ is a symmetric polynomial.
For example, $$\sum_{cyc}a^3b=\frac{1}{2}\sum_{cyc}(a^3b+a^3c)+\frac{1}{2}\sum_{cyc}(a^3c-a^3b)=$$
$$=\frac{1}{2}\sum_{cyc}(a^3b+a^3c)+\frac{1}{2}(a-b)(b-c)(c-a)(a+b+c).$$
| {
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Evaluate: $\lim_{n\to\infty} n^3\int_n^{2n} \frac x{1+x^5}\, dx. $ Is there a way to evaluate: $$\lim_{n\to\infty}n^3\int_n^{2n}\frac x{1+x^5}\, dx,$$ without evaluating the indefinite integral: $$\int \frac x{1+x^5}\, dx\;?$$ Please suggest. Thanks in advance.
| Under $x^{-3}\to x$,
$$ \frac{1}{n^3}\int_{n}^{2n}\frac{x}{1+x^5}dx=\frac{n^3}{3}\int_{\frac1{8n^3}}^{\frac1{n^3}}\frac{dx}{1+x^{\frac{5}{3}}}. $$
By the MVT for integrals, there is $c_n\in (\frac1{8n^3},\frac1{n^3})$ such that
$$ \frac{1}{n^3}\int_{n}^{2n}\frac{x}{1+x^5}dx=\frac1{n^3}\bigg(\frac1{8n^3}-\frac1{n^3}\bigg)\frac{1}{1+c_n^{\frac{5}{3}}}=\frac{7}{24}. $$
So
$$ \lim_{n\to\infty}\frac{1}{n^3}\int_{n}^{2n}\frac{x}{1+x^5}dx=\lim_{n\to\infty}\frac{7}{24}\frac{1}{1+c_n^{\frac{5}{3}}}=\frac{7}{24}. $$
| {
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Determine all pairs of positive integers $(a,b)$ such that $a^2+b^2+ab$ is a perfect square. Consideration via mod $4$ shows that $a,b=0$ mod $4$ or one of between $a,b$ is $=1$ mod $4$ while the other is $=0 $ mod $4$.
Considering $(a+b)^2,a^2+an+b^2,(a+b-1)^2$ as we can deduce that $a,b>2$.
| Just another way to approach the problem.
Since $$a^2+b^2+ab=(a+b)^2-ab $$ is a squared integer, there exists an integer $n$ such that
$$ (a+b)^2-(a+b-n)^2=ab$$
Solving it for $a$, we have
$$a = \frac{n (2 b - n)}{b - 2 n} $$
with the immediate condition $b \neq 2 n$. However, we also have to set two other conditions. Because $a$ is positive, we must have $$ (2b-n)/(b-2n)>0$$ which for positive $n$ leads to
$$b<\frac{n}{2} \,\,\,\text{or} \,\,\, b>2n$$
In addition, because $a$ is integer, we must have
$$(b-2n) |n (2 b - n)$$
Since $n(2b-n)$ can be written as $$n(2b-4n+3n)=2n(b-2n) +3n^2$$ it is divisible by $(b-2n)$ if and only if $(b-2n)$ divides $3n^2$. This occurs only when $b$ is of the form $$b=2n+k$$
where $k$ is a positive integer that divides $3n^2$.
Collecting all results, to find the pairs $a,b$ asked in the OP, we can:
*
*set an arbitrary positive integer $n$;
*take all positive integers $k$ that are divisors of $3n^2$;
*assign to $b$ the values $2n+k$;
*calculate for each $b$ the value $a=n(2b-n)/(b-2n)$.
For example, setting $n=1$, we get that the possible values of $k$, i.e. the divisors of $3 \cdot 1^2=3$, are $1$ and $3$. So we can assign to $b$ the values $2\cdot 1 +1=3$ and $2\cdot 1 +3=5$. The corresponding values of $a$ are $(6-1)/(3-2)=5$ and $(10-1)/(5-2)=3$. So the case $n=1$ yields the symmetric pairs $5,3$ and $3,5$.
As another example, setting $n=6$, we get that the possible values of $k$, i.e. the divisors of $3 \cdot 6^2=108 $, are $1$, $2$, $3$, $4$, $6$, $9$, $12$, $18$, $27$, $36$, $54$ and $108$. Proceeding as above, we can assign to $b$ the values $2\cdot 6 +k$ and calculate the corresponding values of $a$. This yields the $a,b$ pairs $(120,13)$, $(66,14)$, $(48,15)$, $(39,16)$, $(30,18)$, $(24,21)$, plus the other six symmetric pairs.
You can find here a Wolfram confirmation of the $12$ pairs generated by the case $n=6$.
| {
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Permutations of numbers with difference 1 that are lesser than h How many are the permutations of $n$ numbers where the first and last number are $1$, the difference between the numbers is $1$ and the numbers are lesser than h?
For example, for $n = 5$ and $h = 6$ $(h > 4)$ the permutations are $9$:
$ 1-2-3-2-1 $
$ 1-2-2-2-1 $
$ 1-2-2-1-1 $
$ 1-1-2-2-1 $
$ 1-2-1-2-1 $
$ 1-2-1-1-1 $
$ 1-1-2-1-1 $
$ 1-1-1-2-1 $
$ 1-1-1-1-1 $
| These "permutations" are related to Motzkin paths. Compare this picture with your example.
| {
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proving some interesting properties of these matrices let X and Y be two matrices different from I, such that $XY=YX$ and $X^n-Y^n$ is invertible for some natural number n .If
$$X^n-Y^n = X^{n+1}-Y^{n+1} = X^{n+2}-Y^{n+2}$$,
then prove that $I-X,I-Y $ are singular and $X+Y=XY+I$
my approach:
I tried to pre multiply and post multiply by $X and Y$ but it could not produce anything.Kindly help me by providing some suggestions on how to solve this question.
| From $X^n-Y^n=X^{n+1}-Y^{n+1}=X^{n+2}-Y^{n+2}$, we obtain
\begin{align}
X^n(I-X) &= Y^n(I-Y),\tag{1}\\
X^{n+1}(I-X) &= Y^{n+1}(I-Y).\tag{2}
\end{align}
Subtract both sides of $(1)$ by $X^n(I-Y)$, we obtain
\begin{align}
(Y-X)X^n &= (Y^n-X^n)(I-Y).\tag{3}
\end{align}
Substitute the LHS of (1) into the RHS of (2), we get
\begin{align}
X^{n+1}(I-X) &= YX^n(I-X),\\
(X-Y)X^n(I-X) &= 0,\\
(X^n-Y^n)(I-Y)(I-X) &= 0\ \text{ by $(3)$},\\
(I-Y)(I-X) &= 0. \tag{4}
\end{align}
Since $X,Y\ne I$, $(4)$ implies that neither $I-Y$ nor $I-X$ is invertible. Also, by expanding $(4)$, we obtain $X+Y=XY+I$.
| {
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Shuffling the digits of an integer so that the ratio between the resulting numbers is fixed. Suppose we have a 3-digit integer with all digits distinct. If we create its two shift numbers (resulting by rotation of the original digits), the ratio between the two subsequent numbers (if we order them in ascending or descending order) must be fixed.
Find this number.
Suppose the number is xyz and the two resulting numbers, if we shift the digits to the right, are yzx and zxy. Assuming $xyz<yzx<zxy$ (there is no equal, since all digits are distinct), we want $\frac{zxy}{yzx} = \frac{yzx}{xyz} = k$, where k: rational.
Furthermore, it is easy to deduce that 0 is excluded, or one of the resulting numbers would be 2-digit.
I have found 2 such sets of numbers, {243, 324, 432} and {486, 648, 864} and now I am trying to formulate an algebraic solution.
If xyz is the original number, the first rotation is yzx and can be found as follows:
$yzx = xyz*10+x-1000*x$.
Also $zxy= yzx*10+y-1000*y$.
Assuming wlg that $xyz<yzx<zxy$, we must have $\frac{xyz*10+x-1000*x}{xyz} = \frac{yzx*10+y-1000*y}{xyz*10+x-1000*x} = k$
I also noticed that if the 3 numbers are in ascending order $xyz<yzx<zxy$, then their differences
$yzx-xyz$ and $zxy-yzx$ have the same 3 digits in rotation.
but I don't think I can advance it any further...
| We want to find triples of integers $(x,y,z)$ such that
$$0\lt x\le 9, 0\le y\le 9,0\le z\le 9\tag1$$
$$x\not=y,y\not=z,z\not=x\tag2$$
$$\frac{100z+10x+y}{100y+10z+x}=\frac{100y+10z+x}{100x+10y+z}\tag3$$
From $(3)$, we get
$$ yz-x^2=10(xz-y^2)\implies yz\equiv x^2\pmod{10}\tag4$$
and
$$x=-5z+\sqrt{25z^2+10y^2+yz}\implies \sqrt{25z^2+10y^2+yz}\ \in\mathbb Z\tag5$$
Also, if $y\lt z\lt x$, then $yz-x^2=10(xz-y^2)$ does not hold since LHS is negative while RHS is positve.
If $x\lt z\lt y$, then $yz-x^2=10(xz-y^2)$ does not hold since LHS is positive while RHS is negative.
*
*If $x^2\equiv 1\pmod{10}$, then $yz=21$ and so $(y,z)=(3,7),(7,3)$ each of which doesn't satisfy $(5)$.
*If $x^2\equiv 4\pmod{10}$, then $yz=4,14,24,54$ and so $(y,z)=(1,4),(4,1),(7,2),(3,8),(4,6),(6,4),(6,9),(9,6)$ where only $(y,z)=(6,4),(9,6)$ satisfy $(5)$ with $x=8,12$ respectively. (we don't need to consider the case $(y,z)=(2,7)$ since then $x=8$ for which $y\lt z\lt x$ holds. Also, we don't need to consider the case $(y,z)=(8,3)$ since then $x=2$ for which $x\lt z\lt y$ holds.)
*If $x^2\equiv 9\pmod{10}$, then $yz=9$ and so $(y,z)=(1,9),(9,1)$ each of which doesn't satisfy $(5)$.
*If $x^2\equiv 6\pmod{10}$, then $yz=6,16,36,56$ and so $(y,z)=(1,6),(6,1),(3,2),(2,8),(8,2),(4,9),(9,4),(7,8)$ where only $(y,z)=(3,2)$ satisfies $(5)$ with $x=4$. (we don't need to consider the case $(y,z)=(2,3)$ since then $x=4$ or $6$ for which $y\lt z\lt x$ holds. Also, we don't need to consider the case $(y,z)=(8,7)$ since then $x=4$ or $6$ for which $x\lt z\lt y$ holds.)
*If $x^2\equiv 5\pmod{10}$, then $x=5$ and one of $y,z$ is $5$, which does not satisfy $(2)$.
So, the only solutions are
$$\color{red}{(x,y,z)=(4,3,2),(8,6,4)}$$
where we have
$$\frac{243}{324}=\frac{324}{432}=\frac 34=\frac{486}{648}=\frac{648}{864}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
"answer_id": 0
} |
If $f(x)=2|2x-5|+3$, with $x\geq0$, find all $k$ such that $f(x) = kx + 2$ has exactly two roots
If $f(x) = 2|2x-5| + 3 $, $x\geq0$, find the values of $k$ such that the equation $f(x) = kx + 2$ has exactly two roots.
My attempt :-
$2(2x -5)+3 = kx + 2$ ,
$x \geq 2.5$
$x =\frac{9}{4-k}$
Then
$\frac{9}{4-k} \geq \frac{2}{5}$ , $k<4$
Then
$\frac{2}{5} < k < 4$
$2(-2x +5) + 3 = kx +2$ , $ x < \frac{2}{5}$ , $x\geq 0$
$x = \frac{11}{k+4}$
Then
$ \frac{11}{k+4} < \frac{5}{2}$
Then
$k > \frac{2}{5}$
Then
$\frac{2}{5} < k < 4$
Is there any easier procedure to find the values of $k$ ?
| The set of lines $y = kx + 2$ are the non-vertical lines passing through the point $(0,2)$, while the graph of $y = 2|2x - 5| + 3 = 4|x - {5 \over 2}| + 3$ consists of two rays containing $({5 \over 2}, 3)$, one going upwards with slope $4$ and the other going upwards with slope $-4$.
In order for a line containing $(0,2)$ to intersect the graph of $y = 4|x - {5 \over 2}| + 3$ twice, it must (separately) intersect both of the above rays. Thus the minimal value of $k$ corresponding to this situation is when the line $y = kx + 2$ intersects the vertex $({5 \over 2}, 3)$, and the maximal value of $k$ is when the slope $k$ of the line is $4$, the slope of the right-hand ray. In both of these endpoint situations the the line will intersect the graph only once and thus shouldn't be included.
To find the minimum $k$, we simply plug $({5 \over 2}, 3)$ into $y = kx + 2$, obtaining $3 = {5 \over 2}k + 2$ or $k = {2 \over 5}$.
Thus the set of $k$ in question is given by ${2 \over 5} < k < 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\frac{dy}{dx} - {8} -{2}x^2+{4}y^2+y^2x^2 = 0.$ how should I procced from here having the equation $$\frac{dy}{dx} - {8} -{2}x^2+{4}y^2+y^2x^2 = 0.$$
I am getting to the following $$\frac{1}{2^{\frac{3}{2}}}\ln \left(y+\sqrt{2}\right)-\frac{1}{2^{\frac{3}{2}}}\ln \left(y-\sqrt{2}\right)=\frac{x^3}{3}+4x+c$$ from here I can do $\frac{1}{2^{\frac{3}{2}}}(ln\frac{y+\sqrt(2)}{y-\sqrt(2)})=\frac{x^3}{3}+4x+c$
how should I get none implicit function of $y$?
| $$\frac{dy}{dx} = 8+2x^2-4y^2-y^2x^2 = (4+x^2)(2-y^2)$$
$$\frac{dy}{2-y^2}=(4+x^2)dx \iff 2^{-3/2}\ln \Big(\Big \vert\frac{y+\sqrt{2}}{y-\sqrt{2}}\Big \vert\Big) = 4x+\frac{1}{3}x^3+C$$
And this is where you got to. Now we can multiply by $2^{3/2}$ and take $\exp$ of both sides to get
$$\Big \vert \frac{y+\sqrt{2}}{y-\sqrt{2}}\Big \vert= \exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)$$
(considering the positive case of the absolute value)
$$y+\sqrt{2} = y\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big) - \sqrt{2}\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)$$
$$y\Big(1-\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)\Big) = -\sqrt{2}\Big( 1+\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)\Big)$$
$$y=-\sqrt{2} \frac{1+\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)}{1-\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)}$$
if $|y| > \sqrt{2}$, and (by the same method and considering the negative case)
$$y=-\sqrt{2} \frac{1-\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)}{1+\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)}$$
if $|y| < \sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Base conversion problems Let $b$ be an integer greater than 2, and let $N_b = 1_b + 2_b + \cdots + 100_b$ (the sum contains all valid base $b$ numbers up to $100_b$). Compute the number of values of $b$ for which the sum of the squares of the base $b$ digits of $N_b$ is at most 512.
Since we are adding up to $100$ in base $b$, I set up the equation
$$N=\frac{b^2(b^2+1)}{2},$$
and then tried to find the values that satisfy this.
For even values, of $b$, i got 16 values, but I don't know about odd values.
Now I am stuck.
| We have
$$
2N = b^2(b^2+1)=b^4+b^2 = 10100_b
$$
You need to show that
$$
10100_b =
\left\lbrace\begin{array}{c}
2 \cdot \left[ \tfrac{b}{2},0,\tfrac{b}{2},0 \right]_b
&\text{for even $b$}\\
2 \cdot \left[ \tfrac{b-1}{2},\tfrac{b+1}{2},0,0 \right]_b
&\text{for odd $b$}\\
\end{array}\right.
$$
where $[d_3,d_2,d_1,d_0]_b = d_3b^3 + d_2b^2 + d_1b^1 + d_0b^0$.
You can prove the above by addition:
$$
2 \cdot \left[ \tfrac{b}{2},0,\tfrac{b}{2},0 \right]_b
= \left[ \tfrac{b}{2},0,\tfrac{b}{2},0 \right]_b
+ \left[ \tfrac{b}{2},0,\tfrac{b}{2},0 \right]_b \\
2 \cdot \left[ \tfrac{b-1}{2},\tfrac{b+1}{2},0,0 \right]_b
= \left[ \tfrac{b-1}{2},\tfrac{b+1}{2},0,0 \right]_b
+ \left[ \tfrac{b-1}{2},\tfrac{b+1}{2},0,0 \right]_b
$$
The remaining part is easy - just count all even $b$s such that $\frac{b^2}{2} \leq 512$ and all odd $b$s such that $\frac{b^2 + 1}{2} \leq 512$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3722225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Distribution of $\frac{2X_1 - X_2-X_3}{\sqrt{(X_1+X_2+X_3)^2 +\frac{3}{2} (X_2-X_3)^2}}$ when $X_1,X_2,X_3\sim N(0,\sigma^2)$
Given that $X_1, X_2, X_3 $ are independent random variables form $N(0, \sigma^2 )$, I have to indicate that the statistic given below has a $t$ distribution or not.
\begin{equation}
\frac{2X_1 - X_2-X_3}{\sqrt{(X_1+X_2+X_3)^2 +\frac{3}{2} (X_2-X_3)^2}}
\end{equation}
In my attempt of solving this problem:
I start by showing that we can write the numerator as $a^TX$, where $a^T = (2 -1 -1)$ and $X^T= (X_1 X_2 X_3)$. Thus we have that $a^TX \sim N(0, a^T(\sigma^2 I)a)= N(0, 6\sigma^2)$. And so $\frac{1}{\sqrt{6\sigma^2}} a^TX \sim N(0,1)$ or $\frac{1}{\sqrt{6\sigma^2}}(2X_1-X_2-X_3)\sim N(0,1)$.
Next, we know that $(X_1+X_2+X_3) \sim N(0, 3\sigma^2)$. This implies that $\frac{1}{\sqrt{3\sigma^2}}(X_1+X_2+X_3) \sim N(0,1)$ and thus,$\frac{1}{{3\sigma^2}}(X_1+X_2+X_3)^2 \sim \chi^2(1)$. Similarly, $\frac{1}{2\sigma^2}(X_2-X_3)^2 \sim \chi^2(1)$. Therefore, $\frac{1}{{3\sigma^2}}(X_1+X_2+X_3)^2 + \frac{1}{2\sigma^2}(X_2-X_3)^2 \sim \chi^2(2)$ or $\frac{1}{{3\sigma^2}}\left((X_1+X_2+X_3)^2 + \frac{3}{2}(X_2-X_3)^2 \right) \sim \chi^2(2)$.
As a third step I have to show that $\frac{1}{\sqrt{6\sigma^2}}(2X_1-X_2-X_3)$ and $ \frac{1}{{3\sigma^2}}\left((X_1+X_2+X_3)^2 + \frac{3}{2}(X_2-X_3)^2 \right)$ are independent and I am not sure how to show that. Any help would be appreciated.
| Consider the orthogonal transformation
$$\begin{pmatrix}Y_1 \\ Y_2 \\ Y_3\end{pmatrix}=\begin{pmatrix}\frac{2}{\sqrt 6} &-\frac1{\sqrt 6} & -\frac1{\sqrt 6} \\ \frac1{\sqrt 3} & \frac1{\sqrt 3} & \frac1{\sqrt 3}\\ 0 & \frac1{\sqrt 2} & -\frac1{\sqrt 2} \end{pmatrix}\begin{pmatrix}X_1 \\ X_2 \\ X_3\end{pmatrix}$$
So if $Y=(Y_1,Y_2,Y_3)^T$ and $X=(X_1,X_2,X_3)^T$, then $X\sim N(0,\sigma^2 I_3)\implies Y\sim N(0,\sigma^2 I_3)$.
Therefore,
$$T=\frac{2X_1 - X_2-X_3}{\sqrt{(X_1+X_2+X_3)^2 +\frac{3}{2} (X_2-X_3)^2}}=\frac{\sqrt 6Y_1}{\sqrt{3Y_2^2+3Y_3^2}}$$
I think you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3722577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int_0^2 \frac{\ln\left(1+x\right)}{x^2-x+1} \mathop{dx}$ Challenge problem $$\int_0^2 \frac{\ln\left(1+x\right)}{x^2-x+1} \mathop{dx}$$
First thought $u=1+x$, $$ \int_1^3 \frac{\ln{(u)}}{u^2-3u+3} \mathop{du}$$
Here complex analysis or what?
Tips please.
| Via the substitution $u=\sqrt{3}v$ we
get
\begin{equation*}
I=\int_{1}^{3}\dfrac{\ln(u)}{u^2-3u+3}\,\mathrm{d}u =I_1+I_2
\end{equation*}
where
\begin{equation*}
I_1=\dfrac{1}{\sqrt{3}}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\dfrac{\ln(\sqrt{3})}{v^2-\sqrt{3}v+1}\,\mathrm{d}v =\left[\dfrac{\ln 3}{\sqrt{3}}\arctan\left(2v-\sqrt{3}\right)\right]_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}=\dfrac{\pi\ln 3}{2\sqrt{3}}
\end{equation*}
and
\begin{equation*}
I_2=\dfrac{1}{\sqrt{3}}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\dfrac{\ln v}{v^2-\sqrt{3}v+1}\,\mathrm{d}v =[v\mapsto 1/v] =-I_2.
\end{equation*}
Consequently $I_2=0$ and
\begin{equation*}
I=\dfrac{\pi\ln 3}{2\sqrt{3}}.
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3730856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Arrange irrationals in ascending order:$ 2^{\sqrt{\frac{5}{3}}},3^{\sqrt{\frac{3}{5}}},5^{\sqrt{\frac{4}{15}}},29^{\frac{1}{\sqrt{15}}} $ How to arrange following irrational numbers in ascending order:
$$ 2^{\sqrt{\frac{5}{3}}}, 3^{\sqrt{\frac{3}{5}}}, 5^{\sqrt{\frac{4}{15}}}, 29^{\frac{1}{\sqrt{15}}} $$
My try:
Using a calculator I computed the values
$$
2^{\sqrt{\frac{5}{3}}}\approx 2.445, \quad
3^{\sqrt{\frac{3}{5}}}\approx 2.342,\quad
5^{\sqrt{\frac{4}{15}}}\approx 2.296, \quad
29^{\frac{1}{\sqrt{15}}}\approx 2.385.
$$
By checking these values I easily arranged them in ascending order
$$ 5^{\sqrt{\frac{4}{15}}}<3^{\sqrt{\frac{3}{5}}}<
29^{\frac{1}{\sqrt{15}}}<2^{\sqrt{\frac{5}{3}}} $$
This is my final answer I found by calculator.
My question is: Can I arrange these irrational numbers without computing the values because I am not allowed to use calculator.
I have never seen the numbers with irrational powers.
| Equate the powers by taking LCM of denominators and compare the base-numbers as follows
$$ 2^{\sqrt{\dfrac{5}{3}}}, 3^{\sqrt{\dfrac{3}{5}}}, 5^{\sqrt{\dfrac{4}{15}}}, 29^{\dfrac{1}{\sqrt{15}}} $$
$$ 2^{\dfrac{5}{\sqrt{15}}}, 3^{\dfrac{3}{\sqrt{15}}}, 5^{\dfrac{2}{\sqrt{15}}}, 29^{\dfrac{1}{\sqrt{15}}} $$
$$ (2^5)^{\dfrac{1}{\sqrt{15}}}, (3^3)^{\dfrac{1}{\sqrt{15}}}, (5^2)^{\dfrac{1}{\sqrt{15}}}, 29^{\dfrac{1}{\sqrt{15}}} $$
$$(32)^{\dfrac{1}{\sqrt{15}}}, (27)^{\dfrac{1}{\sqrt{15}}}, (25)^{\dfrac{1}{\sqrt{15}}}, 29^{\dfrac{1}{\sqrt{15}}} $$
Now, arrange above base numbers in increasing order as follows $$25<27<29<32$$
$$(25)^{\frac{1}{\sqrt{15}}}<(27)^{\frac{1}{\sqrt{15}}}<(29)^{\frac{1}{\sqrt{15}}}<(32)^{\frac{1}{\sqrt{15}}}$$
$$ \therefore 5^{\sqrt{\dfrac{4}{15}}}<3^{\sqrt{\dfrac{3}{5}}}<
29^{\dfrac{1}{\sqrt{15}}}<2^{\sqrt{\dfrac{5}{3}}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Easier approach to $\int_0^{\infty} \frac{\mathrm{e}^{-x} \cosh(2x/5)}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x$? After playing with this integral for a little bit last night, I eventually resorted to complex analysis to solve it.
Can this be solved without complex analysis? It feels like there ought to be a way. If not, is there an easier way with complex analysis? (I'm still pretty beginner tier at this sort of thing.)
My solution is somewhat involved and it is as follows:
First, get rid of the cosh.
$$
\begin{split}
I &= \int_0^{\infty} \frac{\mathrm{e}^{-x} \cosh(2x/5)}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x \\
&= \frac1{2} \int_0^{\infty} \frac{\mathrm{e}^{-3/5 x} + \mathrm{e}^{-7/5 x}}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x \\
\frac{\mathrm{e}^{-7/5 x}}{1 + \mathrm{e}^{-2x}} &= \frac{\mathrm{e}^{3/5 x}}{1 + \mathrm{e}^{2x}} \\
2I &= \int_0^{\infty} \frac{\mathrm{e}^{-3/5 x}}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x + \int_0^{\infty} \frac{\mathrm{e}^{3/5 x}}{1 + \mathrm{e}^{2x}} \, \mathrm{d}x
\end{split}
$$
Next, do some u-subs to make it nicer.
$$\begin{split}
u = \mathrm{e}^{-x} & \qquad \mathrm{d}u = - \mathrm{e}^{-x} \, \mathrm{d}x \\
\int_0^{\infty} \frac{\mathrm{e}^{-3/5 x}}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x &= \int_{0}^{1} \frac{u^{-2/5}}{1 + u^2} \, \mathrm{d}u \\
\\
u = \mathrm{e}^{x} & \qquad \mathrm{d}u = \mathrm{e}^{x} \, \mathrm{d}x \\
\int_0^{\infty} \frac{\mathrm{e}^{3/5 x}}{1 + \mathrm{e}^{2x}} \, \mathrm{d}x &= \int_{1}^{\infty} \frac{u^{-2/5}}{1 + u^2} \, \mathrm{d}u \\
\\
2I &= \int_{0}^{\infty} \frac{u^{-2/5}}{1 + u^2} \, \mathrm{d}u \\
\end{split}
$$
Our contour is the counterclockwise semicircular arc of radius $R > 1$ in the upper half of the complex plane.
$$
\begin{split}
\oint_C \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \int_{-R}^0 \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z + \int_0^{R} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z + \int_0^{\pi} \frac{{\left(R \mathrm{e}^{i \phi}\right)}^{-2/5}}{1 + {\left(R \mathrm{e}^{i \phi}\right)}^2} \, iR\mathrm{e}^{i \phi} \, \mathrm{d}\phi \\
\lim_{R \rightarrow \infty} \int_0^{\pi} \frac{{\left(R \mathrm{e}^{i \phi}\right)}^{-2/5}}{1 + {\left(R \mathrm{e}^{i \phi}\right)}^2} \, iR\mathrm{e}^{i \phi} \, \mathrm{d}\phi &= 0 \\
\oint_{C, R \rightarrow \infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \int_{-\infty}^0 \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z + \int_0^{\infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z \\
\int_{-\infty}^0 \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= - \int_0^{\infty} \frac{(-z)^{-2/5}}{1 + (-z)^2} \, \mathrm{d}(-z) \\
\oint_{C, R \rightarrow \infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \left(1 + \mathrm{e}^{-2\pi i/5}\right) \int_0^{\infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z \\
\end{split}
$$
Finally, take the residue and solve for the original integral.
$$
\begin{split}
\oint_{C, R \rightarrow \infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= 2 \pi i \operatorname{Res}_{z = i} \left( \frac{z^{-2/5}}{1+z^2} \right) \\
&= 2 \pi i \left( \frac{i^{-2/5}}{2 i} \right) \\
&= \pi i^{-2/5} \\
\int_0^{\infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \pi \left(\frac{i^{-2/5}}{1 + \mathrm{e}^{-2 \pi i / 5}}\right) \\
&= \frac{\pi}{2} \left(\sqrt{5} - 1 \right) \\
2I &= \frac{\pi}{2} \left(\sqrt{5} - 1 \right) \\
I &= \frac{\pi}{4} \left(\sqrt{5} - 1 \right)
\end{split}
$$
| Starting from @Luis Sierra's answer
$$\begin{equation}
I=\frac{1}{2}\int\limits_{0}^{1} \frac{t^{\frac{2}{5}}}{1+t^{2}} \,dt +\frac{1}{2}\int\limits_{0}^{1} \frac{t^{-\frac{2}{5}}}{1+t^{2}}\,dt
\end{equation}$$
Using the quite standard
$$J_a=\int_0^1 \frac {t^a}{1+t^2}\, dt=\frac{1}{4} \left(\psi \left(\frac{a+3}{4}\right)-\psi
\left(\frac{a+1}{4}\right)\right)\qquad \text{if} \qquad \Re(a)>-1$$ So, rearranging,
$$8I=\Big[\psi\left(\frac{17}{20}\right)-\psi\left(\frac{3}{20}\right)\Big]+\Big[\psi \left(\frac{13}{20}\right)-\psi\left(\frac{7}{20}\right)\Big]=\pi \cot \left(\frac{3 \pi }{20}\right)+\pi \tan \left(\frac{3 \pi }{20}\right)$$ that is to say
$$8I=\pi\csc\left(\frac{3 \pi }{20}\right)\,\sec\left(\frac{3 \pi }{20}\right)=2 \left(\sqrt{5}-1\right)\, \pi \implies I=\frac{\sqrt{5}-1}{4} \pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Show that what is the graph of each one of these equations. Given the following equations:
$$1)\;\;x^{2}+2y^{2}+z^{2}-2x+4z-22=0$$
$$2)\;\;5x^{2}+6y^{2}+4z-4x=14$$
$$3)-x^{2}+y^{2}-z^{2}-2x+2z=0$$
$$4) x=z^2$$
Show that what is the graph of each one of these equations.
$$1)$$
$$x^{2}+2y^{2}+z^{2}-2x+4z-22=0$$
$$\frac{\left(x-1\right)^{2}}{27}+\frac{y^{2}}{\frac{27}{2}}+\frac{\left(z+2\right)^{2}}{27}=1$$
Which is an ellipsoid.
$$2)$$
$$5x^{2}+6y^{2}+4z-4x=14$$
$$5x^{2}+6y^{2}+4z-4x=14$$
$$\frac{x^{2}}{12}+\frac{y^{2}}{10}+\frac{z}{15}-\frac{x}{15}=\frac{14}{60}$$
$$\frac{5x^{2}-4x}{60}+\frac{y^{2}}{10}+\frac{z}{15}=\frac{14}{60}$$
$$\frac{\left(x-\frac{2}{5}\right)^{2}}{60}+\frac{y^{2}}{50}+\frac{z}{75}=\frac{\frac{14}{60}-\frac{4}{5\cdot60}}{5}$$
$$3)$$
$$-x^{2}+y^{2}-z^{2}-2x+2z=0$$
$$x^{2}-y^{2}+z^{2}+2x-2z=0$$
$$\left(x+1\right)^{2}-y^{2}+\left(z-1\right)^{2}=2$$
I don't know the last three cases,can someone help me?
|
Equation $1$. Ellipsoid.
The skirt is due to the real hack in the code to deal with complex values.
It shouldn't be there.
Equation $2$. Dome.
Equation $3$. Hyperboloid of one sheet. Note the y axis is vertical.
Octave:
figure 1;
tx = ty = [-5:0.1:5]';
[xx, yy] = meshgrid (tx, ty);
zplus = -2 + real(sqrt(27 - (xx-1).^2 -2*yy.^2)); # real is a hack
zminus = -2 - real(sqrt(27 - (xx-1).^2 -2*yy.^2)); # real is a hack
mesh (tx, ty, zplus);
hold on;
mesh (tx, ty, zminus);
xlabel ("x");
ylabel ("y");
zlabel ("z");
title("[1]: (x-1)^2/27 + 2y^2/27 + (z+2)^2/27 = 1");
figure 2;
tx = ty = [-5:0.1:5]';
[xx, yy] = meshgrid (tx, ty);
z = 75*( (14/60 - 4/(5*60))/5 - (xx-2/5).^2/60 - yy.^2/50);
mesh (tx, ty, z);
xlabel ("x");
ylabel ("y");
zlabel ("z");
title("[2]: (x-2/5)^2/60 + y^2/50 + z/75 = 11/250");
figure 3;
tx = tz = [-5:0.1:5]';
[xx, zz] = meshgrid (tx, tz);
yplus = real(sqrt((xx+1).^2 + (zz-1).^2 - 2)); # real is a hack
mesh (tx, tz, yplus);
hold on;
mesh (tx, tz, -yplus);
xlabel ("x");
ylabel ("z");
zlabel ("y");
title("[3]: (x+1)^2 - y^2 + (z-1)^2 = 2");
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What Is Bigger $100^{100}$or $\sqrt{99^{99} \cdot 101^{101}}$ Hello every what is bigger $100^{100}$or $\sqrt{99^{99} \cdot 101^{101}}$?
I tried to square up and I got $100^{200}$ or $99^{99} \cdot 101^{101}$ and I don't have an idea how to continue.
| It is “well-known” that $\log(u) \le u-1$ for all $u>0$, with strict inequality for $u \ne 1$ (see for example for each $x>1 , \frac{x-1}{x}\ < \ln x < x-1$).
It follows that for all real $x > 1$
$$
\log \left( \frac{x^{2x}}{(x-1)^{x-1} (x+1)^{x+1}}\right)
= (x-1) \log \left( \frac{x}{x-1}\right) + (x+1) \log \left( \frac{x}{x+1}\right) \\
< (x-1) \left( \frac{x}{x-1} - 1 \right) + (x+1) \left( \frac{x}{x+1} - 1\right)
= 1 - 1 = 0
$$
and therefore
$$
x^x < \sqrt{(x-1)^{x-1} (x+1)^{x+1}} \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Three boxes with balls Box $U_1$ contains $1$ white ball and $2$ black balls. Box $U_2$ contains $2$ white balls and $2$ black balls. We extract without reinsertion two balls from every boxes. The four balls are put in a third box $U_3$ initially empty. We randomly extract a ball from $U_3$. Find the probability that the ball is white.
Well, I reasoned in this way. The possible combinations that ensure that $U_3$ contains at least one white ball are BNBB, NBBB, NNBB, BNBN, BNNB, BNNN, NBBN, NBNB, NBNN, NNBN, NNNB. Thus:
*
*$\mathbb{P}$($U_3$ contains $3$ white balls)$=\mathbb{P}($(BNBB)$\cap$(NBBB)$)=(\frac{1}{3}\cdot1 \cdot\frac{1}{2}\cdot\frac{1}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2}\cdot \frac{1}{3})=0,11$
*$\mathbb{P}(U_3$ contains $2$ white balls)$=\mathbb{P}($(NNBB)$\cap$(BNBN)$\cap$(BNNB)$\cap$(NBBN)$\cap$(NBNB)$)=(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})+(\frac{1}{3}\cdot 1\cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{1}{3}\cdot 1 \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{3}\cdot \frac{1}{2} \cdot \frac{2}{3})=0,46$
*$\mathbb{P}(U_3$ contains $1$ white ball)$=\mathbb{P}($(BNNN)$\cap$(NBNN)$\cap$(NNBN)$\cap$(NNNB)$)=(\frac{1}{3}\cdot 1 \cdot \frac{1}{2}\cdot \frac{1}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})+(\frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{2}{3})=0,33$
*$\mathbb{P}(U_3$ doesn't contain any white balls)$=2\mathbb{P}($(NNNN)$)=2(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})=0,1$
Thus $\mathbb{P}($one white ball from $U_3)=\frac{3}{4}\cdot 0,11+\frac{2}{4}\cdot 0,46+\frac{1}{4}\cdot 0,33+\frac{0}{4}\cdot 0,11=0,395$
Is it correct? Particularly I'm interested in reasoning. Thanks in advance.
| In the end, it all drills down to selecting one ball. With probability half, it is a ball from $U_1$ and with probability half, it is a ball from $U_2$.
The probability of choosing a white ball from each box is known, so the total probability is $\tfrac{1}{2}\cdot\tfrac{1}{3}+\tfrac{1}{2}\cdot\tfrac{1}{2}=\tfrac{5}{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $P(x) \bmod (x^2-x-2)$
Let $P(x)$ be a polynomial.
If $P(x) \bmod (x+1)=0$ and $P(x) \bmod (x-2)=6$, then evaluate $P(x) \bmod (x^2-x-2)$.
My attempt:
$$P(x)=k(x)(x-2)+6=k(x)(x+1)-3(k(x)-2)$$
$$\Longrightarrow (k(x))-2) \bmod (x+1)=0$$
$$\Longrightarrow \left(\dfrac{P(x)-6}{x-2}-2\right) \bmod (x+1)=0$$
$$\Longrightarrow (P(x)-2x-2) \bmod ((x+1)(x-2)) =0$$
$$\Longrightarrow P(x) \bmod (x^2-x-2) \equiv (2x+2) \bmod ((x+1)(x-2)) = (2x+2) \bmod (x^2-x-2)=2x+2 $$
Is my solution correct?
| Your solution is correct, but here is a different way to approach it. Though not stated, I'll assume we are working in $\mathbb{Q}[x]$. Notice that $(x^2 - x - 2) = (x-2)(x+1)$, i.e. it is the product of the moduli in your first two congruence equations. A quick check shows that indeed these moduli, $x+1$ and $x-2$, are coprime. This should immediately make you think of Chinese Remainder Theorem.
Let $p(x) = x+1$ and $q(x) = x-2$. We need to find the "inverses" of these, i.e. $p^{-1}(x)$ and $q^{-1}(x)$ such that:
$$
p(x)p^{-1}(x) \equiv 1 \mod q(x), \; \; \; \; \; q(x)q^{-1}(x) \equiv 1 \mod p(x).
$$
With this, the theorem gives the solution to be:
$$
P(x) \equiv a\:q(x)q^{-1}(x) + b\:p(x)p^{-1}(x) \mod (x^2-x-2).
$$
In this case, $a=0$ and $b=6$, so we only need to find $p^{-1}(x)$. A quick computation reveals:
$$
p^{-1}(x) = x-\frac{5}{3}
$$
as:
$$
(x+1)\Big(x-\frac{5}{3}\Big) - 1= x^2 - \frac{2}{3}x - \frac{8}{3} = (x-2)\Big(x+\frac{4}{3}\Big),
$$
i.e. $p\: p^{-1} \equiv 1 \mod q$. So this gives the solution,
$$
P(x) = 6(x+1)\Big(x-\frac{5}{3}\Big) = (x+1)(6x-10) = 6x^2 -4x - 10 = 6(x^2 - x - 2) + 2x+2,
$$
which is to say:
$$
P(x) \equiv 2x+2 \mod (x^2 - x -2).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove $(n + \frac{1}{2})\log(1+\frac{1}{n})$ is increasing in $n$? Is $(n + \frac{1}{2})\log(1+\frac{1}{n})$ increasing in $n$?
I attempted to differentiate $p_n=(n + \frac{1}{2})\log(1+\frac{1}{n})$ with respect to $n$.
$p_n^{'} = \log(1+\frac{1}{n}) + \frac{(n+\frac{1}{2})}{(1+\frac{1}{n})}(-\frac{1}{n^2}) = \log(1+\frac{1}{n}) - \frac{1}{2}(\frac{1}{n}+\frac{1}{n+1})$.
I am stuck at this point and I cannot prove that $\log(1+\frac{1}{n}) - \frac{1}{2}(\frac{1}{n}+\frac{1}{n+1}) > 0$.
I know by the definition $\log (1+\frac{1}{n}) =\int_1^{1+\frac{1}{n}}\frac{1}{x}dx$.
But I can only reach to the point where $ \frac{1}{n+1} < \log (1+\frac{1}{n}) < \frac{1}{n}$.
Thanks for your help in advance.
| As $n\to\infty$, $\log\left(1 + \dfrac1n\right)\to0$. Similarly, as $n\to\infty$, $\dfrac1n\to0$ and $\dfrac1{n+1}\to0$ and, therefore, $\dfrac12\left(\dfrac1n + \dfrac1{n + 1}\right)\to0$. So, $p_n^{'}\to0$ as $n\to\infty$ and, therefore, $p_n$ is not increasing in $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solving $\int_{-1}^{1}8x^3-5x^2+4dx$ I've done the following so far:
$$\left.\int_{-1}^{1}\left(8x^3-5x^2+4\right)dx=\left(\frac{8}{4}x^4-\frac{5}{3}x^3+4x\vphantom{\int}\right)\right|_{-1}^{1}$$
$$=\left(\frac{8}{4}-\frac{5}{3}+4\right)=\frac{13}{3}$$
However, I double-checked on wolfram alpha and the solution is actually $-\dfrac{14}{3}$. Would you know where I went wrong? I have no idea where the negative came from, in the solution, or how it's one value above mine.
| The value of last limit must be $4-(-4)=8$
$$\int_{-1}^1(8x^3-5x^2+4)dx=\left(4x^4-\frac53x^3+4x\right)_{-1}^1$$
$$=\left(4(1)^4-\frac53(1)^3+4(1)-4(-1)^4+\frac53(-1)^3-4(-1)\right) $$
$$=\frac{14}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How prove this $\sum_{i=n+2}^{+\infty}\frac{1}{i^2}>\frac{2n+5}{2(n+2)^2}$ let $n$ be postive integer,show that
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}>\dfrac{2n+5}{2(n+2)^2}\tag{1}$$
I know
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}>\int_{n+2}^{+\infty}\dfrac{1}{x^2}dx=\dfrac{1}{n+2}$$
But $$\dfrac{1}{n+2}-\dfrac{2n+5}{2(n+2)^2}=-\dfrac{1}{2(n+2)^2}<0$$
then this integral method can't solve (1),so How to prove it?Thanks
| At least, for "large" values on $n$, we can show it since
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}=\psi ^{(1)}(n+2)$$
So, using the asymptotics of the digamma function and Taylor expansions, we have
$$\sum_{i=n+2}^{+\infty}\frac{1}{i^2}-\frac{2n+5}{2(n+2)^2}=\frac{1}{6 n^3}+O\left(\frac{1}{n^4}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Derangement related problem-bijective functions $f: A \to A$ such that $f(x) \neq x$ and $f(1) \neq 2$ Let $R$ denote the set of all nonempty relations on the set $A$, where $A = \{1, 2, 3, 4, 5\}$. A function $f(x)$ is chosen at random from set $R$. The probability that $f(x)$ is bijective, $f(x)\ne x$ such that $x\in A$ and $f(1)\ne 2$ is_____
My approach is as follow total number of relations is $5 \cdot 5=25$.
Hence, number of nonempty relations is $2^{25}-1$.
Given the formula of derangement where $f(x)\ne x$ we get $5! \cdot (1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!})=44$, I am not able to insert condition $f(1)\ne 2$ because those case needs to be removed where $f(x)=2$.
| Let us find the number of derangements $\left\{ 1,2,3,4,5\right\} \to\left\{ 1,2,3,4,5\right\} $
that satisfy $f\left(1\right)=2$.
This equals the number of bijections $\left\{ 2,3,4,5\right\} \to\left\{ 1,3,4,5\right\} $
that satisfy $f\left(k\right)\neq k$ for $k\in\left\{ 3,4,5\right\} $.
Let $A_{k}$ denote the set of bijections with $f\left(k\right)=k$.
Then applying the principle of inclusion/exclusion and also symmetry we find: $$\left|A_{3}^{\complement}\cap A_{4}^{\complement}\cap A_{5}^{\complement}\right|=4!-\left|A_{3}\cup A_{4}\cup A_{5}\right|=4!-3\left|A_{3}\right|+3\left|A_{3}\cap A_{4}\right|-\left|A_{3}\cap A_{4}\cap A_{5}\right|=$$$$24-3\times3!+3\times2!-1!=11$$
So if there are indeed $44$ derangements in total then $44-11=33$ of them will satisfy $f(1)\neq2$.
addendum
After a second look I realized that things can be solved a lot easyer (without using PIE).
The set of derangements $\left\{ 1,2,3,4,5\right\} \to\left\{ 1,2,3,4,5\right\} $ can be split up in $4$ disjoint subsets: $D_2,D_3,D_4,D_5$. Here $D_i$ denotes the set of derangements that satisfy $f(1)=i$. By symmetry it is clear that the sets have equal cardinality, so if the sum of these cardinalities equals $44$ then the cardinality of $D_3\cup D_4\cup D_5$ is $33$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
how can I integrate $\int \dfrac{2x+3}{x^3-9x}dx$? How can I integrate this $$\int \dfrac{2x+3}{x^3-9x}dx?$$
My work
$$\dfrac{2x+3}{x^3-9x}=\dfrac{2x+3}{x(x+3)(x-3)}$$
$$=\dfrac{A}{x-3}+\dfrac{B}{x}=\dfrac{C}{x+3}$$
after solving, I got $A=1/2$, $B=-1/3$, $C=-1/6$
partial fractions decompositions:
$$\dfrac{2x+3}{x^3-9x}=\dfrac{1/2}{x-3}+\dfrac{-1/3}{x}=\dfrac{-1/6}{x+3}$$
$$\int \left(\dfrac{1}{2(x-3)}- \dfrac{1}{3x}- \dfrac{1}{6(x+3)}\right)\ dx$$
$$=\dfrac{1}{2}\ln (x-3)- \dfrac{1}{3}\ln (x)- \dfrac{1}{6}\ln (x+3)+C$$
I am not sure if my answer is correct. my question is can I use substitution to solve above integration? if yes please help me solve it by substitution. thanks
| hint for substitution
Write the integral as
$$\int \frac{2dx}{x^2-9}+\int \frac{3xdx}{x^2(x^2-9)}$$
for the first, you can put $$x=3\cosh(t)$$
and for the second, $$x^2=u$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3744234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to show that $J_{n+1} = \frac{3n-1}{3n} J_n$? Let $$J_n := \int_{0}^{\infty} \frac{1}{(x^3 + 1)^n} \, {\rm d} x$$
where $n > 2$ is integer. How to show that $J_{n+1} = \frac{3n-1}{3n} J_n$?
| Hint:
Let $y=\dfrac x{(x^3+1)^m}$
$$\dfrac{dy}{dx} =\dfrac1{(x^3+1)^m}+\dfrac{(-m)x(3x^2)}{(x^3+1)^{m+1}} =\cdots =\dfrac{3m}{(x^3+1)^{m+1}}-\dfrac{3m-1}{(x^3+1)^m}$$
Integrate both sides wrt $$\dfrac x{(x^3+1)^m}=3mI_{m+1}-(3m-1)I_m$$ where $$I_n=\int\dfrac{dx}{(x^3+1)^n}$$
Now $\dfrac x{(x^3+1)^m}\big|_0^\infty=0-0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove The following inequality $(ax+by)^2 \le ax^2+by^2$ for $a+b=1$ Prove The following inequality $(ax+by)^2 \le ax^2+by^2$ for $a+b=1, 0 \le a,b \le 1$
I tried expanding the equation and substituting $b=1-a$
\begin{equation}
(ax+by)^2=a^2x^2+2abxy+b^2y^2=a^2x^2+2axy-2a^2xy+b^2y^2
\end{equation}
The middle member $2axy-2a^2xy$ is negative only for $a>1$, so I'm not sure how to proceed from here.
In the hints I was given it was also said that it can be done by proving that the quadratic form $q(x,y)=ax^2+bx^2-(ax+by)^2$ is always positive. I tried finding the eigenvalues but ended up with huge equation I couldn't make sense of.
| WLOG $a=\sin^2t,b=\cos^2t$
$$ax^2+by^2-(ax+by)^2=\sin^2t\cos^2t(x^2+y^2-2xy)=?$$
Alternatively, $$a(1-a)+b(1-b)=a+b-((a+b)^2-2ab)=\cdots=2ab$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Verifying the period of $f(x)=\sin(x)+\cos(x/2)$ It seems clear from the graph of $f(x)=\sin(x)+\cos(x/2)$ that the period $p$ of the function is equal to $4\pi$.
To verify that $4\pi$ is a period of $f(x)$, note that
\begin{align}
\sin(x + 4\pi) + \cos\left(\frac{x + 4\pi}{2}\right) & =\sin(x)\cos(4\pi)+\cos(x)\sin(4\pi)+\cos(x/2)\cos(4\pi/2)-\sin(x/2)\sin(4\pi/2) \\
& =\sin(x)+\cos(x/2)
\end{align}
Thus $4\pi$ is indeed a period of $f$. My question is, how would one go about trying to prove that $4\pi$ is the smallest $p>0$ such that $f(x+p)=f(x)$?
| Let $$\sin(x + T) + \cos(\frac{x+T}{2}) = \sin(x) + \cos(\frac{x}{2})$$And $T\gt 0$. Then we have $$\sin(x+T) - \sin(x) = \cos(\frac{x}{2}) - \cos(\frac{x+T}{2}) \implies$$
$$2\sin(\frac{T}{2})\cos(\frac{2x+T}{2}) = -2\sin(\frac{2x + T}{4})\sin(\frac{-T}{4}) $$ So then $$\sin(\frac{T}{4}) = 0$$ Or $$2\cos(\frac{T}{4})\cos(\frac{2x+T}{2}) = \sin(\frac{2x + T}{4}) \tag{1}$$ For all $x\in \mathbb{R}$. It can be shown that it's not possible $(1)$ holds for all $x\in \mathbb{R}$. So we have $$T = 4k\pi$$ It implies that the fundamental period is $T = 4\pi$.
One way for proving the mentioned statement is using differentiation. For all $x\in \mathbb{R}$ $$2\cos(\frac{T}{4})\cos(\frac{2x+T}{2}) = \sin(\frac{2x + T}{4}) \implies$$ $$-2\cos(\frac{T}{4})\sin(\frac{2x+T}{2}) = \frac{1}{2}\cos(\frac{2x + T}{4}) \implies$$ $$-2\cos(\frac{T}{4})\cos(\frac{2x+T}{2}) = \frac{-1}{4}\sin(\frac{2x + T}{4}) \implies$$ $$\sin(\frac{2x + T}{4}) = \frac{1}{4}\sin(\frac{2x + T}{4}) \implies$$ $$\sin(\frac{2x + T}{4}) = 0 \tag{2}$$ No matter what's the value of $T$, it's not possible $(2)$ holds for all $x\in \mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Proof that any number is equal to $1$ Before I embark on this bizzare proof, I will quickly evaluate the following infinite square root; this will aid us in future calculations and working:
Consider $$x=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}$$
$$x^2-2=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}=x \implies x^2-x-2=0\implies x=2$$
as $x>0$. Now for the proof:
I was attempting some different infinite expansions/square roots when trying to solve another question of mine (Evaluate $\sqrt{x+\sqrt{{x^2}+\sqrt{{x^3}+\sqrt{{x^4}...}}}}$ ) and I came across this:
$$x+\frac{1}{x}=\sqrt{(x+\frac{1}{x})^2}=\sqrt{2+x^2+\frac{1}{x^2}}=\sqrt{2+\sqrt{(x^2+\frac{1}{x^2}}})^2=\sqrt{2+\sqrt{2+x^4+\frac{1}{x^4}}}=\sqrt{2+\sqrt{2+\sqrt{(x^4+\frac{1}{x^4})^2}}}=\sqrt{2+\sqrt{2+\sqrt{2+x^8+\frac{1}{x^8}}}}=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}=2$$
if you keep on applying this and using the result found at the start of the question. So we have that for any real number $x$ that
$$x+\frac{1}{x}=2\implies x^2-2x+1=0\implies (x-1)^2=0$$
so we finally have:
$$x=1$$
Where have I gone wrong, for surely this cannot be correct?
| Although they look similar at first glance, there is no reason for the two sequences
$$ \sqrt{2+2},\ \sqrt{2+\sqrt{2+2}},\ \sqrt{2+\sqrt{2+\sqrt{2+2}}}, \ldots $$
and
$$\sqrt{2+x^2+1/x^2},\ \sqrt{2+\sqrt{2+x^4+1/x^4}},\ \sqrt{2+\sqrt{2+\sqrt{2+x^8+1/x^8}}}, \ldots $$
to have the same limit unless $x=1$.
| {
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"url": "https://math.stackexchange.com/questions/3750275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Floor function equation $⌊x + 1/2⌋ + ⌊x⌋ = \frac12 x^6$ So in this floor equation $⌊x + 1/2⌋ + ⌊x⌋ = \frac12 x^6$, I've tried putting $x = n + e$, where $0 \le e < 1$, but I didn't get anything useful. What should be an approach in these situations?
| Suppose $x = n + e$ and consider two cases $e < 0.5$ and $e \geq 0.5$.
*
*$0\leq e < 0.5$
$\begin{alignat*}{2}
&2\cdot⌊n + e + 0.5⌋ + 2\cdot⌊n + e⌋ = (n + e)^6\\
&2\cdot n + 2\cdot n = (n + e)^6\\
&4\cdot n = (n + e)^6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\
&2\cdot \sqrt{n} = (n + e)^3\\
&(\sqrt{n} + 1)^2 - (n + 1) = (n + e)^6\\
&-(n + 1) = (n + e)^3 - (\sqrt{n} + 1)^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\end{alignat*}$
Notice, that LHS is negative, while RHS positive for $n \in \mathbb{Z}_{>1}$. Hence, we need to consider only $n \in \{0,1 \}$. Considering $n = 0$ and using $(2)$ yields $x = 0$. While $n = 1$ yields $e = 4^{1/6} - 1 < 0.5$, hence $x = 1 + 4^{1/6} - 1 = 4^{1/6}$.
*
*$0.5\leq e < 1$
The second case can be worked out similarly, but there are no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $\int \sqrt{1-\dfrac{1}{25x^2}}\ dx$? How to integrate the following:
$$\int \sqrt{1-\dfrac{1}{25x^2}}\ dx$$
What I did is:
$$\int \sqrt{1-\dfrac{1}{25x^2}}\ dx=\int \dfrac{\sqrt{25x^2-1}}{5x}\ dx$$
I substituted $5x=\sec\theta$, $dx=\dfrac{1}{5}\sec\theta\tan\theta\ d\theta $
$$=\int \dfrac{\sqrt{\sec^2\theta-1}}{\sec\theta}\ \dfrac{1}{5}\sec\theta\tan\theta\ d\theta$$
$$=\frac15\int \tan^2\theta\ d\theta$$
used $\tan^2\theta=\sec^2\theta-1$
$$=\frac15\int( \sec^2\theta-1)\ d\theta$$
$$=\dfrac15\tan\theta-\frac15\theta+c$$
back to $x$
$$=\dfrac15\sqrt{25x^2-1}-\frac15\sec^{-1}(5x)+c$$
I am not sure whether my answer is correct.
My question: Can I integrate this with other substitutions? If yes, please help me. Thank you
| you can also do it with $t^2=25x^2-1$. To get
\begin{align*}
\int \frac{\sqrt{25x^2-1}}{5x} \, dx&=\frac{1}{5}\int \frac{t^2}{t^2+1} dt\\
&=\frac{1}{5}\left[\int \frac{t^2+1-1}{t^2+1} dt\right]\\
&=\frac{1}{5}\left[t-\int \frac{1}{t^2+1} dt\right]\\
&=\frac{t}{5}-\frac{\arctan t}{5}+c,
\end{align*}
where $t=\sqrt{25x^2-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Evaluating the integral $\int_{f}^{g} \exp \left[-\frac{1}{d}\left(c+b x+a x^{2}\right)\right] d x$ I am trying to evaluate the following definite integral and find an analytic solution:
\begin{equation}
\int_{f}^{g} \exp \left[-\frac{1}{d}\left(c+b x+a x^{2}\right)\right] d x
\end{equation}
I have tried looking here, but this doesn't really help. Any advice would be appreciated on finding the exact form.
| assuming that ad>0
apply linearity
$$\int \ e^{- \frac{ax^2+bx+c}{d} }\, dx =e^{-c/d}\int \ e^{-\frac {ax^2+bx}{d}}dx$$
now solve
$$\int \ e^{-\frac {ax^2+bx}{d}}dx=\int \ e^{-\frac {ax^2}{d}-\frac{bx}{d}}dx$$
complete the square
$$\int\ e^{-\frac {b^2} {4ad}- (\frac{ x\sqrt{a} }{ \sqrt{d} } + \frac{b}{2 \sqrt{ad} } )^2}dx$$
Substitute $$u= \frac{2ax+b}{2 \sqrt{ad} } \longrightarrow \frac{du}{dx} = \frac{ \sqrt{a}}{ \sqrt{d} } $$
$$\Rightarrow = \frac{ \sqrt{d\pi}e^{ \frac{b^2}{4ad} } }{2 \sqrt{a} } \int \frac{2e^{-u^2}}{ \sqrt{\pi} } du$$
This is the Gauss Error Function $=erf(u)$
$$\therefore \frac{ \sqrt{d\pi}e^{ \frac{b^2}{4ad} } }{2 \sqrt{a} } \int \frac{2e^{-u^2}}{ \sqrt{\pi} } du=\frac{ \sqrt{d\pi}e^{ \frac{b^2}{4ad} } erf(u)}{2 \sqrt{a} } $$
undo substitution
$$=\frac{ \sqrt{d\pi}e^{ \frac{b^2}{4ad} } erf(\frac{2ax+b}{2 \sqrt{ad}})}{2 \sqrt{a} } $$
Thus the integral is solved
$$\Rightarrow \int_f^g e^{- \frac{ax^2+bx+c}{d} }\, dx =\frac{ \sqrt{d\pi}e^{ \frac{b^2}{4ad} } erf(\frac{2ag+b}{2 \sqrt{ad}})}{2 \sqrt{a} }-\frac{ \sqrt{d\pi}e^{ \frac{b^2}{4ad} } erf(\frac{2af+b}{2 \sqrt{ad}})}{2 \sqrt{a} }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $n$-th term formula of Fibonacci sequence using generating function I am trying to get the formula $F_n = \frac{\phi^n - \psi^n}{\phi - \psi}$ using generating functions. I managed to find that $G_F(x) = \frac{1}{1 - x - x^2}$ then I used partial fraction decomposition to find that $$G_F(x) = \frac{1}{\phi - \psi} \Biggl(\frac{1}{x - \psi} - \frac{1}{x - \phi}\Biggr)$$
After that I took the following steps to simplify:
$$G_F(x) = \frac{1}{\phi - \psi} \Biggl(\frac{\frac{1}{\psi}}{\frac{x}{\psi} - 1} - \frac{\frac{1}{\phi}}{\frac{x}{\phi} - 1}\Biggr)$$
$$ = \frac{1}{\phi - \psi} \Biggl(\frac{\psi}{\frac{x}{\phi} - 1} - \frac{\phi}{\frac{x}{\psi} - 1}\Biggr), since\ \psi = -\frac{1}{\phi}$$
$$ = \frac{1}{\phi - \psi} \Biggl(\frac{\psi}{-\psi x - 1} - \frac{\phi}{-\phi x - 1}\Biggr)$$
$$ = \frac{1}{\phi - \psi} \Biggl(\frac{\phi}{\phi x + 1} - \frac{\psi}{\psi x + 1}\Biggr) $$
The issue is that this function generates the series
$$a_n = \frac{\phi \cdot (-\phi)^n - \psi \cdot (-\psi)^n}{\phi - \psi}$$
Now, the $n + 1$ as the exponent is probably due to the fact that I started my series with $1$ instead of $0$.But I don't understand why is my series so close yet false.
| Another way is to use exponential generating functions. Start with $F_{n + 2} = F_{n + 1} + F_n$, $F_0 = 0, F_1 = 1$. Define:
$\begin{align*}
\widehat{F}(z)
&= \sum_{n \ge 0} F_n \frac{z^n}{n!}
\end{align*}$
Now you see that:
$\begin{align*}
\frac{d}{d z} \widehat{F}(z)
&= \sum_{n \ge 0} F_{n + 1} \frac{z^n}{n!}
\end{align*}$
Take the recurrence, multiply by $z^n / n!$, sum over $n \ge 0$ and recognize the resulting sums:
$\begin{align*}
\sum_{n \ge 0} F_{n + 2} \frac{z^n}{n!}
&= \sum_{n \ge 0} F_{n + 1} \frac{z^n}{n!}
+ \sum_{n \ge 0} F_n \frac{z^n}{n!} \\
\frac{d^2}{d z^2} \widehat{F}(z)
&= \frac{d}{d z} \widehat{F}(z) + \widehat{F}(z)
\end{align*}$
As initial values you know:
$\begin{align*}
\widehat{F}(0)
&= F_0 = 0 \\
\widehat{F}'(0)
&= F_1 = 1
\end{align*}$
The traditional ODE dance tells you:
$\begin{align*}
\widehat{F}(z)
&= c_1 \exp(\phi z) + c_2 \exp(\psi z)
\end{align*}$
Using the initial conditions gets you:
$\begin{align*}
F_0
&= 0
= c_1 + c_2 \\
F_1
&= 1
= c_1 \phi + c_2 \psi
\end{align*}$
From the first equation we get $c_2 = - c_1$, we also know $\psi = 1 - \phi$:
$\begin{align*}
1
&= c_1 \phi - c_1 (1 - \phi) \\
c_1
&= \frac{1}{2 \phi - 1} \\
&= \frac{1}{\phi - \psi} \\
c_2
&= - \frac{1}{2 \phi - 1} \\
&= \frac{1}{\psi - \phi}
\end{align*}$
Extracting coefficients then gives:
$\begin{align*}
F_n
&= \frac{\phi^n - \psi^n}{\phi - \psi}
\end{align*}$
(Not that world-shattering here, but a useful trick if your recurrence has factors $n$ thrown in).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{x\to 0} \frac{\sqrt{ax+b}-1}{x}=1$ My answer
Let $\sqrt{ax+b}=y$
Then
$$\lim_{y\to \sqrt b} \frac{(y-1)a}{y^2-b}$$
Let $b=1$
Then $$\lim _{y\to 1} \frac{a}{\frac{y^2-1}{y-1}}$$
$$=\frac a2 =1$$
$$a=2$$
The answer is correct, but this relies on assuming $b=1$, and that doesn’t seem appropriate. What is the correct answer for this?
| $$\frac{\sqrt{ax+b}-1}{x}=\frac{ax+b-1}{x(\sqrt{ax+b}+1)}.$$
Now, we see that we need $b=1$ (otherwise, the limit does not exist) and $\frac{a}{\sqrt{b}+1}=1,$
which gives also $a=2$.
If $b\neq1$ for $x\rightarrow0$ we obtain:
$$\frac{\sqrt{ax+b}-1}{x}=\frac{ax+b-1}{x(\sqrt{ax+b}+1)}=\frac{a}{\sqrt{ax+b}+1}+\frac{b-1}{x(\sqrt{ax+b}+1)}.$$
We see that for $b>0$ $$\frac{a}{\sqrt{ax+b}+1}\rightarrow\frac{a}{\sqrt{b}+1},$$ but
$$
\lim_{x\rightarrow0}\frac{b-1}{x(\sqrt{ax+b}+1)}$$ does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving $\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}$ Can we find the solutions for this equation?
$$\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}, \quad x \in \mathbb{R}$$
I tried to amplify the second square root in the $LHS$ with the conjugate and then use AM-GM in order to find where $x$ can be.
Also, the existence conditions imply $x \leq 2$. I obtained $x \leq \frac{4}{3}$.
| \begin{align*}
&\Rightarrow\sqrt{4-2x}\\
&=\sqrt{2+2(1-x)}&(x\in[0,2])\\
&=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\color{red}{(1-x)}}\\
&=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\color{red}{\sqrt{(1-x)^2}}}&(\color{red}{x\in[0,1]})\\
&=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\sqrt{1-(2x-x^2)}}\\
&=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\sqrt{1-\sqrt{2x-x^2}}\sqrt{1+\sqrt{2x-x^2}}}\\
&=\sqrt{\left(\sqrt{1-\sqrt{2x-x^2}}+\sqrt{1+\sqrt{2x-x^2}}\right)^2}\\
&=\sqrt{1-\sqrt{2x-x^2}}+\sqrt{1+\sqrt{2x-x^2}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Simplifying trigonometry function by substitution I want to simplify the following expression: $\frac{{\left( {1 + \sqrt 3 \tan {1^{\circ}}} \right)\left( {1 + \sqrt 3 \tan {2^{\circ}}} \right)\left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right)\left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right)}}{{\left( {1 + {{\tan }^2}{1^{\circ}}} \right)\left( {1 + {{\tan }^2}{2^{\circ}}} \right)}}$
My approach is as follow
$T = \left( {1 + \sqrt 3 \tan {1^{\circ}}} \right)\left( {1 + \sqrt 3 \tan {2^{\circ}}} \right)\left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right)\left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right){\cos ^2}{1^{\circ}}{\cos ^2}{2^{\circ}}$
$\Rightarrow \tan {58^{\circ}}\left( {1 + \sqrt 3 \tan {2^{\circ}}} \right) = \sqrt 3 - \tan {2^{\circ}}$
$\Rightarrow \tan {59^{\circ}}\left( {1 + \sqrt 3 \tan {2^0}} \right) = \sqrt 3 - \tan {1}$
$\tan {60^{\circ}} = \frac{{\left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right)}}{{\left( {1 + \tan {{59}^{\circ}}\tan {1^{\circ}}} \right)}} \Rightarrow \left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right) = \sqrt 3 \left( {1 + \tan {{59}^{\circ}}\tan {1^{\circ}}} \right)$
$\tan {60^{\circ}} = \frac{{\left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right)}}{{\left( {1 + \tan {{58}^{\circ}}\tan {2^{\circ}}} \right)}} \Rightarrow \left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right) = \sqrt 3 \left( {1 + \tan {{58}^{\circ}}\tan {2^{\circ}}} \right)$
$T = \frac{{\sqrt 3 - \tan {1^{\circ}}}}{{\tan {{59}^{\circ}}}} \times \frac{{\sqrt 3 - \tan {2^{\circ}}}}{{\tan {{58}^{\circ}}}} \times \sqrt 3 \left( {1 + \tan {{59}^{\circ}}\tan {1^{\circ}}} \right) \times \sqrt 3 \left( {1 + \tan {{58}^{\circ}}\tan {2^{\circ}}} \right){\cos ^2}{1^{\circ}}{\cos ^2}{2^{\circ}}$
After this step I am confused
| Generalization:
$$\dfrac{(1+\tan y\tan z)(\tan z+\tan(y-z))}{1+\tan^2z} =\cdots =\dfrac{\cos(y-z)\sin(z+y-z)\cos^2z}{\cos y\cos z\cos z\cos(y-z)} =\tan y$$
$$\text{ if }\cos(y-z)\cos z\ne0$$
Here $y=60^\circ$
Then $z=1^\circ, 2^\circ$
| {
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"url": "https://math.stackexchange.com/questions/3764542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove that $|a + b| = |a| + |b| \iff a\overline{b} \ge 0$ I'm reading a complex analysis book. In this book, the author establishes the following statement
If $a,b \in \mathbb{C}$, then $|a + b| = |a| + |b| \iff \left(a\overline{b}\in \mathbb{R}\right) \wedge \left(a\overline{b}\ge 0\right)$
This statement didn't seem intuitive to me, so I decided to try to prove it. I denoted the complex numbers $a$ and $b$ as $a = \alpha + i \beta$ and $b = \gamma + i \delta$. Using this, I get that $a\overline{b} = (\alpha \gamma + \beta\delta) + i(\beta \gamma - \alpha\delta)$, which tells us that
$$
\left(a\overline{b}\in \mathbb{R}\right) \wedge \left(a\overline{b}\ge 0\right) \iff (\alpha \gamma + \beta \delta \ge 0) \ \ \wedge \ \ (\alpha\delta= \beta \gamma )
$$
From here, I do the following
\begin{align}
&2(\alpha\delta)^2 = 2(\alpha\delta)^2 \iff 2(\alpha\delta)(\alpha\delta) = (\alpha\delta)^2 + (\alpha\delta)^2 \iff 2\alpha\delta\beta \gamma = (\alpha\delta)^2 + (\beta \gamma)^2 \notag \\
\iff& (\alpha\gamma)^2 + 2\alpha\gamma\beta \delta + (\beta \delta)^2 =(\alpha\gamma)^2 + (\alpha\delta)^2 + (\beta \gamma)^2 + (\beta \delta)^2 \iff (\alpha\gamma + \beta \delta)^2 = \left(\alpha^2 + \beta^2\right)\left(\gamma^2 + \delta^2\right) \notag \\
\iff& 2(\alpha\gamma + \beta \delta) = 2\sqrt{\left(\alpha^2 + \beta^2\right)\left(\gamma^2 + \delta^2\right)} \qquad \ \text{(here using the hypothesis that $\alpha \gamma + \beta \delta \ge 0$)} \notag \\
\iff & \alpha^2 + \beta^2 + \gamma^2 + \delta^2 + 2(\alpha\gamma + \beta \delta) = \alpha^2 + \beta^2 + \gamma^2 + \delta^2 +2\sqrt{\left(\alpha^2 + \beta^2\right)\left(\gamma^2 + \delta^2\right)}\notag\\
\iff& \left(\alpha^2 +2\alpha\gamma + \gamma^2 \right)+ \left(\beta^2 +2 \beta \delta+ \delta^2\right) = \left(\sqrt{\alpha^2 + \beta^2}\right)^2 +2\sqrt{\alpha^2 + \beta^2}\sqrt{\gamma^2 + \delta^2} + \left(\sqrt{\gamma^2 + \delta^2}\right)^2\notag\\
\iff& (\alpha + \gamma)^2 + (\beta + \delta)^2 = \left(\sqrt{\alpha^2 + \beta^2} +\sqrt{\gamma^2 + \delta^2}\right)^2 \iff |a+ b|^2 = \left(|a| + |b|\right)^2 \iff |a+ b| = |a| + |b|
\end{align}
where in the last equivalence I used the fact that $|z|\ge 0, \ \forall z \in \mathbb{C}$.
Is my proof correct? And also, does anyone know a different (possibly shorter) method of proving the above statement? Any and all help would be greatly appreciated. Thank you!
| First off: Intuitively the statement reads “The distances to the origin of two complex points $a$ and $b$ add up to the distance to the origin of $a + b$ if and only if they lie on the same ray from origin.” This is because $\overline b$ is $b$ reflected on real line, which can be interpreted as “$b$, just with its angle to the positive real ray inverted”.
Next, the absolute value and complex conjugation are related by $|z|^2 = z\overline z$. So maybe it’s easier to prove equivalence to the squared identity. As others already have hinted, $|a + b|^2 = |a|^2 + 2\operatorname{Re} a\overline b + |b|^2$. Hence
\begin{align*}
|a + b|^2 = (|a| + |b|)^2 &\iff 2\operatorname{Re} a\overline b = 2|a||b| \\
&\iff \operatorname{Re} a\overline b = |a\overline b|.
\end{align*}
So this reduces to proving for $z ∈ ℂ$, $\operatorname{Re} z = |z| \iff z ∈ [0..∞)$, which shouldn’t be hard to do.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that in a triangle $\sum\limits_{cyc}\frac{w_bw_c}{w_a}\geq\frac{3}{4}\left(\sum\limits_{cyc}\frac{a^2w_a}{w_bw_c}\right)\geq\sqrt{3}s$ Let $ ABC$ is a triangle, $ w_a, w_b, w_c$ are bisectors of angles, $ h_a, h_b, h_c$ are altitudes respectively, $ r$ is radius of the incircle, prove that:$$ \frac {w_bw_c}{w_a} + \frac {w_cw_a}{w_b} + \frac {w_aw_b}{w_c} \geq \frac {3}{4}\left(\frac {a^2w_a}{w_bw_c} + \frac {b^2w_b}{w_cw_a} + \frac {c^2w_c}{w_aw_b}\right) \geq \sqrt {3}s$$
I found the solution of this inequality
\begin{aligned}
& \frac {w_bw_c}{w_a} + \frac {w_cw_a}{w_b} + \frac {w_aw_b}{w_c} \geq \frac {3}{4}\left(\frac {a^2w_a}{w_bw_c} + \frac {b^2w_b}{w_cw_a} + \frac {c^2w_c}{w_aw_b}\right)\\
\iff & 4w_b^2w_c^2 + 4w_c^2w_a^2 + 4w_a^2w_b^2 - 3a^2w_a^2 - 3b^2w_b^2 - 3c^2w_c^2\geq 0\\
\iff & \sum x^3(y + z)(x - y)(x - z) + 11(x - y)^2(y - z)^2(z - x)^2 + 40\sum y^2z^2(x - y)(x - z))+ 4xyz\sum x(x - y)(x - z) + 9xyz\sum (y + z)(x - y)(x - z) \geq 0
\end{aligned}
In this solution how step $$\sum x^3(y + z)(x - y)(x - z) + 11(x - y)^2(y - z)^2(z - x)^2 + 40\sum y^2z^2(x - y)(x - z))+ 4xyz\sum x(x - y)(x - z) + 9xyz\sum (y + z)(x - y)(x - z) \geq 0$$comes from the step $$4w_b^2w_c^2 + 4w_c^2w_a^2 + 4w_a^2w_b^2 - 3a^2w_a^2 - 3b^2w_b^2 - 3c^2w_c^2\geq 0$$
and also how to prove the right inequality
| The right inequality.
In the standard notation we need to prove that:
$$\sum_{cyc}\frac{a^2\cdot\frac{2bc\cos\frac{\alpha}{2}}{b+c}}{\frac{2ac\cos\frac{\beta}{2}}{a+c}\cdot\frac{2ab\cos\frac{\gamma}{2}}{a+b}}\geq\frac{2(a+b+c)}{\sqrt3}$$ or
$$\sum_{cyc}\frac{a^2\cdot\frac{2bc\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2bc}}{2}}}{b+c}}{\frac{2ac\sqrt{\frac{1+\frac{a^2+c^2-b^2}{2ac}}{2}}}{a+c}\cdot\frac{2ab\sqrt{\frac{1+\frac{a^2+b^2-c^2}{2ab}}{2}}}{a+b}}\geq\frac{2(a+b+c)}{\sqrt3}$$ or
$$\sum_{cyc}\frac{\frac{a^2\sqrt{bc(a+b+c)(b+c-a)}}{b+c}}{\frac{\sqrt{ac(a+b+c)(a+c-b)}}{a+c}\cdot\frac{\sqrt{ab(a+b+c)(a+b-c)}}{a+b}}\geq\frac{2(a+b+c)}{\sqrt3}$$ or
$$\sum_{cyc}\frac{a(a+b)(a+c)}{b+c}\sqrt{\frac{b+c-a}{(a+b-c)(a+c-a)}}\geq\frac{2\sqrt{(a+b+c)^3}}{\sqrt3}$$ or$$\sum_{cyc}\frac{a(a+b)(a+c)(b+c-a)}{b+c}\geq2\sqrt{\frac{(a+b+c)^3\prod\limits_{cyc}(a+b-c)}{3}}.$$
Now, let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that:
$$\sum_{cyc}\frac{x(y+z)(2y+x+z)(2z+x+y)}{2x+y+z}\geq8\sqrt{\frac{(x+y+z)^3xyz}{3}}.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v>0$, and $xyz=w^3$.
Thus, we need to prove that:
$$\sum_{cyc}\frac{(3v^2-yz)(3u+y)^2(3u+z)^2}{\prod\limits_{cyc}(3u+x)}\geq24\sqrt{u^3w^3}$$ or
$$\frac{-w^6+171u^3w^3-9uv^2w^3+405u^4v^2-54u^2v^4}{w^3+54u^3+9uv^2}\geq8\sqrt{u^3w^3}$$ or $f(v^2)\geq0,$ where
$$f(v^2)=-w^6+171u^3w^3-9uv^2w^3+405u^4v^2-54u^2v^4-8(w^3+54u^3+9uv^2)\sqrt{u^3w^3}.$$
But since by Maclaurin $$u\geq v\geq w,$$ we obtain:
$$f'(v^2)=405u^4-108u^2v^2-9uw^3-72u^2\sqrt{uw^3}>0,$$ which says that $f$ increases.
Thus, it's enough to prove our inequality for a minimal value of $v^2$, which by $uvw$ happens for equality case of two variables.
Since our inequality is homogeneous, it's enough to assume that $y=z=1,$ which gives
$$\frac{2x(x+3)^2}{2x+2}+\frac{2(x+1)(x+3)(2x+2)}{x+3}\geq8\sqrt{\frac{(x+2)^3x}{3}}$$ or
$$x(x+3)^2+4(x+1)^3\geq8(x+1)\sqrt{\frac{(x+2)^3x}{3}},$$ which after squaring of the both sides gives
$$(x-1)^2(11x^4+50x^3+91x^2+88x+48)\geq0$$ and we are done!
| {
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"url": "https://math.stackexchange.com/questions/3766609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to evaluate the following limit: $\lim_{x\to 0}\frac{12^x-4^x}{9^x-3^x}$? How can I compute this limit
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}\text{?}$$
My solution is here:
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\dfrac{1-1}{1-1} = \dfrac{0}{0}$$
I used L'H$\hat{\mathrm{o}}$pital's rule:
\begin{align*}
\lim_{x\to 0}\dfrac{12^x\ln12-4^x\ln4}{9^x\ln9-3^x\ln3}&=\dfrac{\ln12-\ln4}{\ln9-\ln3}
\\ &=\dfrac{\ln(12/4)}{\ln(9/3)}
\\ &=\dfrac{\ln(3)}{\ln(3)}
\\ &=1
\end{align*}
My answer comes out to be $1$. Can I evaluate this limit without L'H$\hat{\mathrm{o}}$pital's rule? Thanks.
| If you add and subtract $1$ from numerator and denominator
$$
\lim_{x\to0}\frac{(12^x-1)-(4^x-1)}{(9^x-1)-(3^x-1)}
$$
then dividing each term by $x$
$$
\lim_{x\to0}\frac{\dfrac{12^x-1}{x}-\dfrac{4^x-1}{x}}{\dfrac{9^x-1}{x}-\dfrac{3^x-1}{x}}
$$
Now, all four limits have the form
$$
\lim_{x\to0}\frac{a^x-1}{x}=\log a
$$
so we get
$$
\lim_{x\to0}\frac{\dfrac{12^x-1}{x}-\dfrac{4^x-1}{x}}{\dfrac{9^x-1}{x}-\dfrac{3^x-1}{x}}=\frac{\log12-\log4}{\log9-\log3}=\frac{\log3}{\log3}=1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
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