Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that $\lim_{n\to\infty} (\sqrt{n^2+n}-n) = \frac{1}{2}$ Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon$
$\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| < \epsilon$
$\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} < \epsilon$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} > \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \sqrt{n} > \frac{1-2 \epsilon}{2}$
$\Rightarrow n > \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
| Here is yet another proof. Let $n = u-\frac{1}{2}$. Then we have:
$$\lim_{n \to \infty} \sqrt{n(n+1)} - n$$
$$=\lim_{u \to \infty} \sqrt{\left(u-\frac{1}{2}\right) \left(u+\frac{1}{2}\right)} - u + \frac{1}{2}$$
$$=\lim_{u \to \infty} \sqrt{u^2-\frac{1}{4}} - u + \frac{1}{2}$$
$$=\lim_{u \to \infty} u - u + \frac{1}{2} = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 7
} |
How to find the Square Root of a Polynomial $4x^4 + 4x^3 - 11x^2 -6x + 9$
How do you find the square root of this polynomial? I really don't understand. Please provide an easy-to-understand explanation. Thanks.
| Adjusting the coefficient of $x^3,$
$$4x^4+4x^3+\cdots+x^2=(2x^2)^2+2\cdot2x^2\cdot x+x^2$$
Adjusting the coefficient of $x^2$ and the constants
$$4x^4+4x^3-11x^2+\cdots+3^2=(2x^2)^2+2\cdot2x^2\cdot x+x^2-2\cdot2x^2\cdot3+\cdots+3^2$$
Finally,
$$4x^4+4x^3-11x^2-6x+3^2=(2x^2)^2+2\cdot2x^2\cdot x+x^2-2\cdot2x^2\cdot3+3^2-2\cdot x\cdot3=(2x^2+x-3)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/785070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Two halls 6 and 9 meters perpendicularly intersect. Optimization Two halls 6 and 9 meters perpendicularly intersect. Find the length of the longest straight bar to be passed horizontally from one aisle to another by a corner without deformation.
and this is my try:
How to find the equation to maximize in this problem.Please.
| This superficially appears to be a maximization problem but is really a minimization problem. You've drawn some diagonal lines through the corner that you seem to have labeled $(-6,9)$. You have to figure out which value of the coordinate you've called $x$ makes that diagonal line as short as possible.
You've got
$$
\begin{align}
g(x) = (x+6)^2 + \left( 9 + \frac{54}{x} \right)^2 & = (x+6)^2 + 81\left(\frac{x+6}{x}\right)^2 \\[10pt]
& = (x+6)^2\left( 1 + \frac{81}{x^2} \right).
\end{align}
$$
You need the value of $x$ that minimizes that.
Notice that $g(x)\to\infty$ as $x\downarrow0$ and $g(x)\to\infty$ as $x\to\infty$, and the function is continuous on $(0,\infty)$, so it must have a global minimum somewhere in between. If there's only one place in that interval where $g'=0$, then that must be it.
Alright, in response to comments:
\begin{align}
g'(x) & = (x+6)^2 \frac{d}{dx}\left( 1 + \frac{81}{x^2} \right) + \left( 1 + \frac{81}{x^2} \right) \frac{d}{dx}(x+6)^2 \\[10pt]
& = (x+6)^2 \cdot\frac{-162}{x^3} + \left( 1 + \frac{81}{x^2} \right)2(x+6).
\end{align}
This is $0$ when $x=-6$, and at other points we can divide by $x+6$ both sides of the equation that sets the expression above equal to $0$. We get
$$
(x+6) \cdot\frac{-162}{x^3} + \left( 1 + \frac{81}{x^2} \right)2 = 0.
$$
Multiplying both sides by $x^3$ we get
$$
(x+6)(-162) + (x^3 + 81x)2 = 0
$$
$$
2x^3 - 972 = 0 ,
$$
$$
x^3 - 486 = 0
$$
$$
x = \sqrt[3]{486} = 3\sqrt[3]{18} \approx 7.86.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/785521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral $\int_0^1 \frac{x\log x+1-x}{x \log^2 x}\log(1+x)\, dx=\log\frac{4}{\pi}$ Hi I am trying to prove this
$$
I:=\int_{0}^{1}
{x\log\left(\,x\,\right) + 1 - x \over x\log^{2}\left(\,x\,\right)}\,
\log\left(\,1 + x\,\right)\,{\rm d}x=\log\left(\,4 \over \pi\,\right).
$$
Thanks.
This is just a beautiful integral for many reasons. Logs are everywhere and an inspirational solution!!! I am not sure if breaking it up into three separate pieces is of any use, I tried that by writing
$$
I=\int_0^1\frac{ \log(1+x)}{\log x}dx+\int_0^1\frac{\log(1+x)}{x \log^2 x}dx-\int_0^1\frac{\log(1+x)}{\log^2 x}dx
$$
but wasn't sure how to handle these. Also note that
$$
\int_0^1 \frac{x\log x+1-x}{x \log^2 x}dx=1,
$$
in case that happened to come up anywhere along the calculation.
| Consider
\begin{align*}
\int_0^1 \frac{x\log{x}+1-x}{x}\, x^a\, \log{(1+x)}\, dx &= \int_0^1 \frac{x\log{x}+1-x}{x}\, x^a\, \sum_{n\ge 1} (-1)^{n+1}\frac{x^n}{n}\, dx\\
&=\sum_{n\ge 1} \int_0^1 \, (-1)^{n+1} (x\log{x}+1-x)\, \frac{x^{a+n-1}}{n}\, dx\\
&= \sum_{n\ge 1} - \frac{\left(-1\right)^{n+1}}{{\left(a + n + 1\right)}^{2} n} + \frac{\left(-1\right)^{n+1}}{{\left(a + n\right)} n} -\frac{\left(-1\right)^{n+1}}{{\left(a + n + 1\right)} n}\\
\int_0^1 \frac{x\log{x}+1-x}{x \log{x}}\, x^a\, \log{(1+x)}\, dx &= \sum_{n\ge 1} \frac{\left(-1\right)^{n + 1}}{n} \left(\frac{1}{a + n + 1} + \log\left(\frac{a + n}{a+n+1}\right)\right)\tag{$\int da$}\\
\int_0^1 \frac{x\log{x}+1-x}{x (\log{x})^2}\, x^a\, \log{(1+x)}\, dx &= \sum_{n\ge 1} \frac{\left(-1\right)^{n+1} {\left(a + n\right)} \log\left(\frac{a + n}{a + n + 1}\right) + \left(-1\right)^{n+1}}{n}\tag{$\int da$}\\
\end{align*}
Subst. $a=0$
\begin{align*}
\therefore \int_0^1 \frac{x\log{x}+1-x}{x (\log{x})^2}\, \log{(1+x)}\, dx &= \sum_{n\ge 1} \left(-1\right)^{n+1} \log\left(\frac{ n}{ n + 1}\right) + \frac{\left(-1\right)^{n+1}}{n}\hspace{20pt} \text{(Wallis product and log 2)}\\
&= \log{\left(\frac{2}{\pi}\right)}+\log{2} \\
&= \log{\left(\frac{4}{\pi}\right)}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/785880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
} |
Infinite Product $\prod\limits_{k=1}^\infty\left({1-\frac{x^2}{k^2\pi^2}}\right)$ I've been looking at proofs of Euler's sine expansion, that is
$$\frac{\sin x}{x}=\prod_{k=1}^\infty\left({1-\frac{x^2}{k^2\pi^2}}\right)$$
All the proofs seem to rely on complex analysis and Fourier series.
Is there any more elementary proof?
| I think this is the most motivated elementary proof.
For $z \in \mathbb C$ and $n \in \mathbb N_{> 0}$, let:
$$f_n \left({z}\right) = \frac 1 2 \left[{\left({1 + \frac z n}\right)^n - \left({1 - \frac z n}\right)^n }\right]$$
Then $f_n\left({z}\right) = 0$ if and only if:
\begin{align}
&&\left({1 + \frac z n}\right)^n & = \left({1 - \frac z n}\right)^n \\
& \iff & 1 + \frac z n & = \left({1 - \frac z n}\right) e^{2 \pi i \frac k n} \\
& \iff & z & = n \frac {e^{2 \pi i \frac k n} - 1} {e^{2 \pi i \frac k n} + 1} \\
&&& = n i \tan \left({\frac {k \pi} n }\right)
\end{align}
Let $n = 2 m + 1$.
Then the roots of $f_{2 m + 1} \left({z}\right)$ are $\left({2 m + 1}\right) i \tan \left({\dfrac {k \pi} {2 m + 1}}\right)$ for $- m \le k \le m$.
Observe that $f_{2m + 1} \left({z}\right)$ is a polynomial of degree $2 m + 1$.
Then for some constant $C$, we have:
\begin{align}
f_{2 m + 1} \left({z}\right) & = C z \prod_{\substack {k \mathop = - m \\ k \mathop \ne 0} }^m \left({1 - \frac z {\left({2 m + 1}\right) i \tan \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)\\
& = C z \prod_{k \mathop = 1}^m \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)
\end{align}
It can be seen from the binomial theorem that the coefficient of $z$ in $f_n \left({z}\right)$ is $1$.
Hence $C = 1$, and we obtain:
$$f_{2 m + 1} \left({z}\right) = z \prod_{k \mathop = 1}^m \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)$$
First we consider $z = x$ where $x$ is a non-negative real number.
Let $l < m$.
Then:
$$x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) \le f_{2 m + 1} \left({x}\right)$$
Taking the limit as $m \to \infty$ we have:
\begin{align}
& &\lim_{m \to \infty} x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {k^2 \pi^2} \left({\frac {k \pi / \left({2 m + 1}\right)} {\tan \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)^2 }\right) & \le \frac 1 2 \left({e^x - e^{- x} }\right)\\
& \implies &x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {k^2 \pi^2} }\right) & \le \sinh x
\end{align}
By the inequality $\tan \theta \ge \theta$ for $0 \le \theta < \dfrac {\pi} 2$ we have:
$$f_{2 l + 1} \left({x}\right) \le x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {k^2 \pi^2} }\right) \le \sinh x$$
Taking the limit as $l \to \infty$ we have by Squeeze Theorem:
$$\quad x \prod_{k \mathop = 1}^\infty \left({1 + \frac {x^2} {k^2 \pi^2} }\right) = \sinh x \tag{1}$$
Now let $1 < l < m$.
We have:
\begin{align}
&\left \vert{f_{2 m + 1} \left({z}\right) - z \prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)}\right \vert \\
& =\left \vert{z}\right \vert \left \vert{\prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)}\right \vert \cdot \left \vert{\prod_{k \mathop = l + 1}^m \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) - 1}\right \vert\\
& \le \left \vert{z}\right \vert \left[{\prod_{k \mathop = 1}^l \left({1 + \frac {\left \vert{z}\right \vert^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)}\right] \cdot \left[{\prod_{k \mathop = l + 1}^m \left({1 + \frac {\left \vert{z}\right \vert^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) - 1}\right] \\
& = f_{2 m + 1} \left({\left \vert{z}\right \vert}\right) - \left \vert{z}\right \vert \prod_{k \mathop = 1}^l \left({1 + \frac {\left \vert{z}\right \vert^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)
\end{align}
Taking the limit as $m \to \infty$ we have:
$$\left \vert{\sinh z - z \prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {k^2 \pi^2} }\right)}\right \vert \le \sinh {\left \vert{z}\right \vert} - \left \vert{z}\right \vert \prod_{k \mathop = 1}^l \left({1 + \frac {\left \vert{z}\right \vert^2} {k^2 \pi^2} }\right)$$
Now take the limit as $l \to \infty$.
By $(1)$ and Squeeze Theorem, we have:
$$\sinh z = z \prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {k^2 \pi^2} }\right)$$
Finally, substituting $z \mapsto i z$, we obtain:
$$ \sin z = z \prod_{k \mathop = 1}^l \left({1 - \frac {z^2} {k^2 \pi^2} }\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/786046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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Generalised Integral $I_n=\int_0^{\pi/2} \frac{x^n}{\sin ^n x} \ \mathrm{d}x, \quad n\in \mathbb{Z}^+.$ I have this integral, $$I_n=\displaystyle \int_0^{\pi/2} \frac{x^n}{\sin ^n x} \ \mathrm{d}x, \qquad n\in \mathbb{Z}^+.$$
We have the results
$$
\begin{align}
I_1 & = 2C, \\
I_2 &= \pi\log 2, \\
I_4 & = -\frac{\pi^3}{12} + 2\pi\log 2 + \frac{\pi^3}{3}\log 2-\frac{3\pi}{2}\zeta(3),
\end{align}
$$
where $C$ is Catalan's constant. Can we prove any of these results, or make any progress on $I_3$, or the general case?
| Integrating by parts 3 times,
$$ \begin{align} \int_{0}^{\pi /2} \frac{x^{4}}{\sin^{4} x} \ dx &= - \frac{x^{4}}{3} \cot(x) \left(\csc^{2} (x) +2 \right) \Bigg|^{\pi/2}_{0} + \frac{4}{3} \int_{0}^{\pi /2} x^{3} \cot (x) \left(\csc^{2} (x) +2 \right) \ dx \\ &= \frac{4}{3} \int_{0}^{\pi /2} x^{3} \cot (x) \left(\csc^{2} (x) +2 \right) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx + \frac{4}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \csc^{2}(x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx - \frac{2}{3}x^{3} \cot^{2}(x) \Bigg|^{\pi/2}_{0} + 2 \int_{0}^{\pi /2} x^{2} \cot^{2}(x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx + 2 \int_{0}^{\pi /2} x^{2} \cot^{2} (x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx -2x^{2} \Big( x + \cot(x) \Big) \Bigg|^{\pi/2}_{0} +4 \int_{0}^{\pi /2} x\Big(x+ \cot(x) \Big) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx - \frac{\pi^{3}}{4} + 4 \int_{0}^{\pi /2} x^{2} \ dx + 4 \int_{0}^{\pi /2} x \cot(x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx - \frac{\pi^{3}}{12} + 4 \int_{0}^{\pi /2} x \cot(x) \ dx . \end{align}$$
In general, $$ \int_{a}^{b} f(x) \cot(x) \ dx = 2 \sum_{n=1}^{\infty} \int_{a}^{b} f(x) \sin (2nx) \ dx .$$
See here.
So $$ \begin{align} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx &= 2 \sum_{n=1}^{\infty} \int_{0}^{\pi /2} x^{3} \sin (2nx) \ dx \\ &= 2 \sum_{n=1}^{\infty} \left(\frac{(-1)^{n-1} \pi^{3}}{16n} - \frac{(-1)^{n-1} 3\pi}{8n^{3}} \right) \\ &= \frac{\pi^{3}}{8} \ln (2) - \frac{3 \pi}{4} \eta(3) \\ &= \frac{\pi^{3}}{8} \ln (2) - \frac{9 \pi }{16} \zeta(3). \end{align}$$
And
$$ \begin{align} \int^{\pi /2}_{0} x \cot(x) \ dx &= 2 \sum_{n=1}^{\infty} \int_{0}^{\pi /2} x \sin(2nx) \ dx \\ &= -\frac{\pi}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \\ &= \frac{\pi \ln 2}{2} . \end{align}$$
Therefore,
$$ \begin{align} \int_{0}^{\pi /2} \frac{x^{4}}{\sin^{4} x} \ dx &= \frac{8}{3} \left( \frac{\pi^{3}}{8} \ln (2) - \frac{9 \pi }{16} \zeta(3) \right) - \frac{\pi^{3}}{12} + 4 \left(\frac{\pi \ln 2}{2} \right) \\ &= - \frac{\pi^{3}}{12} + 2 \pi \ln(2) + \frac{\pi^{3}}{3} \ln (2) - \frac{3 \pi}{2} \zeta(3) . \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/786281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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How to estimate the lower bound of a given Toeplitz matrix's eigenvalue? Given the Toeplitz matrix
$$\begin{pmatrix}
1 & a & a^2 & \cdots & a^n \\
a &1 &a & \cdots & a^{n-1} \\
a^2&a & 1 & \cdots& a^{n-2} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a^n & a^{n-1} & a^{n-2} & \cdots & 1\\
\end{pmatrix}$$
where $a \in (0,1)$, can one find the eigenvalues of the matrix? If not, can one find a lower bound?
Any links or reference materials? Thanks.
| Let's call the matrix in question $A$. Its inverse looks quite easy,
$$
A^{-1}
=
\frac{1}{1-a^2}
\begin{bmatrix}
1 & -a & & & & \\
-a & 1 + a^2 & -a & & & \\
& -a & 1+a^2 & -a & & \\
& & \ddots & \ddots & \ddots & \\
& & & -a & 1 + a^2 & -a \\
& & & & -a & 1 \\
\end{bmatrix}
$$
A lower bound on $\lambda_{\min}(A)$ can be estimated using the reciprocal of an upper bound on $\lambda_{\max}(A^{-1})$ for which one can use the Gershgorin theorem:
$$
\lambda_{\max}(A^{-1})\leq\max\left\{\frac{1 + a}{1 - a^2}, \frac{1+2a+a^2}{1 - a^2}\right\}.
$$
We clean up a bit the expressions,
$$\frac{1+a}{1-a^2}=\frac{1}{1-a},\quad \frac{1+2a+a^2}{1-a^2}=\frac{1+a}{1-a},$$
and see that the latter is generally larger. So
$$
\lambda_{\max}(A^{-1})\leq\frac{1+a}{1-a}\quad\Rightarrow\quad\lambda_{\min}(A)\geq\frac{1-a}{1+a}.
$$
Also, the same approach can be used to bound the maximal eigenvalues of $A$:
$$
\lambda_{\max}(A)\leq 1+a+\cdots+a^n=1+\frac{a(1-a^n)}{1-a}\leq \frac{1}{1-a}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $2 \lt f(x) \lt 2.8$? I have a question about a function $f(x)$:
$$f(x) = -x + \frac{x^2}{x-2} - \frac{20}{x^2 + x - 6},\qquad x>2$$
simplifies to
$$f(x) = 2x + \frac{10}{x + 3},\qquad x>2$$
How would you show the range of $f(x)$ is $2 < f(x) < 2.8?$
Thank you.
| I think it should be $$\frac{2x+10}{x+3}=\frac{2(x+3)+4}{x+3}=2+\frac4{x+3}$$
As $\displaystyle x>2,x+3>5\implies0<\frac1{x+3}<\frac15$
| {
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"url": "https://math.stackexchange.com/questions/791705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Verifying Touchard's Identity $$C_{n+1} = \sum_{k=0}^{\lfloor n/2\rfloor}{n\choose 2k}\cdot C_k\cdot 2^{n-2k}$$
where $C_n$ are the Catalan numbers.
I think we start by diving both sides by $2^n$, but unsure of where to go from there
| Notice that $2^{-2k}\binom{n}{2k}\binom{2k}{k}=\binom{n/2}{k}\cdot \binom{n/2-1/2}{k}$. So
$$A_n=\sum_{k=0}^{\lfloor n/2 \rfloor}2^n \cdot 2^{-2k}\cdot \binom{n}{2k}\binom{2k}{k}\cdot\dfrac{1}{k+1}=2^n\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n/2}{k}\cdot \binom{n/2-1/2}{k}\cdot\dfrac{1}{k+1}$$
Without loss of generality, suppose $n/2 \in \mathbb{N}$. By Vandermonde's identity, we have
$$A_n=\dfrac{2^n}{n/2+1} \cdot \binom{n+1/2}{n/2}=\dfrac{1}{n+2}\binom{2(n+1)}{n+1}=C_{n+1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int\limits_0^\pi \frac{x}{1+\sin^2x} \ dx$ How can one evaluate
$$\int_0^\pi \frac{x}{1+\sin^2x} \ dx\ ?$$
| First, make the substitution $x\to \pi - x$ to get $$\int_{0}^{\pi} \frac{x}{1+\sin^2 x} \, dx = \frac{\pi}{2} \int_{0}^{\pi} \frac{1}{1+\sin^2 x} \, dx $$
Now, interpret the integral on the right as the area under a polar curve: set $r(\theta) = \frac{1}{\sqrt{1+\sin^2 \theta}}$. Converting this to Cartesian coordinates gives the ellipse $x^2 + 2y^2 = 1$. The area under this ellipse as the angle varies from $0$ to $\pi$ is $\frac{\pi}{2\sqrt{2}}$. Thus, $$\frac{1}{2} \int_{0}^{\pi} \frac{1}{1+\sin^2 x} \, dx = \frac{1}{2} \int_{0}^{\pi} r^2 \, d\theta = \frac{\pi}{2\sqrt{2}} \implies \int_{0}^{\pi} \frac{1}{1+\sin^2 x} \, dx= \frac{\pi}{\sqrt{2}}$$ Therefore, $$\int_{0}^{\pi} \frac{x}{1+\sin^2 x} \, dx = \frac{\pi^2}{2\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/792692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Calculus Question: $\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx$ Can anyone help me to find $\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx$? Any help would be appreciated. Thanks in advance.
| Let
$$ I(a,b)=\int_0^{\pi/2}(\sin x)^{1-a}(\cos x)^{1-b}dx.$$
Then
$$ I(a,b)=\frac{\Gamma \left(1-\frac{a}{2}\right) \Gamma \left(1-\frac{b}{2}\right)}{2
\Gamma \left(-\frac{a}{2}-\frac{b}{2}+2\right)} $$
and hece
\begin{eqnarray}
\frac{\partial I(a,b)}{\partial a}&=&\frac{\Gamma \left(1-\frac{a}{2}\right) \Gamma \left(1-\frac{b}{2}\right)
\left(\psi ^{(0)}\left(-\frac{a}{2}-\frac{b}{2}+2\right)-\psi
^{(0)}\left(1-\frac{a}{2}\right)\right)}{4 \Gamma
\left(-\frac{a}{2}-\frac{b}{2}+2\right)}.
\end{eqnarray}
Thus
\begin{eqnarray}
\frac{\partial I(0,b)}{\partial a}&=&\frac{\Gamma \left(1-\frac{b}{2}\right) \left(\psi
^{(0)}\left(2-\frac{b}{2}\right)+\gamma \right)}{4 \Gamma
\left(2-\frac{b}{2}\right)}.
\end{eqnarray}
Noting
$$ \Gamma(x)\approx\frac{1}{x} \text{ near }x=0 $$
we have
\begin{eqnarray}
\lim_{b\to2}\frac{\partial I(0,b)}{\partial a}&=&\lim_{b\to2}\frac{\Gamma \left(1-\frac{b}{2}\right) \left(\psi
^{(0)}\left(2-\frac{b}{2}\right)+\gamma \right)}{4 \Gamma
\left(2-\frac{b}{2}\right)}\\
&=&\lim_{b\to2}\frac14 \frac{\psi^{(0)}(2-\frac{b}{2})+\gamma}{1-\frac{b}{2}}\\
&=&\lim_{b\to2}\frac14 \psi^{(1)}(2-\frac{b}{2})\\
&=&\frac{\pi^2}{24}.
\end{eqnarray}
So
$$ \int_0^{\pi/2}\tan x\ln(\sin x)dx=-\frac{\pi^2}{24}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/792950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 6
} |
How to determine the general polar equation of a circle How can you determine that the polar equation $r = a\cos(\theta)$ is a circle?
| Consider
$$
\begin{align}
y^2+(x-a/2)^2
&=\color{#C00000}{(r\sin(\theta))^2}+\color{#00A000}{(r\cos(\theta)-a/2)^2}\\
&=\color{#C00000}{a^2\cos^2(\theta)\sin^2(\theta)}+\color{#00A000}{a^2\cos^4(\theta)-a^2\cos^2(\theta)+a^2/4}\\
&=\color{#0000FF}{a^2\cos^2(\theta)(\sin^2(\theta)+\cos^2(\theta))}-a^2\cos^2(\theta)+a^2/4\\
&=\color{#0000FF}{a^2\cos^2(\theta)}-a^2\cos^2(\theta)+a^2/4\\
&=a^2/4
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/795121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solving cubic equation? I am trying to figure out the following cubic root thing.
$ax^3+bx^2+cx+d=0$
I set up
$x=y-\frac{3}{ba}$
Then I plug in for x
$a(y-\frac{3}{ba})^3+b(y-\frac{3}{3a})^2+c(y-\frac{3}{ba})=0$
The issue I am having trouble with is the simplification
I try to multiply it all but I gets messy.
Maybe the binomial theorem can be used
this is supposed to go down into the depressed cubic which is.
$ay^3+(c-\frac{b^2}{3a})y+(d+\frac{2b^3}{27a^2}-\frac{bc}{2a})$
| I would first consider
$$\begin{align*} 0 &= a(y-t)^3 + b(y-t)^2 + c(y-t) + d \\ &= a(y^3 - 3ty^2 + 3t^2 y - t^3) + b(y^2 - 2yt + t^2) + c(y-t) + d\\ &= ay^3 + (-3at + b)y^2 + (3at^2 - 2bt + c)y + (-at^3 + bt^2 - ct + d). \end{align*}$$
Now, the value of $t$ for which the coefficient of $y^2$ equals zero obviously satisfies $3at-b = 0$, or $t = \frac{b}{3a}$. This is not what you wrote, which is why you are experiencing..."problems." With the correct substitution, the remaining coefficients become
$$\begin{align*}0 &= ay^3 + \bigl(\tfrac{b^2}{3a} - \tfrac{2b^2}{3a} + c\bigr)y + \bigl(-\tfrac{b^3}{27a^2} + \tfrac{b^3}{9a^2} - \tfrac{bc}{3a} + d \bigr) \\ &= ay^3 + \bigl(c - \tfrac{b^2}{3a}\bigr)y + \bigl(\tfrac{2b^3}{27a^2} - \tfrac{bc}{3a} + d\bigr). \end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/796624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Calculating $x^{103} \equiv 2 \pmod{143}$ I need to find x, given that:
$0\leq x \leq 143$ and $x^{103}\equiv 2 \pmod{143}$.
I tried to use Euler's theorem $p(143)=120$, but it didn't help.
| Consider
$$x^{103}\equiv2 \pmod {13}$$
$$x^{103}\equiv2 \pmod {11}$$
en
$$x^7\equiv2 \pmod {13}$$
$$x^3\equiv2 \pmod {11}$$
$7$ is the common primitive root $\bmod {13}$ and $\bmod {11}$
$7^{11}\equiv2 \pmod {13}$, $7^3\equiv2 \pmod {11}$, and
if $(7^t)^7\equiv2 \pmod {13}$, $7t\equiv 11 \pmod {12}$, then $t=5\pmod {12}$
if $(7^s)^3\equiv2 \pmod {13}$, $3s\equiv 3 \pmod {10}$, then $s=1\pmod {10}$
so, the root $x$ is decided by
$$x\equiv 7^5\equiv 11 \pmod {13}$$
$$x\equiv7 \pmod {11}$$
then
$$x\equiv128 \pmod {143}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/798974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to calculate the improper integral $\int_0^\infty\left(\frac{1}{\sqrt{x^2+4}}-\frac{P}{x+2}\right)dx$ This is the first time I've seen a problem like this. I have no idea what to do. Detailed guidance would be of great help.
For which values of P does the integral converge?
$$\int_0^\infty\left(\dfrac{1}{\sqrt{x^2+4}}-\dfrac{P}{x+2}\right)dx$$
Thank You!
| Let our function be $f(x)$. The function $f(x)$ behaves nicely in the interval $[0,1]$, so it is enough to find $p$ such that $\int_1^\infty f(x)\,dx$ converges.
If $p\le 0$, then Comparison shows divergence. So we can assume $p\gt 0$.
Rewrite $f(x)$ as $\frac{x+2-p\sqrt{x^2+4}}{(x+2)\sqrt{x^2+4}}$, and then rationalize the numerator by multiplying top and bottom by $x+2+p\sqrt{x^2+4}$. We get
$$f(x)=\frac{x^2(1-p^2) +4x+4(1-p^2)}{(x+2)\sqrt{x^2+4}\left(x+2+\sqrt{x^2+4}\right)}.$$
If $p=1$, then $f(x)\le \frac{4x}{2x^3}=\frac{2}{x^2}$. Since $\int_1^\infty \frac{2}{x^2}\,dx$ converges, so does $\int_1^\infty f(x)\,dx$.
If $0\lt p\lt 1$ or $p\gt 1$, then our integral diverges. There are two cases to consider.
Suppose that $0\lt p\lt 1$. Let $g(x)=\frac{1}{x}$. One can show that
$$\lim_{x\to\infty} \frac{f(x)}{g(x)}=\frac{1-p^2}{2}.$$
Since $\int_1^\infty g(x)\,dx$ diverges, so does $\int_1^\infty f(x)\,dx$. The argument for $p\gt 1$ is similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/800633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
total least squares derivation with matrices Taken from a computer vision book: "to minimize the sum of the perpendicular distances between points and lines, we need to minimize $$ \sum_i (ax_i + by_i +c)^2$$ subject to $a^2 +b^2 =1$. Now using a Lagrangian multiplier $\lambda$, we have a solution if $$ \left( \begin{array}{ccc}
\overline{x^2} & \overline{xy} & \overline{x} \\
\overline{xy} & \overline{y} & \overline{y} \\
\overline{x} & \overline{y} & 1 \end{array} \right)\left[ \begin{array}{c}
a \\
b \\
c
\end{array} \right] = \lambda \left[ \begin{array}{cc}
2a\\
2b \\
0
\end{array} \right]$$
How is the book getting these matrices?
Also, the notion is that $\overline{u} = \frac{\sum u_i}{k}$. (Yeah, I don't know what $k$ stands for. I can only assume this is an average.)
It goes onto say that $c = -a\overline{x} - b\overline{y}$, and that we can substitute this back to get the eigenvalue problem $$\left[ \begin{array}{cc}
\overline{x^2} -\overline{x}~\overline{x} & \overline{xy} - \overline{x}\overline{y}\\
\overline{xy} - \overline{x}\overline{y} & \overline{y^2} - \overline{y} ~\overline{y} \\
\end{array} \right] \left[\begin{array}{cc}
a\\
b
\end{array} \right] = \mu \left[ \begin{array}{cc}
a\\
b
\end{array} \right].$$
I don't see what they substituted into, and how the answer is derived.
| Take the function
$$
F(a,b,c,\lambda)=\sum_i (ax_i+by_i+c)^2-\lambda(a^2+b^2-1)
$$
so that
\begin{align}
\frac{\partial F}{\partial a}&=2\sum_i (ax_i+by_i+c)x_i-2\lambda a=0\\
\frac{\partial F}{\partial b}&=2\sum_i (ax_i+by_i+c)y_i-2\lambda b=0\\
\frac{\partial F}{\partial c}&=2\sum_i (ax_i+by_i+c)=0\\
\frac{\partial F}{\partial \lambda}&=-(a^2+b^2-1)=0
\end{align}
or better
\begin{align}
&a\sum_i x_i^2 +b\sum_i x_iy_i+c\sum_i x_i=\lambda a\\
&a\sum_i x_iy_i+b\sum_i y_i^2 +c\sum_i y_i=\lambda b\\
&a\sum_i x_i +b\sum_i y_i +nc=0\\
&a^2+b^2=1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/801836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to show $\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}=2\zeta (3)$? How to show this equation is true.
$$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}=2\zeta (3)$$
where $H_{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$
| It can be shown that
\begin{align}
H_{n} = \int_{0}^{1} \frac{1-t^{n}}{1-t} \ dt.
\end{align}
Using this integral form of the Harmonic numbers the series in question becomes
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{1}{n^{2}} \cdot \int_{0}^{1} \frac{1-t^{n}}{1-t} \ dt \\
&= \int_{0}^{1} \left[ \zeta(2) - Li_{2}(t) \right] \frac{dt}{1-t},
\end{align}
where $Li_{2}(x)$ is the dilogarithm. Now using the integral
\begin{align}
\int \frac{Li_{2}(t)}{1-t} \ dt = 2 Li_{3}(1-t) - 2 Li_{2}(1-t) \ \ln(1-t) - Li_{2}(t) \ \ln(1-t) - \ln(t) \ \ln^{2}(1-t)
\end{align}
it is seen that, with the use of $Li_{m}(1) = \zeta(m)$, $Li_{m}(0) = 0$, $\ln(1) = 0$,
\begin{align}
S &= \zeta(2) [ - \ln(1-t)]_{0}^{1} + \zeta(2) \ln(0) + 2 Li_{3}(1) \\
&= - \zeta(2) \ln(o) + \zeta(2) \ln(0) + 2 Li_{3}(1) \\
&= 2 Li_{3}(1)
\end{align}
which yields
\begin{align}
\sum_{n=1}^{\infty} \frac{H_{n}}{n^{2}} = 2 \zeta(3).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/804424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Existence of $x,y$ Satisfying Diophantine Equation Let $a,b$ be positive integers. Prove that there exist positive integers $x,y$ such that
$$ \dbinom{x+y}{2} = ax+by $$
| Solutions of the equation: $\frac{(x+y)(x+y-1)}{2}=ax+by$
Can be written without using Pell's equation:
$x=\frac{(a+b+1)(3b-a+1)+2(b-a)k-k^2}{8(b-a)}$
$y=\frac{(a+b+1)(b-3a-1)+2(b-a)k+k^2}{8(b-a)}$
$k$ - what some integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/804692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove : $0\leq ab+bc+ca-2abc \leq \frac{7}{27}$ $a;b;c\geq 0$ such that : $a^2+b^2+c^2=1$.
Prove : $0\leq ab+bc+ca-2abc \leq \frac{7}{27}$
Thanks :)
P/s : I have no ideas about this problem :(
| This is the incorrect problem. The correct problem is
Given $a,b,c \ge 0, a+b+c = 1,$ prove $0 \le ab+bc+ca-2abc \le \frac{7}{27}$.
For the LHS, $ab \ge abc$ and so on so we are done.
For the RHS, $\sqrt{xy} \le x+y = 1-z $ then $z(x+y)+xy(1-2z) \le z(1-z)+\frac{(1-2z)(1-z)^2}4 \le \frac{7}{27}$ by calculus methods.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/805122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a^4+b^4\in\mathbb Q$ and $a^3+b^3\in\mathbb Q$ and $a^2+b^2\in\mathbb Q$, prove that $a+b\in\mathbb Q$ and $ab\in\mathbb Q$.
If $\begin{cases}a^4+b^4\in\mathbb Q\\ a^3+b^3\in\mathbb Q\\ a^2+b^2\in\mathbb Q\end{cases}$, prove that $a+b\in\mathbb Q$ and $ab\in\mathbb Q$. It is given that $a,b\in\mathbb R$.
The proof of the latter would simply follow from the former, and vice versa. So I think a better question would be:
Prove one of these statements: $a+b\in\mathbb Q$ or $ab\in\mathbb Q$.
The problem is from the selection to IMO.
I've tried a whole lot of things, including the identities: $$a^4+b^4=(a+b)(a^3+b^3)-ab(a^2+b^2)\\ a^3+b^3=(a+b)(a^2-ab+b^2)\\ a^2+b^2=(a+b)^2-2ab\\ (a+b)^3=a^3+b^3+3ab(a+b)\\ \text{etc...}$$
Even if one could solve the problem using these identities, doing it would most likely be quite tedious imho... Any observations would be greatly appreciated. Thanks.
| Hint: $(a^4+b^4)(a^2+b^2)-(a^3+b^3)^2 = a^2b^2(a^2+b^2-2ab)$. Now use @Praphulla's comment.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/805330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 1
} |
Bounds in Pythagorean Triples Let's take a look at this simple task (Pythagorean Triples):
Calculate $A$ and $B$ such that $A^2 + B^2 = C^2$.
$C$ is given.
Is there any way to find an upper bound for $A$, $B$, $A^2$, and $B^2$?
The upper bound will be a function of $C$.
$0 < A < f_1(C)$
$0 < B < f_1(C)$
$0 < A^2 < f_2(C)$
$0 < B^2 < f_2(C)$
$f_1(C) = ?$
$f_2(C) = ?$
Any ideas?
| There are an infinite number of pythagorean triples for which $B + 1 = C$. Let $A = 2n+1$. Let $B = 2n^2 + 2n$ and $C = B + 1 = 2n^2 + 2n + 1$.
$$\begin{align}
\\ A^2 + B^2 &= (2n+1)^2 + (2n^2 + 2n)^2
\\ & = (4n^2 + 4n + 1) + (4n^4 + 8n^3 + 4n^2)
\\ & = 4n^4 + 8n^3 + 8n^2 + 4n + 1
\\ & = (2n^2 + 2n + 1)^2
\\ & = C^2
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/805431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{\sqrt{7}}$
Let $a,b,c>0$ such that
$$\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}=\dfrac{1}{3}.$$
Show that
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}.$$
My try: since
$$\dfrac{1}{2+a^2}=\dfrac{1}{2}\left(1-\dfrac{a^2}{a^2+2}\right)$$
so
$$\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}=\dfrac{7}{3}$$
we only prove
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}$
and I want use Cauchy-Schwarz inequality to prove it,But I can't works,such
$$\left(\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}\right)(a^2+2+b^2+2+c^2+2)\ge (a+b+c)^2$$
$$(a^2+b^2+c^2+6)\ge \dfrac{3}{7}(a^2+b^2+c^2+2ab+2bc+2ac)$$
$$\Longrightarrow 4(a^2+b^2+c^2)+42\ge 6(ab+bc+ac)$$
and let $$p=a+b+c,q=ab+bc+ac,r=abc$$
so
$$2p^2+21\ge 7q$$
and we only prove
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{ab+bc+ac}{abc}=\dfrac{q}{r}\le\dfrac{3}{\sqrt{7}}$$
maybe this is not true.
But this not usefull to solve this problem .
Thank you
| This might not be the "proper" way, but it can be seen that both the expressions are symmetrical in all terms. And we need the maximum of the second eqn. This can happen only when all the terms contribute equally to the sum, if it were not so then, let 1/a<1/b<1/c, then we can make b and c to a, and get higher sum. So, putting it back in the first eqn. we get $ \frac{3}{a^2+2} = 1/3$, solving which we get $ a = \sqrt{7}$ or $-\sqrt{7} $. Choosing the positive root, as we need maximum values, we get $ \frac{1}{a} + \frac{1}{b}+\frac{1}{c} = \frac{3}{\sqrt{7}}$
Missed JJacquelin's answer, by a few minutes. It is more analytical, than mine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/806224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 3
} |
Partial fraction decomposition,how? I need to decompose this fraction:
$${x^2+1\over (x-1)^3\cdot(x+3)}$$
I tried to write it up like this: $${A\over (x-1)}+{B\over (x-1)^2}+{C\over (x-1)^3}+{D\over (x+3)}$$
But now i get $$A\cdot(x-1)^5\cdot (x+3) +B\cdot (x-1)^4\cdot (x+3) +C\cdot(x-1)^3\cdot (x+3) +D\cdot (x-1)^6 =x^2+1$$ which is not correct i think.
What should i do?
| Try this:
$
\frac{x^2+1}{(x-1)^3(x+3)} =
\frac{(x-1)^2+2x}{(x-1)^3(x+3)} =
\frac{(x-1)^2}{(x-1)^3(x+3)}+\frac{2x}{(x-1)^3(x+3)} =
\frac{1}{(x-1)(x+3)}+\frac{2x}{(x-1)^3(x+3)}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/806541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $7\mid(3^{2n+1}+2^{n+2})$ for all $n\in\mathbb{N}$
Prove that the following is true for every $n∈ℕ$:
$$7\mid(3^{2n+1}+2^{n+2}).$$
I've noticed
$$3^{2n+1}+2^{n+2} =3^{2n} \cdot 3+2^{n} \cdot 4.$$
Any suggestions how to continue from there to get something like $7k$ for $k\in\mathbb{N}$.
Thank you in advance!
| Just surprised that the very crude induction method with $u_n=3\cdot 9^n+4\cdot 2^n$ hasn't been posted yet - it goes $$u_{n+1} = 3\cdot 9^{n+1}+4\cdot 2^{n+1}=7\cdot3\cdot 9^n+2\cdot 3\cdot 9^n+2\cdot 4\cdot 2^n=7\cdot 3\cdot 9^n+2\cdot u_n$$ which is divisible by $7$ if $u_n$ is divisible by $7$. Base case is $u_0=7$. It depends on the fact that $9-2$ is divisible by $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How do I differentiate to what they have given? Given that $y = \dfrac{x^3 - 5x}{\sqrt{x}}$, show that $\dfrac{dy}{dx}$= $\dfrac{5(x^2 - 1)}{2 \sqrt{x}}$
(Posted from ``answer'' below: I get as far as $\dfrac{dy}{dx}= \dfrac {5}{2}x^\frac {3}{2} - \dfrac {5}{2}x^\frac{-1}{2}$ but struggling with the first fraction.)
I am just getting to grips with the formatting so hope that is clear!
Many thanks
| $\dfrac{dy}{dx}=\dfrac{(3x^2-5)\sqrt{x}-\frac{x^3-5x}{2\sqrt{x}}}{x}=\frac{3x^2-5}{\sqrt{x}}-\frac{x^2-5}{2\sqrt{x}}=\frac{5(x^2-1)}{2\sqrt{x}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How find this minimum of the $(a-c)^2+(b-d)^2$ let $a,b,c,d\in R$,and such
$$ab=c^2+4d^2=1$$
find the minimum of the value
$$(a-c)^2+(b-d)^2$$
My idea: since
$$(a-c)^2+(b-d)^2=a^2+b^2+c^2+d^2-2ac-2bd$$
I think this inequality can use Cauchy-Schwarz inequality to solve it.Thank you
| $f(a,b,c,d) = (a-c)^2 + (b-d)^2 \geq 0$, and $0$ is the minimum value of $f$. It occurs when:
$a = c$, and $b = d$. Thus:
$ab = cd = 1 = c^2 + 4d^2$. Solving this we get:
$a = c = \dfrac{5 \pm \sqrt{17}}{2}$, and
$b = d = \dfrac{2}{5 \pm \sqrt{17}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/811369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $x^2 + 96=0$ in $\mathbb{Z}_{100}$ I'm trying to find all solutions to $x^2 + 96=0$ in $\mathbb{Z}_{100}$.
$x^2 + 96 \equiv 0 \bmod 100$ implies that $x^2 + 96 \equiv 0 \bmod 2$ and $x^2 + 96 \equiv 0 \bmod 5$.
$$x^2 + 96 \equiv 0 \bmod 2 \iff x^2 \equiv 0 \bmod 2.$$
This means that $x$ must be even, since the square of an odd number is odd.
$$x^2 + 96 \equiv 0 \bmod 5 \iff x^2 \equiv 4 \bmod 5 \implies x=\pm2.$$
Is this correct?
| To fill in a last detail, here's how to work the problem mod $25$: first, solve the equation $x^2-4\equiv0\pmod 5$; you should find $x\equiv\pm2\pmod 5$ as your solutions. Now, we use the technique of Hensel lifting to 'raise' these solutions a degree, from $\pmod 5$ to $\pmod{5^2}$. First look at the $x\equiv 2$ root and write it as $x\equiv 5k+2\pmod{25}$, with $0\leq k\lt 5$. Now, we have $x^2\equiv 25k^2+20k+4\equiv 20k+4\pmod{25}$; so $x^2-4\equiv 20k\pmod{25}$. By easy inspection, the only solution of this in range is $k=0$, so that in fact $x\equiv 2\pmod{25}$. You can use similar algebra to find the root mod $25$ that's $\equiv -2\pmod{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/813177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I prove $2$ is prime (or irreducible) in all $Z[\sqrt{d}]$ with $d < -7$?
How do I prove $2$ is prime (or irreducible) in all $Z[\sqrt{d}]$ with $d < -7$?
I feel like the answer is right under my nose, but I just can't see it.
| For convenience, let's say $-d$ instead of $d$, with $d \in \mathbb{Z}^+$ and $\mu(d) \neq 0$.
If 2 is composite or reducible in $\mathbb{Z}[\sqrt{-d}]$, that means $x^2 + dy^2 = 2$ has a solution in integers, or, if $d \equiv 3 \mod 4$, the equation $\frac{x^2}{4} + \frac{dy^2}{4} = 2$ holds. For the former, the only solutions are $1^2 + 1 \times 1^2$ and $0^2 + 2 \times 1^2$. And for the latter, $\frac{1^2}{4} + \frac{7 \times 1^2}{4}$. These solutions correspond to the factorizations $(1 - i)(1 + i)$, $(-1)(\sqrt{-2})^2$ and $\left(\frac{1}{2} - \frac{\sqrt{-7}}{2}\right)\left(\frac{1}{2} + \frac{\sqrt{-7}}{2}\right)$.
If $d > 9$ and $y \neq 0$, then clearly $x^2 + dy^2 \geq 10 > 2$ or $\frac{x^2}{4} + \frac{dy^2}{4} \geq \frac{5}{2} > 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/815331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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An Inequality Problem $1 \le \frac{a}{1-ab}+\frac{b}{1-bc}+\frac{c}{1-ac} \le \frac{3\sqrt{3}}{2}$ If $a,b,c>0$, are positive real numbers such that $a^2+b^2+c^2=1$ then, the following Inequalities hold:
$\displaystyle 1 \le \frac{a}{1-ab}+\frac{b}{1-bc}+\frac{c}{1-ac} \le \frac{3\sqrt{3}}{2}$
$\displaystyle 1 \le \frac{a}{1+ab}+\frac{b}{1+bc}+\frac{c}{1+ac} \le \frac{3\sqrt{3}}{4}$
Homogenizing the first inequality as,
$\displaystyle \frac{a}{a^2+b^2+c^2-ab}+\frac{b}{a^2+b^2+c^2-bc}+\frac{c}{a^2+b^2+c^2-ac} \le \frac{3\sqrt{3}}{2\sqrt{a^2+b^2+c^2}}$
and noting the cyclic symmtery I tried Rearrangement Inequalities Inequalities.
Using $a^2+b^2+c^2-ab = \frac{1}{4}(a+b)^2+\frac{3}{4}(a-b)^2+c^2 \ge \frac{1}{4}(a+b)^2+c^2 \ge (a+b)c$
$\displaystyle \sum_{cyc} \frac{a}{a^2+b^2+c^2-ab} \le \sum_{cyc} \frac{a}{ac+bc}$ might help, but I couldn't get anywhere with it.
Thank you!
| we can prove
$$\dfrac{x}{1+xy}\le\dfrac{3\sqrt{3}x(x+y+2z)}{4(x^2+y^2+z^2+3(xy+yz+xz))}$$
so
$$\sum_{cyc}\dfrac{x}{1+xy}\le \sum_{cyc}\left(\dfrac{3\sqrt{3}x(x+y+2z))}{4(x^2+y^2+z^2+3(xy+yz+xz)}\right)=\dfrac{3\sqrt{3}}{4}$$
I have an other ugly methods:
this problem is equivalent:$a^2+b^2+c^3=3$,then we have
$$\dfrac{a}{ab+3}+\dfrac{b}{bc+3}+\dfrac{c}{ca+3}\le\dfrac{3}{4}$$
$$\Longleftrightarrow
4abc\sum_{cyc}ab+12\sum_{cyc}ab^2+36abc+36\sum_{cyc}a
\le 3a^2b^2c^2+9abc\sum_{cyc}a+27\sum_{cyc}ab+81$$
use this follow well know reslut
$$ab^2+bc^2+ca^2\le\dfrac{4}{27}(a+b+c)^3-abc$$
so we only
$$4abc\sum_{cyc}ab+12\left(\dfrac{4}{27}(a+b+c)^3-abc\right)
+36abc+36\sum_{cyc}a\le 3a^2b^2c^2+9abc\sum_{cyc}a+27\sum_{cyc}ab+81$$
let
$$p=a+b+c,q=ab+bc+ac,r=abc,p^2-2q=3$$
use AM-GM inequality we have
$$\dfrac{(p^2-3)^2}{4}=q^2\ge 3qr$$
so we consider
$$f(r)=3r^2-(2p^2-9p+18)r-\dfrac{16}{9}p^3+\dfrac{27}{2}p^2-36p
+\dfrac{81}{2}\ge 0$$
so
$$f'(r)=6r-2p^2+9p-18\le\dfrac{(p^2-3)^2}{2p}-2p^2+9p-18=
\dfrac{(p-1)(p-3)(p^2+2)-18}{2p}\le 0$$
so
$$f(r)\ge f\left(\dfrac{(p^2-3)^2}{12}\right)=\dfrac{(p-3)(3p^7-15p^6+27p^5-247p^4
+717p^3-1953p^2+621p-81)}{144p^2}$$
and we can prove
$$3p^7-15p^6+27p^5-247p^4
+717p^3-1953p^2+621p-81\le 0(p\ge \sqrt{3})$$
| {
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"url": "https://math.stackexchange.com/questions/815534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Proving continuity by epsilon-delta proof for a function of two variables. On account of a SE question , I raised the following question.
Let $f:D \to \mathbb R^2$ be a function in two variables. How would we go about setting up an epsilon-delta proof?
Let $f$ for example be given by $f(x,y) = (x+y)^2$ on the domain $D = \mathbb R^2$.
Edit. I am aiming at the kind of continuity that goes like the following for a one-variable function $f$. For every $\varepsilon > 0$, we have an $\delta >0$ such that $|x-a| < \delta$ implies $|f(x)-f(a)| < \varepsilon$.
| Let's show it is continuous at $(a,b)$. Then you calculate:
$|f(x,y) - f(a,b)| = |((x-a) + (y-b) + (a+b))^2 - (a+b)^2| \leq (x-a)^2 + (y-b)^2 + 2|x-a||y-b| + 2|x-a||a+b| + 2|y-b||a+b)| < (x-a)^2 + (y-b)^2 + (x-a)^2 + (y-b)^2 + 4(|a|+|b|)\cdot \sqrt{(x-a)^2 + (y-b)^2} = 2\left((x-a)^2 + (y-b)^2\right) + 4(|a|+|b|)\cdot \sqrt{(x-a)^2 + (y-b)^2} < \left(2 + 4|a| + 4|b|\right)\cdot \sqrt{(x-a)^2 + (y-b)^2}$, if we preset $\sqrt{(x-a)^2 + (y-b)^2} < 1$. From here you can see the answer as you take:
$\delta = \text{min} \{1, \dfrac{\epsilon}{2 + 4|a|+4|b|}\}$ for a given $\epsilon > 0$.
Done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying $\sqrt[4]{161-72 \sqrt{5}}$ $$\sqrt[4]{161-72 \sqrt{5}}$$
I tried to solve this as follows:
the resultant will be in the form of $a+b\sqrt{5}$ since 5 is a prime and has no other factors other than 1 and itself. Taking this expression to the 4th power gives $a^4+4 \sqrt{5} a^3 b+30 a^2 b^2+20 \sqrt{5} a b^3+25 b^4$. The integer parts of this must be equal to $161$ and the coeffecients of the roots must add to $-72$. You thus get the simultaneous system:
$$a^4+30 a^2 b^2+25 b^4=161$$
$$4 a^3 b+20 a b^3=-72$$
In an attempt to solve this, I first tried to factor stuff and rewrite it as:
$$\left(a^2+5 b^2\right)^2+10 (a b)^2=161$$
$$4 a b \left(a^2+5 b^2\right)=-72$$
Then letting $p = a^2 + 5b^2$ and $q = ab$ you get
$$4 p q=-72$$
$$p^2+10 q^2=161$$
However, solving this yields messy roots. Am I going on the right path?
| Denesting $\sqrt w = \sqrt{a+b\sqrt{n}}\,$ can be done by a simple formula that I discovered in my youth.
$ {\bf Simple\ Denesting\ Rule} \ \ \overbrace{\rm \color{#0a0}{subtract\ out}\ \sqrt{norm}^{\phantom .}}^{\textstyle\!\!\! w \to w - \sqrt{ww'} =:\, s\!\!\!\!\!\!}\!\!\!, \ {\rm then}\ \ \overbrace{\color{brown}{\rm divide\ out}\ \sqrt{{\rm trace}}^{\phantom .}}^{\textstyle s\,\to\, s/\sqrt{s+s'}\!\!\!\!\!\!\!}$ from that.
$\!\begin{align}{\rm Recall}\ \ w = a + b\sqrt{n}\rm \ \ has\ \ {\bf norm}\ &=\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2\! - n\: b^2\\[4pt]
{\rm and,\ furthermore,\ }w\rm \ \ has\ \ {\bf trace}\ &=\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\,a\end{align}$
In the norm/trace sqrts either sign works e.g. $\sqrt 1 = \pm1,\,$ so we choose what proves simplest.
Here $\:161-72\sqrt 5\:$ has norm $= 1.\:$ $\rm\ \color{#0a0}{subtracting\ out}\ \sqrt{norm}\ = -1\ $ yields $\ 162-72\sqrt 5\:$
which has $\ {\rm\ \sqrt{trace}}\: =\: \sqrt{324}\ =\ 18.\ \ \ \ \rm \color{brown}{Dividing\ it\ out}\ \,$ of the above yields $\ \ \ 9-4\sqrt 5$
Checking: $\ (9 - 4\sqrt 5)^2 = 9^2\!+\! 4^2(5)- 2(9)4\sqrt 5 = 161-72\sqrt 5 \ \ \large \color{#c00}\checkmark$
Next $\:9-4\sqrt 5\:$ has norm $= 1.\:$ $\rm\ \color{#0a0}{subtracting\ out}\ \sqrt{norm}\ = 1\ $ yields $\ 8-4\sqrt 5\:$
with $\ {\rm\ \sqrt{trace}}\: =\: \sqrt{16}\ =\ 4.\ \ \ \ \rm \color{brown}{Dividing\ it\ out}\,\ $ of the above yields $\,\ \ \ 2-\sqrt 5$
Checking: $\ (2 - \sqrt 5)^2 = 2^2\!+\! 5 - 2\cdot 2\sqrt 5 = 9 - 4\sqrt 5 \ \ \large \color{#c00}\checkmark$
Negating $\,2-\sqrt 5\,$ to get the positive square-root yields the sought result. We chose the signs in $\,\sqrt 1 = \pm 1$ so that arithmetic is simplest. Any choice will work as the proof below shows (e.g. we do both here). For many worked examples see prior posts on denesting. Below is a sketch of a proof.
Lemma $\ \ \sqrt w\, =\, \dfrac{s}t,\ \ \ \begin{align}s &\,=\, w \pm \sqrt{ww'}\\[.1em] t &\,=\: \pm\sqrt{s+s'}\end{align}\ $ when $\ \ \color{#90f}{\sqrt{ww'}\in\Bbb Q}$
Proof $\quad\ s^2 =\, w (w+w' \pm 2\sqrt{ww'})\, =\, w\, t^2$
Necessarily $\ \color{#90f}{\sqrt{ww'}\in \Bbb Q}\,$ if a denesting $\sqrt w = v = c + d\sqrt n\,$ exists, since
$$w = v^2\,\Rightarrow\, w' = v'^2\Rightarrow\, ww' = (vv')^2\in\Bbb Q^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all integers satisfying $m^2=n_1^2+n_1n_2+n_2^2$ I need to solve for $(n_1, n_2, m)$ (where $m, n_1, n_2$ are all integers - may be +ve or -ve) satisfying $m^2=n_1^2+n_1n_2+n_2^2$.
I have found 2 solutions so far - $(3, 5, 7)$ and $(7, 8, 13)$. What is a general solution?
UPDATE: For my purpose, it is further needed that $\gcd(n_1, m) = \gcd(n_2, m) = 1 $.
| This equation is symmetric so many solutions.
Generally speaking, this equation has a lot of formulas for the solution. Because it is symmetrical.
Write the formula can someone come in handy. the equation:
$Y^2+aXY+X^2=Z^2$
Has a solution:
$X=as^2-2ps$
$Y=p^2-s^2$
$Z=p^2-aps+s^2$
more:
$X=(4a+3a^2)s^2-2(2+a)ps-p^2$
$Y=(a^3-8a-8)s^2+2(a^2-2)ps+ap^2$
$Z=(2a^3+a^2-8a-8)s^2+2(a^2-2)ps-p^2$
more:
$X=(a+4)p^2-2ps$
$Y=3p^2-4ps+s^2$
$Z=(2a+5)p^2-(a+4)ps+s^2$
more:
$X=8s^2-4ps$
$Y=p^2-(4-2a)ps+a(a-4)s^2$
$Z=-p^2+4ps+(a^2-8)s^2$
For the particular case: $Y^2+XY+X^2=Z^2$ You can draw more formulas.
$X=3s^2+2ps$
$Y=p^2+2ps$
$Z=p^2+3ps+3s^2$
more:
$X=3s^2+2ps-p^2$
$Y=p^2+2ps-3s^2$
$Z=p^2+3s^2$
In the equation: $X^2+aXY+bY^2=Z^2$ there is always a solution and one of them is quite simple.
$X=s^2-bp^2$
$Y=ap^2+2ps$
$Z=bp^2+aps+s^2$
$p,s$ - integers asked us.
| {
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"timestamp": "2023-03-29T00:00:00",
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Regular pyramid, centre of circumscribed ball=centre of inscribed ball Prove that if the sum of plane angles in the vertex of a regular pyramid equals $180 ^{\circ}$, then in this pyramid the centers of the inscribed and circumscribed balls are equal.
Could you help me a bit?
Thank you!
| To prove it we should easy calculate all.
Let the pyramid's lateral edge is equals to one.
Calculate a length $a$ of a side of the regular polygon M with n sides which is the base of the pyramid:
$$a=2\sin \frac{\phi}{2} = 2\sin \frac{\pi}{2n}$$
Here $\phi=\pi \,/\, n$ is a plane angle at the vertex of the pyramid.
Let R is a radius of the circumscribed circle of the polygon M. Then $a=2R\sin \frac{\pi}{n}$. Hence $$R=\frac{\sin \frac{\pi}{2n}}{\sin \frac{\pi}{n}} = \frac{1}{2\cos\frac{\pi}{2n}}$$
$$\cos\frac{\pi}{2n} = \frac{1}{2R}$$
Calculate a radius r of the inscribed circle of the polygon M:
$$r=R\cos\frac{\pi}{n}=R(2\cos^2\frac{\pi}{2n}-1)=R(\frac{1}{2R^2}-1)=\frac{1-2R^2}{2R}$$
Calculate a height of the pyramid by the Pythagorean theorem:
$$H=\sqrt{ 1 - R^2}$$
Calculate a radius $\rho$ of the inscribed ball of the pyramid by using the formula that a radius of an inscribed circle for a triangle is equals to its area divided by semiperimeter:
$$\rho = \frac{rH}{r+\cos\frac{\pi}{2n}}=H\frac{1}{1+\frac{\cos\frac{\pi}{2n}}{r}} = H\frac{1}{1+\frac{1\,/\,2R}{(1-2R^2)\,/\,2R}}=H\frac{1-2R^2}{2(1-R^2)}$$
Here $\cos\frac{\pi}{2n}$ is a height of a lateral side of the pyramid.
By the formula that a radius of a circumscribed circle of a triangle is equals to a product of lengths of its sides divided by four times the area of a triangle we get a radius of the circumscribed ball of the pyramid:
$$\Omega = \frac{2R \times 1 \times 1}{4HR}=\frac{1}{2H}$$
Finally, to prove our theorem we need to check that $\rho + \Omega = H$:
$$\rho + \Omega= \frac{1}{2H}(2H^2\frac{1-2R^2}{2(1-R^2)} + 1) = \frac{1 - 2R^2 + 1}{2H} = \frac{1-R^2}{H}=H$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate $\int \frac{\cos x}{\sqrt{\sin2x}} \,dx$?
How to integrate $$\int \frac{\cos x}{\sqrt{\sin2x}} \,dx$$ ?
I have:
$$\int \frac{\cos x}{\sqrt{\sin2x}} \,dx = \int \frac{\cos x}{\sqrt{2\sin x\cos x}} \,dx = \frac{1}{\sqrt2}\int \frac{\cos x}{\sqrt{\sin x}\sqrt{\cos x}} \,dx = \frac{1}{\sqrt2}\int \frac{\sqrt{\cos x}}{\sqrt{\sin x}} \,dx = \frac{1}{\sqrt2}\int \sqrt{\frac{\cos x}{\sin x}} \,dx = \frac{1}{\sqrt2}\int \sqrt{\cot x} \,dx \\
t = \sqrt{\cot x} \implies x = \cot^{-1} t^2 \implies \,dx = -\frac{2t\,dt}{1 + t^4}$$
so I have:
$$-\sqrt2 \int \frac{t^2 \,dt}{1 + t^4}$$
I tried partial integration on that but it just gets more complicated. I also tried the substitution $t = \tan \frac{x}{2}$ on this one: $\frac{1}{\sqrt2}\int \sqrt{\frac{\cos x}{\sin x}} \,dx$
$$= \frac{1}{\sqrt2}\int \sqrt{\frac{\frac{1 - t^2}{1 + t^2}}{\frac{2t}{1+t^2}}} \frac{2\,dt}{1+t^2} = \frac{1}{\sqrt2}\int \sqrt{\frac{1 - t^2}{2t}} \frac{2\,dt}{1+t^2} = \int \sqrt{\frac{1 - t^2}{t}} \frac{\,dt}{1+t^2}$$
... which doesn't look very promising.
Any hints are appreciated!
| Having it the form of one polynomial divided by another should suggest one method, ugly though it may be: partial fractions. We have $1+t^4=1+2t^2+t^4-2t^2=(t^2+1-t\sqrt2)(t^2+1+t\sqrt2)$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Find the value of $a$ such that at least one root of the equation $f(x)=x^2 - (a-3)x + a =0$ is greater than $2$. As the title says.
To solve this problem I took two cases and solved them separately:
1. when the $x$ coordinate of the vertex is greater than $2$ and $f(2)>0$;
2. when $f(2)<0$.
However I wasn't able to find the answer as the two conditions have no common solution (obviously they won't have because in one case $f(2)>0$ and in the other one $f(2)<0$.
The inequalities which I got after solving for the two cases separately were:
1. $a<10$ and $a<7$
2. $10 < a$
And help would be appreciated. :)
| $x^2-(a-3)+a=0$
$\therefore \alpha, \beta=\large\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{(a-3)\pm \sqrt{(a-3)^2-4a}}{2}=\frac{(a-3)\pm \sqrt{(a-1)(a-9)}}{2}$
$$\therefore\large \frac{(a-3)\pm \sqrt{(a-1)(a-9)}}{2}>2$$
$$\therefore(a-3)\pm \sqrt{(a-1)(a-9)}>4$$
$$\therefore \pm \sqrt{(a-1)(a-9)} > 4-a+3$$
$$\therefore (a-1)(a-9) > (7-a)^2$$
$$\therefore (a-1)(a-9) \geq 0$$
For this to be true either both brackets on LHS should be positive OR both negative.
$\therefore a \geq9$ OR $a \leq1$
EDIT: substitute $a=9$ and $a=1$ in the original equation and we see that $a=9$ satisfies the constraint while $a=1$ does not, so answer is $a \geq9$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving a separable differential equation Solve the differential equation:
$$y'=\frac{1-y^2}{1-x^2}$$
My book says the solution is: $$y=\frac{x+c}{cx+1},$$ where $c$ is a constant.
It's been ten minutes I tried to verify if it was correct but I'm pretty sure the book is wrong. Can someone confirm it?
| We have
$$
\frac{dy}{dx}=\frac{1-y^2}{1-x^2}.
$$
Using separation variable we obtain
$$
\frac{dy}{1-y^2}=\frac{dx}{1-x^2}.
$$
Integrating both sides yields
$$\eqalign
{
\int\frac{dy}{1-y^2}&=\int\frac{dx}{1-x^2}\\
\frac12\int\left[\frac{1}{1+y}+\frac{1}{1-y}\right]\ dy&=\frac12\int\left[\frac{1}{1+x}+\frac{1}{1-x}\right]\ dx\\
\ln(1+y)-\ln(1-y)&=\ln(1+x)-\ln(1-x)+C_1\\
\ln\frac{1+y}{1-y}&=\ln\frac{1+x}{1-x}+C_1\\
\frac{1+y}{1-y}&=C_2\cdot\frac{1+x}{1-x}\\
y&=\frac{k_1x+k_2}{k_2x+k_1}\quad\Rightarrow\quad\text{dividing by}\ k_1\\
\color{blue}{y(x)}&\color{blue}{=\frac{x+c}{c\, x+1}}.
}
$$
where $C_2=e^{C_1}$, $k_1=C_2+1$, $k_2=C_2-1$, and $c=\dfrac{k_2}{k_1}$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
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} |
Synthetic division of polynomials by factor of the form $(ax+b)$ Find the quotient and remainder when $6x^4-11x^3+5x^2-7x+9$ is divided by $(2x-3)$. I expressed the divisor $(2x-3)$ as $2(x-\frac{3}{2})$ and conducted synthetic division by $\frac{3}{2}$ and obtained the coefficients of the quotients as $6, -2, 2$ and $-4$ and the remainder as $3$. Then I divided the quotients by $2$. Why should I do this? and why should I not divide the remainder by $2$?
| In my opinion, you do not need to change $(2x-3)$ by anything. If you perform the long divition of $6x^4-11x^3+5x^2-7x+9$ by $(2x-3)$, you should directly obtain
$$\frac{6x^4-11x^3+5x^2-7x+9 }{2x-3}=3 x^3-x^2+x-2+\frac{3}{2 x-3}$$ If i May suggest, don't make your life more complex than it is. Cheers.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum the following $\sum_{n=0}^{\infty} \frac {(-1)^n}{4^{4n+1}(4n+1)} $ Evaluate:
$$\sum_{n=0}^{\infty} \frac {(-1)^n}{4^{4n+1}(4n+1)} $$
I rewrote the sum as $$\sum_{n=0}^{\infty} \frac {1}{4^{8n-7}(8n-7)} - \sum_{n=0}^{\infty} \frac {1}{4^{8n-3}(8n-3)}$$
Now, I tried to express this as a Geometric Series and Partial Fraction but was unable to do so. I also tried to use Riemann Sum, but I don't know how to apply it here. Any help will be appreciated. Thanks!
| The series for $\tanh^{-1}(x)$ and $\tan^{-1}(x)$ are
\begin{align}
\tanh^{-1}(x) &= \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1} \\
\tanh^{-1}(x) &= \sum_{n=0}^{\infty} \frac{(-1)^{n} \ x^{2n+1}}{2n+1}.
\end{align}
Adding the series together leads to
\begin{align}
\sum_{n=0}^{\infty} \frac{x^{4n+1}}{4n+1} &= \frac{1}{2} \left( \tanh^{-1}(x) + \tanh^{-1}(x) \right)
\end{align}
and upon setting $x = e^{i \pi/4}/4$ the resulting series is
\begin{align}
\sum_{n=0}^{\infty} \frac{(-1)^{n}}{4^{4n+1} \ (4n+1)} = \frac{1}{2} \left( \tanh^{-1}\left(\frac{e^{i \pi/4}}{4}\right) + \tan^{-1}\left(\frac{e^{i \pi/4}}{4}\right) \right).
\end{align}
After some reductions it can be shown that the series in question is given by
\begin{align}
\sum_{n=0}^{\infty} \frac{(-1)^{n}}{4^{4n+1} \ (4n+1)} &= \frac{1}{4\sqrt{2}} \left[ \ln\left(\frac{17+4\sqrt{2}}{17-4\sqrt{2}}\right) + 2 \tan^{-1}\left(\frac{4\sqrt{2}}{15}\right) \right].
\end{align}
| {
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} |
$\frac{1}{\sin^2(x)}+\frac{1}{\cos^2(x)} = \cos^2(x)+\sin^2(x) ?$ This is a simple question. Since $\cos(\theta)^2 + \sin(\theta)^2 = 1$
Can I take the inverse of this $\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)} = \frac{1}{1}$?
Finally getting $\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)} = \cos(\theta)^2 + \sin(\theta)^2$
If this is incorrect thinking can someone please put me on the right path?
Thanks
| It is not true that the reciprocal of $a+b$ is the reciprocal of $a$ plus the reciprocal of $b$. The reciprocal of $\sin^2\theta+\cos^2\theta$ is not the reciprocal of $\sin^2\theta$ plus the reciprocal of $\cos^2\theta$.
| {
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power series representation in terms of another Hi how to do the following:
Given
$f(z) = \sum c_n z^n$
How to express $\sum c_{3n} z^{3n}$ in terms of $f(z)$
Thanks a lot!
| If you allow the use of complex numbers (numbers of the form $a + bi$ where $i=\sqrt{-1}$ and $a$ and $b$ are ordinary real numbers) then there are three cube roots of 1:
$$
\begin{array}{l}
1 \\
\frac{-1+i\sqrt{3}}{2} \equiv \omega \\
\frac{-1-i\sqrt{3}}{2} = \omega^2 \\
\end{array}
$$
You can easily verify that
$$
\omega^2 = \left(\frac{-1+i\sqrt{3}}{2}\right)^2 =
\frac{1 -2i\sqrt{3} - 3}{4} = \frac{-1-i\sqrt{3}}{2}
$$
and that
$$
\omega^3 = \frac{-1+i\sqrt{3}}{2} \frac{-1-i\sqrt{3}}{2} = \frac{1+3}{4} = 1
$$
Now let's look at what $f(z\omega)$ would be: When the power is a multiple of 3, the term will remain $c_{3n}z^{3n}$ because $\omega^{3n} = 1$. But when the power is $3n+1$ the term is $\omega c_{3n+1}z^{3n+1}$ and when the power is $3n+2$ the term is
$\omega^2 c_{3n+2}z^{3n+2}$.
$$
f(\omega z) = \sum c_{3n} z^{3n} + \omega \sum c_{3n+1}z^{3n+1} + \omega^2 \sum c_{3n+2}z^{3n+2}
$$
Similarly,
$$
f(\omega^2 z) = \sum c_{3n} z^{3n} + \omega^2 \sum c_{3n+1}z^{3n+1} + \omega \sum c_{3n+2}z^{3n+2}
$$ because $\omega^4 = \omega$.
Now here is the cute step: Notice that $\omega + \omega^2 = -1$. So
$$
f(\omega z) + f(\omega^2 z)= 2 \sum c_{3n} z^{3n} -\sum c_{3n+1}z^{3n+1} -\sum c_{3n+2}z^{3n+2}
$$
and the answer to your question is obtained by adding $f(z)$ to get rid of all the $3n+1$ and $3n+2$ powers:
$$
\sum c_{3n} z^{3n} = \frac{f(z)+f(\omega z) + f(\omega^2 z)}{3}
$$
| {
"language": "en",
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} |
Equality of binomial coefficients I have seen that the following equations are equal, but are wondering how this is shown
${n \choose m} \cdot 1 \cdot 3 \cdots (2m-1)\cdot 1 \cdot 3 \cdots (2n-2m-1)
= \frac{n!}{2^n} {2m \choose m} {2n-2m \choose n-m}
=2^n \cdot n! \cdot (-1)^n \cdot {-\frac{1}{2} \choose m}{-\frac{1}{2} \choose n-m}$
I can show that
${n \choose m} \cdot 1 \cdot 3 \cdots (2m-1)\cdot 1 \cdot 3 \cdots (2n-2m-1) =2^n \cdot n! \cdot (-1)^n \cdot {-\frac{1}{2} \choose m}{-\frac{1}{2} \choose n-m} $
But I cannot show one of the two equalities involving $\frac{n!}{2^n} {2m \choose m} {2n-2m \choose n-m}$.
I know it is true due to numerical calculations.
Thanks,
| Cool question. I think the following trick works:
$$ {2m \choose m} = \frac{1 \cdot 3 \cdots (2m-1)}{m!} \cdot \frac{2 \cdot 4 \cdots 2m}{1 \cdot 2 \cdots m} = 1 \cdot 3 \cdots (2m-1) \cdot \frac{2^m}{m!} $$
and
$$ {2n-2m \choose n-m} = \frac{1 \cdot 3 \cdots (2n - 2m -1)}{(n-m)!} \cdot \frac{2 \cdot 4 \cdots (2n-2m)}{1 \cdot 2 \cdots (n-m)} = 1 \cdot 3 \cdots (2n - 2m -1) \cdot \frac{2^{n-m}}{(n-m)!}$$
So
$$ \frac{n!}{2^n} {2m \choose m}{2n-2m \choose n-m} = 1 \cdot 3 \cdots (2m-1) \cdot 1 \cdot 3 \cdots (2n - 2m -1) \cdot \frac{2^{m+n-m}}{2^n} \cdot \frac{n!}{m! (n-m)!} $$
$$ = {n \choose m} \cdot 1 \cdot 3 \cdots (2m-1) \cdot 1 \cdot 3 \cdots (2n - 2m -1) $$
| {
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To Solve an ODE To Solve: $\displaystyle (1+x^2)\frac{d^2y}{dx^2}+1+\left(\frac{dy}{dx}\right)^2=0$
My Attempt: Take $\displaystyle \frac{dy}{dx}=p$
Now we have: $\displaystyle (1+x^2)\frac{dp}{dx}+1+p^2=0$
$\displaystyle \frac{dp}{1+p^2}=-\frac{dx}{(1+x^2)}$
Integrating, $\displaystyle \tan^{-1}p=-\tan^{-1}x$
So now, can we take this as $\displaystyle p = -x$ ?
If I can,we end with: $\displaystyle \frac{dy}{dx}=-x+c_1$
The answer seems different: $\displaystyle y=c_1x+(c_1^2+1)\log(x-c_1)+c_2$
Where am I going wrong?
| Continuing my above comment. We have
$$ y' = p = \tan\bigl(-\tan x + c\bigr) = \frac{c_1-x}{1+c_1x }. $$
Now in case $c_1 = 0$, we have $y' = -x$, hence $y = c_2 - \frac 12 x^2$. Otherwise we have
$$ y' \frac{c_1 - x}{1 + c_1x} = -\frac 1{c_1}\frac{-c_1^2 + c_1x}{1+c_1x}
= -\frac 1{c_1} \cdot \left( 1 - (1+c_1^2)\frac 1{1+c_1 x} \right)
$$
Hence
$$ y = c_2 - \frac 1{c_1} \cdot \left( x - \frac{1+c_1^2}{c_1}\log(1 + c_1 x)\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/831700",
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} |
Circle touching three tangential circles The circles $C_1,C_2$ and $C_3$ with radii $1,2$ and $3$, respectively, touch each other externally. The centres of $C_1$ and $C_2$ lie on the $x$-axis, while $C_3$ touches them from the top. Find the ordinate of the centre of the circle that lies in the region enclosed by the circles $C_1,C_2$ and $C_3$ and touches all of them.
I had been trying to do the sum by assuming the centre of the first circle at $(0,1)$ but it did not help.
| here is my nice answer:
let $C$ be the center of enclosed circle then its radius $r$ is given by set formula by HC Rajpoot
$$r=\frac{abc}{2\sqrt{abc(a+b+c)}+ab+bc+ca}$$
where, $a=1, b=2, c=3$ are radii of three externally touching circles then
$$r=\frac{1\times2\times 3}{2\sqrt{1\times 2\times 3(1+2+3)}+1\times 2+2\times 3+3\times 1}=\frac{6}{23}$$
now, drop a perpendicular $CN$ of length $y$ from center $C$ to the x-axis to get a right triangle $C_1NC$ in which hypotenuse
$C_1C=1+\frac6{23}=\frac{29}{23}$. apply pythagorean
$$C_1N=\sqrt{(C_1C)^2-(CN)^2}=\sqrt{\left(\frac{29}{23}\right)^2-y^2}\ \ \ \ ..........(1)$$
similarly, in right triangle $CNC_2$ in which hypotenuse
$C_2C=2+\frac6{23}=\frac{52}{23}$. apply pythagorean
$$NC_2=\sqrt{(C_2C)^2-(CN)^2}=\sqrt{\left(\frac{52}{23}\right)^2-y^2}\ \ \ \ ........(2)$$
since, circle $C_1$ & $C_2$ are touching hence, $$C_1N+NC_2=C_1C_2=1+2=3$$
$$\sqrt{\left(\frac{29}{23}\right)^2-y^2}+\sqrt{\left(\frac{52}{23}\right)^2-y^2}=3$$
$$\sqrt{\frac{841}{529}-y^2}=3-\sqrt{\frac{2704}{529}-y^2}$$ taking squares on both sides, i get
$$\sqrt{\frac{2704}{529}-y^2}=\frac{1104}{529}$$
again i take square,
$$y^2=\frac{2704}{529}-\left(\frac{1104}{529}\right)^2=\frac{400}{529}$$
$$y=\frac{20}{23}$$
above is the correct value of ordinate of center $C$ of the enclosed circle
| {
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"question_score": "1",
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prove $\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = 4 \cdot \sin(\frac{B - C}{2}) \cdot \sin(\frac{C - A}{2}) \cdot \sin(\frac{A - B}{2})$ If $A + B + C = \pi$,
then prove $\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = 4 \cdot \sin(\frac{B - C}{2}) \cdot \sin(\frac{C - A}{2}) \cdot \sin(\frac{A - B}{2})$
My try:
$\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = \sin(A - C) + \sin(B - A) + \sin(C -B)$
$\sin(A - C) + \sin(B - A) + \sin(C -B) = 2 \cdot \sin(\frac{B -C}{2}) \cdot \cos(A - \frac{C + B}{2}) + \sin(C - B)$
$\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = 2 \cdot \sin(\frac{B -C}{2}) \cdot \sin(\frac{A}{2}) + \sin(C - B)$
I am confused what steps should follow?
| First let $C=\pi -A -B$, and then rewrite both $LHS$ and $RHS$:
*
*$$\color{green}{LHS}=\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B)=\\\sin(B + 2\pi-2A-2B) + \sin(\pi-A-B + 2A) + \sin(A + 2B)=\\\sin(-B-2A) + \sin(\pi+A-B) + \sin(A+2B)=\\\sin(-B-2A)+\sin(A+2B)-\sin(A-B)=\\2\sin\left(\frac{A+2B-B-2A}{2}\right)\cos\left(\frac{A+2B+B+2A}{2}\right)+\sin(B-A)=\\\color{green}{2\sin\left(\frac{B-A}{2}\right)\cos\left(\frac{3A+3B}{2}\right)+\sin(B-A)}$$
*$$\color{blue}{RHS}=4 \cdot \sin\left(\frac{B - C}{2}\right) \cdot \sin\left(\frac{C - A}{2}\right) \cdot \sin\left(\frac{A - B}{2}\right)=\\4 \cdot \sin\left(\frac{B - \pi+A+B}{2}\right) \cdot \sin\left(\frac{\pi-A-B - A}{2}\right) \cdot \sin\left(\frac{A - B}{2}\right)=\\4 \cdot \sin\left(-\left(\frac{\pi}{2}-\frac{A+2B}{2}\right)\right) \cdot \sin\left(\frac{\pi}{2}-\frac{2A+B}{2}\right) \cdot \sin\left(\frac{A - B}{2}\right)=\\-4 \cdot \sin\left(\frac{\pi}{2}-\frac{A+2B}{2}\right) \cdot \cos\left(\frac{2A+B}{2}\right) \cdot \sin\left(\frac{A - B}{2}\right)=\\-4 \cdot \cos\left(\frac{A+2B}{2}\right) \cdot \cos\left(\frac{2A+B}{2}\right) \cdot \sin\left(\frac{A - B}{2}\right)=\\-4\cdot \frac12\sin\left(\frac{A - B}{2}\right)\left[\cos\left(\frac{A+2B}{2}-\frac{2A+B}{2}\right)+\cos\left(\frac{A+2B}{2}+\frac{2A+B}{2}\right)\right]=-2\sin\left(\frac{A - B}{2}\right)\left[\cos\left(\frac{B-A}{2}\right)+\cos\left(\frac{3A+3B}{2}\right)\right]=\\2\sin\left(\frac{B-A}{2}\right)\cos\left(\frac{B-A}{2}\right)+2\sin\left(\frac{B-A}{2}\right)\cos\left(\frac{3A+3B}{2}\right)=\\\color{blue}{\sin(B-A)+2\sin\left(\frac{B-A}{2}\right)\cos\left(\frac{3A+3B}{2}\right)}$$
So if $C=\pi -A -B$,$$LHS=RHS\\\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B)=4\sin\left(\frac{B - C}{2}\right)\sin\left(\frac{C - A}{2}\right)\sin\left(\frac{A - B}{2}\right)\\\color{yellow}{QED}$$
| {
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How to prove the following exercise by using the definition of a determinant? $\begin{align} \begin{vmatrix}
a_{11} & \cdots& a_{1m} & 0 & \cdots & 0 \\
\cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\
a_{m1} & \cdots & a_{mm} & 0 & \cdots & 0 \\
0 & \cdots & 0 & 1 & \cdots & 0 \\
\cdot & \cdots & \cdot & \cdot & \ddots & \cdot \\
0 & \cdots & 0 & 0 & \cdots & 1
\end{vmatrix} =
\begin{vmatrix}
a_{11} & \cdots& a_{1m} \\
\cdot & \cdots & \cdot \\
a_{m1} & \cdots & a_{mm} \\
\end{vmatrix}.
\end{align}$
i.e. Definition of a determinant;
The determinant of the array A is the number
$$ \sum_{(\lambda_1, \cdots , \lambda_n)} {\epsilon (\lambda_1, \cdots , \lambda_n)} a_{1\lambda_1} ,\cdots, a_{n\lambda_n} $$ where the summation extends over all n! arrangements $(\lambda_1,\cdots ,\lambda_n)$ of $(1, \cdots, n)$. This determinant is denoted by
$\begin{align}
\begin{vmatrix}
a_{11} & a_{21} & \cdots & a_{1m}\\
a_{21} & a_{22} & \cdots & a_{2m}\\
\cdot & \cdot & \cdots & \cdot\\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{vmatrix}, \text{or more briefly, by}
\end{align}$
$\begin{align}
{\begin{vmatrix}
a_{ij}\\
\end{vmatrix}_n}.
\end{align}$
$\begin{align}
A =
\begin{matrix}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\cdot & \cdot &\cdots & \cdot\\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{matrix}
\end{align}$
This exercise is taken from the book 'An Introduction to Linear Algebra' by L. Mirsky.
Page number 13, exercise 1.3.1.
| Use the fact that the determinant of a block matrix is equal to the product of the determinants of matrix $A$ and $B$.
So $$\begin{vmatrix} A & 0 \\ 0 & B\end{vmatrix} = \det(A) \det(B)$$
Here, $A$ is an $m\times m$ matrix, $B$ is a $(n - m)\times (n - m)$ matrix.
This can be proven by induction.
| {
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Inequality $\frac1a + \frac1b + \frac1c \leq \frac{a^8+b^8+c^8}{a^3b^3c^3}$
Let $a,b,c$ be positive reals .
Prove that $\displaystyle \frac1a + \frac1b + \frac1c \leq \frac{a^8+b^8+c^8}{a^3b^3c^3}$
I found this one in a book, no hints mentioned, but marked as very hard.
I can't make any progress...
| We might make repeated use of the inequality $x^2+y^2+z^2 \ge xy+yz+zx$, for $x,y,z \in \mathbb{R}$ as well.
We have infact: $$\displaystyle a^8+b^8+c^8 \ge \sum\limits_{cyc} a^4b^4 \ge \sum\limits_{cyc} a^2b^4c^2 \ge \sum\limits_{cyc} ab^2c.bc^2a = \sum\limits_{cyc} a^2b^3c^3 = a^3b^3c^3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)$$
| {
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Local minimum of $f(x) = 4x + \frac{9\pi^2}{x} + \sin x$ What's the minimum value of the function
$$f(x) = 4x + \frac{9\pi^2}{x} + \sin x$$
for $0 < x < +\infty$? The answer should be $12\pi - 1$, but I get stuck with the expression involving both $\cos x$ and $x^2$ in the derivative. Taking the derivative, we have:
$$f'(x) = 4 - \frac{9\pi^2}{x^2} + \cos x.$$
In order to find the local extrema of the function, we set $f'(x) = 0$. Therefore,
\begin{align}
4 - \frac{9\pi^2}{x^2} + \cos x &= 0 \\
4x^2 - 9\pi^2 + x^2 \cos x &= 0 \\
x^2 (4 + \cos x) &= 9\pi^2.
\end{align}
However, I'm not sure what to do from here or if, indeed, I'm doing it right at all. Any help would be appreciated.
| If I may interfer in the discussion, the first derivative being $$f'(x) = 4 - \frac{9\pi^2}{x^2} + \cos x$$ you find the minimum for $x=\frac{3\pi}{2}$ just because, for this specific value, the cosine is zero. I can bet that the problem was set with this in mind.
But, let us change slightly your function to $$f(x) = 4x + \frac{8\pi^2}{x} + \sin x$$ Now, $$f'(x) = 4 - \frac{8\pi^2}{x^2} + \cos x$$ and the problem becomes quite different and you cannot solve anymore $f'(x)=0$ using elementary methods. Instead numerical methods should be used and Newton is probably the simplest root-finder for the solution of non linear equation. Starting from a "reasonable" guess $x_0$, Newton updates are given by $$x_{n+1}=x_n-\frac{f'(x_n)}{f''(x_n)}$$ So, in the case of the modified problem, using $x_0=\frac{3\pi}{2}\simeq 4.71239$, the successive iterates will be $4.53525$, $4.54083$, $4.54084$ which is the solution of $f'(x)=0$ for six significant figures.
| {
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how can I find the period of $f=\sin^4(x)$? How can I find the period of $f=\sin^4(x)$?
$\sin^4(x)=\frac{3-4\cos2x+\cos4x}{8}$. I didn't manage to reduce $\cos 4x$ to $\cos 2x$.
| Note that the equation $$\sin^4 (x+y)=\sin^4 x$$
Implies $$\sin^4 (x+y)-\sin^4 x = \left(\sin (x+y)-\sin x\right)\left(\sin (x+y)+\sin x\right)\left(\sin^2 (x+y)+\sin^2 x\right)=0$$
The final factor is nonnegative, and can only be zero when $\sin x=0$ so contributes no value of $y$ which works for all $x$.
So we have $$\sin (x+y)=\sin x\cos y+\cos x \sin y=\pm \sin x$$ And this has to be true for all $x$ and constant $y$. Set $x=0$ to obtain $\sin y=0$ and $y=n\pi$. So the period is $\pi$ (after checking that these values of $y$ work).
| {
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Matrix with unknown coefficients, finding another basis let $(e_1,e_2,...,e_5)$ canonical basis of $R^5$, $V=(a,b,c,d,e)\in R^5$ with $V\neq(0,0,0,0,0)$. we consider $f:R^5\to R^5$ and its matrix :
$$Mat(f) = M= \begin{pmatrix}
a&a&a&a&a\\
b&b&b&b&b\\
c&c&c&c&c\\
d&d&d&d&d\\
e&e&e&e&e\\
\end{pmatrix}$$
It's rank is $rg(f)=1$ because $V\neq(0,0,0,0,0)$ and all columns are the same. So $Dim(Im(f))=1$ and because of the rank theorem, $Dim(Ker(f))=4$
Basis of $Im(f)=Vect\begin{pmatrix}\begin{pmatrix}a\\b\\c\\d\\e\\\end{pmatrix}\end{pmatrix}$ and by solving with $X=\begin{pmatrix}x\\y\\z\\t\\u\\ \end{pmatrix}$, $ MX=0$ I find : $Ker(M)=Vect\begin{pmatrix}\begin{pmatrix}-1\\1\\0\\0\\0\end{pmatrix},\begin{pmatrix}-1\\0\\1\\0\\0\\\end{pmatrix},\begin{pmatrix}-1\\0\\0\\1\\0\end{pmatrix},\begin{pmatrix}-1\\0\\0\\0\\1\end{pmatrix}\end{pmatrix}$
I was then asked a condition on $a,b,c,d,e$ for $V\in Ker(f)$ and I found $a+b+c+d+e=0$ by solving $MV=0$
And next is a question I need to verify:
Question : let this condition be fullfilled, let $(\epsilon_1,\epsilon_2,\epsilon_3,\epsilon_4)$ basis of $Ker(f)$ chosen so $\epsilon_4=V$
Complete it in a basis of E (here $R^5$) and write the matrix in the new basis
Verification : what I did : Is it okay for me to set $V = \begin{pmatrix}-1\\0\\0\\0\\1\end{pmatrix}$ so it would mean that my matrix $M$ would become $M=\begin{pmatrix}-1&-1&-1&-1&-1\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\1&1&1&1&1\end{pmatrix}$,$\epsilon_1=\begin{pmatrix}-1\\1\\0\\0\\0\end{pmatrix}$, and then respectively for $\epsilon_2,\epsilon_3, ...$
or should I solve with $V=\begin{pmatrix}a\\b\\c\\d\\e\end{pmatrix}$ ? Because I don't understand what they're looking for if it's a general answer or a simple base to create.
because what I could find was my new basis (direcltly made into a base changing matrix because it's hard to type everything) : $P=\begin{pmatrix}-1&-1&-1&-1&-1\\1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\end{pmatrix}$
then I solved and answered $M'=P^-1MP=\begin{pmatrix}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&-1\\0&0&0&0&0\end{pmatrix}$
Then, right after this question, the big problem:
Question: now, $V\notin Ker(f)$ I have to find a basis so B matrix would be : B = \begin{pmatrix}\gamma&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{pmatrix}
and of course I need to precise $\gamma$
So here, I thought I could use $V=\begin{pmatrix}-1\\0\\0\\0\\0\end{pmatrix}$ and keep my $Ker(f)$ basis I found before and use the same formula $B=P^-1M'P$ but it didn't sound good at all ... Because my $\gamma=0$ which is nonsense ... I'm a bit confused and it makes me stuck
Thank you in advance and sorry for the big stuff.
| Hints:
Question 1
I would assume this question requires a general answer, so setting a value for $V$ is incorrect.
Before you can find the matrix of $f$ in the new basis, you need to find $\epsilon_5$. Since $\mathbb{R}^5$ has dimension 5, it is enough that $\{\epsilon_1, \epsilon_2, \epsilon_3, \epsilon_4, \epsilon_5\}$ be linearly independent. How can we choose $\epsilon_5$ to make this true? Remember $\{\epsilon_1, \epsilon_2, \epsilon_3, \epsilon_4\}$ spans the nullspace of $f$. Let's take $\epsilon_5$ to be some arbitrary vector not in $\mathtt{Ker}(f)$. Then, if
$$\lambda_1\epsilon_1 + \lambda_2\epsilon_2 + \lambda_3\epsilon_3 + \lambda_4\epsilon_4 + \lambda_5\epsilon_5 = 0,$$
then
$$f(\lambda_1\epsilon_1 + \lambda_2\epsilon_2 + \lambda_3\epsilon_3 + \lambda_4\epsilon_4 + \lambda_5\epsilon_5) = \lambda_1f(\epsilon_1) + \lambda_2f(\epsilon_2) + \lambda_3f(\epsilon_3) + \lambda_4f(\epsilon_4) + \lambda_5f(\epsilon_5) = 0$$
$$\lambda_1f(\epsilon_5) = 0.$$
Since $\epsilon_5 \not\in \mathtt{Ker}(f)$, we have that $\lambda_5 = 0$. Thus,
$$\lambda_1\epsilon_1 + \lambda_2\epsilon_2 + \lambda_3\epsilon_3 + \lambda_4\epsilon_4 = 0.$$
But the only way for this to happen is if $\lambda_1 = \lambda_2 = \lambda_3 = \lambda_4 = 0$ (why?) so $\{\epsilon_1, \epsilon_2, \epsilon_3, \epsilon_4, \epsilon_5\}$ is linearly independent.
Now, remember that the matrix of $f$ with respect to this basis will be $\left[f(\epsilon_1)|f(\epsilon_2)|f(\epsilon_3)|f(\epsilon_4)|f(\epsilon_5)\right]$.
Question 2
We observe that $f(V) = MV
= \left[\begin{array}{ccccc}
a & a & a & a & a \\
b & b & b & b & b \\
c & c & c & c & c \\
d & d & d & d & d \\
e & e & e & e & e \\
\end{array}\right]
\left[\begin{array}{c}
a \\
b \\
c \\
d \\
e \\
\end{array}\right] = a\left[\begin{array}{c}
a \\
b \\
c \\
d \\
e \\
\end{array}\right] + b\left[\begin{array}{c}
a \\
b \\
c \\
d \\
e \\
\end{array}\right] + c\left[\begin{array}{c}
a \\
b \\
c \\
d \\
e \\
\end{array}\right] + d\left[\begin{array}{c}
a \\
b \\
c \\
d \\
e \\
\end{array}\right] + e\left[\begin{array}{c}
a \\
b \\
c \\
d \\
e \\
\end{array}\right] = (a+b+c+d+e)\left[\begin{array}{c}
a \\
b \\
c \\
d \\
e \\
\end{array}\right]$
What does this tell use about the basis we need and the value of $\gamma$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$A$ is not similar to a diagonal matrix over the reals Let $A = \begin{bmatrix} 6 & -3 & -2 \\ 4 & -1 & -2 \\ 10 & -5 & -3 \end{bmatrix} $ then $A$ is not similar to a diagonal matrix over the reals and it is not similar to a diagonal matrix over the complexes.
We know that $A$ is similar to a diagonal matrix over the reals(complexes) if there exist $D$ diagonal matriz and $P$ invertible matrix both $n \times n$ with real entries (complex entries) such that $A = PDP^{-1}$
I find that the inverse of $A$ is $A^{-1} = \frac{1}{2}\begin{bmatrix} -7 & 1 & 4 \\ -8 & 2 & 4 \\ -10 & 0 & 6 \end{bmatrix}$ and i diagonalize the matrix $A$ and got $A = PBP^{-1}$ with $ P = \begin{bmatrix} 1 & \frac{3}{5}-\frac{i}{2} & \frac{3}{5}+\frac{i}{2} \\ 0 & \frac{3}{5}-\frac{i}{2} & \frac{3}{5}+\frac{i}{2} \\ 2 & 1 & 1 \end{bmatrix} $ , $ P^{-1} = \begin{bmatrix} 1 & -1 & 0 \\ -1 + 3i & 1-\frac{i}{2} & \frac{1}{2}-\frac{3i}{2} \\ -1 - 3i & 1+\frac{i}{2} & \frac{1}{2}+\frac{3i}{2} \end{bmatrix} $ and finally i got $D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -i & 0 \\ 0 & 0 & i \end{bmatrix}$ then A is diagonalizable but my question is how can i conclude that $A$ is not similar to a diagonalize matrix over the reals? , please some help for this.
| The entries on the diagonal matrix are unique, take the characteristic polynomial of $A$ and $D$ which are the same since they are similar matrices (why? $Det(A- y I ) = Det( P D P^{-1} - y I ) = Det ( P ( D - y I ) P^{-1}) = Det (P) Det (D - yI) Det (P) ^{-1}$ ), so the zeros are the same.
Now you notice that the diagonal entries of $D$ are precisely the zeros with multiplicity of the characteristic polynomial, so $p_A$ determine $D$ up to permutation. If there are some $D$ that has complex entries, no $D$ is possible with real entries.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/844724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Series expansion of $\coth x$ using the Fourier transform Hi I have research about the series of coth but all of the solutions emerges from integral on a contour, Could you calculate the fourier transform of coth? Is that possible at all?My goal is to reach : $$\coth x = \frac{1}{x} + 2 \sum_{n=1}^\infty \frac{x}{x^2+\pi^2n^2}$$
through fourier.
| The expansion can be derived from the complex Fourier series of the $2 \pi$-periodic function $f(x) = e^{ax}$ for $- \pi < x < \pi$.
By definition of the complex Fourier series, $$e^{ax} = \lim_{N \to \infty}\sum_{n=-N}^{N} c_{n} e^{inx} $$
where $$ \begin{align} c_{n} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{ax} e^{-inx} \ dx &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{(a-in)x} \ dx \\ &= \frac{1}{2 \pi} \frac{e^{(a-in)x}}{a-in} \Big|^{\pi}_{-\pi} \\ &= \frac{1}{2 \pi} \frac{1}{a-in} \Big( e^{a \pi}e^{-in \pi} - e^{- a \pi}e^{i n \pi} \Big) \\ &= \frac{1}{\pi} \frac{(-1)^{n}}{a-in}\frac{e^{a \pi}-e^{- a \pi}}{2} \\ &= \frac{(-1)^{n}}{\pi} \frac{a+in}{a^{2}+n^{2}} \sinh a \pi \end{align}$$
At $x= \pi$ (a point of discontinuity), $$\frac{e^{a \pi}+e^{- a \pi}}{2} = \cosh a \pi = \lim_{N \to \infty}\sum_{n=-N}^{N} c_{n} (-1)^{n} = \frac{\sinh a \pi}{\pi} \lim_{N \to \infty} \sum_{n=-N}^{N} \frac{a+in}{a^{2}+n^{2}}$$
which implies
$$ \begin{align} \coth a \pi &= \frac{1}{\pi} \lim_{N \to \infty}\sum_{n=-N}^{N} \frac{a+in}{a^{2}+n^{2}} \\ &= \frac{1}{\pi} \lim_{N \to \infty}\sum_{n=-N}^{N} \frac{a}{a^{2}+n^{2}} + \frac{i}{\pi} \lim_{N \to \infty} \sum_{n=-N}^{N} \frac{n}{a^{2}+n^{2}} \\ &= \frac{1}{\pi} \lim_{N \to \infty}\sum_{n=-N}^{N} \frac{a}{a^{2}+n^{2}}+ 0 \\ &= \frac{1}{\pi}\sum_{n=-\infty}^{\infty} \frac{a}{a^{2}+n^{2}} \end{align}$$
And replacing $a$ with $ \displaystyle \frac{x}{\pi}$, we get
$$ \begin{align} \coth x &= \sum_{n=-\infty}^{\infty} \frac{x}{x^{2}+\pi^{2}n^{2}} \\ &= \frac{1}{x} + 2 \sum_{n=1}^{\infty} \frac{x}{x^{2}+\pi^{2}n^{2}} \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/845506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to establish this inequality: $(1-a)(1-b)(1-c) \geq 8abc$ for $a+b+c=1$? Let $a$, $b$, and $c$ be positive real numbers such that $a+b+c = 1$. Then how to establish the following inequality? $$ (1-a)(1-b)(1-c) \geq 8abc.$$
My effort:
Since $a+b+c =1$, we can write $$ (1-a)(1-b)(1-c) = 1 - (a+b+c) +(ab+bc+ca) - abc = 1 - 1 +(ab+bc+ca) - abc = (ab+bc+ca) - abc = abc(a^{-1} + b^{-1} + c^{-1} ) - abc = 3abc (\frac{a^{-1} + b^{-1} + c^{-1}}{3}) - abc.$$ What next?
| @Saaqib Mahmood, I'm expanding on @5xum's answer here. Using $AM\geq HM$,
$$$$${\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\over{3}$$\geq$${3}\over{a+b+c}$
$$$$So, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq9$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of solutions to equation involving floor-function For school I have to solve some problems involving floor-functions, and I have no clue how to solve this one:
for a given $n$, calculate the number of $k$'s, $k\lt n$ such that the number of multiples of 7 between and including $n$ and $n-k$ is bigger then the number of multiples of 7 lower than or equal to k.
I've managed to make it into a formula: $n$ and $k$ are right iff $$ \left \lfloor \frac{n}{7} \right \rfloor-\left\lfloor \frac{n-k}{7}\right \rfloor >\left \lfloor \frac{k}{7} \right \rfloor$$
For $n=17$, these $k$ work (I think): $4,5,6,11,12,13$
Could anyone give me a hint (not the full solution) on how to calculate the number of solutions for a certain n?
| Hint: If $0 \leqslant k \leqslant n-7$, then
$$\left\lfloor \frac{n-(k+7)}{7}\right\rfloor + \left\lfloor \frac{k+7}{7}\right\rfloor = \left\lfloor \frac{n-k}{7}-1\right\rfloor + \left\lfloor \frac{k}{7}+1\right\rfloor = \left\lfloor \frac{n-k}{7}\right\rfloor + \left\lfloor \frac{k}{7}\right\rfloor.$$
Also, $\left\lfloor \frac{k}{7}\right\rfloor = 0$ for $0 \leqslant k < 7$.
For $0 \leqslant k < 7$, we have $\left\lfloor \frac{k}{7}\right\rfloor = 0$, and $\left\lfloor \frac{n}{7}\right\rfloor > \left\lfloor \frac{n-k}{7}\right\rfloor$ if and only if $k > r$, where $r$ is the least non-negative remainder of $n$ modulo $7$. So $\left\lfloor \frac{n}{7}\right\rfloor - \left\lfloor \frac{n-k}{7}\right\rfloor > \left\lfloor \frac{k}{7}\right\rfloor$ if and only if $k = q\cdot 7 + s$, with $r < s < 7$ and $0 \leqslant q < \left\lfloor \frac{n}{7}\right\rfloor$. The number of $k$s is therefore $\left\lfloor\frac{n}{7}\right\rfloor \cdot (6-r)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/846039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How prove this inequality? show that
$$\dfrac{\sqrt{2}}{2}<f(n)=\dfrac{\sqrt{2n+1}-1}{1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\cdots+\dfrac{1}{\sqrt{n}}}<\dfrac{\sqrt{3}}{2}$$
I know this
$$\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}<\dfrac{2}{\sqrt{n}+\sqrt{n-1}}=2(\sqrt{n}-\sqrt{n-1})$$
By the way I can use Stolz lemma find the
$$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}\dfrac{\sqrt{2n+1}-\sqrt{2n-1}}{\dfrac{1}{\sqrt{n}}}=\dfrac{\sqrt{2}}{2}$$
but this can't prove this inequality,Thank you
| The estimate $1/\sqrt{n} < 2(\sqrt{n}-\sqrt{n-1})$ yields
$$\sum_{k=1}^n \frac{1}{\sqrt{k}} < 1 + \sum_{k=2}^n 2(\sqrt{k}-\sqrt{k-1}) = 2\sqrt{n} - 1,$$
and so
$$\frac{\sqrt{2n+1}-1}{\sum_{k=1}^n \frac{1}{\sqrt{k}}} > \frac{\sqrt{2n+1}-1}{2\sqrt{n}-1} > \frac{\sqrt{2}}{2}.$$
For an upper bound, we could look at the analogous $1/\sqrt{n} > 2(\sqrt{n+1}-\sqrt{n})$, which yields
$$\sum_{k=1}^n \frac{1}{\sqrt{k}} > 2(\sqrt{n+1}-1).$$
That would leave
$$\frac{\sqrt{2n+1}-1}{\sqrt{n+1}-1} < \sqrt{3}$$
to be shown, or
$$\sqrt{3}-1 < \sqrt{3(n+1)}-\sqrt{2n+1} = \frac{n+2}{\sqrt{3(n+1)}+\sqrt{2n+1}}.$$
A trivial estimate shows the right hand side is $> \frac{\sqrt{n+2}}{\sqrt{3}+\sqrt{2}}$, which yields the inequality for $n \geqslant 7$.
The cases $1 \leqslant n \leqslant 6$ can be verified by hand.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all the prime integer solutions to $q^2(p-1)=(p+1)(q+1)$ Let $p,q$ be prime numbers. Find all the integer solutions to:
$$q^2(p-1) = (p+1)(q+1)$$
I am almost sure that $q=2$,$p=7$ is the only solution. Thus I assumed that $p$ and $q$ were both odd to reach a contradiction, but so far I haven't made any progress that way.
| $q^2(p−1)=(p+1)(q+1)$
This can be written as:
$\frac{q^2}{q+1} = \frac{p+1}{p-1} = 1 + \frac{2}{p-1}$
In other words, subtracting 1 from both sides,
$\frac{q^2 - q - 1}{q+1} = \frac{2}{p-1}$
Two cases here:
1) p = 2. So $q^2 - q - 1 = q+1$.
Gives: $q^2 - 2q - 2 = 0$. Easy to see this has no integer solutions (just solve the quadratic equation).
2) p is an odd prime. Let p = 2k + 1.
Then $q + 1 = k(q^2 - q - 1)$.
So $(q+1)(k+1) = kq^2$.
Now, q and q+1 are co-prime. So are k+1 and k. Unless k or q is 1, this can only mean:
$q+1 = k$ and $k+1 = q^2$
So $q+2 = q^2$
Solving the quadratic, q = 2 or q = -1. 'q is prime' makes us reject q = -1 case. So q = 2. So p = 2k+1 = 2(q+1) + 1 = 7.
Now, suppose k = 1. Then we get:
$2(q+1) = q^2$. This again gives $q^2 - 2q - 2 = 0$ which has no integer solutions.
q = 1 gives:
$2(k+1) = k$. This gives k = -1/2, which is not an integer.
So p = 7, q = 2 is the only solution to this.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\lim\limits_{x \to 0} \sinh(x)/x =1$.
Prove that $ \lim\limits_{x \to 0} \frac{\sinh x}{x} =1.$
I am having some trouble proving this without derivative. Some help would be much appreciate!
| Rewrite using known limit for $e$:
$$
\lim_{x \rightarrow 0} \frac{\sinh(x)}{x} = \lim_{x \rightarrow 0} \frac{e^x - e^{-x}}{2x} = \lim_{x \rightarrow 0} \frac{e^x - 1 + 1 - e^{-x}}{2x} = \frac{1}{2} \lim \frac{e^x - 1}{x} - \frac{1}{2} \lim_{x \rightarrow 0} \frac{e^{-x}-1}{x}
$$
$$
= \frac{1}{2} \lim \frac{e^x - 1}{x} + \frac{1}{2} \lim_{x \rightarrow 0} \frac{e^x-1}{x} = \lim_{x \rightarrow 0} \frac{e^x - 1}{x} = 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Infinite limit terms under root Suggest me a hint to solve:$$\psi=\lim_{x\to0}{\frac{\sqrt{1-\cos x+\sqrt{1-\cos x+\sqrt{1-\cos x+\cdots}}}-1}{x^2}}$$
My try,
| Let $\displaystyle y=\sqrt{1-\cos x+\sqrt{1-\cos x+\sqrt{1-\cos x+\cdots}}}$
$$\implies y^2=1-\cos x+y\iff y^2-y+\cos x-1=0$$
$$y=\frac{1\pm\sqrt{1-4(\cos x-1)}}2=\frac{1\pm\sqrt{5-4\cos x}}2$$
For $y\ge0$ and $\displaystyle x\to0,x\ne0\cos x<1,\sqrt{5-4\cos x}>1\implies 1-\sqrt{5-4\cos x}<0$ hence can be safely discarded
So, $$\lim_{x\to0}{\frac{\sqrt{1-\cos x+\sqrt{1-\cos x+\sqrt{1-\cos x+\cdots}}}-1}{x^2}}$$
$$=\lim_{x\to0}\frac{\sqrt{5-4\cos x}-1}{2x^2}$$
$$=\lim_{x\to0}\frac{5-4\cos x-1}{2x^2}\cdot\frac1{\lim_{x\to0}\sqrt{5-4\cos x}+1}$$
Now $$\lim_{x\to0}\frac{5-4\cos x-1}{2x^2}=\frac42\lim_{x\to0}\frac{1-\cos x}{x^2}$$
$x=2y\implies$ $$\lim_{x\to0}\frac{1-\cos x}{x^2}=\lim_{y\to0}\frac{1-\cos2y}{(2y)^2}=\frac24\left(\lim_{y\to0}\frac{\sin y}y\right)^2=\cdots$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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finding $ \int_{C(0,2)^+} \frac{z^3}{z^3+z^2-z-1} $ I want to find
$$
I \ = \ \int_{C(0,2)^+} \frac{z^3}{z^3+z^2-z-1}
$$
First of all, I know that $z^3+z^2-z-1 = (z+1)^2(z-1)$. I split up the integral as a sum of residues:
$$
I \ = \
2\pi i \cdot Res_{z=-1}\frac{z^3}{(z+1)^2\cdot(z-1)}
\ + 2 \pi i \cdot Res_{z=1}\frac{z^3}{(z+1)^2\cdot(z-1)}
$$
The rightmost part becomes $2 \pi \cdot 1^3/(1+1)^2 \ = \ \pi i /2$. The same trick can't be applied for the the other part. There was another lemma that was useful though:
$$
Res_{z=1}f(x) \ = \ \frac{1}{(2-1)!} \cdot
\lim_{z \rightarrow -1}\left( (z+1)^2 \cdot \frac{z^3}{(z+1)^2(z-1)}\right)
\ = \ \frac{1}{(1+1)^2} \ = \ \frac{1}{4}
$$
Now I should multiply this by $2 \pi i$, which gives me $\pi \cdot i /2$.
Adding gives us $\pi i $.
Could you please check what I did and tell me if I this is the right way to solve this?
| Your calculation of the residue of $\frac{z^3}{(z+1)^2(z-1)}$ at $z=+1$ is correct but I think there is an error with your calculation at $z=-1$. (I am going to be slightly informal here but this can be easily made formal.)
If you have a Laurent series expansion $\sum_{n=-\infty}^{+\infty} a_n(z-a)^n$, then the residue at $a$ is simply the coefficient $a_{-1}$.
We are interested in calculating the residue of
$$\frac{z^3}{(z+1)^2(z-1)}$$
at $z=\pm 1$.
If $z=-1$, then observe that we can rewrite this expression as a quotient $\frac{\frac{z^3}{(z-1)}}{(z+1)^2}$. The numerator $\frac{z^3}{(z-1)}$ is holomorphic in a neighborhood of $z=-1$ and thus admits a Taylor series expansion $\sum_{n=0}^{\infty} a_n(z+1)^n$. We are interested in the coefficient of $\frac{1}{z+1}$ in the Laurent series expansion of $\frac{\frac{z^3}{(z-1)}}{(z+1)^2}$ - however, this is simply $a_1$!
Now, to fix your calculation, note that $a_1$ is the derivative at $z=-1$ of $\frac{z^3}{(z-1)}$, and this is also the residue at $z=-1$ of $\frac{z^3}{(z+1)^2(z-1)}$. (Exercise)
Your calculation of the residue of $\frac{z^3}{(z+1)^2(z-1)}$ at $z=+1$ is correct, as I stated, and can be calculated using the reasoning above. As you note, we this time rewrite $\frac{z^3}{(z+1)^2(z-1)}$ as $\frac{\frac{z^3}{(z+1)^2}}{z-1}$ where the numerator is holomorphic in a neighborhood of $z=+1$. Thus, it is evident that once we expand this as a Laurent series at $z=+1$, the residue will simply be the value of the numerator at $z=+1$, i.e., the value of $\frac{z^3}{(z+1)^2}$ at $z=+1$, i.e., $\frac{1}{4}$.
The final answer can be calculated (as you already observe) by adding the residues of $\frac{z^3}{(z+1)^2(z-1)}$ at $z=\pm 1$ and multiplying the result by $2\pi i$. (I don't think your final answer is correct as your calculation of the residue at $z=-1$ is not correct so the final answer needs to be recalculated.)
Hope this helps!
| {
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Solving a quadratic trigonometric equation? The equation is $6 \cos^2x+\cos x=1$,
My work:
$6x^2+x-1=0$
$(3x-1)(2x+1)$
$3x-1=0 ∨ 2x+1=0$
$x=\frac{1}{3} ∨ x= \frac{-1}{2}$
But I do not know how to progress further.
| It should be: $$\eqalign{6\cos^2x+\cos x-1=0
&\iff
(3\cos x-1)(2\cos x+1)=0
\\
&\iff 3\cos x-1=0,\;\; 2\cos x+1=0
\\
&\iff\cos x=\tfrac13,\;\; \cos x=-\tfrac12.}$$
Can you take it from here?
| {
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Trigonometry - Finding value of expression Question: Given $\sin x + \sin^2 x = 1$ then find the value of $$\cos^{12} x + 3\cos^{10} x + 3\cos^8 x + 6\cos^6 x + 2\cos^4 x + \cos^2 x -2$$
I have no idea where to start on this question. Please help me!
| $\sin x+\sin ^2x=1$, let's give $\sin x$ another name, $\alpha$.
$\alpha+\alpha ^2 =1$.
$\alpha^2 + \alpha -1=0$.
That gives us either
$\alpha_1=\frac{-1+\sqrt{5}}{2}$ or $\alpha_2=\frac{-1-\sqrt{5}}{2}$.
Notice that $\sin(x) = \frac{-1-\sqrt{5}}{2}$ can't be true. So the only valid solution is $\alpha_1$.
Given that $\sin^2x=(\frac{-1+\sqrt{5}}{2})^2$, this means that $\cos^2x=1-(\frac{-1+\sqrt{5}}{2})^2$
Can you take it from here?
| {
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Evaluate position of first secondary maximum of $\frac{\sin N (x/2)}{\sin (x/2)}$ The function
$$f(x) = \displaystyle \left | \frac{\sin \left( N \displaystyle \frac{x}{2} \right)}{\sin \left( \displaystyle \frac{x}{2} \right)} \right |$$
when evaluated for $x > 0$, has its first zero in
$$x = \frac{2 \pi}{N}$$
But the first secondary maximum is in
$$x = \frac{3 \pi}{N}$$
and not in
$$x = \frac{\pi}{N}$$
because here the function is decreasing from the maximum reached for $x = 0$ (where $f(x) = N$).
My questions are: why? And how can it be seen?
I tried to compute the derivative of $f(x)$:
$$f'(x) = \displaystyle \frac{ \displaystyle \frac{N}{2} \cos \left( \frac{N x}{2} \right) \sin \left( \frac{x}{2} \right) - \frac{1}{2} \sin \left( \frac{N x}{2} \right) \cos \left( \frac{x}{2} \right)}{\sin^2 \left( \displaystyle \frac{x}{2} \right)}$$
for $0 \leq x < \displaystyle \frac{2 \pi}{N}$. It is not immediately visible that it is always negative in that range. I tried to rearrange the numerator with the Werner formulas, but I did not obtain a more clear expression.
Are there any other methods or assumptions that can be done in order to prove that between $x = 0$ and $x = \displaystyle \frac{2 \pi}{N}$ the function $f(x)$ is monothonically decreasing?
| By using your formula for $f'(x)$ we have to show that
$$N \cos\left(\frac{Nx}{2}\right) \sin\left(\frac{x}{2}\right) \le \sin\left(\frac{Nx}{2}\right) \cos\left(\frac{x}{2}\right)$$
when $0 \le x < \frac{2\pi}{N}$. This clearly holds if $N = 1$. Assume thus that $N \ge 2$ and write $y = x/2$.
If $\frac{\pi}{2} \le Ny < \pi$, then $\cos(Ny) \le 0$, and the inequality is trivial.
Assume therefore that $0 \le Ny < \frac{\pi}{2}$. The inequality is then equivalent to
$$N\tan(y) \le \tan(N y).$$
This is true because $\tan(y)$ is a superadditive function in that range. (It is enough to check that it is convex and $0$ at $0$.)
| {
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deriving $y=\sqrt{x+\sqrt{x+\sqrt{x}}\cdots} $ How to derive $y=\sqrt{x+\sqrt{x+\sqrt{x}}\cdots}$ at $x=6$ ?
| We have:
$$\begin{equation}
\begin{split}
y=\sqrt{6+\sqrt{6+\sqrt{6}}\cdots}\Leftrightarrow
y^2&=6+\sqrt{6+\sqrt{6+\sqrt{6}}\cdots}, \\&=6+y.
\end{split}
\end{equation}$$
Hence:
Find $y$ in $$y^2-y-6=0.$$
Use the following: $y^2-y-6=(y-3)(y+2).$
Hence: $y=3$ or $y=-2$.
Since, $\sqrt{6+\sqrt{6+\sqrt{6}}\cdots}\geqslant0$ then $\sqrt{6+\sqrt{6+\sqrt{6}}\cdots}=3.$
| {
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Combination of quadratic and arithmetic series Problem:
Calculate $\dfrac{1^2+2^2+3^2+4^2+\cdots+23333330^2}{1+2+3+4+\cdots+23333330}$.
Attempt:
I know the denominator is arithmetic series and equals
$$\frac{n}{2}(T_1+T_n)=\frac{23333330}{2}(1+23333330)=272222156111115,$$
but how do I calculate the numerator without using a calculator?
| Intuitively,
\begin{align}
S_1&=\frac{1^2}{1}=1=\frac{3}{3}\\
S_2&=\frac{1^2+2^2}{1+2}=\frac{5}{3}\\
S_3&=\frac{1^2+2^2+3^2}{1+2+3}=\frac{7}{3}\\
S_4&=\frac{1^2+2^2+3^2+4^2}{1+2+3+4}=3=\frac{9}{3}\\
\vdots\\
\large\color{blue}{S_n}&\color{blue}{=\frac{2n+1}{3}}.
\end{align}
| {
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The points $(0,0)$, $(a, 11)$, and $(b,37)$ are vertices of an equilateral triangle. Find the product $ab$. "The points $(0,0),\;(a, 11), \text{ and } (b,37)$ are vertices of an equilateral triangle. Find the product $ab$."
I'm not sure how to start this problem. I of course drew out an equilateral triangle with those points, but i'm not sure what information I can draw from them. We know that the sides all have equal length obviously. I'm not sure what to do.
| I guess we obtain the following equations:
$a^2+11^2 = b^2+37^2$
$a^2+11^2 = (a-b)^2 + (37-11)^2$
These simplify to:
$a^2-b^2 = 1248$
$2ab = b^2 + 555$
We can solve the second equation for $a$, and obtain $a=\frac{b^2+555}{2b}$. Substituting this into the first equation, we obtain:
$\left(\frac{b^2+555}{2b}\right)^2 - b^2 = 1248$
Multiplying by $4b^2$, this becomes:
$(b^2+555)^2 - 4b^4 = 4992b^2$,
or:
$3b^4+3882b^2-308025=0$
Now we can use the quadratic formula to obtain:
$b^2 = \frac{-3882\pm\sqrt{3882^2+4\cdot 3\cdot 308025}}{6}$
We need a positive answer, so we choose the plus sign, and get:
$b^2= 75 \\
b = 5\sqrt{3} \\
a^2 = 1323 \\
a=21\sqrt{3}$
Finally, we want the product $ab$, which equals $5\sqrt{3}\cdot 21\sqrt{3} = 315$
| {
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solve $3\sin\theta+4\cos\theta=0$ Solve for $0 < \theta < 360$
Question
$3 \sin \theta + 4 \cos \theta = 0$
Please help. I really can't figure this out
Thanks :)
What I have tried
I tried using the a $\sin \theta + b \cos \theta = r \sin (\theta + a)$ rule but didn't work out (or I might have done it wrong)
| Divide both sides by 5, and observe that the coefficients satisfy
$\left(\frac{3}{5}\right)^2+ \left(\frac{4}{5}\right)^2=1.$ Consequently there is some angle $\alpha$ such that $\cos \alpha = 3/5, \sin\alpha =4/5$. (For comparison with other answers, note that $\tan\alpha = 4/3$). With this in mind we can use the sum-to-product identity to write
$$ \sin(\alpha+\theta) = \cos\alpha \sin \theta+\sin\alpha \cos\theta = \frac{3}{5}\sin\theta+\frac{4}{5}\cos\theta = 0.$$ So all that's left is to find the solutions of this equation.
| {
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Find the value of $\frac{S_{5}S_{2}}{S_{7}}$ If $a$, $b$, $c$ $\in \mathbb R$, we define $S_{k}=\frac{a^k+b^k+c^k}{k}$ (where $k$ is a non-negative integer). Given that $S_{1}=0$, find the value of $$\frac{S_{5}S_{2}}{S_{7}}$$
I tried:
We are given that $S_{1}=0$ i.e. $a + b + c=0$ $\implies a^3+b^3+c^3=3abc$ or $S_{3}=abc$. Similarly, for getting a relation for $S_{2}$ I squared the given condition. However, for higher powers, finding a condition becomes tedious. Even after raising the given equation to power $5$ and $7$ and then substituting for $S_{5}$ and $S_{2}$, I'm not arriving at a particular answer. However, I think that the fact that the degree of numerator and denominator in $\frac{S_{5}S_{2}}{S_{7}}$ is equal can somehow be used but I'm not getting it. Please Help!
Thanks!
| Observe that
$$a^{n+3}+b^{n+3}+c^{n+3}=\\(a+b+c)(a^{n+2}+b^{n+2}+c^{n+2})-(ab+ac+bc)(a^{n+1}+b^{n+1}+c^{n+1})+abc(a^n+b^n+c^n)$$
So since $a+b+c=0$ we have
$$a^2+b^2+c^2=-2(ab+ac+bc)\\ a^3+b^3+c^3=3abc\\
a^4+b^4+c^4=2(ab+ac+bc)^2\\
a^5+b^5+c^5=-5abc(ab+ac+bc)\\
a^6+b^6+c^6=-2(ab+ac+bc)^3+3(abc)^2\\
a^7+b^7+c^7=7abc(ab+ac+bc)^2$$
From where
$S_5S_2-S_7=\frac{1}{10}10abc(ab+ac+bc)^2-\frac{1}{7}7abc(ab+ac+bc)^2=0$
| {
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Prove Basis for symmetric matrix. **Let V be the vector subspace of M$_{2}$ ($\mathbb{R})$ consisting
of all symmetric matrices, That is
A$^{t}$ = A.
1) Show that
$\clubsuit$= $\left\{ \left(\begin{array}{cc}
1 & -2\\
-2 & 1
\end{array}\right),\left(\begin{array}{cc}
2 & 1\\
1 & 3
\end{array}\right),\left(\begin{array}{cc}
4 & -1\\
-1 & -5
\end{array}\right)\right\} $ is a basis for V.
2) Find the co-ordinates of Z = $\left(\begin{array}{cc}
4 & -11\\
-11 & -7
\end{array}\right)$ with respect to this Basis.**
For Part 1) could we argue that the generalised form of a symmetric
matrix in M$_{2}$ ($\mathbb{R})$ would be
something like $\left(\begin{array}{cc}
x & z\\
z & y
\end{array}\right)$: x,y,z $\epsilon\mathbb{R}$. Then if $\left(\begin{array}{cc}
1 & -2\\
-2 & 1
\end{array}\right),$$\left(\begin{array}{cc}
2 & 1\\
1 & 3
\end{array}\right)$, $\left(\begin{array}{cc}
4 & -1\\
-1 & -5
\end{array}\right)$ are linearly independent and spanning then it is
a basis for V , so could we get a set of equations like
$\left(\begin{array}{cc}
x & z\\
z & y
\end{array}\right)$= A$\left(\begin{array}{cc}
1 & -2\\
-2 & 1
\end{array}\right)$+ B $\left(\begin{array}{cc}
2 & 1\\
1 & 3
\end{array}\right)$+ C$\left(\begin{array}{cc}
4 & -1\\
-1 & -5
\end{array}\right)$ - (1)
Giving
x= A + 2B +4C
y= A + 3B -5C
z = -2A+B -C
So if this system of equations has a solution not all zero, then $\clubsuit$
is a basis for V.
2) For part 2) would we have a similar approach to part one solving a set of values in (1) but replacing the LHS of (1) with Z?.
| For 1) it is enough to show that they are spanning all the space or independent, since as if they are independent and the set of $2$ by $2$ symmetric matrices is a vector space of dimension $3$, then in-dependency results in spanning also. Note that if
$$
T =\left(
\begin{matrix}
1 & 2 &4\\ 1& 3& -5\\ -2 &1 &-1\\
\end{matrix}\right)
$$
then $\det T \ne 0$
For 2) your trend is correct and you should find the coefficients in linear representation, i.e.
$$
\left(
\begin{matrix}
1 & 2 &4\\ 1& 3& -5\\ -2 &1 &-1\\
\end{matrix}\right)
\left(
\begin{matrix}
x\\y\\z
\end{matrix}\right)=
\left(
\begin{matrix}
4\\-7\\-11
\end{matrix}\right).
$$
| {
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Integration problem $\displaystyle \int \frac{dx}{x(x^3+8)}$ $$\int \frac{dx}{x(x^3+8)}$$
I think I'm supposed to use partial fractions, but I am unsure of how to start the problem. Any help would be appreciated.
| Hint: Note that $\displaystyle x^3+8=(x+2)(x^2-2x+4)$. It's because you have $a^3+b^3=(a+b)(a^2-ab+b^2)$. For $x^2-2x+4$ you have $\Delta<0$ so you should find partial fractions:
$$\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x^2-2x+4}+\frac{Dx}{x^2-2x+4}$$
| {
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Use an appropriate Half-Angle Formula to find the exact value of $\sin 105^\circ$ Is this correct? $$\sin 105^\circ = \frac{\sqrt{2+\sqrt{3}}}{4}$$
Because $105^\circ$ is in the second quadrant the square root after the equal sign is $+$. However, because $\cos 210^\circ$ is negative, it value is negative $-\frac{\sqrt{3}}{2}$.
Is my thought process correct?
Also, can I factor out $\frac{1}{4}$?
| This is an equivalent way to get the same answer as above:
$\displaystyle\sin 105^{\circ}=\sqrt{\frac{1-\cos210^{\circ}}{2}}
=\sqrt{\frac{1-(-\frac{\sqrt{3}}{2})}{2}}=\sqrt{\frac{2+\sqrt{3}}{4}}=\frac{\sqrt{2+\sqrt{3}}}{2}$.
Notice that using the addition formula for the sine gives a different-looking answer:
$\sin 105^{\circ}=\sin(60^{\circ}+45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$.
| {
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$\ \sqrt{x+39}-\sqrt{x+7}=4 $ So I tried to solve this problem for x
$\ \sqrt{x+39}-\sqrt{x+7}=4 $
I multiplied both sides ($\ \sqrt{m}\cdot\sqrt{n}=\sqrt{mn} $)
$\ (\sqrt{x+39}-\sqrt{x+7})^2=16 $
$\ (x+39)-2(x^2+46x+273)-(x+7) $
$\ 0x+32+(-2x^2-92x-546) $
$\ -2x^2-92x-514 $
divide the 2 out
$\ x^2+46x+257=-8 $
$\ x^2+46x+265=0 $
Use the quadratic formula (or scientific calculator) and the answers are -6.752 and -39.248. I know the answer is exactly -3. What went wrong?
| You made two mistakes when squaring. The correct equation is
$$(x + 39) - 2\sqrt{(x+39)(x+7)} + (x + 7) = 16$$
This can be rearranged as
$$2x + 30 = 2\sqrt{(x+39)(x+7)}$$
Dividing by 2:
$$x + 15 = \sqrt{(x+39)(x+7)}$$
Square again
$$x^2 + 30x + 225 = (x+39)(x + 7)$$
$$= x^2 + 46x + 273$$
So your equation is the same as
$$30x + 225 = 46x + 273$$
So $16x = -48$ and therefore $x = -3$.
When squaring equations always plug in your answer to make sure it's the actual solution. And in this case it is.
| {
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Solution of the Legendre's ODE using Frobenius Method This is the Legendre's differential equation given in my book:
$(1-x)^{2}\ddot{y}-2x\dot{y}+k(k+1)y=0$
I solved this equation by taking:
$y=x^{c}\{a_{0}+a_{1}x+a_{2}x^{2}+.....+a_{r}x^{r}+.....\}$
Therefore, each term in the equation becomes (I deliberately didn't use the sum notation to see it clearly):
$k(k+1)y=k(k+1)a_{0}x^{c}+k(k+1)a_{1}x^{c+1}+k(k+1)a_{2}x^{c+2}+.....+k(k+1)a_{r}x^{c+r}+.....$
$-2x\dot{y}=-2ca_{0}x^{c}-2(c+1)a_{1}x^{c+1}-2(c+2)a_{2}x^{c+2}-.....-2(c+r)a_{r}x^{c+r}-.....$
$\ddot{y}=(c-1)ca_{0}x^{c-2}+c(c+1)a_{1}x^{c-1}+(c+1)(c+2)a_{2}x^{c}+.....+(c+r-1)(c+r)a_{r}x^{c+r-2}+.....$
$-x^{2}\ddot{y}=-(c-1)ca_{0}x^{c}-c(c+1)a_{1}x^{c+1}-(c+1)(c+2)a_{2}x^{c+2}-.....-(c+r-1)(c+r)a_{r}x^{c+r}-.....$
The indicial equation is $(c-1)ca_{0}x^{c-2}=0$. Therefore, there are two solutions for $c=0$ and $c=1$. Since,
$(c+r+2)(c+r+1)a_{r+2}+k(k+1)a_{r}-2(c+r)a_{r}-(c+r)(c+r-1)a_{r}=0$
general recurrence equation is as follows:
$a_{r+2}=\frac{[(c+r-k)(c+r+k+1)]a_{r}}{(c+r+2)(c+r+1)}$
There is no problem so far. But my book gives the following answer for $c=0$:
$y=a_{0}\left\{ 1-\frac{k(k+1)}{2}x^{2}+\frac{k(k-2)(k+1)(k+3)}{4!}x^{4}-.....\right\}$
This is only possible for $a_1=0$. But I can't see that. I will be very glad if anyone can show me why $a_1$ should be equal to zero when $c=0$.
Answer:
I solved both for $c=0$ and $c=1$ as @Semiclassical suggested. For $c=0$, $a_1$ is indeterminate because of the term: $c(c+1)a_1x^{c-1}$ (It can be any value). Therefore, the solution is:
$u=a_{0}\left\{ 1-\frac{k(k+1)}{2!}x^{2}+\frac{k(k-2)(k+1)(k+3)}{4!}x^{4}-.....\right\} +a_{1}\left\{ x-\frac{(k-1)(k+2)}{3!}x^{3}+\frac{(k-3)(k-1)(k+2)(k+4)}{5!}x^{5}-.....\right\}$
For $c=1$, $a_1=0$ because of the term: $c(c+1)a_1x^{c-1}$, thus the solution is:
$w=a_{0}\left\{ x-\frac{(k-1)(k+2)}{3!}x^{3}+\frac{(k-3)(k-1)(k+2)(k+4)}{5!}x^{5}-.....\right\}$
As it is seen this is not a distinct solution. It is already available in the solution for $c=0$. Therefore, the solution for $c=0$ in my book is incomplete. Actually, it should give the complete solution as above. I learned that this happens when indicial roots differ by an integer value.
| When $c=0$, the recurrence doesn't determine $a_0$ or $a_1$. There is a two-parameter family of solutions, and you can use $a_0$ and $a_1$ as the parameters. So you can choose $a_0$ and $a_1$ to be whatever you like.
It's convenient to take the two basic solutions to be the ones with $(a_0,a_1) = (1,0)$ and $(0,1)$; every solution is then a linear combination of these two.
When you have a regular singular point and the indicial roots differ by an integer, there can be a logarithmic term that complicates things, but that doesn't happen in this case ($x=0$ is not a singular point).
| {
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Does this sequence consist of squares of integers? Question:
let sequence $\{x_{n}\}$ such $$x_{0}=0,x_{1}=1,x_{2}=0,x_{3}=1$$
and such
$$x_{n+3}=\dfrac{(n^2+n+1)(n+1)}{n}x_{n+2}+(n^2+n+1)x_{n+1}-\dfrac{n+1}{n}x_{n}$$
show that:$ x_{n}$ are all square number?
My idea: I have
$$x_{4}=3\cdot 2-2\cdot 1=4=2^2$$
$$x_{5}=\dfrac{21}{2}\cdot 4+(2^2+2+1)\cdot 1=49=7^2$$
and so on,but for all $n$,How prove it?
Thank you
| In fact $x_{n+1}=a_n^2$, where $\{a_n\}$ are determined by the recurrence relation
$$a_{n+1}=n a_n+a_{n-1}$$
with the initial conditions $a_0=1$, $a_1=0$. This can be exprerimentally discovered using the online encyclopedia of integer sequences, and then the proof can be easily carried out by induction.
Proof: The only nontrivial induction step:
\begin{align}
x_{n+3}=\frac{(n^2+n+1)(n+1)}{n}a_{n+1}^2+(n^2+n+1)a_{n}^2-\frac{n+1}{n}a_{n-1}^2=\ldots=a_{n+2}^2,
\end{align}
where the dots denote expressing $a_n$ and $a_{n-1}$ in terms of $a_{n+2}$ and $a_{n+1}$. $\quad \blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/874873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluation of $ \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ Evaluation of $\displaystyle \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$
$\bf{My\; Try::}$ Given $\displaystyle \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx = \int \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx$
Now Let $\displaystyle \tan x = t^2\;,$ Then $\sec^2 xdx = 2tdt$ or $\displaystyle dx = \frac{2t}{1+t^4}dt$
So Integral is $\displaystyle \int\frac{t}{1+t}\cdot \frac{2t}{1+t^4}dt = 2\int\frac{t^2}{(1+t)\cdot (1+t^4)}dt$
Now How can I solve after that
Help me
Thanks
| By collecting all the suggestions, you should be able to prove that:
$$\frac{2t}{(t+1)(t^4+1)}=\frac{t}{1+t^4}+\frac{t^3}{1+t^4}+\frac{1-t^2}{1+t^4}-\frac{1}{1+t},$$
and since $(1+t^4)=(1+\sqrt{2}t+t^2)(1-\sqrt{2}t+t^2)$, it follows that:
$$\int\frac{2t\,dt}{(t+1)(t^4+1)}=\frac{1}{2}\arctan t^2+\frac{1}{4}\log(1+t^4)+\frac{1}{2\sqrt{2}}\log\frac{1+\sqrt{2}t+t^2}{1-\sqrt{2}t+t^2}-\log(1+t).$$
$\\$
Addendum
Suggestion for evaluating $\displaystyle{\int \frac{1 - t^2}{t^4 + 1}\,\mathrm{d}t}$:
Write
$$
\begin{aligned}
\int- \frac{t^2 - 1}{t^4 + 1}\,\mathrm{d}t &= -\int \frac{1 - \dfrac{1}{t^2}}{t^2 + \dfrac{1}{t^2}}\,\mathrm{d}t \\
&=-\int \frac{1 - \dfrac{1}{t^2}}{\left(t + \dfrac{1}{t}\right)^2 - 2}\,\mathrm{d}t
\end{aligned}
$$
Now, set $\displaystyle{u = t + \frac{1}{t}}$ and $\mathrm{d}u = \left(1 - \dfrac{1}{t^2}\right)\,\mathrm{d}t$:
$$
\begin{aligned}
-\int \frac{\mathrm{d}u}{u^2 - 2} &= \dfrac{1}{2\sqrt{2}}\ln\left|\frac{u + \sqrt{2}}{u - \sqrt{2}} \right| + C \\
&=\frac{1}{2\sqrt{2}}\ln\left|\frac{t^2 + t\sqrt{2} +1}{t^2 - t\sqrt{2}+1} \right| + C
\end{aligned}
$$
Note that $t^2 + t\sqrt{2}+1 = \left(t + \dfrac{1}{\sqrt{2}}\right)^{\!2} + \dfrac{1}{2} >0$ and $t^2 - t\sqrt{2}+1 = \left(t - \dfrac{1}{\sqrt{2}}\right)^{\!2} + \dfrac{1}{2} >0$, then we can get rid of the absolute value bars:
$$\int \frac{1-t^2}{t^4 + 1}\,\mathrm{d}t = \frac{1}{2\sqrt{2}}\ln\!\left(\dfrac{t^2 + t\sqrt{2}+1}{t^2 - t\sqrt{2}+1}\right) + C $$
This is a very famous technique.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Which one is correct for $\sqrt{-16} \times \sqrt{-1}$? $4$ or $-4$? As we can find in order to evaluate $\sqrt{-16} \times \sqrt{-1}$, we can do it in two ways.
FIRST
\begin{align*}
\sqrt{-16} \times \sqrt{-1} &= \sqrt{(-16) \times (-1)}\\
&= \sqrt{16}\\
&=4
\end{align*}
SECOND
\begin{align*}
\sqrt{-16} \times \sqrt{-1} &= \sqrt{16i^2} \times \sqrt{i^2}\\
&= 4i \times i\\
&=4i^2\\
&=-4
\end{align*}
Incidentally if the above is input in complex mode of Casio scientific calculator, the result comes out as $-4$.
Which of the above solutions is correct?
| The 'square root function' (the function we represent by $\sqrt{\cdot}$) is actually a branch cut of the multiple valued 'square root relation'. The typical square root function returns a complex value with an argument in $[0, \pi)$.
So $\sqrt{-16}\cdot\sqrt{-1}=4i\cdot i=-4$ because $i$ has an argument of $\pi/2$. The other value that we would reasonably see as being "$\sqrt{-16}$" is $-4i$, but this complex number has an argument of $3\pi/2$.
This pattern also persists in the real numbers. We have that $\sqrt{4}=2$. But we could also reasonably say that $\sqrt{4}=-2$. But $2$ has an argument of $0$ while $-2$ has an argument of $\pi$.
| {
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"question_score": "3",
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How to have this equation $s^2-2(p+q+r+2pqr)s+(p^2+q^2+r^2-2(pq+qr+rp)-4)=0$？ Old Question:
For $x,y,z\in N^{+}$, if such $(xy+1)(yz+1)(zx+1)$ is a perfect square ,show that
$$(xy+1),(yz+1),(xz+1)$$ are all perfect square .
and I konw this PDF have solution, http://math.elinkage.net/attachment.php?aid=18
also this have post solution:http://math.elinkage.net/showthread.php?tid=49&my_post_key=7753a3538fd25446d120928362d64e03&language=chs
and I read this paper and this link.
My question: this key solve this problem why consider this follow equation
$$s^2-2(p+q+r+2pqr)s+(p^2+q^2+r^2-2(pq+qr+rp)-4)=0$$
How to have this equation?Thank you
| The inventor of this solution had the idea to consider $(xy+1)(yz+1)(zy+1)$ as the discriminant of a certain quadratic equation. The question is which equation.
Let's say that it has the form $s^2+bs+c=0$, implying that $b^2-4c=(xy+1)(yz+1)(zy+1)$. Suppose we want to keep the symmetry in $x,y,z$, which means $b$ and $c$ should be symmetric functions of $x,y,z$. Expanding and rewriting, we find
$$(xy+1)(yz+1)(zy+1)=x^2y^2z^2+xyz(x+y+z)+xy+yz+zx+1.$$
The first two terms are of the form $g^2+gh$. In number theory it is often useful to rewrite such expressions as $\frac14(4g^2+4gh)=\frac14((2g+h)^2-h^2)$. So far, we have
$$(xy+1)(yz+1)(zy+1)=\frac14((2xyz+x+y+z)^2-(x+y+z)^2)+xy+yz+zx+1.$$
This can be rewritten as
$$\begin{align*}4(xy+1)(yz+1)(zy+1)
&=(2xyz+x+y+z)^2-(x+y+z)^2+4(xy+yz+zx+1)\\
&=(2xyz+x+y+z)^2-(x^2+y^2+z^2)+2(xy+yz+zx)+4.\end{align*}$$
In order to get something of the form $b^2-4c$ we multiply by $4$:
$$16(xy+1)(yz+1)(zy+1)=(2(2xyz+x+y+z))^2-4(x^2+y^2+z^2-2xy-2yz-2zx-4).$$
This means it is a good idea to consider $b=\pm\,2(2xyz+x+y+z)$ and $c=x^2+y^2+z^2-2xy-2yz-2zx-4$, i.e. the equation
$$s^2\pm\,2(2xyz+x+y+z)s+(x^2+y^2+z^2-2(xy+yz+zx)-4)=0.$$
Note that it has discriminant $16(xy+1)(yz+1)(zy+1)$, but since $16$ is a square this is practically what we intended.
Finally, observe that this is almost symmetric in $x,y,z$ and $s$ too: If we take the $-$ sign, this can be rewritten as
$$s^2+x^2+y^2+z^2-2(xy+yz+zx+sx+sy+sz)-4xyz-4=0.$$
This equation is even more symmetric than we hoped, so certainly pointing towards a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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minimal polynomial given an algebraic number I am trying to find the minimal polynomial for the algebraic number $1+\sqrt{2}+\sqrt{3}$. My original thought was just let $\alpha=1+\sqrt{2}+\sqrt{3}$. The method I use though seems very complicated. For example,
$$\alpha^2=(1+\sqrt{2}+\sqrt{3})^2=1+\sqrt{2}+\sqrt{3}+\sqrt{2}+2+\sqrt{2}\sqrt{3}+\sqrt{3}+\sqrt{2}\sqrt{3}+3$$
$$\alpha^2=1+\sqrt{2}+\sqrt{3}+1+\sqrt{2}+\sqrt{3}+4+2\sqrt{2}\sqrt{3}$$
$$\alpha^2=4+2\alpha+2\sqrt{2}\sqrt{3}$$
If I substitute for $\sqrt{2}=\alpha-1-\sqrt{3}$ and $\sqrt{3}=\alpha-1-\sqrt{2}$, I'm still going to end up with $\sqrt{2}$ and $\sqrt{3}$. So I figured I'd square again. Moving the 4 over makes the calculation easier since binomials are easier than trinomials....
$$(\alpha^2-4)^2=(2\alpha+2\sqrt{2}\sqrt{3})^2$$
$$\alpha^4-8\alpha^2+16=4\alpha^2 +8\alpha\sqrt{2}\sqrt{3}+24$$
$$\alpha^4-12a^2-8=4\alpha(2\sqrt{2}\sqrt{3})$$
From the above calculation I see that $2\sqrt{2}\sqrt{3}=\alpha^2-2\alpha-4$
$$\alpha^4-12a^2-8=4\alpha(\alpha^2-2\alpha-4)$$
$$\alpha^4-4\alpha^3-4\alpha^2+16\alpha-8=0$$
But my question is, how do I KNOW this is the minimal polynomial? Yes, it is true now, since I constructed it so, that if $f(x)=x^4-4x^3-4x^2+16x-8$, then $f(\alpha)=0$ Do I simply attempt to factor out an $(x-\alpha)$. In that case, $f(x)=(x-\alpha)g(x)$. Then I show that $g(\alpha)\neq 0$?
| Consider the polynomial
$$(y-\sqrt{2}-\sqrt{3})(y-\sqrt{2}+\sqrt{3})(y+\sqrt{2}-\sqrt{3})(y+\sqrt{2}+\sqrt{3}).$$
The first two terms have product $y^2-2\sqrt{2}y-1$ and the next two have product $y^2+2\sqrt{2}y-1$. Multiply. We get $y^4-10y^2+1$.
We can now see that $\alpha$ is a zero of the polynomial $(x-1)^4-10(x-1)^2+1$.
To show this is minimal, some algebra is useful. One can show that the field $\mathbb{Q}(\sqrt{2})$ has degree $2$ over the rationals, since $\sqrt{2}$ is irrational. And $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree $2$ over $\mathbb{Q}(\sqrt{2})$, since $\sqrt{3}$ cannot be expressed as $s+t\sqrt{2}$ with $s$ and $t$ rational. Thus $\mathbb{Q}(\alpha)$ has degree $4$ over the rationals, so the minimal polynomial has degree $4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Convergence of $a_n=1+1/5+1/9+\ldots+\frac{1}{4n-3}$ Show that the sequence $$a_n=1+1/5+1/9+\ldots+\frac{1}{4n-3}$$ does not converge but the sequence $b_n=\frac{a_n}{n}$ converges.
I can show the first part. For the second part, will it be sufficient to just show that $$\lim \frac{a_n}{n}=0?$$
| note that $a_{{n}}=\sum _{k=1}^{n}{ \frac{1}{4k-3}}$, then
$\sum _{k=1}^{n}{ \frac{1}{4k-3}}=\frac{1}{4}\sum _{k=1}^{n}{ \frac{1}{k-\frac{3}{4}}}$
$>\frac{1}{4}\sum _{k=1}^{n}{ \frac{1}{k}}$
and the last series is the harmonic series, which does not converge. Therefore $a_{{n}}$ does not converge. Similarily, $b_{{n}}=\frac{1}{4}\sum _{k=1}^{n}{ \frac{1}{n(k-\frac{3}{4})}}$, and
$\frac{1}{4}\sum _{k=1}^{n}{ \frac{1}{n(k-\frac{3}{4})}}<\frac{1}{4}\sum _{k=1}^{n}{ \frac{1}{(k-\frac{3}{4})^2}}=\frac{1}{4}\sum _{k=2}^{n}{ \frac{1}{(k-\frac{3}{4})^2}}+4<\frac{1}{4}\sum_{k=1}^{n}{\frac{1}{k^2}}+4$
and the last series converges. Therefore $b_{{n}}$ also converges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the Value of Trigonometric Expression Given $$\begin{align} \frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1 \end{align} \tag{1}$$ Find the value of
$$\begin{align} \frac{\cos^3 \beta}{\cos \alpha}+\frac{\sin ^3\beta}{\sin \alpha} \end{align} \tag{2} $$
I Tried like this: From $(1)$ we have$$\sin\alpha \cos\beta+\cos\alpha \sin\beta=-\sin\beta \cos\beta$$ $\implies$
$$\sin(\alpha+\beta)=-\sin\beta \cos\beta \tag{3}$$
Eq $(2)$ is $$\frac{\frac{\sin\alpha}{4}\left(3\cos\beta+\cos3\beta\right)+\frac{\cos\alpha}{4}\left(3\sin\beta-\sin3\beta\right)}{\sin\alpha \cos\alpha}=\frac{\frac{3}{4}\sin(\alpha+\beta)+\frac{1}{4}\sin(\alpha-3\beta)}{\sin\alpha \cos\alpha}=\frac{\frac{-3}{4}\sin\beta \cos\beta+\frac{1}{4}\sin(\alpha-3\beta)}{\sin\alpha \cos\alpha}$$
I cannot proceed any further..please help me
| Let $x = \dfrac{\cos \alpha}{\cos \beta}$, and $y = \dfrac{\sin \alpha}{\sin \beta}$, then we have: $x + y = - 1$,and we need to find: $S = \dfrac{\cos ^2\beta}{x} + \dfrac{\sin^2\beta}{y}$.
But $(x\cdot \cos \beta)^2 + (y\cdot \sin \beta)^2 = \cos ^2\alpha + \sin ^2\alpha = 1 \to x^2(1-\sin^2\beta) + y^2(1-\cos^2 \beta) = 1 \to x^2\sin^2\beta + y^2\cos^2\beta = x^2 + y^2 - 1$. Thus:
$S = 1\cdot S = -(x+y)\cdot S = -\left(1 +\dfrac{x}{y}\cdot \sin^2\beta + \dfrac{y}{x}\cdot \cos^2\beta\right) = -\left(1+ \dfrac{x^2+y^2-1}{xy}\right) = -\left(1+\dfrac{1-2xy-1}{xy}\right) = 1$.
| {
"language": "en",
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"source": "stackexchange",
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Maximum value of $f(x) = \left|\sqrt{\sin^2 x+2a^2} - \sqrt{2a^2-1-\cos^2 x}\right|\;\;,$ Where $a,x\in \mathbb{R}$ Calculation of Maximum value of $\displaystyle f(x) = \left|\sqrt{\sin^2 x+2a^2} - \sqrt{2a^2-1-\cos^2 x}\right|\;\;,$ Where $a,x\in \mathbb{R}$
$\bf{My\; Try::}$ We Can Write It as $f(x) = \left|\sqrt{2a^2-\cos^2 x+1}-\sqrt{2a^2-\cos^2 x-1}\right|$
Now Let $\left(2a^2-\cos^2 x\right) = b\in \mathbb{R}\;,$ Then Expression is $f(b) = \left|\sqrt{b+1}-\sqrt{b-1}\right|\;,$ where $b\in \mathbb{R}.$
Now How Can I find max. value of $f(b)$
Help me
Thanks
| Hint: Let $g(b)=\sqrt{b+1}-\sqrt{b-1}$ where $b\in [2a^2-1,2a^2]$ by using derivative find max and min of $g$ then result follows.
| {
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How does $-\sqrt {\frac{{2 - \sqrt 2 }}{{2 + \sqrt 2 }}} $ simplify to $1 - \sqrt 2 $? I've the answer for a question in my textbook to be:
$-\sqrt {\frac{{2 - \sqrt 2 }}{{2 + \sqrt 2 }}} $ which i've then simplifed to: $-\sqrt {3 - 2\sqrt 2 } $
However my textbook states $-\sqrt {\frac{{2 - \sqrt 2 }}{{2 + \sqrt 2 }}} = 1 - \sqrt 2 $
How is this?
| $$\frac{2-\sqrt{2}}{2+\sqrt{2}}=\frac{2-\sqrt{2}}{2+\sqrt{2}}\frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{\left(2-\sqrt{2}\right)^{2}}{2}$$
so $$\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}=\frac{2-\sqrt{2}}{\sqrt{2}}=\sqrt{2}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/879840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Number of possible patterns? Using the following rule: Each column and each row must contain at least one point, how many patterns can a 4x4 grid (thus with 16 possible point positions) generate? (this rule would thus make the answer different than the traditional one given by combinations)
More specifically, how many patterns can be made with 8 points on this 4x4 grid?
| There are no positions such that the total number of row and column that does not contains any point is at least $3$ (we will say that those row and column are "avoided").
Number of position that avoid at least $1$ row or column is $8\times\begin{pmatrix}12\\8\end{pmatrix}$. This is because there are $8$ choice of which column or row to avoid, and once that is chosen, there are $12$ position left to put $8$ points.
Intersection of a set of position that avoid a particular row or column and another of such set is always a set of position that avoid at least 2 row and column.
Number of position that avoid at least $2$ row and column is $2\times\begin{pmatrix}4\\2\end{pmatrix}\begin{pmatrix}8\\8\end{pmatrix}+4\times 4\times\begin{pmatrix}9\\8\end{pmatrix}$. The first term is for when we avoid 2 row or avoid both column. Since the situation is the same for the case of both row and the case of both column, there is a factor of 2, and we only need to consider the case of both row: in that case, we have 2 among the 4 rows we can choose to avoid, which leave us with $8$ position left to put $8$ points. The second term is for when we pick 1 row and 1 column to avoid, then there are 4 choice of column and 4 choice of row, and after that is picked, we are left with $9$ position to put $8$ points.
Now use inclusion-exclusion to get number of position that avoid at least 1 row or column to be $8\times\begin{pmatrix}12\\8\end{pmatrix}-2\times\begin{pmatrix}4\\2\end{pmatrix}\begin{pmatrix}8\\8\end{pmatrix}-4\times 4\times\begin{pmatrix}9\\8\end{pmatrix}$.
Number of position ignoring the restriction is $\begin{pmatrix}16\\8\end{pmatrix}$.
Hence total number of allowed position is $\begin{pmatrix}16\\8\end{pmatrix}-8\times\begin{pmatrix}12\\8\end{pmatrix}+2\times\begin{pmatrix}4\\2\end{pmatrix}\begin{pmatrix}8\\8\end{pmatrix}+4\times 4\times\begin{pmatrix}9\\8\end{pmatrix}=\frac{16!}{8!8!}-\frac{12!}{4!7!}+12+144=9066$.
| {
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If the numerator of a fraction is increased by $2$ and the denominator by $1$, it becomes $5/8$.... If the numerator of a fraction is increased by $2$ and the denominator by $1$, it becomes $\displaystyle \frac{5}{8}$ and if the numerator and the denominator of the same fraction are each increased by $1$, the fraction becomes equal to $\displaystyle \frac{1}{2}$. Find the fraction.
I tried,
Let the numerator of the fraction be $x$,
Let the denominator of the fraction be $y$
Therefore, Original fraction is $\displaystyle \frac{x}{y}$.
$$
\left\{
\begin{eqnarray*}
\frac{x+2}{y+1}=\frac{5}{8} \\[2mm]
\frac{x+1}{y+1}=\frac{1}{2} \\
\end{eqnarray*}
\right.
$$
What should I do now?
| We have the equations
$$\frac{x+2}{y+1}=\frac{5}{8}, \ \frac{x+1}{y+1}=\frac{1}{2}=\frac{4}{8}$$
Hence
$$x+1=4, \ y+1=8 \ \implies x=3, \ y=7$$
We are allowed to do so since $(x+2)-(x+1)=5-4$
| {
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Why $ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$? Why $ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$?
I am learning trigonometric identities one identity I have to proof is the next:
$ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$
so I tried to resolve the identity for the left:
$ 1 + \sin^2 \alpha+ \cos^2\alpha - 2\sin\alpha + 2\cos\alpha - 2\sin\alpha\cos\alpha $
$= 1 + 1 - 2\sin\alpha + 2\cos\alpha - 2\sin\alpha\cos\alpha $
$= 2 (1 - \sin\alpha + \cos\alpha - \sin\alpha\cos\alpha)$
And I got stuck, I did not know what to do, so I went to see the problem's answer and I was going fine, the part I was not able to resolve is the next one:
$ 2 (1 - \sin\alpha + \cos\alpha - \sin\alpha\cos\alpha)$
$= 2 (1 + \cos\alpha - \sin\alpha(1+\cos\alpha))$
$= 2 (1 - \sin \alpha)(1+ \cos \alpha)$
So the question is how did the teacher do the last three steps? I cant figure it out.
| Indeed
$$
(1 - \sin \alpha + \cos \alpha)^2 = (1 - \sin \alpha)^2 + \cos^2 \alpha+ 2(1 - \sin \alpha)\cos \alpha
$$
$$
= (1 - 2\sin \alpha + \sin^2\alpha + \cos^2\alpha) + 2(1 - \sin \alpha)\cos \alpha
$$
$$
= 2(1 - \sin \alpha) + 2(1 - \sin \alpha)\cos \alpha = 2(1 - \sin \alpha)(1 + \cos \alpha)
$$
| {
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"url": "https://math.stackexchange.com/questions/881273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\int x^2 \sqrt{x^2-1} dx$ How do I evaluate the following indefinite integral?
$$\int x^2 \sqrt{x^2-1} dx$$
Through integration of parts, I have obtained
$$ \frac{x}{3}(x^2-1)^{3/2} - \frac{1}{3} \int (x^2-1)^{3/2} dx $$
I've attempted evaluating the second term through substitution, where
$$ x = \sec(u)$$
However, I am stuck with
$$ \frac{x}{3}(x^2-1)^{3/2} - \frac{1}{3} \int \tan^4(u) \sec (u) du $$
What would be my next step?
| Let $x = \sec \theta$. Hence, $dx = \sec \theta \tan \theta d\theta$.
$$
\int x^2\sqrt{x^2 - 1}dx = \int \sec^2\theta \tan \theta \sec \theta \tan \theta d\theta = \int \sec^3 \theta \tan^2 \theta d\theta
$$
$$
= \int \sec^3d\theta(\sec^2\theta - 1)d\theta = \int \sec^5\theta d\theta - \int \sec^3 \theta d\theta
$$
In this link
$$
K_n(\theta) := \int \sec^n \theta d\theta = \frac{1}{n-1}\sec^{n-2}\theta\tan \theta + \frac{n-2}{n-1}K_{n-2}(\theta)
$$
Thus,
$$
\int \sec^5\theta d\theta = \dfrac{1}{4}\sec^3d\theta \tan \theta d\theta + \dfrac{3}{4}K_3(\theta)
$$
and
$$
K_3(\theta) = \dfrac{1}{2}\sec \theta \tan \theta + \dfrac{1}{2}\ln(\sec \theta + \tan \theta)
$$
| {
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Factoring $x^4+x^2+2x+6$ We have to factor $x^4+x^2+2x+6$.Factoring through factor theorem is not helpful in this question as the question does not follow the integral root theorem i.e. the root of this expression is not any of the factors of 6. I am totally confused.Please help in factoring it!
| $\begin{eqnarray}{\bf Hint}\quad x^4\!+x^2\!+2x+6\, &=& \qquad x^4 + 4 &+& x^2\!+2x+2\\
&=&\!\! (x^2\!+\!2)^2\!\!-(2x)^2 &+& x^2\!+2x+2\\
&=&\!\! (x^2\!+\!2 -2x)(\color{#c00}{x^2\!+2 + 2x})\!\! &+& \color{#c00}{x^2\!+2x+2}\\
&=&\!\! (x^2\!-2x+3)(\color{#c00}{x^2\!+ 2x\! + 2})\!\!\!\
\end{eqnarray}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
prove that : $\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c}+ \frac{c^2+a^2}{c+a} \geq 3$
For $a^2+b^2+c^2 =3$, with $a$, $b$ and $c$ are positive real numbers, prove that:
$$\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c}+ \frac{c^2+a^2}{c+a} \geq 3.$$
Can any one help me with this problem?
| Applying AM-GM, we get:
$\dfrac{\dfrac{a^2 + b^2}{a+b} + \dfrac{a^2 + c^2}{a+c} + \dfrac{c^2 + b^2}{c+b}}{3} \geq \Bigg( \Bigg(\dfrac{a^2 + b^2}{a+b} \Bigg) \Bigg(\dfrac{a^2 + c^2}{a+c} \Bigg) \Bigg(\dfrac{c^2 + b^2}{c+b} \Bigg)\Bigg)^{\frac{1}{3}}$
$\geq \Bigg( \Bigg(\dfrac{a+ b}{2} \Bigg) \Bigg(\dfrac{a+ c}{2} \Bigg) \Bigg(\dfrac{c + b}{2} \Bigg)\Bigg)^{\frac{1}{3}}$
(Follows from AM-GM again.)
$\geq \frac{1}{2} ((a+b)(b+c)(a+c))^{\frac{1}{3}}$
$\geq \frac{1}{2} ((a+b+c)(ab+bc+ca) -abc)^{\frac{1}{3}} \quad \ldots (1)$
Clearly $(abc)_{max} = \bigg( \dfrac{a^2 + b^2 +c^2}{3}\bigg)^{\frac{3}{2}} = 1$
Now by Lagrange's method of undetermined multipliers it isn't hard to show that subject to the constraint $a^2 +b^2 +c^2 = 3$, the maximum values of $a+b+c$ and $ab +bc +ca$ are both 3.
Together we get:
$\dfrac{\dfrac{a^2 + b^2}{a+b} + \dfrac{a^2 + c^2}{a+c} + \dfrac{c^2 + b^2}{c+b}}{3} \geq \dfrac{1}{2}((3 \times 3) -1)^{\frac{1}{3}}$
And the required inequality follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
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Is this formula for the harmonic numbers true? Is this formula for the harmonic numbers true?
$$H_n = \lim_{s\to 0} \, \int \frac{(s+1)^{(-n-1)}+s-1}{s} \, ds$$
Mathematica:
Clear[n, s]
Monitor[Table[
Limit[Integrate[((s + 1)^(-n - 1) + s - 1)/s, s], s -> 0], {n, 1,
12}], n]
Differences[%]
| I claim that
$$\frac{1}{(s+1)^n s} = \frac{1}{s} - \sum_{j=1}^{n} \frac{1}{(s+1)^j}.$$
Let us prove this by induction. If $n=1$ then one may check that
$$\frac{1}{s (s+1)} = \frac{1}{s} - \frac{1}{s+1}$$
as desired. Now assume the statement is true for $n=k$. Using the inductive hypothesis we get
$$\frac{1}{(s+1)^{k+1} s} = \frac{\frac{1}{(s+1)^k s}}{s+1} = \frac{1}{s (s+1)} - \sum_{j=2}^{k+1} \frac{1}{(s+1)^j}.$$
Performing the same decomposition again gives
$$\frac{1}{(s+1)^{k+1} s} = \frac{1}{s} - \sum_{j=1}^{k+1} \frac{1}{(s+1)^j}$$
as desired.
Using the claim, we have
$$\frac{1}{(s+1)^{n+1} s} + 1 - \frac{1}{s} = 1 - \sum_{j=1}^{n+1} \frac{1}{(s+1)^j}$$
Now for the original problem we have
$$\int \frac{1}{(s+1)^n s} + 1 - \frac{1}{s} ds = s - \ln(s+1) + \sum_{j=2}^{n+1} \frac{1}{(j-1) (s+1)^{j-1}} + C$$
so the limit is
$$\sum_{j=2}^{n+1} \frac{1}{j-1} + C = H_n + C$$
Note that the second $s$ in the numerator is in some sense superfluous, since the result of the limit is the same with or without it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/884896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration of a second degree polynomial underneath a radical. I have a question about how to apply the substitution rule in an arc length problem. The problem asks for the arc length of $x^2$ on the interval $[-1,2]$.
Here is what I have come up with:
$$y=x^2
\frac {dy}{dx} = 2x
(\frac {dy}{dx}^2 = 4x^2$$
So, $$L = \int_{-1}^2 \sqrt {1+ \frac {dy}{dx}^2}dx
= \int_{-1}^2 \sqrt {1+4x^2}dx$$
So the question is, how do I solve the last step, i.e. $\int_{-1}^2 \sqrt {1+4x^2}dx$?
(If the mathjax is bad, please forgive me, I am just learning)
| I shall give you the indefinite integral using IBP:
$$I=\int \sqrt{4x^2+1}\ dx=x\sqrt{4x^2+1}-\int \frac {(4x^2+1)-1}{\sqrt{4x^2+1}}\ dx$$
This means, $$2I=x\sqrt {4x^2+1}+\frac 12 \int \frac {dx}{\sqrt{x^2+(\frac 12)^2}}$$
Now, $$\int \frac {dx}{\sqrt {x^2+a^2}}=\int \frac {1+\frac {x}{\sqrt {x^2+a^2}}}{x+\sqrt {x^2+a^2}}\ dx=\ln|x+\sqrt {x^2+a^2}|+C$$
Hence we get:
$$I=\frac {x\sqrt {4x^2+1}}{2}+\frac 14 \ln|x+\sqrt {x^2+\frac 14}|+C$$
Now you can put the required limits to obtain the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
let $a,b,c >0 $ and $abc=1$,prove that $\sqrt{1+8a^2}+ \sqrt{1+8b^2}+ \sqrt{1+8c^2}\leq 3(a+b+c )$ let $a,b,c >0 $ and $abc=1$,prove that $\sqrt{1+8a^2}+ \sqrt{1+8b^2}+ \sqrt{1+8c^2}\leq 3(a+b+c )$
can anyone help me with this question.
i've tried to assume that $a\geq b \geq c $ as my teacher said,however i couldn't solve it
| Let
$$ f(a,b,c)=3\sum_{cyc}a-\sum_{cyc}\sqrt{1+8a^2}.$$
Assuming that $a\geq b\geq c$, the Cauchy-Schwarz inequality gives:
$$ f(a,b,c)\geq f(\sqrt{ab},\sqrt{ab},c)\tag{1}$$
hence it is sufficient to prove the inequality in the case $(a,b,c)=\left(x,x,\frac{1}{x^2}\right):$
$$ 3\left(2x+\frac{1}{x^2}\right)-2\sqrt{1+8x^2}-\sqrt{1+\frac{8}{x^4}}\geq 0,$$
$$ 3(2x^3+1)\geq 2x^2\sqrt{1+8x^2}+\sqrt{x^4+8}$$
$$ 1+36x^3-5x^4+4x^6 \geq 4x^2\sqrt{(1+8x^2)(8+x^4)}.\tag{2}$$
To prove $(2)$ it is sufficient to prove that for any $x\in\mathbb{R}^+$ we have:
$$ 1+72x^3 - 138x^4 + 2328 x^6 - 360 x^7 \geq 0 \tag{3}$$
or the still weaker:
$$ 12-23x+328x^3 \geq 0,$$
$$ f(x)=1-2x+27x^3 \geq 0,\tag{4}$$
that is trivial since that cubic polynomial has a negative discriminant, hence only one real root, and since $f(0)>0$ while $f(-1)<0$ that root is between $-1$ and $0$, so the cubic polynomial is positive over $\mathbb{R}^+$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Inductive proof of a formula for Fibonacci numbers May someone help me? I am trying to use induction to prove that the formula for finding the $n$-th term of the Fibonacci sequence is:
$$F_n=\frac{1}{\sqrt{5}}⋅\left(\frac{1+\sqrt{5}}{2}\right)^n-\frac{1}{\sqrt{5}}⋅\left(\frac{1-\sqrt{5}}{2}\right)^n.$$
I tried to put $n=1$ into the equation and prove that if $n=1$ works then $n=2$ works and it should work for any number, but it didn't work. I need to prove that this formula gives the $n$th Fibonacci number.
| Let $\phi=\dfrac{\sqrt{5}+1}2$ and note that $\phi^{-1} =\dfrac 1\phi= \dfrac{\sqrt{5}-1}2$.
Note also that $1+\dfrac 1\phi=\phi$ and $1-\phi=-\dfrac 1\phi$.
From your formula,
$$F_n = \frac 1{\sqrt{5}}\left[\phi^n-(-\frac 1\phi)^n \right]$$
For $n=k$ and $n=k-1$,
$$\begin{align}
F_k &= \frac 1{\sqrt{5}}\left[\phi^k-(-\frac 1\phi)^k \right]\\
F_{k-1} &= \frac 1{\sqrt{5}}\left[\phi^{k-1}-(-\frac 1\phi)^{k-1} \right]\\
&=\frac 1{\sqrt{5}} \left[\phi^k \cdot \frac 1\phi -(-\frac 1\phi)^k \cdot (-\phi)\right]\\
\end{align}$$
Hence,
$$\begin{align}
F_{k+1}&=F_{k}+F_{k-1}\\
&=\frac 1{\sqrt{5}} \left[\phi^k \cdot \left( 1+\frac 1\phi \right)
-(-\frac 1\phi)^k \cdot \left( 1-\phi \right)\right]\\
&=\frac 1{\sqrt{5}} \left[\phi^k \cdot \phi
-(-\frac 1\phi)^k \cdot \left( -\frac 1\phi \right)\right]\\
&=\frac 1{\sqrt{5}} \left[\phi^{k+1}-(-\frac 1\phi)^{k+1} \right]
\end{align}$$
i.e. if formula is true for $n=k-1$ and $n=k$, it is also true for $n=k+1$.
For $n=0$ and $n=1$, $F_0=0$ and $F_1=1$ respectively.
Hence $F_2=F_0+F_1=1$.
It can easily be shown that the formula is true for $n=2$.
Hence, by induction, formula is true for all positive integer $n\geq 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Simplifying expressions with algebraic indices I am having a ridiculous amount of trouble solving questions of this style - this one in particular, I know the answer is $6^{2n-4}$ but I can't get there!!!!
$ \frac {36^{2n}.6^{n+2}} {216^{n+2}}$
Could someone please help me out with some steps involved? I would be extremely grateful!
I understand the rules $a^x b^x = (ab)^x $ and $x^a x^b = x^{a+b} $ I just don't understand why I'm not able to get this.
These are my steps I have been taking:
$\frac {3^{2n}.12^{2n}.2^{n+2}.3^{n+2}} {12^{n+2}.18^{n+2}}$
$ \frac {3^{2n}.3^{2n}.4^{2n}.2^{n+2}.3^{n+2}} {4^{n+2}.3^{n+2}.3^{n+2}.3^{n+2}.2^{n+2}} $
I've simplified in this manner and then have tried cancelling and am coming out with completely wrong answers every time.
| $$ \frac {36^{2n}.6^{n+2}} {216^{n+2}}=\frac {(6^2)^{2n}.6^{n+2}} {(6^3)^{n+2}}=\frac{6^{2\cdot2n+(n+2)}}{6^{3\cdot(n+2)}}=\frac{6^{5n+2}}{6^{3n+6}}=6^{(5n+2)-(3n+6)}=6^{2n-4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/887551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How show that $AB>BR$ in rhombus? Let $A,B$ be points on the side $QR$ and $C,D$ be points on the side $RS$ of the rhombus $PQRS$, such that $B$ is closer to $R$ than $A$ is and that $C$ is closer to $R$ than $D$ is. Suppose that the segments $PA$, $PB$, $PC$, $PD$ divide the angle $\angle SPQ$ into five equal angles.
How show that $AB>BR$?
| connect $SR$, let $\angle BSR=x \implies \angle ASB=2x, \angle BRS=5x<\dfrac{\pi}{2} $
let $BR=p,AB=q,SR=a,SB=b$, apply sin law:
$\dfrac{p}{\sin{x}}=\dfrac{a}{\sin{6x}},\dfrac{q}{\sin{2x}}=\dfrac{b}{\sin{8x}},\dfrac{a}{\sin{6x}}=\dfrac{b}{\sin{5x}} \implies \dfrac{q}{p}=\dfrac{\sin{2x}\sin{5x}}{\sin{x}\sin{8x}}=f(x)$
$f(x)=\dfrac{2\sin{5x}\cos{x}}{2\sin{4x}\cos{4x}}=\dfrac{\sin{6x}+\sin{4x}}{2\sin{4x}\cos{4x}}=\dfrac{\sin{4x}\cos{2x}+\cos{4x}\sin{2x}}{2\sin{4x}\cos{4x}}+\dfrac{1}{2\cos{4x}}=\dfrac{\cos{2x}}{2\cos{4x}}+\dfrac{\sin{2x}}{2\sin{4x}}+\dfrac{1}{2\cos{4x}}=\dfrac{\cos{2x}}{2\cos{4x}}+\dfrac{1}{4\cos{2x}}+\dfrac{1}{2\cos{4x}}=\dfrac{2\cos^2{2x}+1}{4\cos{4x}\cos{2x}}+\dfrac{1}{2\cos{4x}}=\dfrac{\cos{4x}+2}{4\cos{4x}\cos{2x}}+\dfrac{1}{2\cos{4x}}=\dfrac{1}{4\cos{2x}}+\dfrac{1}{2\cos{4x}\cos{2x}}+\dfrac{1}{2\cos{4x}}$
note $2x<4x <5x <\dfrac{\pi}{2}, \dfrac{1}{\cos{2x}}$ and $\dfrac{1}{\cos{4x}}$ are mono increasing funtion and positive $\implies f(x)$ is mono increasing function $ \implies f_{min}=f(0)=\dfrac{5}{4} \implies \dfrac{q}{p}>\dfrac{5}{4}>1 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/887832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Equality of two iterated square roots Solve for $x$:
$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}}=\sqrt{x\sqrt{x\sqrt{x\sqrt{x\dots}}}}$
My attempt:
The L.H.S is equal to $\dfrac{1+\sqrt{4x+1}}{2}$ and R.H.S equals $x^2$
Equating both sides:
$\implies 4x+1=(2x^2-1)^2$
$\implies 4x+1=4x^4-4x^2+1$
$\implies 4x^4-4x^2-4x=0$
$\implies x(x^3-x-1)=0$
Disregarding the complex roots,
$\implies x=0$ 0r $\dfrac{1}{3}\sqrt[3]{\dfrac{27-3\sqrt{69}}{2}}+\dfrac{\sqrt[3]{\dfrac{9+\sqrt{69}}{2}}}{3^{2/3}}$
Is my solution correct? By the way I would like to see other methods to solve it. Thanks!
| I get a different solution:
Let $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+..}}}}=\sqrt{x\sqrt{x\sqrt{x\sqrt{x...}}}}$.
Then $y^2 = x+ y$ and $y^2 = xy$.
From the second equation we see that $x = y = 0$ is a possible solution, otherwise, $x = y \neq 0$.
So from the first equation: $x^2 = 2x \Rightarrow x = y = 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/888130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Distance between triangle's centroid and incenter, given coordinates of vertices
If $G$ is the centroid and $I$ is the in-center of the triangle, with vertices $A(-36,7)$, $B(20,7)$, and $C(0,-8)$, then find the length of $GI$.
Well the obvious way to approach this problem would be to centroid of the triangle and then the incenter of the triangle, and then find the distance. Is there an easier method to do this problem? Doing it that way would get a bit lengthy.
| The side lengths are $a=25,b=39,c=56$ by the pythagorean theorem and the area is $\Delta=420$ by the Heron's formula or the shoelace formula, hence $r=7$. Since $h_c=\frac{2\Delta}{c}$, the length of the height relative to $C$ is $15$, hence the distance of $G$ from $AB$ is just $5$.
It is well known that $I=\frac{aA+bB+cC}{a+b+c}$ and $AI^2=bc-4rR=bc-\frac{2abc}{a+b+c}$, hence $IG^2$ can also be computed through the parallel axis theorem. Let we assume that $A$ has mass $a$, $B$ has mass $b$ and $C$ has mass $c$. With such assumptions, $I$ is the center of mass of $S=\{A,B,C\}$ and the moment of inertia of $S$ around $I$ is given by:
$$ M_I = a AI^2 + b BI^2 + c CI^2 = 3abc - 4(a+b+c) rR = abc $$
while the moment of inertia of $S$ around $G$ is given by:
$$ M_G = a AG^2 + b BG^2 + c CG^2 = \frac{1}{9}\sum_{cyc}a(2b^2+2c^2-a^2) $$
and by the parallel axis theorem $ M_G = (a+b+c) IG^2 + M_I$, hence:
$$ IG^2 = \frac{M_G-M_I}{a+b+c} = \boxed{\frac{\frac{2}{3}(a+b+c)^3-\frac{5}{3}(a^3+b^3+c^3)-13abc}{9(a+b+c)}}.$$
A difficult problem has just born:
Prove that if $a,b,c$ are the side lengths of a triangle,
$$ 2(a+b+c)^3 \geq 5 (a^3+b^3+c^3) + 39abc. $$
Tricky solution: $IG^2\geq 0$, and equality is achieved only when $I\equiv G$, i.e. when $a=b=c$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $(1-a)(1-b)(1-c)(1-d)\geq abcd$ if $a^2+b^2+c^2+d^2=1$ Let $a,b,c,d\geq0$, $a^2+b^2+c^2+d^2=1$
Prove
$\displaystyle (1-a)(1-b)(1-c)(1-d)\geq abcd$
I mutiplied both with $\displaystyle (1+a)(1+b)(1+c)(1+d)$ to use $1-a^2=b^2+c^2+d^2$ and try using the cauchy-schwarz and holder but it is doesn't work.
| $a^2+b^2+c^2+d^2=1$ => $a,b,c,d\in[0,1]$.
If $\displaystyle abcd=0$ so the inequality is true.
If $\displaystyle abcd>0$, set :
$x=\frac{1-a}{a},
y=\frac{1-b}{b},
z=\frac{1-c}{c},
w=\frac{1-d}{d}$
We have
$\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+w)^2}=1$
And the inequality become
$\displaystyle xyzw\geq1$
We will prove that with $x,y,z,w\geq0$ and $xyzw=1$:
$\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+w)^2}\geq1$
(1)
Using Cauchy-Schwarz we have
$\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+w)^2}\geq
\frac{1}{(\frac{x}{y}+1)(xy+1)}+\frac{1}{(\frac{y}{x}+1)(xy+1)}
+\frac{1}{(\frac{z}{w}+1)(zw+1)}+\frac{1}{(\frac{w}{z}+1)(zw+1)}
=\frac{1}{xy+1}+\frac{1}{zw+1}
=\frac{1}{xy+1}+\frac{1}{\frac{1}{xy}+1}
=\frac{1}{xy+1}+\frac{xy}{xy+1}=1$
Suppose that $xyzw<1$. Set $t=\frac{1}{xyz}$ so $xyzt=1$ and $t>w$. Using (1), we have
$1\le\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+t)^2}
<\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+w)^2}=1$
So the suppose is false
=>$xyzw\geq1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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How to get the two eigen vectors for eigen =1 I have to find the eigen vectors for this matrix.
\begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 0\\
0 & 0 & 0
\end{pmatrix}
I end up with this matrix to plug in the eigen values.
\begin{pmatrix}
1-\lambda & 0 & 1\\
0 & 1-\lambda & 0\\
0 & 0 & -\lambda
\end{pmatrix}
Eigen values are 1,0
When I plug in zero then I get
\begin{pmatrix}
-1\\
0\\
1
\end{pmatrix}
But I'm not sure how the answer key is getting two vectors when using eigen = 1
The vectors should be:
\begin{pmatrix}
1\\
0\\
0
\end{pmatrix}
\begin{pmatrix}
0\\
1\\
0
\end{pmatrix}
| For each eigenvalue $\lambda$, use the definition of "eigenvector" and just solve
$\begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 0\\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix} a \\ b \\c
\end{pmatrix}=\lambda \begin{pmatrix} a \\ b \\c
\end{pmatrix}$.
This will furnish you with eigenvectors for eigenvalues. It turns out in this case that the eigenspace for the value $1$ is two dimensional, so you can produce two linearly independent eigenvectors for $1$.
From $\begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 0\\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix} a \\ b \\c
\end{pmatrix}=\begin{pmatrix} a \\ b \\c
\end{pmatrix}$
we learn that $\begin{pmatrix} a +c \\ b \\0
\end{pmatrix}=\begin{pmatrix} a \\ b \\c
\end{pmatrix}$, so in other words $c=0$, and an eigenvector for $1$ must look like this:
$\begin{pmatrix} a \\ b \\0
\end{pmatrix}$. Now you just have to find two different sets of coefficients satisfying this such that the vectors are linearly independent: pretty easy to do in this case. You already have an answer in front of you, so see how it fits in.
This is equivalent to substituting the eigenvalue into the lambda matrix, and then computing generators of the nullspace of that matrix.
In this case, for $\lambda=1$, you would be looking for solutions to this equation:
$\begin{pmatrix}
0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & -1
\end{pmatrix}\begin{pmatrix} a \\ b \\c
\end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\0
\end{pmatrix}$.
The resulting thing you learn is that $ \begin{pmatrix} c \\ 0 \\-c
\end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\0
\end{pmatrix}$, which is just saying $c=0$.
That means you are free to choose $a,b$ in the vector $ \begin{pmatrix} a \\ b \\0
\end{pmatrix}$ as you wish, just as in my previous paragraphs.
| {
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Compute $\iint_S \mathbf{F}\cdot d\mathbf{S}$ where $S$ is the surface that bounds the sphere $x^2+y^2+z^2=16$ and $\mathbf{F}=\langle z,y,x \rangle$ The problem is actually to verify the divergence theorem by computing both $\iiint_E \text{div } \mathbf{F\space} dV$, which was relatively easy to compute and gives $\frac{256\pi}{3}$. To find $\iint_S \mathbf{F}\cdot d\mathbf{S}$, I parametrized the surface with spherical coordinates: $\mathbf{r}=\langle 4\sin\phi\cos\theta,4\sin\phi\sin\theta,4\cos\phi\rangle$ with $0\leq\phi\leq\pi,0\leq\theta\leq2\pi$. Now, noting that $\iint_S\mathbf{F}\cdot d\mathbf{S}=\iint_S \mathbf{F}\cdot \mathbf{n}\space dS$ where $\mathbf{n}$ is the normal vector to the $S$, and since $S$ is a sphere, we have $\mathbf{n}=\langle\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi\rangle$, and $$\mathbf{F\cdot n}=4\sin\phi\cos\phi\cos\theta+4\sin^2\phi\sin\theta\cos\theta+4\sin\phi\cos\phi\cos\theta$$
But the integral of this is 0.
Where am I going wrong?
| On the surface of the sphere $\langle x,y,z \rangle = \langle \sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi \rangle$.
Hence, $F = \langle z,y,x \rangle = \langle 4\cos\phi, 4\sin\phi\sin\theta, 4\sin\phi\cos\theta \rangle$.
Also, $n = \frac{1}{4}\langle x,y,z \rangle = \langle \sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi \rangle$.
Therefore, $F \cdot n = 4\cos\phi\sin\phi\cos\theta + 4\sin^2\phi\sin^2\theta + 4\cos\phi\sin\phi\cos\theta$.
This shouldn't integrate to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Radical Inequality $\sqrt{2x-1}$ + $\sqrt{3x-2}$ > $\sqrt{4x-3}$ + $\sqrt{5x-4}$
I have attempted to solve this by squaring each side, resulting in
$5x + 2\sqrt{2x-1}\sqrt{3x-2} - 3 > 9x + 2\sqrt{(4x-3)(5x-4)} - 7 $
$4 + 2\sqrt{2x-1}\sqrt{3x-2} > 4x + 2\sqrt{(4x-3)(5x-4)}$
$1 + 1/2\sqrt{2x-1}\sqrt{3x-2} > x + 1/2\sqrt{(4x-3)(5x-4)}$
After this my thought was to square again, but I don't think that'd help too much - and I don't even know if this is correct anymore.
| Assuming that all the arguments of the square roots are non negative, hence $x\geq\frac{4}{5}$, we get by squaring the following equivalent inequality:
$$\sqrt{(4x-3)(5x-4)}-\sqrt{(2x-1)(3x-2)}<2x-2,$$
that turns into:
$$\frac{2(x-1)(7x-5)}{\sqrt{(4x-3)(5x-4)}+\sqrt{(2x-1)(3x-2)}}<2(x-1).$$
Dividing by $x-1$ (and being careful in switching the direction of the inequality when needed) this boils down to proving:
$$f(x)=\sqrt{(4x-3)(5x-4)}+\sqrt{(2x-1)(3x-2)} \lessgtr 7x-5=g(x). \tag{1}$$
$g(x)$ is just the equation of the tangent line to the graphics of $f(x)$ in $x=1$, but since the LHS is a concave function as the sum of two concave functions, the initial inequality holds for any $x\in[4/5,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/897029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Expansion of Logarithms with Cube Roots Does the following expand to the following
$$ \log_6(11^6\sqrt[3]{12}) $$
= $ 6\log_6(11) + \log_6 (\sqrt[3]{12})$
| First note that
\[ \sqrt[n]{x}=x^{\frac{1}{n}} \]
\[ \log_{b}(xy)=\log_{b}(x)+\log_{b}(y) \]
\[ \log_{b}(x^{n})=n\log_{b}(x) \]
\[ \log_{b}(x)=\frac{\log_{10}(x)}{\log_{10}(b)}=\frac{\log(x)}{\log(b)} \]
So then
\[
\log_{6}(11^{6}\cdot\sqrt[3]{12})=\log_{6}(11^{6}\cdot 12^{\frac{1}{3}})=\log_{6}(11^{6})+\log_{6}(12^{\frac{1}{3}})=6\log_{6}(11)+\frac{1}{3}\log_{6}(12)
\]
Or perhaps
\[
\frac{6\log(11)}{\log(6)}+\frac{\log(12)}{3\log(6)}
\]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/897484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove by contradiction that $(n+1)^3 \not= n^3 +(n-1)^3$ for $3$ consecutive positive integers Prove by contradiction that if $n-1$, $n$, $n+1$ are consecutive positive integers, then the cube of the largest cannot be equal to the sum of the cubes of the other two.
Assume that:
$$ (n+1)^3 = (n-1)^3+n^3 $$
$$n^3+3n^2+3n+1=n^3-3n^2+3n-1+n^3$$
$$3n^2+1=-3n^2-1+n^3$$
$$n^3-6n^2-2=0$$
I don't know how to move from here. Instead of this solution considered that two of the integers are either odd or even but this idea didn't help much.
| $n^3-6n^2-2=0 \implies n^2(n-6)=2$
*
*$n<6 \implies n^2(n-6)<0$
*$n=6 \implies n^2(n-6)=0$
*$n>6 \implies n^2(n-6)\geq49$
Therefore, $\forall{n}\in\mathbb{N}:n^2(n-6)\neq2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/900405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.