Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Cyclic inequality How can we prove that:
$$a^{60} c^{10} +b^{60}a^{10}+c^{60}b^{10}+a^{50} c^{20} +b^{50}a^{20}+c^{50}b^{20}\geq 2(a^{51}b^{9}c^{10}+b^{51}c^9 a^{10}+c^{51}a^9 b^{10}), \ \forall\ a,b,c\geq 0.$$
I proved only that $32S_1+18S_2\geq 2 S$, using AM-GM inequality, where
$S_1=a^{60} c^{10} +b^{60}a^{10}+c^{60}b^{10},$
$S_2=a^{50} c^{20} +b^{50}a^{20}+c^{50}b^{20},$
$S=a^{51}b^{9}c^{10}+b^{51}c^9 a^{10}+c^{51}a^9 b^{10}$
Note that Muirhead's inequality is not working there...
| Update 2019/10/05 (Simplify some proof.)
With computer, here is a solution:
Clearly, we only need to prove the case when $a, b, c > 0$.
Let $x=a^{10}, \ y = b^{10}, \ z=c^{10}$. The inequality is written as
$$xy^6 + yz^6 + zx^6 + x^2y^5 + y^2z^5 + z^2 x^5 \ge
2x^5yz\sqrt[10]{\frac{x}{y}} + 2xy^5z\sqrt[10]{\frac{y}{z}} + 2xyz^5\sqrt[10]{\frac{z}{x}}.$$
We will use the following bound:
$$\sqrt[10]{u} \le f(u) = \frac{11 u^2 + 418 u + 171}{20(19 u + 11)}, \quad \forall u > 0$$
which follows from (letting $u = v^{10}$)
\begin{align}
&11 v^{20} + 418 v^{10} + 171 - 20(19v^{10}+11)v\\
=\ & (v-1)^4\\
&\times \left(11 v^{16}+44 v^{15}+110 v^{14}+220 v^{13}+385 v^{12}+616 v^{11}+924 v^{10}+1320 v^9+1815 v^8\right.\\
&\qquad \left. +2040 v^7+2044 v^6+1876 v^5+1585 v^4+1220 v^3+830 v^2+464 v+171\right).
\end{align}
With the bound above, it suffices to prove that
$$xy^6 + yz^6 + zx^6 + x^2y^5 + y^2z^5 + z^2 x^5 \ge
2x^5yzf(x/y) + 2xy^5zf(y/z) + 2xyz^5f(z/x)$$ or $F(x,y,z)\ge 0$
where $F(x,y,z)$ is a homogeneous polynomial of degree $10$.
We use the Buffalo Way. WLOG, assume that $z = \min(x,y,z) $. There are two possible cases:
1) $z \le y \le x$: Let $z=1, \ y=1+s, \ x = 1+s + t; \ s,t \ge 0$. Then, $F(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients.
2) $z \le x \le y$: Let $z=1, \ x=1+s, \ y=1+s+t; \ s,t\ge 0$. Then, $F(1+s, 1+s+t, 1)$ is a polynomial in $s, t$ with non-negative coefficients.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
For positive integers $m,n$ if $\sqrt 7 - \frac{m}{n} > 0$ then prove that $\sqrt 7 - \frac{m}{n} > \frac{1}{{mn}}$ For positive integers $m,n$ if $\sqrt 7 - \frac{m}{n} > 0$ then prove that $\sqrt 7 - \frac{m}{n} > \frac{1}{{mn}}$.
| By condition, $7n^2 > m^2$. The square of an integer when divided by 7 moiety can be given in only 0, 1, 2 and 4. Therefore, none of the numbers $m^2+1$, $m^2+2$ is not divisible by 7, where $7n^2\ge m^2+3$. Then $n\sqrt7\ge\sqrt{m^2+3}\ge\sqrt{m^2+2+\frac1{m^2}} > m+\frac1m$ at $m > 1$, so $\sqrt7-\frac{m}n > \frac1{mn}$.
Case $m=1$ immediately obvious, but you can also notice that in this case, true strict inequality $7n^2 > m^2+3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\cos(x)\cos(2x)\cos(4x)=1/8$ I have to solve $\cos(x)\cos(2x)\cos(4x)=1/8$.
I can express it for $x$ only with $\cos(2x)=\cos^2(x)-\sin^2(x)$ and $\cos(4x)=\cos(2x+2x)$, but it only seems to become a really big expression and I have no clue how to proceed after... Any suggestions?
| You can use the double angle identity
$$ \sin 2x = 2\sin x \cos x
$$
By multiplying $\sin x$,
$$\begin{align*}
\cos(x)\cos(2x)\cos(4x) &= \frac{1}{8} \\
\frac{1}{2} \sin (2x) \cos(2x)\cos(4x) &= \frac{1}{8}\sin x \\
\frac{1}{4} \sin (4x)\cos(4x) &= \frac{1}{8}\sin x \\
\frac{1}{8} \sin (8x) &= \frac{1}{8}\sin x \\
\sin (8x) &= \sin x
\end{align*}$$
The last equation will have $16$ roots in $[0,2\pi)$, but $0$ and $\pi$ do not solve the original equation since they are introduced by $\sin x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proving that $2\sqrt 3+3\sqrt[3] 2-1$ is irrational
Prove that $2\sqrt 3+3\sqrt[3] 2-1$ is irrational
My attempt:
$$k=2\sqrt 3+3\sqrt[3] 2-1$$
Suppose $k\in \mathbb Q$, then $k-1\in \mathbb Q$.
$$2\sqrt 3+3\sqrt[3] 2=p/q$$
I'm stuck here and don't know how to procced. I tried to do this:
$$\sqrt 3=\frac{p/q-3\sqrt[3] 2}{2}$$
contradiction, but I'm not at all sure about that. How should I proceed?
| Assuming $$\frac{p}{q}=2\sqrt{3}+3\sqrt[3]{2}$$
You will have :
$$54 = (\frac{p}{q} - 2\sqrt{3})^3$$
$$= \frac{p^3}{q^3}-6\frac{p^2}{q^2}\sqrt{3}+18\frac{p}{q}-24\sqrt{3}$$
So finally :
$$54 -\frac{p^3}{q^3}-18\frac{p}{q}= -6\frac{p^2}{q^2}\sqrt{3}-24\sqrt{3} = -6 (\frac{p^2}{q^2}+4)\sqrt{3}$$
which is impossible because $\sqrt{3}$ is irrational and $54 -\frac{p^3}{q^3}-18\frac{p}{q}$ and $-6 (\frac{p^2}{q^2}+4)$ are both rational
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Find the minimum and maximum of $h(x) = \dfrac{1}{1+|x|}+\dfrac{1}{1+|x-a_1|}$
Let $a_1 \in \mathbb{R}$. Find the minimum and maximum of $h(x) = \dfrac{1}{1+|x|}+\dfrac{1}{1+|x-a_1|}$.
This question seems hard to solve since we don't know what $a_1$ is. We want both $|x|,|x-a_1|$ to be as small as possible or as large as possible to find the maximum or minimum, so how do I find where that is possible? Maybe it might occur at the median or some other point.
| Set $b=a_1/2$ and the translation $x\mapsto x-b$, $y\mapsto y$. This leads to considering the function
$$
f(x)=\frac{1}{1+|x-b|}+\frac{1}{1+|x+b|}
$$
and it's not restrictive to assume $b\ge0$. Leave out the case $b=0$, for the moment. The function is even, so we just need to study it for $x\ge0$ and we can write it as
$$
f(x)=\begin{cases}
\dfrac{1}{1+b-x}+\dfrac{1}{1+x+b} & \text{if $0\le x<b$}\\[6px]
\dfrac{1}{1+x-b}+\dfrac{1}{1+x+b} & \text{if $x\ge b$}
\end{cases}
$$
You can see that $f(0)=2/(1+b)$ and that
$$
\lim_{x\to\infty}f(x)=0
$$
Thus there is no absolute minimum, because $f(x)>0$ for all $x$. We also have
$$
f(b)=1+\frac{1}{1+2b}
$$
The derivative is
$$
f'(x)=\begin{cases}
\dfrac{1}{(1+b-x)^2}-\dfrac{1}{(1+b+x)^2} & \text{if $0\le x<b$}\\[6px]
-\dfrac{1}{(1-b+x)^2}-\dfrac{1}{(1+b+x)^2} & \text{if $x>b$}
\end{cases}
$$
Note that $f$ is decreasing in the interval $(b,\infty)$ (and not differentiable at $b$). In the interval $[0,b)$ we can write
$$
f'(x)=\frac{4x(1+b)}{(1+b-x)^2(1+b+x)^2}
$$
so the function has zero derivative at $0$ and is increasing in the interval $(0,b)$.
This should be enough to finish up.
The case $b=0$ is quite easy:
$$
\frac{2}{1+|x|}\le 2
$$
for all $x$, equality only for $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How do I show $\int_0^\infty \frac{x}{(x+1)^2}dx$ diverges I did $$\int_0^\infty \frac{x}{(x+1)^2}dx=\int_0^\infty \frac{1}{(x+2+\frac1x)}dx\le\int_0^\infty \frac{1}{x}dx$$ but $\int_0^\infty \frac{1}{x}dx$ diverges. So my logic fails.
| Let $c$ be the upper limit of the integral and take the limit as $c$ goes to infinity.
$$I = \int_0^\infty \frac{x}{(x+1)^2}dx = \lim_{c \rightarrow \infty} \int_0^c \frac{x}{(x+1)^2}dx.$$
Then compute the definite integral as normal;
$$I = \lim_{c \rightarrow \infty}\int_0^c \frac{x}{(x+1)^2} = \lim_{c\rightarrow\infty}\left[\log(x + 1) - \frac{1}{x+1}\right]_0^c = 1 + \lim_{c \rightarrow \infty} \left(\log(c+1) - \frac{1}{c+1}\right).$$
The logarithm diverges as its argument gets arbitrarily large and the $\frac{1}{c+1}$ term vanishes. Hence,
$$\lim_{c \rightarrow \infty} \int_0^c \frac{x}{(x+1)^2}dx = \infty$$
and we conclude that $I$ diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1699135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Find the minimum polynomial of $\sqrt{i+\sqrt{2}}$ over $ \mathbb{R}$ So here's what I know about the minimum polynomial $p(x)$ of $c=\sqrt{i+\sqrt{2}}$ over $\mathbb{R}$...
*
*$p(x)$ has to be degree 2
*$p(x)$ must be a multiple of $q(x)=x-\sqrt{i+\sqrt{2}}$ because $q(x)$ is the generator of the kernel of the substitution function $\sigma_c$ that maps polynomials over $\mathbb{C}$ to $\mathbb{C}$.
*$p(x)$ must be a divisor of $r(x) = (x^2-i-\sqrt{2})(x^2+i-\sqrt{2})$, because $r(c)=0$.
So I tried multiplying $q(x)$ by each degree-1 factor of $x^2+i-\sqrt{2}=(x+\sqrt{\sqrt{2}-i})(x-\sqrt{\sqrt{2}-i})$, but (I think) neither product is over $\mathbb{R}$... I must be doing something wrong. Any tips?
| Robert Israel's answer is excellent and complete, but I thought I might offer some supporting computations. Let $\alpha = i+\sqrt{2}$, and let $\beta = \sqrt{\alpha}$. Then $|\alpha| = \sqrt{3}$, so we can rewrite $\alpha$ as
$$\alpha = \sqrt{3}(\cos(\theta)+i\sin(\theta))$$
where $\cos(\theta) = \frac{\sqrt{2}}{\sqrt{3}}$ and $\sin(\theta) = \frac{1}{\sqrt{3}}$. Then using the principal branch of the square root, we see
$$\beta = \sqrt{\alpha} = \sqrt[4]{3}\left(\cos\left(\frac{\theta}{2}\right)+i\sin\left(\frac{\theta}{2}\right)\right)$$
The half-angle formulas give us
$$\cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1+\cos(\theta)}{2}} = \sqrt{\frac{\sqrt{3}+\sqrt{2}}{2\sqrt{3}}}$$
and
$$\sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1-\cos(\theta)}{2}} = \sqrt{\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}}$$
whence
$$\beta = \sqrt{\frac{\sqrt{3}+\sqrt{2}}{2}} + \sqrt{\frac{\sqrt{3}-\sqrt{2}}{2}}i$$
Then it is easy to see that the trace of $\beta$ is
$$\mathrm{Tr}(\beta) = 2\sqrt{\frac{\sqrt{3}+\sqrt{2}}{2}} = \sqrt{2\sqrt{3}+2\sqrt{2}}$$
and the norm of $\beta$ is
$$|\beta| = (\sqrt[4]{3})^{2} = \sqrt{3}$$
Then the minimal polynomial of $\beta$ over $\mathbb{R}$ is
$$(X-\beta)(X-\overline{\beta}) = X^{2} - \mathrm{Tr}(\beta) + |\beta|^{2} = X^{2} - \sqrt{2\sqrt{3}+2\sqrt{2}}X + \sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find Nth formula of recursive formula $a_n=a_{n-1}+n(n-1)a_{n-2}$ $$a_n=a_{n-1}+n(n-1)a_{n-2}$$
$$a_0=1, a_1=-\frac{1}{2}$$
Is it possible to find explicit formula for $a_n$ just by using $a_0$ and $a_1$?
I know how to solve this problem if $a_n=Aa_{n-1}+Ba_{n-2}$ where $A$ and $B$ is some real number(constant), but in this problem that is not a case.
| Let $\displaystyle\;b_n = \frac{a_n}{n!}$, the recurrence relation at hand can be transformed to
$$n ( b_n - b_{n-2} ) = b_{n-1}\quad\text{ with }\quad \begin{cases} b_0 = 1\\b_1 = -\frac12\end{cases}\tag{*1}$$
Let $f(z) = \sum_{n=0}^\infty b_n z^n$ be the OGF for $b_n$. If we multiple $(*1)$
by $z^n$ and start to sum from $n = 2$. We get
$$\begin{align}
& z\frac{d}{dz}\left[ (1-z^2) f(z) - 1 + \frac{z}{2} \right] = z(f(z)-1)\\
\iff & (1-z^2) \frac{df(z)}{dz} - (1+2z) f(z) = -\frac{3}{2}
\end{align}
$$
Solving the ODE gives us
$$f(z) = \frac{5 - 3\sin^{-1}(z)}{2(1-z)\sqrt{1-z^2}} - \frac{3}{2(1-z)}$$
Expanding $f(z)$ as a power series in $z$, we obtain following
ugly expression of $a_n$.
$$a_n = \frac{n!}{2}\left[
5\sum_{p=0}^{\lfloor \frac{n}{2}\rfloor} \frac{\binom{2p}{p}}{4^p}
- 3\sum_{s=0}^{\lfloor\frac{n-1}{2}\rfloor} \sum_{p=0}^s \frac{\binom{2p}{p}\binom{2s-2p}{s-p}}{4^s(2p+1)} - 3
\right]$$
As a double check, the first few values of $a_n$ according this formula are
$$(2a_0,2a_1,\ldots) = 2, -1, 3, -3, 33, -27, 963, -171, 53757, 41445,4879575, 9438525, \ldots$$
and they do satisfy the recurrence relation.
Update
It turns out we can simplify the mess a little bit.
The recurrence relation on $b_n$ can be rewritten as $b_n = (n+1)(b_{n+1} - b_{n-1})$. The RHS has the form of a finite difference. We can use it to get
rid of one level of summation. The end result is for all $n \ge 0$,
$$a_n = \frac{(n+1)!}{2}\left[
5\frac{\binom{2r}{r}}{4^r}
- 3\sum_{p=0}^s \frac{\binom{2p}{p}\binom{2s-2p}{s-p}}{4^s(2p+1)}
\right]
\quad\text{ with }\quad
\begin{cases}
r = \lfloor \frac{n+1}{2}\rfloor\\
s = \lfloor \frac{n}{2}\rfloor
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Sum of Consecutive Perfect Squares If $n\in\mathbb{Z}$ and $2n+1$ is a perfect square, then is it true that $n+1$ is a sum of two consecutive perfect squares?
| Since $2n + 1$ is odd, if it is a perfect square it is the square of an odd number. So write $2n+1 = (2k+1)^2$. Then we can solve to get $n+1 = \frac{(2k+1)^2 - 1}{2} + 1$, and further simplify this to $2k^2 + 2k + 1 = k^2 + k^2 + 2k + 1 = k^2 + (k+1)^2$, showing that $n+1$ is indeed the sum of two consecutive squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$ $$\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$$
I don't think L'hospital's rule will make the problem easy. (I am afraid to differentiate the numerator). The given limit has a $\frac{0}{0}$ form. I tried using taylor series but the it made the problem more complicated.
| This is exactly the kind of question which "looks very difficult" and at the same time is "extremely easy to answer". It appears that it is specially crafted to generate a complicated looking expression in order to intimidate a casual reader/student. The following evaluation shows that the difficulty is only superficial:
\begin{align}
L &= \lim_{x \to 0}\frac{\log(1 + \sin^{3}x \cos^{2}x)\cot(\log^{3}(1 + x))\tan^{4}x}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\log(1 + x^{2})}\tag{1}\\
&= \lim_{x \to 0}\dfrac{\log(1 + \sin^{3}x \cos^{2}x)\cot(\log^{3}(1 + x))\cdot\dfrac{\tan^{4}x}{x^{4}}\cdot x^{4}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\cdot\dfrac{\log(1 + x^{2})}{x^{2}}\cdot x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\log(1 + \sin^{3}x \cos^{2}x)\cot(\log^{3}(1 + x))\cdot 1 \cdot x^{4}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\cdot 1 \cdot x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\log(1 + \sin^{3}x \cos^{2}x)x^{2}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\tan(\log^{3}(1 + x))}\notag\\
&= \lim_{x \to 0}\dfrac{\dfrac{\log(1 + \sin^{3}x \cos^{2}x)}{\sin^{3}x \cos^{2}x}\cdot \sin^{3}x \cos^{2}x\cdot x^{2}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\cdot\dfrac{\tan(\log^{3}(1 + x))}{\log^{3}(1 + x)}\cdot \dfrac{\log^{3}(1 + x)}{x^{3}}\cdot x^{3}}\notag\\
&= \lim_{x \to 0}\dfrac{1\cdot \sin^{3}x \cos^{2}x\cdot x^{2}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\cdot1\cdot 1\cdot x^{3}}\notag\\
&= \lim_{x \to 0}\dfrac{\sin^{3}x \cdot 1}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})x}\notag\\
&= \lim_{x \to 0}\dfrac{\dfrac{\sin^{3}x}{x^{3}} \cdot x^{3}}{\dfrac{\sin(\sqrt{x^{2} + 2} -\sqrt{2})}{\sqrt{x^{2} + 2} -\sqrt{2}}\cdot(\sqrt{x^{2} + 2} -\sqrt{2})x}\notag\\
&= \lim_{x \to 0}\dfrac{1 \cdot x^{3}}{1\cdot(\sqrt{x^{2} + 2} -\sqrt{2})x}\notag\\
&= \lim_{x \to 0}\frac{x^{2}}{\sqrt{x^{2} + 2} - \sqrt{2}}\tag{2}\\
&= \lim_{x \to 0}\frac{x^{2}(\sqrt{x^{2} + 2} + \sqrt{2})}{(x^{2} + 2) - 2}\notag\\
&= \lim_{x \to 0}(\sqrt{x^{2} + 2} + \sqrt{2})\notag\\
&= \sqrt{2} + \sqrt{2}\notag\\
&= 2\sqrt{2}\notag
\end{align}
The simplification of the complicated expression in equation $(1)$ to a very simple expression in equation $(2)$ is done via the use of standard limits $$\lim_{x \to 0}\cos x = \lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{\tan x}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$ In evaluation of limits it is important to figure out the use of standard limits to simplify a complicated expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Showing $ | 1 - \frac{x^2}{2} - \cos x | \leq \frac{1}{24}$ using Taylor Use Taylor's theorem to show that for $x \in [-1, 1] $ we have that $$ \left| 1 - \frac{x^2}{2} - \cos x \right| \leq \frac{1}{24}. $$
Attempt: Since $\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \ldots $, I see that $1 - \frac{x^2}{2} = P_2 (x)$ is the second order Taylor polynomial of $f(x) = \cos(x)$ around $x_0 = 0$. So we have that $\cos(x) = P_2(x) + r(x)$, where $r(x)$ is the remainder term. But I'm not sure how to use this remainder term to get a proper upper bound for $ | 1 - \frac{x^2}{2} - \cos x | \leq \frac{1}{24}$.
| Since the term in $x^3$ is $0$, your approximation $P_2(x)$ can also be thought of as $P_3(x)$.
By the Lagrange form of the remainder, the absolute value of the error in $P_3(x)$ is $\frac{1}{4!}|\cos(c_x)||x|^4$ for some $c_x$ between $0$ and $x$. We have $|\cos(c_x)|\le 1$, and $|x|\le 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit calculation with Riemann integral Help me calculating the limit with Riemann integral:
$$a_n=\frac{1}{n\sqrt{n^2+1}}+\frac{2}{n\sqrt{n^2+4}}+\frac{3}{n\sqrt{n^2+9}}+...+\frac{n-1}{n\sqrt{n^2+(n-1)^2}}$$
| Hint: This can be written as
$$\lim_{n \to \infty}\frac{1}{n} \sum_{r=1}^{n-1} \frac{r}{\sqrt{(n^2+r^2)}}$$
$$=\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n-1} \frac{\frac{r}{n}}{\sqrt{1+(\frac{r}{n})^2}}$$
$$=\int_0^1 \frac {xdx}{\sqrt{1+x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Partial derivative of $f(x,y) = \frac{x^3+y^3}{x^2+y^2}$ at $f_1(0,0)$ To be more precise than the title, the function is actually piecewise
$$
f(x,y) = \begin{cases}
\frac{x^3+y^3}{x^2+y^2} & (x,y) \ne (0,0) \\
0 & (x,y) = (0,0) \\
\end{cases}
$$
I checked that the function is continuous at $(0,0)$, so I then calculated the partial derivative with respect to $x$ as
$$
f_1(x,y) = \frac{x^4-2xy^3+3x^2y^2}{(x^2+y^2)^2} \tag{1}
$$
This is undefined at $(0,0)$, so I then tried to find the limit around accumulation points. Let $S_1$ be the points on the $x$ axis
$$
\lim_{x \to 0} f_1(x,0) = \frac{x^4}{(x^2)^2} = 1 \tag{2}
$$
Let $S_2$ be the points on the line $y = x$
$$
\lim_{x \to 0} f_1(x,x) = \frac{x^4-2x^4+3x^4}{(x^2+x^2)^2} = \frac{2x^4}{(2x^2)^2} = \frac{1}{2} \tag{3}
$$
So, the limits are different around different accumulation points. That's where I'm confused because the answer should be $1$.
| Hint: Compute the limit (as $h$ goes to $0$) of $(f(0+h,0)-f(0,0))/h$ to get the value of the partial derivative at $(0,0)$.
What you are looking at with those path limits is the question of continuity of this partial derivative function at $(0,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Confusion determining eigenvalues Matrix $A$:
$$A = \begin{bmatrix}
-5 &14& -8 \\
-9 &16& -6 \\
-9 &11& -1
\end{bmatrix} $$
Determining eigenvalue of Matrix $A - 1I$ hence $\lambda = 1$ (other eigenvalues are 4 and 5).
Matrix $A - 1I:$
$$\begin{bmatrix}
-6 &14& -8 \\
-9 &15& -6 \\
-9 &11& -2
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}$$
1. Figuring out each $x_i$ so the above equation complies will give me the eigenvector for eigenvalue $\lambda = 1$ right?
2. the sum of all $x_i$ must be $0$?
3. How would you go about determining $x_i$, do you just play around till you get a right combination?
| Using gaussian elimination, your system is equivalent to the following, writing only the matrix, not the $x_i$ nor the right hand side since there are only zeros in the RHS anyway:
$$\left(\begin{matrix}
-6 &14& -8 \\
-9 &15& -6 \\
-9 &11& -2
\end{matrix}\right)$$
$$\left(\begin{matrix}
-3 &7& -4 \\
-9 &15& -6 \\
-9 &11& -2
\end{matrix}\right)$$
$$\left(\begin{matrix}
-3 &7& -4 \\
0 &-6& 6 \\
0 &-10& 10
\end{matrix}\right)$$
$$\left(\begin{matrix}
-3 &7& -4 \\
0 &-6& 6 \\
0 &0& 0
\end{matrix}\right)$$
$$\left(\begin{matrix}
-3 &7& -4 \\
0 &-6& 6 \\
0 &0& 0
\end{matrix}\right)
\left(\begin{matrix}
x_1 \\ x_2 \\ x_3
\end{matrix}\right)
=\left(\begin{matrix}
0 \\ 0 \\ 0
\end{matrix}\right)$$
That is, $x_3$ is arbitrary, then $x_2=x_3$ from the second equation, and finally $x_1=-\frac13(4x_3-7x_2)=x_3$.
Therefore, $(1,1,1)^T$ is an eigenvector of $A$ for the eigenvalue $1$.
When solving such system, if the eigenvalue is correct, you know for sure there is at least one arbitrary component. There are more if the eigenspace has dimension greater than $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Existence of rational roots in a quadratic equation Consider the quadratic equation $(a+c-b)x^2 + 2cx + b+c-a = 0 $ , where a,b,c are distinct real numbers and a+c-b is not equal to 0. Suppose that both the roots of the equation are rational . Then
a) a,b, and c are rational
b)$c/(a-b)$ is rational
c)$b/(c-a)$ is rational
d)$a/(b-c)$ is rational
My attempt - I used the discriminant method to find out the possible roots of the given equation.
Which for me came out to be $(-2c + a-b)/(2a -2b +2c)$ and $ ( -2c + b - a)/(2c + 2a - 2b )$
Hence according to me option a is the correct answer , while the correct answer which is given is b. Please tell me where am I going wrong , or what am I missing ?
| The roots are
$$\frac{-2c\pm \sqrt{4c^2 - 4(c+(a-b))(c-(a-b))}}{2(a+c-b)}$$
$$=\frac{-c\pm (a-b)}{a+c-b}$$
$=\frac{a-b-c}{a-b+c}$ or $\frac{-a+b-c}{a-b+c}=-1$
For the first root,
$$=\frac{a-b+c-2c}{a-b+c}=1-\frac{2c}{a-b+c}$$
So $\frac{2c}{a-b+c}$ is rational, so its inverse is also rational
$\frac{a-b+c}{2c}=\frac{a-b}{2c}+\frac 1 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Evaluation of $\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$ Evaluate :
$\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$
My approach : I multiplied both sides by $\sqrt{a+x}$ and after simplification it comes down to :
$\int_{0}^{a} \frac{a}{\sqrt{a^{2}-x^{2}}} dx + \int_{0}^{a} \frac{x}{\sqrt{a^{2}-x^{2}}} dx$, let's denote them by $I_1$ and $I_2$ respectively.
$I_1$ can be easily solved to $a \sin ^{-1} \frac{x}{a}$.
If I write $I_2$ as $ \frac{-1}{2}\int_{0}^{a} \frac{-2x}{\sqrt{a^{2}-x^{2}}} dx$, I get the solution as $\frac{a}{2}(\pi+1)$ but the answer given is $\frac{a}{2}(\pi+2)$. Where did I go wrong?
| So for your solution:
$$I_2 = \int_{0}^{a} \frac{x}{\sqrt{a^2 - x^2}} dx$$
Let $u = a^2 - x^2, du = -2x dx, dx = -\frac{1}{2x} du$ then it is clear that
$$ I_2 = - \frac{1}{2} \int \frac{1}{\sqrt{u}} du = - \sqrt{u} $$
Now evaluating back at $x$,
$$ = - \sqrt{a^2 - x^2} | _0^a = - \sqrt{0} + \sqrt{a^2} = a$$
I would recommend against, the expression
$$ - \frac{1}{2} \int_{0}^{a} \frac{-2x}{\sqrt{a^2 - x^2 }} dx $$
Since it doesn't really make it any easier to integrate.But if we insist, again making the substitution $u = a^2 - x^2, dx = -\frac{1}{2x} du$ we get:
$$ - \frac{1}{2} \int \frac{1}{\sqrt{u}} dx $$
Which is evaluated the same way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Residue of $f(z) = \frac{1}{z-\sin z}$ at $z=0$ My attempt:
$$ f(z) = \frac{1}{z-\sin z}$$
$$\frac{1}{z-(z-\frac{z^3}{6}+\frac{z^5}{120}-...)}$$
$$\frac{1}{z(1-(1-\frac{z^2}{6}+\frac{z^4}{120}-...))}$$
$$Res(f(z),0) = \lim_{z \to 0} z \cdot \frac{1}{z(1-(1-\frac{z^2}{6}+\frac{z^4}{120}-...))}$$
$$ = \lim_{z \to 0} \frac{1}{(1-(1-\frac{z^2}{6}+\frac{z^4}{120}-...))}$$
$$=\frac{1}{0}$$
This is where I'm stuck. Using L'Hopital's rule doesn't help. The answer should be $\frac{3}{10}$.
| Two small hints in addition to the nice answer of @C.Dubussy
*
*If $f$ has a pole of order $n$ in $z_0$, the $Res$ of $f$ at $z_0$ is
\begin{align*}
Res(f,z_0)=\lim_{z\rightarrow z_0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}\left[(z-z_0)^nf(z)\right]
\end{align*}
*From your series expansion
\begin{align*}
f(z)&=\frac{1}{z-\sin z}\\
&=\frac{1}{z-(z-\frac{z^3}{6}+\frac{z^5}{120}-\cdots)}\\
&=\frac{1}{z^3\left(\frac{1}{6}-\frac{z^2}{120}-\cdots\right)}
\end{align*}
you can immediately deduce that $z_0=0$ is a pole of order $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the integral $\int \frac{2x^3+3x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$ $$\int \frac{2x^3+3x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$$$$=2\int \frac{x^3}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx+3\int \frac{x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$$
For these two integrals, I have tried Euler substitutions and various types of substitutions of irrational functions, but it seems that those substitutions don't simplify the integral.
What substitution is useful for this type of integrals?
| Since $x^2 + 2x -3 = (x+1)^2 -4$, you can use the substitution $x+1 = 2 \cosh t$. This way you are going to get a rational function in $\cosh t$ and $\sinh t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Why is $x^3-x^2+x$ injective? The function $$ f(x) = x^3-x^2+x$$ is injective (as seen on the graph), but by doing f(a) = f(b) I can't get to the point where a = b.
Can this be indicated in an analytical way (without graph) ?
| Well, you should not give up hope on that $f(a)=f(b)$ approach :-):
$$f(a)=f(b)\\a^3-a^2+a=b^3-b^2+b\\a^3-b^3-(a^2-b^2)+(a-b)=0\\(a-b)(a^2+ab+b^2-a-b+1)=0$$
Which means that either $a=b$ or $a^2+ab+b^2-a-b+1=0$
Considering this as a quadratic with respect to a, we calculate it's discriminant as $$\Delta=(b-1)^2-4(b^2-b+1)=\\-3b^2+2b-3$$Which can be showed to be $<0$.
Thus we must have $a=b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 3
} |
Integral through $u$-substitution v. multiplying out I have the integral:
$ \int (x^2)(x^3-1)^2 \, dx $ Through $u$-substitution, I can write this is equal to $ \frac {1}{3} \int (3x^2)(x^3-1)^2 dx $ which equals $ \frac{1}{3}\cdot\frac{(x^3-1)^3}{3}$.
However, if rather than using $u$-substitution, I multiply within $ \int (x^2)(x^3-1)^2 \, dx $ from the beginning, I don't get the same answer. For example: $ \int (x^2)(x^3-1)^2 \, dx $ = $ \int (x^2)(x^6 -2x^3 + 1) \, dx = \int (x^8 -2x^5 + x^2) \, dx$. The anti derivative of this becomes $ \frac{x^9}{9} - \frac{2x^6}{6} + \frac{x^3}{3} $.
However, these anti-derivatives are not equal. Through u-substitution, when $x=1$, the anti derivative evaluates to zero. In the second route, when I plug in $1$, I get $1/9$.
Where did I go wrong? From what I see, the $u$-substitution version is correct.
| $$
\underbrace{\frac{x^9}{9} - \frac{2x^6}{6} + \frac{x^3}{3} + \text{a constant}} = \underbrace{ \frac 1 3\cdot \frac{(x^3-1)^3} 3 + \text{a different constant}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
partial fractions
$$\frac{1}{1-x^2}$$
$$\frac{1}{1-x^2}=\frac{a}{1-x}+\frac{b}{1+x}$$
$$1=a+ax+b-bx$$
$$1=a+b+x(a-b)$$
$a+b=1$ and $x(a-b)=0\Rightarrow a-b=0\Rightarrow a=b$
$$2a=1\Rightarrow a=\frac{1}{2}$$
$b=\frac{1}{2}$
$$\frac{1}{1-x^2}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}$$
Where I went wrong?
| You didn't do anything wrong.
$$\frac{1}{1-x^2} = \frac{1}{2(x+1)} - \frac{1}{2(x-1)}$$
is equivalent to your answer of
$$\frac{1}{1-x^2} = \frac{1}{2(1-x)} + \frac{1}{2(1+x)}.$$
I think we can agree that both answers have a common term of $\dfrac{1}{2(1+x)}$. Now, notice:
$$\frac{1}{2(1-x)} = \frac{1}{2 \cdot [-1(x-1)]} = \frac{1}{-2(x-1)} = -\frac{1}{2(x-1)}$$
Therefore $-\dfrac{1}{2(x-1)} = \dfrac{1}{2(1-x)}$, and so the answers are the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Loop invariants in logic I am working on some questions about hoare calculus/logic. The given program $\pi$ is:
$ x:=0; y:=1; WHILE \; \lnot x=n \; DO (y:=2y+x + 1; \; x:= x + 1) $
The hoare-logic rules that I can use are in the list below)
H1 {[$t$\x]$\varphi$} x := $t$ {$\varphi$}, provided that $t$ is free for $x$ in $\varphi$
H2 from {$\varphi$} $\pi_1$ {$\psi$} and {$\psi$} $\pi_2$ {$\chi$} follows {$\varphi$} ($\pi_1;\pi_2$) {$\chi$}
H3 from {$\varphi \wedge \epsilon$} $\pi_1$ {$\psi$} and {$\varphi \wedge \neg \epsilon$} $\pi_2$ {$\psi$} follows {$\varphi$} IF $\epsilon$ THEN $\pi_1$ ELSE $\pi_2$ {$\psi$}
H4 from {$\varphi \wedge \epsilon$} $\pi$ {$\varphi$} follows {$\varphi$} WHILE $\epsilon$ DO $\pi$ {$\neg \epsilon \wedge \varphi$}
H5 if $\varphi \rightarrow \varphi`$ and $\psi \rightarrow \psi`$ are true in M = (D, I), then: from {$\varphi`$} $\pi$ {$\psi`$} follows {$\varphi$} $\pi$ {$\psi$}
The questions are:
1) Show that $y = 3 \cdot 2 ^x - x - 2$ is a loop variant for the while-loop from the program $\pi$.
2) Prove $\{true\} \; \pi \{y = 3 \cdot 2^n -n - 2 \} $ .
An update for my solution for the first question, so far. Made some edits.
1) $ \{y = 3 \cdot 2 ^x - x - 2\} \; y:=2y+x + 1; \; x:= x + 1 \; \{y = 3 \cdot 2 ^x - x - 2\} $ Start
2) $ \{y=3⋅2x^{+1}−(x+1)−2\} \; x:= x + 1 \; \{y = 3 \cdot 2 ^x - x - 2\} $ H1. substitute $[x+1/x]$ in the precondition, starting from the right.
3) $ \{2y +x+1 = 3 \cdot 2 ^{x+1} - x - 3\} \; y:=2y+x + 1; \; \{y = 3 \cdot 2 ^x - x - 2\}$ H1. Substitution: $[y:=2y+x + 1;/y]$ in the precondition, starting from the right.
4) $ \{2y +x+1 = 3 \cdot 2 ^{x+1} - x - 3\} y:=2y+x + 1; \; x:= x + 1 \{y = 3 \cdot 2 ^x - x - 2\} $ H2 on 2,3
5) $ \{y = 3 \cdot 2 ^x - x - 2 \land \lnot x = n\} y:=2y+x + 1; \; x:= x + 1 \{y = 3 \cdot 2 ^x - x - 2\} $ H5 on 4
6) $\{y = 3 \cdot 2 ^x - x - 2\} WHILE \lnot x = n\ DO \; y:=2y+x + 1; \; x:= x + 1 \{y = 3 \cdot 2 ^x - x - 2\} $ H4, apply the loop
Question 2. I have continued the numbering:
7) $ \{y=1\} y:=1 \; \{y:=1\} \; $ H1
8) $ \{0=0\} x:=0 \{x:=0\} $ H1
9) $ \{0=0\} x:=0; \; y:=1 \{ x:=0 \} $ H2, 7,8
10) $ \{TRUE\} \; x:= 0 \; ;y:=1; \; \{y = 3 \cdot 2 ^x - n - 2\} \; $ H5,9
11) $ \{TRUE\} \; x:= 0 \; ;y:=1; \; WHILE \lnot x = n\ DO \; y:=2y+x + 1; \; x:= x+1 \{y = 3 \cdot 2 ^x - x - 2\} $ H2 on 6 and 10
| Here's what we want to prove:
{y = 3*2^x - x - 2}
y := 2*y + x + 1;
x := x + 1;
{y = 3*2^x - x - 2}
Indeed, this follows easily after sticking another assertion between the two assignments:
{y = 3*2^x - x - 2}
y := 2*y + x + 1;
{y = 3*2^(x + 1) - x - 3}
x := x + 1;
{y = 3*2^x - x - 2}
To see why the first 3 lines hold, observe that:
\begin{align*}
&y = 3 \cdot 2^x - x - 2 \\
&\implies 2y + x + 1 = 2(3 \cdot 2^x - x - 2) + x + 1 \\
&\implies 2y + x + 1 = (3 \cdot 2^{x + 1} - 2x - 4) + x + 1 \\
&\implies 2y + x + 1 = 3 \cdot 2^{x + 1} - x - 3
\end{align*}
Likewise, to see why the last $3$ lines hold, observe that:
\begin{align*}
&y = 3 \cdot 2^{x + 1} - x - 3 \\
&\implies y = 3 \cdot 2^{x + 1} - (x + 1) - 2
\end{align*}
For the second question, combine H4 with your loop invariant (after noting that the initial conditions satisfy the loop invariant) so that you can replace $x$ with $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve: $\int x^3\sqrt{4-x^2}dx$ $$\int x^3\sqrt{4-x^2}dx$$
I tried changing the expression like this:
$$\int x^3\sqrt{2^2-x^2}dx=\int x^3\sqrt{(2-x)(2+x)}dx$$
And that's where I got stuck. I tried integration by parts but it doesn't simplify the expression.
| Substituting $x=2\sin t$, we get:
$$ \int x^3\sqrt{2^2-x^2}dx=32\int \sin^3 t \cos^2 t dt=32\int(1-\cos^2 t)\cos ^2 t\sin t dt$$
$$=32\int (\cos^2 t-\cos ^4 t)\sin tdt$$
$$=-32\int z^2 dz + 32\int z^4 dz$$ Where $z=\cos t.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Trigonometric polynom Prove that $$\cos\frac{\pi}{7},\cos\frac{3\pi}{7},\cos\frac{5\pi}{7}$$ roots of polynomial $8x^3-4x^2-4x+1=0$
I'm confused, what can i do with $\frac{\pi}{7}$
| Let $z=e^{\frac{i\pi}{7}}$. Then $\cos (\frac{\pi}{7})=\frac{z+z^{-1}}{2}$, $\cos (\frac{2\pi}{7})=\frac{z^2+z^{-2}}{2}$, $\cos (\frac{3\pi}{7})=\frac{z^3+z^{-3}}{2}$. Also $0=\frac{z^7-1}{z-1}=1+z+z^2+z^3+z^4+z^5+z^6$. As $z\ne0,$ divide both sides by $z^3$ to get $$z^3+\frac1{z^3}+z^2+\frac1{z^2}+z+\frac1z+1=0.$$
Now use $$\displaystyle z^2+\frac1{z^2}=\left(z+\frac1z\right)^2-2,\quad z^3+\frac1{z^3}=\left(z+\frac1z\right)^3-3\left(z+\frac1z\right)$$
to obtain a cubic polynomial with the above roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Definite integral with trigononmetric functions I have arrived at definite integral with trigonometric functions
$I(a, b) = \int_0^{2 \pi} \frac{1 - a \sin(\theta) - b \cos(\theta)}{(1 +a^2 +b^2 - 2 a \sin(\theta) - 2 b \cos(\theta) )^{3/2}} \mathrm{d} \theta, \quad 0 \le a, b < 1$
I was expecting result with elliptic integrals, but Mathematica could not evaluate it :-(
Any suggestions to evaluate such integral analytically?
| Using a CAS, $$\alpha\,I(a,b)=\left(a^2+b^2+1-2 \sqrt{a^2+b^2}\right) K(z(a,b))+\left(1-a^2-b^2\right)
E(z(a,b))$$ where $$z(a,b)=\frac{4 \sqrt{a^2+b^2}}{a^2+b^2+1+2 \sqrt{a^2+b^2}}$$ $$\alpha=\frac 12 \sqrt{a^2+b^2+1+2 \sqrt{a^2+b^2}}\,\left(a^2+b^2+1-2 \sqrt{a^2+b^2}\right)\,$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \lfloor\frac{\lfloor x\rfloor}{n}\rfloor=\lfloor\frac{x}{n}\rfloor$. How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor=\left\lfloor\frac{x}{n}\right\rfloor$.
So we want to prove $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\ge\left\lfloor\frac{x}{n}\right\rfloor$ and $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\le\left\lfloor\frac{x}{n}\right\rfloor$
Since $\lfloor x\rfloor\le x$, we can just start from here and prove
$\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\le\left\lfloor\frac{x}{n}\right\rfloor$
But for $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\ge\left\lfloor\frac{x}{n}\right\rfloor$, I have no idea how to start.
| Since $n$ is a natural number we can divide $x$ (with remainder) by $n$ in order to express $x = nb +r_1$ for some $b \in \mathbb{N}$ and $r_1 < n$. Now $\lfloor \frac{x}{n} \rfloor = \lfloor \frac{nb+r}{n} \rfloor = \lfloor b + \frac{r}{n} \rfloor = b$ since $r<n$. On the other hand we have that $\lfloor x \rfloor = bn + r_2$ for the $\textbf{same}$ b and with $r_2<n$. This is because n cannot divide the fractional part of x since it is a natural number. Then $\lfloor \frac{\lfloor x \rfloor}{n} \rfloor = \lfloor \frac{ bn + r_2}{n} \rfloor = b$. Therefore, $\lfloor \frac{\lfloor x \rfloor}{n} \rfloor = \lfloor \frac{x}{n} \rfloor$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
In a right triangle, can $a+b=c?$ I understand that due to the Pythagorean Theorem, $a^2+b^2=c^2$, given that $a$ and $b$ are legs of a right triangle and $c$ is the hypotenuse of the same right triangle. However, most of the time, $a+b\neq c$. What I have been wondering is, is there any set of values for $a$, $b$, and $c$ that make the statement $a+b=c$ true?
| Substituting $c = a + b$ into $a^2+b^2=c^2$ gives us $a^2+b^2=(a+b)^2$. Multiplying that out gives us $a^2+b^2=a^2+2ab+b^2$ which means $2ab=0$
So to satisfy both $c = a + b$ and $a^2+b^2=c^2$ either $a=0$ or $b=0$.
The question then becomes a bit more philosophical. We have a result that satisfies the Pythagorean equation but can we really consider it a right angle triangle? Is a "side" of length zero really a side at all?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1735126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 11,
"answer_id": 4
} |
Derive identities for $\cos(4x)$ and $\sin(4x)$ using following fact So I need to use the fact that: $$\cos(4x) + i\sin(4x) = \left(\cos(x) + i\sin(x)\right)^4$$ to derive identities for $\cos(4x)$ and $\sin(4x)$ in terms of $\cos(x)$ and $\sin(x)$. I'm not sure how to go about this, could I please get some help.
| $$(\cos x +i \sin x)^4=\cos 4x+i \sin 4x$$
$$(\cos x +i \sin x)^4=\cos^4x+4\cos ^3x \cdot (i \sin x)+6\cos ^2x \cdot (i \sin x)^2+4\cos x \cdot (i \sin x)^3+(i \sin x)^4=$$
$$=\cos^4x-6 \cos ^2x \sin^2x+\sin^4x+$$
$$+i(4 \cos ^3 x \sin x-4\cos x\sin^3x)$$
Then $$\cos 4x=\cos^4x-6 \cos ^2x \sin^2x+\sin^4x$$
and $$\sin 4x=4 \cos ^3 x \sin x-4\cos x\sin^3x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1735653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sin n\theta=n\sin \theta-\frac{n(n^2-1)}{3!}\sin^3\theta+\frac{n(n^2-1)(n^2-3^2)}{5!}\sin^5\theta+\cdots$
Prove that
$$\sin n\theta=n\sin \theta-\frac{n(n^2-1)}{3!}\sin^3\theta+\frac{n(n^2-1)(n^2-3^2)}{5!}\sin^5\theta+-\cdots$$
If I am not mistaken, this identity was either proven by Newton or known to him, so if possible I would really like to see the way he approached it, though any solution will suffice.
My brief efforts involved induction on $n$ which failed since I ended up with having to manipulate $\sin( n+1)\theta=\sin( n\theta +\theta)=\sin n\theta \cos\theta+\cos n\theta\sin\theta $, which involves $\cos n\theta$.
I tried the "familiar" method of expansion of $$\sin n\theta=n\theta-\frac{(n\theta)^3}{3!}+\frac{(n\theta)^5}{5!}-+\cdots$$
but this only made it more complicated$$\sin n\theta=n\theta-\frac{(n\theta)^3}{3!}+\frac{(n\theta)^5}{5!}-+\cdots=\\n\Big(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-+\cdots\Big)-\frac{n(n^2-1)}{3!}\Big(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-+\cdots\Big)^3+-\cdots$$
Another attempt would be to use the identity $$\sin n\theta=\frac{1}{2i}(e^{in\theta}-e^{-in\theta})$$
though this was obviously not known to Newton.
In any case, any thougths, ideas are welcome..
EDITA thorough answer has been provided below, but since it involves the use of complex numbers, I deem that the search for another answer, one based solely on the mathematical knowledge up to Newton's time is open.
After a bit more research, it appears that Newton came up with the formula after reading a book by Vieta, but I have been unable to gather further info on whether the formula was known to Vieta as well.
| The series can be retrieved by expressing the Maclaurin expansion of $\sin(nx)$ considered as a function of $\sin x$, avoiding then the use of complex numbers. Moreover, it can be extended to the case of non-integer $n$.
Let $u$ a real number, we introduce the notations $y=\sin x$ et $f(y)=\sin ux$. Successive derivations of $f(y)$ with respect to $x$ give
\begin{align*}
\cos xf^{\prime}(y)&=u\cos ux\\
-\sin xf^{\prime}(y)+(1-\sin^{2}x)f^{\prime\prime}(y)&=-u^{2}\sin ux %
\end{align*}
The latter expression can be written as
\begin{equation}
(1-y^{2})f^{\prime\prime}(y)=yf^{\prime}(y)-u^{2}f
\end{equation}
Now, successive derivation with respect to $y$ are
\begin{align*}
(1-y^{2})f^{(3)}(y) & =3yf^{\prime\prime}(y)+(1-u^{2})f^{\prime}(y)\\
(1-y^{2})f^{(4)}(y) & =5yf^{(3)}(y)+(4-u^{2})f^{\prime\prime}(y)
\end{align*}
This form suggest the recurrence relation:
\begin{equation}
(1-y^{2})f^{(n+2)}(y)=(2n+1)yf^{(n+1)}(y)+(n^{2}-u^{2})f^{(n)}(y)
\end{equation}
which is easily established.
As $f(0)=0$ and $f^{\prime}(0)=u$, we deduce for $p\geq1$
\begin{align}
f^{(2p)}(0)&=0\\
f^{(2p+1)}(0)&=u(1-u^{2})(3^{2}-u^{2})\dots\left[ (2p-1)^{2}-u^{2}\right]
\end{align}
The Maclaurin expansion of $f(y)$ gives:
\begin{equation}
\sin ux=u\sum\limits_{p=0}^{\infty}(1^{2}-u^{2})(3^{2}-u^{2})\dots\left[
(2p-1)^{2}-u^{2}\right] \frac{\sin^{2p+1}x}{(2p+1)!}
\end{equation}
(For $p=0$, the product of the factors in the summation is taken to be equal to 1 by definition).
Additional expansions can be obtained. By taking $y=\sin x$ and $g(y)=\cos ux$, and using the same method, we obtain the same recurrence expression. Now, $g(0)=1$ and $g^{\prime}(0)=0.$ Then
\begin{align}
g^{(2p+1)}(0)&=0\\
g^{(2p)}(0)&=u^{2}(2^{2}-u^{2})(4^{2}-u^{2})\dots\left[ (2p-2)^{2}-u^{2}\right]
\end{align}
Thus,
\begin{equation}
\cos ux=1+\sum\limits_{p=0}^{\infty}(0-u^{2})(2^{2}-u^{2})(4^{2}%
-u^{2})\dots\left[ (2p)^{2}-u^{2}\right] \frac{\sin^{2p+2}x}{(2p+2)!}
\label{cosux}%
\end{equation}
By changing $u\to2u$ in the previous expression, we obtain also
\begin{equation*}
\sin^2ux=\sum\limits_{p=0}^{\infty}(u^{2}-0)(u^{2}-1^{2})(%
u^{2}-2^{2})\dots\left( u^{2}-p^{2}\right) \frac{\left( -1 \right)^p2^{2p+1}\sin^{2p+2}x}{(2p+2)!}
\end{equation*}
From derivation of the above expressions, we can also find the following expansions
\begin{align}
\frac{\cos ux}{\cos x}&=\sum\limits_{p=0}^{\infty}(1^{2}-u^{2})(3^{2}%
-u^{2})\dots\left[ (2p-1)^{2}-u^{2}\right] \frac{\sin^{2p}x}{(2p)!}\\
\frac{\sin ux}{\cos x}&=\frac{1}{u}\sum\limits_{p=0}^{\infty}(0-u^{2}%
)(2^{2}-u^{2})(4^{2}-u^{2})\dots\left[ (2p)^{2}-u^{2}\right] \frac{\sin
^{2p+1}x}{(2p+1)!} \label{Dercosux}%
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
} |
Find the number of solutions of the equation $\sin^22x-\cos^28x=\frac{1}{2}\cos10x$ Find the number of solutions of the equation $\sin^22x-\cos^28x=\frac{1}{2}\cos10x$ lying in the interval $(0,\frac{\pi}{2})$
I found the period of $\sin^22x-\cos^28x$ as $\pi$ and the period of $\frac{1}{2}\cos10x$ is $\frac{\pi}{5}$
I do not know how to solve it further.
| $$\sin^2 2x-\cos^2 8x=\frac{1}{2} \cos10x$$
Using the formulae, $$\sin^2\alpha=\frac{1-\cos 2 \alpha}{2}; \cos^2\alpha=\frac{1+\cos 2 \alpha}{2}$$
$$\frac{1-\cos 4x}{2}-\frac{1+\cos 16x}{2}=\frac{1}{2} \cos10x$$
$$-\cos 4x-\cos 16 x=\cos 10 x$$
$$2 \cos 10x \cos 6x + \cos 10x=0$$
$\cos 10x =0$ or $\cos 6x= -\frac 12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integration of $\int \frac{x^2+20}{(x \sin x+5 \cos x)^2}dx$ How do we integrate
$$\int \frac{x^2+20}{(x \sin x+5 \cos x)^2}dx$$
Could someone give me some hint for this question?
| \begin{align}
I=\int\frac{x^2+20}{(x\sin x+5\cos x)^2}dx&=\int\frac{x^2+20}{(x^2+5^2)(\frac{x}{\sqrt{x^2+5^2}}\sin x+\frac{5}{\sqrt{x^2+5^2}}\cos x)^2}dx\\
&=\int\frac{\left(1-\frac{5}{x^2+5^2}\right)dx}{\cos ^2\left(x-\arctan (\frac{x}{5})\right)}\\
&=\int\frac{d\left(x-\arctan\left(\frac{x}{5}\right)\right)}{\cos ^2\left(x-\arctan (\frac{x}{5})\right)}\\
&=\tan \left(x-\arctan (\frac{x}{5})\right)+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to evaluate $\lim _{x\to \infty }\:\frac{\left(\sqrt{1+\frac{x^3}{x+1}}-x\right)\ln x}{x\left(x^{\frac{1}{x}}-1\right)+\sqrt{x}\ln^2x}$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\:\frac{\left(\sqrt{1+\frac{x^3}{x+1}}-x\right)\ln x}{x\left(x^{\frac{1}{x}}-1\right)+\sqrt{x}\ln^2x}$$
The result should be $-\frac{1}{2}$, but wolfram says that is $0$
| As usual I prefer the elementary way. We have
\begin{align}
L &= \lim_{x \to \infty}\dfrac{\left(\sqrt{1 + \dfrac{x^{3}}{x + 1}} - x\right)\log x}{x(x^{1/x} - 1) + \sqrt{x}\log^{2}x}\notag\\
&= \lim_{x \to \infty}\dfrac{\left(1 + \dfrac{x^{3}}{x + 1} - x^{2}\right)\log x}{\{x(x^{1/x} - 1) + \sqrt{x}\log^{2}x\}\left(\sqrt{1 + \dfrac{x^{3}}{x + 1}} + x\right)}\notag\\
&= \lim_{x \to \infty}\dfrac{(1 + x - x^{2})\log x}{x(x + 1)\{x(x^{1/x} - 1) + \sqrt{x}\log^{2}x\}\left(\sqrt{\dfrac{1}{x^{2}} + \dfrac{x}{x + 1}} + 1\right)}\notag\\
&= \frac{1}{2}\lim_{x \to \infty}\dfrac{(1 + x - x^{2})\log x}{x(x + 1)\{x(x^{1/x} - 1) + \sqrt{x}\log^{2}x\}}\notag\\
&= \frac{1}{2}\lim_{x \to \infty}\dfrac{\left(\dfrac{1}{x^{2}} + \dfrac{1}{x} - 1\right)\log x}{\left(1 + \dfrac{1}{x}\right)\{x(x^{1/x} - 1) + \sqrt{x}\log^{2}x\}}\notag\\
&= -\frac{1}{2}\lim_{x \to \infty}\dfrac{\log x}{x(\exp((\log x)/x) - 1) + \sqrt{x}\log^{2}x}\notag\\
&= -\frac{1}{2}\lim_{x \to \infty}\dfrac{1}{\dfrac{\exp((\log x)/x) - 1}{(\log x)/x} + \sqrt{x}\log x}\notag\\
&= -\frac{1}{2}\lim_{x \to \infty}\dfrac{\dfrac{1}{\sqrt{x}\log x}}{\dfrac{1}{\sqrt{x}\log x}\cdot\dfrac{\exp((\log x)/x) - 1}{(\log x)/x} + 1}\notag\\
&= -\frac{1}{2}\cdot\dfrac{0}{0\cdot 1 + 1}\notag\\
&= 0\notag
\end{align}
In the last step we have used the fact that $\sqrt{x}\log x \to \infty, (\log x)/x \to 0$ as $x \to \infty$ and $(\exp(t) - 1)/t \to 1$ as $t \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Eigenvalues of symmetric block matrices related How to find eigenvalues of following block matrices?
$M=\begin{bmatrix}
A & B & O & O & O & O & O & \cdots & O & B\\
B & A & B & O & O & O & O & \cdots & O & O\\
O & B & A & B & O & O & O & \cdots & O & O\\
O & O & B & A & B & O & O & \cdots & O & O\\
O & O & O & B & A & B & O & \cdots & O & O\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots\\
O & O & O & O & O & O & O & O & A & B\\
B & O & O & O & O & O & O & O & B & A\\
\end{bmatrix} $
$N=\begin{bmatrix}
A & B & O & O & O & O & O & \cdots & O & O\\
B & A & B & O & O & O & O & \cdots & O & O\\
O & B & A & B & O & O & O & \cdots & O & O\\
O & O & B & A & B & O & O & \cdots & O & O\\
O & O & O & B & A & B & O & \cdots & O & O\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots\\
O & O & O & O & O & O & O & O & A & B\\
O & O & O & O & O & O & O & O & B & A\\
\end{bmatrix} $
Where,
$A=\begin{bmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{bmatrix}$
$B=\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}$
$O=\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}$
| Let $n$ be the number of $3 \times 3$ blocks in each row and column of $M$.
Let $P$ be the $3n \times 3n$ permutation matrix such that $P_{k,3k-2} = P_{n+k,3k-1} = P_{2n+k,3k} = 1$ for $k = 1,2,\ldots,n$ and all other entries are $0$.
Then, we have $$PMP^{-1} = \begin{bmatrix}X&I_{n \times n}&I_{n \times n}\\I_{n \times n}&O_{n \times n}&I_{n \times n}\\I_{n \times n}&I_{n \times n}&O_{n \times n}\end{bmatrix}$$ where $X$ is the $n \times n$ matrix defined by $$X = \begin{bmatrix}0 & 1 & & & 1 \\ 1 & 0 & \ddots & & \\ & \ddots & \ddots & \ddots & \\ & & \ddots & 0 & 1 \\ 1 & & & 1 & 0\end{bmatrix}.$$
The eigenvalues $\mu_k$ and corresponding eigenvectors $u_k$ of $X$ are $$\mu_k = 2\cos \dfrac{2k\pi}{n} \ \text{and} \ u_k = \begin{bmatrix}1\\e^{i2\pi k/n} \\ e^{i4\pi k/n} \\ \vdots \\ e^{i(2n-2)\pi/n}\end{bmatrix}, \ \text{for} \ k = 1,2,\ldots,n.$$
If $\lambda$ is an eigenvalue of $PMP^{-1}$ with corresponding eigenvector $\begin{bmatrix}a\\b\\c\end{bmatrix} \neq 0$, then \begin{align*}Xa+b+c = \lambda a \\ a+c = \lambda b \\ a+b = \lambda c \end{align*}
Adding equations 2 and 3 gives $2a = (\lambda-1)(b+c)$, so $b+c = \dfrac{2}{\lambda-1}a$ (I'll let you show that $\lambda = 1$ isn't possible). Substituting this into equation 1 gives $Xa = \left(\lambda - \dfrac{2}{\lambda-1}\right)a$.
There are 2 possibilities:
1) $a$ is a scalar multiple of $u_k$ and $\lambda-\dfrac{2}{\lambda-1} = \mu_k \leadsto \lambda = \dfrac{1}{2}\left(\mu_k+1 \pm \sqrt{\mu_k^2-2\mu_k+9}\right)$
2) $a = 0$, which yields $b+c = 0$, and thus, $c = -b$.
Combining the two cases, we get $3n$ eigenvalues and linearly independent eigenvectors of $PMP^{-1}$:
$$\lambda_{+}^{(k)} = \dfrac{1}{2}\left(\mu_k+1+\sqrt{\mu_k^2-2\mu_k+9}\right) \ \text{and} \ w_{+}^{(k)} = \begin{bmatrix}u_k \\ \tfrac{1}{\lambda_{+}^{(k)}-1}u_k \\ \tfrac{1}{\lambda_{+}^{(k)}-1}u_k \end{bmatrix}, \ \text{for} \ k = 1,2,\ldots,n$$
$$\lambda_{-}^{(k)} = \dfrac{1}{2}\left(\mu_k+1-\sqrt{\mu_k^2-2\mu_k+9}\right) \ \text{and} \ w_{-}^{(k)} = \begin{bmatrix}u_k \\ \tfrac{1}{\lambda_{-}^{(k)}-1}u_k \\ \tfrac{1}{\lambda_{-}^{(k)}-1}u_k \end{bmatrix}, \ \text{for} \ k = 1,2,\ldots,n$$
$$\lambda^{(k)} = -1 \ \text{and} \ w^{(k)} = \begin{bmatrix}0\\\hat{e}_k \\ -\hat{e}_k\end{bmatrix}, \ \text{for} \ k = 1,2,\ldots,n.$$
The eigenvalues of $M$ are the same as the eigenvalues of $PMP^{-1}$.
We can do a similar calculation to get the eigenvalues of $N$. The only difference is that the eigenvalues for the matrix $X$ in this case are $\mu_k = 2\cos\dfrac{k\pi}{n+1}$, and the eigenvectors $u_k$ are different.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola $x^2 -y^2 = 2000^2$ I found this answer here on AoPS. I agree with the answer till it multiplies $49$ by $2$. I think it should be multiplied by $4$ since there are $4$ possible cases:
1) $x+y, x-y$ is positive.
2) $x+y$ is neg and $x-y$ is pos.
3) $x+y$ is pos and $x-y$ is neg.
4) $x+y,x-y$ are neg.
So, I think the answer must be $49 \cdot 4 = 196$.
Consider, the case $x+y=4000$ and $x-y=1000$. There are 4 solutions. $(2500, 1500), (-2500, 1500), (2500, -1500), (-2500, -1500)$.
Am I right? If not, can you highlight my fault and an intuitive understanding to the question?
| Take the simpler case $x^2-y^2=(x-y)(x+y)=20^2=2^4\cdot 5^2$ We give a factor $2$ to each of $x,y$ then distribute the factors of $2^2 \cdot 5^2$ to $x+y,x-y$. There are $9$ of them: $1,2,4,5,10,20,25,50,100$ The case $x-y=8, x+y=50$ gives $x=29, y=21$. You cannot separately pick the signs of $x+y$ and $x-y$, because they need to be of the same sign. There are only two cases, not four.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $\gcd(a,b)=d\Rightarrow\gcd(a^2,b^2)=d^2\ $ Show that if $\gcd(a,b)=d\Rightarrow\gcd(a^2,b^2)=d^2\ $
$\gcd(a,b)=d\Rightarrow\ d\mid a,b\Rightarrow\ \ d^2\mid a^2,b^2\Rightarrow\ d^2\mid\gcd(a^2,b^2)$.
But to complete the proof we must show that: $\gcd(a^2,b^2)\mid d^2$
How can I achieve this?
| Simply, write $a=da'$ and $b=db'$ such that $$\gcd(a', b') =1$$ Hence, $a^2= d^2(a')^2$ and $b^2=d^2(b')^2$.
Since, $\gcd(a', b') =1$, $\gcd((a')^2, (b')^2) =1$ and thus $$gcd(a^2, b^2) = d^2\gcd((a')^2, (b')^2) = d^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1742419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find all continuous functions over reals such that $f(x)+f(y) = f(x+y)-xy-1$ for all $x,y \in \mathbb{R}$
Find all continuous functions over reals such that $f(x)+f(y) = f(x+y)-xy-1$ for all $x,y \in \mathbb{R}$.
I saw first that $f(0) = -1$ but then I am struggling to see how to get a formula for $f(x)$. If I do $x = 0$ we get $f(0) + f(x) = f(x)-1$ which doesn't really help. Is there a better way to get a formula for $f(x)$ here?
| Let $g(x) = f(x) + 1-x^2/2$. Then $g$ is continuous and
$$g(x+y) = g(x) + g(y)$$
The only functions with those properties (continuous and additive) are those of the form $g(x) = ax$, for $a \in \mathbb{R}$. (See Continuous and additive implies linear )
Hence $f(x) = -1+ax + \frac{x^2}{2}$ for some constant $a$.
EDIT: Why that $g$? First, $xy = \frac{1}{2} (x+y)^2 - \frac{1}{2}x^2 - \frac{1}{2} y^2$, so the relation can be written as
$$f(x) - \frac{1}{2}x^2 + f(y) - \frac{1}{2} y^2 = f(x+y) - \frac{1}{2}(x+y)^2 -1. $$
If $h(x) = f(x) -\frac{1}{2} x^2$, then
$$h(x)+h(y) = h(x+y) -1.$$
There are two $h$ on the left and only one on the right, and the balance is $-1$. So if we define $g(x) = h(x) +1$, then
$$g(x)-1 + g(y) -1 = g(x+y)-1 -1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Doubt in solving $\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}+\cot^{-1}\frac{1}{x}=\pi$ Find the value of $x$ if $$\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}+\cot^{-1}\frac{1}{x}=\pi$$ First i tried to calculate the value of
$$\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}=\sin^{-1}\frac{2}{\sqrt{5}}+\sin^{-1}\frac{3}{\sqrt{10}}$$ Letting
$$\theta=\sin^{-1}\frac{2}{\sqrt{5}}+\sin^{-1}\frac{3}{\sqrt{10}}$$ taking $\sin$ both sides and using $sin(A+B)=sinAcosB+cosAsinB$ we get
$$\sin\theta=\sin\left(\sin^{-1}\frac{2}{\sqrt{5}}\right)\cos\left(\sin^{-1}\frac{3}{\sqrt{10}}\right)+\cos\left(\sin^{-1}\frac{2}{\sqrt{5}}\right)\sin\left(\sin^{-1}\frac{3}{\sqrt{10}}\right)$$ so
$$\sin\theta=\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}}$$
Now my doubt is will $\theta=\frac{\pi}{4}$ or $\theta=\frac{3\pi}{4}$ ?
My book has taken $\theta=\frac{3\pi}{4}$
| Use the definition of Principal Value,
$$\sec^{-1}\sqrt5=\tan^{-1}\sqrt{5-1}=\tan^{-1}2$$
If $\csc^{-1}\dfrac{\sqrt{10}}3=y,\csc y=\dfrac{\sqrt{10}}3$ and $0<y<\dfrac\pi2$
$\cot y=+\sqrt{\left(\dfrac{\sqrt{10}}3\right)^2-1}=\dfrac13\iff\tan y=3$
$\implies\csc^{-1}\dfrac{\sqrt{10}}3=\tan^{-1}3$
Like my answer in showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$,
$\tan^{-1}2+\tan^{-1}3=\pi+\tan^{-1}\dfrac{2+3}{1-2\cdot3}=\pi+\left(-\dfrac\pi4\right)$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding equation of a circle given three non - collinear points A circle is given which passes through three non collinear points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ then prove that equation of this circle is given by
$$\begin{vmatrix}
x^2+y^2&x&y&1\\
x_1^2+y_1^2&x_1&y_1&1\\
x_2^2+y_2^2&x_2&y_2&1\\
x_3^2+y_3^2&x_3&y_3&1\\
\end{vmatrix} = 0$$
It is just like one of answers to this question Finding an equation of circle which passes through three points
I would like to know how we get to this determinant and I have no idea from where I can start?
| The quadratic form $Ax^2+By^2+Cxy+Dx+Ey+F=0$ represents always a conic (eventualy degenerate in two right lines if it is a product of two linear equations) and it represents a circle if and only if $A=B\ne0$ and $C=0$. Then for a circle we have the equation
$$A(x^2+y^2)+Dx+Ey+F=0$$ When we want a circle passing through three non collinear points $(x_i,y_i)$ we have a linear system of three equations whose solution gives the three values $ a=\dfrac{D}{A}, b=\dfrac{E}{A}, c=\dfrac{F}{A}$ of the required circle.
But when we want to have the equation of the circle without getting explicitely $a,b,c$ we consider the linear system of four equations with three unknowns $a,b,c$
$$(x^2+y^2)+ax+by+c=0\\(x_1^2+y_1^2)+ax_1+by_1+c=0\\(x_2^2+y_2^2)+ax_2+by_2+c=0\\(x_3^2+y_3^2)+ax_3+by_3+c=0$$
and it is well known that the condition of compatibility of this four equations is that its determinant be equal to zero. In other words, making more explicitely, we have the matricial equation
$$\begin{pmatrix}
x^2+y^2&x&y&1\\
x_1^2+y_1^2&x_1&y_1&1\\
x_2^2+y_2^2&x_2&y_2&1\\
x_3^2+y_3^2&x_3&y_3&1\\
\end{pmatrix}\cdot\begin{pmatrix}1\\a\\b\\c\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}$$
There are two evident solutions, $(0,0,0,0)$ and $(1,a,b,c)$ so the determinant must be equal to $0$ because if not then the solution should be unique.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find maximum of $f(x)=\frac{x^m(1-x)^n}{x(1-x)} $ on $0< x< 1$ I would appreciate if somebody could help me with the following problem:
Q: Find maximum $f(x)$ on $0< x< 1$ ($n,m \in \mathbb{N}$)
$$f(x)=\frac{x^m(1-x)^n}{x(1-x)} $$
I try
$$f'(x)=(m-1) x^{m-2} (1-x)^{n-1}-(n-1) x^{m-1} (1-x)^{n-2}$$
...
| If $m=n=1$, then $f(x)=1$ is a constant function. Thus suppose $m>1$ or $n>1$.
Factoring $f'(x)$,
\begin{align}
f'(x)&=x^{m-2}(1-x)^{n-2}((m-1)(1-x)-(n-1)x)\\
&=x^{m-2}(1-x)^{n-2}((m-1)-(m+n-2)x)
\end{align}
and so $x=\frac{m-1}{m+n-2}$ satisfies $f'(x)=0$. Is $\frac{m-1}{m+n-2}$ between $0$ and $1$? If $m>1$ and $n=1$, then it is $1$ and not in $(0,1)$. If $m=1$ and $n>1$, then it is $0$ and not in $(0,1)$. In fact, $f(x)=(1-x)^{n-1}$ and $f(x)=x^{m-1}$ do not have the maximum in $(0,1)$. However, if $n\ge 2$ and $m\ge 2$, $0<\frac{m-1}{m+n-2}<1$, so $f(x)$ has a extreme value at $x=\frac{m-1}{m+n-2}$, and there exists a maximum value $f(\frac{m-1}{m+n-2})$.
$$
\therefore(\text{maximum of $f(x)$})=\begin{cases}
1,&\text{if }m=1\text{ and } n=1\\
\text{doesn't exist},&\text{if only one of $m$ and $n$ is $1$}\\
\left(\frac{m-1}{m+n-2}\right)^{m-1}\left(\frac{n-1}{m+n-2}\right)^{n-1},&\text{if }m\ge 2\text{ and }n\ge 2\\
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simplifying radicals inside radicals: $\sqrt{24+8\sqrt{5}}$ Simplify: $\sqrt{24+8\sqrt{5}}$
I removed the common factor 4 out of the square root to obtain $2\sqrt{6+2\sqrt{5}}$, but the answer key says it is $2+2\sqrt{5}$.
Am I missing out on some general rule here?
| $$\sqrt{24+8\sqrt{5}}=\sqrt{20+4+2\cdot 2\cdot 2\sqrt{5}}=\sqrt{(2\sqrt5)^2+2^2+2 \cdot 2 \cdot 2\sqrt{5}}=\sqrt{(2+2\sqrt5)^2}=2+2\sqrt5$$
or, to continue on your start,
$$\sqrt{6+2\sqrt5}=\sqrt{1^2+(\sqrt5)^2+2\sqrt5}=1+\sqrt5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
How many elements are in the set $ \{ \left( \frac{2+i}{2-i} \right) ^n : n \in \mathbb N\}$ How many elements are in the set $$ \left\{ \left( \frac{2+i}{2-i} \right) ^n : n \in \mathbb N\right\}$$
My attempt:
$\left( \frac{2+i}{2-i} \right) ^n = \left( \frac{3}{5} + \frac{4}{5}i \right)^n=e^{\arctan(4/3)ni}$
So if for $n_1 < n_2$ it is true that $e^{\arctan(4/3)n_1 i}=e^{\arctan(4/3)n_2 i}$, then $z = n_2-n_1 = \frac{2\pi}{\arctan(4/3)}k, k\in \mathbb Z$
Because it is not really clear what $\arctan(4/3)$ is, I don't know whether we can find $z,k$ to satisfy the equation above.
| I was recently given off line a solution to this problem which to me seems a bit down to earth but probably easier to grasp.
If we have $a+bi$ such that $a = 3 \mod 5$ and $b = 4 \mod 5 $, i.e. $a = 5n + 3, b = 5m + 4$ for $m,n \in \mathbb Z$, then $(a+bi)^2 = x + yi$ and $x = 3 \mod 5$ and $y = 4 \mod 5 $ (can be easily checked by a direct calculation).
By induction $\forall n \in \mathbb N, (a+bi)^n = x + yi$ and $x = 3 \mod 5$ and $y = 4 \mod 5 $.
Now if we have $n_1<n_2$ such that $$\left( \frac{3}{5} + \frac{4}{5}i \right)^{n_1}=\left( \frac{3}{5} + \frac{4}{5}i \right)^{n_2}$$ then there should be $n_0 \in \mathbb N$ such that $$\frac{1}{5^{n_0}}\left( 3 + 4i \right)^{n_0}=\frac{1}{5^{n_0}}\left( x + yi \right)=1$$
So $x=5^{n_0}$, i.e. $x = 0 \mod 5$, but we just showed that $x = 3 \mod 5$. So all elements in the set are different.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\int \frac{1-\sin x}{(1+\sin x)\cos x}dx$ Evaluate $$I=\int \frac{(1-\sin x) dx}{(1+\sin x)\cos x}$$
I tried in the following way:
$$1-\sin x=1-\cos\left(\frac{\pi}{2}-x\right)=2 \sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$$
Similarly $$1+\sin x=2 \cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$$
So
$$I=\int \tan^2\left(\frac{\pi}{4}-\frac{x}{2}\right) \sec x \: dx=\int \sec^2\left(\frac{\pi}{4}-\frac{x}{2}\right) \sec x \: dx-\int \sec x \:dx$$
If $$J=\int \sec^2\left(\frac{\pi}{4}-\frac{x}{2}\right) \sec x \: dx$$ Applying parts for $J$ we get
$$J=-2\sec x \tan\left(\frac{\pi}{4}-\frac{x}{2}\right)+2\int \sec x \tan x \tan\left(\frac{\pi}{4}-\frac{x}{2}\right)dx $$
But i am clueless from here
| One may write
$$
\begin{align}
I&=\int \frac{(1-\sin x) dx}{(1+\sin x)\cos x}
\\\\&=\int \frac{(1-\sin x)\cos x dx}{(1+\sin x)\cos^2 x}
\\\\&=\int \frac{(1-\sin x)\cos x dx}{(1+\sin x)(1-\sin^2 x)}
\\\\&=\int \frac{(1-u)du}{(1+u)(1-u^2)}\quad (u=\sin x)
\\\\&=\int \frac{du}{(1+u)^2}
\\\\&=-\frac{1}{1+\sin x}+C.
\end{align}
$$
Edit. (from @Bernard) The Bioche rules are rules that give hints on the useful substitutions for the integration of rational functions in $\sin$ and $\cos$. Namely you consider the differential form $f(\sin x,\cos x)dx$, and substitute successively $−x$ to $x$, then $π−x$, and finally $π+x$. If the differential form is invariant by one of these substitutions, set $u$ equal to the function which is invariant by the same substitution – which is $u=\cos x$, $u=\sin x$ or $u=\tan x$, respectively.
If it happens the differential form is invariant by two of these substitutions, one may set $u$ equal to a trigonometric function of $2x$.
The worst case is when none of these substitutions work. As a last resort, one sets $u=\tan \frac x2$, since there are standard formulae which express $\sin x,\cos x$ and $\tan x$ as functions of $u$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Rational Expression equivalent form EDIT: I know how to find the answer, but does anyone know why plugging in numbers for x does not work?
The Question: If the rational expression $\frac {3x^2}{3x-1}$ is rewritten in the equivalent form $\frac {\frac 13}{3x-1}+A$, what must expression A be in terms of x?
The four answer choices: A) $x+ \frac 13$ B) $x+1$ C) $x-1$ D) $x-3$
The Answer:
A) $x+ \frac 13$
I don't really know the way this question is "supposed" to be solved, so I just tried plugging in some numbers.
*
*Plugging in 0 for x, I would get $0=\frac {-1}3+ A$
*Plugging in 1 for x, I would get $\frac 32=\frac 23 + A$.
*Plugging in 2 for x, I would get $\frac {12}{5}=\frac 53 +A$.
The equation for $x=0$ makes it seem like $A$ really is the correct answer, but plugging in 1 and 2, I wasn't sure which one, if any were correct. Pretty sure I'm just missing something really obvious here, but asking anyway. Answer Key contains the answer listed above.
| You can solve the equation for A and get
$A = \frac {3x^2-\frac 13}{3x-1}$.
In the numerator 3 can be factored out.
$A = \frac {3(x^2-\frac 19)}{3x-1}$
$x^2-\frac 19$ is equivalent to the third binomial formula $a^2-b^2=(a-b)\cdot (a+b)$.
Therfore $ x^2-\frac 19=(x-\frac 13)\cdot (x+\frac 13)$
$A= \frac {3\cdot (x-\frac 13)\cdot (x+\frac 13)}{3x-1}=\frac { (3x-1)\cdot (x+\frac 13)}{3x-1}$
Numerical example: $x=2$
$\frac {3x^2}{3x-1}=\frac {3\cdot 2^2}{3\cdot 2-1}=\frac{12}{5}$
This result has to be the same like
$\frac{\frac13}{3x-1}+x+\frac 13=\frac{1}{9x-3}+x+\frac13$
$=\frac{1}{15}+2+\frac 13=\frac{1}{15}+\frac{30}{15}+\frac{5}{15}=\frac{36}{15}=\frac{12}{5}\quad \checkmark$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 1
} |
order $a$ = 5, $a^3b = ba^3$. show that that $ab = ba$. Let $a, b$ be elements of a group $G$. Suppose that a has order $5$ and that $a^3b = ba^3$. I want to show that that $ab = ba$.
Here is what I think:
We know that we have $a^1, a^2, a^3, a^4, a^5 = 1$. So, $a^4 = a^{-1}$ and $a^3 = a^{-2}$ and $a^{2} = a^{-4}$. Now,
$a^3b = ba^3$ then $a^{-2}b = ba^{-2}$ then $ba^2 = a^2b$ then $ba^{-4} = a^{-4}b$ then $a^{-1}b = ba^{-1}$ then multiply on the left and right by $a$ to get $ba=ab$. Is this correct or is there a simpler approach?
| Since $a$ has order 5, we have $a = a^6$. Then:
$$ ab = a^6b = a^3(a^3b) = a^3(ba^3) = (a^3b)a^3 = (ba^3)a^3 = ba^6 = ba \;\;\;\;\square $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$ Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$.
I am not asked to use any form, so I am going to use $k\cos(x-\alpha)$.
$8\cos x - 6\sin x = k\cos(x-\alpha)$
$$=k(\cos x\cos\alpha + \sin x\sin\alpha)$$
$$=k\cos\alpha\cos x + k\sin\alpha\sin x$$
equating coefficients:
$k\cos\alpha = 8$
$k\sin\alpha = 6$, or is it $k\sin\alpha = -6$? I find this really confusing
$k = \sqrt{8^2 + 6^2} = 10$
$\alpha$ is in the 1st quadrant where both sin and cos are positive.
$\alpha = \arctan\frac{6}8 = 36.8^{\circ}$
$\therefore10\cos(x - 36.8) = 5$
I have a maximum and minimum value of 10
From here I do not know how to finish solving for x.
| converting the given equation into $$\tan(x/2)$$ we get
$$8\,{\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left(
\tan \left( x/2 \right) \right) ^{2}}}-12\,{\frac {\tan \left( x/2
\right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}}-5
=0$$
setting $$\tan(x/2)=t$$ you have to solve this equation
$$-{\frac {13\,{t}^{2}+12\,t-3}{{t}^{2}+1}}=0$$ for $$t$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Dimension about space of matrices of order 3 over the field of the real. Consider the vector space of the matrices of order 3 over the field of the real $M_{3}\left(\Re\right)$ numbers. and let S be the subspace such that is spanned by the matrices of the form $AB-BA$. What is the dimension of S?
I was thinking in... the form of that matrix, and i know... If $X=AB−BA$ , then $trX=tr(AB−BA)=tr(AB)−tr(BA)=tr(AB)−tr(AB)=0$
well, i worked in an basis of that space... this:
$\left\{ \begin{pmatrix}0 & 1 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
1 & 0 & 0\\
0 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{pmatrix},\\ \begin{pmatrix}0 & 0 & 0\\
0 & 0 & 0\\
1 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
0 & 0 & 0\\
0 & 1 & 0
\end{pmatrix},\begin{pmatrix}1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & -1
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & 1
\end{pmatrix}\right\} $,
it's fine the basis?
| Let $X=AB-BA$ then $tr(X)=tr(AB-BA) = tr(AB) - tr(BA) = 0 $ we have the subspace of matrices with trace = 0, and I found the basis for the space:
$\left\{ \begin{pmatrix}0 & 1 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
1 & 0 & 0\\
0 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{pmatrix},\\ \begin{pmatrix}0 & 0 & 0\\
0 & 0 & 0\\
1 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
0 & 0 & 0\\
0 & 1 & 0
\end{pmatrix},\begin{pmatrix}1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & -1
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1
\end{pmatrix},\right\}
$
then the dimension of the space is 8.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
prove $\sum_{k=1}^n(-1)^{k+1}\cos^{2n+1}\left(\dfrac{k\pi}{2n+1}\right)=\frac{1}{2}-\frac{2n+1}{2^{2n+1}}$ Today I found the identity : $$\sum_{k=1}^n(-1)^{k+1}\cos^{2n+1}\left(\dfrac{k\pi}{2n+1}\right)=\frac{1}{2}-\frac{2n+1}{2^{2n+1}}$$.
How to prove or disprove this?
Thank you.
| Hint : $$\cos^{2n+1}\left(\dfrac{k\pi}{2n+1}\right)=\frac{1}{2^{2n+1}}(e^{\frac{ik\pi}{2n+1}}+e^{-\frac{ik\pi}{2n+1}})^{2n+1}.$$ Then use the binomial theorem and switch the sums to get $$\frac{1}{2^{2n+1}}\sum_{j=0}^{2n+1}\sum_{k=1}^n (-1)^{k+1} \binom{2n+1}{j}e^{\frac{ikj\pi}{2n+1}}e^{-\frac{ik\pi(2n+1-j)}{2n+1}}$$ which is equal to $$-\frac{1}{2^{2n+1}}\sum_{j=0}^{2n+1} \binom{2n+1}{j} \sum_{k=1}^n e^{\frac{2ikj\pi}{2n+1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
equation of an ellipse given its center and two tangent lines There exists an ellipse centered at (0,0) with two tangent lines given by $y=-\frac{1}{2}x + \frac{\sqrt{39}}{2}$ and $y=\frac{1}{3}x + \frac{7}{3}$. Find the ellipse.
So, I used the equation of one line and substituted it into the equation of an ellipse:
$\frac{x^2}{a^2} + \frac{(\frac{1}{3}x + \frac{7}{3})^2}{b^2}$=1.
Then I get a quadratic equation in $x$:
$(9b^2+a^2)x^2 + (14a^2)x + (49a^2 - 9a^2b^2)$ = 0
When I solve for $x$ using the quadratic formula, I get a very complicated expression in terms of $a$ and $b$:
$x = \frac{-14a^2 \pm \sqrt{196a^2 - 1764a^2b^2 + 324a^2b^4 - 98a^4 + 36a^4b^2}}{18b^2 + 2a^2} $
At this point, I figure that there must be a simpler way of solving this problem. Otherwise, I have to plug this expression into the original equation of an ellipse to get an equation in terms of a and b. Then, repeat the entire process with the second line to get a second equation in terms of a and b, and then use that system to solve for both a and b.
I've also thought about doing this with derivatives:
The slope of the second line is $\frac{7}{3}$, so the derivative at the $(x,y)$ coordinate where that line intersects the ellipse, of the ellipse, using implicit differentiation, yields (after solving):
$\frac{dy}{dx} = -\frac{xb^2}{ya^2} = \frac{7}{3}$
This, again, seems like it would lead to a lot of messy algebra if I were to plug this back into the original equation after solving for x:
$x=-\frac{7ya^2}{3b^2}$
Then plugging in:
$\frac{(-\frac{7ya^2}{3b^2})^2}{a^2} + \frac{y^2}{b^2} = 1$
As you can see, the algebra gets messy. Is there a more simplistic way of going about this?
| You can prove that the tangent line to an ellipse of equation
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$$
at a point $(s,t)$ on the ellipse is
$$
\frac{sx}{a^2}+\frac{ty}{b^2}=1
$$
You can rewrite the equations of your tangent lines as
$$
\frac{x}{\sqrt{39}}+\frac{2y}{\sqrt{39}}=1,
\qquad
\frac{-x}{7}+\frac{3y}{7}=1
$$
Thus there must exist $(s,t)$ and $(u,v)$ on the ellipse such that
$$
\frac{s}{a^2}=\frac{1}{\sqrt{39}},
\quad
\frac{t}{b^2}=\frac{2}{\sqrt{39}},
\qquad
\frac{u}{a^2}=-\frac{1}{7},
\quad
\frac{v}{b^2}=\frac{3}{7}
$$
With the assumption that $\dfrac{s^2}{a^2}+\dfrac{t^2}{b^2}=1$ we get
$$
\frac{s}{\sqrt{39}}+\frac{2t}{\sqrt{39}}=1
$$
and, similarly,
$$
-\frac{u}{7}+\frac{3v}{7}=1
$$
Now $\dfrac{s}{\sqrt{39}}=\dfrac{a^2}{39}$ and $\dfrac{2t}{\sqrt{39}}=\dfrac{4b^2}{39}$, so we get
$a^2+4b^2=39$.
Similarly $-\dfrac{u}{7}=\dfrac{a^2}{49}$ and $\dfrac{3v}{7}=\dfrac{9b^2}{49}$, so we get $a^2+9b^2=49$.
This implies $b^2=2$ and $a^2=31$, so your ellipse has equation
$$
\frac{x^2}{31}+\frac{y^2}{2}=1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1757284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the system of equations: $a+b+c=2$, $a^2+b^2+c^2=6$, $a^3+b^3+c^3=8$ If we have \begin{cases} a+b+c=2 \\ a^2+b^2+c^2=6 \\ a^3+b^3+c^3=8\end{cases} then what is the value of $a,b,c$?
| Since the expressions you know the value of are homogeneous and symmetric in $a,b,c$, we know (or should know) they are expressible in terms of homogeneous polynomials in the elementary symmetric functions $E_1=a+b+c$, $E_2=ab+ac+bc$, $E_3=abc$ and in polynomial terms.
In order to find these expressions I prefer a more systematic approach, which can be as follows, rather than plugging in the values and going by attempts.
Degree considerations tell us that
$$
a^2+b^2+c^2=\alpha E_1^2+\beta E_2
$$
With $(a,b,c)=(1,0,0)$ we obtain $1=\alpha$; with $(a,b,c)=(1,1,0)$ we get $2=4\alpha+\beta$. Thus $\beta=-2$.
Also
$$
a^3+b^3+c^3=\alpha E_1^3+\beta E_1E_2+\gamma E_3
$$
With $(a,b,c)=(1,0,0)$ we obtain $1=\alpha$; with $(a,b,c)=(1,1,0)$ we get $2=8\alpha+2\beta=0$; with $(a,b,c)=(1,1,1)$ we get $3=27\alpha+9\beta+\gamma$. Therefore $\beta=-3$ and $\gamma=3$.
The relations are then
$$
\begin{cases}
a^2+b^2+c^2=E_1^2-2E_2\\[4px]
a^3+b^3+c^3=E_1^3-3E_1E_2+3E_3
\end{cases}
$$
Since you're given $E_1=2$, $a^2+b^2+c^2=6$ and $a^3+b^3+c^3=8$, the relations become
$$
E_2=-1,\qquad
E_3=2E_2=-2
$$
Why is this important? Because given $E_1$, $E_2$ and $E_3$, the polynomial having $a,b,c$ as roots is
$$
x^3-E_1x^2+E_2x-E_3
$$
(Viète's formulas). Thus the polynomial is
$$
x^3-2x^2-x+2=x^2(x-2)-(x-2)=(x^2-1)(x-2)=(x-1)(x+2)(x-2)
$$
and you found the roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
How to show that $f(x) = x \arctan \left(x \sin^{2} \left(\frac{1}{x}\right)\right)$ is strictly increasing for $x \geq 1$? I am trying to prove that $f(x) = x \arctan \left(x \sin^{2} \left(\frac{1}{x}\right)\right)$ is a strictly increasing function for $x \geq 1$.
I try to do this by showing that $f'(x)>0$ for all $x \geq 1$.
We have $$f'(x)= \arctan(x \sin^{2} \frac{1}{x}) + \frac{x}{1+x^{2} \sin^{4} \frac{1}{x}} \left(\frac{-2\cos\frac{1}{x} \sin \frac {1}{x}}{x} + \sin^{2} \frac{1}{x}\right),$$
so it's not clear if this quantity is positive.
Now I can show that $x \arctan \left(\frac{1}{x}\right)$ is strictly increasing (by computing its derivative and using simple trig inequalities to show that it is positive), and that $$x \sin^{2} \left(\frac{1}{x}\right)- \frac{1}{x}>0$$ So my aim is somehow to compare $f(x)$ with $x\arctan \frac{1}{x}$ and use the above inequality to prove that $f'(x)>0$.
But now I am stuck. Any suggestions as to how to proceed? Or any other approach to the problem would be appreciated!
| Simple, use derivative, not approximations. You get:
$(x\arctan(x\sin^2(\frac{1}{x})))'=\arctan(x\sin^2(\frac{1}{x}) + \frac{x}{1+x^2\sin^4(\frac{1}{x})}(sin^2(\frac{1}{x}) + \frac{x}{1+x^2\sin^4(\frac{1}{x})}(-2x\sin\frac{1}{x}\cos\frac{1}{x}(-\frac{1}{x^2}))$
so all are positive for $x \geq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Hard summation involving binomial and quadratic What is $$\sum \frac{2r^2-98r+1}{(100-r)({100\choose r})}$$ Where $r\in [1,99]$I have reduced it to $$\frac{(2r^2-98r+1)}{(100){99\choose r}}$$ what to do further? Partial fractions don't seem to help.
| Using that
$$2r^2-98r+1=(r+1)^2-r(100-r)$$
yields a telescopic sum :
$$\begin{align}\sum_{r=1}^{99}\frac{2r^2-98r+1}{(100-r)\binom{100}{r}}&=\sum_{r=1}^{99}\frac{(2r^2-98r+1)\cdot r!\cdot (100-r)!}{(100-r)\cdot 100!}\\\\&=\frac{1}{100!}\sum_{r=1}^{99}(2r^2-98r+1)\cdot r!\cdot (99-r)!\\\\&=\frac{1}{100!}\sum_{r=1}^{99}(r^2+2r+1-100r+r^2)\cdot r!\cdot (99-r)!\\\\&=\frac{1}{100!}\sum_{r=1}^{99}((r+1)^2-r(100-r))\cdot r!\cdot (99-r)!\\\\&=\frac{1}{100!}\sum_{r=1}^{99}((r+1)\cdot (r+1)!\cdot (99-r)!-r\cdot r!\cdot (100-r)!)\\\\&=\frac{1}{100!}(100\cdot 100!\cdot 0!-1\cdot 1!\cdot 99!)\\\\&=\color{red}{\frac{9999}{100}}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$x^2+y^2=2z^2$, positive integer solutions Determine all positive integer solutions of the equation $x^2+y^2=2z^2$.
First I assume $x \geq y$, and I have $x^2-z^2=z^2-y^2$. Then I have $(x-z)(x+z)=(z-y)(z+y)$, but from here, I don't know how it can help me to describe solutions (I know that there are infinitely many).
| Just a start:
Note that $$\left(\frac{x-y}{2}\right)^2 + \left(\frac{x+y}{2}\right)^2=\frac{x^2+y^2}{2}=z^2.$$
And $\frac{x-y}{2}$ and $\frac{x+y}{2}$ are integers (why?)
So you need to find solutions to $u^2+v^2=z^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Show that this limit yields $\gamma=\lim_{n \to \infty}\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k= n}^{(n-1)^2}\frac{1}{k}$ $\gamma =0.5772156...$ is Euler's constant
Show that this limit yields
$$\gamma=\lim_{n \to \infty}\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k=n}^{(n-1)^2}\frac{1}{k}$$
| Since
$$\frac{1}{k} \leqslant \int_{k-1}^k \frac{dx}{x} = \log k - \log(k-1) \leqslant \frac{1}{k-1},$$
it follows that
$$\sum_{k = n}^{(n-1)^2}\frac{1}{k} \leqslant \log(n-1)^2 - \log(n-1) = \log(n-1) \leqslant \sum_{k = n}^{(n-1)^2}\frac{1}{k} + \frac{1}{n-1} - \frac{1}{(n-1)^2},$$
and
$$ \sum_{k = 1}^{n-1}\frac{1}{k} - \log(n-1) + \frac{1}{n-1} - \frac{1}{(n-1)^2}\leqslant \sum_{k = 1}^{n-1}\frac{1}{k} - \sum_{k = n}^{(n-1)^2}\frac{1}{k} \leqslant \sum_{k = 1}^{n-1}\frac{1}{k} - \log(n-1).$$
By the squeeze theorem
$$\lim_{n \to \infty}\left(\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k= n}^{(n-1)^2}\frac{1}{k}\right) = \lim_{n \to \infty}\left(\sum_{k=1}^{n-1}\frac{1}{k} - \log(n-1)\right) = \gamma$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Game theory probability I have a variant of two-finger morra game, where the winner is determined by the parity of the sum of the two numbers thrown, but the amount won or lost is the product of the two numbers. There are two player - Alice and Bob
If Alice plays:
1 finger with probability $p$
2 fingers with probability $1 − p$
Alice wins when outcome is even, and Bob wins if it's odd.
What’s the expected payoff that Bob can achieve if he knows $p$?
How should Alice choose $p$ such that Bob’s payoff is minimized?
| I suppose that Alice wins if the sum of the numbers is even. If $a \{1,2\}$ and $b \in \{1,2\}$ are the strategies played by Alice and Bob, respectively. The payoff of Bob is:
$$B(a,b) = \begin{cases}
-1 & \text{if}~a = b = 1 \\
2 & \text{if}~a = 1, b = 2 ~\text{or}~ a=2, b=1\\
-4 & \text{if}~a = b = 2 \\
\end{cases}.$$
The expected payoff of Bob knowing $p$ is:
$$\mathbb{E}[B(a, b) | b] = \begin{cases}
B(1,1)p + B(2,1)(1-p) & \text{if}~b = 1 \\
B(1,2)p + B(2,2)(1-p) & \text{if}~b = 2
\end{cases}\\= \begin{cases}
-1p + 2(1-p) & \text{if}~b = 1 \\
2p -4(1-p) & \text{if}~b = 2
\end{cases}\\ = \begin{cases}
2 -3p & \text{if}~b = 1 \\
6p-4 & \text{if}~b = 2
\end{cases}$$
Bob will choose strategy $b = 1$ if
$$2-3p > 6p-4 \Rightarrow p<\frac{2}{3}.$$
Of course, he will choose $b=2$ if $p>\frac{2}{3}$. If $p=\frac{2}{3}$, there is no difference between $a$ and $b$.
Now, if $p < \frac{2}{3}$, then Bob plays $b=1$ and his average payoff is $2-3p \in [0, 2]$. On the other hand, if Bob plays $b=2$, then his average payoff is $6p-4 \in [0, 2]$. The minimum of the Bob's payoff is $0$ and it is attained for $p = \frac{2}{3}$. This means that Alice must choose $p = \frac{2}{3}$ to minimize Bob's payoff.
Notice that $p = \frac{2}{3}$ is the component of Nash equilibrium for Alice in the space of mixed strategies.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $a, b$ so that $\lim_{n \to \infty}((1-n^3)^{\frac{1}{3}}-an-b)=0$ If we go by definition, then we have to find $a,b$, so that for every $\varepsilon >0, \exists N(\varepsilon), $ so that $\forall n>N(\varepsilon)$
$$|(1-n^3)^{\frac{1}{3}}-an-b| < \varepsilon$$
My attempt,
$$ \left| \frac{1-n^3-an\left(\sqrt[3]{1-n^3}^2\right)-b\left(\sqrt[3]{1-n^3}\right)^2}{\left(\sqrt[3]{1-n^3}\right)^2} \right|< \varepsilon$$
but I'm stuck and I'm not sure if this is how you find $a$ and $b$.
| You want $(1-n^3)^{\frac 13} \rightarrow an+b$
If $(1-n^3)^{\frac 13} = an+b + \delta$
Then $(1-n^3) = (an+b + \delta)^3$
$1-n^3 = (an+b + \delta)(a^2n^2+b^2 + \delta^2+2abn+2an \delta + 2b \delta)$
$1-n^3 = a^3n^3+ab^2n + an\delta^2+2a^2bn^2+2a^2n^2 \delta + 2abn \delta$
$+a^2bn^2+b^3 + b\delta^2+2ab^2n+2abn \delta + 2b^2 \delta$
$+a^2n^2\delta+b^2\delta + \delta^3+2abn\delta+2an \delta^2 + 2b \delta^2$
$1-n^3 = a^3n^3+(3a^2b+3a^2 \delta)n^2 +(3ab^2 + 3a\delta^2+ 6ab \delta )n + 3b\delta^2 + 3b^2 \delta+b^3 + \delta^3$
Compare coefficients of $n^3$ to get $a=-1$
$1 = (3b+3 \delta)n^2 -(3b^2 + 3\delta^2+ 6b \delta )n + 3b\delta^2 + 3b^2 \delta+b^3 + \delta^3$
Coefficient of $n^2$ is $3b+3\delta$, where $\delta \rightarrow 0$. Therefore set $b=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Conditional Probabilities multi coloured balls from a bag I have two bags
Bag1 : Contains 4 red and 5 black balls
Bag2 : Contains 3 red and 7 black balls
One ball is drawn from bag1 and 2 balls are drawn form bag2. Find the probability of 2black balls and 1 red ball.
How do I go about solving this problem?.
This is what I have come up with so far:
$$P(B,RR) + P(R,BB) = \frac{5}{9} \cdot \frac{7}{10} \cdot \frac{3}{9} + \frac{4}{9} \cdot \frac{7}{10} \cdot \frac{6}{9}$$
Thanks.
|
This is what I have come up with so far:
$$P(B,RR) + P(R,BB) = \frac{5}{9} \cdot \frac{7}{10} \cdot \frac{3}{9} + \frac{4}{9} \cdot \frac{7}{10} \cdot \frac{6}{9}$$
That's not quite okay. The probability of drawing the red from bag-2 is twice that. It is the old "two ways to draw the same thing" problem: red then black, black then red.
$$P(B,RR) + P(R,BB) = \frac{5}{9} \cdot \frac{7}{10} \cdot \frac{3}{9} + \frac{5}{9} \cdot \frac{3}{10} \cdot \frac{7}{9} + \frac{4}{9} \cdot \frac{7}{10} \cdot \frac{6}{9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of three consecutive cubes equals a perfect square I have found this problem in an old German textbook:
Find all sets of three consecutive integers such that the sum of their cubes is a perfect square.
We can write
$$S = (x-1)^3 + x^3 + (x+1)^3 = (x-1+x+1)((x-1)^2 - (x-1)(x-1) + (x+1)^2) + x^3$$
which reduces to
$$S = 3x(x^2 + 2).$$
If we set $x^2 + 2 = 3x$, we get
$$x^2 - 3x + 2 = 0 \iff (x-1)(x-2) = 0$$
and we thus obtain the solutions $(0,1,2)$ and $(1,2,3)$.
At first I conjectured that these are the only solutions, but I couldn't prove this. However, I was wrong: $(23,24,25)$ also satisfies the relationship.
It is worth noting that these are the only solutions for $x \leq 100000$. Can anyboy help me prove that these are the only ones? Or otherwise, help me find all other triples?
| What you can do (but not ask from seventh graders) is to turn the equation into an elliptic curve in Weierstrass normal form, by multiplying both sides by $9$ and putting $9S=Y^2$ and $X=3x$. Then we get
$$Y^2=X^3+18X$$
which indeed denotes an elliptic curve and we can use for instance SageMath to return its integral points: $(0, 0), (3, 9), (6, 18), (72, 612)$.
Then we find $x=0,1,2,24$.
In fact, something similar can be done if we have an odd number, say $2t+1$ of consecutive cubes whose sum must be a perfect square. If we let $n^3$ be the central element of the cubes, the sum is $(n-t)^3+\cdots +n^3+\cdots +(n+t)^3=(2t+1)n^3+6n(1+4+\cdots +t^2)=(2t+1)n^3+t(t+1)(2t+1)n$
We know this is a square when $n=t,t+1$ (since the sum of the first $k$ cubes is a perfect square), and less obvious, but easy to verify, is that $n=4t(t+1)(2t+1)$ (which is $24$ times the sum of the first $t$ squares) is also a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
Is $\frac{1010103010101}9$ prime or composite? $1010103010101$ obviously divisible by $9$.
Is $\frac{1010103010101}9$ prime or composite?
The answer would be obtained without using WolframAlpha
| The polynomial $ x^6 + x^5 + x^4 + 3x^3 + x^2 + x + 1 $ can be factored through the following manipulations:
$$x^3 (x^3 + x^2 + 1) + x^4 + 2x^3 + x^2 + x + 1 = (x^3 + 1)(x^3 + x^2 + 1) + x^4 + x^3 + x = (x^3 + x + 1)(x^3 + x^2 + 1) $$
Now, plug in $ x = 10^2 $ to get a factorization of your number, which easily confirms that your number is composite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
$x,y,z >0$, prove
$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$
Note:
Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don't believe that any students can solve this problem in 3 hour time frame.
Update 1: In this forum, somebody said that BW is the only solution for this problem, which to the best of my knowledge is wrong. This problem is listed as "coffin problems" in my country. The official solution is very elementary and elegant.
Update 2: Although there are some solutions (or partial solution) based on numerical method, I am more interested in the approach with "pencil and papers." I think the approach by Peter Scholze in here may help.
Update 3: Michael has tried to apply Peter Scholze's method but not found the solution yet.
Update 4: Symbolic expanding with computer is employed and verify the inequality. However, detail solution that not involved computer has not been found. Whoever can solve this inequality using high school math knowledge will be considered as the "King of Inequality".
| Too long for a comment.
The Engel form of Cauchy-Schwarz is not the right way:
$$\frac{(x^2)^2}{8x^3+5y^3}+\frac{(y^2)^2}{8y^3+5z^3}+\frac{(z^2)^2}{8z^3+5x^3} \geq \frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}$$
So we should prove that
$$\frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}\geq\frac{x+y+z}{13}$$
which is equivalent to
$$\frac{(x^2+y^2+z^2)^2}{(x^3+y^3+z^3)}\geq x+y+z$$
but by Cauchy-Schwarz again we have $$x+y+z=\frac{(x^2)^2}{x^3}
+\frac{(y^2)^2}{y^3}+\frac{(z^2)^2}{z^3}
\geq \frac{(x^2+y^2+z^2)^2}{(x^3+y^3+z^3)}$$
and the inequalities are in the wrong way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "124",
"answer_count": 8,
"answer_id": 7
} |
Cubic Spline Interpolation My problem is to find a interpolating cubic spline to the points
$$\left\{(0,0), \left(\frac{\pi}{2}, 1\right), \left(\pi,0\right), \left(\frac{3\pi}{2}, -1\right),(2\pi,0)\right\}$$
I did as follows:
I set up the system
$$ \begin{cases}
& hg_0+2(h+h)g_1+hg_{2} = 6\left(\dfrac{y_{2}-y_1}{h} - \dfrac{y_1-y_{0}}{h}\right)\\
& hg_{1}+2(h+h)g_2+hg_{3} = 6\left(\dfrac{y_{3}-y_2}{h} - \dfrac{y_2-y_{1}}{h}\right)\\
& hg_{2}+2(h+h)g_3+hg_{4} = 6\left(\dfrac{y_{4}-y_3}{h} - \dfrac{y_3-y_{2}}{h}\right)
\end{cases} $$
Naturally cubic spline, used $ g_0 = g_4 = 0 $.
$$ \begin{cases}
& \phantom{hg_0+ }4hg_1+hg_{2} = -\dfrac{12}h\\
& hg_{1}+4hg_2+hg_{3} = 0\\
& hg_{2}+4hg_3\phantom{ +hg_{4}} = \dfrac{12}{h}
\end{cases} $$
I solved the system and found $ g_1 = -12/\pi^2 $, $ g_2 = 0 $ e $ g_3 = 12/\pi^2 $.
These formulas used to calculate the coefficients.
\begin{equation}
a_k = \frac{g_k-g_{k-1}}{6h_k};\;\;\;\;\;\;b_k = \frac{g_k}{2};\;\;\;\;\;\;
c_k = \frac{y_k-y_{k-1}}{h_k} + \frac{2h_kg_k+g_{k-1}h_k}{6};\;\;\;\;\;\;
d_k = y_k.
\end{equation}
Calculate the values of $a_k$ and of the values of $b_k$.
When I calculate the value of $c_1$, got
$$ c_1 = \frac{y_1-y_{0}}{h} + \frac{2hg_1+g_0h}{6} = \frac{1-0}{\pi/2} + \frac{2(\pi/2)(-12/\pi^2)+0(\pi/2)}{6} = 0 $$
But my book says that the result is $c_1 = 3/\pi$
Where you are wrong?
| Find the Cubic Spline for the points:
$$\left\{(0,0), \left(\frac{\pi}{2}, 1\right), \left(\pi,0\right), \left(\frac{3\pi}{2}, -1\right),(2\pi,0)\right\}$$
We will make use of this algorithm, which is certainly equivalent to the one you are using.
Since we have equal spacing, all of the $h_i = \dfrac{\pi}{2}$.
We have:
$$a = \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \\ a_4 \end{bmatrix} = \begin{bmatrix} y_0 \\ y_1 \\ y_2 \\ y_3 \\ y_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ -1 \\ 0 \end{bmatrix}$$
Forming $Ac = v$ and solving for the $c_i$ terms, yields:
$$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ \dfrac{\pi}{2} & 2 \pi & \dfrac{\pi}{2} & 0 & 0 \\ 0 & \dfrac{\pi}{2} & 2 \pi & \dfrac{\pi}{2} & 0 \\ 0 & 0 & \dfrac{\pi}{2} & 2 \pi & \dfrac{\pi}{2} \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix} = \begin{bmatrix} 0 \\ -\dfrac{12}{\pi} \\ 0 \\ \dfrac{12}{\pi} \\ 0 \end{bmatrix} \implies \begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix} = \begin{bmatrix} 0 \\ -\dfrac{6}{\pi^2} \\ 0 \\ \dfrac{6}{\pi^2} \\ 0 \end{bmatrix}$$
We now form the $b_i$ terms as:
$$b_i = \dfrac{1}{h_i}(a_{i+1}-a_i) - \dfrac{h_i}{3}(2 c_i + c_{i+1}), i= 0 \ldots 3$$
This yields:
$$b = \begin{bmatrix} b_0 \\ b_1 \\ b_2 \\ b_3 \end{bmatrix} = \begin{bmatrix} \dfrac{3}{\pi} \\ 0 \\ -\dfrac{3}{\pi} \\ 0 \end{bmatrix}$$
We now form the $d_i$ terms as:
$$d_i = \dfrac{1}{3h_i}(c_{i+1}-c_i) , i= 0 \ldots 3$$
This yields:
$$d = \begin{bmatrix} d_0 \\ d_1 \\ d_2 \\ d_3 \end{bmatrix} = \begin{bmatrix} -\dfrac{4}{\pi^3} \\ \dfrac{4}{\pi^3} \\ \dfrac{4}{\pi^3} \\-\dfrac{4}{\pi^3} \end{bmatrix}$$
We are now ready to form the splines as:
$$\begin {align} S_0(x) &= a_0 + b_0(x-x_0)+c_0(x-x_0)^2 + d_0(x-x_0)^3 \\
S_1(x) &= a_1 + b_1(x-x_1)+c_1(x-x_1)^2 + d_1(x-x_1)^3 \\ S_2(x) &= a_2 + b_2(x-x_2)+c_2(x-x_2)^2 + d_2(x-x_2)^3 \\ S_3(x) &= a_3 + b_3(x-x_3)+c_3(x-x_3)^2 + d_3(x-x_3)^3\end{align} $$
The final result is:
$$\begin {align} S_0(x) &= -\dfrac{4 x^3}{\pi ^3} + \dfrac{3 x}{\pi }, x \in \left[0, \dfrac{\pi}{2} \right] \\
S_1(x)& = \dfrac{4 x^3}{\pi ^3}-\dfrac{12 x^2}{\pi ^2}+\dfrac{9 x}{\pi }-1, x \in \left[\dfrac{\pi}{2}, \pi \right] \\
S_2(x)& = \dfrac{4 x^3}{\pi ^3}-\dfrac{12 x^2}{\pi ^2}+\dfrac{9 x}{\pi }-1, x \in \left[\pi, \dfrac{3 \pi}{2} \right] \\
S_3(x) &= -\dfrac{4 x^3}{\pi ^3}+\dfrac{24 x^2}{\pi ^2}-\dfrac{45 x}{\pi }+26,x \in \left[\dfrac{3 \pi}{2}, 2 \pi \right] \end {align}$$
If we plot the splines over their respective ranges, we get:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplifying the result formula for depressed Cubic After understanding the Cardano's formula for solving the depressed cubic (of the form $x^3+mx=n$, of course), I tried to find the solution of the equation $$x^3+6x=20.$$
After plugging into the formula
$$x=(n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}+(-n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}$$
where $m=6$ and $n=20$, we get
$$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}.$$
However, we notice that, without using Cardano's formula, that $x=2$ is the solution for the equation $x^3+6x=20.$
My question is: how does the equation $$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}$$ get simplified to $x=2$?
P.S. I understand that it was Niccolo Fontana who first figured out how to solve depressed cubic, to give one the proper credit.
| With the benefit of hindsight we notice that $10+\sqrt{108}=10+6\sqrt{3}$ and
$$10+6\sqrt{3}=(1+\sqrt{3})^3.$$
Similarly,
$$-10+6\sqrt{3}=(-1+\sqrt{3})^3.$$
Take the (real) cube roots and subtract.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Evaluating the rational integral $\int \frac{x^2+3}{x^6(x^2+1)}dx $
Evaluate $$\int \frac{x^2+3}{x^6(x^2+1)}dx .$$
I am unable to break into partial fractions so I don't think it is the way to go. Neither is $x=\tan \theta$ substitution. Please give some hints. Thanks.
| Let's try
$$\frac{x^2+3}{x^6(x^2+1)}=\frac ax+\frac b{x^2}+\frac c{x^3}+\frac d{x^4}+\frac e{x^5}+\frac f{x^6}+\frac{gx+h}{x^2+1}\implies$$
$$x^2+3=ax^5(x^2+1)+bx^4(x^2+1)+cx^3(x^2+1)+dx^2(x^2+1)+ex(x^2+1)+f(x^2+1)+(gx+h)x^6$$
and now:
$$x=0\implies 3=f\;,\;\;coef. \;x^7\implies 0=a+g\;,\;\;coef.\;x^6\implies 0=b+h$$
$$coef.\;x^5\implies 0=a+c\;,\;\;coef.\;x^4\implies0=b+d\;,\;\;coef.\;x^3\implies 0=c+e$$
$$coef.\;x^2\implies1=d+f\;,\;\;coef.\;x\implies0=e\;$$
From here:
$$f=3\;,\;\;d=-2\;,\;\;b=2\;,\;\;h=-2\;,\;\;c=e=a=g=\implies0$$
$$\frac{x^2+3}{x^6(x^2+1)}=\frac2{x^2}-\frac2{x^4}+\frac3{x^6}-\frac2{x^2+1}\implies$$
$$\int\frac{x^2+3}{x^6(x^2+1)}dx=-\frac2x+\frac2{3x^3}-\frac3{5x^5}-2\arctan x+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
} |
If $x^2+ax-3x-(a+2)=0\;,$ Then $ \min\left(\frac{a^2+1}{a^2+2}\right)$
If $x^2+ax-3x-(a+2)=0\;,$ Then $\displaystyle \min\left(\frac{a^2+1}{a^2+2}\right)$
$\bf{My\; Try::}$ Given $x^2+ax-3x-(a+2)=0\Leftrightarrow ax-a = -(x^2-3x-2)$
So we get $$a=\frac{x^2-3x-2}{1-x} = \frac{x^2-2x+1+1-x-4}{1-x} = \left[1-x-\frac{4}{1-x}+1\right]$$
Now $$f(a) = \frac{a^2+1}{a^2+2} = \frac{a^2+2-1}{a^2+2} = 1-\frac{1}{a^2+2}$$
So $$f(x) = 1-\frac{1}{\left[(1-x)-\frac{4}{1-x}+1\right]^2+2}$$
Now put $1-x=t\;,$ Then we get $$f(t) =1- \frac{1}{\left(t-\frac{4}{t}+1\right)^2+2}$$
Now How can I maximize $\displaystyle \frac{1}{\left[(1-x)-\frac{4}{1-x}+1\right]^2+2, }\;,$ Help Required, Thanks
| there is a high-school method.
since $\frac{a^2+1}{a^2+2}$ is an even function, and it's decreasing when $x \lt 0$, and increasing when $x \gt 0$, all we need to do is to find the min value of $\mid a \mid$.
since $x^2 + (a - 3) x - (a - 2) = 0$ has root(s) as real number(s), it follows that
$$
\Delta = (a - 3)^2 + 4 \cdot 1 \cdot (a - 2) \geq 0
$$
so we have $a^2 - 2a + 1 \geq 0$, and the min value of $a$ is $0$
so the min value of $\frac{a^2+1}{a^2+2}$ is $\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$
This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 2005.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2}\geq\frac{a+b+c}{3}.$$
My proof of this inequality you can see here: https://artofproblemsolving.com/community/c6h22937p427220
But this way does not help for the starting inequality.
A big problem we have around the point $(a,b,c)=(0.785, 1.25, 1.861)$ because the difference between the LHS and the RHS in this point is $0.0000158...$.
I tried also to use Cauchy-Schwarz, but without success.
Also, I think the BW (see here https://math.stackexchange.com/tags/buffalo-way/info I tryed!) does not help.
| Let $f,g,p,u,k,d,x,y,z>0$ such that $u\geq k$:
$$y=\left(\frac{x\left(g+u\right)p}{g+k}\right),x=\left(\frac{z\left(d+u\right)p}{d+k}\right),z=\left(\frac{y\left(f+u\right)p}{f+k}\right)$$
Then we have the functions :
$$h\left(x\right)=\frac{x}{13+\frac{5p^{2}\left(x+u\right)^{2}}{\left(x+k\right)^{2}}}$$
Wich is convex on $(0,\infty)$ and $u\ge k>0$ and $p> 0$ :
Then using Jensen's inequality we have :
$$\frac{z}{f}h(f)+\frac{x}{d}h(d)+\frac{y}{g}h(g)\geq \left(\frac{z}{f}+\frac{x}{d}+\frac{y}{g}\right)h\left(\frac{\left(z+x+y\right)}{\frac{z}{f}+\frac{x}{d}+\frac{y}{g}}\right)$$
Now the problem is :
$$\left(\frac{z}{f}+\frac{x}{d}+\frac{y}{g}\right)h\left(\frac{\left(z+x+y\right)}{\frac{z}{f}+\frac{x}{d}+\frac{y}{g}}\right)\geq^{?}\frac{\left(x+y+z\right)}{18}\tag{I}$$
After simplification we have the constraint :
$$p(\frac{\left(z+x+y\right)}{\frac{z}{f}+\frac{x}{d}+\frac{y}{g}}+u)\leq \frac{\left(z+x+y\right)}{\frac{z}{f}+\frac{x}{d}+\frac{y}{g}}+k$$
we are done .
Last edit 23/04/2022 :
We have the inequality for $x\in[0,1]$ :
$$f(x)=\left(\frac{1}{13+5x^{-2}}\right)\geq g(x)=\left(\frac{1}{6}\left(\frac{1}{2+x^{-2}}+\frac{x^{2}\left(1-x\right)^{2}}{5}\right)\right)\left(1+\frac{\left(x\left(1-x\right)\right)^{2}}{16.75}\right)^{3}$$
Now we need to show under some assumptions given below:
$$p(a,b,c)=-\frac{\left(a+b+c\right)}{18}+ag\left(\frac{a}{b}\right)+bf\left(\frac{b}{c}\right)+\frac{c^{3}}{13c^{2}+5a^{2}}\geq 0$$
For that we use Buffalo's way and a constraint we have all the coefficient positives in ($x\ge 0$ and $n\geq 1$ a natural number) :
$$p(0.785,1.25+x,1.861+x^n)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 5,
"answer_id": 3
} |
How to solve $\cos(\frac{\alpha }{2} )=\frac{a}{\sqrt{a^{2}+b^{2} } }$ for $\cos(\alpha)$ using half-angle formula. I have $\cos(\frac{\alpha }{2} )=\frac{a}{\sqrt{a^{2}+b^{2} } } $
How can I get $\cos(\alpha ) $ from this?
I know this identitiy.
$\cos(\frac{\alpha }{2} )=\sqrt{\frac{1+\cos(\alpha ) }{2} } $
But just cant figure out, how to do it.
| $\cos(\frac{\alpha }{2} )=\frac{a}{\sqrt{a^{2}+b^{2} } } $
$\cos^2(\frac{\alpha }{2} )=\frac{a^2}{a^2+b^2}$
$\frac{1}{2}(1 + \cos(\alpha))=\frac{a^2}{a^2+b^2}$
$\cos(\alpha)=\frac{2a^2}{a^2+b^2} - 1$
half-angle formula
$\cos^2(\frac{\alpha }{2} ) = \frac{1}{2}(1 + \cos(\alpha))$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to compute $\int_0^1\frac{\ln(x)}{1+x^5}dx$?
Let $\phi$ denote the golden ratio $\phi=\frac{1+\sqrt5}{2}$.
How can I prove this sum?
$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\phi}{(5n+1)^2}+\frac{\phi^{-1}}{(5n+2)^2}-\frac{\phi^{-1}}{(5n+3)^2}-\frac{\phi}{(5n+4)^2}\right]=\left(\frac{2\pi}{5}\right)^2$$
My try:
Change the sums into an integrals:
$\sum_{n=0}^{\infty}\frac{(-1)^n}{(5n+1)^2}=\int_0^1\frac{-\ln(x)}{1+x^5}dx$
Can somebody give a hint how on to integrate this integral.
Try substitution by letting $u=\ln(x)$ is not working and integration by part is making it more complicated than before. What kind of substitution should I be using?
| \begin{align*}
\sum_{n=0}^{\infty}(-1)^{n}\left[\frac{\phi}{(5n+1)^{2}}-\frac{\phi}{(5n+4)^{2}}\right] & =\frac{2}{125}\left(5+2\sqrt{5}\right)\pi^{2}\\
\sum_{n=0}^{\infty}(-1)^{n}\left[\frac{\phi^{-1}}{(5n+2)^{2}}-\frac{\phi^{-1}}{(5n+3)^{2}}\right] & =\frac{2}{125}\left(5-2\sqrt{5}\right)\pi^{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Show that $a^{13} \equiv a \pmod{3 \cdot 7 \cdot 13}$.
Show that $a^{13} \equiv a \pmod{3 \cdot 7 \cdot 13}$.
I want to know if my attempt is correct.
First $a^{13} \equiv (a^3)^4 \cdot a \equiv a^4 \cdot a \equiv a^3 \cdot a^2 \equiv a \cdot a^2 \equiv a^3 \equiv a \pmod 3$.
Second, $(a^7)^2 \cdot a^{-1} \equiv a^2 \cdot a^{-1} \equiv a \pmod 7$.
Third, $a^{12} \equiv 1 \pmod{13} \implies a^{13} \equiv a \pmod{13}$.
So, $3, 7, 13 | a^{13} - a$. Since they are relatively prime in pairs, $$3 \cdot 7 \cdot 13 | a^{13} - a \implies a^{13} \equiv a \pmod{ 3 \cdot 7 \cdot 13}$$
Is it correct? Could you suggest an easier or a better proof?
Thanks.
| By Euler's Theorem (or you may even use Fermat's theorem),
$a^2 \equiv 1 \pmod {3} \implies a^{12} \equiv 1 \pmod 3 \implies a^{13} \equiv a \pmod 3$
$a^6 \equiv 1 \pmod {7} \implies a^{12} \equiv 1 \pmod 7 \implies a^{13} \equiv a \pmod 7$
$a^{12} \equiv 1 \pmod {13} \implies a^{13} \equiv a \pmod {13} $
$\therefore a^{13} \equiv a \pmod {lcm[3,7,13]} \implies a^{13} \equiv a \pmod {3\times7\times13}$
It is quit trivial if $a$ is not co-prime to at least one of $3, 7, 13.$ Then you will not even need Euler or Fermat's theorem.
Remember, if
$a\equiv b \pmod {m_1}$
$a\equiv b \pmod {m_2}$
.
.
.
$a\equiv b \pmod {m_n}$
then, $a\equiv b \pmod {lcm[m_1,m_2,...,m_n]}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$ The question
Prove that:
$$\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$$
What I've tried
Knowing that:
$$\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right)$$
evaluating at $z=i$ gives
$$ \frac{e^π - e^{-π}}{2i} = \sin(πi) = πi \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right)$$
so:
$$ \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right) = \frac{e^π - e^{-π}}{2π}$$
I'm stucked up and don't know how to continue, any help?
| I'll reproduce the answer @C.Dubussy have just deleted:
$$ \prod_{n=2} \left( 1 - \frac{1}{n^4} \right) = \prod_{n=2}^∞ \left( 1 + \frac{1}{n^2}\right) \prod_{n=2}^∞ \left( 1 - \frac{1}{n^2}\right) = \frac{\sin{iπ}}{iπ} \prod_{n=2}^∞ \frac{n-1}{n} \prod_{n=2}^∞ \frac{n+1}{n} $$
And because the last product gives $\frac{1}{2}$, we have it!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
What is $\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n}$ for $x > 4$.
What is $\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n}$ for $x > 4$.
Here is what I got so far (using Cauchy's integral formula) :
$$\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n} ~=~ \sum\limits_{n=0}^\infty \left(\frac{1}{2\pi i}\int_\mathbb{T} \frac{(1+z)^{2n}}{z^{n+1}}\; dz\right) \cdot \frac{1}{x^n} \\=~ \frac{1}{2\pi i}\sum\limits_{n=0}^\infty \frac{1}{x^n}\int_{\mathbb{T}} \frac{1}{z}\cdot \left( \frac{(1+z)^2}{z}\right)^n \; dz \\=~ \frac{1}{2\pi i} \int_{\mathbb{T}} \frac{1}{z} \sum\limits_{n=0}^\infty \left( \frac{(1+z)^2}{xz}\right)^n \; dz \\=~ \frac{1}{2\pi i}\int_\mathbb{T} \frac{1}{z} \cdot \left( \frac{1}{1- \left(\frac{(1+z)^2}{xz}\right)}\right) \; dz \\=~ \frac{1}{2\pi i}\int_\mathbb{T} \frac{1}{z} \cdot \left(\frac{xz}{xz- (1+z)^2}\right) \; dz \\=~ \frac{1}{2\pi i}\int_\mathbb{T} \frac{x}{xz- (1+z)^2} \; dz \\=~ \frac{1}{2\pi i}\int_\mathbb{T} \frac{x}{-1 + (x-2)z- z^2} \; dz$$
According to Wolframalpha I should get $\sqrt{\frac{x}{x-4}}$. Where did I commit a sin ?
I'm expecting to use the Residue theorem or Cauchy's integral formula at some point but I feel something is fishy.
| An ODE-based proof. If we set
$$ A(t) = \sum_{n\geq 0}\binom{2n}{n}t^n $$
we have an analytic function in a neighbourhood of the origin with radius of convergence $\rho\geq\frac{1}{4}$.
We may notice that:
$$ A'(t) = \sum_{n\geq 1}\binom{2n}{n} n t^{n-1} = \sum_{n\geq 0}\binom{2n}{n}(4n+2) t^n $$
while:
$$ \frac{d}{dt}\left( t A(t)\right) = \sum_{n\geq 0}\binom{2n}{n}(n+1) t^n $$
hence:
$$ A'(t) = 4\frac{d}{dt}\left( t A(t)\right) - 2A(t) = 2A(t)+4t A'(t) $$
or:
$$ \frac{A'(t)}{A(t)} = \frac{2}{1-4t}$$
that leads to:
$$ A(t) = \sum_{n\geq 0}\binom{2n}{n}t^n = \frac{1}{\sqrt{1-4t}}. \tag{1}$$
By replacing $t$ with $\frac{1}{x}$ we get:
$$ \forall x>4,\qquad \sum_{n\geq 0}\binom{2n}{n}\frac{1}{x^n}=\color{red}{\sqrt{\frac{x}{x-4}}}.\tag{2} $$
$(1)$ also follows from Lagrange's inversion theorem (that leads to Bring radical).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$
Evaluation of $$\int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$$
$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$$
Put $x=\cos 2 \theta\;,$ Then $dx = -2\sin 2 \theta d\theta$ and Changing Limit, We get
$$I = \int_{0}^{\frac{\pi}{4}}\frac{2\sin 2 \theta}{\sqrt{2}\cos \theta+\sqrt{2}\sin \theta+2}d\theta$$
So $$I = \int_{0}^{\frac{\pi}{4}}\frac{\sin 2 \theta}{\cos\left(\theta-\frac{\pi}{4}\right)+1}d\theta$$
Now Put $\displaystyle \theta-\frac{\pi}{4}=t\;,$ Then $d\theta = dt$ and changing limits
So $$I = \int_{-\frac{\pi}{4}}^{0}\frac{\cos 2t}{\cos t+1}dt=2\int_{-\frac{\pi}{4}}^{0}\frac{2\cos^2 t-2+1}{\cos t+1}dt$$
after that we can solve easily,
My question is can we solve it without Using Trig substution,
Plz explain me
Thanks
| multiply the integrand by
$$
1=\frac{\left(-\sqrt{1-x}-\sqrt{x+1}+2\right) \left(\sqrt{1-x^2}+1\right)}{\left(-\sqrt{1-x}-\sqrt{x+1}+2\right) \left(\sqrt{1-x^2}+1\right)}
$$
doing the alegbra correctly this indeed eliminates the roots in the denominator and we end up with
$$
I=\frac{1}{2}\int_{0}^{1}\frac{-\sqrt{1-x} x+\sqrt{1+x} x-2 \sqrt{1-x}+2 \sqrt{1-x^2}-2 \sqrt{x+1}+2}{x^2}dx=\\
$$
taking into account a proper limiting procedure for $x\rightarrow 0,1$, this integral can be split in elementary pieces (solvable by integration by parts and/or subs of the form $y\rightarrow y\pm 1$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Using $\epsilon-\delta$ proof to prove continuity
Use an $\epsilon-\delta$ proof to show that
$f : R \setminus \left \{ \frac{-3}{2} \right \} \rightarrow R$ , $$f(x) = \frac{3x^2-2x-5}{2x+3}$$
is continuous at $x = -1$
Hello there. Can anyone here help me with this? I know i need to show that $|x-l|<\delta$ implies $|f(x) - f(l)| < \epsilon$ but I don't know how to do it for this specific example.
| Fix $\varepsilon>0$ and let $x\in\mathbb{R}\setminus\{-\frac{3}{2}\}$ such that $|x-(-1)|=|x+1|<\delta$ for $\delta>0$ to be chosen later. As $f(-1)=0$, we have :
\begin{align}
\left|f(x)-f(-1)\right|&=\left|\frac{3x^2-2x-5}{2x+3}-0\right|\\
&=\left|\frac{3x^2-2x-5}{2x+3}\right|.
\end{align}
Now observe that
\begin{align}
3x^2-2x-5&=3(x+1)\left(x-\frac{5}{3}\right),\quad\quad\quad\quad(*)
\end{align}
\begin{align}
\left|x-\frac{5}{3}\right|&=\left|x+1-1-\frac{5}{3}\right|\\
&\leq\left|x+1\right|+\left|1+\frac{5}{3}\right|\\
&<\delta+\frac{8}{3},\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(**)
\end{align}
and that (by the inverse triangle inequality)
\begin{align}
|2x+3|&\geq|2|x|-3|\\
&\geq3-2|x|\\
&>3-2(1+\delta)\\
&=1-2\delta\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(***)
\end{align}
so that :
\begin{align}
\left|\frac{3x^2-2x-5}{2x+3}\right|&=\frac{3|x+1|\left|x-\frac{5}{3}\right|}{|2x+3|}\\
&<\frac{3\delta\left(\delta+\frac{8}{3}\right)}{|1-2\delta|}.
\end{align}
We now choose $\delta<1/4$ for $1-2\delta>1/2>0$, and then
\begin{align}
\left|f(x)-f(-1)\right|&<6\delta\left(\delta+\frac{8}{3}\right).
\end{align}
Finally, if we choose $\varepsilon=6\delta\left(\delta+\frac{8}{3}\right)$ for any $\delta<1/4$, then we get
$$|x-(-1)|<\delta\quad\Longrightarrow\quad|f(x)-f(-1)|<\varepsilon$$
whence $f$ is continuous at $x=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Modulo Equations I am trying to solve a problem involving modulo arithmetic but I am not sure what method to use as I have never done this style of question before nor do I have any examples to work from.
The question is:
Solve for $x$ where $x^5 = 11 \mod (35)$
So I though Eculid's GCD algorithm would help but I just get circular equations and get nowhere.
Thanks
| Let $x\in\mathbb{Z}$ be such that $x^5\equiv 11\pmod{35}$. Then, $x^{15}=\left(x^5\right)^3\equiv 11^3\equiv 1\pmod{35}$. Since $\gcd(x,35)=1$, $x^{12}=x^{\text{lcm}(5-1,7-1)}=x^{\lambda(35)}\equiv 1\pmod{35}$, where $\lambda$ is the Carmichael function. Thus, $x^3=x^{\gcd(15,12)}=1\pmod{35}$, whence $1\equiv x^6\equiv x^5\cdot x\equiv 11\pmod{35}$, so that $x\equiv 11^{-1}\equiv 16\pmod{35}$, which is obviously a solution to this congruence.
Alternatively, $x^5\equiv 11\pmod{35}$ iff $x^5\equiv 1\pmod{5}$ and $x^5\equiv 4\equiv 3^4\pmod{7}$, which is equivalent to $x\equiv 1\pmod{5}$ and $x\equiv 3^2\equiv 2\pmod{7}$. Consequently, $x^5\equiv 11\pmod{35}$ if and only if $x\equiv 16\pmod{35}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7} -\frac{1}{8}-\frac{1}{9}-\frac{1}{10} -\cdots $ converge $$1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}
-\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{10} ... $$
I added parentheses for each sub-sequence with the same sing.
so i got :
$$1-(\dfrac{1}{2}+\dfrac{1}{3})+(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})-(\dfrac{1}{7}
+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}) ... $$
I want to show that the new sequence is a leibniz sequence and by that conclude that is converge.
I managed to show that each pair of parentheses is greater than:
$$\dfrac{2}{n+1} $$
Cant find a way to proceed.
Thanks for helping.
| Compare each group, term by term, with the next group.
$$\frac14-\frac17+\frac15-\frac18+\frac16-\frac19=\frac3{4\cdot7}+\frac3{5\cdot8}+\frac3{6\cdot9}>\frac{3^2}{6\cdot9}>\frac19>\frac1{10}$$
Each group is greater than the next group, so the alternating series test applies.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Prove that $1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$
Prove that $$1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$$
Attempt:
We can easily show that an eighth power can be expressed as a fourth power since $x^8 = (x^2)^4$. Conversely, by Fermat's Little Theorem, $x^{\phi(25)} = x^{20} \equiv 1 \pmod{25}$ if $\gcd(x,25) = 1$ and thus $x^4 = x \cdot x^3 \equiv x^{21} \cdot x^3 \equiv x^{24} \equiv (x^3)^8 \pmod{25}$. $\square$
I am not sure if the above proves the result, but it does show that if we have an $8$th power of an integer modulo $25$ and it is relatively prime to $25$, then it is equal to a fourth power of an integer modulo $25$ and vice-versa. Does that therefore mean they are equivalent?
| $100 = 2^2 \times 5^2$. It suffices to show that the two sums are equal mod $4$ and mod $25$.
If $x$ is even, $x^4 \equiv x^8 \equiv 0 \mod 2^2$, while if $x$ is odd,
$x^4 \equiv x^8 \equiv 1 \mod 2^2$. So $\sum_{i=1}^{99} x^4 \equiv \sum_{i=1}^{99} x^8 \mod 2^2$.
Similarly, if $x$ is divisible by $5$, $x^4 \equiv x^8 \equiv 0 \mod 5^2$.
On the other hand, the multiplicative group $U$ of units in $\mathbb Z / 25 \mathbb Z$ (i.e. the numbers mod $25$ that are not divisible by $5$) is cyclic of order $\phi(5^2) = 20$. The maps $t \to t^4$ and $t \to t^8$ are homomorphisms of this group, with the same range (namely the $5$ elements $y$ such that $y^5 = 1$). Thus $x^4$ for $x \in U$ and $x^8$ for $x \in U$ run over this same set, each element occurring $4$ times. If we take the
numbers from $1$ to $99$ not divisible by $5$, each element occurs $20$ times. Thus the two sums mod $25$ are the same.
EDIT: Somewhat more generally, $\sum_{x=1}^{N-1} x^4 \equiv \sum_{x=1}^{N-1} x^8 \mod N$ for all positive integers $N$. One way to see this is that
$$ \sum_{x=1}^{N-1} (x^8 - x^4) = \dfrac{N\; (2\,{N}^{8}-9\,{N}^{7}+12\,{N}^{6}-12\,{N}^{4}+9\,{N}^{3}-2\,{N}^{2})}{18}$$ and $2\,{N}^{8}-9\,{N}^{7}+12\,{N}^{6}-12\,{N}^{4}+9\,{N}^{3}-2\,{N}^{2} \equiv 0 \mod 18$ for all $N$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Is there any integer $n>1$ such that $3^n - 1$ is divisible by $2^n - 1$? Is there any integer $n>1$ such that $3^n - 1$ is divisible by $2^n - 1$?
I guess not. For every even integer $n$, we can show that $3^n - 1$ is not divisible by $2^n - 1$ because $2^n -1$ is a multiple of $3$, but $3^n-1$ is not. Would anyone give some tips for the other case (case for $n$ odd integer)?
| Here is a proof by contradiction. Assume there exists such $n$.
If $n$ is even, notice $2^n-1 \equiv 0 \pmod 3$. Thus, as $3^n-1$ is not dividible by $3$, $n$ is odd. $$2^n-1 \equiv 7 \pmod {12}$$So there exists such a $p$ that $$p \equiv \pm 7 \pmod {12}, 2^n-1 \equiv 0 \pmod p \Rightarrow 3^n-1 \equiv 0 \pmod p$$ So notice if $x=3^{\frac{n-1}{2}}$, $$3^{n} \equiv 3x^2 \equiv 1 \pmod {p} \Rightarrow (3x)^2 \equiv 3 \pmod {p}$$ This implies $3$ is a quadratic residue mod $p$. A contradiction, as $p \equiv \pm 7 \pmod {12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the remainder of $\sum_{i=0}^{99} 2^{i^2}$ when dividing by 7 and determine if the quotient is even or odd I've recently had this problem in an exam and couldn't solve it.
Find the remainder of the following sum when dividing by 7 and determine if the quotient is even or odd:
$$\sum_{i=0}^{99} 2^{i^2}$$
I know the basic modular arithmetic properties but this escapes my capabilities. In our algebra course we've seen congruence, divisibility, division algorithm... how could I approach it?
| By Fermat's little theorem
$$ 2^{i^2}\!\!\!\pmod{7}=\left\{\begin{array}{ll}\color{green}{1}&\text{if } i\equiv 0\pmod{6}\\\color{blue}{2}&\text{if } i\equiv 1\pmod{6}\\\color{blue}{2}&\text{if } i\equiv 2\pmod{6}\\\color{green}{1}&\text{if } i\equiv 3\pmod{6}\\\color{blue}{2}&\text{if } i\equiv 4\pmod{6}\\\color{blue}{2}&\text{if } i\equiv 5\pmod{6}\\\end{array}\right.$$
hence:
$$ \sum_{i=0}^{99}2^{i^2} \equiv \color{blue}{2\sum_{i=0}^{99}1}-\color{green}{\sum_{k=0}^{33}1} \equiv 2\cdot 100-34 \equiv \color{red}{5}\!\!\pmod{7}.$$
In a similar way
$$ 2^{i^2}\!\!\!\pmod{14}=\left\{\begin{array}{ll}\color{green}{1}&\text{if } i\equiv 0\pmod{6}\\\color{blue}{2}&\text{if } i\equiv 1\pmod{6}\\\color{blue}{2}&\text{if } i\equiv 2\pmod{6}\\\color{purple}{8}&\text{if } i\equiv 3\pmod{6}\\\color{blue}{2}&\text{if } i\equiv 4\pmod{6}\\\color{blue}{2}&\text{if } i\equiv 5\pmod{6}\\\end{array}\right.$$
hence:
$$ \sum_{i=0}^{99}2^{i^2}\equiv 2\cdot 100-34+\color{purple}{7\cdot 17} \equiv \color{red}{5}\pmod{14}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $276 x\equiv 90\pmod {666}$
Solve $276 x\equiv 90\pmod {666}$
I found using Euclidean algorithm that $\gcd (276,666)=6$
then I divided by $6$ and I got:
$$46x\equiv 15\pmod {111}$$
and I found that $\gcd(46,111)=1$ using Euclidean algorithm
I am stuck here and don't know what to do
| You correctly reduced the problem to $46x \equiv 15 \pmod{111}$. Applying the Euclidean algorithm to find $\gcd(46, 111)$ yields
\begin{align*}
111 & = 2 \cdot 46 + 19\\
46 & = 2 \cdot 19 + 8\\
19 & = 2 \cdot 8 + 3\\
8 & = 2 \cdot 3 + 2\\
3 & = 1 \cdot 2 + 1\\
2 & = 2 \cdot 1
\end{align*}
so $\gcd(46, 111) = 1$ as you found.
We now follow E. Girgin's advice and apply the extended Euclidean algorithm to find the multiplicative inverse of $46$ modulo $111$. Working backwards to write $1$ as a linear combination of $46$ and $111$ yields
\begin{align*}
1 & = 3 - 2\\
& = 3 - (8 - 2 \cdot 3)\\
& = 3 \cdot 3 - 8\\
& = 3(19 - 2 \cdot 8) - 8\\
& = 3 \cdot 19 - 7 \cdot 8\\
& = 3 \cdot 19 - 7(46 - 2 \cdot 19)\\
& = 17 \cdot 19 - 7 \cdot 46\\
& = 17(111 - 2 \cdot 46) - 7 \cdot 46\\
& = 17 \cdot 111 - 41 \cdot 46
\end{align*}
Hence, $-41 \cdot 46 = 1 - 17 \cdot 111 \implies 46^{-1} \equiv -41 \equiv -41 + 111 \equiv 70 \pmod{111}$. Multiplying both sides of the equation $46x \equiv 15 \pmod{111}$ by $70$ yields
\begin{align*}
70 \cdot 46x & \equiv 70 \cdot 15 \pmod{111}\\
x & \equiv 1050 \pmod{111}\\
x & \equiv 9 \cdot 111 + 51 \pmod{111}\\
x & \equiv 51 \pmod{111}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplify $\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$ Simplify $$\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$$
I know there is another easier method except the one I answered. I cannot find it. Can you please help? Thanks in advance.
| For a general approach related to lab's hint: We want to (hopefully) simplify $\sqrt{4 \pm 2\sqrt{3}}$. Let's look at $\sqrt{4 + 2\sqrt{3}}.$
$$
\sqrt{4 + 2\sqrt{3}} = a + b\sqrt{3}
$$
Let's find $a$ and $b$. First square both sides.
$$
4 + 2\sqrt{3} = a^2 + 3b^2 + 2ab\sqrt{3}.
$$
This gives us two equations:
$$
\left\{
\begin{array}{rcl}
a^2 + 3b^2 &= & 4\\
2ab &= & 2
\end{array}
\right.
$$
From the second equation we can say $a = 1/b$. Plugging this into the first equation gives us
$$
\frac{1}{b^2} + 3b^2 = 4.
$$
This can be rewritten as $3b^4 - 4b^2 + 1 = 0$. This is an equation of "quadratic type" (because it's quadratic in $b^2$) and when we solve it (by factoring or by using the quadratic formula) we get $b^2 = 1$ and $b^2 = 1/3$. Therefore $b = \pm 1$ or $b = \pm 1/\sqrt 3$. Since $a = 1/b$, this yields four possibilities for $a + b\sqrt{3}$:
$$
\underbrace{1 + \sqrt{3}}_{b=1}, \qquad
\underbrace{-1 - \sqrt{3}}_{b=-1}, \qquad
\underbrace{\sqrt{3} + 1}_{b = 1/\sqrt{3}}, \qquad
\underbrace{-\sqrt{3} - 1}_{b = -1/\sqrt{3}}
$$
Notice that this is actually only two distinct possibilities: $1 + \sqrt{3}$ and $-1 - \sqrt{3}$. Because, by convention, $\sqrt{4 + 2\sqrt{3}}$ must be positive, we can throw out the extraneous $-1 - \sqrt{3}$ solution and we end up with
$$ \sqrt{4 + 2\sqrt{3}} = 1 + \sqrt{3}.$$
A similar analysis will show that
$$ \sqrt{4 - 2 \sqrt{3}} = -1 + \sqrt{3}.$$
Therefore we have:
$$
\frac{1}{\sqrt{4 + 2\sqrt{3}} - \sqrt{4 - 2\sqrt{3}}} = \frac{1}{(1+\sqrt{3}) - (-1 + \sqrt{3})} = \frac{1}{2}
$$
It's subjective whether or not this is really better/easier than what you did.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Prove $\frac{1^5}{1^5}+\frac{1^5-2^5}{1^5+2^5}+\frac{1^5-2^5+3^5}{1^5+2^5+3^5}+\frac{1^5-2^5+3^5-4^5}{1^5+2^5+3^5+4^5}+\cdots?$ On my previous page I proposed,
$$\frac{1^3}{1^3}+\frac{1^3-2^3}{1^3+2^3}+\frac{1^3-2^3+3^3}{1^3+2^3+3^3}+\frac{1^3-2^3+3^3-4^3}{1^3+2^3+3^3+4^3}+\cdots=3-\frac{\pi^2}{4}$$
Jack D'Aurizio proved the above identity in a simple manner.
He proposed the changing of cubes into the fifth power and he also offered the closed form of the identity.
$$\frac{1^5}{1^5}+\frac{1^5-2^5}{1^5+2^5}+\frac{1^5-2^5+3^5}{1^5+2^5+3^5}+\frac{1^5-2^5+3^5-4^5}{1^5+2^5+3^5+4^5}+\cdots=\\
15-\frac{\pi}{2}\left[3\pi+3\sec\left(\frac{\pi\sqrt3}{2}\right)-4\sqrt{3}\tan\left(\frac{\pi\sqrt3}{2}\right)\right]$$
Can anybody provide a proof for it?
| We need some preliminary lemma.
$$ S_N=\sum_{n=1}^{N}n^5 = \frac{1}{12}N^2(N+1)^2 (2N^2+2N-1) \tag{1} $$
$$ P_N=\sum_{n=1}^{2N}(-1)^{n+1} n^5 = N^2(5-20 N^2-16 N^3) \tag{2}$$
$$ D_N=\sum_{n=1}^{2N-1}(-1)^{n+1} n^5 = N^2(5-20 N^2+16 N^3) \tag{3}$$
They give:
$$ S = \sum_{N\geq 1}\left(\frac{D_N}{S_{2N-1}}+\frac{P_N}{S_{2N}}\right)=\\\sum_{N\geq 1}\left(\frac{15-12 N^2 (5+4 N)}{(1-2 N)^2 \left(-1-4 N+8 N^2\right)}+\frac{15+12 N^2 (-5+4 N)}{(1+2 N)^2 \left(-1+4 N+8 N^2\right)}\right)\tag{4}$$
and such a series can be computed from partial fraction decomposition, the identity:
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{5}$$
and the reflection formula for the $\psi=\frac{d}{dx}\log\Gamma$ function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods and without gamma functions? I know that
$$\int_0^\infty \sin x^2\, dx = \sqrt{\frac{\pi}{8}}$$
but all of the methods I've found seem to be too complicated for an early calculus student. Is there any method of calculating this with real methods and without gamma functions?
| We first transform the integral by putting $u=x^2$, $$
\int_0^{\infty} \sin \left(x^2\right) d x=\frac{1}{2} \int_0^{\infty} \frac{\sin u}{\sqrt{u}} d u
$$
Using the Gaussian integral: $\int_0^{\infty} e^{-u v^2} d v=\frac{\sqrt{\pi}}{2 \sqrt{u}} $, where $u$ is a constant, we have
$$
\begin{aligned}
I &=\frac{1}{2} \int_0^{\infty} \frac{2 \sin u}{\sqrt{\pi}} \int_0^{\infty} e^{-u v^2} d v d u\\&=\frac{1}{\sqrt{\pi}} \int_0^{\infty}\left(\int_0^{\infty} \sin u \cdot e^{-u v^2} d u\right) d v \\
&=\frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{v^4+1} d v
\end{aligned}
$$
Playing a little on the integrand, we obtain
$$
\begin{aligned}
I&=\frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{\frac{1}{v^2} }{v^2+\frac{1}{v^2}} d v \\
&=\frac{1}{2 \sqrt{\pi}} \int_0^{\infty} \frac{\left(1+\frac{1}{v^2}\right)-\left(1-\frac{1}{v^2}\right)}{v^2+\frac{1}{v^2}} d v \\
&=\frac{1}{2 \sqrt{\pi}}\left[\int_0^{\infty} \frac{d\left(v-\frac{1}{v}\right)}{\left(v-\frac{1}{v}\right)^2+2}- \underbrace{\int_0^{\infty} \frac{d\left(v+\frac{1}{v}\right)}{\left(v+\frac{1}{v}\right)^2-2}}_{=0}\right] \\
&=\frac{1}{2 \sqrt{2} \cdot \sqrt{\pi}}\left[\tan ^{-1}\left(\frac{v-\frac{1}{v}}{\sqrt{2}}\right)\right]_0^{\infty} \\
&=\frac{\pi}{2 \sqrt{2} \sqrt{\pi}} \\
&=\sqrt{\frac{\pi}{8}} \\
&
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding centre of sphere inscribed in tetrahedron Given the tetrahedron with vertices defined by vectors $a=(-4, -3, 1)$, $b=(8,3,1)$, $c= (2, 6, 1)$, $d=(4,3,3)$, find the centre of the sphere inscribed in the tetrahedron.
My train of thought: consider the intersection of the four bisectors of the vertices of the tetrahedron. The centre of the sphere will be in the intersection of the four angle-bisecting planes.
Is this correct?
If so, then we need to find four normal vectors, equations of the four planes, and equate them all together, then find $x$, the centre of the sphere. Is this correct?
Is there a simpler way?
| The four vertices are $A=(4,3,3),B=(2,6,1),C=(-4,-3,1),D=(8,3,1)$.
The four faces of the tetrahedron are $BCD:z-1=0$; $ACD:x-2y+2z-4=0$; $ABD:x+2y+2z-16=0$; and $ABC:3x-2y-6z+12=0$. So the distance of a point $(a,b,c)$ from the three faces is $|c-1|,\ |\frac{1}{3}|a-2b+2c-4|,\ \frac{1}{3}|a+2b+2c-16|,\ \frac{1}{7}|3a-2b-6c+12|$. These must all be equal.
That gives multiple solutions because we also get all the exspheres.
So we use JeanMarie's idea that the signs of the distances must be same for the incentre and the centroid which is $G=(\frac{5}{2},\frac{9}{4},\frac{3}{2})$.
The signs of its distances from $BCD,ACD,ABD,ABC$ are +,-,-,+.
So the distances $c-1,-(a-2b+2c-4)/3,-(a+2b+2c-16)/3,(3a-2b-6c+12)/7$ must all be equal.
Solving, we find that $(\frac{26}{7},3,\frac{13}{7})$ is the incentre.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1791289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Eigenvalue of block matrix of order $2n$ How to find eigenvalues of following block matrix?
$$P=\begin{bmatrix} A & B \\ B & A \end{bmatrix}$$
Where,
$A=\begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 & \cdots & 1 \\ 1 & 0 & 1 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 0 & \ddots & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \end{bmatrix}_n$
$B=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \end{bmatrix}_n$
I had read one result for block matrix which says that eigenvalues of matrix $P$ is union of eigenvalues of $A+B$ and $A-B$
Here,
$A+B=\begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0 & \cdots & 1 \\ 1 & 0 & 1 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 0 & \ddots & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \end{bmatrix}_n$
$A-B=\begin{bmatrix} -1 & 1 & 0 & 0 & 0 & 0 & \cdots & 1 \\ 1 & 0 & 1 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 0 & \ddots & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \end{bmatrix}_n$
How to find eigenvalues of $A+B$ and $A-B$?
I know that eigenvalues of $A$ are $2\cos\frac{2\pi j}{n},j=1,2,\cdots,n$
| Note that the eigenvalues of $P$ are real and in $[-3,3]$.
Let $U_n$ be the matrix that is derived from $A_n$ by putting the entries $[1,n],[n,1]$ equal to $0$.
Then $\det(A_n\pm B_n-\lambda I_n)=\det(A_n-\lambda I_n)\pm \det(U_{n-1}-\lambda I_{n-1})=p_n(\lambda)\pm q_{n-1}(\lambda)$.
The roots of $p_n$ are $2\cos(\frac{2\pi j}{n}),j=1,\cdots,n$.
The roots of $q_{n-1}$ are $2\cos(\frac{\pi j}{n}),j=1,\cdots,n-1$.
Remark 1. When $n$ is even, $p_n$ is even and $q_{n-1}$ is odd. Thus $\det(P-\lambda I_{2n})=p_n^2(\lambda)-q_{n-1}^2(\lambda)$ is even.
Remark 2. The roots of $p_n,q_{n-1}$, for $1\leq j\leq n-1$ and even, are the same. Then we explicitly know half of the eigenvalues of $A\pm B$. Finally, we know half of the eigenvalues of $P$ (they have multiplicity $2$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Need help with Differential equation concept Having trouble with this differential equation.
Find the general solution to the differential equation:
$$ \frac {dy}{dx}= \frac {√y(x^2 + x − 4)} {(x^2 + 1)(x − 1)} $$
I don't know where to start.
| $$\text{y}'\left(x\right)=\frac{\sqrt{\text{y}(x)}(x^2+x-4)}{(x^2+1)(x-1)}\space\Longleftrightarrow\space\int\frac{\text{y}'\left(x\right)}{\sqrt{\text{y}\left(x\right)}}\space\text{d}x=\int\frac{x^2+x-4}{(x^2+1)(x-1)}\space\text{d}x\tag1$$
Substitute $\text{u}=\text{y}\left(x\right)$ and $\text{d}\text{u}=\text{y}'\left(x\right)\space\text{d}x$ to find the integral $\int\frac{\text{y}'\left(x\right)}{\sqrt{\text{y}\left(x\right)}}\space\text{d}x$:
$$\int\frac{\text{y}'\left(x\right)}{\sqrt{\text{y}\left(x\right)}}\space\text{d}x=2\sqrt{\text{y}\left(x\right)}+\text{C}_1\tag2$$
For the RHS use partial fractions:
$$\frac{x^2+x-4}{(x^2+1)(x-1)}=\frac{2x}{x^2+1}+\frac{3}{x^2+1}-\frac{1}{x-1}\tag3$$
So, you get that:
$$\int\frac{x^2+x-4}{(x^2+1)(x-1)}\space\text{d}x=\ln\left|x^2+1\right|+3\arctan(x)-\ln\left|x-1\right|+\text{C}_2\tag4$$
So, we know that:
$$2\sqrt{\text{y}\left(x\right)}=\ln\left|x^2+1\right|+3\arctan(x)-\ln\left|x-1\right|+\text{C}\tag5$$
Solving for $\text{y}\left(x\right)$:
$$\text{y}\left(x\right)=\left\{\frac{\ln\left|\frac{x^2+1}{x-1}\right|+3\arctan(x)}{2}+\text{C}\right\}^2\tag6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Need help solving $x^4-3x^3-11x^2+3x+10=0$ Solve $x^4-3x^3-11x^2+3x+10=0$
I have tried to solve this equation using 'general formula from roots' from https://en.wikipedia.org/wiki/Quartic_function.
$$ax^4+bx^3+cx^2+dx+e=0$$
$$x_{1,2}=-\frac b{4a}-S \pm 0.5\sqrt{-4S^2-2P+ \frac q S}$$
$$x_{3,4}=-\frac b{4a} + S \pm 0.5\sqrt{-4S^2-2P-\frac q S}$$
$$p=\frac{8ac-3b^2}{8a^2}$$
$$q= \frac{b^3-4abc+8a^2d}{8a^3}$$
$$S=0.5\sqrt{-\frac{2p} 3 + \frac{Q+\Delta_0} Q}{3a}$$
$$Q=\left(\Delta_1+\sqrt{\Delta_1^2-4\Delta_0^3} \cdot 0.5\right)^{1/3}$$
$$\Delta_0=c^2-3bd+12ae$$
$$\Delta_1=2c^3-9bcd+27eb^2+27ad^2-72ace$$
Those formula does not work. I ended up with complex roots.
Please help me to use those formulas to solve quartic equation. Any help will be very much appreciated.
| One can check that $1$ and $-1$ are both roots of this polynomial. After factoring out $x-1$ and $x+1$, you're left with a quadratic polynomial.
As a general rule, the rational root theorem is a good place to start for questions like these.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove $\frac{ab}{4-d}+ \frac{bc}{4-a}+\frac{cd}{4-b}+\frac{da}{4-c} \leqslant \frac43$ for $a^2 + b^2 + c^2 + d^2 = 4$
Let $a,b,c,d \geqslant 0$ and $a^2+b^2+c^2+d^2=4$. Prove that
$$\frac{ab}{4-d}+ \frac{bc}{4-a}+\frac{cd}{4-b}+\frac{da}{4-c} \leqslant \frac43.$$
I try some reverse AM-GM techniques but fail. I don't think rearrangement inequality works because of cyclic nature of this inequality. I try to homogenize this inequality to remove the constrain, but the square root in the denominator kills me.
| Alternative proof:
We have
$$\frac{5}{18} + \frac{1}{18}d^2
- \frac{1}{4 - d} = \frac{(2 - d)(d - 1)^2}{18(4 - d)} \ge 0.$$
Thus, we have
\begin{align*}
&\frac{ab}{4-d}+ \frac{bc}{4-a}+\frac{cd}{4-b}+\frac{da}{4-c}\\
\le\,& \frac{5}{18}\sum_{\mathrm{cyc}} ab + \frac{1}{18}\sum_{\mathrm{cyc}}abd^2 \\
\le\,& \frac{5}{18}\sum_{\mathrm{cyc}} \frac{a^2 + b^2}{2} + \frac{1}{18}\sum_{\mathrm{cyc}}\frac{a^2 + b^2}{2}\cdot d^2\\
=\,& \frac{5}{18} (a^2 + b^2 + c^2 + d^2) + \frac{1}{18}
\left(\frac{(a^2+c^2)(b^2+d^2)}{2} + a^2c^2 + b^2d^2\right)\\
\le\,& \frac{5}{18}(a^2 + b^2 + c^2 + d^2) + \frac{1}{18}
\left(\frac{(a^2+c^2)(b^2+d^2)}{2} + \frac{(a^2+c^2)^2}{4} + \frac{(b^2+d^2)^2}{4}\right)\\
=\,& \frac{5}{18}(a^2 + b^2 + c^2 + d^2) + \frac{1}{18}
\frac{(a^2 + c^2 + b^2 + d^2)^2}{4}\\
=\,& \frac{4}{3}.
\end{align*}
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solution of $x^y=y^x$ and $x^2=y^3$ Solve the given set of equations:
$x^y=y^x$ and $x^2=y^3$ where $x,y \in \mathbb{R}$
Would any other solution exist other that $x=y=1$ because I think $x^2=y^3$
will only be true for $x=y=1$ or $x=y=0$
| Note that if $x^2=y^3$, then $y=x^{2/3}$. Therefore, we find that
$$x^{x^{2/3}}=\left(x^{2/3}\right)^x \tag 1$$
Taking the logarithm of both sides of $(1)$ yields
$$x^{2/3}\log(x)=\frac23 x\log(x) \tag2$$
Solutions to $(2)$ are $x=1$ and $x=27/8$. For $x=1$, $y=1$ and for $x=27/8$, $y=9/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
How can I compute $\tan(.5\arctan(x))$? The plot for this function appears to be in the form of $\alpha*\arctan(\beta*x)$ but I've no clue how to go about simplifying the expression.
| Recall that $\tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan a \tan b}$. Also let $c = \tan({\frac{\arctan x}{2}})$.
\begin{align}
c &= \tan(\arctan x - \frac{\arctan x}{2})\\
&= \frac{\tan(\arctan x) - \tan(\frac{\arctan x}{2})}{1 + \tan(\arctan x)\tan(\frac{\arctan}{2})}\\
&= \frac{x - c}{1 + xc}\\
c^2x + c &= x- c\\
c^2x + 2c - x & = 0
\end{align}
So by the quadratic formula(note that we take the positive root since due to the domain of the inverse tangent function)
\begin{equation}
c = \frac{-2 + \sqrt{4 - 4(x)(-x)}}{2x} = -\frac{1}{x} + \sqrt{1 + \frac{1}{x^2}}
\end{equation}
Therefore
\begin{equation}
\tan(\frac{\arctan x}{2}) = -\frac{1}{x} + \sqrt{1 + \frac{1}{x^2}}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Permutations conjugated
Show that the permutations:
$\alpha=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 5 & 3 & 6 & 1 & 4 \\
\end{pmatrix}
$
and
$\beta=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 3 & 4 & 2 & 1 & 6 \\
\end{pmatrix}
$
Are conjugated in $S_6$ and you hrite all permutations $\gamma\in S_6$ such that $\gamma\alpha\gamma^{-1}=\beta$
I know that $\alpha=(1,2,5)(4,6)(3)$ and $\beta=(2,3,4)(1,5)(6)$ also if $\sigma_1=\bigl( \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 1 & 5 & 4 & 6 \end{smallmatrix} \bigr)$ then $\sigma_1(1,2,5)\sigma_1^{-1}=(2,3,4)$; if $\sigma_2=\bigl( \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 4 & 1 & 6 & 5 \end{smallmatrix} \bigr)$ then $\sigma_2(4,6)\sigma_2^{-1}=(1,5)$ and if $\sigma_3=(3,6)$ then $\sigma_3(3)\sigma_3^{-1}=(6)$.
But I don't know how to build $\gamma$ such that $\gamma\alpha\gamma^{-1}=\beta$. Can you help me, please?
| One solution is to rewrite $\gamma\alpha\gamma^{-1}=\beta$ as $\gamma\alpha=\beta\gamma$.
Then $\gamma\alpha(1)=\beta\gamma(1)$, and from $\alpha(1)=2$ we get $\gamma(2)=\beta\gamma(1)$.
Similarly, $\gamma(5)=\beta\gamma(2)$ and $\gamma(1)=\beta\gamma(5)$.
Combining these, we get $\gamma(2)=\beta\gamma(1)=\beta\beta\beta\gamma(2)$, so $\gamma(2)$ has order 3 under $\beta$, and the same is true for $\gamma(5)$ and $\gamma(1)$.
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Is there a simple, intuitive way to see that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$ Is there a simple intuitive way to show that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$?
I sense it could be done more simple than this:
1 - take the derivative
$f'(x)=1-\frac{x}{\sqrt{x^2-1}}<0$ if $x>1$ so the slope of $f(x)$ in $(1,\infty)$ is negative.
2 - conclude
Since $f(1)=1$ and the derivative is negative we have showed that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$
Question: is there a simple, intuitive way to see that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$?
| Notice that $(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})=x^2-(x^2-1)=1$, and that $x+\sqrt{x^2-1}>x>1$, and therefore, we have that $x-\sqrt{x^2-1}=\frac{1}{x+\sqrt{x^2-1}}<\frac{1}{1}=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 0
} |
What are all the uses of the determinant? I've learned how to calculate the determinant but what is the determinant used for? So far, I only know that there is no inverse if the determinant is 0.
| *
*It allows you to evaluate cross products and find the general equation of the plane if given $3$ points. For instance: $A(1,1,0),\, B =(1,0,1),\,C=(0,1,2)$ $$B-A=(1,0,1)-(1,1,0)=(0,-1,1)$$ and $$C-A=(0,1,2)-(1,1,0)=(-1,0,2)$$ You now use the cross-product of $$(B-A)\times(C-A)=\begin{bmatrix}i & j & k \\0 & -1 & 1 \\-1 & 0 & 2 \end{bmatrix}=(-2,-1,-1)= \vec{n}$$ $i$, $j$ and $k$ are unit vectors that are directed along the $x$, $y$ and $z$ axis respectively. $\vec{n}$ is known as the normal vector $\vec{n}$ and is perpendicular to the equation of the plane.
*For square matrices formulae such as $$\det(AB) = \det (A) \det (B)$$ and $$\det (A^{-1}) = \frac 1 {\det (A)}$$ are applicable
*Allows you to check if your matrix has an inverse: $$\det=\begin{vmatrix}0 & 0 & 2 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & -1 & 3 & 0\\ 2 & 1 & 5 & -3 \end{vmatrix}$$
$$=0\times\begin{vmatrix} 0 & 0 & 1 \\ -1 & 3 & 0 \\ 1 & 5 & -3 \end{vmatrix}+0\times\begin{vmatrix} 1 & 0 & 1 \\ 0 & 3 & 0 \\ 2 & 5 & -3 \end{vmatrix}+2\times\begin{vmatrix} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 2 & 1 & -3 \end{vmatrix}+0\times\begin{vmatrix} 1 & 0 & 0 \\ 0 & -1 & 3 \\ 2 & 1 & 5 \end{vmatrix}$$$$=2\times\begin{vmatrix} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 2 & 1 & -3 \end{vmatrix}$$ $$=2\left( 1\times\begin{vmatrix}-1 & 0 \\ 1 & -3 \\ \end{vmatrix}+0\times\begin{vmatrix}0 & 0 \\ 2 & -3 \\ \end{vmatrix}+ 1\times\begin{vmatrix}0 & -1 \\ 2 & 1 \\ \end{vmatrix}\right)$$$$=2\times \left(1\times(3-0)+1\times(0--2)\right)= 2(3+2)=10 \ne 0$$ So your matrix is invertible (or has an inverse).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Strong Pseudoprime to base square? Question:
Let n>1 be an odd composite integer and let a be an integer with (a,n)=1.
Show that, if n is a strong pseudoprime to base a, then n is also a strong pseudoprime to base $a^2$
What I have done is that : let $n-1=2^{s}t$ then assumption implies that
$1. a^t \equiv 1 \pmod n$ or
$a^t \equiv -1 \pmod n$ or
$2.a^{2t} \equiv -1 \pmod n$
or... $a^{2^{s-1}t} \equiv -1 \pmod n$
so by part 1, $a^{2t} \equiv 1 \pmod n$ or
by part 2, $a^{2t} \equiv -1 \pmod n$
or... $a^{2^{s-1}t} \equiv -1 \pmod n$
these two results means that $n$ is strong pseudoprime to base $a^2$
when(if) we have also that $a^{2^{s}t} \equiv -1 \pmod n$
but the last term always be false because $a^{2^{s}t} \equiv a^{n-1}\equiv 1 \pmod n$ by that (strong pseudoprime is fermat pseudoprime) so we don't need that $a^{2^{s}t} \equiv -1 \pmod n$
so n is also a strong pseudoprime to base $a^2$ .. is my opinion right?
| Yes, the strong pseudoprime test doesn't require that you test $a^{n-1} \equiv 1 \bmod n$, but in any case that also works. Your proof basically works OK but gets a little unclear, perhaps because of the confusion between $a^{2k}$ and $(a^2)^k$.
So to restate the strong pseudoprime rule:
Define $t,d$ with $t\cdot 2^d = n-1$ with $t$ odd. $n$ is a pseudoprime base $a$ if either:
*
*$a^t \equiv 1 \pmod n$
*$a^{t\cdot2^k} \equiv -1 \pmod n$ for some $0 \le k \le (d-1)$
Note that these conditions also assure that $a^{n-1} \equiv 1 \pmod n$ without further evaluation.
Now we assume that $a$ meets one of these conditions, and check that $b= a^2$ also meets them:
*
*$a^t \equiv 1 \pmod n \implies b^t = a^{2t} \equiv 1 \pmod n $ (condition 1 applies for $b$)
*$a^t = a^{t\cdot2^0} \equiv -1 \pmod n \implies b^t = a^{2t} \equiv 1 \pmod n $ (condition 1 applies for $b$)
*$a^{t\cdot2^k} \equiv -1 \pmod n$ for some $1 \le k \le (d-1) \implies b^{t\cdot2^(k-1)} = a^{t\cdot2^k} \equiv -1 \pmod n$ (condition 2 applies for $b$)
Hence $n$ is also a strong pseudoprime base $b=a^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Asymptotic expansion of ratio function I want to expand the following function:
$$
f(x)=\frac{1}{(1-e^{-x})}
$$
$f(x)$ can be rewritten as
$$
f(x) \sim \frac{1}{x-x^2/2 + x^3/2/3}
$$
But I want to express big-oh notation such that
$$
f(x) = \frac{1}{x} + .... +O(x^2)
$$
up to $x^2$ order.
How to do it?
| Around $0$:
$$
f(x) = \frac{1}{x-\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)}
= \frac{1}{x}\frac{1}{1-\frac{x}{2}+\frac{x^2}{6}+o(x^2)}
$$
(Advice: when doing Taylor expansions, avoid the equivalents $\sim$, which are made to capture the first term only. Confusions and mistake may arise very quickly otherwise.)
Now, use the fact that $\frac{x}{2}+\frac{x^2}{6}+o(x^2) \xrightarrow[x\to0]{}0$, and that $$\frac{1}{1+u} = 1-u+u^2 + o(u^2)$$ when $u\to 0$ (your $u$ will be $-\frac{x}{2}+\frac{x^2}{6}+o(x^2)$):
$$\begin{align}
f(x) &= \frac{1}{x}\frac{1}{1-\frac{x}{2}+\frac{x^2}{6}+o(x^2)}
= \frac{1}{x} \left(1+\frac{x}{2}-\frac{x^2}{6}+o(x^2)+\left(\frac{x}{2}+\frac{x^2}{6}+o(x^2)\right)^2\right)\\
&= \frac{1}{x} \left(1+\frac{x}{2}-\frac{x^2}{6}+o(x^2)+\left(\frac{x^2}{4}+o(x^2)\right)\right)
= \frac{1}{x} \left(1+\frac{x}{2}-\frac{x^2}{6}+o(x^2)+\frac{x^2}{4}\right)\\
&= \frac{1}{x} \left(1+\frac{x}{2}+\frac{x^2}{12}+o(x^2)\right)
= \frac{1}{x}+\frac{1}{2}+\frac{x}{12}+o(x)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the hypotenuse ( Trigonometric Identities!) If the sides of a right-angled triangle are $\cos 2a + \cos 2b + 2\cos(a+b)$ and $\sin 2a + \sin 2b + 2\sin(a+b)$, find the hypotenuse-
I can simplify this but I end up having a lot of terms in the end :/
| $$\cos 2a+\cos 2b+2\cos(a+b)=\cos^2 a-\sin^2 a + \cos^2b-\sin^2b+2\cos a\cos b-2\sin a\sin b=(\cos a+\cos b)^2-(\sin a+\sin b)^2$$
$$\sin2a+\sin2b+2\sin(a+b)=2\sin a\cos a+2\sin b\cos b+2\sin a\cos b+2\cos a\sin b=2(\sin a+\sin b)(\cos a+\cos b)$$
Let $\cos a+\cos b=u$, $\sin a+\sin b=v$.
$$\text{Hypotenuse}=\sqrt{(u^2-v^2)^2+(2uv)^2}=u^2+v^2=(\cos a+\cos b)^2+(\sin a+\sin b)^2=2(1+\cos(a-b))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a,b$ or $c$ is divisible by $7$.
Show that if $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a$, $b$ or $c$ is divisible by $7$.
Here it seems Fermat's theorem has no use. We could consider many different cases of remainders of $a,b,c$ modulo $7$ but that's tedious.
Any ideas for a much shorter solution?
| You want to prove the statement:
$[a^3+b^3+c^3\equiv0\pmod7]\implies[a\equiv0\pmod7]\vee[b\equiv0\pmod7]\vee[c\equiv0\pmod7]$
Instead, prove the equivalent statement:
$[a\not\equiv0\pmod7]\wedge[b\not\equiv0\pmod7]\wedge[c\not\equiv0\pmod7]\implies[a^3+b^3+c^3\not\equiv0\pmod7]$
Now, observe the following:
*
*$n\equiv1\pmod7 \implies n^3\equiv1^3\equiv 1\equiv1\pmod7$
*$n\equiv2\pmod7 \implies n^3\equiv2^3\equiv 8\equiv1\pmod7$
*$n\equiv3\pmod7 \implies n^3\equiv3^3\equiv 27\equiv6\pmod7$
*$n\equiv4\pmod7 \implies n^3\equiv4^3\equiv 64\equiv1\pmod7$
*$n\equiv5\pmod7 \implies n^3\equiv5^3\equiv125\equiv6\pmod7$
*$n\equiv6\pmod7 \implies n^3\equiv6^3\equiv216\equiv6\pmod7$
Therefore, $[a\not\equiv0\pmod7]\wedge[b\not\equiv0\pmod7]\wedge[c\not\equiv0\pmod7]\implies$ one of the following:
*
*$a^3+b^3+c^3\equiv1+1+1\equiv 3\equiv3\not\equiv0\pmod7$
*$a^3+b^3+c^3\equiv1+1+6\equiv 8\equiv1\not\equiv0\pmod7$
*$a^3+b^3+c^3\equiv1+6+1\equiv 8\equiv1\not\equiv0\pmod7$
*$a^3+b^3+c^3\equiv1+6+6\equiv13\equiv6\not\equiv0\pmod7$
*$a^3+b^3+c^3\equiv6+1+1\equiv 8\equiv1\not\equiv0\pmod7$
*$a^3+b^3+c^3\equiv6+1+6\equiv13\equiv6\not\equiv0\pmod7$
*$a^3+b^3+c^3\equiv6+6+1\equiv13\equiv6\not\equiv0\pmod7$
*$a^3+b^3+c^3\equiv6+6+6\equiv18\equiv4\not\equiv0\pmod7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Bilinear transformation which maps $z=(\infty, i, 0)$ and $w= (-1, -i, 1)$ I have three equations after simplifying this a bit
$a+c=0$
$ai+b-c=0$
$b-d=0$
How do I proceed further?
If you care to know this is from the chapter Complex Variables
| We want $w(z)=\frac{az+b}{cz+d}$, such as $w(\infty)=-1,w(0)=1,w(i)=-i$.
This lead to the system :
$$\begin{align}\begin{cases} \frac{a}{c}=-1\\ \frac{b}{d}=1 \\ \frac{ai+b}{ci+d}=-i \end{cases} &\Rightarrow \begin{cases} c=-a\\ b=d \\ \frac{ai+b}{-ai+b}=-i \end{cases} \\&\Rightarrow \begin{cases} c=-a\\ b=d \\ \frac{(ai+b)^2}{b^2+a^2}=-i \end{cases} \\&\Rightarrow \begin{cases} c=-a\\ b=d \\ \frac{b^2-a^2+i2ab}{b^2+a^2}=-i \end{cases} \\&\Rightarrow \begin{cases} c=-a\\ b=d \\ b^2-a^2+i2ab=-ib^2-ia^2 \end{cases} \\&\Rightarrow \begin{cases} c=-a\\ b=d \\ b^2-a^2=0 \\2ab=-b^2-a^2 \end{cases} \\&\Rightarrow \begin{cases} c=-a\\ b=d \\ b^2=a^2 \\2ab=-2a^2 \end{cases} \\&\Rightarrow \begin{cases} c=-a\\ b=d \\ b^2=a^2 \\ab=-a^2 \end{cases} \end{align}$$
Consider the two cases $a=b$ and $a=-b$.
If $a=b$ :
$$\begin{cases} c=-a\\ b=d \\ a=b \\a^2=-a^2 \end{cases} \Rightarrow a=b=c=d=0$$
But such a transformation is not defined, so this case is not possible.
If $a=-b$ :
$$\begin{cases} c=-a\\ b=d \\ a=-b \\-a^2=-a^2 \end{cases} \Rightarrow -a=b=c=d$$
So $w(z)$ is of the form $\frac{-az+a}{az+a}=\frac{-z+1}{z+1}$ (if $a \neq 0$).
Let's verify that such an $w$ is solution.
$$w(\infty)=-1,w(0)=1,w(i)=\frac{-i+1}{i+1}=-i$$
So $w(z)=\frac{-z+1}{z+1}$ is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving Chinese Remainder Theorem Algebraically I am doing a practice problem for my final which asks:
Solve the following Chinese Remainder Theorem:
$$
x \equiv 2 \pmod{3}, \\
x \equiv 3 \pmod{5}, \\
x \equiv 5 \pmod{7}, \\
x \equiv 7 \pmod{11} \\
x \equiv 11 \pmod{13}
$$
From the first I can conclude that $x = 3k + 2$ for some $k \in \mathbb{Z}$.
Now I can apply that to the second equation which gives $ 3k+2 \equiv 3 \pmod{5}.$
Then I get lost here. Do I subtract $2$ and solve $ 3k \equiv 1 \pmod{5}$?
I don't have a solid understanding of solving the Chinese Remainder Theorem algebraically in general.
| We have x= 2 (mod 3) and x= 4 (mod 5).
From the first x= 2+ 3i for some integer i.
From the second x= 4+ 5j for some integer j.
2+ 3i= 4+ 5j so 3i- 5j= 2.
Now use the "Chinese remainder theorem to solve that 'Diophantine equation'.
3 divides into 5 once with remainder 2: 5- 3= 2
2 divides into 3 once with remainder 1: 3- 2= 1
3- 2= 3- (5- 3)= 2(3)- 5= 1.
Multiply by 2: 4(3)- 2(5)= 2.
So i= 4+ 5m and j= 2+ 3m for any integer m:
3(4+ 5m)- 5(2+ 3m)= 12+ 15m- 10- 15m= 2.
x= 2+ 3i so x= 2+ 3(4+ 5m)= 14+ 15m.
Now add "x= 5 (mod 7)" which means x= 5+ 7n for some integer n.
x= 14+ 15m= 5+ 7n so 7n- 15m= 14- 5= 9.
7 divides into 15 twice with remainder 1: 15- 2(7)= 1.
Multiply by 9: 9(15)- 18(7)= 9. n= -18+ 15p and m= -9+ 7p.
If we want positive integer answers, replace p with p+ 2:
n= -18+ 15p+ 30= 12+ 15p and m= -9+ 7p+ 14= 5+ 7p.
We now have x= 5+ 7n= 5+ 7(12+ 15p)= 5+ 84+ 105p= 89+ 105p.
I'll let you finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Express the function $f(x)= (1-\sin x)/(1+\sin x)$ as the sum of an even and odd function. Express the function $f(x)= (1-\sin x)/(1+\sin x)$ as the sum of an even and odd function.
|
Hint: In order to answer such questions the systematic approach provided by @coffeemath is very helpful, since it provides you a method to always find the odd and even part of a function $f$ regardless of its shape.
You want to split the function $f$ into the sum of an even function $g$ and and odd function $h$
\begin{align*}
f(x)=g(x)+h(x)
\end{align*}
Recall a function $g$ is even if $g(x)=g(-x)$.
If we put
\begin{align*}
g(x)=\frac{f(x)+f(-x)}{2}\tag{1}
\end{align*}
we obtain
\begin{align*}
g(-x)=\frac{f(-x)+f(x)}{2}=g(x)
\end{align*}
Recall a function $h$ is odd if $h(x)=-h(-x)$.
If we put
\begin{align*}
h(x)=\frac{f(x)-f(-x)}{2}\tag{2}
\end{align*}
we obtain
\begin{align*}
-h(-x)=-\frac{f(-x)-f(x)}{2}=\frac{f(x)-f(-x)}{2}=h(x)
\end{align*}
Putting all together results in
\begin{align*}
g(x)+h(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}=f(x)
\end{align*}
and we are done.
$$ $$
The even part of $f(x)$
With the help of (1) we systematically find the even part
\begin{align*}
g(x)&=\frac{1}{2}\left(f(x)+f(-x)\right)\\
&=\frac{1}{2}\left(\frac{1-\sin(x)}{1+\sin(x)}+\frac{1-\sin(-x)}{1+\sin(-x)}\right)\\
&=\frac{1}{2}\left(\frac{1-\sin(x)}{1+\sin(x)}+\frac{1+\sin(x)}{1-\sin(x)}\right)\\
&=\frac{1}{2}\cdot\frac{\left(1-\sin(x)\right)^2+\left(1+\sin(x)\right)^2}{1-\sin^2(x)}\\
&=\frac{1}{2}\cdot\frac{2+2\sin^2(x)}{1-\sin^2(x)}\\
&=\frac{1+\sin^2(x)}{\cos^2(x)}\\
\end{align*}
The odd part of $f(x)$
Similarly with the help of (2) we find the odd part
\begin{align*}
h(x)&=\frac{1}{2}\left(f(x)-f(-x)\right)\\
&=\frac{1}{2}\left(\frac{1-\sin(x)}{1+\sin(x)}-\frac{1-\sin(-x)}{1+\sin(-x)}\right)\\
&=\frac{1}{2}\left(\frac{1-\sin(x)}{1+\sin(x)}-\frac{1+\sin(x)}{1-\sin(x)}\right)\\
&=\frac{1}{2}\cdot\frac{\left(1-\sin(x)\right)^2-\left(1+\sin(x)\right)^2}{1-\sin^2(x)}\\
&=\frac{1}{2}\cdot\frac{-4\sin(x)}{1-\sin^2(x)}\\
&=-\frac{2\sin(x)}{\cos^2(x)}\\
\end{align*}
We conclude the representation by even and odd part of the function $f$ is
\begin{align*}
f(x)=\frac{1+\sin^2(x)}{\cos^2(x)}-\frac{2\sin(x)}{\cos^2(x)}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
how many answers do this equation have$3^{2x}-34(15^{x-1})+5^{2x}=0$ How many answers do this equation have?
$3^{2x}-34(15^{x-1})+5^{2x}=0$
My Attempt:$3^{2x}+5^{2x}=34(15^{x-1})$.Now what to do?
| $$3^{2x}-34(15^{x-1})+5^{2x}=0$$
$$3^{2x}-\frac{34}{15}(15^{x})+5^{2x}=0$$
$$15\cdot3^{2x}-34\cdot 3^x \cdot 5^x+15\cdot5^{2x}=0$$
$$15\cdot\left(\frac{3}{5}\right)^{2x}-34\cdot \left(\frac{3}{5}\right)^{x}+15=0$$
$\left(\frac{3}{5}\right)^{x}=t$
$$15\cdot t^{2}-34\cdot t+15=0$$
$t=\frac35$ or $t=\frac53$
$x=1$ or $x=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How can I use Bessel's equation to solve the Lengthening Pendulum differential equation? Taking a small extract of this previous bounty question of mine:
It can be shown that the differential equation: $$\fbox{$y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^{\prime}+\left[\left(bcx^{c-1}\right)^2+\frac{a^2-p^2c^2}{x^2}\right]y=0$}\tag{1}$$ has the solution $$\fbox{$y=x^aZ_p\left(bx^c\right)$}\tag{2}$$ where $Z_p$ stands for $J_p$ or $N_p$ or any linear combination of them, and $a,b,c,p$ are
constants.
To see how to use this, let us “solve” the differential equation: $$y^{\prime\prime}+9xy=0\tag{3}$$ If $(3)$ is of the type $(1)$, then we must have $$1-2a=0$$ $$2(c-1)=1$$ $$(bc)^2=9$$ $$a^2-p^2c^2=0$$ from these $4$ equations we find
$$a=\dfrac12$$ $$c=\dfrac32$$ $$b=2$$ $$p=\dfrac{a}{c}=\dfrac13$$
Then the solution of $(3)$ is $$y=x^{1/2}Z_{1/3}\left(2x^{3/2}\right)$$ This means that the general solution of $(3)$ is $$y=x^{1/2}\left[AJ_{1/3}\left(2x^{3/2}\right)+BN_{1/3}\left(2x^{3/2}\right)\right]$$ where $A$ and $B$ are arbitrary constants.
The differential equation of a lengthening pendulum is $$l\dfrac{d^2\theta}{dl^2}+2\dfrac{d\theta}{dl}+\dfrac{g}{v^2}\theta=0\qquad\quad\tag{4}\longleftarrow\text{proved in this former question}$$
I have to solve this differential equation for $\theta$ by comparing $(4)$ with $(1)$ in the same manner as used to find the solution to $(3)$
So here is my attempt:
First I begin by writing $(1)$ in terms of the new variables $\theta$ and $l$ $$\frac{\mathrm{d}^2\theta}{\mathrm{d}l^2}+\left(\frac{1-2a}{l}\right)\frac{\mathrm{d}\theta}{\mathrm{d}l}+\left[\left(b\,c\,l^{c-1}\right)^2+\frac{a^2-p^2c^2}{l^2}\right]\theta=0\tag{5}$$
Now I compare $(4)$ with $(5)$ to obtain
$$l=1$$
$$\frac{1-2a}{l}=2\implies a=-\frac12\qquad\text{since $l=1$}$$
$$\left(b\,c\,l^{c-1}\right)^2+\frac{a^2-p^2c^2}{l^2}=\frac{g}{v^2}\tag{6}$$
Substituting $a=\frac12$ and $l=1$ into $(6)$
$$\implies b^2c^2+\frac14 -p^2c^2=\frac{g}{v^2}\tag{7}$$
Now if $l=1$ we must have $$c=1$$
Substituting $c=1$ into $(7)$
$$\implies b^2+\frac14 -p^2=\frac{g}{v^2}\tag{8}$$
But I am unable to proceed from here and I have made a mistake anyway as I can tell you that the correct answer is $$\theta=l^{-1/2}Z_1\left(\frac{2{g}^{1/2}}{v}l^{1/2}\right)$$ Comparison with
$$\fbox{$y=x^aZ_p\left(bx^c\right)$}\tag{2}$$
shows that $$a=-\frac12, \quad c=\frac12, \quad b=\frac{2{g}^{1/2}}{v},\quad p=1$$
Could someone please help me find the correct values of $a,b,c,p$?
Any hints or advice is well appreciated.
| Hint: In the pendulum example the variable $l$ plays the role of $x$ and $\theta$ the role of $y$.
Keeping this in mind and comparing
\begin{align*}
\frac{d^2\theta}{dl^2}+\frac{2}{l}\frac{d\theta}{dl}+\frac{g}{v^2}\frac{\theta}{l}=0
\end{align*}
with
\begin{align*}
\frac{d^2\theta}{dl^2}+\left(\frac{1-2a}{l}\right)\frac{d\theta}{dl}+\left[(bcl^{c-1})^2+\frac{a^2-p^2c^2}{l^2}\right]\theta=0
\end{align*}
we get the following conditions:
\begin{align*}
1-2a&=2\\
2(c-1)&=-1\\
b^2c^2&=\frac{g}{v^2}\\
a^2-p^2c^2&=0
\end{align*}
and conclude
\begin{align*}
a=-\frac{1}{2},\quad c=\frac{1}{2},\quad b=\sqrt{\frac{4g}{v^2}}=\frac{2}{v}\sqrt{g},\quad p=\sqrt{\frac{a^2}{c^2}}=1
\end{align*}
We finally obtain
\begin{align*}
\theta=l^aZ_p\left(bl^c\right)=\frac{1}{\sqrt{l}}Z_1\left(\frac{2}{v}\sqrt{gl}\right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} \leq \frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)$ for $x,y,z \geq 0$ This inequality is wrong - see the accepted answer (it appears there is no general inequality for these two expressions).
On the left we have harmonic mean of pairwise geometric means, which obeys:
$$\sqrt[3]{xyz} \geq \color{blue}{ \frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}} \geq 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}} \geq \frac{3xyz}{xy+yz+zx}$$
On the right we have arithmetic means of pairwise harmonic means, obeying the same inequality (the first inequality here may not be true as well, waiting for proof in another question):
$$\sqrt[3]{xyz} \geq \color{blue}{ \frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)} \geq 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}} \geq \frac{3xyz}{xy+yz+zx}$$
According to Wolfram Alpha the following is true:
$$\frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} \leq \frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)$$
But I have not been able to prove it so far.
It may help to transform the RHS in the following way:
$$\frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)=\frac{2}{3} \frac{xyz(x+y+z)+(xy+yz+zx)^2}{(x+y+z)(xy+yz+zx)-xyz}$$
| Wrong Hint:
$\frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)^{-1} \leq \frac{2}{3} $
$\to 2.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \leq \frac{2}{3}.\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3\sqrt{xyz}}=\frac{2}{9}.\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}} \right) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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.