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Where have I made my mistake in my integrating $\int x \sqrt{2x - 1} \, dx$? $$\int x \sqrt{2x - 1} \,dx$$
Let $u = 2x - 1$
$$\int x \sqrt{u}\, dx$$
$$\frac{du}{dx} = 2 \implies \frac{1}{2}du = dx.$$
So the integral is written as
$$\int \frac{1}{2} u^{\frac{1}{2}} \, du$$
$$ = \frac{1}{2} \left(\frac{2}{3} u^{\frac{3}{2}} \right)$$
$$ = \frac{1}{3} (2x - 1)^{\frac{3}{2}} + c$$
But apparently this is wrong.
Wolfram says it is something completely different. Where have I made the mistake?
EDIT: OOOPS!! Sorry. Silly mistake.
| $$\int x \sqrt{u} dx\tag{1}$$
$$\int \frac{1}{2} u^{\frac{1}{2}} du\tag{2?}$$
What happened to the factor of $x$ going from $(1) \to (2)$?
We have that $u = 2x-1$, so $x = \dfrac{u+1}{2},\;$ and as you know, $dx = \dfrac 12 du.\,$ This gives us:
$$\int x \sqrt{2x - 1} \,dx \quad = \quad \frac{1}{2} \int \frac{(u+1)}{2} u^{1/2} \, du\tag{2} \quad =\quad \frac 14 \int \left(u^{3/2} + u^{1/2}\right) \,du$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/345179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove that $f \circ f (x) = x$ $$f(x) = \pi - \sqrt{1-a^2}\int_{0}^{x} \dfrac{dt}{1+\cos\,t}$$
Using that if $u = \tan \frac t2$, then $\cos\,t = \dfrac{1-u^2}{1+u^2}$
$u = \tan \frac t2 \Rightarrow t = 2 \arctan u \Rightarrow dt = \dfrac{2}{1+u^2} du$
substituting in the integral
$$\int_0^{\tan \frac x2} \dfrac{2\,du}{(1+u^2)\left(1+\dfrac{1-u^2}{1+u^2}\right)}$$
$$= \int_0^{\tan \frac x2} du $$
Then: $$f(x) = \pi - \sqrt{1-a^2}\int_0^{\tan\frac x2}du = \pi - \sqrt{1-a^2}\tan \frac x2$$
but I can't see what I can do to prove that $f\circ f(x) = x$ ( I have tried with trignometric identities and get this ugly expression:
$$ \pi - \sqrt{1-a^2} \, \cot \left(\dfrac{\sqrt{1-a^2}}{2}\, \tan \frac x2 \right) $$
| Something is wrong with your
definition of $f(x)$,
because,
if $a=1$ then $f(x) = \pi$ for all $x$,
so $f(f(x)) = \pi$ for all $x$.
In particular, it seems that the $a$ is out of place.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$
Prove that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$.
I know that for proving the $\gcd(a,b) = c$ you need to prove
*
*$c|a$ and $c|b$
*$c$ is the greatest number that divides $a$ and $b$
Number 2 is what I'm struggling with. Does anybody have any ideas?
| First of all, $$5\cdot(5^{98}+3)-1\cdot(5^{99}+1)=14$$
$$\implies (5^{98}+3,5^{99}+1)\mid 14$$
Now both the numbers are even $\implies 2\mid (5^{98}+3,5^{99}+1) $
Using Fermat’s Little theorem $5^6\equiv1\pmod 7$
$\implies 5^{98}=(5^6)^{16}\cdot5^2\equiv 25\pmod 7\equiv-3\implies 7\mid (5^{98}+3)$
and $5^{99}\equiv(5^6)^{16}\cdot5^3\equiv125\pmod 7\equiv-1 \implies 7\mid (5^{99}+1)$
$\implies 7\mid (5^{98}+3,5^{99}+1) $
$\implies (5^{98}+3,5^{99}+1)$ is divisible by lcm$(2,7)=14$
| {
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"url": "https://math.stackexchange.com/questions/346524",
"timestamp": "2023-03-29T00:00:00",
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What does it mean to eliminate $\theta$ from these equations? How should I do it?
Eliminate $\theta$ from the following pairs of equations:
A) $x=\sin \theta$, $y=\sin 2\theta$
B) $x=3\cos 2\theta +1$, $y=2\sin \theta$
My problem is I really don't understand what the question is asking? The answer in the book seems like it is asking me to put everything in terms of $x$ and $y$? And if it is, how do I go about that?
Thank you!
EDIT: My attempt:
$$y = 2\sin \theta \cos \theta $$
$${y \over {2\sin \theta }} = \cos \theta $$
so as $$x = \sin \theta $$
$$\cos \theta = {y \over {2x}}$$
what do i do from here?
| You've done the hard part. Of A):
$$
\begin{align}
\sin\theta & = x \\ \\
\cos\theta & = {\frac y{2x}}
\end{align}$$
This is where the Pythagorean Identity (as you call it) comes in very handy, indeed:
$$\quad\quad\quad\sin^2(\theta) + \cos^2(\theta) = 1$$
$$\iff \left(\sin\theta \right)^2 + \left(\cos\theta \right)^2 = 1$$
$$x^2 + \left(\dfrac{y}{2x}\right)^2 = 1$$
Then just simplify!
Part B:
$$x = 3 \cos 2\theta, \quad x = 2\sin \theta$$
$$x = 3(1-2\sin^2\theta) \tag{$\cos(2\theta) = 1 - 2\sin^2\theta$}$$
$$x= \frac 32(2 - (1 - \cos^2\theta)) \tag{$\sin^2\theta = 1 - \cos^2 \theta$}$$
$$x = \frac 32 (\cos^2 \theta + 1) \iff \frac 23 x - 1 = \cos^2\theta$$
$$ $$
$$ y = 2 \sin \theta \iff \sin \theta = y/2$$
$$ $$
$$\sin^2 \theta + \cos ^2 \theta = 1\tag{Pythagorean identity again}$$
$$\left(\frac y2\right)^2 + \left(\frac 23 x - 1\right) = 1$$
Simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/347737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the limit $\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|$ Calculate
$$\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|$$
| If you know a little bit about Taylor expansions and big O notation, you can do that easily. We will use
$$
\sqrt{1+u}=1+\frac{u}{2}+O(u^2)\qquad \mbox{when } u\longrightarrow 0.
$$
Now
$$
\pi\sqrt{n^2+n+1}=\pi n\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}=\pi n\left(1+\frac{1}{2n}+O\left(\frac{1}{n^2} \right)\right)=\pi n+\frac{\pi}{2}+O\left(\frac{1}{n} \right)
$$
So
$$
\sin (\pi\sqrt{n^2+n+1})=\sin\left(\pi n+\frac{\pi}{2}+O\left(\frac{1}{n} \right) \right)=(-1)^n\sin\left(\frac{\pi}{2}+O\left(\frac{1}{n} \right) \right).
$$
It follows that
$$
|\sin (\pi\sqrt{n^2+n+1})|=\left| \sin\left(\frac{\pi}{2}+O\left(\frac{1}{n} \right) \right)\right|\longrightarrow \left| \sin\left(\frac{\pi}{2}\right)\right|=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/349522",
"timestamp": "2023-03-29T00:00:00",
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Proving $x$ is divisible by $20$ I need to prove that $x$ divisible by $20$ if and only if $x=0\pmod4$ and $x=0 \pmod 5$
proving that if $x=0 \pmod 4$ and $x=0 \pmod 5$ than $x$ divisible by $20$ is by the Chinese theorem (am I right??)
But the other way - I dont understand why its not enough to be divisible only by one of them ($5$ OR $4$)?
P.S. How does it help me prove $(7^n+4*2^n+8^n-3^n)|20$?? How do I open this?
| $20 = 4*5$ so if $x$ is divisible by $20$ then it is divisible by $4$ and $5$. And hence by definition $x \equiv 0 \pmod 4$ and $x \equiv 0 \pmod 5$
Conversely, if $x \equiv 0 \pmod 4$ and $x \equiv 0 \pmod 5$ then $x = 4n$ and $x= 5m$ for some integers $n$ and $m$.
So $4n$ = $5m$.
Now since $5$ does not divide $4$, it must divide $n$. Hence $n = 5l$ for some integer $l$.
Hence $x = 4n = 4*5*l$.
Hence $x$ is divisible by $20$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/350670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find minimum of $\frac{a+3c}{a+2b+c}+\frac{7a+6b+3c}{a+b+2c}+\frac{c-a}{2a+b+c}$ for non-negative reals
Let $a, b, c\ge 0$, not all zero. Find the minimum of
$$N = \frac{a+3c}{a+2b+c}+\frac{7a+6b+3c}{a+b+2c}+\frac{c-a}{2a+b+c}. $$
| (This initial part is not a solution. It is too long to be a comment. This demonstrates that the suggested approach of substituting the denominator doesn't work out because equality is not achieved as the condition $ a \geq 0$ is not met.)
We use the substitution $x = a + 2b + c, y = a+b+2c, z = 2a + b + c$.
This system gives us $ a = (-x-y+3z)/4, b = (3x-y-z)/4, c = (-x+3y-z)/4$.
The expression becomes:
$$\frac{ -x + 2y } { x} + \frac{ 2 x - y + 3z } { y} + \frac{ y - z } { z} = \frac{ 2y}{x} + \frac{ 2x}{y} + \frac{3z}{y} + \frac{ y}{z} -3. $$
Since $ \frac{ 2y}{x} + \frac{ 2x}{y} \geq 2\sqrt{4} = 4$ and $ \frac{ 3z}{y} + \frac{y}{z} \geq 2 \sqrt{3} $, hence a lower bound is $ 2 \sqrt{3} +1 \approx 4.46 $.
Equality is achieved when $ y = x, y = \sqrt{3} z$, or equivalently when
$ a : b : c = 3 - 2 \sqrt{3} : 2 \sqrt{3} -1 : 2 \sqrt{3} - 1 $, which we have to verify have the same sign.
However, they do not have the same sign, so equality CANNOT be achieved. The minimum, or infimum, is higher.
We need the extra condition that (symmetric) $3z \geq x + y$ in order to get (symmetric) $a \geq 0$.
This condition was violated when we tried to find the equality case.
Since the expression is symmetric, WLOG we may assume that $ y = 1$. If so, we want to minimize
$$ \frac{2}{x} + 2x + 3z + \frac{1}{z} - 3 $$
subject to $3z \geq x + 1, 3 \geq x + z, 3x \geq z + 1$.
Even doing Lagrangian, this is horrendous and has very ugly solutions. We can show that the minimum happens on the boundary $3z=x+1$, but the actual solution ($\approx 4.493$) is pretty ugly.
I'm doubtful there's a contest-math solution.
My guess is that the problem setters made a mistake by not checking $ a : b : c$.
I do wish there was a nice way to complete this problem though, since it would illustrate the importance of checking conditions instead of just assuming as most of us did.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Existence and value of $\lim_{n\to\infty} (\ln\frac{x}{n}+\sum_{k=1}^n \frac{1}{k+x})$ for $x>0$ Does the limit
$$W(x)=\lim_{n\to\infty} \left(\ln\frac{x}{n}+\sum_{k=1}^n \frac{1}{k+x} \right)$$
exist for all $x>0$? If so, what is the limit
$$\lim_{x\to\infty}W(x)?$$
| We have
$$\sum_{k=1}^n \dfrac1{k+x} = \int_{1^-}^{n^+} \dfrac{d \lfloor t \rfloor}{t+x} = \left. \dfrac{\lfloor t \rfloor}{t+x} \right\vert_{t=1^-}^{t=n^+} + \int_{1^-}^{n^+} \dfrac{\lfloor t \rfloor}{(t+x)^2} dt = \dfrac{n}{n+x} + \int_{1^-}^{n^+} \dfrac{\lfloor t \rfloor}{(t+x)^2} dt$$
Now
$$\int_{1^-}^{n^+} \dfrac{\lfloor t \rfloor}{(t+x)^2} dt = \int_1^{n^+} \dfrac{t}{(t+x)^2} dt - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt$$
$$\int_1^{n^+} \dfrac{t}{(t+x)^2} dt = \int_1^{n} \dfrac{dt}{t+x} - x \int_1^n \dfrac{dt}{(t+x)^2} = \log(n+x) - \log(1+x) -x \left(\dfrac1{1+x} - \dfrac1{n+x}\right)$$
Hence, we get that
\begin{align}
\log(x/n) + \sum_{k=1}^n \dfrac1{k+x} & = \log\left(\dfrac{x}n \right) + \dfrac{n}{n+x} + \log \left(\dfrac{n+x}{1+x}\right) -\dfrac{x}{1+x} + \dfrac{x}{n+x} - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt\\
& = - \dfrac{x}{1+x} + 1 + \log\left(\dfrac{x}n \cdot \dfrac{n+x}{1+x}\right) - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt\\
& = \dfrac1{1+x} + \log\left(\dfrac{x}n \cdot \dfrac{n+x}{1+x}\right) - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt
\end{align}
Now letting $n \to \infty$, we get that
$$W(x) = \dfrac1{1+x} + \log \left(\dfrac{x}{1+x}\right) - \underbrace{\int_1^{\infty} \dfrac{\{t\}}{(t+x)^2} dt}_{\text{Converges since }\{t\} \in [0,1)}$$
$$0 \leq \overbrace{\int_1^{\infty} \dfrac{\{t\}}{(t+x)^2} dt}^{g(x)} \leq \int_1^{\infty} \dfrac1{(t+x)^2} dt = \dfrac1{1+x}$$ There might be some name for $g(x)$ (Probably some of the number theorists on this website might be able to identity this). For instance, $g(0) = 1-\gamma$, where $\gamma \approx 0.57721$ is the Euler Mascheroni constant.
Now $$\lim_{x \to \infty} W(x) = 0 + \log(1) + 0 = 0$$
Another method is as follows. From here, we have
\begin{align}
\sum_{k=1}^n \left(\dfrac1k - \dfrac1{x+k} \right) & = \sum_{k=1}^n \int_0^1 (y^{k-1} - y^{x+k-1})dy\\
& = \int_0^1 (1-y^x) \sum_{k=1}^n y^{k-1} dy\\
& = \int_0^1 (1-y^x) \dfrac{1-y^n}{1-y} dy
\end{align}
Hence, we have
\begin{align}
\log(x/n) + \sum_{k=1}^n \dfrac1{k+x} & = \log(x/n) + \sum_{k=1}^n \left(\dfrac1{k+x} - \dfrac1k \right) + \sum_{k=1}^n \dfrac1k\\
& = \log(x) + \sum_{k=1}^n \dfrac1k - \log(n) + \sum_{k=1}^n \left(\dfrac1{k+x} - \dfrac1k \right)\\
& = \log(x) + \sum_{k=1}^n \dfrac1k - \log(n) - \int_0^1 (1-y^x) \dfrac{1-y^n}{1-y} dy
\end{align}
Now letting $n \to \infty$, we get that
$$W(x) = \log(x) + \gamma - \int_0^1 \dfrac{1-y^x}{1-y} dy$$
Now as $x \to \infty$, we have $$\int_0^1 \dfrac{1-y^x}{1-y} dy = \log(x) + \gamma + \mathcal{O}(1/x)$$ Hence, we get that
$$\lim_{x \to \infty} W(x) = 0$$
Let us prove why, as $x \to \infty$, we have $$\int_0^1 \dfrac{1-y^x}{1-y} dy = \log(x) + \gamma + \mathcal{O}(1/x)$$
The proof is the same as before. We have
$$\sum_{k=1}^n \dfrac1k = \int_0^1 \sum_{k=1}^n y^{k-1} dy = \int_0^1 \dfrac{1-y^n}{1-y} dy$$
But we know that $\displaystyle \sum_{k=1}^n \dfrac1k = \log(n) + \gamma + \mathcal{O}(1/n)$. Hence, we get that
$$\int_0^1 \dfrac{1-y^n}{1-y} dy = \log(n) + \gamma + \mathcal{O}(1/n)$$
Replacing $n$ by $x$ and because the integral is a smooth function of $x$, we can conclude that
$$\int_0^1 \dfrac{1-y^x}{1-y} dy = \log(x) + \gamma + \mathcal{O}(1/x)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Reasoning about the Chebyshev functions: How does one check an upper bound based on the second Chebyshev function? In Ramanujan's proof of Bertrand's Postulate, Ramanujan states:
$\log([x]!) - 2\log([\frac{1}{2}x]!) \le \psi(x) - \psi(\frac{1}{2}x) + \psi(\frac{1}{3}x)$
where:
$\vartheta(x) = \sum_{p \le x} \log p$
$\psi(x) = \sum_{n\ge1} \vartheta(x^{\frac{1}{n}})$
$\log([x]!) = \sum_{n\ge1} \psi(\frac{1}{n}x)$
$\log([x!] - 2\log([\frac{1}{2}x]!) = \psi(x) - \psi(\frac{1}{2}x) + \psi(\frac{1}{3}x) - \psi(\frac{1}{4}x) + \ldots$
How does one go about proving that $\log([x]!) - 2\log([2x]!) \le \psi(x) - \psi(\frac{1}{2}x) + \psi(\frac{1}{3}x)$ is correct and something like $\log([2x!]) - \log([x!]) \le \psi(2x)$ is not correct?
Is $\log([3x!]) - \log([2x]!) \le \psi(3x) - \psi(2x) + \psi(\frac{3}{2}x)$?
| I found the answer to my question in M. Ram Murty, Problems in Analytic Number Theory.
If $a_0 \ge a_1 \ge a_2 \ge \ldots$ is a decreasing sequence of real numbers tending to $0$, then:
$a_0 - a_1 \le \sum_{n=0}^{\infty}(-1)^{n}a_n \le a_0 - a_1 + a_2$.
This result is straight forward to show using induction.
This can be applied to the Chebyshev second function since:
$\log([x]!) - 2\log([\frac{x}{2}]!) = \sum_{n \le x}(-1)^{n-1}\psi(\frac{x}{n})$
It is now straight forward to resolve $\log([2x]!) - \log([x]!)$ and $\log([3x!]) - \log([2x]!)$.
In both cases, the sequence may not be strictly decreasing so we cannot apply the upper bound.
In the case of $\log([2x]!) - \log([x]!)$, we have:
$\psi(2x) - \psi(x) + \psi(x) - \psi(\frac{1}{2}x) + \psi(\frac{2}{3}) - \ldots$ where $\psi(\frac{2}{3}x)$ may be greater than $\psi(\frac{1}{2}x)$.
In the case of $\log([3x]!) - \log([2x]!)$, we have:
$\psi(3x) - \psi(2x) + \psi(\frac{3}{2}x) - \psi(x) + \psi(x) - \psi(\frac{2}{3}x) + \psi(\frac{3}{4}x) - \ldots$ where $\psi(\frac{3}{4}x)$ may be greater than $\psi(\frac{2}{3}x)$.
| {
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"url": "https://math.stackexchange.com/questions/353794",
"timestamp": "2023-03-29T00:00:00",
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If $\frac{1}{2}1-\left(a_1+\frac{a_2}{2}+\cdots+\frac{a_n}{2^{n-1}}\right)$
Let $n>1$ be a positive integer and $\frac{1}{2}<a_{j}<1$ for $j=1,2,\ldots,n$. Show that $$(1-a_{1})(1-a_{2})\cdots (1-a_{n})>1-\left(a_{1}+\frac{a_{2}}{2}+\cdots+\frac{a_{n}}{2^{n-1}}\right).$$
I have no ideas.
| For another approach, we re-write as:
$$(1-a_1)(1-a_2) \dots (1-a_n) + \left(a_1 + \frac{a_2}{2} + \dots + \frac{a_n}{2^{n-1}}\right) > 1$$
Let $P = (1-a_1) (1-a_2)\dots (1-a_n)$ and $Q = \left(a_1 + \frac{a_2}{2} + \dots + \frac{a_n}{2^{n-1}}\right) $. It may be noted that for any $a_k$ we can define $p = \dfrac{P}{1-a_k}$ which does not depend on $a_k$ and $0 < p < \dfrac{1}{2^{n-1}}$. Similarly $q = Q - \frac{a_k}{2^{k-1}}$ does not depend on $a_k$.
Now $LHS = f(a_k) = P + Q = p(1-a_k) + q + \dfrac{a_k}{2^{k-1}} = \left(\dfrac{1}{2^{k-1}}-p\right)a_k + p+q$
is a linear function of $a_k$, with positive slope $\frac{1}{2^{k-1}}-p$.
Hence $f(a_k)$ achieves its lower bound at the lower limit of its domain, i.e. when $a_k \rightarrow \frac{1}{2}$. As the reasoning was for a general $a_k$, the minimum of the LHS must be when all $a_k \rightarrow \frac{1}{2}$. Hence $LHS > \dfrac{1}{2^n} + (1 - \dfrac{1}{2^n}) = 1$.
EDITED to explain the general case and clarify the notations used.
| {
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Fast way to calculate determinant for a block matrix I have a block matrix
$$Q_{(n+m-1)\times(n+m-1)} = \begin{pmatrix} A & -J\\-J^t & B \end{pmatrix}$$
where
$$A_{(m-1)\times(m-1)} = n*I_{(m-1)\times(m-1)} \text{ and } B_{n\times n} = m*I_{n\times n}$$
where $I$ is identity matrix. $J$ is then $(m-1)\times n$ matrix with all entries $1$.
For example, when $m = 4, n = 2$ we have
$$Q_{5\times 5} = \begin{pmatrix} 2 & 0 & 0 & -1 & -1\\ 0 & 2 & 0 & -1 & -1\\0&0&2&-1&-1\\-1&-1&-1&4&0\\-1&-1&-1&0&4\end{pmatrix}$$
Answer is $n^{m-1}\cdot m^{n-1}$ but I don't know how to show that. Please help. Thank you.
| See in wikipedia the page of determinat in section 3.3 on the Block matrices.
When $A$ is Invertible matrix, we have
$$
\det\begin{pmatrix}A& B\\ C& D\end{pmatrix} = \det(A) \det(D - C A^{-1} B).
$$
When $D$ is invertible, a similar identity with $\det(D)$ factored out can be derived analogously,
$$
\det\begin{pmatrix}A& B\\ C& D\end{pmatrix} = \det(D) \det(A - B D^{-1} C).
$$
In this case
\begin{align}
\det\begin{pmatrix}A & -J \\ -J^T & B \end{pmatrix}=
&
\det A \cdot \det\big( B-JA^{-1}J^T \big)
\\
=
&
n^{m-1}\det
\left[
\begin{array}{c}
m & m-1 & \ldots & m-1 & m-1\\
m-1 & m & \ldots & m-1 & m-1\\
\vdots & \vdots & \ddots &\vdots & \vdots\\
m-1 & m-1 & \ldots & m & m-1\\
m-1 & m-1 & \ldots & m-1 & m\\
\end{array}
\right]_{n\times n}
\end{align}
Now, note that
$$
\left[
\begin{array}{c}
m & \ldots & m-1 & \ldots & m-1\\
m-1 &\ldots & m-1 & \ldots & m-1\\
\vdots & & \vdots & & \vdots\\
m-1 & & m & & m-1\\
\vdots & & \vdots & & \vdots\\
m-1 &\ldots & m-1 & \ldots & m-1\\
m-1 &\ldots & m-1 & \ldots & m\\
\end{array}
\right]_{n\times n}
\left[
\begin{array}{c}
1\\
1\\
\vdots\\
0\\
\vdots\\
1\\
1\\
\end{array}
\right]
=
(m-1)
\left[
\begin{array}{c}
1\\
1\\
\vdots\\
0\\
\vdots\\
1\\
1\\
\end{array}
\right]
$$
implies
$$
\det
\left[
\begin{array}{c}
m & \ldots & m-1 & \ldots & m-1\\
m-1 &\ldots & m-1 & \ldots & m-1\\
\vdots & & \vdots & & \vdots\\
m-1 & & m & & m-1\\
\vdots & & \vdots & & \vdots\\
m-1 &\ldots & m-1 & \ldots & m-1\\
m-1 &\ldots & m-1 & \ldots & m\\
\end{array}
\right]_{n\times n}
=
(m-1)^n
$$
and then
$$
\det\begin{pmatrix}A & -J \\ -J^T & B \end{pmatrix}=n^{m-1}\cdot (m-1)^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/355886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Choosing teams with minimum number of boys and girls. I need to find different number of teams I can make with 6 people that needs to have at least 2 girls and 2 boys. There are 8 girls and 12 boys. So the way I think is I need to find total number of different teams that I can make first and subtract all boys team, all girls team, one girl 5 boys team, 5 girls one boy team.
$$\binom{20}{6} - \binom{12}{6} - \binom{8}{6} - \left(\binom{8}{5} \binom{12}{1}\right) - \left(\binom{8}{1} \binom{12}{5}\right)$$
$$\binom{20}{6} - \binom{12}{6} - \binom{8}{6} - \left( \binom{8}{5} \binom{12}{1}\right) - \left( \binom{8}{1} \binom{12}{5}\right) $$
| Just calculate the 3 cases instead.
(b,g)=(4,2), (3,3) and (2,4)
Anyways you r also right.
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Bounds for the exponential integral In Abramowitz and Stegun: Handbook of Mathematical Functions
(on page 229, property 5.1.20) it is found that
$$
\frac{1}{2} \log \left(1 + \frac{2}{x} \right) < \exp(x) E_1(x) < \log \left(1 + \frac{1}{x} \right) \qquad (x > 0)
$$
where
$$
E_1(x) = \int_x^\infty \frac{\exp(-t)}{t} dt
$$
How does the proof go?
| Here is a proof following from the theorem that, if for all $x$, $g'(x) < f'(x) < h'(x)$, and the inequality $g(x)<f(x)<h(x)$ holds at some point, then this inequality holds true for all $x$.
First we can rearrange the inequality as
$$e^{-x}\frac{\ln(x+2)-\ln(x)}{2}< \int_x^\infty \frac{e^{-t}}{t}dt<e^{-x} \big(\ln(x+1)-\ln(x)\big)$$
Then we can take the derivative of each side and divide by $e^{-t}$ to obtain another inequality to prove.
$$\frac{\ln(x)-\ln(x+2) +\frac{1}{x+2} - \frac{1}{x}}{2} <\frac{-1}{x}, < \ln(x)-\ln(x+1) +\frac{1}{x+1}-\frac{1}{x}$$
Now we can continue down the line, taking the derivative again.
$$\frac{\frac{1}{x}-\frac{1}{x+2}-\frac{1}{(x+2)^2}+\frac{1}{x^2}}{2} < \frac{1}{x^2} < \frac{1}{x}-\frac{1}{x+1}-\frac{1}{(x+1)^2}+\frac{1}{x^2}$$Finally, we have reduced the problem to something easier to prove. For the LHS, we can multiply out $2x^2(x+2)^2$ (a positive value since $x>0$) to obtain the equivalent inequality,
$$x(x+2)^2-x^2(x+2) - x^2 + (x+2)^2<2(x+2)^2$$
Now we can expand and everything cancels out nicely to get
$$x^3+4x^2+4x-x^3-2x^2 - x^2-x^2-4x-4 = -4<0$$
Now for the RHS, we can similarly multiply by $x^2(x+1)^2$ to obtain another equivalent inequality
$$(x+1)^2<x(x+1)^2-x^2(x+1) -x^2+(x+1)^2$$
Expanding and canceling we get
$$0<x^3+2x^2+x-x^3-x^2-x^2=x$$
Which is true since we are only considering $x>0$. Now that we've finally proved that $g''(x)<f''(x)<h''(x)$, the last step is to show that $g'(x)<f'(x)<h'(x)$ and $g(x)<f(x)<h(x)$. These can be verified at $x=1$, and if anyone would like more on this verification, just ask.
| {
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"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
How to find the minimum of $f(x)=(\sin(x)+\cos(x)+\tan(x)+\cot(x)+\sec(x)+\csc(x))^2$? I need to find the minimum of $f(x)$ with
$$f(x)=(\sin(x)+\cos(x)+\tan(x)+\cot(x)+\sec(x)+\csc(x))^2$$
Could you help me with some clues?
| To built upon @Queayiouer if you split the function as
$$ f(x) = \left( g(x) \right)^2 = \left( \left(\sin x+\frac{1}{\sin x}\right) + \left(\cos x+\frac{1}{\cos x}\right) + \left(\tan x+\frac{1}{\tan x}\right) \right)^2 $$
where the minimum occurs if $g(x)=0$ or $g'(x)=0$. The first is not going to happen.
So now we have
$$ g(x) = g_1(x) + g_2(x) + g_3(x) = \left(\sin x+\frac{1}{\sin x}\right) + \left(\cos x+\frac{1}{\cos x}\right) + \left(\tan x+\frac{1}{\tan x}\right) $$
Let find if they three functions have a common minimum since
$$ g_1'(x) = \cos(x) \left(1-\frac{1}{\sin^2 x} \right) = 0 $$
$$ g_2'(x) = \sin(x) \left(\frac{1}{\cos^2 x}-1 \right) = 0 $$
$$ g_3'(x) = \tan^2 x - \frac{1}{\tan^2 x} = 0$$
Unfortunately those are never all zero at the same time. So my guess it to proceed numerically with Newton-Raphson method where
$$ x \rightarrow x - \frac{ g_1(x) + g_2(x) + g_3(x) }{ g_1'(x) + g_2'(x) + g_3'(x) } $$
starting from $x=1$ or something. I get between $2.5<x<2.7$.
| {
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} |
How to find the eigenvalues and Jordan canonical form of this matrix Question:
let $a_{i,j}\in R,A=(a_{i,j})_{n\times n} $,and
$a_{i,j}=\begin{cases}
1&i+j\in\{n,n+1\}\\
0&i+j\notin\{n,n+1\}
\end{cases}$
that's meaning:
$$A=\begin{bmatrix}
0&0&0&\cdots&0&1&1\\
0&0&0&\cdots&1&1&0\\
0&0&\cdots&1&1&0&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
0&1&1&\cdots&0&0&0\\
1&1&0&\cdots&0&0&0\\
1&0&0&\cdots&0&0&0
\end{bmatrix}_{n\times n}$$
Problem (1): Find the Jordan canonical form of $A$.
I know this matrix Jordan is
$$diag(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})$$
where $\lambda_{i}$ is eigenvalue
But this problem key find the eigenvalue is hard,
Thank you.maybe this problem is not easy,But I hope see someone can solve it.Thank you very much!
| I show below that the eigenvalues of $A$ are exactly the numbers
$-2\cos\big(j\frac{2\pi}{2n+1}\big)$ for $1\leq j\leq n$.
What makes everything work is the
identity $2\cos(\theta)\cos(k\theta)=\cos((k-1)\theta)+\cos((k+1)\theta)$.
Unfortunately the eigenvectors are a little complicated to express directly,
the best presentation I could find so far was by successively applying
two reasonably simple change of bases.
Denote by $\alpha$ the endomorphism of ${\mathbb R}^n$ canonically associated
to $A$, and by $(d_1,d_2,d_3, \ldots,d_n)$ the canonical basis of ${\mathbb R}^n$. We thus
have
$$
\alpha(d_j)=
\left\lbrace\begin{array}{lcl}
d_{n-j}+d_{n+1-j}, & \text{for} & j<n, \\
d_1, & \text{for} & j=n.
\end{array}\right. \tag{1}
$$
Let us put $z_1=d_1,z_2=d_2$, and $z_j=d_{j}-d_{j-2}$ for $j>2$. We have thus
defined a new basis ${\mathcal Z}=(z_1,z_2,z_3, \ldots,z_n)$, and for any $j$
we have $d_j=\sum_{k\leq j, k \equiv j \ ({\sf mod} \ 2)} z_k$. We deduce that
$$
\alpha(z_j)=
\left\lbrace\begin{array}{lcl}
\sum_{k=1}^n z_k, & \text{for} & j=1, \\
\sum_{k=1}^{n-1} z_k, & \text{for} & j=2, \\
-z_{n+2-j}-z_{n+3-j}, & \text{for} & j>2. \\
\end{array}\right. \tag{2}
$$
In other words, the matrix of $\alpha$ relative to the basis $\mathcal Z$ is
$$
Z=\left(\begin{array}{ccccccccc}
1 & 1 & & & & \ldots & & & \\
1 & 1 & & & & \ldots & & & -1 \\
1 & 1 & & & & \ldots & & -1 & -1 \\
1 & 1 & & & & \ldots & -1 & -1 & \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1 & 1 & & & -1 & \ldots & & & \\
1 & 1 & & -1 & -1 & \ldots & & & \\
1 & 1 & -1 & -1 & & \ldots & & & \\
1 & & -1 & & & \ldots & & & \\
\end{array}\right) \tag{3}
$$
Next, let us define a new basis ${\mathcal W}=(w_1,w_2,w_3, \ldots,w_n)$, by putting
$$
w_j=
\left\lbrace\begin{array}{lcl}
z_{l+1}, & \text{for} & j=n-2l, l\geq 0 \\
z_{n+1-l}, & \text{for} & j=n-(2l+1),l\geq 1 \\
-(\sum_{k=1}^n z_k), & \text{for} & j=n-1. \\
\end{array}\right. \tag{4}
$$
By construction, we have
$$
z_j=
\left\lbrace\begin{array}{lcl}
w_{n+2-2j}, & \text{when} & n+2-2j\geq 1 \\
w_{2j-(n+3)}, & \text{when} & 2j-(n+3)\geq 1 \\
-(\sum_{k=1}^n w_k), & \text{when} & j \text{ is the unique integer in } [\frac{n+2}{2},\frac{n+3}{2}]. \\
\end{array}\right. \tag{5}
$$
We deduce that
$$
\alpha(w_j)=
\left\lbrace\begin{array}{lcl}
w_1+\sum_{k=3}^n w_k, & \text{for} & j=1, \\
-w_{j-1}-w_j, & \text{for} & 1 < j < n-1, \\
-w_{n-1}-2w_n, & \text{for} & j=n-1, \\
-w_{n-1}, & \text{for} & j=n.
\end{array}\right. \tag{6}
$$
In other words, the matrix of $\alpha$ relative to the basis $\mathcal W$ is
$$
W=\left(\begin{array}{cccccccccc}
1 & -1 & & & & \ldots & & & & \\
0 & & -1 & & & \ldots & & & & \\
1 & -1 & & -1 & & \ldots & & & & \\
1 & & -1 & & -1 & \ldots & & & & \\
1 & & & -1 & & \ldots & & & & \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots \\
1 & & & & & \ldots & -1 & & & \\
1 & & & & & \ldots & & -1 & & \\
1 & & & & & \ldots & -1 & & -1 & \\
1 & & & & & \ldots & & -1 & & -1 \\
1 & & & & & \ldots & & & 2 & \\
\end{array}\right) \tag{7}
$$
From the identity
$2\cos(\theta)\cos(k\theta)=\cos((k-1)\theta)+\cos((k+1)\theta)$, (combined
with $\cos(-k\theta)=\cos(k\theta)$ and $\displaystyle\sum_{k=1}^{n} \cos(k\theta)=-\frac{1}{2}$ when
$\theta$ is a multiple of $\frac{2\pi}{2n+1}$), we see that for $1\leq j \leq n$, the vector
$$
v_j=\sum_{k=1}^{n} \cos\bigg(jk\frac{2\pi}{2n+1}\bigg)w_j
$$
is an eigenvector for $W^T$, associated to the eigenvalue $-2\cos\big(j\frac{2\pi}{2n+1}\big)$. This finishes the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/363237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
The minimum and the maximum of $y=\sin^2x/(1+\cos^2x)$ I was asked to find the minimum and maximum values of the functions:
*
*$y=\sin^2x/(1+\cos^2x)$;
*$y=\sin^2x-\cos^4x$.
What I did so far:
*
*$y' = 2\sin(2x)/(1+\cos^2x)^2$
How do I check if they are suspicious extrema points? After this function is cyclical and therefore only section that is not $(-\infty,\infty)$ can there be a local minimum/maximum.
*$y' = \sin(2x)+4\cos^3(x)\cdot\sin(x)$
Any suggestions?
| If you don't really need to use derivatives,
$1.$
Clearly, $y=\frac{\sin^2x}{1+\cos^2x}\ge 0$ in fact $=0$ if $\sin x=0$
$y-1=\frac{\sin^2x}{1+\cos^2x}-1=\frac{\sin^2x-(1+\cos^2x)}{1+\cos^2x}=-2\frac{\cos^2x}{1+\cos^2x}\le 0$ in fact $=0$ if $\cos x=0$
$\implies y-1\le 0\iff y\le 1\implies 0\le y\le 1$
$2.$ $$y=\sin^2x-\cos^4x=1-\cos^2x-\cos^4x=1-\cos^2x(1+\cos^2x)$$
$$\text{Now, }0\le \cos^2x\le1\implies 1\le\cos^2x+1\le2 \implies 0\le \cos^2x(\cos^2x+1)\le2$$
$$\implies 0\ge-\cos^2x(\cos^2x+1)\ge-2\iff 1\ge1-\cos^2x(\cos^2x+1)\ge-1$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
} |
Inequality: $\frac{a^2+b^2}{(a+b)^4}+\frac{b^2+c^2}{(b+c)^4}+\frac{c^2+d^2}{(c+d)^4}+\frac{d^2+a^2}{(d+a)^4} \leq \frac{1}{abcd}$ If $ab+bc+cd+da\leq 8$ and $a,b,c,d \in \mathbb{R}_{+},$how can I prove the following inequality :
$$\frac{a^2+b^2}{(a+b)^4}+\frac{b^2+c^2}{(b+c)^4}+\frac{c^2+d^2}{(c+d)^4}+\frac{d^2+a^2}{(d+a)^4} \leq \frac{1}{abcd}$$
I use : $$\frac{a+b}{2}\geq \frac{2ab}{a+b}$$ or $$\frac{1}{(a+b)^2} \leq \frac{1}{4ab}$$
$$\frac{1}{(a+b)^4} \leq \frac{1}{16a^2b^2}$$
$$\frac{a^2+b^2}{(a+b)^4}\leq \frac{a^2+b^2}{16a^2b^2}$$
but from here I have no idea.
thanks :)
| Multiplying $(ab+bc+cd+da)/8$ to the RHS, it suffices to show that
$$\sum \frac{(a^2+b^2)}{(a+b)^4}\le \sum\frac{1}{8ab}.$$
Very luckily we have
$$\frac{(a^2+b^2)}{(a+b)^4} \le \frac{1}{8ab},$$
which is equivalent to $(a+b)^4 \ge 8ab(a^2+b^2)$ or $(a-b)^4 \ge 0$. Q.E.D.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Sequence and equation I am having a problem finding an equation for this sequence: 2, 2+4 , 2+4+6, 2+4+6+8 .
Can someone show me the steps in solving this?
| We have $a_1 = 2 \times 1$, $a_2 = 2 \times 1 + 2 \times 2$, $a_3 = 2 \times 1 + 2 \times 2 + 2 \times 3$.
In general, we have
$$a_n = 2 \times 1 + 2 \times 2 + 2 \times 3 + \cdots + 2 \times n = 2 \times (1+2+\cdots+n) = 2 \times\dfrac{n(n+1)}2 = n(n+1)$$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $3\sum\limits_{i=0}^k\binom{n}{3i}\leq2^n+2$ If $n\in \mathbb{Z^+}$ and $k$ is the largest integer for which $3k\leq n$, then is it true that $\sum_{i=0}^k\binom{n}{3i}\leq \frac{1}{3}(2^n+2)$?
My work: We can break this into two cases: $n=3k+1$ and $n=3k+2$.If $n=3k+1$, then we need to prove that $$\sum_{i=0}^k\binom{3k+1}{3k}\leq \frac{1}{3}(2^{3k+1}+2)$$ The first thing which strikes me is the fact that $2^{3k+1}=\sum_{i=0}^{3k+1}\binom{3k+1}{i}$ and the inequality can now be written after simplification as $$2\sum_{i=0}^k\binom{3k+1}{3i}\leq 2+\sum_{i=0}^k\binom{3k+1}{3i+1}+\sum_{i=0}^k\binom{3k+1}{3i+2}$$ but it does not seem to be helpful.Besides, there is another case($n=3k+2$) left to be tackled as well. Any ideas or hints will be useful.
| Here’s a completely elementary solution. Let
$$\begin{align*}
a_n&=\sum_k\binom{n}{3k}\;,\\
b_n&=\sum_k\binom{n}{3k+1}\;,\text{ and}\\
c_n&=\sum_k\binom{n}{3k+2}\;.
\end{align*}$$
Clearly $a_n+b_n+c_n=2^n$. Moreover, it follows from Pascal’s identity that
$$\begin{align*}
a_n&=a_{n-1}+c_{n-1}\;,\\
b_n&=b_{n-1}+a_{n-1}\;,\text{ and}\\
c_n&=c_{n-1}+b_{n-1}\;.
\end{align*}$$
Finally, $a_0=1$, and $b_0=c_0=0$, and the recurrences above make it very easy to calculate the first few values:
$$\begin{array}{c|ccc}
n&a_n&b_n&c_n\\ \hline
0&\color{red}1&0&0\\
1&1&1&\color{blue}0\\
2&1&\color{red}2&1\\
3&\color{blue}2&3&3\\
4&5&5&\color{red}6\\
5&11&\color{blue}{10}&11\\
6&\color{red}{22}&21&21
\end{array}$$
The pattern is clear:
$$\begin{align*}
a_n&=\begin{cases}
\left\lceil\frac{2^n}3\right\rceil,&\text{if }n\equiv 0,1,5\pmod6\\\\
\left\lfloor\frac{2^n}3\right\rfloor,&\text{if }n\equiv 2,3,4\pmod6
\end{cases}\\\\
b_n&=\begin{cases}
\left\lceil\frac{2^n}3\right\rceil,&\text{if }n\equiv 1,2,3\pmod6\\\\
\left\lfloor\frac{2^n}3\right\rfloor,&\text{if }n\equiv 0,4,5\pmod6
\end{cases}\\\\
c_n&=\begin{cases}
\left\lceil\frac{2^n}3\right\rceil,&\text{if }n\equiv 3,4,5\pmod6\\\\
\left\lfloor\frac{2^n}3\right\rfloor,&\text{if }n\equiv 0,1,2\pmod6\;.
\end{cases}
\end{align*}\tag{1}$$
It’s a straightforward (if slightly tedious) matter to use the recurrences to prove $(1)$ by induction on $n$. Then merely note that for all $n$ we have
$$a_n\le\left\lceil\frac{2^n}3\right\rceil\le\frac{2^n+2}3\;.$$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding generating function for the recurrence $a_0 = 1$, $a_n = {n \choose 2} + 3a_{n - 1}$ I am trying to find generating function for the recurrence:
*
*$a_0 = 1$,
*$a_n = {n \choose 2} + 3a_{n - 1}$ for every $n \ge 1$.
It looks like this:
*
*$a_0 = 1$
*$a_1 = {1 \choose 2} + 3$
*$a_2 = {2 \choose 2} + 3{1 \choose 2} + 9$
*$a_3 = {3 \choose 2} + 3{2 \choose 2} + 9{1 \choose 2} + 27$
*$a_4 = {4 \choose 2} + 3{3 \choose 2} + 9{2 \choose 2} + 27 {1 \choose 2} + 81$
I know what the generating function of the sequence $3 ^n = (1, 3, 9, 27, 81, \dots)$ is, as well as what the generating functions for some sequences of combinatorial numbers are, but how do I split the sequence up into these pieces I know?
(The problem is those combinatorial numbers "move right" every time. If they were growing left-to-right along with their coefficients, it would be much easier. And there is no constant difference between $a_i$ and $a_{i + 1}$.)
| Sneaky. Write your recurrence without subtractions in indices, i.e.:
$$
a_{n + 1} = 3 a_n + \binom{n + 1}{2}
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply by $z^n$, sum over $n \ge 0$ and recognize the resulting sums, particularly:
\begin{align}
\sum_{n \ge 0} \binom{n + 1}{2} z^n
&= z \sum_{n \ge 0} \binom{n + 1}{2} z^{n - 1} \\
&= z \sum_{n \ge 0} \binom{n + 2}{2} z^n \\
&= \frac{z}{(1 - z)^3}
\end{align}
so that:
$$
\frac{A(z) - a_0}{z} = 3 A(z) + \frac{z}{(1 - z)^3}
$$
Using $a_0 = 1$ and solving as partial fractions:
$$
A(z) = \frac{11}{8 (1 - 3 z)}
- \frac{1}{2 (1 - z)^3}
+ \frac{1}{4 (1 - z)^2}
- \frac{1}{8 (1 - z)}
$$
We can read off the coefficients here:
\begin{align}
a_n &= \frac{11}{8} \cdot 3^n
- \frac{1}{2} \binom{-3}{n} (-1)^n
+ \frac{1}{4} \binom{-2}{n} (-1)^n
- \frac{1}{8} \\
&= \frac{11 \cdot 3^n - 1}{8}
- \frac{1}{2} \binom{n + 3 - 1}{3 - 1}
+ \frac{1}{4} \binom{n + 2 - 1}{2 - 1} \\
&= \frac{11 \cdot 3^n - 1}{8}
- \frac{1}{2} \cdot \frac{(n + 2) (n + 1)}{2!}
+ \frac{1}{4} \cdot \frac{n + 1}{1!} \\
&= \frac{11 \cdot 3^n - 2 n^2 - 4 n - 3}{8}
\end{align}
| {
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} |
cauchy schwarz inequality problemes I have to prove that for all $x,y,z>0$,
$$\left(\frac{x+y}{x+y+z}\right)^{0.5} + \left(\frac{x+z}{x+y+z}\right)^{0.5} + \left(\frac{z+y}{x+y+z}\right)^{0.5} \leq 6^{0.5}$$
using Cauchy-Schwarz inequality? How do I do that ?
I have to define an inner product but I do not what, and what are the vectors?
| Something to notice is that $\left(\frac{x+y}{x+y+z}\right)+ \left(\frac{x+z}{x+y+z}\right) + \left(\frac{z+y}{x+y+z}\right)= 2 $. This suggests writing $\left(\frac{x+y}{x+y+z}\right)^{0.5} + \left(\frac{x+z}{x+y+z}\right)^{0.5} + \left(\frac{z+y}{x+y+z}\right)^{0.5} $ as the dot product of $\left(\left(\frac{x+y}{x+y+z}\right)^{0.5},\left(\frac{x+z}{x+y+z}\right)^{0.5},\left(\frac{z+y}{x+y+z}\right)^{0.5}\right)$ with $(1,1,1)$, since the sum of the squares of the three terms will be on the right-hand side. And sure enough, using the Cauchy-Schwarz inequality will immediately give the bound you seek.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/373649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
(4x^2+2kx-5)/(x+2) remainder is 3 find value of k? 2 methods - first is long division by $(x+2)$, 2nd is to use remainder theorem
let $f(x) = 4x^2+2kx-5$ and $g(x) = x+2$
to find the remainder of $\frac{f(x)}{g(x)}$ where $g(x) = (x+c)$ we need to evaluate $f(-c)$
$f(-2) = 4(-2)^2+2k(-2) -5$
Because the remainder is 3, we know that $f(-2)=3$
so $$16 - 4k -5 = 3$$
$$4k = 8$$
$$k=2$$
is that correct?
| We have
\begin{align}
\dfrac{4x^2 + 2kx - 5}{x+2} & = \dfrac{4(x+2)^2 - 4 \cdot(4x) - 4 \cdot 4 + 2kx-5}{x+2}\\
& = 4(x+2) + \dfrac{(2k-16)x - 21}{x+2}\\
& = 4(x+2) + \dfrac{(2k-16)x - 21}{x+2}\\
& = 4(x+2) + \dfrac{(2k-16)(x+2) - 21-2(2k-16)}{x+2}\\
& = 4(x+2) + (2k-16) + \dfrac{11-4k}{x+2}
\end{align}
This means $11-4k = 3 \implies k = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/375469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
$ax+by+cz=d$ is the equation of a plane in space. Show that $d'$ is the distance from the plane to the origin. This is a 3 part practice question I would like to get some feedback on. I think I have solved the 1st two parts, but I need a little direction for part (c) (the title is Part (a) ) which is repeated here with more detail,
a) $ax+by+cz=d$ is the equation of a plane in space. Divide this equation by a real number to get $a'x+b'y+c'z=d'$.
Show that $d'$ is the distance from the plane to the origin.
It is required that $ a'^2 + b'^2 + c'^2 = 1 $ and $d' >= 0$
Here is my solution to part (a)
Let $p$ be a real number $ (p \neq 0)$, such that
$\frac{1}{p}(ax + by + cz) = \frac{d}{p} = a'x + b'y + c'z = d'$
It is required that
$ a'^2 + b'^2 + c'^2 = 1 $ so,
$ (\frac{a}{p})^2 + (\frac{b}{p})^2 + (\frac{c}{p})^2 = 1 $, this implies
$ a^2 + b^2 + c^2 = p^2 $
$ \pm\sqrt{a^2 + b^2 + c^2} = p $
Since $d = ax_0 + by_0 + cz_0 $ (where $ Q(x_0,y_0,z_0) $ is a point on the plane),
$d' = \frac{1}{p}(ax_0 + by_0 + cz_0)$
Substituting $ p $ into $d'$, we get
$d' = \frac{1}{\sqrt{a^2 + b^2 + c^2}}(ax_0 + by_0 + cz_0) = \frac{ax_0 + by_0 + cz_0}{\sqrt{a^2 + b^2 + c^2}}$
(taking the positive since it is required that $d' >= o$)
Since
$ ai + bj + ck $ is a normal vector to the plane and,
$ x_0i + y_0j + z_0k $ is the position vector of point $Q$
$d' = \frac{ax_0 + by_0 + cz_0}{\sqrt{a^2 + b^2 + c^2}} = \frac{|n \cdot \vec{\mathbf{OQ}|} }{|n|}$
This is the formula from a point to a plane (in this case the origin).
b) Find the position vector of the point $S$ that is closest to the origin on the plane, with equation in the primed form of part (a).
My solution to part (b):
From part (a) it was determined that
$d' = \frac{ax_0 + by_0 + cz_0}{\sqrt{a^2 + b^2 + c^2}} = \frac{|n \cdot \vec{\mathbf{OQ}|} }{|n|}$ , is the distance from the plane to the origin.
And, in the primed form
$ \vec{\mathbf{n}} = a'i + b'j + c'k $ is a normal vector to the plane
Since $ \vec{\mathbf{n}} \cdot \vec{\mathbf{n}} = |\vec{\mathbf{n}}|^2 = a'^2 + b'^2 + c'^2 = 1 $ so $ |\vec{\mathbf{n}}| = 1$
$ \vec{\mathbf{n}} $ is the unit normal vector to $ a'x + b'y + c'z = d'$
The position vector $ \vec{\mathbf{S}} $ is then:
$ d'\vec{\mathbf{n}} = d' (a'i + b'j + c'k) $
Is this a sufficient answer? Or should I expand d'?
c) Consider the plane given by the equation in the primed form of part (b). Why is each plane containing the origin described by two distinct equations of this
form? Why does each plane that does not contain the origin have a unique equation
of this form?
This question, I need a little direction.
If the equation of a plane in form $ a'x + b'y + c'z = d'$ contains the origin, then
$d' = 0$, so
$ a'x + b'y + c'z = \frac{a}{p}x + \frac{b}{p}x + \frac{c}{p}x = 0$
What would be a 2nd distinct equation of this form? Could someone point me in the right direction?
| Your answer looks good. Here is an alternate method to find the minimum point.
To minimize
$$
x^2+y^2+z^2\tag{1}
$$
while maintining
$$
ax+by+cz=d\tag{2}
$$
we want to have
$$
2x\,\delta x+2y\,\delta y+2z\,\delta z=0\tag{3}
$$
for all $(\delta x,\delta y,\delta z)$ so that
$$
a\,\delta x+b\,\delta y+c\,\delta z=0\tag{4}
$$
By orthogonality, this means we need $(a,b,c)\|(x,y,z)$; that is,
$$
(x,y,z)=k(a,b,c)\tag{5}
$$
Substituting $(5)$ into $(2)$ yields
$$
k(a^2+b^2+c^2)=d\tag{6}
$$
Plugging $(5)$ and $(6)$ into $(1)$, we get the square of the minimum distance is
$$
\left(\frac{d}{a^2+b^2+c^2}\right)^2(a^2+b^2+c^2)\tag{7}
$$
Thus, the minimum distance is
$$
\frac{|d|}{\sqrt{a^2+b^2+c^2}}\tag{8}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/378903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How prove this $T_{n}<\dfrac{5}{4}|b_{1}|$ let $\{b_{n}\}$ such that $b^2_{n}=b_{n+1}(1+b^2_{n})$,
and
$$T_{n}=\sum_{k=1}^{n}\dfrac{(-1)^k(k+2)}{(k+1)^2}b_{k}$$
show that
$T_{n}<\dfrac{5}{4}|b_{1}|$
my idea
$b^2_{n}=b_{n+1}(1+b^2_{n})$ then $b_{n}>0,n\ge 2$
and
$$b^2_{n}\ge 2b_{n+1}b_{n},\Longleftrightarrow b_{n+1}\le \dfrac{1}{2}b_{n}$$
then $$b_{n}\le \dfrac{1}{2^{n-2}}b_{2}=\dfrac{1}{2^{n-2}}\dfrac{b^2_{1}}{1+b^2_{1}},n\ge 2$$
But is follow very ugly. Thank you veryone
| Suppose $b_1\ne 0$. (If $b_1=0$, then $b_n=0$ for all $n$ and the inequality does not hold.) And I also assume $n\ge 1$ in $b_n^2=b_{n+1}(1+b_n^2)$.
Lemma. $0<b_{n+1}\leq b_n/2$ for $n\ge 2$, and $0<b_2\leq |b_1|/2$.
Proof of lemma. (From the OP's argument.) $b_{n+1}=\frac{b_n^2}{1+b_n^2}>0$ for $n\ge 1$. Using the fact that $1+b_n^2\ge 2|b_n|$, we find $|b_n|^2=b_n^2=b_{n+1}(1+b_n^2)\ge 2b_{n+1}|b_n|$, hence $|b_{n}|\ge 2b_{n+1}$.
Proof. The $(2l-1)$-th and $2l$-th terms of $T_n$ ($l\ge 1$) are
\begin{align*}
-\frac{2l+1}{(2l)^2}b_{2l-1}+\frac{2l+2}{(2l+1)^2}b_{2l}.
\end{align*}
For $l\ge 2$, this is negative, because $\dfrac{2l+1}{(2l)^2}\geq \dfrac{2l+2}{(2l+1)^2}>0$ ($\because\dfrac{k+2}{(k+1)^2}=\dfrac1{k+1}+\dfrac1{(k+1)^2}$ is a decreasing function of $k\ge 1$) and $0<b_{2l}\leq b_{2l-1}/2$ (from the lemma).
For $l=1$, this is equal to
\begin{align*}
-\frac34b_1+\frac49b_2
&\leq\frac34|b_1|+\frac49\cdot\frac12|b_1|\\
&=\frac{35}{36}|b_1|
\end{align*}
If $n$ is even, the above argument suggests $T_n\le\frac{35}{36}|b_1|<\frac54|b_1|$. If $n$ is odd ($\ge 3$), the last term remains, which is negative, so still $T_n\le\frac{35}{36}|b_1|<\frac54|b_1|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Is $u_n\le(1-a)^n\forall n\in\mathbb{N}$? Consider the sequence $\{u_n\}$ where $u_0=1,u_1=1-a$ for some $0< a < 1/4$, and $u_{n+2} = u_{n+1}-au_n$. Is $u_n\le(1-a)^n\forall n\in\mathbb{N}$?
| Let $u_n = (1-a)^n b_n$
Then we have that $b_0 = 1$ and $b_1 = 1$.
Now $$b_{n+2} = \frac{b_{n+1}}{1-a} - \frac{ab_n}{(1-a)^2}$$
And so
$$b_{n+2} - b_{n+1} = \frac{ab_{n+1}}{1-a} - \frac{ab_n}{(1-a)^2} = \frac{a}{1-a}\left(b_{n+1} - \frac{b_n}{1-a}\right)$$
Now if $b_{n+1} \le b_n \le \frac{b_n}{1-a}$, we must have that $b_{n+2} \le b_{n+1}$
And so $b_n$ is a decreasing sequence and thus $b_n \le 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
How to prove that an integral doesn't exist? $$\int_{0}^{\infty}\sin^2\left(\pi \left(x + \frac{1}{x} \right) \right) dx $$
Should I use any test for convergence?
| Since $\left|\sin^2(a)-\sin^2(b)\right|=|\sin(a)+\sin(b)|\,|\sin(a)-\sin(b)|\le2|a-b|$, we have
$$
\begin{align}
&\left|\int_a^b\sin^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x
-\int_{a+1/2}^{b+1/2}\cos^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x\right|\\
&=\left|\int_a^b\sin^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x
-\int_a^b\sin^2\left(\pi x+\frac\pi{x+1/2}\right)\,\mathrm{d}x\right|\\
&\le\int_a^b2\left(\frac\pi{x}-\frac\pi{x+1/2}\right)\,\mathrm{d}x\\
&\le\frac\pi{a}\tag{1}
\end{align}
$$
Since $\sin^2(a)+\cos^2(a)=1$, the sum of the integrands is $1$ on $[a+1/2,b]$, therefore,
$$
\begin{align}
&\int_a^b\sin^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x
+\int_{a+1/2}^{b+1/2}\cos^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x\\
&\ge b-a-1/2\tag{2}
\end{align}
$$
The triangle inequality applied to $(1)$ and $(2)$ yields
$$
\int_a^b\sin^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x
\ge\frac12\left(b-a-1/2-\frac\pi{a}\right)\tag{3}
$$
Inequality $(3)$ shows that $\int_0^\infty\sin^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x$ diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/387245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Evaluate the infinite series $\sum_{r=1}^{\infty} \frac{2^{-r}}{r^{2}}$ My teacher posed an infinite series question to me today and I'm not quite sure how to start to go about it.
$$\sum_{r=1}^{\infty} \dfrac{2^{-r}}{r^{2}}$$
Any hints would be much appreciated.
| Claim: $$\color{blue}{\boxed{\displaystyle \sum_{k=1}^{\infty} \dfrac1{2^k k^2} = \dfrac{\pi^2}{12} - \dfrac{\log^2(2)}2}}$$
Proof
What you want to evaluate is the PolyLogarithm function, $\text{Li}_2(x)$, at $x=1/2$. The PolyLogarithm function $\text{Li}_s(x)$ is defined as
$$\text{Li}_s(x) = \sum_{k=1}^{\infty} \dfrac{x^k}{k^s}$$
We are lucky here to evaluate this sum, since in general, we cannot obtain such a nice answer for $\text{Li}_s(x)$. For instance, there is no nice answer to $$\text{Li}_{2}(1/3) = \sum_{k=1}^{\infty} \dfrac1{3^k k^2}$$
Below is the procedure on how to obtain the solution $\displaystyle \sum_{k=1}^{\infty} \dfrac1{2^k k^2} = \dfrac{\pi^2}{12} - \dfrac{\log^2(2)}2$.
From geometric series, we have
$$\sum_{r=1}^{\infty} x^{r-1} = \dfrac1{1-x}$$
Integrating from 0 to $x$, we get that
$$\sum_{r=1}^{\infty} \dfrac{x^r}r = -\log(1-x) \implies \sum_{r=1}^{\infty} \dfrac{x^{r-1}}r = -\dfrac{\log(1-x)}x$$
Again integrating from $0$ to $1/2$, we get that
$$S = \sum_{r=1}^{\infty} \dfrac1{2^rr^2} = - \int_0^{1/2} \dfrac{\log(1-t)}t dt = -\int_1^{1/2} \dfrac{\log(t)}{1-t}(-dt) =-\int_{1/2}^1 \dfrac{\log(t)}{1-t}dt$$
Now we have
\begin{align}
\int_{1/2}^1 \dfrac{\log(t)}{1-t}dt & = \int_{1/2}^1 \sum_{k=0}^{\infty} t^k \log(t) dt = \sum_{k=0}^{\infty} \int_{1/2}^1 t^k \log(t) dt\\
& = \sum_{k=0}^{\infty}\dfrac{-1 + 2^{-(1+k)} + (k+1) \log(2)2^{-(1+k)}}{(1+k)^2}\\
& = - \dfrac{\pi^2}6 + S + \log(2) \cdot \overbrace{\sum_{k=0}^{\infty} \dfrac{2^{-(k+1)}}{k+1}}^{-\log(1-1/2) = \log(2)}
\end{align}
Hence, we get that
$$S = -\left(-\dfrac{\pi^2}6 + S + \log^2(2)\right) \implies 2S = \dfrac{\pi^2}6 - \log^2(2) \implies S = \dfrac{\pi^2}{12} - \dfrac{\log^2(2)}2$$
Hence,
$$\color{red}{\boxed{\displaystyle \sum_{k=1}^{\infty} \dfrac1{2^k k^2} = \dfrac{\pi^2}{12} - \dfrac{\log^2(2)}2}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Explanation of why $\int \sin^2 x\cos^2 x\; dx = 1/3 \sin^3 x - 2/5 \sin^5 x + 1/7 \sin^7 x +c$
verify the solution
$$\int \sin^2 x\cos^5 x\; dx = 1/3 \sin^3 x - 2/5 \sin^5 x + 1/7 \sin^7 x +c$$
I have hit this in my book and can't work it out. Does anyone have any ideas or a walk-through that might help?
| It's easier to start from the right hand side. Derive to get:
$$ (\sin^2 x - 2\sin^4 x + \sin^6 x)\cos x$$
Note that this is just:
$$(\sin x - \sin^3 x)^2\cos x=\sin^2 x(1-\sin^2x)^2\cos x=\sin^2 x\cos^5x$$
So, it seems you are missing a couple of powers of $\cos x$ in the integral..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Examine the continuity of $f(x)=x^2+\frac{x^2}{(1+x^2)}+\frac{x^2}{(1+x^2)^2}+...+ \frac{x^2}{(1+x^2)^n}+....$ at $x=0$
Examine the continuity of $$f(x)=x^2+\frac{x^2}{(1+x^2)}+\frac{x^2}{(1+x^2)^2}+...+ \frac{x^2}{(1+x^2)^n}+....$$ at $x=0$
I tried to solve the problem using $$\lim_{x \to0^+}f(x)=\lim_{x \to0^-}f(x)=f(0)$$ and I got both limits to be $0$. I think there is an error somewhere since my book says it is discontinuous.
Can you help me out?
Thankyou
| If the dots imply an infinite series, then you can first consider
$$
g(x)=\sum_{n\ge0}\frac{1}{(1+x^2)^n}
$$
that converges to
$$
g(x)=\frac{1}{1-\dfrac{1}{1+x^2}}=\frac{1+x^2}{x^2}
$$
whenever
$$
\frac{1}{1+x^2}<1
$$
that is, for all $x\ne0$.
Thus your function is
$$
f(x)=
\begin{cases}
1+x^2 &\text{if $x\ne0$}\\
0&\text{if $x=0$}
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determine the $n$-th power of matrix Determine the $n$-th power of the matrix. $$\pmatrix{2 & 2 & 0 \\ 1 & 2 & 1 \\1 & 2 & 1 \\ }$$ May you help me with the answer since the book states : $${1\over 6}\pmatrix{4+2\cdot4^n & 3\cdot4^n & -4+4^n \\ -2+2\cdot4^n& 3\cdot4^n & 2+4^n\\-2+2\cdot4^n & 3\cdot4^n & 2+4^n \\ }$$ while mine misses the ${1\over 6}$. So who is right?
$$$$Thanks in advance for any help you are able to provide.
| I think about $$\pmatrix{\left(2^{2n-1}-\frac{2}{3}(4^n-1)\right) & 2^{2n-1} & \frac{2}{3}(4^n-1) \\ \left(2^{2n-1}-\frac{2}{3}(4^n-1)+1\right) & 2^{2n-1} & \frac{2}{3}(4^n-1)+1 \\\left(2^{2n-1}-\frac{2}{3}(4^n-1)+1\right) & 2^{2n-1} & \frac{2}{3}(4^n-1)+1 \\ }$$
| {
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"url": "https://math.stackexchange.com/questions/389560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Pell's type equation in sum I have an another observation on concatenation in sum. For instance, take $12^2 + 33^2 = 1233$. Find as many such pairs (x, y) with $x^2 + y^2 = xy\,$(here xy is concatenation)is possible. Also, discuss how this is happening and is there any connection between this kind of problem and pell's equations?
I got $10^2 + 1^2 = 101$;
$0^2 + 1^2 = 01$;
$88^2 + 33^2 = 8833$;
$12^2 + 33^2 = 1233$;
such pairs.
| This is equivalent to $x^2 + y^2 = 10^n x + y$ with size constraints on $y$ (preferably $10^{n-1} \le y < 10^n$, though we will shortly see that the right-hand inequality is superfluous).
Completing the square gives $(2x - 10^n)^2 + (2y - 1)^2 = 10^{2n} + 1$, so we are looking for a way to write $10^{2n}+1$ as the sum of two squares $A^2+B^2$, where $A = 2x-10^n$ is even and $B=2y-1$ is odd. Since clearly $B \le 10^n$, we needn't worry about $y$ being too large, but we'd like $B \ge \tfrac15\cdot10^n - 1$ so that $y$ doesn't need to be zero-padded before concatenation (otherwise we could use the trivial solution $A=10^n$, $B=1$ giving $x=10^n$, $y=1$).
A simple way to construct solutions is to choose $n$ in certain congruence classes so that $10^{2n}+1$ has a particular prime factor that yields a non-trivial decomposition as the sum of two squares. For instance if $n = 16k+12$ we have $10^n \equiv -4 \pmod{17}$ and $17 \mid 10^{2n}+1$, which gives the decomposition
$$10^{2n}+1 = [\tfrac1{17}(15\cdot 10^{n}-8)]^2 + [\tfrac1{17}(8\cdot 10^n +15 )]^2.$$
Here, $B = \tfrac1{17}(8\cdot 10^n + 15)$ so we may take $y = \tfrac4{17}(10^n+4)$. Similarly, $x = \tfrac4{17}(4\cdot 10^n-1)$. Taking $k=0$ this gives
$$ \color{red}{941176470588}^2 + \color{blue}{235294117648}^2 = \color{red}{941176470588}\color{blue}{235294117648}.$$
I don't see a strong connection to Pell equations but it certainly is possible when working with conics like these.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$
Prove that $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$
This should be fairly straightforward but the proof seems to be alluding me.
I want to show $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$. I recognize this shouldn't be too difficult but perhaps finals have fried my brain.
| Recursive integration We know that $$0\le\cos a\le 1\implies \sin t = \int_0^t\cos s ds < t$$ for $0\lt t\lt z\lt x$. Integrating over $\color{blue}{(0,z)}$ we get
$$1-\cos z=\int_0^z\sin tdt < \int_0^ztdt= \frac{z^2}{2}$$
that is for all $0<z<x$ we have,
$$\color{blue}{1-\frac{z^2}{2}< \cos z\le 1}$$
integrating again over $\color{blue}{(0,x)}$ we get
$$\color{red}{x-\frac{x^3}{6} = \int_0^x 1-\frac{z^2}{2} dz< \int_0^x\cos z dz=\sin x}$$
that is
$$\color{blue}{x-\frac{x^3}{6} <\sin x< x}$$
continuing with this process you get,
$$\color{blue}{1-\frac{x^2}{2}< \cos x< 1-\frac{x^2}{2}+\frac{x^4}{24} }$$
$$\color{blue}{x-\frac{x^3}{6} <\sin x< x-\frac{x^3}{6} +\frac{x^5}{5!}}$$
More generally for $n\geq1$, by induction we get
$$\color{blue}{\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k}}{(2k)!}<\cos x<\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k}}{(2k)!}+\frac{x^{4n}}{(4n)!} }$$
$$\color{blue}{\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k+1}}{(2k+1)!} <\sin x<\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k+1}}{(2k+1)!}+\frac{x^{4n+1}}{(4n+1)!}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\sqrt[3]{x+10}-\sqrt[3]{x-10}=2$.
Solve $\sqrt[3]{x+10}-\sqrt[3]{x-10}=2$.
I tried cubing the both sides but things then go very ugly. Are there simpler way to solve it? Thanks.
p.s. The answers are $\pm 6\sqrt 3$.
| Set $x+10=t^3$ and $x-10 = u^3$. So we want to solve $t - u = 2$.
However $t^3 - u^3 = (t-u)^3 + 3tu(t-u) = 20$.
That is $2^3 + 6tu = 8 + 6tu =20$.
Now we have $tu = 2$. This means $x^2 -100 = 8$ and thus $x = \pm6\sqrt{3}.$
| {
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Evaluating $\sum_{n=0}^\infty \frac 1 {(2n+1)^4}$ using Mittag-Leffler's expansion I am trying to evaluate the following series using Mittag-Leffler's expansion theorem. What function would be useful?
$$\sum_{n=0}^\infty \frac 1 {(2n+1)^4} = \frac{\pi^4}{96}$$
I considered differentiating the following relation twice but it did not help. And I managed to prove $0 = 0$ :D
$$\text{tanh}(z) = 2z \sum_{n=0}^\infty \frac{1}{z^2 + ((2n+1)\pi/2)^2}$$
The above expansion gives $\displaystyle \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$. I need something of $4-th$ order. Which function's expansion will give me something like $\displaystyle F(z) = \sum_{n=0}^\infty \frac{1}{(2n+1)^4 + f(z)}$
\begin{align*}
\frac 1z \frac{d}{dz}\left( \frac{\text{tanh}(z)}{z} \right )
&= \frac{z \; \text{sech}(z)^2 -\text{tanh}(z)}{z^3} \\
&= \frac{z (1 - z^2 + O(z^3)) - (z - z^3/3 + O(z^2))}{z^3} \\
&= \frac{-\frac 2 3 z^3 + O(z^5)}{z^3 } \\
&= -2/3 \\
&= \frac 1z \frac{d}{dz}\left( 2 \sum_{n=0}^\infty \frac{1}{z^2 + ((2n+1)\pi/2)^2}\right )\\
&= \frac 1z 2 \sum_{n=0}^\infty - \frac{2z}{(z^2 + ((2n+1)\pi/2)^2)^2} \\
&= - 4\sum_{n=0}^\infty \frac{1}{(z^2 + ((2n+1)\pi/2)^2)^2}
\end{align*}
| You should evaluate:
$$
\lim_{z \to 0} -\frac{(\pi/2)^4}{2 z} \frac{\mathrm{d}}{\mathrm{d}z} \frac{\tanh(z)}{2z}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving two equations involving the greatest common divisor
Show or prove that $$\gcd \left(\frac{a^{2m}-1}{a+1} ,a + 1\right )=\gcd(a + 1 , 2m),$$
and that
$$\gcd \left(\frac{a^{2m + 1}+1}{a+1} , a + 1\right)=\gcd(a + 1 , 2m + 1).$$
| I will show that $$\gcd \left(\frac{a^{2m}-1}{a+1} ,a + 1\right )=\gcd(a + 1 , 2m)$$
First note that $$a^{2m}-1=(a^2)^m-1=(a^2-1)(a^{2(m-1)}+
a^{2(m-2)}+\cdots+1)$$
so that $$\frac{a^{2m}-1}{a+1}=(a-1)(a^{2(m-1)}+\cdots+1)$$
But $a\equiv -1 \mod a+1$ so we have
$$\frac{a^{2m}-1}{a+1}\equiv -2((-1)^{2(m-1)}+\cdots+1)=-2m$$
which means $$\gcd \left(\frac{a^{2m}-1}{a+1} ,a + 1\right )=\gcd(a + 1 , 2m)$$
Can you do something similar with the other?
| {
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Why is this derivative incorrect? We have to find the derivative of $$f(x) = \dfrac{\tan(2x)}{\sin(x)}$$
I would like to know why my approach is incorrect:
$$f'(x) = \dfrac{\sin(x) \cdot \dfrac{2}{\cos^2(2x)} - \tan(2x) \cdot \cos(x)}{\sin^2(x)}$$
$$ = \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)}{\cos^2(2x) \cdot \sin^2(x)}$$
$$ = \dfrac {2 \sin(x) - \sin(2x) \cdot \cos(x)}{\cos^3(2x) \cdot \sin^2(x)}$$
p.s. - To avoid confusion ; I wanted to get rid of the $\tan$. I'm sure there is a shorter method than this but I don't want it; I just want to know why this is wrong.
| $$f'(x) = \dfrac{\sin(x) \cdot \dfrac{2}{\cos^2(2x)} - \tan(2x) \cdot \cos(x)}{\sin^2(x)}\tag{1}$$
So far so good...
But your following line is where you made an algebraic error: you forgot to multiply both (the entire numerator (both terms)) and (the denominator) by $\cos^2(2x)$. Doing this gives us:
$$f'(x) = \dfrac{\color{blue}{\bf \cos^2(2x)}\left(\sin(x) \cdot \dfrac{2}{\cos^2(2x)} - \tan(2x) \cdot \cos(x)\right)}{\color{blue}{\bf \cos^2(2x)}\sin^2(x)}$$
NOW we distribute across the numerator, canceling the denominator of the first term, and multiplying the second term by $\cos^2{2x}$:
$$ = \dfrac{2\sin(x) - \tan(2x) \cdot \cos(x)\cdot \cos^2(2x)}{\cos^2(2x)\sin^2(x)}\tag{2}$$
$$ = \dfrac{2\sin(x) - \sin(2x) \cdot \cos^2(2x)\cdot \cos(x)}{\cos^3(2x)\sin^2(x)}\tag{3}$$
| {
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Evaluate $\int \frac{1+\cos(x)}{\sin^2(x)}\,\operatorname d\!x$ I`m trying to solve this integral and I did the following steps to solve it but don't know how to continue.
$$\int \dfrac{1+\cos(x)}{\sin^2(x)}\,\operatorname d\!x$$
$$\begin{align}\int \dfrac{\operatorname d\!x}{\sin^2(x)}+\int \frac{\cos(x)}{\sin^2(x)}\,\operatorname d\!x &= \int \dfrac{\operatorname d\!x}{\sin^2(x)}+\int \frac{\cos(x)}{1-\cos^2(x)} \\
&=\int \sin^{-2}(x)\,\operatorname d\!x + \int \cos(x)\,\operatorname d\!x - \int \frac{\operatorname d\!x}{\cos(x)}\end{align}$$
Any suggestions how to continue?
Thanks!
| Hints:
$1+\cos x=2 \cos^2 \dfrac{x}{2}$
$\sin^2 x=(2\sin\dfrac{x}{2} \cos \dfrac{x}{2})^2$
You expression will be $\dfrac{1}{2} \int \dfrac{1}{\sin^2\dfrac{x}{2}}$
| {
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What would be the value of $a$ and $b$ in following rational expression? If $(5 + 2\sqrt{3})/(7 + \sqrt{3}) = (a - \sqrt{3b})$,
How do I find the value of $a$ and $b$ where $a$ and $b$ are rational numbers?
| First rewrite $5+2\sqrt{3} = (a-\sqrt{3b})(7+\sqrt{3})$.
Now suppose $b=3n^2$, then we get $5+2\sqrt{3} = (a-3|n|)(7+\sqrt{3})$, which can not be solved for rational $a$ and $n$.
Next, suppose $b=n^2$, then we get $5+2\sqrt{3} = (a-|n|\sqrt{3})(7+\sqrt{3}) = 7a-3|n|+(a-7|n|)\sqrt{3}$ and hence $5=7a-3|n|$ and $2=a-7|n|$. This gives $9=-46|n|$, which has no solutions either for rational $a$ and $n$.
Other $b$ won't work either, since you'll be left with a $\sqrt{b}$. Thus your equation has no rational solutions.
| {
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Matrix Equation, Solving for Variables. I'm going through my exercises, and came across a problem that wasn't covered in our lectures. Here's the question:
$
\begin{align}
\begin{bmatrix}
a-b & b+c\\
3d+c & 2a-4d
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
8 & 1\\
7 & 6
\end{bmatrix}
\end{align}
$
What I have done so far is:
$
\begin{align}
\begin{bmatrix}
a-b-8 & b+c-1\\
3d+c-7 & 2a-4d-6
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix}
\end{align}
$
And the solving for variables,
$$
a-b-8 = 0
$$
$$
a-b = 8
$$
$$
a = \frac{8}{-b}
$$
$$
b+c-1=0
$$
$$
b+c=1
$$
$$
b=\frac{1}{c}
$$
$$
3d+c-7=0
$$
$$
3d+c=7
$$
$$
3d=\frac{7}{c}
$$
$$
d=\frac{7}{3c}
$$
$$
2a-4d-6=0
$$
$$
2a-4d=6
$$
$$
\frac{16}{-2b}-\frac{28}{12c}=6
$$
Am I going about this correctly? Or am I just doing this completely incorrect?
| The best way is follow the solution given by @Alex Wertheim. Let me show you a tedious substitution method to solve this. From $a_{11}$ we have $$b=a-8$$ (dividing by $a$ or $b$ is invalid as we may have zeros). Now from $a_{12}$ entry we have $$c=-b+1=-a+9,$$ and using $a_{21}$ $$d=\dfrac{-c+7}{3}=\dfrac{a-2}{3}.$$ Finally from the last entry $a_{22}$ we have $$a-2d=a-\dfrac23(a-2)=3.$$ Find $a$ from this simple linear equation and substitute back to find $b, c$ and $d.$
| {
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Find the modular residue of this product.. Please help me solve this and please tell me how to do it..
$12345234 \times 23123345 \pmod {31} = $?
edit: please show me how to do it on a calculator not a computer thanks:)
| As $10^1\equiv 10\pmod{31},$
$10^2=100\equiv7,$
$10^3\equiv10\cdot7\equiv8,$
$10^4\equiv49\equiv18,$
$10^5=10^2\cdot10^3\equiv 7\cdot8\equiv25,$
$10^6=(10^3)^2\equiv8^2\equiv2,$
$10^7=10^4\cdot10^3\equiv18\cdot8\equiv20,$
$$12345234=4+3\cdot10+2\cdot10^2+5\cdot10^3+4\cdot10^4+3\cdot10^5+2\cdot10^6+1\cdot10^7$$
$$\equiv4+3\cdot10+2\cdot7+5\cdot8+4\cdot18+3\cdot25+2\cdot2+1\cdot20\pmod{31}$$
| {
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How to solve this simultaneous equation of $3$ variables. I've stuck in this equation system. No clue how to start ?
$$\begin{eqnarray}
x+y+z &=&a+b+c\tag{1} \\
ax+by+cz &=&a^{2}+b^{2}+c^{2}\tag{2} \\
ax^{2}+by^{2}+cz^{2} &=&a^{3}+b^{3}+c^{3}\tag{3}
\end{eqnarray}$$
Find the value of $x,y,z$ is in the form of $a,b$ and $c$.
I want to know steps of solution.
| This is a slightly interesting question. Let me try and provide some motivation for the ugly values that Americo calculated.
Let me assume that $a\neq b, b \neq c, c \neq a $. We'd deal with this issue separately.
Note that the first 2 equations give the equation on a plane. Hence, their intersection is a line, which we can calculate. The first 2 equations are equivalent to
$$ \begin{array} {llll}
&&(x-a) &+ &(y-b) &+ &(z-c) &= 0\\
&a&(x-a) &+b&(y-b) & +c&(z-c) & = 0 \\ \end{array}$$
Since
$$ \begin{pmatrix} 1\\1\\1\\ \end{pmatrix} \times \begin{pmatrix} a\\b\\c\\ \end{pmatrix} = \begin{pmatrix} c-b\\a-c\\b-a\\ \end{pmatrix}, $$
we know that the line is given by $\frac{x-a}{c-b} = \frac{ y-b}{a-c} = \frac{z-c}{b-a} = k $. (This is valid because the denominator is never 0.) This is equivalent to $x=a + k(c-b), y = b+k(a-c) , z = c+k(b-a)$.
Next, we are interested in when this line intersects the ellipsoid $ax^2 + by^2 + cz^2 = a^3 + b^3 + z^3 $, which happens in at most 2 places. (Of course, one place is $k=0$, which leads to the solution $x=a, y=b, z=c$.) Using the equation of the line above, we get that
$$\begin{array}{l l l l}
ax^2 &= a^3 & + 2a^2k(c-b) &+k^2 a(c-b)^2 \\
by^2 &= b^3 & + 2b^2k(a-c) & + k^2 b(a-c)^2 \\
cz^2 &= c^3 & + 2c^2k(b-a) & + k^2 c(b-a)^2 \\
\end{array} $$
Summing up these three equations, and using the last equation given, we get that
$$-2k(a-b)(b-c)(c-a) + k^2 (a^2b+a^2c+b^2a+b^2c+c^2a+c^2b - 6abc) = 0. $$
One solution is $k=0$ as mentioned, and the other is
$$ k = \frac{ 2 (a-b)(b-c)(c-a) } { a^2b+a^2c+b^2a+b^2c+c^2a+c^2b - 6abc} .$$
With this value of $k$, we get the answer that Americo calculated above.
Now, what happens when $a=b$ or $b=c$ or $c=a$? Interestingly, this forces there to be only 1 solution, namely $k=0$. You can also get this result by substituting $a=b$ into the value of $k$ and showing that the numerator is 0. The geometric interpretation is that the line becomes tangential to the ellipsoid.
Similarly, if $a=b=c$, then there is only 1 solution.
| {
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Let $k \geq 3$; prove $2^k$ can be written as $(2m+1)^2+7(2n+1)^2$ Prove:
If $k \geq 3$, then $2^k$ can be written as $(2m+1)^2+7(2n+1)^2$, where $k, m, n \in \mathbb{N}$.
| You need to be careful about just one thing in the induction step. Note that the first case is $$ 1^2 + 7 \cdot 1^2 = 8 = 2^3. $$
The important thing is that
$$ 1 \equiv 1 \pmod 4. $$
So, we begin with $$ a^2 + 7 b^2 = 2^k $$
with $k \geq 3$ AND
$$ a \equiv b \pmod 4, $$ then we do the induction step.
What the formula of Brahmagupta does for you is to allow us to multiply the value $2^k$ by 8, as in
$$ (a - 7 b)^2 + 7 (a+b)^2 = 8 \cdot 2^k = 2^{k+3}. $$
At this stage, both numbers $a-7b,a+b$ are even, which is bad. On the other hand,
$$ a - 7 b \equiv a+b \pmod 8. $$ As a result, when we divide by 2, we get
$$ \frac{ a - 7 b}{2} \equiv \frac{a+b}{2} \pmod 4. $$
Furthermore, as $ a \equiv b \pmod 4, $ we know that $ a + b \equiv 2 \pmod 4, $ so that $\frac{a+b}{2} $ is once again ODD.
We have completed the induction step, as
$$ \left( \frac{ a - 7 b}{2} \right)^2 + 7 \left( \frac{a+b}{2} \right)^2 = 2 \cdot 2^k = 2^{k+1}, $$ with both numbers odd and the extra property we need to continue the induction,$$ \frac{ a - 7 b}{2} \equiv \frac{a+b}{2} \pmod 4. $$
It may be worth mentioning that we need to allow negative values for $(a,b)$ in order to keep the mod 4 thing going. So the pairs go
$$ (1,1), \; \; (-3,1), \; \; (-5,-1), \; \; (1,-3), \; \; (11,-1), \; \; (9,5), $$
$$ (-13,7), \; \; (-31,-3), \; \; (-5,-17), \; \; (57,-11), \; \; (67,23), \; \; (-47,45), $$
$$ (-181,-1), \; \; (-87,-91), \; \; (275,-89), \; \; (449,93), \; \; (-101,271), \; \; (-999,85), \ldots $$
| {
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Is there a simpler approach to these system of equations? I recently came across the following system of equations:
$$x + y + z = 1 \\
x^2 + y^2 + z^2 = 2 \\
x^3 + y ^3 + z^3 = 3$$
And I have two questions:
One, is there a way to prove or disprove whether there is a solution for this particular set of equations? Furthermore, is there a way to expand the proof for a more generalized set of equations, that is for this set :
$$x + y + z = 1 \\ x^2 + y^2 + z^2 = 2 \\ ...\\x^n + y^n + z^n = n$$
Two, is there a simpler approach for the prior solution set than substitution? As of now, I'm not getting anywhere with this method. I end up getting into a long-winded series of substitution and isolation that yields something like $z = z$ or $1 = 1$.
Sorry if it's a repeat question; I couldn't exactly find an accurate way to search for the equations.
| We will make this into a cubic equation in terms of $z$. We have $x+y = 1-z$ and $x^2 + y^2 = 2-z^2$.
Factor: $$x^3 + y^3 = ( x + y ) ( x^2 - xy + y^2 ) = ( 1-z ) ( 2-z^2 - xy )$$
$$2xy = ( x + y )^2 - ( x^2 + y^2 ) = ( 1-z )^2 - ( 2 - z^2 ) = 2z^2 - 2z - 1$$
So our third equation becomes (multiplying by $2$):
$$( 1 - z ) ( -4z^2 + 2z + 5 ) + 2z^3 = 6$$
Multiplying both sides by $2$ and expanding the polynomial,
$$6z^3 - 6z^2 - 3z -1 = 0$$
This will have $3$ roots (in the complex numbers), which will be solutions for $x$, $y$, and $z$ (by symmetry).
| {
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what values of b_2 result in solving this system? Solve the system of equations:
$$ \begin{pmatrix}
0 & 1 & 0 & 3 \\
0 & 2 & 0 & 6 \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3\\ x_4 \end{pmatrix}
= \begin{pmatrix} b_1 \\ b_2\end{pmatrix}$$
What value of $b_2$ results in this system having a solution?
I did my row reduction and came up with:
$$\begin{pmatrix}
0 & 1 & 0 & 3 & b_1\\
0 & 0 & 0 & 0 & 2b_1 - b_2\\
\end{pmatrix} $$
therefore, $2b_1=b_2$
Is it really that simple?
Any help is greatly appreciated!
| Your answer is correct and your reasoning is good (although I think most people would come up with $b_2-2b_1$ rather than $2b_1-b_2$ in the reduced row echelon form). However, in this particular case, there is a conceptually simpler solution (I emphasize "particular" because in general, your way of solving the problem is preferred). Note that
$$
\underbrace{\begin{pmatrix}0&1&0&3\\ 0&2&0&6\end{pmatrix}}_{A}
\begin{pmatrix}x_1 \\ x_2 \\ x_3\\ x_4\end{pmatrix}
=x_1\begin{pmatrix}0\\ 0\end{pmatrix}
+x_2\begin{pmatrix}1\\ 2\end{pmatrix}
+x_3\begin{pmatrix}0\\ 0\end{pmatrix}
+x_4\begin{pmatrix}3\\ 6\end{pmatrix}
=(x_2+3x_4)\begin{pmatrix}1\\ 2\end{pmatrix}.
$$
Therefore the system of linear equations is solvable if and only if $(b_1,b_2)^T$ is a scalar multiple of $(1,2)^T$, i.e. iff $b_2=2b_1$.
Using the language of linear algebra, $A\mathbf{x}=\mathbf{b}$ is solvable iff $\mathbf{b}$ lies in the column space of $A$. Now the column space of $A$ is the line spanned by $(1,2)^T$. Hence the result.
| {
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Finding the rational values of constant for which these constants are roots of equation Problem :
Determine all rational values for which $a,b,c$ are the roots of $x^3+ax^2+bx+c=0$
Solution :
Sum of the roots $a+b+c = -a$ ........(i) ( Since , as per question $a,b,c$ are roots of equation and we have to find values )
$\sum_{a,b,c} ab = b ........(ii)$
$abc= -c \Rightarrow ab = -1 \Rightarrow a =\frac{-1}{b} .....(iii)$
Solving (i),(ii) and (iii) simultaneously we get an equation which is biquadratic in "a" which is :
$2a^4-2a^2-a+1=0$ Here we can find that $a = 1$ is a factor of this equation and by long division of polynomial we can divide $a-1$ with the given biquadratic we get cubic polynomial in a which is $2a^3+2a^2-1$ ...
Please guide further and suggest whether this is correct or not.. Thanks...
| Neither $-1$ nor $1$ is a solution to your final cubic, so this calls for the use of the cubic formula.
$$a = 2, b = 2, c = 0, d = -1$$
$$\Delta_0 = 2^2 - 3(2)(0) = 4$$
$$\Delta_1 = 2(2^3) - 9(2)(2)(0) + 27(2^2)(-1) = 16 - 108 = -92$$
$$C = \sqrt[3]{\frac{-92 + \sqrt{(-92)^2 - 4(4^3)}}{2}} = \sqrt[3]{\frac{-92 + \sqrt{8208}}{2}} = \sqrt[3]{\frac{12\sqrt{57} - 92}{2}} = \sqrt[3]{6\sqrt{57} - 46}$$
We can divide $8208$ by $-27a^2$ (or $-108$) to get the equation's discriminant, $-76$. This means the equation has one real root and two complex roots. That real root is given by:
$$-\frac{1}{6}\left(2 + \sqrt[3]{6\sqrt{57} - 46} + \frac{4}{\sqrt[3]{6\sqrt{57} - 46}}\right)$$
A number which is definitely not rational. So you have found the only rational root, 1.
| {
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} |
Find non-singular matrices P and Q such that PAQ is in the normal form for the matrix A. $A= \left[ \begin{array}{ccc}
1 & 2 & 3 & -2 \\
2 & -2 & 1 & 3 \\
3 & 0 & 4 & 1 \end{array} \right]$
$A=IAI$
$\left[ \begin{array}{ccc}
1 & 2 & 3 & -2 \\
2 & -2 & 1 & 3 \\
3 & 0 & 4 & 1 \end{array} \right]$ = $\left[ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right] A \left[ \begin{array}{ccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right]$
| If I understand what you mean by normal form here, we would do the following series of operations ($R$ is row, $C$ is Column, and note the row operations are duplicated on the RHS $3x3$ and Column Operations are duplicated on the $4x4$, while performing the operations listed below on the LHS, that is, the matrix $A$ itself):
*
*$R2: R2 - 2R1$
*$R3: R3 - 3R1$
*$R3: R3 - R2$
*$C4: C4 + C2$
*$C3: C3 + 5 C4$
*$C2: C2 + 6C4$
*$C2: C2 - 2C1$
*$C3: C3 - 3C1$
*$ C2 \leftrightarrow C4$ (that is, swap these two columns)
So, we start with:
$\left[ \begin{array}{ccc}
1 & 2 & 3 & -2 \\
2 & -2 & 1 & 3 \\
3 & 0 & 4 & 1 \end{array} \right]$ = $\left[ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right] A \left[ \begin{array}{ccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right]$
and after performing the operations outlined above, end up with:
$\left[ \begin{array}{ccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 \end{array} \right]$ = $\left[ \begin{array}{ccc}
1 & 0 & 0 \\
-2 & 1 & 0 \\
-1 & -1 & 1 \end{array} \right] A \left[ \begin{array}{ccc}
1 & 0 & -3 & -2 \\
0 & 1 & 5 & 7 \\
0 & 0 & 1 & 0 \\
0 & 1 & 5 & 6 \end{array} \right]$
Some observations:
*
*$P$ and $Q$ are nonsingular (full rank) matrices
*$P$ and $Q$ are NOT unique. A different sequence of operations to obtain the normal form gives different $P$ and $Q$ (you should verify this by finding a different sequence).
*$A = P^{-1} \cdot \begin{bmatrix}I_r & 0\\0 & 0\end{bmatrix} \cdot Q^{-1}$, where $I_r$ is the $I_2$ matrix listed in the final form of the LHS reductions above.
*If $A$ is a nonsingular matrix, $A = P^{-1}Q^{-1}$ (In this example, rank $A = 2$, so this does not apply).
| {
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} |
Stupid question about $1 - \frac{1}{2}-\frac{1}{4}+\frac{1}{3}- \frac{1}{6}-\frac{1}{8}+\frac{1}{5}\dots$ I have
$$1 - \frac{1}{2}-\frac{1}{4}+\frac{1}{3}- \frac{1}{6}-\frac{1}{8}+\frac{1}{5}\dots$$
Partial sum $S_{3n}$ of the above is:
$$(1 - \frac{1}{2}-\frac{1}{4})+(\frac{1}{3}- \frac{1}{6}-\frac{1}{8})+(\frac{1}{5}-\dots$$
But what is $S_{3n-1}$ and $S_{3n-2}$ ?
| By guessing the pattern hidden in the ellipsis, it seems that you consider the series
$\sum_{k=1}^\infty a_k $
where $a_{3n-2}=\frac1{2n-1}$, $a_{3n-1}=-\frac1{4n-2}$, $a_{3n}=-\frac1{4n}$ for $n\ge 1$.
Thus $a_{3n-2}+a_{3n-1}+a_{3n}=\frac1{2n-1}-\frac1{4n-2}-\frac1{4n}=\frac1{4n(2n-1)}$ and we have the partial sums
$$S_{3n}=\sum_{k=1}^{3n} a_k =\sum_{k=1}^n\frac1{4n(2n-1)}$$
$$S_{3n-1}=\sum_{k=1}^{3n-1} a_k =\sum_{k=1}^n\frac1{4n(2n-1)}+\frac1{4n}$$
$$S_{3n-2}=\sum_{k=1}^{3n-2} a_k =\sum_{k=1}^n\frac1{4n(2n-1)}+\frac1{4n}+\frac1{4n-2}$$
| {
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Prove inequality $(x+y+z-2xyz)^2 \le 2$
Problem: Prove inequality $(x+y+z-2xyz)^2 \le 2\ (1)$ with
$x^2+y^2+z^2 = 1 \land x,y,z \in \mathbb R$
I tried expand $LHS$ and have:
$$(1)\iff 1 - 2 (xy+yz+xz) + 4 xyz(x+y+z)-(2xyz)^2 \ge 0$$
Denote: $xy = a, yz = b,xz=c \implies (1) \iff1-2\sum a+ 4 \sum ab - 2abc \ge0$
But stuck. Please help me.
| we let $x\le y\le z$,then $$z^2\ge\dfrac{1}{3},2xy\le x^2+y^2\le\dfrac{2}{3}$$
use Cauchy-Schwarz inequality we have
$$(x+y+z-2xyz)^2=[(x+y)+z(1-2xy)]^2\le [(x+y)^2+z^2][1+(1-2xy)^2]$$
we only prove
$$[(x+y)^2+z^2][1+(1-2xy)^2]\le 2$$
since:
$$ [(x+y)^2+z^2][1+(1-2xy)^2]=(1+2xy)[1+(1-2xy)^2]=2+4x^2y^2(2xy-1)\le 2$$
By done!
| {
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How do I find the exact value of $\cos\frac{\pi}{12}\cos\frac{5\pi}{12}\cos\frac{7\pi}{12}\cos\frac{11\pi}{12}$? I know that $\cos(6\phi)\equiv32c^6-48c^4+18c^2-1$ where $c=\cos\phi$.
I also know that when $\cos(6\phi)=0$, then $\phi=\frac{k\pi}{12}$ ($k = 1,3,5,7,9,11$).
How do I find the exact value of:
$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \cos\left(\frac{11\pi}{12}\right)$$
| Mosquito-nuking linear algebraic solution: it can be shown that the tridiagonal matrix
$$\mathbf T=\frac12\begin{pmatrix}-1&1&&&&\\1&0&1&&&\\&1&0&1&&\\&&1&0&1&\\&&&1&0&1\\&&&&1&1\end{pmatrix}$$
has the eigenvalues $\cos\dfrac{(2k-1)\pi}{12},\quad k=1,\dots,6$. Since the product of the eigenvalues is equal to the determinant of the matrix,
$$\cos\frac{\pi}{12}\cos\frac{5\pi}{12}\cos\frac{7\pi}{12}\cos\frac{11\pi}{12}=\frac{\det\mathbf T}{\cos\tfrac{\pi}{4}\cos\tfrac{3\pi}{4}}=\frac{-\tfrac1{32}}{-\tfrac12}=\frac1{16}$$
| {
"language": "en",
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} |
Correlation Coefficient - $\rho(X,Y)$. If I have two aleatory variables
$$X=\begin{pmatrix}2&3&4&5&6&7&8&9&10&11&12\\ \frac{1}{36}&\frac{2}{36}&\frac{3}{36}&\frac{4}{36}&\frac{5}{36}&\frac{6}{36}&\frac{5}{36}&\frac{4}{36}&\frac{3}{36}&\frac{2}{36}&\frac{1}{36}\end{pmatrix}$$ and $$Y=\begin{pmatrix}2&3&4&5&6&7&8&9&10&11&12\\ \frac{1}{36}&\frac{2}{36}&\frac{3}{36}&\frac{4}{36}&\frac{5}{36}&\frac{6}{36}&\frac{5}{36}&\frac{4}{36}&\frac{3}{36}&\frac{2}{36}&\frac{1}{36}\end{pmatrix}$$ how can I calculate $$\rho(X,Y) \mbox{?}$$
$\rho$ -correlation coefficeint.
thanks:)
| You cannot find the correlation without additional information. $X$ and $Y$ could be independent, in which case the correlation is $0$, or (in this instance) $X$ might always be equal to $Y$ so the correlation would be $1$, or $X$ might (again, in this instance) be equal to $14-Y$ (as $X$ goes from $2$ to $12$, then $14-X$ goes from $12$ to $2$), in which case the correlation is $-1$, or they could be related in more complicated ways, in which case the correlation could have any of a large (but in this case finite) number of possible other values between $1$ and $-1$.
Since the number of possible values of the correlation is in this case finite, one could say that there's enough information given to deduce at least something about the correlation.
| {
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Find volume using double integrals? Question: Use double integral to find the volume of the solid enclosed by the spheres $x^2+y^2+z^2=1$ and $x^2+y^2+(z-1)^2=1$
Alright so I tried to doing this by myself and I'm not sure if this is right. Could someone check over my work?
Curve of intersection:
\begin{align*}
x^2 + y^2 + z^2 &= x^2 + y^2 + (z - 1)^2\\
\implies z^2 &= z^2 - 2z + 1\\
\implies z &= 1/2.
\end{align*}
So, the curve of intersection is $x^2 + y^2 = 3/4$ with $z = 1/2$.
By symmetry, it suffices to double the volume between $z = 1/2$ and $z = 1$.
$x^2 + y^2 + (z-1)^2 = 1$ is above $x^2 + y^2 + z^2 = 1$ when $z > 1/2$.
Solving for (positive) $z$ yields
$z = \sqrt{1 - x^2 - y^2}$ and $z = 1 + \sqrt{1 - x^2 - y^2}$.
Hence, the volume equals
\begin{align*}
V &= 2 \iint \left[(1 + \sqrt{1 - x^2 - y^2}) - \sqrt{1 - x^2 - y^2}\right]\, dA\\ \\
&= 2 \int_0^{2\pi}\!\int_0^{\sqrt{3/4}}\!r\,dr \,d\theta, \qquad\textrm{via polar coordinates}\\ \\
&= 2 \int_0^{2\pi} \!\left[(1/2)r^2\right]_{r = 0}^{\sqrt{3/4}}\, d\theta\\ \\
&= \int_0^{2\pi}\!3/4\,d\theta\\ \\
&= 3\pi/2.
\end{align*}
| By the word "enclosed", I am assuming you want to find the volume bounded by these two spheres, which means the solid formed by their intersection. You claimed:
$x^2 + y^2 + (z-1)^2 = 1$ is above $x^2 + y^2 + z^2 = 1$ when $z > 1/2$.
If we are truely like to find the volume of the intersection, then this is not accurate. First sphere is centered at $(0,0,1)$ with radius $1$, second has origin as its center with the same radius. If you imagine a viewer sitting on the $x$-axis ($y=0$), this viewer will see the $x^2 + y^2 + (z-1)^2 = 1$'s bottom lies beneath the top of $x^2 + y^2 + z^2 = 1$, the integral should be set as:
$$
V = \int^{\sqrt{3}/2}_{-\sqrt{3}/2} \int^{\sqrt{3/4 - y^2}}_{-\sqrt{3/4 - y^2}}
\int^{\sqrt{1 - x^2 - y^2}}_{1- \sqrt{1 - x^2 - y^2}} 1\,dz\,dx\,dy.
$$
Changing to cylindrical coordinates:
$$
V = \int^{2\pi}_{0} \int^{\sqrt{3}/2}_{0}
\int^{\sqrt{1 - \rho^2}}_{1- \sqrt{1 - \rho^2}} \rho\,dz\,d\rho\,d\theta.
$$
Integrate above (for you to try):
$$
V = \frac{5\pi}{12},
$$
which coincides with the spherical cap formula. (EDIT: computation error earlier, now fixed)
| {
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How many compositions of n are there with same parity Let n be a non-negative integer. How many compositions of n are there where the i-th part has the same parity as i? For example, compositions of 7 that satisfy this condition are (7), (5,2), (3,4), (1,6), (1,2,1,2,1). Express your answer as the coefficient of a simplified rational expression.
can someone help me please :(
| Hint: The number of solutions where there is exactly 1 part is the coefficient of $x^n$ in $x + x^3 + x^5 + \ldots = \frac{x}{1-x^2}$.
Hint: The number of solutions where there are exactly 2 parts is the coefficients of $x^n$ in $(x+x^3+x^5 + \ldots ) \times (x^2 + x^4 + x^6 + \ldots ) = \frac{ x } { 1-x^2} \times \frac{x^2} { 1-x^2} $.
Hence, the number of solutions is (summing over all possible number of parts) is the coefficient of $x^n$ in
$$ \sum_{i=1}^\infty \left(\frac{x}{1-x^2} + \frac{ x } { 1-x^2} \times \frac{x^2} { 1-x^2} \right) \left(\frac{ x } { 1-x^2} \times \frac{x^2} { 1-x^2} \right)^i $$
Simplify this further.
| {
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Similarity and Scale Factor If I have two triangles ABC and ADE and I know that each of the triangles is equilateral, each of the angles is 60 degrees. Is there a way to determine the scale factor of the two triangles, for example:
The ratio:
Area of ADE $:$Area of ABC
Thanks in advance.
| If side $DC$$ = $ $x$, then $AC$ must be $2x$ because both triangles are equilateral. By Pythagorean theorem, $AD$ is $3^\frac{1}{2}x$. Then $AD:AC=3^\frac{1}{2}:2=\frac{3^\frac{1}{2}{}}{2}:1$. So the scale factor $k$, is $\frac{3^\frac{1}{2}}{2}$. Using the formula:
$\frac{\text{Area of }AED}{\text{Area of }ABC}=k^2$
$\frac{\text{Area of }AED}{\text{Area of }ABC}=(\frac{3^\frac{1}{2}}{2})^2$
$\frac{\text{Area of }AED}{\text{Area of }ABC}=\frac{3}{4}$
Therefore $k^2= \frac{3}{4}$ and so the ratio between triangle $ADE$ and $ABC$ is $1:\frac{3}{4}$
| {
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simplifying $\min(\max(A,B),C) $ In a larger problem, I have to make use of the following
$$\min(\max(A,\ B),\ C)$$
Please how do I simplify?
| For $a, b ∈ \Bbb R:$
$$\max\left(a,b\right) = \frac{a+b+\left|a-b\right|}{2}$$
$$\min\left(a,b\right) = \frac{a+b-\left|a-b\right|}{2}$$
Therefore, if $a,b,c ∈ \Bbb R:$
$$\min\left(\max\left(a,b\right),c\right) =
\min\left(\frac{a+b+\left| a-b\right|}{2},c\right) = $$
$$=\frac{\frac{a+b+\left| a-b\right|}{2}+c-\left|\frac{a+b+\left|a-b\right|}{2}-c\right|}{2}=
\frac{a+b+\lvert a-b\rvert}{4}+\frac{c}{2}-\left|\frac{a+b+|a-b|}{4}-\frac{c}{2}\right|=$$
$$=\frac{a+b+\left|a-b\right|+2c}{4}-\left|\frac{a+b+|a-b|-2c}{4}\right|=$$
$$=\frac{a+b+\left|a-b\right|+2c-\Big|a+b+\left|a-b\right|-2c\Big|}{4}$$
| {
"language": "en",
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Calculating $\sum^{10}_{k=1}\left(\sin\frac{2k\pi}{11}+i\cos\frac{2k\pi}{11}\right)$
Find the value of $$\sum^{10}_{k=1}\left (\sin\left (\frac{2k\pi}{11} \right )+i\cos\left (\frac{2k\pi}{11}\right ) \right)$$
My approach:
Since $\cos\theta + i\sin\theta = e^{i\theta}$, we can write the given equation as:
$$\begin{align*}
&i \left \{\sum^{10}_{k=1} \left (\cos\frac{2k\pi}{11} -i\sin\frac{2k\pi}{11} \right ) \right \}\\
= &i \left \{\sum^{10}_{k=1}\left (e^{-i\frac{2k\pi}{11}} \right ) \right \} \tag{i}
\end{align*}$$
Solving the index part only which is
$$\begin{align*}
-i\frac{2k\pi}{11} &= -i\frac{2\pi}{11}(1+2+3+\cdots+10) \quad (\text{putting the values of } k)\\
&= -i\frac{2\pi}{11}( 55) \quad \left(\text{By applying sum of first $n$ natural numbers} = \frac{n(n+1)}{2}\right )\\
&=-i10\pi
\end{align*}$$
Putting this value in $(\text{i})$ we get:
$e^{-i10\pi} = i\cos10\pi = i.$
But the answer is $-i$. Please suggest where I went wrong… Thanks..
| I am not sure where your steps came from. You are right that you are summing
$$i \sum_{k=1}^{10} e^{-i 2 \pi k/11}$$
This is a geometric series and has value
$$i\frac{e^{-i 2 \pi/11}-e^{-i 2 \pi}}{1-e^{-i 2 \pi/11}} = i\frac{e^{-i 2 \pi/11}-1}{1-e^{-i 2 \pi/11}}= -i$$
| {
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approximate $\int_{u=0}^b e^{-\frac{u^2 + ac}{au}}du.$ I'm trying to find an approximation (or exact solution if possible) for an integral of the form:
$$\int_{u=0}^b e^{-\frac{u^2 + ac}{au}}du.$$
I was thinking of somehow applying a Gauss Hermite Quadriture Expansion, but I'm not sure how I would do this. Does anyone know the best way I should go about this?
| I do not know much about numerical integration, but the following trick may help you avoid exponents diverge ad infinitum.
We are dealing with the integral
$$ I = \int_{0}^{b} \exp\left( - \frac{u}{a} - \frac{c}{u} \right) \, du. $$
With the substitution $u = \sqrt{a c} x^{2}$, for $\beta = \left( b / \sqrt{ac} \right)^{1/2}$ we have
\begin{align*}
I
&= 2\sqrt{ac} \int_{0}^{\beta} x \exp\left\{ - \sqrt{\frac{c}{a}} \left( x^{2} + x^{-2} \right) \right\} \, dx \\
&= 2\sqrt{ac} e^{-2\sqrt{c/a}} \int_{0}^{\beta} x \exp\left\{ - \sqrt{\frac{c}{a}} \left( x - x^{-1} \right)^{2} \right\} \, dx.
\end{align*}
With the substitution $y = x - x^{-1}$ and
$$ \alpha := \frac{\beta + \sqrt{\beta^{2} + 4}}{2} = \frac{\sqrt{b} + \sqrt{b + 4\sqrt{ac}}}{2\sqrt[4]{ac}}, \qquad \gamma := \sqrt{\frac{c}{a}}, $$
we find that
\begin{align*}
I &= 2a\gamma e^{-2\gamma} \int_{-\infty}^{\alpha} \frac{\left(\sqrt{y^{2}+4}+y\right)^{2}}{4 \sqrt{y^{2}+4}} e^{-\gamma\,y^{2}} \, dy.
\end{align*}
Now with aid of Mathematica, we can check that
$$\int_{-\infty}^{0} \frac{\left(\sqrt{y^{2}+4}+y\right)^{2}}{4 \sqrt{y^{2}+4}} e^{-\gamma y^{2}} \, dy = \frac{1}{2} e^{2 \gamma} K_1(2 \gamma)-\frac{1}{4 \gamma}, $$
where $K_{n}(z)$ is the modified Bessel function of the second kind. This allows us to write
\begin{align*}
I &= a\gamma e^{-2\gamma} \left[ e^{2\gamma} K_1\left(2\gamma \right)-\frac{1}{2\gamma} + 2 \int_{0}^{\alpha} \frac{\left(\sqrt{y^{2}+4}+y\right)^{2}}{4 \sqrt{y^{2}+4}} e^{-\gamma\,y^{2}} \, dy \right],
\end{align*}
with each term being much tractable.
| {
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why is this Markov Chain aperiodic I have this Matrix:
$$P=\begin{pmatrix} 0 & 1 \\ 0.3 & 0.7 \end{pmatrix}$$
this markov chain is said to be aperiodic, I dont understand how it comes to it. Period $\delta$ is the gcd of the set of all diagonal elements, right? if $\delta>1$, $P$ is periodic, if $\delta=1$, then aperiodic.
but here it is not $\delta=1$, is it? or do i have to transit the matrix to some certain form?
| Since the Jordan Normal Form of the matrix is
$$
\begin{bmatrix}
0&1\\
0.3&0.7
\end{bmatrix}
=\begin{bmatrix}
-10&3\\3&3
\end{bmatrix}
\begin{bmatrix}
-0.3&0\\0&1
\end{bmatrix}
\begin{bmatrix}
-10&3\\3&3
\end{bmatrix}^{-1}
$$
we have
$$
\begin{align}
\begin{bmatrix}
0&1\\
0.3&0.7
\end{bmatrix}^n
&=\begin{bmatrix}
-10&3\\3&3
\end{bmatrix}
\begin{bmatrix}
-0.3&0\\0&1
\end{bmatrix}^n
\begin{bmatrix}
-10&3\\3&3
\end{bmatrix}^{-1}\\
&\to
\begin{bmatrix}
-10&3\\3&3
\end{bmatrix}
\begin{bmatrix}
0&0\\0&1
\end{bmatrix}
\begin{bmatrix}
-10&3\\3&3
\end{bmatrix}^{-1}\\
&=\frac1{13}\begin{bmatrix}
3&10\\3&10
\end{bmatrix}
\end{align}
$$
Thus, the initial state $\begin{bmatrix}x\\y\end{bmatrix}$ tends toward the fixed point state
$$
\left(\frac3{13}x+\frac{10}{13}y\right)\begin{bmatrix}1\\1\end{bmatrix}
$$
| {
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How to prove $(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$ without calculations I read somewhere that I can prove this identity below with abstract algebra in a simpler and faster way without any calculations, is that true or am I wrong?
$$(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$$
Thanks
| Notice that this is invarient if the same constant is added to all three variables, so one of them may be set equal to zero. Set $c=0$. Then the expression becomes $(a-b)^3 -a^3 +b^3+3ab(a-b)$, easily seen to be zero.
BTW, apparently the original expression is equivalent to saying $x^3 +y^3+z^3 =3xyz$ if $x+y+z=0$.
EDIT:Yes, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$
| {
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Prove $\frac{1}{2 \pi} \int_0^{2\pi} \frac{1 - r^2}{1 + r^2 - 2r \cos{t}} dt = 1$ using contour integration The question is to solve the integral using concepts of contour integrals:
$$\frac{1}{2 \pi} \int_0^{2\pi} \frac{1 - r^2}{1 + r^2 - 2r \cos{t}} dt = 1$$
| Let $$I(r) = \dfrac{1}{2 \pi} \int_0^{2\pi} \frac{1 - r^2}{1 + r^2 - 2r \cos{t}} dt \tag{$\spadesuit$}$$
Splitting the integral from $0$ to $\pi$ and $\pi$ to $2\pi$, changing $t \to \pi-t$, in the second integral, we get
$$I(r) = \dfrac{1}{\pi} \int_0^{\pi} \frac{1 - r^2}{1 + r^2 - 2r \cos{t}} dt = \dfrac{1}{\pi} \int_0^{\pi} \frac{1 - r^2}{1 + r^2 + 2r \cos{t}} dt$$
Replacing $t$ by $2t$ in $(\spadesuit)$, we get that
$$I(r) = \dfrac1{\pi} \int_0^{\pi} \frac{1 - r^2}{1 + r^2 - 2r \cos{2t}} dt = \dfrac1{\pi} \int_0^{\pi} \dfrac{1 - r^2}{(1+r)^2-4r \cos^2(t)} dt$$
\begin{align}
I(r^2) & = \dfrac1{\pi} \int_0^{\pi} \dfrac{1 - r^4}{(1+r^2)^2-4r^2 \cos^2(t)} dt\\
& = \dfrac1{2\pi}\left(\int_0^{\pi} \dfrac{1 - r^2}{1+r^2-2r \cos(t)} dt + \int_0^{\pi} \dfrac{1 - r^2}{1+r^2+2r \cos(t)} dt\right)\\
& = I(r)
\end{align}
Now note that $I(r)$ is continuous for $r \in [0,1)$ and $r \in (1,\infty)$. Now note that
$$I(0) = 1 \text{ and } \lim_{r \to \infty} I(r) = -1$$
Since $I(r) = I(r^2)$, we have $I(r) = \cdots I(r^{2^n})$. For $r\in [0,1)$, we have
$$I(r) = \lim_{n \to \infty} I(r^{2^n}) = I(\lim_{n \to \infty} r^{2^n}) = I(0) = 1$$
For $r\in (1,\infty)$, we have
$$I(r) = \lim_{n \to \infty} I(r^{2^n}) = I(\lim_{n \to \infty} r^{2^n}) = -1$$
Hence,
$$I(r) = \begin{cases} 1 & \text{ if }r\in[0,1)\\ 0 & \text{ if } r = 1\\ -1 & \text{ if }r \in (1,\infty)\end{cases}$$
| {
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"url": "https://math.stackexchange.com/questions/414832",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Is my proof correct? $2^n=x^2+23$ has an infinite number of (integer) solutions. This is how I tried to prove it. Is it correct? Thanks!!
$2^n = x^2+23$
$x^2$ must be odd, therefore $x^2 = 4k+1$, where $k \in \mathbb{N}$.
$2^n=4k+24$
$k=2(2^{n-3}-3)$
Since $x^2=4k+1$,$ \ \ \ \ $ $k_1 = \frac{x_1^2-1}{4}$
and $k_2=\frac{(x_1+2)^2-1}{4}=\frac{x_1^2+4x_1+3}{4}$
If we substitute $x_1^2=4k_1 + 1$, we end up with:
$k_2=k_1 + \sqrt{4k_1+1} + 1$
Therefore, finding solutions of $k=2(2^{n-3} - 3)$ is comparable to
finding the solutions of $k+\sqrt{4k+1}+1=2(2^{n-3}-3)$
Let $p=2(2^{n-3}-3)$
Therefore, $(\sqrt{4k+1})^2=(p-k-1)^2$, so
$(p-k)^2 = 2(p+k)$
Since $p$ has an infinite number of solutions $(p-k)^2=2(p+k)$ also has an infinite number of solutions, which implies the the original does also.
| There are only finitely many solutions. Pillai's conjecture is overkill because it allows for all possible exponents simultaneously, whereas the solutions of $2^n = x^2 + 23$ must lie on one of the three curves $y^3 = x^2+23$, $2y^3 = x^2+23$, $4y^3= x^2+23$, each of which has finitely many integer points.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to integrate $\frac{x+2}{x^2+2x+2}$ with the substitution method This is what I got so far:
$$\frac{x+2}{x^2+2x+2} = \frac{x+2}{(x+1)^2+1} = \int \frac{x}{(x+1)^2+1} + 2 \times \int \frac{1}{(x+1)^2+1}$$
I know the last integral $= \arctan(x+1) + c$, where $c$ is a constant, but I don't know how to integrate the first one, i.e $\int \frac{x}{(x+1)^2+1}$.
Thanks for your help.
EDIT: MY SOLUTION
I split the fraction: $\frac{x+1}{x^2+2x+2} + \frac{1}{x^2+2x+2}$
$$\int \frac{x+1}{(x+1)^2+1}\;dx \implies (x+1)^2+1 = u$$
$$\implies \int\frac{(x+1)}{u} \text{ and } \frac{du}{2(x+1)}$$
$$=\frac{1}{2}\int \frac{1}{u}\,du$$
$$=\frac{1}{2}\ln(u)$$
$$=\frac{1}{2}\ln(x^2+2x+2)$$
so finally the last integral $\int \frac{1}{x^2+2x+2}\,dx = \int \frac{1}{(x+1)^2+1}\,dx = \arctan(x+1)$
and now the solution = $\frac{1}{2}\ln(x^2+2x+2) + \arctan(x+1) + C$
| HINT:
$$\int\frac{x+2}{x^2+2x+2}dx=\int\frac{x+2}{(x+1)^2+1^2}dx$$
Put $x+1=\tan \theta$
Alternatively, put $x+2= A\frac{d(x^2+2x+2)}{dx}+B$
i.e., $x+2=A\cdot2(x+1)+B=2Ax+2A+B$
Comparing the coefficients of $x,1=2A\implies A=\frac12$
Comparing the constants, $2A+B=2\implies B=2-2A=1$
So, $$\int\frac{x+2}{x^2+2x+2}dx$$
$$=\frac12\int \frac{d(x^2+2x+2)}{dx}\cdot\frac1{x^2+2x+2} dx+\int\frac{dx}{x^2+2x+2}$$
$$=\frac12\int \frac{d(x^2+2x+2)}{x^2+2x+2}+\int \frac{dx}{(x+1)^2+1^2}$$
Can you take it from here?
| {
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The system of equations $x^2 + y^2 - x - 2y = 0$ and $x + 2y = c$ I have
$(1.) \quad x^2 + y^2 - x - 2y = 0 \\
(2.) \quad x + 2y = c$
Solving for $y$ in $(2.)$ gives
$y = (c - x) / 2$
Is there a way to simplify equation $(1.)$?
Because at the end I arrive at
$c^2 - 2x - c = 0$
and can't proceed. Need to get typical form of square equation $Ax^2 + Bx + C = 0$.
The solution for $c$ is $0$ or $5$.
EDIT 1:
For what real values of the parameter do the common solutions of the following pairs of simultaneous equations became identical?
(g) $ \quad x^2 + y^2 - x - 2y = 0, \quad x + 2y = c$, Ans. c = 0 or 5.
The process is to solve for $y$, then to substitute that into second equation.
From that we get A, B and C. Delta being $B^2 - 4AC$ we can get parameter.
So
$y = (c - x) / 2$
$x^2 + y^2 - x -2y = 0$
$x^2 + ((c-x)/2)^2 - x - 2 * ((c-x)/2) = 0$
and i got lost...
| If you substitute correctly, you will get the equation $5\,{x}^{2}-2\,c\,x+{c}^{2}-4\,c=0$.
If you solve this, you will get $x = \frac{1}{5}(c \pm 2\sqrt{c(5-c)})$. The other value is given by $y=\frac{1}{2}(c-x)$, but you don't need this to answer your question.
If the two sets of solutions have the same values, then we must have $c(5-c)=0$, hence $c=0$ or $c=5$.
Addendum: To get the above equation, replace $y$ in $x^2+y^2-x-2y=0$ by $y = \frac{1}{2}(c-x)$ (from the second equation). That is, the equation
\begin{eqnarray}
x^2+\frac{1}{4}(c-x)^2 -x -(c-x) &=& \frac{1}{4}(4x^2+(c-x)^2 -4c) \\
&=& \frac{1}{4}(5 x^2-2 c x +c^2-4c)
\end{eqnarray}
Then the solutions are (ignoring the $\frac{1}{4}$, of course):
\begin{eqnarray}
x &=& \frac{1}{10}\left(2c \pm \sqrt{4 c^2-20(c^2-4c)} \right) \\
&=& \frac{1}{10}\left(2c \pm \sqrt{16 c(5-c)} \right) \\
&=& \frac{1}{5}\left(c \pm 2\sqrt{ c(5-c)} \right)
\end{eqnarray}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How find $a$ such that $x^2-\sqrt{a-x}=a$ has exactly two real solutions Consider the equation
$$ x^2-\sqrt{a-x}=a.$$ I wish to determine the values of $a$ for which the above equation has exactly two real solutions (for $x$).
My idea:
$$a-x=(x^2-a)^2=x^4-2ax^2+a^2\Longrightarrow f(x)=x^4-2ax^2+x+a^2-a=0$$ and we must have
$$a-x\ge 0$$
$$x^2= a+\sqrt{a-x}\ge a,$$
so $f(x)=x^4-2ax^2+x+a^2-a=0$ has only real solution $x$, and this solution $x^2\ge a, x\le a.$ But can I use this to find the possible values of $a$?
then we have $f''(x)=0 ?$
$$\Longrightarrow f'(x)=4x^3-4ax+1,\Longrightarrow f''(x)=12x^2-4a=0$$
then $$12x^2=4a\Longrightarrow a=3x^2>0$$
so $$ a^2\le \sqrt{\dfrac{a}{3}}\le a$$
$$\Longrightarrow \dfrac{1}{3}\le a\le \sqrt[3]{\dfrac{1}{3}}$$
my reslut is true?and I think this problem have other nice methods.
I hope someone can write the final results，because I don't know the correct result.Thank you everyone
| Another idea is:
Equation is equivalent to the system
1) $\sqrt{a-x}=y$
2) $\sqrt{a+y}=|x|$
with conditions
3) $x\leq a$ and $y\geq0$, $y \geq -a$.
Are obtained by squaring
1') $ a-x = y^{2}$
2') $ a+y = x^{2}$.
Subtracting the last equation is obtained:
$(x+y)(1-x+y)=0$ si cazurile $y= -x$ si $y =x-1$.
Replace $ y $ in 2 ') etc...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to determine the rank and determinant of $A$? let $A$ be $$A_{a} = \begin{pmatrix} a & 1 & 1 & 1 \\ 1 & a & 1 & 1\\ 1 & 1 & a & 1\\ 1 & 1 & 1 & a \end{pmatrix}$$
How can I calculate the rank of $A$ by the Gauss' methode and $\det A$?
| $\det A_{a} =\det \begin{pmatrix} a & 1 & 1 & 1 \\ 1 & a & 1 & 1\\ 1 & 1 & a & 1\\ 1 & 1 & 1 & a \end{pmatrix}$
Adding all the columns to the 1st column we have,
$\det A_{a} = \det \begin{pmatrix} a+1+1+1 & 1 & 1 & 1 \\ a+1+1+1 & a & 1 & 1\\ a+1+1+1 & 1 & a & 1\\ a+1+1+1 & 1 & 1 & a \end{pmatrix}$
Taking $a+1+1+1$ common from row 1 we have,
$\det A_{a} = (a+1+1+1)\det \begin{pmatrix}1 & 1 & 1 & 1 \\ 1 & a & 1 & 1\\ 1 & 1 & a & 1\\ 1 & 1 & 1 & a \end{pmatrix}$
Now subtract the 1st row from the 2nd row 1st from the third and so on to get
$\det A_{a} = (a+1+1+1)\det \begin{pmatrix}1 & 1 & 1 & 1 \\ 0 & a & 1 & 1\\ 0 & 1 & a & 1\\ 0 & 1 & 1 & a \end{pmatrix}$
Now it will be easy.
After finding determinant of A find out the values of A for which the determinant is $0$.For those values of A find the rank of the matrix by noting the number of independent row of $A$ otherwise the rank is $4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$, given $x+y+z=3$ and $x,y,z\ge0$
Let $x+y+z=3,x,y,z\ge 0$,show that
$$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$$
Additional information
I have seen the following problem:
$x,y,z>0,x+y+z=3$, prove that
$$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge 6.$$
Without loss of generality we can let $x=\max{\{x,y,z\}}$
Proof: case 1
$x\ge y\ge z$
we can easily prove
$$\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge\sqrt{y^2+x+2}+\sqrt{z^2+z+2}$$
and
$$\sqrt{x^2+y+2}+\sqrt{y^2+x+2}\ge\sqrt{x^2+x+2}+\sqrt{y^2+y+2}$$
so we have
$$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge \sqrt{x^2+x+2}+\sqrt{y^2+y+2}+\sqrt{z^2+z+2}.$$
Then use
$$\sqrt{x^2+x+2}\ge\dfrac{3}{4}x+\dfrac{5}{4}$$
$$\sqrt{y^2+y+2}\ge\dfrac{3}{4}y+\dfrac{5}{4}$$
$$\sqrt{z^2+z+2}\ge\dfrac{3}{4}z+\dfrac{5}{4}$$
to get the result. Whereas the case 2 when $x\ge z\ge y$ can be proved using the same methods.
Now,I have another idea: using Holder inequality
we have
$$\left(\sum\sqrt{x^2+yz+2}\right)^2\left(\sum\dfrac{x^2+2yz+9}{x^2+yz+2}\right)\ge 36^3$$
$$\Longleftrightarrow \sum\dfrac{x^2+2yz+9}{x^2+yz+2}\le 1296$$
and the following link has some discussion about this problem
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=538230
and
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=538752&p=3097872#p3097872
and Vasc gave the hint:
$$\sum\sqrt{8(a^2+bc+2)}\ge \sum\sqrt{(3a+b+c)^2+7}\ge 12\sqrt 2$$
How prove this hint?Thank you everyone.
and my other idea is as follows:
let $a=\min(a,b,c)$
we can prove
$$\sqrt{b^2+ca+2}+\sqrt{c^2+ab+2}\geq \sqrt{(b+c)^2+2a(b+c)+8-(b-c)^2}\tag{1}$$
\begin{align*}
&\sqrt{a^2+bc+2}+\sqrt{(b+c)^2+2a(b+c)+8-(b-c)^2}\\
\geq &\sqrt{a^2+\frac{(b+c)^2}{4}+2}+\sqrt{(b+c)^2+2a(b+c)+8} \tag{2}
\end{align*}
Summing up
\begin{align*}
&\sum_{cyc}{\sqrt{a^2+bc+2}}\\
\geq &\sqrt{a^2+\frac{(b+c)^2}{4}+2}+\sqrt{(b+c)^2+2a(b+c)+8}\\
=&\sqrt{a^2+\frac{(3-a)^2}{4}+2}+\sqrt{(3-a)^2+2a(3-a)+8}
\end{align*}
By the way: someone said $(1)$ is wrong? why? can anyone give an example? And hopefully someone can use this method to prove this inequality? Thank you very much!
| Here is a sketch of another alternative approach.
Substitute variables $x=a+1,y=b+1,z=c+1$; then $$f(a,b,c)=\sqrt{4+a^2+bc+a}+\sqrt{4+b^2+ca+b}+\sqrt{4+c^2+ab+c}$$ — I exploited the fact that $a+b+c=x+y+z-3=0$; also note, that $-1\le a,b,c \le 2$
Use Taylor formula $\sqrt{4+t}=2\sqrt{1+\frac{t}{4}}=2+\frac{t}{4}-\frac{t^2}{16}+R(t)$ to get Taylor expansion for $f$ up to second degree; I got $$64f=64\cdot6+16(a+b+c)+15(a^2+b^2+c^2)+16(ab+bc+ca)+R(a,b,c)=64\cdot6+7(a^2+b^2+c^2)+R(a,b,c)$$ — I again exploit that $(a+b+c)^2=0$
So, it is local minimum; after getting some bound for $R(a,b,c)$ one may get a value for $\varepsilon$ to conclude that for any $a,b,c$ that $|a|<\varepsilon,|b|<\varepsilon,|c|<\varepsilon$ we have $f(a,b,c)\ge 6$
Next, we remember that $a^2+bc+a=x^2+yz\ge0,...$, so $$|f'_a(a,b,c)|\le |\frac{2a+1}{2\sqrt{2}}+\frac{c}{2\sqrt{2}}+\frac{b}{2\sqrt{2}}|\le\frac{9}{2\sqrt{2}}$$ and so $|f'_b(a,b,c)|\le \frac{9}{2\sqrt{2}}$, $|f'_c(a,b,c)|\le \frac{9}{2\sqrt{2}}$
Last step is just brute force calculation of values of $f$ in network of points $a=\frac{k}{N}$, $b=\frac{l}{N}$, $c=-\frac{k+l}{N}$ (except such values that lies in $|a|<\varepsilon,|b|<\varepsilon,|c|<\varepsilon$ where $f\approx 6$) where $N$ sufficiently large and $-N\le (k,l,-k-l)\le 2N$; after such calculation one may observe that $f(\frac{k}{N}, \frac{l}{N}, -\frac{k+l}{N})>6+\frac{3\cdot 9}{2\sqrt{2}N}$ so there is no possibility that $f=6$, because $f(a,b,c)\ge f(\frac{k}{N}, \frac{l}{N}, -\frac{k+l}{N})-\frac{1}{N}\frac{9}{2\sqrt{2}}(|a-\frac{k}{N}|+|b-\frac{l}{N}|+|c-\frac{-k-l}{N}|)>6$
If current $N$ is not enough to observe that $f(\frac{k}{N}, \frac{l}{N}, -\frac{k+l}{N})>6+\frac{3\cdot 9}{2\sqrt{2}N}$, then one have to do brute force calculation with, say $N'=2N$; this will lead to success very soon, because $f''$ is small. I suspect that such calculation may be limited to <100 values, if one get good estimates for $R$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
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Determinant of symmetric matrix Given the following matrix, is there a way to compute the determinant other than using laplace till there're $3\times3$ determinants?
\begin{pmatrix}
2 & 1 &1 &1&1 \\
1 & 2 & 1& 1 &1\\
1& 1 & 2 & 1 &1\\
1&1 &1 &2&1\\
1&1&1&1&-2
\end{pmatrix}
| You can substract the first row from every other rows and get matrix of form:
$$\begin{pmatrix}
2 & 1 &1 &1&1 \\
-1 & 1 & 0& 0 &0\\
-1& 0 & 1 & 0 &0\\
-1&0 &0 &1&0\\
-1&0&0&0&-3
\end{pmatrix}.$$
Computing the determinant is now much easier.
| {
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Evaluating $\lim_{x\to1}\frac{\sqrt{x^2+3}-2}{x^2-1}$? I tried to calculate, but couldn't get out of this:
$$\lim_{x\to1}\frac{x^2+5}{x^2 (\sqrt{x^2 +3}+2)-\sqrt{x^2 +3}}$$
then multiply by the conjugate.
$$\lim_{x\to1}\frac{\sqrt{x^2 +3}-2}{x^2 -1}$$
Thanks!
| You were right to multiply "top" and "bottom" by the conjugate of the numerator. I suspect you simply made a few algebra mistakes that got you stuck with the limit you first posted:
So we start from the beginning:
$$\lim_{x\to1}\frac{\sqrt{x^2 +3}-2}{x^2 -1}$$
and multiply top and bottom by the conjugate, $\;\sqrt{x^2 + 3} + 2$:
$$\lim_{x \to1} \, \frac{\sqrt{x^2 + 3} -2 }{x^2 - 1} \cdot \frac{\sqrt{x ^ 2 + 3} +2}{\sqrt{x^2+3}+2}$$
You were correct to do that. You just miss-calculated and didn't actually need to expand the denominator, that's all.
In the numerator, we have a difference of squares (which is the reason we multiplied top and bottom by the conjugate), and expanding the factors gives us: $$(\sqrt{x^2+3}-2)(\sqrt{x^2 + 3} + 2) = (\sqrt{x^2+3})^2 - (2)^2 = (x^2 + 3) - 4 = x^2 - 1$$ And now there's no reason to waste time trying to simplify the denominator*, since we can now cancel the factor $(x^2 - 1)$ from top and bottom:
$$\lim_{x\to1}\frac{\color{blue}{\bf (x^2- 1)}}{\color{blue}{\bf (x^2 - 1)}(\sqrt{x^2 +3}+2)}\; = \;\lim_{x \to 1} \dfrac{1}{\sqrt{x^2 + 3}+2}\; = \;\frac{1}{\sqrt{1+3} + 2} \;=\; \dfrac 14$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the generating function for this set of strings Let $a(n)$ be the number of $\{0,1\}$-strings of length $n$ which contain no $4$ consecutive $1$'s and no $4$ consecutive $0$'s (don't contain "$0000$" or "$1111$"). Find the generating function for this.
Thank you.
| Let $a(n)$ denote the number of admissible binary strings of length $n$. First, note that precisely half of the strings of any length end in $\color{red}{0}$ (or $\color{green}{1}$). Now, for a string of length $n>4$, there are $8$ possible endings with the following numbers of occurrences:
$$\begin{align*}
&\overbrace{\ldots???\color{green}{1}}^{\text{length }n-3}\color{red}{0}\color{red}{0}\color{red}{0}\quad\tfrac{1}{2}a(n-3)\\\
&\ldots????\color{red}{0}\color{red}{0}\color{green}{1}\quad\tfrac{1}{2}a(n-4)+\tfrac{1}{2}a(n-3)\\
&\ldots????\color{red}{0}\color{green}{1}\color{red}{0}\quad\tfrac{1}{2}a(n-2)\\
&\ldots????\color{red}{0}\color{green}{1}\color{green}{1}\quad\tfrac{1}{2}a(n-2)\\
&\ldots????\color{green}{1}\color{red}{0}\color{red}{0}\quad\tfrac{1}{2}a(n-2)\\
&\ldots????\color{green}{1}\color{red}{0}\color{green}{1}\quad\tfrac{1}{2}a(n-2)\\
&\ldots????\color{green}{1}\color{green}{1}\color{red}{0}\quad\tfrac{1}{2}a(n-4)+\tfrac{1}{2}a(n-3)\\
&\underbrace{\ldots???\color{red}{0}}_{\text{length }n-3}\color{green}{1}\color{green}{1}\color{green}{1}\quad\tfrac{1}{2}a(n-3)
\end{align*}$$
Summing up, we obtain the following recursion formula:
$$a(n)=2a(n-2)+2a(n-3)+a(n-4)$$
Finally, we determine $a(n)$ from the initial values
$$a(1)=2, a(2)=4, a(3)=8, a(4)=14.$$
Reassuring note:
Substituting $a(n-1)=a(n-2)+a(n-3)+a(n-4)$ in $\textbf{Calvin Lin}$'s recurrence relation $a(n)=a(n-1)+a(n-2)+a(n-3)$, we obtain agreement.
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Find all $n$ such that $\sigma(n)=12$ Let $ \sigma (n) = \sum_{k|n}^{}{k} $. I need to solve $\sigma(n)=12$.
Probably the following might be of use: if $n={p_1}^{a_1}{p_2}^{a_2}...{p_s}^{a_s}$ then $\sigma(n)=\frac{{p_1}^{a_1+1}-1}{p_1-1}\frac{{p_2}^{a_2+1}-1}{p_2-1}...\frac{{p_s}^{a_s+1}-1}{p_s-1}$.
| If RHS $=12,$ as $1+p+p^2=7$ or $\ge 1+3+3^3=13,n$ must be square-free
So, $\sigma(n)=\prod(1+p_i)$ where $p_i$s are distinct prime divisors $(\ge2)$ of $n$
Also, $\sigma(n)\ge 1+n \ \ \ \ (1)$
The equality occurs if $n$ is prime i.e, here $n+1=12\implies n=11$ which is prime.
$(1)\implies n\le 11$
So, other the values of $n$ must be non-prime & square-free
As $2\le n\le 11,$ the values can be $6,10$
$\sigma(6)=\sigma(2\cdot3)=(1+2)(1+3)=12$
$\sigma(10)=\sigma(2\cdot5)=(1+2)(1+5)=18$
Alternatively,
As $n$ is square-free, the values of $1+p_i$s are $3,4,6,12$ as $\ge2$
If $1+p_1=3,1+p_2=4\implies n=p_1\cdot p_2=2\cdot 3=6$
If $1+p_1=6,1+p_2=2\implies p_2=1<2$ so this case does not arise
If $1+p_1=12, n=p_1=11$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An identity involving the Beta function I'm trying to show that
$$ \int _0^1 \frac{x^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}}dx = \frac{B(a,b)}{(1+c)^ac^b}$$
Where $$B(a,b) := \int _0^1 x^{a-1}(1-x)^{b-1}dx $$ is the "Beta function". I am supposed to use a substitution but I'm pretty much stuck. I am familiar with the basic properties of the Beta function, its relation to the gamma function etc. Any hints or advice you care to offer would be super cool.
| Let $\displaystyle \int _0^1 \frac{x^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}}dx = I$
We have ,
$(1+c)^ac^bI=\displaystyle \int _0^1 \frac{(1+c)^ac^bx^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}}dx $
$=\displaystyle \int _0^1 \frac{(1+c)c((1+c)x)^{a-1}(c(1-x))^{b-1}}{(x+c)^{a+b}}dx $
$=\displaystyle \int _0^1 \left(\frac{(1+c)c}{(x+c)^2}\right)\left(\frac{(1+c)x}{(x+c)}\right)^{a-1}\left(\frac{c(1-x)}{(x+c)}\right)^{b-1}dx$
Let $\displaystyle y=\left(\frac{c(1-x)}{(x+c)}\right)$ and we have $\displaystyle dy=-\left(\frac{(1+c)c}{(x+c)^2}\right)dx$
Then we have the above $=-\displaystyle\int _1^0y^{b-1}(1-y)^{a-1}dy=\int_0^1y^{b-1}(1-y)^{a-1}dy=B(b,a)=B(a,b)$
So ultimately we have,
$\displaystyle(1+c)^ac^bI=B(a,b)\Rightarrow I=\frac{B(a,b)}{(1+c)^ac^b}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/420024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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About rationalizing expressions For example, rationalizing expressions like
$$\frac{1}{\pm \sqrt{a} \pm \sqrt{b}}$$
Is straightforward. Moreover cases like
$$\frac{1}{\pm \sqrt{a} \pm \sqrt{b} \pm \sqrt{c}}$$
and
$$\frac{1}{\pm \sqrt{a} \pm \sqrt{b} \pm \sqrt{c} \pm \sqrt{d}}$$
Are still easy to rationalize. But my question is in the more general case
$$\frac{1}{\pm \sqrt{a_1} \pm \sqrt{a_2} \cdots \pm \sqrt{a_n}}$$
Where $n \ge 5$
Are they always rationalizable? If so, how would be an algorithm to rationalize them. If not, then a proof must exist.
From my point of view, I can't find an obvious way to rationalize the case $n=5$, since grouping the radicals in a group of 3 and a group of 2 radicals and then applying the identity
$$(a-b)*(a+b)=a^2-b^2$$
Just modifies the denominator from
$$\pm \sqrt{a} \pm \sqrt{b} \pm \sqrt{c} \pm \sqrt{d} \pm \sqrt{e}$$
to
$$\pm v \pm \sqrt{w} \pm \sqrt{x} \pm \sqrt{y} \pm \sqrt{z}$$
Will this help in something? Or a different method or identity is needed?
| Yes, it's always possible to rationalize such expressions.
Without loss of generality, it's enough to consider the first square root having positive sign in front of it. Then, see what happens if you multiply together terms with all possible combinations of the remaining signs:
$$\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b$$
$$\begin{eqnarray}
\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)
\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)\\
\left(\sqrt{a}-\sqrt{b}+\sqrt{c}\right)
\left(\sqrt{a}-\sqrt{b}-\sqrt{c}\right) & = & a^2 + b^2 + c^2 - 2ab - 2ac - 2bc \\
\end{eqnarray}$$
The resulting expressions get progressively more complicated (the resulting polynomial will be homogeneous, symmetric polynomial of degree $2^{n-2}$ -- it'll contain all possible combinations of the variables with all possible combinations of exponents), but it won't contain any of the square roots anymore.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Are these rational curves? I have to find the singular points of the following curves and tell if they are rational. The curves are $C=Z(x^2+y^2+x^2y^2)$ and $C=Z(x^3+y^3-1)$, and the base field is the complex one.
I think I found the singular points but I'm stuck in finding the rational morphism. Can some one help me?
| A plane affine curve is rational iff its projectivization is rational, so let's study these projectivizations since we have more tools at our disposal in the projective plane.
2) The projectivization of the second curve is the curve $x^3+y^3-z^3=0$.
This is a smooth curve of genus $\frac {(3-1)(3-2)}{2}=1$, so it is not rational since rational curves have genus $0$ .
1) The projectivization of the first curve is the curve $C$ with equation $x^2z^2+y^2z^2+x^2y^2=0$, which has three nodal singularities at $(1:0:0),(0:1:0),(0:0:1)$.
The birational transformation of the plane (traditionally "quadratic transformation") $x=\frac {1}{X}, y=\frac {1}{Y}, z=\frac {1}{Z}$ transforms (after multiplying by $X^2Y^2Z^2$) the curve $C$ into the conic $C'$ of equation $Y^2+X^2+Z^2=0$.
So the curve $C$ is indeed rational, since the conic $C'$ is rational .
Another argument for the rationality of $C$ is that its geometric genus is $\frac {(4-1)(4-2)}{2}-3\cdot 1=0$ and a curve of geometric genus $0$ is rational (the subtracted term in the formula comes from the the three nodal singularities of the curve).
| {
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"url": "https://math.stackexchange.com/questions/422815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Derivation of Pythagorean Triple General Solution Starting Point: I was reading on proof wiki about the derivation of the general solution to the pythagorean triple diophantine equation:
$$
x^2 + y^2 = z^2,
$$
where $x,y,z > 0$ are integers.
I came across the following general solution to the primitive function:
\begin{align*}
x &= 2mn\\
y &= (m^2 - n^2)\\
z &= (m^2 + n^2)\\
\end{align*}
for coprime $m,n$.
I looked at the proof of it working (if you square $x$ and $y$ and add it it does indeed equal $z^2$)
My one qualm was, how the hell did they start with that? For example, is there a natural way by starting with the original problem that you end up with the expression above?
I noticed that when attempting to derive the general solution myself, from start to finish,
I would begin by noting I can find all pairs such that $z$ and $y$ differ by a constant $k$... but I cannot make that final leap to end up with the equation above so that for any given $k$ you can find a solution.
Thanks ahead of time!
| Here is the way to generate all relatively prime pythagorean triples:
Theorem: Let $m$ and $n$ be positive integers so that
$$
\begin{align}
&m\gt n\tag{1}\\
&m+n\text{ is odd}\tag{2}\\
&m\text{ and }n\text{ are relatively prime}\tag{3}
\end{align}
$$
Then,
$$
\begin{align}
a &= m^2 - n^2\\
b &= 2mn\\
c &= m^2 + n^2
\end{align}\tag{4}
$$
gives all positive, relatively prime $a$, $b$, and $c$ so that
$$
a^2 + b^2 = c^2\tag{5}
$$
Proof: $(5)\Rightarrow(4):$
Suppose $a$, $b$, and $c$ are positive, relatively prime, and $a^2 + b^2 = c^2$.
Because $(2k)^2 = 4k^2$ and $(2k+1)^2 = 4(k+1)k + 1$, the square of an even
integer must be $0 \bmod{4}$ and the square of an odd integer must be $1 \bmod{4}$.
At least one of $a$ and $b$ must be odd; otherwise $a$, $b$, and $c$ would share a
common factor of $2$. If both are odd, then $c^2$ would need to be $2 \bmod{4}$,
which is impossible. Thus, one must be even and the other must be odd. This means that $c$ must be odd. Without loss of generality, let $b$ be even.
Let $M = (c+a)/2$ and $N = (c-a)/2$. Then
$$
\begin{align}
a &= M - N\tag{6}\\
c &= M + N\tag{7}\\
b^2 &= 4MN\tag{8}
\end{align}
$$
Thus, we have that $M \gt N \gt 0$ and one of $M$ and $N$ must be even and the othermust be odd. Furthermore, $\gcd(M,N)$ divides $a$, $b$, and $c$; thus, $\gcd(M,N) = 1$. Since $b^2 = 4MN$ and $\gcd(M,N) = 1$, both $M$ and $N$ must be perfect squares. Let $M = m^2$ and $N = n^2$, where $m$ and $n$ are positive; then, $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied.
$(4)\Rightarrow(5):$
Suppose $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied. Then $(5)$ is satisfied:
$$
\begin{align}
a^2 + b^2
&= (m^2 - n^2)^2 + (2mn)^2\\
&= m^4 - 2 m^2 n^2 + n^4 + 4 m^2 n^2\\
&= m^4 + 2 m^2 n^2 + n^4\\
&= (m^2 + n^2)^2\\
&= c^2\tag{9}
\end{align}
$$
Furthermore, $a$ and $b$ are relatively prime since
$$
\begin{align}
\gcd(a,b)
&= \gcd(m^2-n^2,2mn)\\
&\:\mid\:\gcd(m-n,2) \gcd(m-n,m) \gcd(m-n,n)\\
&\times\gcd(m+n,2) \gcd(m+n,m) \gcd(m+n,n)\\
&=\gcd(m+n,2)^2 \gcd(n,m)^4\\
&= 1\tag{10}
\end{align}\\
$$
$\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/423196",
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"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
How to determine whether this series convergent or divergent? Does
$$
\dfrac{7}{19}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}+\cdots+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}\cdots\sqrt[n]{\dfrac{7}{19}}+\cdots
$$
converge or diverge?
The following is my idea:
use
$1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}>\ln{n}$
$$
\sum_{n=1}^{\infty}\left(\dfrac{7}{19}\right)^{(1+1/2+\cdots+1/n)}
< \sum_{n=1}^{\infty}\left(\dfrac{7}{19}\right)^{\ln{n}}
= \sum_{n=1}^{\infty}n^{-\ln(19/7)}
$$
But $p=\ln{\dfrac{19}{7}}<1$, becasue $\dfrac{19}{7}\approx 2.71428<e=2.71828$
I guess the series is divergent because I use
$$1+1/2+\cdots+1/n\approx \ln{n}, n\to\infty$$
to find $$\sum_{n=1}^{\infty} (1/x)^{1+1/2+1/3+\cdots+1/n}$$ is convergent only if $x>e$.
So, my question is: how do I determine whether
$$
\dfrac{7}{19}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}+\cdots+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}\cdots\sqrt[n]{\dfrac{7}{19}}+\cdots
$$
converges or diverges? Thank you
| Since
$$
H_n = \sum_{k=1}^{n} \frac{1}{k} = \log n + \gamma + o(1),
$$
we have, for $x > 0$,
$$
x^{H_n} \sim x^{\log n + \gamma} = x^\gamma n^{\log x}
$$
As $n \to \infty$. We may then apply the limit comparison test to conclude that
$$
\sum_{n=1}^{\infty} \frac{1}{x^{H_n}}
$$
converges when $\log x > 1 \Leftrightarrow x > e$ and diverges when $\log x \leq 1 \Leftrightarrow x \leq e$.
Since $19/7 < e$ the series in your question diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How can evaluate $\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$ I don't know if I apply for this case sin (a-b), or if it is the case of another type of resolution, someone with some idea without using derivation or L'Hôpital's rule? Thank you.
$$\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$$
| Using $\sin\left(a\right)-\sin\left(b\right)=2\sin\left(\frac{a-b}{2}\right)\cos\left(\frac{a+b}{2}\right)$ and $\cos\left(a+b\right)=\cos\left(a\right)\cos\left(b\right)-\sin\left(a\right)\sin\left(b\right)$: $$L=\lim_{x\rightarrow 0}\frac{2\sin\left(x^2/2\right)\cos\left(x^{-1}+x^2/2\right)}{x}\\=\lim_{x\rightarrow 0}\frac{2\sin\left(x^2/2\right)\left(\cos\left(x^2/2\right)\cos\left(x^{-1}\right)-\sin\left(x^2/2\right)\sin\left(x^{-1}\right)\right)}{x}\\=\lim_{x\rightarrow 0}\frac{\sin\left(x^2\right)\cos\left(x^{-1}\right)-2\sin^2\left(x^2/2\right)\sin\left(x^{-1}\right)}{x}\\=\lim_{x\rightarrow 0}x\cos\left(1/x\right)-\lim_{x\rightarrow 0}\frac{x^3\sin\left(1/x\right)}{2}=0.$$
| {
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"source": "stackexchange",
"question_score": "5",
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Prove that $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b} \ge a+b+c$ If $a,b,c$ are positive , show that $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge a+b+c$$
Trial: Here I proceed in this way $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge \dfrac{2bc}{b+c}+\dfrac{2ca}{c+a}+\dfrac{2ab}{a+b}$$ then how I proceed. Please help.
| It suffices to show, by symmetry, that
$$\frac{a^2+b^2}{a+b}\ge \frac{a+b}{2}.$$
This is equivalent to
$$2a^2+2b^2\ge a^2+b^2+2ab\iff a^2+b^2\ge 2ab.$$
The last inequality is just the standard AM-GM inequality.
| {
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 2
} |
Linear Independence by row space.
Let $(1,1,-1), (2,1,0)$ and $(-1,0,1)$ be vectors, show if they are independent.
I wrote each vector on the rows of the matrix $A$.
$A=\begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & 0\\ -1 & 0 & 1 \end{pmatrix}$
Then I put $A$ on an echelon form.
$A=\begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & 0\\ -1 & 0 & 1 \end{pmatrix}\overset{L2-2L1} {\rightarrow}\begin{pmatrix} 1 & 1 & -1 \\ 0 & -1 & 2\\ -1 & 0 & 1 \end{pmatrix}\overset{L3+L1} {\rightarrow}\begin{pmatrix} 1 & 1 & -1 \\ 0 & -1 & 2\\ 0 & 1 & 0 \end{pmatrix}\overset{L3+L2}{\rightarrow}\begin{pmatrix} 1 & 1 & -1 \\ 0 & -1 & 2\\ 0 & 0 & 2 \end{pmatrix}$
So, $r(A)=3$. If $\space r(A)=3$ that is equal to the number of rows$(n)$ of $A$, then I can conclued that the $3$ vectors are independent.
If $r(A)<n$ , then the set of the $3$ vectors would be dependent. Is this right? Thanks.
| Yes, this is correct.
You could also just compute the determinant of $A$ and conclude that the rows (and columns too) are linear independent.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
A question on differentiation : Let $$f(x)=\sin^{-1}(2x\sqrt{1-x^2})$$
I found out $f'(x)$ in three methods and got three different answers !
1) Putting $x=\cos\theta$, we get $f(x)=2\cos^{-1}x$, on differentiating this we get
$$f'(x)=\frac{-2}{\sqrt{1-x^2}}$$
2) Putting $x=\sin\theta$, we get $f(x)=2\sin^{-1}x$, on differentiating we get
$$f'(x)=\frac{2}{\sqrt{1-x^2}}$$
This is contradicting with the previous result that we arrived at by substituting $x=\cos\theta$. Can someone help me out with the correct solution for this problem!
| One should, before starting differentiating, check where the expression is defined. The function $f$ is defined where $-1\le 2x\sqrt{1-x^2}\le1$, that is,
$$
4x^2(1-x^2)\le 1
$$
which is satisfied for all $x$. Thus the function $f$ is defined in $[-1,1]$ because of the square root.
Apply the chain rule; the derivative of $t\mapsto\arcsin t$ is $1/\sqrt{1-t^2}$; the derivative of $2x\sqrt{1-x^2}$ is
$$
2\left(\sqrt{1-x^2}+x\cdot\frac{-x}{\sqrt{1-x^2}}\right)
=\frac{2(1-2x^2)}{\sqrt{1-x^2}}
$$
For $t=2x\sqrt{1-x^2}$ we have
$$
1-t^2=1-4x^2(1-x^2)=1-4x^2+4x^4=(1-2x^2)^2
$$
so
$$
\frac{1}{\sqrt{1-t^2}}=\frac{1}{|1-2x^2|}
$$
and doing the product gives
$$
f'(x)=\frac{1}{|1-2x^2|}\frac{2(1-2x^2)}{\sqrt{1-x^2}}
=\begin{cases}
\frac{2}{\sqrt{1-x^2}} &\text{if $1-2x^2>0$}\\[2ex]
-\frac{2}{\sqrt{1-x^2}} &\text{if $1-2x^2<0$}
\end{cases}
$$
Thus your seemingly contradictory results are “explained”. The function is not differentiable where $x^2=1/2$.
You can use the derivative for simplifying the expression of $f$. For instance, when $-\frac{1}{\sqrt{2}}<x<-\frac{1}{\sqrt{2}}$, you know that
$$
f(x)=C_1+2\arcsin x
$$
and you can determine that $C_1=0$ by computing $f(0)=0$ and $2\arcsin0=0$. For $-1<x<-\frac{1}{\sqrt{2}}$, you have
$$
f(x)=C_2-2\arcsin x
$$
and computing in $-1$ tells $f(-1)=0$, $-2\arcsin(-1)=\pi$, so $C_2=-\pi$. For $\frac{1}{\sqrt{2}}<x<1$, again
$$
f(x)=C_3-2\arcsin x
$$
and $f(1)=0$, $-2\arcsin1=-\pi$, so $C_3=\pi$. Hence
$$
f(x)=\begin{cases}
-\pi-2\arcsin x & \text{for }x\in[-1,-1/\sqrt{2})\\
2\arcsin x & \text{for }x\in[-1/\sqrt{2},1/\sqrt{2}]\\
\pi-2\arcsin x & \text{for }x\in(1/\sqrt{2},1]
\end{cases}
$$
You can check that
$$
-\pi-2\arcsin\frac{-1}{\sqrt{2}}=-\pi+\frac{\pi}{2}
=-\frac{\pi}{2}=2\arcsin\frac{-1}{\sqrt{2}}
$$
and
$$
\pi-2\arcsin\frac{1}{\sqrt{2}}=\pi-\frac{\pi}{2}=\frac{\pi}{2}
=2\arcsin\frac{1}{\sqrt{2}}
$$
confirming that $f$ is everywhere continuous.
| {
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"source": "stackexchange",
"question_score": "6",
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Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $ Could you, please, check if I solved it right.
\begin{align*}
\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
&= \lim_{n \rightarrow \infty} \sqrt{n^2(1 + \frac1n)}
- \sqrt[3]{n^3(1 + \frac1n)})\\
&= \lim_{n \rightarrow \infty} (\sqrt{n^2} - \sqrt[3]{n^3})\\
&= \lim_{n \rightarrow \infty} (n - n) = 0.
\end{align*}
| You want
$\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
$.
More generally,
consider
$f_{a, c}(n)
=\sqrt[a]{n^a+n^{a-c}}
$
where
$a> 0$
and
$a > c > 0$.
I will use the
generalized binomial theorem
and generalized binomial coefficients.
We have
$\begin{array}\\
f_{a, c}(n)
&=\sqrt[a]{n^a+n^{a-c}}\\
&=n\sqrt[a]{1+n^{-c}}\\
&=n(1+n^{-c})^{1/a}\\
&=n\sum_{k=0}^{\infty} \binom{1/a}{k}n^{-ck}
\qquad\text{(generalized binomial theorem)}\\
&=n\sum_{k=0}^{\infty} \dfrac{\prod_{j=0}^{k-1}(1-aj)}{a^kk!}n^{-ck}
\qquad\text{(see (**) below)}\\
&=n\left(1+\sum_{k=1}^{\infty} \dfrac{\prod_{j=0}^{k-1}(1-aj)}{a^kk!n^{ck}}\right)\\
&=n+\sum_{k=1}^{\infty} \dfrac{\prod_{j=0}^{k-1}(1-aj)}{a^{k}k!n^{ck-1}}\\
&=n+\dfrac{1}{an^{c-1}}+\dfrac{1-a}{2a^2n^{2c-1}}+\dfrac{(1-a)(1-2a)}{6a^3n^{3c-1}}+...\\
\end{array}
$
If $c=1$,
$f_{a, 1}(n)
=\sqrt[a]{n^a+n^{a-1}}
=n+\dfrac{1}{a}+\dfrac{1-a}{2a^2n}+\dfrac{(1-a)(1-2a)}{6a^3n^{2}}+...
$.
In particular
$f_{2, 1}(n)
=\sqrt{n^2+n}
=n+\dfrac{1}{2}-\dfrac{1}{8n}+\dfrac{3}{48n^{2}}+...
=n+\dfrac{1}{2}-\dfrac{1}{8n}+\dfrac{1}{16n^{2}}+...
$
and
$f_{3, 1}(n)
=\sqrt[3]{n^3+n^{2}}
=n+\dfrac{1}{3}-\dfrac{2}{18n}+\dfrac{10}{162n^{2}}+...
=n+\dfrac{1}{3}-\dfrac{1}{9n}+\dfrac{5}{81n^{2}}+...
$.
Therefore the limit is
$\lim_{n \to \infty} ((n+\dfrac{1}{2}+O(1/n))-(n+\dfrac{1}{3}+O(1/n)))
=\dfrac16
$.
Note:
In this kind of computation,
I compute first and
worry about convergence later
(or never).
**
We have,
since
$\binom{r}{k}
=\dfrac{\prod_{j=0}^{k-1}(r-j)}{k!}
$,
$\begin{array}\\
\binom{1/a}{k}
&=\dfrac{\prod_{j=0}^{k-1}(\frac1{a}-j)}{k!}\\
&=\dfrac{\prod_{j=0}^{k-1}(1-aj)}{a^kk!}\\
\end{array}
$
| {
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"url": "https://math.stackexchange.com/questions/433527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 7
} |
Finding the value of $\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$ Is it possible to find the value of
$$\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
Does it help if I set it equal to $x$? Or I mean what can I possibly do?
$$x=\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
$$x^2=1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$
$$x^2-1=2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$
$$\frac{x^2-1}{2}=\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$
$$\left(\frac{x^2-1}{2}\right)^2=2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}$$
$$\left(\frac{x^2-1}{2}\right)^2-2=3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}$$
$$\vdots$$
I don't see it's going anywhere. Help appreciated!
| Not an answer of a closed form, but we can use Ramanujan's formula to approximate:
For any $n\in \mathbb{N}$ $$f(n) = 1+ n = \sqrt{1+n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)\sqrt{1+\dots}}}}}.$$
Let the sum be $x$.
$$
\begin{aligned}
x &> \sqrt{1+2\sqrt{\color{blue}{1}+3\sqrt{\color{blue}{1}+4\sqrt{\color{blue}{1}+5\sqrt{\color{blue}{1}+\dots}}}}} \\&= \sqrt{1+2f(3)}:=a_1 = 3.
\\
x &> \sqrt{1+2\sqrt{2+3\sqrt{\color{blue}{1}+4\sqrt{\color{blue}{1}+5\sqrt{\color{blue}{1}+\dots}}}}} \\&= \sqrt{1+2\sqrt{2+3f(4)}} :=a_2\approx 3.040758335
\\
x &> \sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{\color{blue}{1}+5\sqrt{\color{blue}{1}+\dots}}}}} \\&= \sqrt{1+2\sqrt{2+3\sqrt{3+4f(5)}}} :=a_3\approx 3.063938469
\\
x &> \sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{\color{blue}{1}+\dots}}}}} \\&= \sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5f(6)}}}} :=a_4\approx 3.074786007.
\end{aligned}
$$
So the nested radical does not go on infinitely in this sequence $\{a_n\}$, where
$$
a_n = \sqrt{1+2\sqrt{2+3\sqrt{3+\dots\sqrt{\dots\sqrt{(n-1)+n\sqrt{n+(n+1)(n+3)}}}}}}.
$$
And if we compute a few more terms:
$$
a_5 \approx 3.079604451993
\\
a_6 \approx 3.081712705722
\\
a_7 \approx 3.082633123037
\\
a_8 \approx 3.083036100688
\\
a_9 \approx 3.083213386604
\\
a_{10} \approx 3.083291812809
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/435778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "51",
"answer_count": 5,
"answer_id": 0
} |
show that the function $z = 2x^2 + y^2 +2xy -2x +2y +2$ is greater than $-3$ Show that the function
$$z = 2x^2 + y^2 +2xy -2x +2y +2$$
is greater than $-3$
I tried to factorize but couldn't get more than $(x-1)^2 + (x+y)^2 +(y-1)^2 - (y)^2$.
Is there any another way to factorize or another method??
| The equation $2x^2 + y^2 +2xy -2x +2y +2=0$ is a conic in the plane; so you can use the good old methods to put it in canonical form; set $x=X+a$ and $y=Y+b$, to get
$$
2X^2+4aX+2a^2+
Y^2+2bY+b^2+
2XY+2bX+2aY+2ab
-2X-2a+
2Y+2b+
2
=2X^2+2XY+Y^2+(4a+2b-2)X+(2b+2a+2)Y+2a^2+2ab+b^2-2a+2b+2
$$
and you get zero coefficients for $X$ and $Y$ by solving
$$
\begin{cases}
2a+b=1\\
a+b=-1
\end{cases}
$$
that is $a=2$ and $b=-3$. Now the polynomial becomes
$$
2X^2+2XY+Y^2-3
$$
and now it is clear that its values are $\ge-3$, because $2X^2+2XY+Y^2$ is positive definite (the conic is an ellipse), having negative discriminant $4-8=-4$. Or you can write it as
$$
X^2+(X+Y)^2-3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/436679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find $\cos(2\alpha)$ given $\cos(\theta -\alpha)$ and $\sin(\theta +\alpha)$ My question is:
If $\cos(\theta -\alpha) = \frac{3}{5}$ and $\sin(\theta +\alpha) =\frac{12}{13}$, find $\cos(2\alpha)$.
Attempt I:
\begin{align*}
&\cos^2(\theta -\alpha)+\sin^2(\theta +\alpha)
= \frac{9}{25} + \frac{144}{169}\\
\Rightarrow &\cos^2(\theta -\alpha)- \cos^2(\theta +\alpha)
= \frac{9}{25} + \frac{144}{169} - 1 = \frac{896}{4225}.
\end{align*}
But I thought it won't work.
Then I tried this:
Attempt II:
\begin{align*}
&\begin{cases}
\cos\theta \cos\alpha + \sin\theta \sin\alpha = \cos(\theta -\alpha) = \frac{3}{5},\\
\sin\theta \cos\alpha + \cos\theta \sin\alpha = \sin(\theta +\alpha) =\frac{12}{13},
\end{cases}\\
&\cos\theta (\sin\alpha + \cos\alpha) + \sin\theta (\sin\alpha + \cos\alpha) = \frac{99}{65},\\
&\sqrt{2}\left[\sin\left(\alpha+\frac{\pi}{4}\right) \cos\left(\alpha -\frac{\pi}{4}\right)\right] = \frac{99}{65}.
\end{align*}
But I thought here its better to convert both parts to cosines, so I did:
\begin{align*}
&\sqrt{2}\left[\cos\left(\alpha-\frac{\pi}{4}\right) \cos\left(\alpha -\frac{\pi}{4}\right)\right] = \frac{99}{65},\\
&\sqrt{2} \cos^2\left(\alpha-\frac{\pi}{4}\right) = \frac{99}{65}.
\end{align*}
But I think it also didn't work ....
Please guide. Thanks.
| HINT:
$$\cos(2\alpha)=\cos\{\theta+\alpha-(\theta-\alpha)\}=\cos(\theta+\alpha)\cos(\theta-\alpha)+\sin(\theta+\alpha)\sin(\theta-\alpha)$$
As $\cos(\theta-\alpha)=\frac35,$
$\sin(\theta-\alpha)=\pm \sqrt{1-\left(\frac35\right)^2}=\pm\frac45$
Similarly, as $\sin(\theta+\alpha)=\frac{12}{13},$ $\cos(\theta+\alpha)=\pm\sqrt{1-\left(\frac{12}{13}\right)^2}=\pm \frac5{13}$
If we assume $0<\theta \pm \alpha<\frac\pi2, $ we can consider the positive values only.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality $\frac{1-3ab}{1-2ac}+\frac{1-3bc}{1-2ba}+\frac{1-3ca}{1-2cb}\geq 0$
Let $a\ne 0$, $b\ne 0$ and $c\ne 0$ such that $a^2+b^2+c^2=1$. Prove that: $$\dfrac{1-3ab}{1-2ac}+\dfrac{1-3bc}{1-2ba}+\dfrac{1-3ca}{1-2cb}\geq 0.$$
My attempt to the solution: We get that $ab +bc+ca$ lies between $-0.5$ and $1$. We can use this. But I don't know how?
| I will use the notation $\sum_{cyc} f(x,y,z)$ to mean that the sum is over the so-called cyclic permutations of $(x,y,z)$ (or the even ones if you prefer),
so $\sum_{cyc} f(x,y,z):=f(x,y,z)+f(y,z,x)+f(z,x,y)$ (in place of $x,y,z$ there will also be other variables).
Lemma. If $x,y,z\ge 0$ then $\sum_{cyc} (x+y-2z)(z+x)(x+y)\ge 0$.
Proof. We have $\sum_{cyc}(x+z-2y)(z+x)(x+y)$
$=\sum_{cyc}(x^2z+xyz+x^3+x^2y+xz^2+yz^2+x^2z+xyz-2xyz-2y^2z-2x^2y-2xy^2)$
$=\sum_{cyc} x^3-2\sum_{cyc} x^2y+\sum_{cyc} xy^2=\sum_{cyc} x(x-y)^2\ge 0$. $\blacksquare$
Your inequality (using the hypothesis $a^2+b^2+c^2=1$) can be written as $\sum_{cyc}\frac{(a-b)^2+c^2-ab}{(a-c)^2+b^2}\ge 0$.
Let us call $A=(b-c)^2+a^2$ and $B$, $C$ similarly (by cyclically permuting the variables).
The inequality becomes $\sum_{cyc} \frac{C}{B}\ge\sum_{cyc}\frac{ab}{B}$.
Putting $S=a^2+b^2+c^2$ we clearly have $ab=\frac{S-C}{2}$ (and the cyclic ones), so the inequality is equivalent to
$3\sum_{cyc}\frac{C}{B}\ge \sum_{cyc}\frac{S}{B}$ or (which is the same) $3\sum_{cyc}\frac{B}{A}\ge \sum_{cyc}\frac{S}{A}$.
We will prove something slightly stronger, namely that $3\sum_{cyc}\frac{B}{A}\ge\sum_{cyc}\frac{A+B+C}{A}$ (this implies your inequality since $A+B+C\ge S$).
Now observe that $A-B=2c(a-b)\le (a-b)^2+c^2=C$ (and by symmetry $B-C\le A$, $C-A\le B$), so that $A,B,C$ satisfy the triangle inequalities
$A\le B+C$, $B\le C+A$, $C\le A+B$.
We can thus perform the substitution $A=y+z,B=z+x,C=x+y$ with $x,y,z\ge 0$ (if you didn't know this fact, check it!).
So we are left to prove that $\sum_{cyc}\frac{3(z+x)-2(x+y+z)}{y+z}\ge 0$, which is the Lemma (upon clearing the positive denominators).
| {
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"url": "https://math.stackexchange.com/questions/438017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Factor Equation Help me with this,
Question: factor $x^3y-x^3z+y^3z-xy^3+xz^3-yz^3$.
Solution:
$$\begin{eqnarray}&=&x^3y-x^3z+y^3z-xy^3+xz^3-yz^3\\
&=&x\left(z^3-y^3\right)+y\left(x^3-z^3\right)+z\left(y^3-x^3\right)\\
&=&x\left[(z-y)\left(z^2+zy+y^2\right)\right]+y\left[(x-z)\left(x^2+xz+z^2\right)\right]+z\left[(y-x)\left(y^2+xy+x^2\right)\right]\end{eqnarray}$$
This expression is quite simple at first glance, but I stuck up again in that line. I appreciate any help.
| $x^3y-x^3z+y^3z-xy^3+xz^3-yz^3$
$=x^3(y-z)+yz(y^2-z^2)-x(y^3-z^3)$
$=x^3(y-z)+yz(y+z)(y-z)-x(y-z)(y^2+yz+z^2)$
$=(y-z)\{x^3+yz(y+z)-x(y^2+yz+z^2)\}$
Now, $x^3+yz(y+z)-x(y^2+yz+z^2)$
$=x^3+y^2z+yz^2-xy^2-xyz-z^2x$
$=x(x^2-y^2)-yz(x-y)-z^2(x-y)$
$=(x-y)\{x(x+y)-yz-z^2\}$
Now, $x(x+y)-yz-z^2$
$=x^2+xy-yz-z^2=(x+z)(x-z)+y(x-z)=-(z-x)(x+y+z)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/438088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
General solution for trigonometry equation How should I state the general solution for the equation $\sin(4\phi)=\cos(2\phi)$.
The angles are $15$, $45$, $75$ and $135$ if I restrict myself within the range $[0,360]$
| As $\cos2\phi=\sin4\phi=\cos(90^\circ-4\phi)$
$\implies 2\phi=n360^\circ\pm(90^\circ-4\phi)$ where $n$ is any integer
Taking '+' sign, $2\phi=n360^\circ+90^\circ-4\phi$
$\implies 6\phi=n360^\circ+90^\circ \implies \phi=n60^\circ+15^\circ$
As $0\le \phi<360^\circ, 0\le n60^\circ+15^\circ<360^\circ\implies 0\le n\le 5$
Taking '-' sign, $2\phi=n360^\circ-90^\circ+4\phi$
$\implies 2\phi=90^\circ-n360^\circ\implies \phi=45^\circ-n180^\circ$
As $0\le \phi<360^\circ, 0\le 45^\circ-n180^\circ<360^\circ \implies -1\le n\le0$
So, there are $6+2=8$ solutions in $\in[0, 360^\circ)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/438231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $a^3+b^3+c^3 \geq a^2b+b^2c+c^2a$
Let $a,b,c$ be positive real numbers. Prove that $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$.
My (strange) proof:
$$
\begin{align*}
a^3+b^3+c^3 &\geq a^2b+b^2c+c^2a\\
\sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\
\sum\limits_{a,b,c} a^2 &\geq \sum\limits_{a,b,c} ab\\
a^2+b^2+c^2 &\geq ab+bc+ca\\
2a^2+2b^2+2c^2-2ab-2bc-2ca &\geq 0\\
\left( a-b \right)^2 + \left( b-c \right)^2 + \left( c-a \right)^2 &\geq 0
\end{align*}
$$
Which is obviously true.
However, this is not a valid proof, is it? Because I could just as well have divided by $a^2$ rather than $a$:
$$
\begin{align*}
\sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\
\sum\limits_{a,b,c} a &\geq \sum\limits_{a,b,c} b\\
a+b+c &\geq a+b+c
\end{align*}
$$
Which is true, but it would imply that equality always holds, which is obviously false. So why can't I just divide in a cycling sum?
Edit: Please don't help me with the original inequality, I'll figure it out.
| WOLG, Let $c$=Max{$a,b,c$}, then there is 2 cases:
case I: $0<a \le b \le c$, we want to prove $c^2(c-a) \ge a^2(b-a)+b^2(c-b)$
we have $c^2\ge b^2, c^2\ge a^2 \to $,RHS $\le c^2(b-a)+c^2(c-b)=c^2(c-a)$
case II: $0<b \le a \le c$, we want to prove $a^2(a-b)+c^2(c-a) \ge b^2(c-b)$
we have $a^2\ge b^2,c^2 \ge b^2, \to$LHS $ \ge b^2(a-b)+b^2(c-a)=b^2(c-b)$
to summary 2 cases, we have $a^2(a-b)+b^2(b-c)+c^2(c-a) \ge 0$
QED
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
How do I write a sum of cosines as a product of sines? I am trying to prove that
$$\cos A+\cos B+\cos C=4\sin\frac A2\sin\frac B2\sin\frac C2$$ for ABC is a triangle.
I tried up to the stage of
$$-2\sin^2 C+2\cos\frac{180-C}2 \cos\frac{A+B}2$$
but how do I proceed from here?
| PROOF: There is something wrong in the question
TO PROVE :$\cos A+\cos B+\cos C -1=4\sin\frac A2\sin\frac B2\sin\frac C2$
PROOF: You can proceed from the right hand side too
$4\sin\frac A2\sin\frac B2\sin\frac C2 = 2\sin\frac A2\sin\frac B2 2\sin\frac C2$
NOW, $2\sin\frac A2\sin\frac B2 = cos\frac{A-B}2 - \cos\frac{A+B}2 = cos\frac{A-B}2 - \sin\frac C2$
The expression becomes,
$2\sin\frac C2\cos\frac{A-B}2 - 2\sin^2\frac C2 = \sin\frac{A-B+C}2 - \sin\frac{A-B-C}2 + cosC - 1$ = $cosB + cosA+ cosC- 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/439523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove this inequality $a^{\frac{a}{b}}b^{\frac{b}{c}}c\geq1$ Please help me to prove this inequality.
Assume $a,b,c>0$ and $abc\geq1$ then $a^{\frac{a}{b}}b^{\frac{b}{c}}c\geq1$.
Thanks.
| I think the right inequality is:
$a^{\frac{a}{b}}b^{\frac{b}{c}}c^{\frac{c}{a}} \ge 1$
to prove it is to prove $\dfrac{a}{b}\log{a}+\dfrac{b}{c}\log{b}+\dfrac{c}{a}\log{c} \ge 0$
let $abc=1 \to a=\dfrac{1}{bc}$
LHS=$(b^3-1)\log{b}+(b^3c^3-1)\log{c}$
it is trivial when $b>1$ and $c>1$, $b<1$ and $c<1$, LHS $> 0$
when $b<1, c>1$, if $b^3c^3 >1$, LHS $>0$
so we only check $b^3c^3<1$, ie ,$1<c<\dfrac{1}{b}$ , in this case,
$0<\log{c} <-\log{b},0<1-b^3c^3<1-b^3 \implies (b^3c^3-1)\log{c}>(1-b^3)\log{b} \implies $ LHS$>0$
with same mehtod, we can also get when $b>1, 1>c>\dfrac{1}{b}$, LHS$>0$, when $b>1, c<\dfrac{1}{b}$, it is trivial LHS$>0$
since $(b^3-1)\log{b}\ge 0$, when and only when $b=1$ the "$=0$" holds, if LHS $=0 \implies b=1, c=1 \implies a=1$
it is not difficult to prove when $abc>1,a^{\frac{a}{b}}b^{\frac{b}{c}}c^{\frac{c}{a}}>x^{\frac{x}{y}}y^{\frac{y}{z}}z^{\frac{z}{x}}$,where $xyz=1 $ and $a=px,b=py,c=pz, abc=p^3xyz,p>1$,
| {
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"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_2^\infty{\frac{3x-2}{x^2(x-1)}}$ To be shown that $\int_2^\infty{\dfrac{3x-2}{x^2(x-1)}}=1-\ln2$
My thought: $\dfrac{3x-2}{x^2(x-1)}=\dfrac{3x}{x^2(x-1)}-\dfrac{2}{x^2(x-1)}$
• $\dfrac{3x}{x^2(x-1)}=\dfrac{3}{x(x-1)}=\ldots=-\dfrac{3}{x}+\dfrac{3}{x-1}$
• $\dfrac{2}{x^2(x-1)}=\ldots=-\dfrac{2}{x^2}+\dfrac{2}{x-1}$
But developing the sum of the integrals of the above two gives ln of infinite in my results. I don't know what I am doing wrong. Can you please confirm the dots above? Although I have several times. Thanks a lot
| Hint
Write the integrand as: $$\frac{3x-2}{x^2(x-1)}=\frac{Ax+B}{x^2}+\frac{C}{x-1}$$ in which $A,B$ and $C$ is unknown constants and so find the proper values for them. I think via this way you can find them easier than the way you noted. Indeed, $$\frac{Ax+B}{x^2}+\frac{C}{x-1}=\frac{(A+C)x^2+x(B-A)-B}{x^2(x-1)}$$ and therefore you get $$A+C=0, ~~B-A=3,~~-B=-2$$ and so...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/442900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Maximize $(a-1)(b-1)(c-1)$ knowing that : $a+b+c=abc$. If : $a,b,c>0$, and : $a+b+c=abc$, then find the maximum of $(a-1)(b-1)(c-1)$.
I noted that : $a+b+c\geq 3\sqrt{3}$, I believe that the maximum is at : $a=b=c=\sqrt{3}$. (Can you give hints).
| Hints:
Lagrange multipliers. Define
$$H(x,y,z,\lambda):=(x-1)(y-1)(z-1)+\lambda(x+y+z-xyz)$$
$$\begin{align*}H'_x&=(y-1)(z-1)+\lambda(1-yz)=0\iff \lambda=\frac{(y-1)(z-1)}{yz-1}\\
H'_y&=(x-1)(z-1)+\lambda(1-xz)=0\iff \lambda=\frac{(x-1)(z-1)}{xz-1}\\
H'_z&=(y-1)(x-1)+\lambda(1-yx)=0\iff \lambda=\frac{(y-1)(x-1)}{yx-1}\\
H'_\lambda&=x+y+z-xyz=0\end{align*}$$
From the first three equations we get (using symmetry)
$$\frac{x-1}{xy-1}=\frac{z-1}{yz-1}\implies (x-z)(1-y)=0=(x-y)(1-z)=(y-z)(1-x)$$
Check that $\,x=y=z=1\,$ is not a good option...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $(n+1)(n+2)(n+3)$ is $O(n^3)$ Problem
*
*Prove that $(n+1)(n+2)(n+3)$ is $O(n^3)$
Attempt at Solution
*
*$f(n) = (n+1)(n+2)(n+3)$
*$g(n) = n^3$
*Show that there exists an $n_0$ and $C > 0$ such that $f(n) \le Cg(n)$ whenever $n > n_0$
*$f(n) = n^3+6n^2+11n+6 = n^3(1 + 6/n + 11/n^2 + 6/n^3)$
*$f(n) \le C*g(n)$ is
*$n^3(1 + 6/n + 11/n^2 + 6/n^3) \le C*n^3$ is
*$(1 + 6/n + 11/n^2 + 6/n^3) < C$
That's as far as I got. Should I plug in a value for n to find C? And then, would that value I plugged in for n be $n_0$?
Any help is appreciated.
Thank you in advance.
| You are almost there.
Remember that big-O (and little-O) statements
apply for all large enough $n$,
not for all $n$.
So, in
$(3 + 9/n + 6/n^2) < C$,
give some lower bound for $n$,
say $n > 3$.
Then $9/n < 3$ and
$6/n^2 < 2/3$,
so
$3 + 9/n + 6/n^2 < 3+3+2/3 < 7$,
so $C=7 $ works for $n > 3$.
Note that, as the lower bound for $n$ gets larger
(for example, see what you get for $n = 100$),
the bound on $C$ gets closer to $3$,
its best value.
But it never reaches $3$.
But you only need $a$ bound, not the best bound.
| {
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"url": "https://math.stackexchange.com/questions/447696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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For what natural numbers is $n^3 < 2^n$? Prove by induction Problem
For what natural numbers is $n^3 < 2^n$?
Attempt @ Solution
*
*For $n=1$, $1 < 2$
*Suppose $n^3 < 2^n$ for some $n = k \ge 1$
*It looks like the inequality is true for $n = 0$, $n = 1$ and $n\ge10$
*But, how can I prove this through induction?
| You need to gather more data first:
$$\begin{array}{rcc}
n:&1&2&3&4&5&6&7&8&9&10&11\\
n^3:&1&8&27&64&125&216&343&512&729&1000&1331\\
2^n:&2&4&8&16&32&64&128&256&512&1024&2048
\end{array}$$
Notice that $n^3$ is not less than $2^n$ for $n=2,3,\ldots,9$; the fact that $1^3<2^1$ is an isolated success. The simplest reasonable guess at this point is that $n^3<2^n$ if $n\ge 10$. Thus, the basis step for your induction will be the calculation that $10^3<2^{10}$.
For your induction step you’ll assume as induction hypothesis that $k^3<2^k$ for some integer $k\ge 10$. Your goal, then, will be to show that $(k+1)^3<2^{k+1}$. Note that
$$(k+1)^3=k^3\cdot\frac{(k+1)^3}{k^3}=k^3\left(\frac{k+1}k\right)^3\;.$$
Since $k\ge 10$, $\frac{k+1}k=1+\frac1k\le 1+\frac1{10}=\frac{11}{10}$, and therefore
$$\left(\frac{k+1}k\right)^3\le\left(\frac{11}{10}\right)^3=\frac{1331}{1000}<2\;,$$
and therefore $(k+1)^3<2k^3$. By the induction hypothesis $k^3<2^k$, so ... ?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
find correlation coefficient of $f(x,y)=2$ for $0
Find the correlation coefficient for the random variables $X$ and $Y$ having joint density $f(x,y)=2$ for $0 < x \leq y<1$.
Seem like a simple problem but I'm stuck.
Since $Corr(X,Y) = \frac{Cov(X,Y)}{\sqrt{Var(X)}\sqrt{Var(Y)}}$, I figure I start with $Var(X)$ and $Var(Y)$.
$E(X) = \int_0^1 \int_0^y 2xdxdy = \int_0^1y^2dy = \frac{1}{3}$
$E(X^2) = \int_0^1 \int_0^y 2x^2dxdy = \int_0^1\frac{2}{3}y^3dy = \frac{1}{6}$
$Var(X) = E(X^2) - E(X)^2 = \frac{1}{6} - \frac{1}{9} = \frac{1}{18}$
$E(Y) = \int_0^1 \int_x^1 2ydydx = \int_0^1 1-x^2 dx = \frac{2}{3}$
$E(Y^2) = \int_0^1 \int_x^1 2y^2dydx = \frac{2}{3}\int_0^1 1-x^3dx = \frac{1}{2}$
$Var(Y) = E(Y^2) - E(Y)^2 = \frac{1}{2} - \frac{4}{9} = \frac{1}{18}$
$\frac{1}{\sqrt{Var(X)}\sqrt{Var(Y)}} = \frac{1}{\frac{1}{18}} = 18$
It seems alright up to this point.
$Cov(X,Y) = E(XY) - E(X)E(Y)$
$E(XY) = \int_0^1 \int_0^y 2xy dxdy = \int_0^1 y^3 dy = \frac{1}{4}$
$Cov(X,Y) = \frac{1}{4} - (\frac{1}{18})^2$
And this is where I'm stuck.
$Corr(X,Y) = (\frac{1}{4} - (\frac{1}{18})^2) \times 18 = \frac{18}{4} - \frac{1}{18} > 1$
wolfram says my calculation are correct. so this must be a concept problem.
The $0 < x \leq y<1$ is really throwing me off but Devore doesn't really provide a clear explanation.
Thanks for your hints.
| Your integrals look right. Note that $E(X)E(Y)=\frac{2}{9}$, so the covariance calculation is not right.
The covariance is $\frac{1}{4}-\frac{2}{9}=\frac{1}{36}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Existence of linear mapping I am studying for an exam in linear algebra and I am having trouble solving the following:
Do linear mappings $\phi : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ with the following properties exist?
$1)$ $\phi_1 \begin{pmatrix} 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $, $\phi_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix} $, $\phi_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\3 \end{pmatrix} $
$2)$ $\phi_2 \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\0\end{pmatrix} $, $\phi_2 \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2\\ 3 \end{pmatrix} $
$3)$ $\phi_3 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1\\ 0 \end{pmatrix} $, $\phi_3 \begin{pmatrix} -2 \\ 5 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \end{pmatrix} $, $\phi_3 \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\1 \end{pmatrix} $
I know that the following properties have to hold for a linear mapping:
*
*$f(\mathbf{x}+\mathbf{y}) = f(\mathbf{x})+f(\mathbf{y})$
*$f(\alpha \mathbf{x}) = \alpha f(\mathbf{x})$
*$f(0) = 0 $
I conclude that $2)$ is a not a linear mapping since $\phi(0)$ is not mapped to $0$.
But how shall I proceed with the others?
| 1) You are given the image of $3$ vectors whereas you only need $2$ to define the mapping on $\Bbb R^2$ (because $\Bbb R^2$ is of dimension $2$). So you can assume you defined $\phi$ with the first two and then check if the $3^\text{rd}$ one is true. Now $\begin{pmatrix}1\\2\end{pmatrix}=2\begin{pmatrix}1\\1\end{pmatrix}-\cfrac{1}{2}\begin{pmatrix}2\\0\end{pmatrix}$ so we should have $\begin{pmatrix}0\\1\end{pmatrix}=2\begin{pmatrix}5\\2\end{pmatrix}-\cfrac{1}{2}\begin{pmatrix}2\\3\end{pmatrix}$ which is not the case so the answer is No.
2) As you said, $0$ should be mapped to $0$, and it is not so No.
3) It's the same thing as 1), you need the image of two vectors foming a basis to define your mapping but you have $3$ so you have to check whether the $3^{\text{rd}}$ one is consistent with the two others.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to search quadratic function If a graph of the quadratic function $f(x)=ax^2+bx+c$, where $a$, $b$ and $c$ are constants. If this function vertex is $(13,−169)$ and the distance between the two intersection points with the $x$-axis is $26$, what is the value of $a$, $b$, and $c$?
| $$y=ax^2+bx+c \implies \left(x+\frac b{2a}\right)^2=4\cdot\frac14\left(y-\left(c-\frac{b^2}{4a}\right)\right)$$
Comparing with $(x-h)^2=4a(y-k)$ the vertex will be $\left(-\frac b{2a},c-\frac{b^2}{4a}\right)$
So, $-\frac b{2a}=13\implies b=-26a\ \ \ \ (1)$
and $c-\frac{b^2}{4a}=-169\implies b^2-4ac=169\cdot4a\ \ \ \ (2)$
$\implies 4ac=b^2-169\cdot4a=(-26a)^2-169\cdot4a$
$\implies c=169(a-1) \ \ \ \ (3)$ assuming $a\ne0$
For the two intersection points with the x-axis, $y=f(x)=0$
If $(x_1,0),(x_2,0)$ are the two intersection points, $x_1,x_2$ are the roots of $ax^2+bx+c=0$
$$\implies x_1+x_2=-\frac ba,x_1x_2=\frac ca$$
So, the distance will be $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=\sqrt{(x_1+x_2)^2-4x_1x_2}=\frac{\sqrt{b^2-4ac}}{|a|}=\frac{\sqrt{676a}}{|a|}\text{ using }(2)$
So, the distance $=26\sqrt a$ assuming $a>0$ otherwise the distance will be complex
Then we have $26\sqrt a=26$
Now use $(1),(3)$ to find $b,c$ respectively
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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closed form of $\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^n}$ closed form of $$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^n}$$
for $a,b>0$
n=1 we get
$$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^1}=\frac{2\pi}{ab}$$
n=2 we get
$$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2}=\frac{\pi(a^2+b^2)}{a^3b^3}$$
but what the integral for n ????
I hope to be there two different solution one by contour integration and the other by real analysis and thanks for all
| In the following we assume that $n\geq 1$.
Thanks to parity, the integral can be written as
\begin{align}
I_n=\int_0^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2 x)^n}&=2\int_0^{\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2 x)^n}=\\
&=2\int_0^{\pi}\frac{dx}{(\frac{a^2+b^2}{2}+\frac{a^2-b^2}{2}\cos 2x)^n}=\\
&=\frac{1}{i}\oint_{|z|=1}\frac{z^{-1}dz}{(\frac{a^2+b^2}{2}+\frac{a^2-b^2}{2}\frac{z+z^{-1}}{2})^n}=\\
&=\frac{2^{2n}}{(a^2-b^2)^n}\frac{1}{i}\oint_{|z|=1}\frac{z^{-1}dz}{(z+z^{-1}-c-c^{-1})^n}=\\
&=\frac{2^{2n}}{(a^2-b^2)^n}\frac{1}{i}\oint_{|z|=1}\frac{z^{n-1}dz}{(z-c)^n(z-c^{-1})^n}
\end{align}
where the parameter $c$ is defined by the equation
$$c+c^{-1}=2\frac{b^2+a^2}{b^2-a^2}.$$
In particular, we can (and will) set $\displaystyle c=\frac{b-a}{b+a}$. Now for $b>a>0$ the point $z=c$ is inside, and $z=c^{-1}$ outside the unit circle. Therefore, by residues,
\begin{align}
I_n&=2\pi\cdot\frac{2^{2n}}{(a^2-b^2)^n}\cdot\mathrm{res}_{z=c}\frac{z^{n-1}}{(z-c)^n(z-c^{-1})^n}=\\
&=\frac{2\pi}{(n-1)!}\cdot\frac{2^{2n}}{(a^2-b^2)^n}\cdot\left[\frac{d^{n-1}}{dz^{n-1}}
\frac{z^{n-1}}{(z-\frac{b+a}{b-a})^n}\right]_{z=\frac{b-a}{b+a}}
\end{align}
In particular,
\begin{align}
I_1&=\frac{2\pi}{ab},\\
I_2&=\frac{\pi(a^2+b^2)}{a^3b^3},\\
I_3&=\frac{\pi(3a^4+2a^2b^2+3b^4)}{4a^5b^5},\\
&\ldots
\end{align}
Explicit computation (by RGB)
We can compute the residue explicitly by using the binomial formula.
We have that $$z^{n-1}=c^{n-1}\left(1+\frac{z-c}{c}\right)^{n-1}=\sum_{k=0}^{\infty}\binom{n-1}{k}\frac{(z-c)^k}{c^{k-n+1}},$$
and that $$(z-c^{-1})^{-n}=(c-c^{-1})^{-n}\left(1+\frac{z-c}{c-c^{-1}}\right)^{-n}=\sum_{k=0}^{\infty}\binom{-n}{k}\frac{(z-c)^k}{\left(c-c^{-1}\right)^{k+n}}.$$
We multiply these two to get a series $\sum_{k=0}^{\infty}A_k(z-c)^k$. We need then to divide by $(z-c)^n$ and get the coefficient of the term with degree $-1$. This will be the coefficient $A_{n-1}$.
Using the formula for the product of two series we get
$$A_{n-1}=\sum_{r=0}^{n-1}\frac{\binom{n-1}{r}}{c^{r-n+1}}\frac{\binom{-n}{n-1-r}}{\left(c-c^{-1}\right)^{2n-r-1}}.$$
Finally we get $$I_n=2\pi\cdot\frac{2^{2n}}{(a^2-b^2)^n}\cdot\sum_{r=0}^{n-1}\frac{\binom{n-1}{r}}{c^{r-n+1}}\frac{\binom{-n}{n-1-r}}{\left(c-c^{-1}\right)^{2n-r-1}}.$$
Explicit computation II (by O.L.)
Let us write
$$\frac{z^{n-1}}{(z-c^{-1})^n}=\frac{((z-c^{-1})+c^{-1})^{n-1}}{(z-c^{-1})^n}=\sum_{k=0}^{n-1}{n-1\choose k}c^{k+1-n}(z-c^{-1})^{k-n}$$
We have to compute $(n-1)$th derivative of this expression and then evaluate it at $z=c$. This gives the formula
$$I_n=2\pi\cdot\frac{2^{2n}}{(a^2-b^2)^n}\cdot\sum_{k=0}^{n-1}(-1)^{n-1}{n-1\choose k}{2n-2-k\choose n-1}c^{k+1-n}(c-c^{-1})^{k-2n+1}$$
which is equivalent to that of RGB.
| {
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"timestamp": "2023-03-29T00:00:00",
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Summation and proof by induction question: $\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$ I can't figure this out based on examples in textbooks, etc.
Show via induction that $\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$
So far, I have:
(a) base case
$P(1)= 1(1+1)(1+2) = \frac{1(1+1)(1+2)(1+3)}{4} = 6 = 6$
(b) inductive step $P(k)\rightarrow P(k+1)$
If $P(k)$ is true, then
$P(k+1) = (2)(3)+(2)(3)(4)+(3)(4)(5)+\cdots+(k+1)(k+2)(k+3)= \frac{(k+1)(k+2)(k+3)(k+4)}{4}$
But this doesn't seem to be going anywhere. Any ideas?
| First, to improve the logic of your proof, you don't want to put the thing that you are trying to prove, $P(k+1) = (2)(3)+(2)(3)(4)+(3)(4)(5)+\cdots+(k+1)(k+2)(k+3)= \frac{(k+1)(k+2)(k+3)(k+4)}{4}$ at the beginning of (the second part of) your proof. Same goes for the base case where you you are trying to show that the left and right side both equal 6, but you are doing them in parallel. Try either:
$P(1)= 1(1+1)(1+2) = 6 = \frac{1(1+1)(1+2)(1+3)}{4}$
or
$P(1)= 1(1+1)(1+2) = 6$
and
$\frac{1(1+1)(1+2)(1+3)}{4} = 6$
Now for the actual proof:
We have $\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$. Now take $\sum_{j=1}^{n+1}j(j+1)(j+2)=\sum_{j=1}^{n}j(j+1)(j+2) + (n+1)(n+2)(n+3)$
and by the inductive hypothesis
$\sum_{j=1}^{n}j(j+1)(j+2) + (n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3)}{4} + (n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3)+4(n+1)(n+2)(n+3)}{4}=\frac{(n+4)(n+1)(n+2)(n+3)}{4}$ which concludes our proof.
It is generally not a good idea to expand your sums like this $(2)(3)+(2)(3)(4)+(3)(4)(5)+\cdots+(k+1)(k+2)(k+3)$. It is sometimes helpful for spotting patterns like telescoping sums, but it generally turns into a mess.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the limit $\lim \limits_{n \to \infty}\ (\cos \frac x 2 \cdot\cos \frac x 4\cdot \cos \frac x 8\cdots \cos \frac x {2^n}) $ This limit seemed quite unusual to me as there aren't any intermediate forms or series expansions which are generally used in limits. Stuck on this for a while now .Here's how it goes :
$$
\lim \limits_{n \to \infty}
\left[\cos\left(x \over 2\right)\cos\left(x \over 4\right)
\cos\left(x \over 8\right)\ \cdots\ \cos\left(x \over 2^{n}\right)\right]
$$
| Put $z=e^{i x/2^n}.$ Then the product becomes
$$\prod_{k=1}^n \cos\left(\frac{x}{2^k}\right)
= \frac{1}{2^n} \prod_{k=0}^{n-1} \left(z^{2^k}+z^{-2^k}\right)
= \frac{1}{2^n} z^{-\sum_{k=0}^{n-1} 2^k}
\prod_{k=0}^{n-1} \left(z^{2^{k+1}}+1\right)\\
= \frac{1}{2^n} z^{1-2^n} \frac{1}{z+1} \prod_{k=0}^n \left(z^{2^k}+1\right).$$
Now the product in this last formula is easily seen to be the generating function of the positive integers less than $2^{n+1}-1$ (consider the binary representation of an integer $q$ from this interval), so that we may continue with
$$\frac{1}{2^n} z^{1-2^n} \frac{1}{z+1} \left(1+z+z^2+\cdots+z^{2^{n+1}-1}\right)
= \frac{1}{2^n} z^{1-2^n} \frac{1}{z+1} \frac{1-z^{2^{n+1}}}{1-z}\\
= \frac{1}{2^n} \frac{1}{z+1} \frac{z^{1-2^n}-z^{1+2^n}}{1-z}
= \frac{1}{2^n} \frac{z}{z+1} \frac{z^{-2^n}-z^{2^n}}{1-z}
= \frac{1}{2^n} z \frac{z^{-2^n}-z^{2^n}}{1-z^2} \\
= \frac{1}{2^n} \frac{z^{-2^n}-z^{2^n}}{1/z-z}
= \frac{1}{2^n} \frac{\sin(x)}{\sin(x/2^n)}
= \frac{x/2^n}{\sin(x/2^n)} \frac{\sin(x)}{x}.$$
Finally recall that $\sin(x) \sim x$ in a neighborhood of zero, so that the limit of the first term is one, resulting in a final answer of
$$\frac{\sin(x)}{x}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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generating function Homework Question 1 This is a HW question
I am asked to find a closed form generating function for $1,1,0,1,1,0,1,1,0....$
so then $f(x)=x^0+x^1+0x^2+x^3+x^4+0x^5+x^6+x^7+0x^8$
could use some hint or help.
| Hint: See if writing it as follows helps:
$$f(x) = 1 + x + x^{3} +x^{4} + \cdots = ( 1 + x + x^{2} + x^{3} + x^{4} + x^{5} + \cdots ) - (x^{2} + x^{5} + x^{8} + \cdots) = \frac{1}{1-x} - x^{2}(1+x^{3}+x^{6}+\cdots)$$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\frac92\right)^2} +\frac92\\
&= \sqrt {5^2 -2\times5\times\frac92 + \left(\frac92\right) ^2} + \frac92\\
&= \sqrt {\left(5-\frac92\right)^2} +\frac92\\
&= 5 + \frac92 - \frac92 \\
&= 5\end{align}$$
Where did I go wrong
| How did $\sqrt{16 -2\times4\times\frac92 +(\frac92)^2}$ turn into $\left(\sqrt{16 -36 + (\frac92)^2}\right)^2$?
Then later, you seem to assume that since $\left(4-\frac92\right)^2$ is the same as $\left(5-\frac92\right)^2$, it follows that $4-\frac92=5-\frac92$. Like saying that since $3^2=(-3)^2$, it follows that $3=-3$. A well known mistake.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.