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Why is $ \frac{5}{64}((161+72\sqrt{5})^{-n}+(161+72\sqrt{5})^{n}-2)$ always a perfect square? I'm working on a puzzle, and the solution requires me somehow establishing that
$$ f(n):=\frac{5}{64}\Big(\big(161+72\sqrt{5}\big)^{-n}+\big(161+72\sqrt{5}\big)^{n}-2\Big)$$
is a perfect square for $n\in \mathbb{Z}_{\geq 0}$.
I've done a lot of simplification to get to this point, and am stuck here. I can provide the context of the puzzle if necessary, but it's pretty far removed from what I have here. The goal is basically to show that a formula generates solutions to a given equation.
Any tips on how to proceed?
Here's the first few values:
$$\begin{array}{|c|c|}
\hline
n&\text{value}\\ \hline
0&0\\ \hline
1& 5^2 \\ \hline
2&90^2 \\ \hline
3& 1615^2\\ \hline
4& 28980^2\\ \hline
\end{array}$$
| Note that $161+72\sqrt{5} = (161-72\sqrt{5})^{-1}$ and $161+72\sqrt{5}=(9+4\sqrt{5})^2,$ and therefore
\begin{align*}
\big(161+72\sqrt{5}\big)^{-n}+\big(161+72\sqrt{5}\big)^{n}-2 &= \big( (161+72\sqrt{5})^{n/2} - (161-72\sqrt{5})^{n/2} \big)^2 \\
&= \big( (9+4\sqrt{5})^n - (9-4\sqrt{5})^n \big)^2.
\end{align*}
It therefore suffices to prove that $(9+4\sqrt{5})^n - (9-4\sqrt{5})^n$ is always $\sqrt5$ times an integer that is a multiple of $8$, which can be done using the binomial expansions of $(9\pm 4\sqrt{5})^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find $L=\lim_{n\to \infty }\frac{1}{n}\sum_{k=1}^{n}\left\lfloor 2\sqrt{\frac{n}{k}} \right\rfloor -2\left\lfloor \sqrt{\frac{n}{k}} \right\rfloor$ Question:- Find Limit $$L=\lim_{n\to \infty }\frac{1}{n}\sum_{k=1}^{n}\left\lfloor 2\sqrt{\frac{n}{k}} \right\rfloor -2\left\lfloor\sqrt{\frac{n}{k}} \right\rfloor \text , $$ where $\lfloor x \rfloor$ represents greatest integer function.
Yesterday, my friend sent me this limit question.Greatest integer function is the biggest problem here.I don't know how to evaluate the summation to find the given limit.
Can anybody help me!!
| As this is a Riemann sum, you can convert it into an integral.
This becomes:
$$\int_0^1 \left \lfloor \frac2{\sqrt x} \right \rfloor -2\left \lfloor\frac1{\sqrt x} \right \rfloor\,dx$$
Put $\sqrt x \rightarrow 1/t$ to get:
$$ = 2\int_1^\infty \frac{\left \lfloor 2t \right \rfloor}{t^3} -2\frac{\left \lfloor t \right \rfloor}{t^3}\,dt$$
$$ = 2\left(\sum_{r=1}^\infty\int_{(r+1)/2}^{r/2 + 1}\frac{r+1}{t^3}\,dt - 2\sum_{r=1}^\infty\int_{r}^{r + 1}\frac{r}{t^3}\,dt\right)$$
$$ = 2\sum_{r=1}^\infty\left(\frac{2(2r+3)}{(1+r)(2+r)^2} - \frac{2r+1}{r(1+r)^2}\right)$$
$$ = 2\sum_{r=1}^\infty\left(\frac{4}{(r+1)(r+2)} - \frac{2}{(r+1)(r+2)^2} - \frac{2}{r(r+1)} + \frac{1}{r(1+r)^2}\right)$$
$$ = 2\sum_{r=1}^\infty\left(\frac{1}{r(1+r)^2}-\frac{2}{(r+1)(r+2)^2}\right)$$
$$ = 1 - 2\sum_{r=1}^\infty\left(\frac{1}{r(1+r)^2}\right)$$
$$= \boxed{\frac{\pi^2}3 - 3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}$ given that $\sin\alpha-\cos\alpha=\frac12$ Given that $\sin\alpha-\cos\alpha=\frac12$. What is the value of $$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}?$$
My work:
$$\sin\alpha-\cos\alpha=\frac12$$
$$\sin\alpha\frac1{\sqrt2}-\cos\alpha\frac1{\sqrt2}=\frac1{2\sqrt2}$$
$$\sin\left(\alpha-\frac{\pi}{4}\right)=\frac1{2\sqrt2}$$
$$\alpha-\frac{\pi}{4}=\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)$$
I calculated the value of $\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)\approx 20.705^\circ$,
so I got $\alpha\approx 45^\circ+20.705^\circ=65.705^\circ$
I calculated
$$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}=\frac{1}{\sin^365.705^\circ}-\frac{1}{\cos^3 65.705^\circ}\approx -13.0373576$$
My question: Can I find the value of above trigonometric expression without using calculator? Please help me solve it by simpler method without solving for $\alpha$. Thanks
| You can easily evaluate it without calculator as follows
$$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}=\frac{\cos^3\alpha-\sin^3\alpha}{\sin^3\alpha\cos^3\alpha}$$
$$=\frac{(\cos\alpha-\sin\alpha)(\cos^2\alpha+\sin^2\alpha+\cos\alpha\sin\alpha)}{\sin^3\alpha\cos^3\alpha}$$
$$=\frac{-(\sin\alpha-\cos\alpha)(1+\cos\alpha\sin\alpha)}{\frac18(2\sin\alpha\cos\alpha)^3}$$
$$=\frac{-4(\sin\alpha-\cos\alpha)(3-(\sin\alpha-\cos\alpha)^2)}{(1-(\sin\alpha-\cos\alpha)^2)^3}$$
$$=\frac{-4(\frac12)(3-(\frac12)^2)}{(1-(\frac12)^2)^3}$$
$$=-\frac{352}{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Another way to solve $\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$ without the substitution $y=\tan\left(\frac{x}{2}\right)$? Is there another way to solve an integral $$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$$ without the substitution $y=\tan\left(\frac{x}{2}\right)$?
$\large \int \frac{\sin^3(x)}{1+\cos^2(x)}\ dx$ is easily solved using the substitution $y=\cos(x)$.
What if the power of sine is even?
| Hint:
Bioche's rules suggest to use the substitution
$$t=\tan x,\quad \mathrm d x=\frac{\mathrm dt}{1+t^2}.$$
Indeed, as $\cos^2x=\frac 1{1+t^2}$, $\:\sin ^2x=\frac{t^2}{1+t^2}$, one obtains
$$\int\frac{\sin^4(x)}{1+\cos^2(x)}\,\mathrm dx=\int\frac{\frac{t^4}{(1+t^2)^2}}{1+\frac1{1+t^2}}\,\frac{\mathrm dt}{1+t^2}= \int\frac{t^4\,\mathrm dt}{(1+t^2)^2(2+t^2)}$$
which is comptuted with a decomposition into partial fractions:
$$\frac{t^4}{(1+t^2)^2(2+t^2)}=\frac{At+B}{1+t^2}+\frac{Ct+D}{(1+t^2)^2}+\frac{Et+F}{2+t^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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How to evaluate $\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}$ directly? The rational zeta series
$$
\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}=\frac{3}{2}-\ln \pi \tag1
$$
can be derived from other well known rational zeta series.
$$
\sum_{n=2}^{\infty}\frac{\left ( -1 \right )^{n}\left ( \zeta (n)-1 \right )}{n+1}=\frac{3}{2}+\frac{\gamma }{2}-\frac{\ln 8\pi}{2} \tag2
$$
$$
\sum_{n=2}^{\infty}\frac{\zeta (n)-1}{n+1}=\frac{3}{2}-\frac{\gamma }{2}-\frac{\ln 2\pi}{2} \tag3
$$
Zeta series (2) and (3) can be derived by integrating the Taylor series of logarithm of gamma function. Zeta series (2)+(3) gives
$$
\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{2n+1}=\frac{3}{2}-\frac{\ln 4\pi}{2} \tag4
$$
The zeta series below can be derived directly with the integral definition of $\zeta(2n)$.
$$
\sum_{n=1}^{\infty}\frac{\zeta (2n)}{(n+1)(2n+1)}=\frac{1}{2} \tag5
$$
From zeta series (5) we get
$$
\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{(2n+1)(2n+2)}=\frac{3}{4}-\ln 2 \tag6
$$
Zeta series (6) can be rewritten as
$$
\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{2n+1}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}=\frac{3}{4}-\ln 2 \tag7
$$
Tegether with zeta series (4), we get the result of zeta series (1).
Other than using known results of rational zeta series, how to evaluate zeta series (1) directly with elementary sum of series and integral?
I have tried several ways without success. One of my attempts:
$$
\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}x^{n+1}=\sum_{n=1}^{\infty}\sum_{k=2}^{\infty}\frac{1}{k^{2n}}\int_{0}^{x}t^{n}dt=\sum_{k=2}^{\infty}\int_{0}^{x}\sum_{n=1}^{\infty}\left ( \frac{t}{k^{2}} \right )^{n}dt \\
=\sum_{k=2}^{\infty}\int_{0}^{x}\frac{t}{k^{2}-t}dt=\sum_{k=2}^{\infty}\left ( k^{2}\ln\frac{k^{2}}{k^{2}-x}-x\right )
$$
It seems this attempt won't give any useful result for a closed form, although the sum of this series does converge to $(3/2-\ln\pi)$ slowly when setting $x=1$.
| We write
$$ S := \sum_{n=1}^{\infty} \frac{\zeta(2n)-1}{n+1} $$
for the sum to be computed.
1st Solution. We have
\begin{align*}
S
= \sum_{n=1}^{\infty} \frac{1}{n+1} \sum_{k=2}^{\infty} \frac{1}{k^{2n}}
= \sum_{k=2}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n+1} \frac{1}{k^{2n}}
= \sum_{k=2}^{\infty} \left( - k^2 \log \left( 1 - \frac{1}{k^2} \right) - 1 \right).
\end{align*}
In order to compute this, we write $S_K$ for the partial sums of the last step. Then
\begin{align*}
S_K
&= -K + 1 + \sum_{k=2}^{K} k^2 \log \left( \frac{k^2}{(k+1)(k-1)} \right) \\
&= -K + 1 + \sum_{k=2}^{K} 2 k^2 \log k - \sum_{k=3}^{K+1} (k-1)^2 \log k - \sum_{k=1}^{K-1} (k+1)^2 \log k \\
&= -K + 1 + \log 2 - K^2 \log(K+1) + (K+1)^2 \log K \\
&\quad + \sum_{k=2}^{K} (2 k^2 - (k-1)^2 - (k+1)^2 ) \log k \\
&= -K + 1 + \log 2 - K^2 \log\left(1 + K^{-1}\right) + (2K+1)\log K - 2 \log (K!).
\end{align*}
Now by the Stirling's approximation and the Taylor series of $\log(1+x)$,
$$ 2\log (K!) = \left(2K + 1\right) \log K - 2 K + \log(2\pi) + \mathcal{O}(K^{-1}) $$
and
$$ K^2 \log\left(1 + K^{-1}\right) = K - \frac{1}{2} + \mathcal{O}(K^{-1}) $$
as $K \to \infty$. Plugging this back to $S_K$, we get
$$ S_K = \frac{3}{2} - \log \pi + \mathcal{O}(K^{-1}) $$
and the desired identity follows by letting $K\to\infty$.
2nd Solution. We begin by noting that the Taylor expansion of the digamma function
\begin{align*}
\psi(1+z)
&= -\gamma + \sum_{k=1}^{\infty} (-1)^{k-1} \zeta(k+1) z^{k} \\
&= -\gamma + \zeta(2) z - \zeta(3) z^2 + \zeta(4) z^3 - \dots,
\end{align*}
holds for $|z| < 1$. Then by the Abel's Theorem,
\begin{align*}
S
&= \int_{0}^{1} \sum_{n=1}^{\infty} 2 (\zeta(2n)-1) x^{2n+1} \, \mathrm{d}x \\
&= \int_{0}^{1} x^2 \left( \psi(1+x) - \psi(1-x) - \frac{2x}{1-x^2} \right) \, \mathrm{d}x \\
&= \int_{0}^{1} x^2 \left( \psi(1+x) - \psi(2-x) + \frac{1}{1+x} \right) \, \mathrm{d}x, \tag{1}
\end{align*}
where the identity
$$ \psi(1+z) = \psi(z) + \frac{1}{z} \tag{2} $$
is used in the last step. Then by using the substitution $x\mapsto 1-x$, we get
$$ \int_{0}^{1} x^2 \psi(2-x) \, \mathrm{d}x
= \int_{0}^{1} (1-x)^2 \psi(1+x) \, \mathrm{d}x. $$
Plugging this back to $\text{(1)}$ and performing integration by parts,
\begin{align*}
S
&= \int_{0}^{1} (2x-1) \psi(1+x) \, \mathrm{d}x + \int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x \\
&= -2 \int_{0}^{1} \log\Gamma(1+x) \, \mathrm{d}x + \int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x.
\end{align*}
Now the integrals in the last step can be computed as
$$ \int_{0}^{1} \log\Gamma(1+x) \, \mathrm{d}x = -1 + \frac{1}{2}\log(2\pi)
\qquad \text{and} \qquad
\int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x = -\frac{1}{2} + \log 2. $$
For instance, the first integral can be computed by writing $\log\Gamma(x+1) = \log\Gamma(x) + \log x$ and applying the Euler's reflection formula. For a more detail, check this posting. Finally, plugging these back to $S$ proves the desired identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{u}{(u^2+1)^2}-\dfrac{u}{(u^2+1)^3}$$
Substitute $u^2+1=t$, $2u\ du=dt$, $u\ du=dt/2$
$$\int\frac{u^3}{(u^2+1)^3}du=\int \frac{dt/2}{t^2}-\int \frac{dt/2}{t^3}$$
$$=\frac12\dfrac{-1}{t}-\frac{1}{2}\dfrac{-1}{2t^2}$$
$$=-\dfrac{1}{2t}+\dfrac{1}{4t^2}$$
$$=-\dfrac{1}{2(u^2+1)}+\dfrac{1}{4(u^2+1)^2}+c$$
My question: Can I integrate this with suitable substitution? Thank you
| Hint: $\frac {u^{3}} {(u^{2}+1)^{3}}=u\frac {(u^{2}+1)-1} {(u^{2}+1)^{3}}=\frac u {(u^{2}+1)^{2}}-\frac u {(u^{2}+1)^{3}}$. Split then integral into two parts and use the substitution $x=1+u^{2}$ in both . The answer is $-\frac 1 {2(u^{2}+1)} -\frac 1 {4(u^{2}+1)^{2}}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 3
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For $x≠y$ and $2005(x+y) = 1$; Show that $\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$ Problem:
Let $x$ and $y$ two real numbers such that $x≠0$ ; $y≠0$ ; $x≠y$ and $2005(x+y) = 1$
*
*Show that $$\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$$
*Calculate $l$:
$$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$
For the first question, I tried to work it out with algebra; I solved for x through the equation given, then multiplied it by y and I got the value of $\frac{1}{xy} = 2005\left(\frac{1}{y-2005y^2}\right) $. Then I tried proving that $\frac{1}{y-2005y^2} =\frac{1}{x} + \frac{1}{y} $ but I failed at this.
| *
*$$\frac{1}{xy} = 2005(\frac{1}{x}+\frac{1}{y}) \iff \frac{1}{xy}=\frac{2005(x+y)}{xy}$$
which follows immediately from the condition
*$$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2=$$$$= \frac{y}{y-x}-({1}-\frac{x}{y})-\frac{x}{y-x}-(-1+\frac{y}{x})+\frac{y}{x} - \frac{x}{y}+2=$$$$=\frac{y-x}{y-x}+2=3$$
Explanation:
*
*First divide the fraction into two fractions (like $\frac{y-x}{y}=1-\frac{x}{y}$)
*Cancel out the opposite terms
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$ What I attempted thus far:
Multiply by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} \cdot \frac{\sqrt{1 + x\sin x} + \sqrt{\cos x}}{\sqrt{1 + x\sin x} + \sqrt{\cos x}} = \lim_{x \to 0} \frac{1 + x\sin x - \cos x}{x\tan x \cdot(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$$
From here I can’t see any useful direction to go in, if I even went in an useful direction in the first place, I have no idea.
| Similar to Varun Vejalla's answer, but without L'Hopital: Multiplying by the conjugate and replacing $\tan x$ by $\sin x/\cos x$, we have
$$\begin{align}
{\sqrt{1+x\sin x}-\sqrt\cos x\over x\tan x}
&={\cos x\over\sqrt{1+x\sin x}+\sqrt\cos x}\cdot{1+x\sin x-\cos x\over x\sin x}\\
&={\cos x\over\sqrt{1+x\sin x}+\sqrt\cos x}\left(1+{1-\cos x\over x\sin x}\right)\\
&={\cos x\over\sqrt{1+x\sin x}+\sqrt\cos x}\left(1+{1-\cos^2x\over x\sin x(1+\cos x)}\right)\\
&={\cos x\over\sqrt{1+x\sin x}+\sqrt\cos x}\left(1+{\sin x\over x}\cdot{1\over1+\cos x} \right)\\
&\to{1\over\sqrt{1+0}+\sqrt1}\left(1+1\cdot{1\over1+1}\right)={1\over2}\left(1+{1\over2}\right)={3\over4}
\end{align}$$
Note, we also multiplied by the conjugate of $1-\cos x$, made use of the trig identity $1-\cos^2x=\sin^2x$, and, finally, assumed the familiar limit ${\sin x\over x}\to1$ as $x\to0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding units of $\mathbb{Z}[\delta]$ with $\delta=\frac{1+\sqrt{d}}{2}$ and $d\equiv 1\pmod{4}.$ Let $d$ be a square free integer congruent to 1 modulo 4. Let
$$\delta=\frac{1+\sqrt{d}}{2}$$
(a) Prove that the subset $\mathbb{Z}[\delta]$ of the complex numbers defined here
$$\mathbb{Z}[\delta]:=\left\{a+b\delta:a,b\in\mathbb{Z}\right\}$$
is a subring.
(b) Assume that $d$ is negative. Prove that the units of $\mathbb{Z}[\delta]$ are only $\left\{\pm 1\right\}$, unless $d=-1$ or $-3$ dwhere the units are $\left\{\pm 1,\pm i\right\}$ and $\left\{\pm 1,\pm w\pm w^2\right\}$, respectively, where $w=\frac{-1+\sqrt{-3}}{2}$ is a non-trivial third root of unity.
For (a), I can't prove that $\mathbb{Z}[\delta]$ is closed under multiplication...
My solution for (b)
(b) Let $d\equiv 1\pmod{4}$ with $d\neq -3,-1$. Let $N:\mathbb{Z}[\delta]-\left\{0\right\}\to\mathbb{N}$ norm by $$N(a+b\delta)=(a+b\delta)(a+b\overline{\delta})$$
where $\overline{\delta}=\frac{1-\sqrt{d}}{2}$.
We will use it in a way that is in the domain of integrity $R$ it holds that $\forall r\in R^{\times}\Leftrightarrow N(r)=1$.
Let $a+b\delta\in\mathbb{Z}[\delta]^{\times}$. Then
\begin{align*}1&=N(a+b\delta)\\
&=(a+b\delta)(a+b\overline{\delta})\\
&=a^2+ab\overline{\delta}+ab\delta+b^2\delta\overline{\delta}\\
&=a^2+ab+b^2(\frac{1-d}{4})
\end{align*}
If $d\equiv 1\pmod{4}$ then $1-d=4q$ some $q\in\mathbb{N}$ because $d$ is negative. Therefore we have to
$$1=a^2+ab+b^2q.$$
If $ab\geq 0$, $1=a^2+ab+b^2q\geq a^2+b^2$ then $a=\pm 1\wedge b=0\vee a=0\wedge b=\pm 1$. But, if $a=0\wedge b=\pm 1$ then $1=q$. This implies that $1-d=4$ i.e. $d=-3$ a contradiction.
If $ab<0$ then $1=a^2+ab+b^2q=a^2+2ab+b^2+b^2(q-1)-ab=(a+b)^2-ab+b^2(q-1)$ then $1+ab=(a+b)^2+b^2(q-1)$. Since $(a+b)^2+b^2(q-1)\geq 0$ and $1+ab\leq 0$ it must be fulfilled that
$(a+b)^2+b^2(q-1)=0$ then $a+b=0\wedge b^2(q-1)=0$. If $b=0$, $a=0$ a contradiction because $a+b\delta\neq 0$.
Therefore $a=-b\wedge q=1\wedge b\neq 0$. Then $1-a^2=0$ implies $a=\pm 1$.
Therefeore $a+b\delta=\pm 1$. It is also evident that
$\pm 1\in\mathbb{Z}[\delta]^{\times}$. It is concluded that
$\mathbb{Z}[\delta]^{\times}=\left\{\pm 1\right\}$.
How would it be with the case $d = -1$ or $d = -3$? I can't do something similar.
Actualization 2. If $d=-1$ then $1=a^2+ab+\frac{b^2}{4}$. Now, I want proves that
$a = \pm 1, b = 0$ or $a = -1, b = 2$ or $a = 1, b = -2$
Actualization 3. If $d=-1$ in an earlier step we got the equation
$$1=a^2+ab+b^2(\frac{1-d}{4})$$
If $d=-1$ we have that
$$1=a^2+ab+\frac{b^2}{2}$$
From the latter, if
$b=0$ is obtained
$a=\pm 1$.
If $b\neq 0$, as $b=\pm \sqrt{2-a^2}-a$. and $2-a^2\geq 0$ you should have to
$a=1$ and $b=-1$ or $a=-1$ and $b=1$.
Therefore , $\mathbb{Z}[\delta]^{\times}\subset \left\{\pm 1,\pm 1\mp 2\delta\right\}$ moreover $\pm 1\mp 2\delta=\pm 1\mp 2(\frac{1+i}{2})=\pm 1\mp(1+i)=\pm i$. Finally, with $N(\pm 1), N(\pm i)=1$ the equality holds.
Actualization 3. If $d=-3$ I have this:
If $b=0$ then $a=\pm 1$
If $b\neq 0$, $b=\frac{1}{2}(\pm \sqrt{4-3a^2}-a)$ then $4-3a^2\geq 0$, so $a\in \left\{0,\pm 1\right\}$
If $a=0$ then $b=\pm 1$
If $a=\pm 1$ then $b=\mp 1$, respectively.
Therefore, $\pm \delta$ and $\pm 1\mp \delta\in (\mathbb{Z}[\delta])^{\times}$ with $\delta=\frac{1+\sqrt{-3}}{2}.$
The problem is that in the statement, it is mentioned that
$(\mathbb{Z}[\delta])^{\times}\left\{\pm 1,\pm w\pm w^2\right\}$ different from what I found.
| Hint for (a): $$\delta^2={d+1+2\sqrt d\over4}={d-1\over4}+\delta$$
| {
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How to prove that if $x>0$ and $y>0$, then $\sqrt{x}+\sqrt{y}>\sqrt{x+y}$, using the relation of arithmetic and geometric means?
How to prove that if $x>0$ and $y>0$, then $$\sqrt{x}+\sqrt{y}>\sqrt{x+y}\,,$$ using the relation of arithmetic and geometric means?
I started by showing that if $x>0$ and $y>0$, based on the relation of arithmetic and geometric means, $\dfrac{x+y}{2}\ge\sqrt{xy}$.
Hence, $x+y\ge2\sqrt{xy}$.
I am now stuck here and don't know what must be the next step.
Any suggestions or comments will be much appreciated.
|
Theorem (A Baby Version of the Triangle Inequality). Let $a$, $b$, $c$, and $d$ be real numbers. We have
$$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\geq \sqrt{(a+c)^2+(b+d)^2}\,.$$
The equality holds if and only if
*
*$(a,b)=(0,0)$, or
*there exists $\lambda\geq 0$ such that $(c,d)=(\lambda a,\lambda b)$.
Let $x$ and $y$ be nonnegative real numbers. Note that
$$\sqrt{x}+\sqrt{y}=\sqrt{\sqrt{x}^2+0^2}+\sqrt{0^2+\sqrt{y}^2}\,.$$
By the Baby Triangle Inequality above,
$$\sqrt{\sqrt{x}^2+0^2}+\sqrt{0^2+\sqrt{y}^2}\geq \sqrt{(\sqrt{x}+0)^2+(0+\sqrt{y})^2}\,.$$
That is,
$$\sqrt{x}+\sqrt{y}\geq \sqrt{x+y}\,.\tag{#}$$
By the equality conditions of the Baby Triangle Inequality, (#) is an equality if and only if $x=0$ or $y=0$.
Now, I shall prove the Baby Triangle Inequality using the AM-GM Inequality to fulfill the OP's request that the AM-GM Inequality must be used. By squaring the required inequality, what we need to prove is equivalent to
$$\sqrt{a^2+b^2}\sqrt{c^2+d^2}\geq ac+bd\,.$$
We, in fact, have a stronger inequality:
$$\sqrt{a^2+b^2}\sqrt{c^2+d^2}\geq |a|\,|c|+|b|\,|d|\,. \tag{*}$$
By squaring the inequality above, we know that (*) is equivalent to
$$a^2d^2+b^2c^2\geq 2\,|a|\,|b|\,|c|\,|d|\,,$$ which is true by the AM-GM Inequality:
$$\frac{a^2d^2+b^2c^2}{2}=\frac{|ad|^2+|bc|^2}{2}\geq \sqrt{|ad|^2\cdot |bc|^2}=|a|\,|b|\,|c|\,|d|\,.$$
| {
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Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\\
&=\int \dfrac{3\sec^2\theta d\theta}{81\sec^4\theta}\\
&=\dfrac{1}{27}\int \cos^2\theta d\theta\\
&=\dfrac{1}{27}\int \frac{1+\cos2\theta}{2} d\theta\\
&=\dfrac{1}{54}\left(\theta+\frac{\sin2\theta}{2}\right)+C
\end{align*}
This is where I got stuck. How can I get the answer in terms of $x$?
Can I solve it by other methods?
| substitute $\theta=\tan^{-1}\left(\frac{x-2}{3}\right)$ & $$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{2\left(\frac{x-2}{3}\right)}{1+\left(\frac{x-2}{3}\right)^2}=\frac{6(x-2)}{x^2-4x+13}$$
After substituting $\theta$ and $\sin2\theta$, you will get final answer
$$I=\frac{1}{54}\tan^{-1}\left(\frac{x-2}{3}\right)+\frac{x-2}{18(x^2-4x+13)}+C$$
| {
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How to solve polynomial rational relations for $y$ (e.g $\sqrt{4-3y-y^2} = x(y+4)$)? From time to time, I struggle to solve polynomial relations for $y$.
A trivial example is :
$$ \frac{y}{x} = x \iff y = x^2$$
Easy.
But consider this relation:
$$ \sqrt{4-3y-y^2} = x(y+4)$$
No matter how much I mess around it, seems impossible to bring it in $y = f(x)$ form.
*
*$ \frac{\sqrt{4-3y-y^2}}{(y+4)}= x $
*$ 4 - 3y - y^2 = x^2(y^2 + 8y + 16) \iff (x^2-1)y^2+y(-8x-3)+4(1-4x) = 0$
Is there a trivial methodology that I am missing or is it indeed impossible to inverse some relations?
| $$ \sqrt{4-3y-y^2} = x(y+4)$$
$$ {4-3y-y^2} = x^2(y+4)^2$$
$${ y^2+3y-4} = -x^2(y+4)^2$$
$$(y+4)(y-1) = -x^2(y+4)^2$$
$$\frac{y-1}{y+4} = -x^2$$
$$\frac{1-y}{y+4} = x^2$$
$$\sqrt{\frac{1-y}{y+4}} = x$$
$$ \frac{1-y}{y+4} = x^2$$
$$1-y = x^2y + 4x^2$$
$$1-y = x^2y + 4x^2$$
$$-y = x^2y + 4x^2-1$$
$$y = 1-x^2y - 4x^2 $$
$$y + x^2y = 1-4x^2$$
$$y(1+x^2) = 1-4x^2$$
$$y = \frac{1-4x^2}{1+x^2}$$
| {
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Newton's evaluation of $1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots$ How might have Newton evaluated the following series?
$$\sqrt{2} \, \frac{\pi}{4} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots$$
The method of the this thread applies by setting $x=\pi/4$ in the Fourier series for $f(x) = \pi/2 - x/2$ and then subtracting the extraneous terms (which are a multiple of the Gregory-Leibniz series for $\pi/4$).
I read that this series appears in a letter from Newton to Leibniz. However, I do not have access the letter which appears in this volume.
| Nick Mackinnon gives what appears to be the story in an article that appeared in the Mathematical Gazette in March 1992 (Vol. 76, No. 475), entitled "Newton's Teaser." He writes that Newton conjured up the poser, in fits and starts, in response to Leibniz's series
$$
1 - \frac13 + \frac15 - \frac17 + \cdots = \frac\pi4
$$
The other answers here are not off the mark, really. Newton was able to evaluate—determine areas for—among other things, expressions of the form
$$
\int \frac{dx^{\eta-1}}{e+fx^\eta+gx^{2\eta}}
$$
(Actually, he used $z$ instead of $x$, but I'll use the more usual $x$, because that's what Mackinnon does in most of his exposition other than direct cites of Newton.) He pointed out that by letting $\eta = 1$, $e = g = 1$, and $f = 0$, then the result can be used to evaluate Leibniz's series, and he further suggested that setting $\eta = 1$, $e = g = 1$, and $f^2 = 2eg$ (i.e., $f = \sqrt2$) enables the evaluation of the series in question:
$$
1 + \frac13 - \frac15 - \frac17 + \frac19 + \frac{1}{11} - \cdots
$$
Following Newton's suggestion, we have
\begin{align}
\int_{x=-1}^1 \frac{dx}{1+\sqrt2x+x^2}
& = \int_{x=-1}^1 \frac{dx}{\left(x+\frac{1}{\sqrt2}\right)^2
+\left(\frac{1}{\sqrt2}\right)^2} \\
& = \left. \sqrt2 \arctan \left(
\frac{x+\frac{1}{\sqrt2}}{\sqrt2}\right) \right]_{x=-1}^1 \\
& = \sqrt2 \left[ \arctan \left( \frac12 + \frac{1}{\sqrt2} \right)
- \arctan \left( \frac12 - \frac{1}{\sqrt2} \right) \right] \\
& = \sqrt2 \left( \frac{3\pi}{8} + \frac\pi8 \right) \\
& = \frac{\pi}{\sqrt2}
\end{align}
Newton apparently recorded in his worksheets the factorization
$$
1+x^4 = (1+\sqrt2x+x^2)(1-\sqrt2x+x^2)
$$
so evidently he means to evaluate the integral alternatively as
\begin{align}
\require{cancel}
\int_{x=-1}^1 \frac{dx}{1+\sqrt2x+x^2}
& = \int_{x=-1}^1 \frac{1-\sqrt2x+x^2}{1+x^4} \, dx \\
& = \int_{x=-1}^1 \frac{1+x^2}{1+x^4} \, dx
- \cancel{\int_{x=-1}^1 \frac{\sqrt2x}{1+x^4} \, dx}
\qquad \text{because $\sqrt2x$ is odd} \\
& = 2\int_{x=0}^1 \frac{1+x^2}{1+x^4} \, dx \qquad \text{because this is even}
\end{align}
Combining these gives us
\begin{align}
\frac{\pi}{2\sqrt2}
& = \int_{x=0}^1 \frac{1+x^2}{1+x^4} \, dx \\
& = \int_{x=0}^1 \frac{dx}{1+x^4} + \int_{x=0}^1 \frac{x^2\,dx}{1+x^4} \\
& = \int_{x=0}^1 1-x^4+x^8-x^{12}+\cdots \, dx
+ \int_{x=0}^1 x^2-x^6+x^{10}-x^{14}+\cdots \, dx \\
& = \left. x-\frac{x^5}{5}+\frac{x^9}{9}
-\frac{x^{13}}{13}+\cdots \right]_{x=0}^1
+ \left. \frac{x^3}{3}-\frac{x^7}{7}+\frac{x^{11}}{11}
-\frac{x^{15}}{15}+\cdots \right]_{x=0}^1 \\
& = 1+\frac13-\frac15-\frac17+\frac19+\frac{1}{11}-\frac{1}{13}-\frac{1}{15}
+ \cdots
\end{align}
Mackinnon adduces some circumstantial evidence that strongly suggests Leibniz never cracked Newton's little chestnut.
| {
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Proving $\frac1{2\pi} \int_0^{2\pi} \frac{R^2-r^2}{R^2-2Rr\cos\theta+r^2} d\theta =1$ by integrating $\frac{R+z}{z(R-z)}$ without residue theorem. I was given the function:
$$ \frac{R+z}{z(R-z)} $$
And I was asked to integrate it around a closed contour to prove:
$$\frac1{2\pi} \int_0^{2\pi} \frac{R^2-r^2}{R^2-2Rr\cos\theta+r^2} d\theta =1$$
I've seen people get a proof quite easily by using the residue theorem, but I have not studied it yet so I am not supposed to do it.
My attempt:
Let $\gamma = re^{it}$,
$$\int_\gamma f dz = \int_\gamma \frac1z + \frac2{R-z} dz$$
$$\Rightarrow \int_\gamma f dz = \int_0^{2\pi} \frac{ire^{it}}{re^{it}}dt + \int_0^{2\pi} \frac{2ire^{it}}{R-re^{it}}dt$$
$$ = 2\pi i + \int_0^{2\pi} \frac{2Rr\cos t + 2r}{R^2+2Rr\cos t + r^2} dt$$
But I don't know what else should I do. Any ideas?
Edit: Sorry I had a typo, the function to integrate was $ \frac{R+z}{z(R-z)} $ and not $ \frac{R-z}{z(R-z)} $
| We could avoid complex analysis altogether
$$I = \frac{1}{\pi}\int_0^\pi \frac{R^2-r^2}{R^2+r^2 - 2Rr\cos\theta}\:d\theta$$
$$ = \frac{1}{\pi}\int_0^\pi \frac{R^2-r^2}{R^2+r^2\left(\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}\right) - 2Rr\left(\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}\right)}\:d\theta$$
$$\frac{1}{\pi}\int_0^\pi \frac{(R-r)(R+r)}{(R-r)^2\cos^2\frac{\theta}{2}+(R+r)^2\sin^2\frac{\theta}{2}}\:d\theta = \frac{2}{\pi}\int_0^\pi \frac{\left(\frac{R-r}{R+r}\right)\cdot\frac{1}{2}\sec^2\theta\:d\theta}{\left(\frac{R-r}{R+r}\right)^2+\tan^2\frac{\theta}{2}}$$
$$= \frac{2}{\pi}\tan^{-1}\left[\left(\frac{R-r}{R+r}\right)\tan\frac{\theta}{2}\right]\Biggr|_0^{\pi^-} = 1$$
with the assumption that $|R|\neq|r|$
| {
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Rolling Dice Game, Probability of Ending on an Even Roll The game is described as follows. $A$ and $B$ take turns rolling a fair six sided die. Say $A$ rolls first. Then if $A$ rolls {1,2} they win. If not, then $B$ rolls. If $B$ rolls {3,4,5,6} then they win. This process repeats until $A$ or $B$ wins, and the game stops.
What is the probability that the game ends on an even turn when $A$ rolls first?
Now the book gives the answer as $\frac{4}{7}$, however, when try to calculate I end up with $\frac{2}{11}$.
Below is my work:
To calculate this probability, we decompose the event into two disjoint events, (a) the event where $A$ wins on an even roll, and (b) the event where $B$ wins on an even roll.
(a) Now, the probability $A$ wins can be calculated as follows
\begin{align*}
\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{1}{3}\biggr) + \biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{1}{3}\biggr) + \dots = \sum_{k=0}^\infty \biggr(\frac{2}{9}\biggr)^{2k+1}\frac{1}{3}\\
= \sum_{k=0}^\infty \frac{2}{27}\biggr(\frac{2}{9}\biggr)^{2k} = \sum_{k=0}^\infty \frac{2}{27}\biggr(\frac{4}{81}\biggr)^k = \frac{2}{27}\cdot \frac{1}{1- \frac{4}{81}} = \frac{6}{77}.
\end{align*}
(b) Similarly we calculate the probability $B$ wins on an even roll as
\begin{align*}
\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3}\cdot \frac{2}{3}\biggr) + \biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3}\cdot\frac{2}{3}\biggr) + \dots = \sum_{k=0}^\infty \biggr(\frac{2}{9}\biggr)^{2k+1}\frac{4}{9}\\
= \sum_{k=0}^\infty \frac{8}{81}\biggr(\frac{2}{9}\biggr)^{2k} = \sum_{k=0}^\infty \frac{8}{81}\biggr(\frac{4}{81}\biggr)^k = \frac{8}{81}\cdot \frac{1}{1- \frac{4}{81}} = \frac{8}{77}.
\end{align*}
Therefore, it follows that the probability of the game ending on an even number of rolls is
\begin{equation*}
\frac{6}{77} + \frac{8}{77} = \frac{2}{11}.
\end{equation*}
Am I missing something?
| The problem is not clear as stated.
Interpretation $\#1$: If you interpret it as "find the probability that the game end in an evenly numbered round" you can reason recursively.
Let $P$ denote the answer. The probability that the game ends in the first round is $\frac 26+\frac 46\times \frac 46=\frac 79$. If you don't end in the first round, the probability is now $1-P$. Thus $$P=\frac 79\times 0 +\frac 29\times (1-P)\implies \boxed{P=\frac 2{11}}$$
as in your solution.
Interpretation $\#2$: If the problem meant "find the probability that $B$ wins given that $A$ starts" that too can be solved recursively. Let $\Psi$ denote that answer and let $\Phi$ be the probability that $B$ wins given that $B$ starts. Then $$\Psi=\frac 46\times \Phi$$ and $$\Phi=\frac 46 +\frac 26\times \Psi$$ This system is easily solved and yields $$\boxed {\Psi=\frac 47}$$ as desired.
| {
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Evaluating $\iint dx\,dy$ over the region bounded by $y^2=x$ and $x^2+y^2=2x$ in the first quadrant
Identify the region bounded by the curves $y^2=x$ and $x^2+y^2=2x$, that lies in the first quadrant and evaluate $\iint dx\,dy$ over this region.
In my book the solution is like:
$$\begin{align}\\
\iint dx\,dy &=\int_{x=0}^1\int_{y=\sqrt x}^{\sqrt{2x-x^2}} \, dx \, dy\\
&=\int_{x=0}^1 \big[y\big]_{\sqrt x}^{\sqrt{2x-x^2}}\,dx\\
&=\int_0^1\left(\sqrt{2x-x^2}-\sqrt{x}\right)\,dx\\
&{\begin{aligned}\\
=\int_0^1\sqrt{1-x^2}\,dx-\int_0^1\sqrt{x}&\,dx\text{(applying} \int_0^af(x)\,dx=&\int_0^af(a-x)\,dx \text{ in the first part)}\\
\end{aligned}\\}\\
&=\left[\frac{\sqrt{1-x^2}}{2}+\sin^{-1}x\right]_0^1-\left[\frac{x^{\frac{3}{2}}}{\frac32}\right]_0^1\\
&=\frac{\pi}{2}-\frac12-\frac23(1-0)\\
&=\frac{\pi}{2}-\frac76\\
\end{align}\\
$$
And I did it like:
$$\begin{align}\\
\iint dx\,dy &=\int_{x=0}^1\int_{y=\sqrt x}^{\sqrt{2x-x^2}}dx\,dy\\
&=\int_0^1\left(\sqrt{2x-x^2}-\sqrt{x}\right)\,dx\\
&=\int_0^1\sqrt{1-(x-1)^2}\,dx-\int_0^1\sqrt{x}\,dx\\
&{\begin{aligned}\\
=&\left[\frac{x-1}{2}\sqrt{1-(x-1)^2}+\frac12\sin^{-1}(x-1)\right]_0^1&-\left[\frac23x^{\frac32}\right]_0^1\\
\end{aligned}\\}\\
&=-\frac{\pi}{4}-\frac23\\
\end{align}\\
$$
Which one is correct?
| $$\int\limits_0^1\left(\sqrt{2x-x^2}-\sqrt{x}\right)\,dx =\\=
\int_\limits0^1\sqrt{2x-x^2}\,dx-\int\limits_0^1\sqrt{x}\,dx = \frac{\pi}{4} - \frac{2}{3}
$$
For first one:
$$\int_\limits0^1\sqrt{2x-x^2}\,dx = \int_\limits0^1\sqrt{1-(x-1)^2}\,dx=\\=\int\limits_{-1}^{0}\sqrt{1-y^2}\,dy=\int\limits_{-\frac{\pi}{2}}^{0}\cos^2t\,dt = \int\limits_{-\frac{\pi}{2}}^{0}\frac{1+\cos 2t}{2}\,dt=\\=\frac{1}{2}\left(\int\limits_{-\frac{\pi}{2}}^{0}\,dt+ \int\limits_{-\frac{\pi}{2}}^{0}\cos 2t\,dt \right) =\frac{1}{2}\left(\frac{\pi}{2} +0\right)= \frac{\pi}{4}
$$
| {
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Evaluate $\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$
Evaluate:
$$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$$
I could find the integral by setting it equal to $$\frac{ax+b}{(1+x^3)^{1/2}}$$
and differentiating both sides w.r.t.$x$ as
$$\frac{2-x^3}{(1+x^3)^{3/2}}=\frac{a(1+x^3)^{3/2}-(1/2)(ax+b)3x^2(1+x^3)^{-1/2}}{(1+x^3)}$$$$=\frac{a-ax^3/2-3bx^2}{(1+x^3)^{3/2}}$$
Finally by setting $a=2,b=0$, we get $$I(x)=\frac{2x}{(1+x^3)^{1/2}}+C$$
The question is: How to do it otherswise?
| $$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx=\int \frac{\frac{1}{x^3}(2-x^3)}{\frac1{x^3}\left(1+x^3\right)^{3/2}} \ dx$$
$$=\int \frac{\left(\frac{2}{x^3}-1\right)dx}{\left(\frac{1}{x^2}+x\right)^{3/2}}$$
$$=-\int \frac{d\left(\frac{1}{x^2}+x\right)}{\left(\frac{1}{x^2}+x\right)^{3/2}}$$
$$=- \frac{\left(\frac{1}{x^2}+x\right)^{-\frac32+1}}{-\frac32+1}+C$$
$$=\frac{2}{\sqrt{\frac1{x^2}+x}}+C$$
$$=\bbox[15px, #ffd, border:1px solid green]{\frac{2x}{\sqrt{1+x^3}}+C}$$
| {
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Show that the solution of the equation $x^5-2x^3-3=0$ are all less than 2 (using proof by contradiction).
The question is "Show that the solution of the equation $x^5-2x^3-3=0$ are all less than 2."
I have attempted to answer this question using proof by contradiction and I think my answer is either wrong or not a well written solution. I would like to know if I solved it right and if I did, I would like some advice on how I can improve writing proofs.
My attempt:
Assume to the contrary that solution of this equation is greater then or equal to 2.
Let $x=\frac 2p$.
Then we have
$ (\frac 2p)^5-2(\frac 2p)^3-3=0 $
$ \frac {2^5}{p^5} - \frac {2^4}{p^3} - 3 = 0$
We now consider two case: when $p=1$ and $p<1$.
when $p=1$:
$ 2^5 - 2^4 - 3 = 32 - 16 - 3 = 13$.
Since $x = 2$ is not a solution this is a contradiction.
When $p<1$:
If $p<1$, then we know $1/p>1.$ This implies $\frac {1}{p^{n+1}} > \frac {1}{p^n}.$
Since $\frac 1{p^3}(2^5-2^4)-3 > 2^5-2^4-3 = 13 > 0$, it is clear that the inequality
$ \frac 1{p^5}(2^5)- \frac 1{p^3} 2^4-3 > \frac 1{p^3}(2^5-2^4)-3 > 2^5-2^4-3 = 13 > 0$ holds
Since for all $x > 2$ is not a solution this is a contradiction.
Thus, solution $x$ is less then 2.
| Here is another way of doing this:
If $x\gt 2$ then $x^5\gt 4x^3\left(=2x^3+2x^3\right) \gt 2x^3+16\gt 2x^3+3$
I have added this because it clicked into my head immediately as a way of estimating and thinking though different approaches gives you a range of tools to solve such problems.
| {
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If the largest positive integer is n such that $\sqrt{n - 100} + \sqrt{n + 100}$ is a rational no. , find the value of $\sqrt{n - 1}$ . So here is the Problem :-
If the largest positive integer is n such that $\sqrt{n - 100} + \sqrt{n + 100}$ is a rational no. , find the value of $\sqrt{n - 1}$ .
What I tried :- I think that for $\sqrt{n - 100} + \sqrt{n + 100}$ to be a rational no. , both $(n - 100)$ and $(n + 100)$ have to be squares. Suppose :- $(n - 100)$ = $k^2$ and $(n + 100)$ = $m^2$ for some positive integers $k,m$ , and in the end I could only deduce that $(m + 10)(m - 10) = k^2 + 100$ , but then I couldn't proceed .
Also by guesswork, I could deduce that for $n = 125$, both nos. do become squares, although I don't know whether $n = 125$ is the highest or not.
Any hints or explanations to this problem will be greatly appreciated !
| Let $\sqrt{n-100} + \sqrt{n+100} = p$, where $p$ is rational.
$$\implies 2n + 2\sqrt{n^2 - 10000} = p^2$$
But that must mean that $2\sqrt{n^2 - 10000}$ is rational.
Which must mean that $\sqrt{n^2 - 10000}$ is rational.
$$\implies n^2 - 10000 = k^2$$
$$\implies (n+k)(n-k) = 10000$$ The problem requires us to maximize $n$, notice that we'll get the maximum value of $n$ if we split $10000 = 5000 \times 2$ and set $n+k = 5000$ and $n-k = 2$ to get $n = 2501$.
Hence, $\boxed{\sqrt{n-1} = 50}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Proofs by Induction: are my two proofs correct? I have been trying to understand how proof by mathematical induction works, and I am struggling a bit. But, I think I am understanding it and I just want to verify that what I am doing is correct (and if not, why?)
I have attached a screenshot (as a link) of my problem (black ink) and my work (red ink). My main issue is understanding what the final conclusion should be. What I did was check to see if the left and right side of the problem were equal after assuming $k + 1$ is true, and adding the appropriate terms to both sides, and simplifying.
So, in my final steps of the induction phase, my question is, did I reach the right result?
Prove: $1 + 3 + 6 + \cdots + \dfrac{n(n + 1)}{2} = \dfrac{n(n + 1)(n + 2)}{6}$.
Base: $P(1) = 1$.
Induction:
\begin{align*}
\underbrace{1 + 3 + 6 + \cdots + \frac{k(k + 1)}{2}}_{\dfrac{k(k + 1)(k + 2)}{6}} + \frac{(k + 1)(k + 2)}{2} & = \frac{(k + 1)(k + 2)(k + 3)}{6}\\
\frac{k(k + 1)(k + 2)}{6} + \frac{(k + 1)(k + 2)}{2} & = \frac{(k + 1)(k + 2)(k + 3)}{6}\\
\frac{k(k + 1)(k + 2) + 3(k + 1)(k + 2)}{6} & = \frac{(k + 1)(k + 2)(k + 3)}{6}\\
\frac{(k + 1)(k + 2)(k + 3)}{6} & = \frac{(k + 1)(k + 2)(k + 3)}{6}
\end{align*}
Prove: $5 + 10 + 15 + \cdots + 5n = \dfrac{5n(n + 1)}{2}$
Base: $P(1) = 5$
Induction:
\begin{align*}
5 + 10 + 15 + \cdots + 5k + 5(k + 1) & = \frac{5k(k + 1)}{2} + 5(k + 1)\\
\frac{5k(k + 1)}{2} + 5(k + 1) & = \frac{5k(k + 1)}{2} + 5(k + 1)
\end{align*}
My problem and my work
| Yes.... it is okay but it needs a bit of exposition-- that is an explanation of what you are doing and why you are doing it and why it proves what you want.
I would add after you write the word "Induction:" I would expound words to the effect
Induction step: Supose $P(k)$.
Suppose that for some $k$ that $1+ 3+ 6 + ... +\frac {k(k+1)}2 = \frac {k(k+1)(k+2)}6$. Then we must prove $P(k+1)$ or in other words that $1+3 +.....+\frac {(k+1)(k+2)}2 = \frac {(k+1)(k+2)(k+3)}6$.
Then I'd not that as that is what we are attempt to prove I wouldn't write is an "$= \frac {(k+1)(k+2)(k+3)}6$" every step of the way. It's not clear you are attempting to verify a result and looks as though you are stating a bunch of things without cause.
$\require{cancel}$
$\begin{align*}
\underbrace{1 + 3 + 6 + \cdots + \frac{k(k + 1)}{2}}_{\dfrac{k(k + 1)(k + 2)}{6}} + \frac{(k + 1)(k + 2)}{2} =& \color{red}{\cancel{= \frac{(k + 1)(k + 2)(k + 3)}{6}}}\\
\frac{k(k + 1)(k + 2)}{6} + \frac{(k + 1)(k + 2)}{2} =&\color{red}{\cancel{= \frac{(k + 1)(k + 2)(k + 3)}{6}}}\\
\frac{k(k + 1)(k + 2) + 3(k + 1)(k + 2)}{6} =& \color{red}{\cancel{= \frac{(k + 1)(k + 2)(k + 3)}{6}}}\\
\frac{(k + 1)(k + 2)(k + 3)}{6} & \color{red}{\cancel{= \frac{(k + 1)(k + 2)(k + 3)}{6}}}
\end{align*}$
And we'd be done.
But words like "And therefore we have shown that $P(k) \implies P(k+1)$, and thus the induction step is valid" wouldn't hurt.
| {
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"answer_count": 2,
"answer_id": 1
} |
Find $\nabla_x\left\langle L,Z-\begin{bmatrix}Tu&x\\ x^H&t\end{bmatrix}\right\rangle+p/2\left\|Z-\begin{bmatrix}Tu&x\\ x^H&t\end{bmatrix}\right\|^2$ I am currently reading Atomic norm denoising with applications to line spectral estimation by Bhaskar et al.
In appendix E, an ADMM algorithm is presented to solve the SDP program
\begin{equation*}
\min_{t, u, x, Z} \frac{1}{2} \| x - y \|_2^2 + \frac{\tau}{2}(t + u_1)
\quad \text{s.t.} \quad
Z = \begin{bmatrix} T(u) & x \\
x^{\mathsf{H}} & t \end{bmatrix}, \
Z \succeq 0,
\end{equation*}
where $\tau > 0$ is a regularisation parameter and $T(u)$ is the Hermitian Toeplitz matrix, whose first row is $u$.
According to the paper
augmented Lagrangian is
\begin{equation*}
L_{\rho}(t, u, x, Z, \Lambda)
:= \frac{1}{2} \| x - y \|_2^2
+ \frac{\tau}{2}(t + u_1)
+ \left\langle \Lambda, Z - \begin{bmatrix} T(u) & x \\
x^{\mathsf{H}} & t \end{bmatrix} \right\rangle_F
+ \frac{\rho}{2} \left\| Z - \begin{bmatrix} T(u) & x \\
x^{\mathsf{H}} & t \end{bmatrix} \right\|_F^2,
\end{equation*}
where $\rho > 0$.
The ADMM algorithm consists of the update steps
\begin{align*}
(t^{k + 1}, u^{k + 1}, x^{k + 1})
& \leftarrow \text{argmin}_{t, u, x} L_{\rho}(t, u, x, Z^k, \Lambda^k) \\
Z^{k + 1}
& \leftarrow \text{argmin}_{Z \succeq 0} L_{\rho}(t^{k + 1}, u^{k + 1}, x^{k + 1}, Z, \Lambda^k) \\
\Lambda^{k + 1}
& \leftarrow \Lambda^k + \rho\left( Z^{k + 1} - \begin{bmatrix} T(u^{k + 1}) & x^{k + 1} \\ \overline{x^{k + 1}} & t^{k + 1} \end{bmatrix}\right).
\end{align*}
These updates have a closed form:
\begin{gather*}
t^{k + 1}
= Z_{n + 1, n + 1}^{k} + \frac{1}{\rho} \left( \Lambda_{n + 1, n + 1}^{k} - \frac{\tau}{2}\right) \\
x^{k + 1}
= \frac{1}{2 \rho + 1}\left(y + 2 \rho z_1^k + 2\lambda_1^k\right) \\
u^{k + 1}
= W\left(T^*\left (Z_0^k + \frac{1}{\rho} \Lambda_0^k\right) - \frac{\tau}{2 \rho} e_1\right),
\end{gather*}
where $W$ is a diagonal $n \times n$-matrix with the entries
\begin{equation*}
W_{i i}
:= \begin{cases}
\frac{1}{n}, & i = 1, \\
\frac{1}{2(n - i + 1)}, & i > 1.
\end{cases}
\end{equation*}
and we partition each $Z$ as
\begin{equation}
Z =
\begin{bmatrix}
Z_0 & z_1 \\
z_1^{\mathsf{H}} & Z_{n + 1, n + 1}
\end{bmatrix} \tag{1}
\end{equation}
and $\Lambda$ in the same manner.
My Question
I can't find the closed form for the $x$ update.
What I've tried
I got, dropping all terms independent of $x$,
$$
\frac{\partial}{\partial x} \frac{1}{2} \| x - y \|_2^2 + \frac{\tau}{2}(t + u_1)
= x - y
$$
and using the bilinearity of the inner product and linearity of the trace,
\begin{align*}
\frac{\partial}{\partial x} \left\| Z - \begin{bmatrix} T(u) & x \\ x^{\mathsf{H}} & t \end{bmatrix} \right\|_F^2
& = \frac{\partial}{\partial x}
\left(\text{Tr}\left(\begin{bmatrix} T(u) & x \\ x^{\mathsf{H}} & t \end{bmatrix}^{\mathsf{H}} \begin{bmatrix} T(u) & x \\ x^{\mathsf{H}} & t \end{bmatrix}\right)
- 2 \Re\left( \text{Tr}\left(\begin{bmatrix} T(u) & x \\ x^{\mathsf{H}} & t \end{bmatrix}^{\mathsf{H}} Z \right) \right)\right) \\
& = \frac{\partial}{\partial x}
\left(\text{Tr}\left(\begin{bmatrix} T(\bar{u}) & x^{\mathsf{H}} \\ x & t \end{bmatrix} \begin{bmatrix} T(u) & x \\ x^{\mathsf{H}} & t \end{bmatrix}\right)
- 2 \Re\left( \text{Tr}\left(\begin{bmatrix} T(\bar{u}) & x^{\mathsf{H}} \\ x & t \end{bmatrix} Z \right) \right)\right) \\
& = \frac{\partial}{\partial x} \left(\sum_{k = 1}^{n} \overline{x_k}^2 + x_k^2
- 2 \Re\left( 2 \sum_{k = 1}^{d} z_{d + 1, k} \Re(x_k) \right)\right) \\
& = \frac{\partial}{\partial x} \left(\sum_{k = 1}^{n} \overline{x_k}^2 + x_k^2 \right)
- 4 \frac{\partial}{\partial x} \left(\sum_{k = 1}^{d} \Re(z_{d + 1, k}) \Re(x_k) \right).
\end{align*}
Using Wirtinger calculus as described here, I got $\frac{\partial}{\partial x_k} x_k^2 + \overline{x_k}^2 = x_k$ and $\frac{\partial}{\partial x_k} \Re(x_k) = \frac{1}{2}$ and thus the above expression reduces to
\begin{align}
x - 2 z_{1}.
\end{align}
In conclusion we have
\begin{align}
\frac{\partial}{\partial x} L_{\rho}(t, u, x, Z, \Lambda)
= x - y - 2 \lambda_1 + \frac{\rho}{2} \cdot (x - 2 z_1)
\end{align}
and setting this to zero yields
\begin{align}
\rho z_1 + y + 2 \lambda_1 = \left(1 + \frac{\rho}{2}\right) x,
\end{align}
which is equivalent to
\begin{align}
x = \frac{2}{\rho + 2}\left(\rho z_1 + y + 2 \lambda_1\right),
\end{align}
which is different from the $x$-update in the paper.
Where have I gone wrong?.
| The error is in the derivative of the last term. That should be $\rho/2 \cdot4(x-z_1) = 2\rho(x-z_1)$.
The easiest way to see this is by writing the Frobenius norm as the sum of the squared components (or squared lenghts of the matrix component for the complex plane). The only places where $x$ occurs are the first column where you have $z_1^H - x^H$ and the first row where you have $z_1-x$, from which the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving the system $\sqrt{x} + y = 7$, $x + \sqrt{y} = 11$ I want to solve the following nonlinear system of algebraic equations. Indeed, I am curious about a step by step solution for pedagogical purposes. I am wondering if you can come up with anything. I tried but to no avail.
\begin{align*}
\sqrt{x} + y &= 7 \\
x + \sqrt{y} &= 11
\end{align*}
The answer is $x=9,\,y=4$. A geometrical investigation can give us better insights as depicted below.
$\hspace{2cm}$
| Here is another elegant solution that one of my friends suggested. Although it is so simple and straight forward, it just works for these specific numbers $7$ and $11$! Let us start by rewriting the equations in the following forms
\begin{align*}
(\sqrt{x}-3)+(y-4)&=0,\\
(x-9)+(\sqrt{y}-2)&=0.
\end{align*}
Next, using the famous identity $a^2-b^2=(a-b)(a+b)$, we rearrange the above equations into the following form
\begin{align*}
(\sqrt{x}-3)+(\sqrt{y}-2)(\sqrt{y}+2)&=0,\\
(\sqrt{x}-3)(\sqrt{x}+3)+(\sqrt{y}-2)&=0.
\end{align*}
Solving for $\sqrt{y}-2$ from the second equation and substituting the result into the first leaves us with
\begin{align*}
(\sqrt{x}-3)-(\sqrt{x}-3)(\sqrt{x}+3)(\sqrt{y}+2)&=0.
\end{align*}
Now, it is easy to see what is going to happen. We just factor out the term $\sqrt{x}-3$, and then we will get a product which should be zero.
\begin{align*}
(\sqrt{x}-3)(1-(\sqrt{x}+3)(\sqrt{y}+2))&=0.
\end{align*}
At least one of these terms must vanish and it turns out that the only possibility is $\sqrt{x}-3=0$. This clearly implies that $x=9$ and substituting back into the original equations gives us $y=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
Show that $(a^3+a+1)(b^3+b+1)(c^3+c+1)\le 27$ Let $a,b,c\ge 0$ be such that $a^2+b^2+c^2=3$. Show that
$$(a^3+a+1)(b^3+b+1)(c^3+c+1)\le 27$$
I want to consider the function
$$f(x)=\ln{(x^{3/2}+x^{1/2}+1)}$$ Maybe it isn't the case $f''(x)\le 0$, so I can't use Jensen's inequality.
| Your idea works, but in another writing:
$$3\ln3-\prod_{cyc}(a^3+a+1)=\sum_{cyc}\left(\ln3-\ln(a^3+a+1)\right)=$$
$$=\sum_{cyc}\left(\ln3-\ln(a^3+a+1)+\frac{2}{3}(a^2-1)\right)\geq0$$
because easy to see that $f(x)\geq0$ for any $x\geq0$, where $$f(x)=\ln3-\ln(x^3+x+1)+\frac{2}{3}(x^2-1).$$
Indeed, $$f'(x)=\frac{4(x-1)(2x+3)(2x^2-x+1)}{3(x^3+x+1)},$$ which gives $x_{min}=1$, $f(1)=0$ and we are done!
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving a result for $\prod_{k=0}^{\infty}\Bigl(1-\frac{4}{(4k+a)^2}\Bigr)$ $$\prod_{k=0}^{\infty}\Bigl(1-\frac{4}{(4k+a)^2}\Bigr)=\frac{(a^2-4)\Gamma^2\bigl(\frac{a+4}{4}\bigr)}{a^2\Gamma\bigl(\frac{a+2}{4}\bigr)\Gamma\bigl(\frac{a+6}{4}\bigr)}$$
According to WA. I attempted using
$$\prod_{k=0}^{\infty}\Bigl(1-\frac{x^2}{\pi^2k^2}\Bigr)=\frac{\sin x}{x}$$
But I couldn’t reindex the product appropriately to use it the way I wanted to (factoring out a 4 and then continuing from there). I’d like to have at least a direction to go in or an idea on how to do the product.
| Use (proved by induction)
$$
\prod_{k=0}^{n-1} (a+4k) = \frac{4^n\Gamma(n+\frac{a}{4})}{\Gamma(\frac{a}{4})}
$$
together with
$$
1-\frac{4}{(4\,k+a)^2}={\frac { \left( a+4\,k+2 \right)
\left( a+4\,k-2 \right) }{ \left( 4\,k+a \right) ^{2}}}
$$
to get
$$
\prod _{k=0}^{n-1}{\frac { \left( a+4\,k+2 \right) \left( a+4\,k-2
\right) }{ \left( 4\,k+a \right) ^{2}}}
={\frac { \left( a-2
\right) \left( a+4\,n-2 \right) \Gamma \left(\frac{a-2}{4}+n
\right) ^{2} \;\Gamma \left( \frac{a-2}{4} \right) ^{2}
}{16 \; \Gamma \left( \frac{a}{4} +n\right) ^{2} \; \Gamma
\left( \frac{a+2}{4} \right) ^{2}}}
$$
Than take limit as $n \to \infty$. My result is
$$
\frac{\displaystyle\Gamma\left(\frac{a}{4}\right)^2}{\displaystyle \Gamma\left(\frac{a-2}{4}\right)\Gamma\left(\frac{a+2}{4}\right)}
$$
which agrees with Polygon.
| {
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Is this a valid proof for $I(n^2) \geq \frac{5}{3}$, if $q^k n^2$ is an odd perfect number with special prime $q$? Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$, and let $I(x)=\sigma(x)/x$ be the abundancy index of $x$.
Note that both $\sigma$ and $I$ are multiplicative functions.
A number $m$ is said to be perfect if $\sigma(m)=2m$. Equivalently, $I(m)=2$.
Euler proved that an odd perfect number, if one exists, must have the form
$$m = q^k n^2$$
where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since $q$ is prime, we have
$$\frac{q+1}{q} = I(q) \leq I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{q^{k+1} - 1}{q^k (q - 1)} < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1}$$
from which it follows that
$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$
Note that we then have the lower bound
$$I(n^2) > \frac{2(q-1)}{q} \geq \frac{8}{5}$$
since $q$ is a prime satisfying $q \equiv 1 \pmod 4$.
Here is my initial question:
Can we improve the lower bound for $I(n^2)$ to
$$I(n^2) \geq \frac{5}{3}$$
using the following argument?
$$\bigg(\frac{2q}{q+1} \geq I(n^2) > \frac{5}{3}\bigg) \implies q > 5 \implies q \geq 13 \implies \bigg(I(n^2) > \frac{2(q-1)}{q} \geq \frac{24}{13} > \frac{5}{3}\bigg)$$
Thus, we have the biconditional
$$I(n^2) > \frac{5}{3} \iff q > 5.$$
Next, we have the implication
$$I(n^2) = \frac{5}{3} \implies q = 5.$$
It then suffices to prove the implication
$$q = 5 \implies I(n^2) = \frac{5}{3}$$
to finally show that
$$I(n^2) \geq \frac{5}{3},$$
since $q \geq 5$ holds.
But note that, if $q=5$, then
$$\frac{5}{3} = I(n^2) = \frac{2}{I(5^k)} = \frac{2\cdot{5^k}(5-1)}{5^{k+1}-1}$$
which implies that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.
Still, notice that we have
$$k=1 \implies I(q^k) = I(q) = \frac{q+1}{q} = 1 + \frac{1}{q} \leq \frac{6}{5} \implies I(n^2) = \frac{2}{I(q^k)} = \frac{2}{I(q)} \geq \frac{2\cdot{5}}{6} = \frac{5}{3},$$
which is what we set out to prove.
Here is my final question:
Would it be possible to remove the reliance of the proof on the Descartes-Frenicle-Sorli Conjecture?
| I think that the answer for your initial question is yes. I've found no errors in the argument.
I think that the answer for your final question is no since under the condition that $q=5$, we see that $I(n^2)\ge \dfrac 53$ is equivalent to $k=1$ as follows :
$$\begin{align}I(n^2)\ge\frac 53&\iff \frac{8\cdot 5^k}{5^{k+1}-1}\ge\frac 53
\\\\&\iff 24\cdot 5^k\ge 5(5^{k+1}-1)
\\\\&\iff 5^k\le 5
\\\\&\iff k\le 1
\\\\&\iff k=1\end{align}$$
| {
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Special value of hypergeometric function $\, _2F_1\left(a,a+\frac{1}{3};\frac{4}{3}-a;-\frac{1}{8}\right)$ How can one prove $$\, _2F_1\left(a,a+\frac{1}{3};\frac{4}{3}-a;-\frac{1}{8}\right)=\frac{\left(\frac{2}{3}\right)^{3 a} \Gamma \left(\frac{2}{3}-a\right) \Gamma \left(\frac{4}{3}-a\right)}{\Gamma \left(\frac{2}{3}\right) \Gamma \left(\frac{4}{3}-2 a\right)}$$
This originate from an integral of algebraic functions. What exact transformation can be used to prove this identity? I'd like you to give some suggestions. Thank you!
Update:
The following Mathematica commands verify the quartic transformation given in @pisco's answer:
DifferentialRootReduce[Hypergeometric2F1[4 b/3, (4 b + 1)/3, (4 b + 5)/6, x], x]
DifferentialRootReduce[(1 + 8 x)^(-b) Hypergeometric2F1[b/3, (b + 1)/3, (4 b + 5)/6, 64 x (1 - x)^3/(1 + 8 x)^3], x]
Series[Hypergeometric2F1[4 b/3, (4 b + 1)/3, (4 b + 5)/6, x], {x, 0, 2}]
Series[(1 + 8 x)^(-b) Hypergeometric2F1[b/3, (b + 1)/3, (4 b + 5)/6, 64 x (1 - x)^3/(1 + 8 x)^3], {x, 0, 2}]
Limit[(1 + 8 x)^(-b) Hypergeometric2F1[b/3, (b + 1)/3, (4 b + 5)/6, 64 x (1 - x)^3/(1 + 8 x)^3], x -> -1/8, Direction -> -1]
| From the transformation ${_2F_1}(a,b,c,z) = (1-z)^{c-a-b}{_2F_1}(c-a,c-b,c,z)$, it is equivalent to find
$${_2F_1}(\frac{4}{3}-2a,1-2a,\frac{4}{3}-a,-\frac{1}{8}) = {_2F_1}(\frac{4b}{3},\frac{4b+1}{3},\frac{4b+5}{6},-\frac{1}{8})\qquad b=\frac{3}{4}-\frac{3a}{2}$$
We have the quartic transformation:
$$\tag{1}{_2F_1}(\frac{{4b}}{3},\frac{{4b + 1}}{3},\frac{{4b + 5}}{6},x) = {(1 + 8x)^{ - b}}{_2F_1}(\frac{b}{3},\frac{{b + 1}}{3},\frac{{4b + 5}}{6},\frac{{64x{{(1 - x)}^3}}}{{{{(1 + 8x)}^3}}})$$
It holds in a neighborhood of $x=0$ (more precisely $-1/8<x<\frac{3 \sqrt{3}-5}{4} \approx 0.049$). As $x\to -1/8^-$, the argument of $_2F_1$ on RHS tends to $-\infty$. Invoking the asymptotic expansion of ${_2F_1}$ gives
$${_2F_1}(\frac{{4b}}{3},\frac{{4b + 1}}{3},\frac{{4b + 5}}{6},-\frac{1}{8})= (\frac{4}{9})^b \frac{\Gamma (\frac{1}{3}) \Gamma (\frac{4b+5}{6})}{\Gamma (\frac{b}{3}+\frac{5}{6}) \Gamma (\frac{b+1}{3})}$$
so OP's original formula is proved.
Similar to what I said here, $(1)$ is trivial to prove after it has been explicitly conjectured.
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Inverse Laplace Transform, need help with this $\frac{s^2}{s^2+\sqrt{2}s+1}$ Inverse Laplace Transform of $\frac{s^2}{s^2+\sqrt{2}s+1}$
I transformed the denominator as $(s+\frac{\sqrt{2}}{2})^{2}$ + $\frac{1}{2}$
$\frac{s^2}{(s+\frac{\sqrt{2}}{2})^{2} + \frac{1}{2}}$ and I have no idea how to move forward because partial fraction decomposition fails.
| Substitute $u=s+\dfrac{\sqrt{2}}{2}$:
$$\dfrac{s^2}{(s+\dfrac{\sqrt{2}}{2})^{2} + \frac{1}{2}}=\dfrac{(u-\dfrac{\sqrt{2}}{2})^2}{u^{2} + \dfrac{1}{2}}$$
Then expand...And apply the first Shifting Theorem.
$$\mathcal{L^{-1}}\dfrac {\left(s+\frac{\sqrt2}2\right)}{{\left(s+\frac{\sqrt2}2\right)^2+\frac12}}=e^{-\frac {\sqrt 2}{2}t}\cos \left(\dfrac {\sqrt 2}{2}t \right)$$
| {
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Evalution of a function where $t = x + \frac{1}{x}$ Consider a function $$y=(x^3+\frac{1}{x^3})-6(x^2+\frac{1}{x^2})+3(x+\frac{1}{x})$$ defined for real $x>0$. Letting $t=x+\frac{1}{x}$ gives: $$y=t^3-6t^2+12$$
Here it holds that $$t=x+\frac{1}{x}\geq2$$
My question is: how do I know that $t=x+\frac{1}{x}\geq2$ ?
I want to know how to get to this point without previouly knowing that $t=x+\frac{1}{x}\geq2$
| We have that for $x>0$
$$x+\frac{1}{x}\geq2 \iff x\cdot x+x\cdot \frac{1}{x}\geq x\cdot 2 \iff x^2-2x+1\ge 0 \iff (x-1)^2 \ge 0$$
and the equality holds if and only if $x=1$.
| {
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Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$ Question:
Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$.
My try:
$$\begin{align}\\
&x^2+y^2-4x=0\\
&\implies (x-2)^2+y^2=4\\
&\implies |z-2|=2\\
\end{align}\\
$$
Now, $w=\frac{2z+3}{z-4}$
$$\begin{align}\\
&\frac w2=\frac{2z+3}{2z-8}\\
&\implies\require{cancel}\frac{w}{w-2}=\frac{2z+3}{\cancel{2z}+3-\cancel{2z}+8}\\
&\implies\frac{w}{w-2}=\frac{2z+3}{11}\\
&\implies\frac{2z}{11}=\frac{w}{w-2}-\frac{3}{11}\\
&\implies\frac{2z}{\cancel{11}}=\frac{8w+6}{\cancel{11}(w-2)}\\
&\implies z=\frac{4w+3}{w-2}\\
&\implies z-2=\frac{2w+7}{w-2}\\
\end{align}\\
$$
$$\therefore\left|\frac{2w+7}{w-2}\right|=2\\
\implies 2w+7=2w-4
$$
Now, what to do? Where is my fault? Or how to do it? Is there any other possible ways?
| This is equivalent to showing that $4w+3$ is pure imaginary i.e. $\arg(4w+3)=\pm\frac{\pi}{2}$.
Since $0$ and $4$ are end points of the circle diameter and $4w+3=11\frac{z-0}{z-4}$, then $\arg(4w+3)=\pm\frac{\pi}{2}$.
| {
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} |
Let $\frac{\tan A}{1-\tan^2A}=\sin^220^\circ-\sin160^\circ\sin220^\circ+\sin^2320^\circ$, find $\tan6A$ Let $\dfrac{\tan A}{1-\tan^2A}=\sin^220^\circ-\sin160^\circ\sin220^\circ+\sin^2320^\circ$, find $\tan6A$
My attempt :
\begin{align*}
\dfrac{\tan2A}{2}=\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ\\
\tan2A=2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ)
\end{align*}
and
\begin{align*}
\tan6A&=\tan(2A-60^\circ)\tan2A\tan(2A+60^\circ)\\
&=(\dfrac{\tan2A-\sqrt{3}}{1+\sqrt{3}\tan60^\circ})(\tan2A)(\dfrac{\tan2A+\sqrt{3}}{1-\sqrt{3}\tan60^\circ})
\end{align*}
give
$$\tan6A=(\dfrac{2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ)-\sqrt{3}}{1+\sqrt{3}\tan60^\circ})(2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ))(\dfrac{2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ)+\sqrt{3}}{1-\sqrt{3}\tan60^\circ})$$
This method is incredibly long, there may be better way to deal with this problem.
| Note that:
\begin{array}{}
\dfrac{\tan2A}{2}&=&\sin^2160^\circ-\sin160^\circ\sin220^\circ+\sin^2220^\circ\\
\dfrac{\tan2A}{2}&=&(\sin160^\circ-\sin220^\circ)^2+\sin160^\circ\sin220^\circ\\
\dfrac{\tan2A}{2}&=&(\sin20^\circ+\sin40^\circ)^2-\sin20^\circ\sin40^\circ\end{array}
Apply product-sum and sum-product
\begin{array}{}\dfrac{\tan2A}{2}&=&(2\sin30^\circ\cos10^\circ)^2-\dfrac{\cos20^\circ-\cos60^\circ}{2}\\
\dfrac{\tan2A}{2}&=&\cos^210^\circ-\dfrac{1-2\sin^210^\circ}{2}+\dfrac{1}{4}\\
\dfrac{\tan2A}{2}&=&\cos^210^\circ+\sin^210^\circ-\dfrac{1}{4}\\
\dfrac{\tan2A}{2}&=&\dfrac{3}{4}\\
\tan(2A)&=&\dfrac{3}{2}
\end{array}
| {
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Is $g(x) = \frac{x^3+9}{x^2}$ one-to-one? Let $g(x) =\frac{x^3+9}{x^2}$ restricted to $D(g) = (-\infty,0)$. Is it 1-1?
My approach
$g(x) \text{ 1-1 }: g(x_1)=g(x_2) \iff x_1 = x_2$
Let $x_1, x_2 \in D(g)$ then,
$$\frac{x_1^3+9}{x_1^2} = \frac{x_2^3+9}{x_2^2} \iff x_2^2 \cdot x_1^3 + 9x_2^2 - x_1^2 \cdot x_2^3 + 9x_1^2 =0 \iff\\
\iff x_2^2(x_1^3+9)-x_1^2(x_2^3+9) = 0 \quad (1)$$
In order for $(1)$ to hold,
*
*$x_2 = x_1 = 0$, which can't be because $0 \notin D(g)$ or
*$x_1 = x_2 = \sqrt[3]{-9} \in D(g)$
Therefore $g(x_1)=g(x_2) \iff x_1 = x_2$ holds only for $\sqrt[3]{-9}$ (and not for every $x \in D(g)$)
$$\boxed{ \text{Hence, }g(x)\text{ is NOT 1-1 }}$$
Is this correct?
| The step "In order for (1) to hold..." is incorrect. Just before it, you had an expression of the form $A-B=0$ ($A$ happens to be $x_2^2(x_1^3+9)$ and $B$ happens to be $x_1^2(x_2^3+9)$.) From it you have concluded that $A=0$ and $B=0$. This is not correct - the difference of two numbers can be $0$ even if neither of them is zero (as long as they are equal, that is!).
Now, maybe it will help to write the original function as:
$$g(x)=x+\frac{9}{x^2}$$
You can see that both addends here are strictly increasing. Let's assume $x_1<x_2<0$, you have $x_1^2>x_2^2$ and so $\frac{9}{x_1^2}<\frac{9}{x_2^2}$, and finally, $g(x_1)=x_1+\frac{9}{x_1^2}\lt x_2+\frac{9}{x_2^2}=g(x_2)$
Thus $g(x)$ on $(-\infty, 0)$ is "1-1".
| {
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If $a^2+a b+b^2=40$ and $a^2-\sqrt{a b}+b=5$, then find $a^2+\sqrt{a b}+b$ I was given this problem to solve with elementary methods (High School level).
Knowing that
$$\begin{align}
a^2+a b+b^2 &=40 \\
a^2-\sqrt{a b}+b &=\phantom{0}5
\end{align}$$
find
$$a^2+\sqrt{a b}+b$$
I tried to look for $\sqrt{a b}$, since the requested quantity is
$$a^2+\sqrt{a b}+b=(a^2-\sqrt{a b}+b)+2\sqrt{ab}=5 + 2\sqrt{ab}$$
So I set
$$x=a^2;\;y=\sqrt{ab};\;b=z$$
and the system became
$$x+y^2+z^2=40;\;x-y-z=5$$
subtracting the two equations I got
$$z^2-z+y^2+y-35=0$$
which has one real solution when the discriminant is zero.
That is
$1-4(y^2+y-35)=0$
and then
$y=\frac{1}{2} \left(-1\pm\sqrt{142}\right)$
and finally
$$a^2+\sqrt{a b}+b=4\pm\sqrt{142}$$
I know that there are other solutions because I've found them with Wolfram Mathematica, but I couldn't find them with elementary methods.
Any help will be appreciated.
| We could try to "brute force" it at the high school level. Start with the second equation
$$a^2 - \sqrt{ab} + b = 5$$
$$a^2 + b -5 = \sqrt{ab}$$
$$a^4 + b^2 + 25 + 2a^2 b -10a^2 -10b = ab$$
$$b^2 + ( 2a^2 -10 -a) b + (a^4 -10a^2 +25) =0$$
$$b = \frac{-(2a^2 -10 -a) \pm \sqrt{( 2a^2 -10 -a)^2 - 4(a^4 -10a^2 +25))}}{2}$$
$$b = \frac{-(2a^2 -10 -a) \pm \sqrt{a (-4 a^2 + a + 20)}}{2}$$
Now let's just call $b=x \pm \sqrt{y}.$
$$b^2 = x^2 \pm 2x\sqrt{y} + y$$
$$ab = ax \pm a\sqrt{y}$$
Plug this into the second equation:
$$a^2 +ab + b^2 =40$$
$$a^2 + ax \pm a\sqrt{y} + x^2 \pm 2x\sqrt{y} + y =40 $$
$$(\pm a \pm 2x)\sqrt{y}=40 -a^2 - ax - x^2 -y$$
$$(a^2 \pm 2xa + 4x^2)y=(40 -a^2 - ax - x^2 -y)^2$$
Now plug in for $x, y$ and solve for $a$. What you end up with is:
$$-225 + 950 a - 290 a^2 - (255 a^3)/2 - (53 a^4)/4 - (45 a^5)/2 +
14 a^6 + 2 a^7 - a^8 =0$$ This is not tractable, and it is likely the problem has a typo.
If you type in Mathematica
Solve[{a^2 + a*b + b^2 == 40, \[Sqrt](a*b) + 5 == a^2 + b}, {a, b}]
It will spit out 3 gross looking expressions possible answers. If you then evaluate these for $a^2 + b + \sqrt{ab}$
a^2 + b + Sqrt[a b] /. %OUTPUT NUMBER
Then the resulting expression doesn't simplify nicely. (Which is not a guarantee; Mathematica's simplify function is far from perfect).
| {
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Sum of Series S =?
Proving:
$$\displaystyle\sum_{n=0}^{\infty}\frac{(n^2+4n-1)(\pi-\phi)^{3n}}{n!(n+2)}=(\pi-\phi)^3e^{(\pi-\phi)^3}+2\left(e^{(\pi-\phi)^3}-1\right)-5\left(\frac{e^{(\pi-\phi)^3}((\pi-\phi)^3-1)+(\pi-\phi)^4-(\pi-\phi)^3+1}{(\pi-\phi)^6}\right) $$
Where .$ \phi $ is golden ration
Note:this series is convergent
Proof :
we put $$S=\displaystyle\sum_{n=0}^{\infty}\frac{n^2+4n-1}{n!(n+2)}x^{3n}$$
we have :
\begin{align*}
\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|
&=\lim_{n\to\infty}|\frac{\frac{(n+1)^2+4(n+1)-1}{(n+1)!(n+3)}}{\frac{n^2+4n-1}{n!(n+2)}}|\\
&=\lim_{n\to\infty}\frac{(n+1)^2+4(n+1)-1}{n^2+4n-1}\frac{(n+2)}{(n+1)(n+3)}=0\\
&\implies R=\infty
\end{align*}
So This series is convergent for each $x\in \mathrm R$
| Note that for $n\geq 1$,
$$
\frac{{n^2 + 4n - 1}}{{n!(n + 2)}}x^{3n} = \frac{{x^{3n} }}{{(n - 1)!}} + 2\frac{{x^{3n} }}{{n!}} - 5\frac{{x^{3n} }}{{n!(n + 2)}}.
$$
Hence,
$$
\sum\limits_{n = 0}^\infty {\frac{{n^2 + 4n - 1}}{{n!(n + 2)}}x^{3n} } = e^{x^3 } (x^3 + 2) - 5\sum\limits_{n = 0}^\infty {\frac{{x^{3n} }}{{n!(n + 2)}}} .
$$
But
\begin{align*}
\sum\limits_{n = 0}^\infty {\frac{{x^{3n} }}{{n!(n + 2)}}} & = \frac{1}{{x^6 }}\sum\limits_{n = 0}^\infty {\frac{{x^{3(n + 2)} }}{{n!(n + 2)}}} = \frac{1}{{x^6 }}\sum\limits_{n = 0}^\infty {\frac{{(n + 1)x^{3(n + 2)} }}{{(n + 2)!}}} \\ & = \frac{1}{{x^3 }}\sum\limits_{n = 0}^\infty {\frac{{x^{3(n + 1)} }}{{(n + 1)!}}} - \frac{1}{{x^6 }}\sum\limits_{n = 0}^\infty {\frac{{x^{3(n + 2)} }}{{(n + 2)!}}} \\ & = \frac{1}{{x^3 }}(e^{x^3 } - 1) - \frac{1}{{x^6 }}(e^{x^3 } - 1 - x^3 ) = \frac{{e^{x^3 } (x^3 - 1) + 1}}{{x^6 }}.
\end{align*}
Accordingly,
$$
\sum\limits_{n = 0}^\infty {\frac{{n^2 + 4n - 1}}{{n!(n + 2)}}x^{3n} } = e^{x^3 } (x^3 + 2) - 5\frac{{e^{x^3 } (x^3 - 1) + 1}}{{x^6 }}.
$$
Note that this expression differs from your result. The term $(\pi-\phi)^4$ in your answer is suspicious to me. Please check it numercially again.
| {
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Quasilinear PDE $u_t + (u^2)_x = 0$ cauchy problem The problem I am trying to solve is:
\begin{equation}\label{eq:3.1}
\begin{cases}
\partial_t u + \partial_x(u^2)=0 & x\in \mathbb{R}, t \in (0,\infty]\\
u(x,0)=
\begin{cases}
0 & x\leq 0\\
x & 0<x\leq 1\\
1 & x>1
\end{cases}
\end{cases}
\end{equation}
What I have done is:
We will try to reduce the problem to ODEs on a curve $x(t)$ on the $(t,x)$ plane. The equation can be compared with the canonical form,
\begin{equation}
a\frac{\partial u}{\partial x} +b\frac{\partial }{\partial t} = c,
\end{equation}
where $a = 2u$, $b= 1$ and $c=0$. From the Lagrange-Charpit equations, we have,
\begin{align}\label{eq:3.2}
&\frac{dx}{a}=\frac{dt}{b}=\frac{du}{c} & \text{ substituting we have,}\nonumber\\
\implies &\frac{dx}{2u}=\frac{dt}{1}=\frac{du}{0}&
\end{align}
using second and the third ratio from the equation we have,
\begin{align}\label{eq:3.3}
&\frac{du}{dt}=0 & \text{integrating we have,} \nonumber\\
\implies&u=B,&
\end{align}
where $B$ is a arbitrary constant. Using the initial conditions,
\begin{equation}\label{eq:3.4}
u(x,0)=
\begin{cases}
0 & x\leq 0\\
x & 0<x\leq 1\\
1 & x>1
\end{cases}
\end{equation}
where the characteristic curve $x(t)$, passes through $(c,0)$. By substitution we have,
\begin{equation}
B=
\begin{cases}
0 & x\leq 0\\
c & 0<x\leq 1\\
1 & x>1.
\end{cases}
\end{equation}
Therefore solution can be written as
\begin{equation}\label{eq:3.5}
u=
\begin{cases}
0 & x\leq 0\\
c & 0<x\leq 1\\
1 & x>1.
\end{cases}
\end{equation}
using first and the second ratios from the equation we have,
\begin{align}\label{eq:3.6}
&\frac{dx}{dt}=2u & \text{substituting we have,} \nonumber\\
\implies&\frac{dx}{dt}=
\begin{cases}
0 & x\leq 0\\
2c & 0<x\leq 1\\
2 & x>1.
\end{cases}
&\text{integrating we have,}\nonumber\\
\implies&x=
\begin{cases}
B & x\leq 0\\
2ct+B & 0<x\leq 1\\
2t+B & x>1.
\end{cases}
&\nonumber\\
\end{align}
where $B$ is a arbitrary constant. Using the initial conditions , and that the characteristic curve $x(t)$ passes through $(c,0)$ we have,
\begin{equation}
x=
\begin{cases}
c & x\leq 0\\
2ct+c & 0<x\leq 1\\
2t+c & x>1.
\end{cases}
\end{equation}
Therefore $u$ becomes,
\begin{equation}
u(x,t)=
\begin{cases}
0 & x\leq 0\\
\frac{x}{2t+1} & 0<x\leq 1\\
1 & x>1.
\end{cases}
\end{equation}
I think I am missing something. The solution should have $t$ dependence in the intervals. Thanks.
| The main part that you did seems correct. Except the limits at the end.
$$\begin{equation}
\frac{\partial u}{\partial x} +2u\frac{\partial u}{\partial t} = 0
\end{equation}$$
Your Charpit-lagrange characteristic ODEs are correct :
$$\frac{dx}{2u}=\frac{dt}{1}=\frac{du}{0}$$
A first characteristic equation comes from $du=0$ :
$$u=c_1$$
A second characteristic equation comes from $\frac{dx}{2c_1}=\frac{dt}{1}$ :
$$x-2c_1t=c_2$$
The general solution of the PDE expressed on implicite form $c_1=F(c_2)$ is :
$$u=F(x-2ut)$$
where $F$ is an arbitrary function to be determined according to the initial condition.
\begin{equation}
u(x,0)=F(x)=
\begin{cases}
0 & x\leq 0\\
x & 0<x\leq 1\\
1 & x>1
\end{cases}
\end{equation}
So, the function $F$ is determined any variable $\chi$ :
\begin{equation}
F(\chi)=
\begin{cases}
0 & \chi\leq 0\\
\chi & 0<\chi\leq 1\\
1 & \chi>1
\end{cases}
\end{equation}
We put this function $F(\chi)$ into the above general solution where $\chi=x-2ut$
\begin{equation}
u=F(x-2ut)=
\begin{cases}
0 & x-2ut\leq 0\\
x-2ut & 0<x-2ut\leq 1\\
1 & x-2ut>1
\end{cases}
\end{equation}
Case $u=0$ and $x-2ut\leq 0\quad\to\quad x\leq 0$ .
Case $u=x-2ut$ and $0<x-2ut\leq 1\quad\to\quad u=\frac{x}{1+2t}$ and $0<x\leq 1+2t$
Case $u=1$ and $x-2ut>1 \quad\to\quad x>1+2t$
The solution is :
\begin{equation}
u(x,t)=
\begin{cases}
0 & x\leq 0\\
\frac{x}{1+2t} & 0<x\leq 1+2t\\
1 & x>1+2t
\end{cases}
\end{equation}
| {
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What kind of curve do these lines make? Our teacher has given us a question to be solved.
What curve does the intersection points of the given lines make? A parabola, hyperbola, or none of them? (please look at the image I've posted. I am not speaking about all of the intersection points. Just the tangent curve beneath the lines)
the lines are as follows:
$y=\frac{4\sqrt{3}}{5}x+\frac{41\sqrt{3}}{10}$
$y=\frac{3\sqrt{3}}{5}x+\frac{17\sqrt{3}}{5}$
$y=\frac{2\sqrt{3}}{5}x+\frac{29\sqrt{3}}{10}$
$y=\frac{\sqrt{3}}{5}x+\frac{13\sqrt{3}}{5}$
$y=\frac{5\sqrt{3}}{2}$
$y=-\frac{\sqrt{3}}{5}x+\frac{13\sqrt{3}}{5}$
$y=-\frac{2\sqrt{3}}{5}x+\frac{29\sqrt{3}}{10}$
$y=-\frac{3\sqrt{3}}{5}x+\frac{17\sqrt{3}}{5}$
$y=-\frac{4\sqrt{3}}{5}x+\frac{41\sqrt{3}}{10}$
$y=\sqrt{3}x+5\sqrt{3}$
$y=-\sqrt{3}x+5\sqrt{3}$
my solution: I tried to find if it is a parabola first. Assuming the maximum point to be $(0,\frac{13\sqrt{3}}{5})$ (the intersection point of $4$th and $6$th lines), I found the equation to be $y=-\frac{13\sqrt{3}}{125}x^2+\frac{13\sqrt{3}}{5}$. But after plotting it in desmos, the curve didn't meet the intersection points.
Then I tried to assume $(0,\frac{5\sqrt{3}}{2})$ as the maximum; I yielded $y=-\frac{\sqrt{3}}{10}x^2+\frac{5\sqrt{3}}{2}$; but it again failed in plotting.
I then switched to hyperbola: After a long effort I found the equations to be $$\frac{(y-5\sqrt{3})^2}{(\frac{5\sqrt{3}}{2})^2}-\frac{x^2}{(\frac{5\sqrt{3}}{3})^2}=1$$ and $$\frac{(y-5\sqrt{3})^2}{(\frac{12\sqrt{3}}{5})^2}-\frac{x^2}{(\frac{5\sqrt{3}}{3})^2}=1$$ but I hink both of them are incorrect; because they are irrelevant to their asymptotes ($10$th & $11$th lines in the set of lines).
line sets image
Sorry, I had to stretch the picture a little so that it better be displayed.
| Inspection shows that the given lines can be produced as follows:
$$y_k(x)={\sqrt{3}\over10}\bigl(2kx+(k^2+25)\bigr)\qquad(-5\leq k\leq 5)\ .\tag{1}$$One now should look for the envelope of the family $\bigl(\ell_c\bigr)_{c\in{\mathbb R}}$ of lines
$$\ell_c:\quad y-y_c(x)=0\ ,$$where the integer $k$ in $(1)$ has been replaced by the real parameter variable $c$.
Doing the standard calculations for finding the envelope one finds that the envelope $\epsilon$ of the line family $\bigl(\ell_c\bigr)_{c\in{\mathbb R}}$ is the parabola
$$\epsilon:\qquad y={\sqrt{3}\over10}(25-x^2)\ ,$$
plotted in red in the following figure.
Referring to these "standard calculations": When a family of curves in the $(x,y)$-plane is given by an equation of the form
$$F(x,y,c):=y-{\sqrt{3}\over10}\bigl(2c x+(c^2+25)\bigr)=0$$ then most points $(x,y)$ are "ordinary" points, meaning that there is no curve going through them, or that in the neighborhood of $(x,y)$ the curves are lined up like nice "parallel" stream lines. When $$F(x_0,y_0,c_0)=0\qquad\wedge\qquad F_c(x_0,y_0,c_0)\ne0$$ then $(x_0,y_0)$ is an "ordinary" point. When $$F(x_0,y_0,c_0)=0\qquad\wedge\qquad F_c(x_0,y_0,c_0)=0\tag{1}$$ then $(x_0,y_0)$ is a "special" point, since $c$, defined by $F(x,y,c)=0$, is no longer a good function of $(x,y)$ in the neighborhood of $(x_0,y_0)$.
Some of the "special" points are indeed lying on an envelope $\epsilon$ of the given curve family. You obtain these "special" points by eliminating $c$ from the two equations $(1)$.
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Evaluate $\frac{1+3}{3}+\frac{1+3+5}{3^2}+\frac{1+3+5+7}{3^3}+\cdots$ It can be rewritten as
$$S = \frac{2^2}{3}+\frac{3^2}{3^2}+\frac{4^2}{3^3}+\cdots$$
When $k$ approaches infinity, the term $\frac{(k+1)^2}{3^k}$ approaches zero. But, i wonder if it can be used to determine the value of $S$. Any idea?
Note: By using a programming language, i found that the value of $S$ is $3.5$
| Hint: Let $f(t)=\sum t^{k}$. Then $tf'(t)=\sum kt^{k}$ and $t^{2}f''(t)=\sum (k^{2}-k)t^{k}$. This gives $\sum kt^{k}$ and $\sum k^{2}t^{k}$ in terms of $f,f'$ and $f''$. Put $t=1/3$ and use the fact that $(k+1)^{2}=k^{2}+2k+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show $ \prod_{k=1}^{n} 4^k = 2^{n*(n+1)}$, where did I go wrong in my induction step? Can someone help me out with the induction step?
Show $ \prod_{k=1}^{n} 4^k = 2^{n*(n+1)}$
Base case n=1: $$4^1 = 2^{1*(1+1)} = 2^2$$
Induction step (to show: $2^{(n+1)*(n+2)} = 2^{n^2+3n+2}$ ) :
$$ \prod_{k=1}^{n+1} 4^k = 4^{n+1} * 2^{n*(n+1)} = \frac{1}{2} * 4^{n+1} * 4^{n*(n+1)} = \frac{1}{2} 4^{n^2+n+n+1}= \frac{1}{2} 4^{n^2+2n+1} = 2^{n^2+2n+1}$$
but now we have
$$ 2^{n^2+3n+2} \neq 2^{n^2+2n+1} $$
Where did I go wrong?
| Your mistake was writing $2^{n(n+1)}$ as$\tfrac124^{n(n+1)}$ rather than $4^{\tfrac12n(n+1)}$. The inductive step should be$$4^{\tfrac12n(n+1)}4^{n+1}=4^{\tfrac12n(n+1)+n+1}=4^{(\tfrac12n+1)(n+1)}=4^{(n+1)(n+2)}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove this equality of the determinant of matrix?
Prove that
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
=(a^2-bc)(b^2-ca)(c^2-ab)\end{equation*}
My attempt:
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
=\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
a(b-a) & b(c-b) & c(a-c) \\
(b+a)(b-a) & (c+b)(c-b) & (a+c)(a-c)
\end{pmatrix}
\end{equation*}
But I think my direction is incorrect. Can anyone give me some hints or the solution of this question?
| It has to be a degree-$6$ homogeneous polynomial in $a,\,b,\,c$ that vanishes if $a^2=bc$, because if $a/b=b/c=k$ the determinant is $c^6$ times$$\left|\begin{array}{ccc} k^{4} & k^{2} & 1\\ k^{3} & k & k^{2}\\ k^{2} & 1 & k^{4} \end{array}\right|=k^{4}\left(k^{5}-k^{2}\right)-k^{2}\left(k^{7}-k^{4}\right)=0.$$Repeating this logic for $2$ other factors, the determinant must be proportional to $(a^2-bc)(b^2-ca)(c^2-ab)$. The case $a=1,\,b=2,\,c=3$ lets you verify no proportionality constant is needed.
| {
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"url": "https://math.stackexchange.com/questions/3793112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The ratio of the area of two regular polygons The polygons in the figure below are all regular polygons(regular heptagon), share a vertex and the orange line crosses the three vertices of the two regular polygons, the area of the small regular polygon and the large regular polygon is denoted as $S_1$, $S_2$, what is $\frac{S_1}{S_2}$?
Additional question (regular nine-sided polygon)
|
Won't go through the calculation, but this is the idea.
First since $\triangle ADE$ and $\triangle BDF$ are similar, we know $AE$ pass through $G$.
Now we can calculate $DG$,$GC$,$AG$ based on the left heptagon and since $AD\parallel CE$ we can calculate $GE=GC\cdot {AD\over DG}$. Also we know $\angle DGE=180^{\circ}-\angle AGD={5\over 7}180^{\circ}$.
Therefore $DE^2=DG^2+GE^2-2\cos({5\over 7}180^{\circ})DG\cdot GE$.
If you let $a=DG,b=DA,c=DB$, there are some identity here
Using the identity, $\cos({5\over 7}180^{\circ})=-{a^2+c^2-b^2\over 2ac}=-{a+b\over 2c}$
New edit:
Actually just realized $\angle GEB=\angle GAD=\angle GBE$ so $GE$ is actually just $b$.
Now the calculation is really simple:
$$ED^2=a^2+b^2+ab\cdot{(a+b)\over c}$$
$$=a^2+b^2+{bc(c-b)+c(c+a)(c-b)\over c}$$
$$=a^2+b^2+bc-b^2+c^2+ac-bc-ab$$
$$=a^2+c^2+ac-ab$$
$$=a^2+c^2+b^2-a^2-c^2+b^2$$
$$=2b^2$$
So the area is exactly twice.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $a,b,c$ :- $a + \frac{1}{b} = \frac{7}{3}$ , $b + \frac{1}{c} = 4$ , $c + \frac{1}{a} = 1$
Solve for $a,b,c$ :- $a + \frac{1}{b} = \frac{7}{3}$ , $b + \frac{1}{c} = 4$ , $c + \frac{1}{a} = 1$
What I Tried :- We have $3$ equations , $3$ variables, should be easy .
I did everything, solved these equations but in the end I am getting $3a^2 - 10a + 9 = 0$ . (My work is a bit long so can't show it)
This has complex roots , so this problem just got a little tough .
Can anyone help me?
| From the last equation we have $c=1-\frac{1}{a}=\frac{a-1}{a}$ and from the second $b=4-\frac{1}{c}=4-\frac{a}{a-1}==\frac{3a-4}{a-1}$ so the first equation becomes $a+\frac{1}{b}=a+\frac{a-1}{3a-4}=\frac{7}{3}$ or $3a^{2}-4a+a-1=\frac{7}{3}(3a-4)$ so $9a^{2}-9a-3=21a-28$ and hence we have $9a^{2}-30a+25=(3a-5)^{2}=0.$
| {
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"timestamp": "2023-03-29T00:00:00",
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$(a+1)(b+1)(c+1)\leq4$ for triangle sides $a,b,c$ with $ab+bc+ac=1$
Given that $a,b,c$ are the lengths of the three sides of a triangle, and $ab+bc+ac=1$, the question is to prove $$(a+1)(b+1)(c+1)\leq4\,.$$
Any idea or hint would be appreciated.
This is Problem 6 of Round 1 of the BMO (British Mathematical Olympiad) 2010/2011, as can be seen here.
Remark. This question has been self-answered. Nevertheless, any new approach is always welcome!
| We need to prove $$abc+a+b+c\leq2$$ or
$$(abc+(a+b+c)(ab+ac+bc))^2\leq4(ab+ac+bc)^3$$ or
$$\prod_{cyc}(a(b+c-a)+bc)\geq0$$ and we are done!
We can get a last factoring by the following way.
For $ab+ac+bc=a^2$ we obtain:
$$(abc+(a+b+c)(ab+ac+bc))^2=(abc+(a+b+c)a^2)^2=a^2(a^2+ab+ac+bc)^2=$$
$$=(ab+ac+bc)(2(ab+ac+bc))^2=4(ab+ac+bc)^3$$ and since we work with symmetric polynomials, we got the needed factorization.
| {
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"timestamp": "2023-03-29T00:00:00",
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Check the convergence of the series $\displaystyle{\sum_{n=1}^{+\infty}\frac{\left (n!\right )^2}{\left (2n+1\right )!}4^n}$ I want to check if the following series converge or not.
*
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{\left (n!\right )^2}{\left (2n+1\right )!}4^n}$
I suppose we have to find here an upper bound and apply then the comparison test. But I don’t really have an idea which bound we could take. Could you give me a hint?
*
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2n}}$
We have a term that is a product of the form $\frac{2i-1}{2i}=1-\frac{1}{2i}$. To apply the comparison test we have to find an upper bound. Does it holds that $1-\frac{1}{2i}\leq \frac{1}{2}$ and so $$\prod_{i=1}^n\left (1-\frac{1}{2i}\right )\leq \prod_{i=1}^n \frac{1}{2}=\frac{1}{2^n}$$ Then taking the sum we get $$\sum_{n=1}^{+\infty}\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2n}\leq \sum_{n=1}^{+\infty} \frac{1}{2^n}=1$$ So from the comparison test the original sum must converge also.
Is everything correct?
*
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2n\cdot (2n+2)}}$
We have a term that is a product of the form $\frac{2i-1}{2i+2}$. Which upper bound could we use in this case?
| The Stirling approximation implies $\binom{2n}{n}\sim\frac{4^n}{\sqrt{n\pi}}$, so $\frac{n!^24^n}{(2n+1)!}\sim\frac{\sqrt{\pi}/2}{\sqrt{n}}$, so the series diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
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Two inequalities with parameters $a,b,c>0$ such that $ca+ab+bc+abc\leq 4$
Let $a,b,c>0$ be such that $bc+ca+ab+abc\leq 4$. Prove the following inequalities:
(a) $8(a^2+b^2+c^2)\geq 3(b+c)(c+a)(a+b)$, and
(b) $\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{a^2b}+\dfrac{2}{b^2c}+\dfrac{2}{c^2a}\geq 9$.
Prove also that the unique equality case for both inequalities is given by $a=b=c=1$.
Below are some probably useful or relevant results.
*
*https://artofproblemsolving.com/community/c6h1241430p6342224
*https://artofproblemsolving.com/community/c6h284290p1535893
*https://artofproblemsolving.com/community/c6h608971p3619202
*https://artofproblemsolving.com/community/c6h1804479p11995588
*If $ab+bc+ca+abc=4$, then $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c$
Techniques used in solving the inequalities in these links may prove useful in proving our inequalities.
Attempt. In the simplest case, $a=b=c=:t$, we have $t^3+3t^2-4\leq 0$, whence $0<t\leq 1$. Therefore, the inequalities (a) and (b) become
$$24t^2\geq 24t^3$$
and
$$\frac{3}{t^2}+\frac{6}{t^3}\geq 9\,,$$
which are obviously true. How to prove these inequalities in general?
| The first inequality.
Let $a=kx$, $b=ky$ and $c=kz$ such that $k>0$ and $xy+xz+yz+xyz=4.$
Thus, the condition gives $$k^2(xy+xz+yz)+k^3xyz\leq xy+xz+yz+xyz$$ or
$$(k-1)((k+1)(xy+xz+yz)+(k^2+k+1)xyz)\leq0$$ or $$k\leq1.$$
Thus, we need to prove that $$8(x^2+y^2+z^2)\geq3k(x+y)(x+z)(y+z)$$ and since $0<k\leq1$, it's enough to prove that $$8(x^2+y^2+z^2)\geq3(x+y)(x+z)(y+z).$$
Now, rewrite the new condition in the following form:
$$\sum_{cyc}\frac{1}{x+2}=4$$ and let $x=\frac{2p}{q+r}$ and $y=\frac{2q}{p+r},$ where $p$, $q$ and $r$ are positives.
Thus, $z=\frac{2r}{p+q}$ and after this substitution we obtain something obvious.
But it's better to prove before that $$x+y+z\geq xy+xz+yz,$$ for which we need to prove that:
$$\sum_{cyc}\frac{2p}{q+r}\geq\sum_{cyc}\frac{4pq}{(p+r)(q+r)}$$ or
$$\sum_{cyc}p(p+q)(p+r)\geq2\sum_{cyc}pq(p+q)$$ or
$$\sum_{cyc}(p^3-p^2q-p^2r+pqr)\geq0,$$ which is true by Schur.
Now, since $$1\geq\frac{xy+xz+yz}{x+y+z},$$ it's enough to prove that $$8(x^2+y^2+z^2)(xy+xz+yz)\geq3(x+y+z)(x+y)(x+z)(y+z)$$ or
$$\sum_{cyc}(5x^3y+5x^3z-6x^2y^2-4x^2yz)\geq0,$$ which is true by Muirhead.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $f(x)=\sin^{-1} (\frac{2x}{1+x^2})+\tan^{-1} (\frac{2x}{1-x^2})$, then find $f(-10)$ Let $x=\tan y$, then
$$
\begin{align*}\sin^{-1} (\sin 2y )+\tan^{-1} \tan 2y
&=4y\\
&=4\tan^{-1} (-10)\\\end{align*}$$
Given answer is $0$
What’s wrong here?
| $$\arcsin(\sin(x))=\left\{
\begin{array}{ll}
x-2\pi n&\hbox{for }-\frac{\pi}{2}+2\pi n\le x\le \frac{\pi}{2}+2\pi n\\
\pi-x-2\pi n&\hbox{for }\frac{\pi}{2}+2\pi n\le x\le \frac{3\pi}{2}+2\pi n\\
\end{array}\right.,$$
$$\arctan(\tan(x))=x-\left\lfloor\frac{x+\pi/2}{\pi}\right\rfloor\pi$$
| {
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"timestamp": "2023-03-29T00:00:00",
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solve for $x$, $(\sqrt{a+ \sqrt{a^2-1}})^x+(\sqrt{a- \sqrt{a^2-1}})^x=2a$ Find the value of $x$ when $$(\sqrt{a+ \sqrt{a^2-1}})^x+(\sqrt{a- \sqrt{a^2-1}})^x=2a.$$See, by hit and trial method it is clear that $x=2$ is a solution. But I failed to solve this explicitly to get the solutions.
My Attempt: \begin{align*} &(\sqrt{a+ \sqrt{a^2-1}})^x+(\sqrt{a- \sqrt{a^2-1}})^x=2a \\ \implies \> & (a+ \sqrt{a^2-1})^{x/2}+(a- \sqrt{a^2-1})^{x/2}=2a\\ \implies \>& (a+ \sqrt{a^2-1})^x+(a- \sqrt{a^2-1})^x+2(a+ \sqrt{a^2-1})^{x/2}(a- \sqrt{a^2-1})^{x/2} = 4a^2.\end{align*} I have no idea how to proceed after this. Also I tried by multiplying conjugate up and down, but I failed. Please help me to solve this.
Thanks in advance.
| put $u=(\sqrt{a+ \sqrt{a^2-1}})^x$
then taking conjugate,$\frac{1}{u}$=$(\sqrt{a- \sqrt{a^2-1}})^x$
or we have to solve
$u+\frac{1}{u}=2a$ which can be easily solved.
| {
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Find all $x\in \mathbb{R}$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$
Find all $x\in \mathbb{R}$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$
Letting $a=2^x$ and $b=3^x$ we get $$\frac{a^3+b^3}{a^2b+ab^2} = \frac{7}{6}$$
from the numerator we have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=7$$
since $7$ is a prime we can say that $$a+b=1, a^2-ab+b^2=7.$$
It follows that $$a=1-b$$
from where $$(1-b)^2-(1-b)b+b²=7$$
this quadratic has solutions $b=1, b=0.$
what I now did was consider cases. Firstly $a=b=0$ which has no solutions for $x$.
case $a=b=1$ has the solution $x=0$.
However the actual solutions for this were $x=1, x=-1$ which I don't see how they came up with. What is wrong with my approach?
| The problem here is that $a+b$ and $a^2 - ab + b^2$ are not necessarily integers.
@lab bhattacharjee gives a complete answer for the same problem. In short, observe that both $a^3+b^3$ and $a^2b + ab^2$ are homogeneous of degree $3$. Dividing by $a^3$ yields $$\frac{7}{6}=\frac{a^3+b^3}{a^2b+ab^2}=\frac{1+t^3}{t+t^2}=\frac{1-t+t^2}{t}$$ where $t=\frac{b}{a}=\left( \frac 3 2 \right)^x$. You can proceed from here.
| {
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Integrate $ \int \frac{1}{\sin^{4}x+\cos^{4}x}dx $ Show that$$ \int \frac{1}{\sin^{4}(x)+\cos^{4}(x)}dx \ = \frac{1}{\sqrt{2}}\arctan\left(\frac{\tan2x}{\sqrt{2}}\right)+C$$
I have tried using Weierstrass substitution but I can't seem to get to the answer... Should I be using the said method or is there another way I can approach the question? Since the integrand evaluates into an arctangent function I am assuming there is some trickery in the manipulation that can get me there. But I just can't seem to see it...
| Original post was to evaluate $$\int\frac{\,dx}{\sin^4(x)\cos^4(x)}$$ but I believe that the intended integral was $$\int\frac{\,dx}{\sin^4(x)+\cos^4(x)}$$ which I evaluate after the original request.
$$\begin{align}I=&\int\frac{\,dx}{\sin^4(x)\cos^4(x)}\\&=2^4\int\frac{\,dx}{\bigr[2\sin(x)\cos(x)\bigr]^4}\\&=2^4\int\frac{\,dx}{\sin^4(2x)}\\&=2^4\int\csc^4(2x)\,dx\\&=2^3\int \csc^4(u)\,du\end{align}$$
Using the reduction formula for cosecant, $$\int\csc^{m}(x)\,dx=-\frac{\csc^{m-1}(x)\cos(x)}{m-1}+\frac{m-2}{m-1}\int\csc^{m-2}(x)\,dx$$ or more concisely, $$J_m= -\frac{\csc^{m-1}(x)\cos(x)}{m-1}+\frac{m-2}{m-1}J_{m-2} $$ then we can obtain $$J_4=-\frac{\csc^{3}(u)\cos(u)}{3}+\frac{2}{3}J_2$$ where $J_2=-\cot(u)+C$
Let $$J_4=\int\csc^4{u}\,du$$ so that $$\begin{align}I&=2^3J_4\\&=2^3\bigg[-\frac{\csc^{3}(u)\cos(u)}{3}-\frac{2}{3}\cot(u)\bigg] +C\\&=-\frac{2^3}{3}\bigg[\csc^{3}(2x)\cos(2x)+2\cot(2x)\bigg] +C\\&=-\frac{2^3}{3}\cot(2x)\bigg[\csc^2(2x)+2\bigg]+C\end{align}$$
If you don't know the reduction formula off-hand or don't feel like deriving it, expand the integrand as $$\csc^4(u)=\csc^2(u)\csc^2(u)=\csc^2(u)(1+\cot^2(u))=\csc^2(u)+\csc^2(u)\cot^2(u)$$ which is fairly simple to integrate.
Now to evaluate $$I=\int\frac{\,dx}{\sin^4(x)+\cos^4(x)}$$
$$\begin{align}I&=\int\frac{\,dx}{\cos^4(x)-2\sin^2(x)\cos^2(x)+\sin^4(x)+2\sin^2(x)\cos^2(x)}\\&=\int\frac{\,dx}{[\cos^2(x)-\sin^2(x)]^2+2\sin^2(x)\cos^2(x)}\\&=\int\frac{\,dx}{\cos^2(2x)+\frac{\sin^2(2x)}{2}}\\&=\int\frac{\sec^2(2x)}{1+\frac{\tan^2(2x)}{2}}\,dx\\&=\int\frac{\sec^2(2x)}{1+\bigg(\frac{\tan(2x)}{\sqrt{2}}\bigg)^2}\,dx \end{align} $$
Let $u=\frac{\tan(2x)}{\sqrt{2}}$ and $\,du=\frac{2}{\sqrt{2}}\sec^2(2x)\,dx$. Then $$\begin{align}I&=\frac{\sqrt{2}}{2}\int\frac{\,du}{1+u^2}\\&=\frac{1}{\sqrt{2}}\arctan(u)+C\\&=\frac{1}{\sqrt{2}}\arctan\bigg(\frac{\tan(2x)}{\sqrt{2}}\bigg)+C \end{align} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $a^{2020}+b^{2020}+c^{2020}$.
Question: Let $f(x)=x^3+ax^2+bx+c$ and $g(x)=x^3+bx^2+cx+a$ where $a,b,c\in\mathbb{Z}$, $c\neq 0$. Suppose $f(1)=0$ and roots of $g(x)=0$ are squares of the roots of $f(x)=0$. Find the value of $a^{2020}+b^{2020}+c^{2020}$.
My approach: Let the other roots of $f(x)$ except $1$ be $\alpha$ and $\beta$. This implies that the roots of $g(x)$ are $1,\alpha^2$ and $\beta^2$. Thus, we have $1+\alpha+\beta=-a,$ $\alpha+\beta+\alpha\beta=b,$ $\alpha\beta=-c$ and $1+\alpha^2+\beta^2=-b,$ $\alpha^2+\beta^2+\alpha^2\beta^2=c,$ $\alpha^2\beta^2=-a.$ Note that thus we have $c^2=-a$. Again, $$a^2=(1+\alpha+\beta)^2=1+\alpha^2+\beta^2+2(\alpha+\beta+\alpha\beta)=-b+2b=b.$$ Also, since $f(1)=0\implies a+b+c=-1$. Now since we have $a=-c^2$ and $b=a^2\implies b=c^4,$ which in turn implies that $$-c^2+c^4+c+1=0\implies c^4-c^3+c^3-c^2+c-1=-2\\\implies (c^3+c^2+1)(c-1)=-2.$$ Thus, we either have $$\begin{cases}c^3+c^2+1=1\\ c-1=-2\end{cases} \text{ or }\begin{cases}c^3+c^2+1=-1\\ c-1=2\end{cases} \text{ or }\begin{cases}c^3+c^2+1=2\\ c-1=-1\end{cases} \text{ or }\begin{cases}c^3+c^2+1=-2\\ c-1=1\end{cases}.$$ Note that only the first diophantine equation yields a solution, that is $c=-1$. Thus, $a=-1$ and $b=1$. Therefore, $$a^{2020}+b^{2020}+c^{2020}=1+1+1=3.$$
Is this solution correct and rigorous enough and is there any other way to solve the problem? Also, does anyone know the original source of this problem?
| Another way to reach the solution is this:
Since $a=-c^2$ and $b=c^4$ you have $f(x) =x^3-c^2x^2+c^4x+c$ and since $f(1)=0$ you obtain $1-c^2+c^4+c=0$ i.e. $c^2(c^2-1)+c+1=(c+1)(c^3-c^2+1)=0$, by observing that $c^3-c^2+1=0$ can't have neither odd or even solution, you obtain $c=-1$ and your solution.
| {
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Inconsistent solution to differential equation $Q.$
Let f(x) be a continuous function satisfying the following differential equation-
$$f(x)=(1+x^2)(1+\int_0^x\frac{f^2(t)dt}{1+t^2})$$
$$\text{Find f(1)}$$
My Work-
1)Putting $x=0$ in the equation we get $f(0)=1$
2)Dividing by $1+x^2$ and differentiating w.r.t. $x$ we get-
$$(\frac{y}{1+x^2})'=\frac{y^2}{1+x^2}\qquad\text{∴ y=f(x)}$$
Simplifying-
$$y'(1+x^2)-2xy=y^2(1+x^2)$$
so either $f(x)=0$ or
$$\frac{-dy(1+x^2)}{y^2}+\frac{2xdx}{y}=(-1-x^2)dx$$
$$\frac{d}{dx}(\frac{1+x^2}{y})=\frac{d}{dx}(-x-\frac{x^3}{3})$$
$$\frac{1+x^2}{y}=-x-\frac{x^3}{3}+c$$
Using $y(0)=1$ we get $c=1$ and hence $$f(1)=-6$$
My issue- looking at the question, no matter the value of f(x), the R.H.S. of the equation must always be positive (squared) however the answer is coming to be negative. Am I missing something? Or is there any error in the question? Or is $f(x)=0$ the only acceptable solution?
| $$\frac{f(x)}{1+x^2}=1+\int_{0}^{x} \frac{f^2(t)}{1+t^2}dt \implies f(0)=1$$
D. w.r.t. $x$, using Lebnitz, to get
$$f'(x)-\frac{2x}{1+x^2}f(x)=f^2(x)\implies f^{-2}f'-\frac{2x}{1+x^2}\frac{1}{f}=1$$
This is Bernoulli equation. Let $1/f=v$, then we get
$$\frac{dv}{dx}+\frac{2x}{1+x^2}v=-1$$
This is linear equation, whose integrating factor is $I=\exp[\int \frac{2x}{1+x^2}dx]\implies I= (1+x^2)$
$$v=(1+x^2)^{-1} \int -1(1+x^2) dx + C (1+x^2)^{-1}.$$
$$\implies v=\frac{-x-x^3/3}{1+x^2}+C(1+x^2)^{-1}=\frac{1}{f}$$
Used $f(0)=1 \implies C=1.$ Finally, $$f(x)=\frac{1+x^2}{1-x-x^3/3}.$$
So $f(1)=-6.$ The plot of $f(x)$ is given below wherein $f(x)>0$only in (0,0.8177.. ), then there is a singularity at $x=0.8177..$ and it becomes negative. The answer to OP's interesting question could lie in the non-linearity of this ODE.
| {
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"url": "https://math.stackexchange.com/questions/3813224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of non negative integer solutions of $x+y+2z=20$
The number of non negative integer solutions of $x+y+2z=20$ is
Finding coefficient of $x^{20}$ in
$$\begin{align}
&\left(x^0+x^1+\dots+x^{20}\right)^2\left(x^0+x^1+\dots+x^{10}\right)\\
=&\left(\frac{1-x^{21}}{1-x}\right)^2\left(\frac{1-x^{11}}{1-x}\right)\\
=&\left(1-x^{21}\right)^2(1-x)^{-3}\left(1-x^{11}\right)
\end{align}$$
i.e.finding coefficient of $x^{20}$ in $$(1-x)^{-3}-x^{11}(1-x)^{-3}$$
Or, coefficient of $x^{20}$ in $(1-x)^{-3}-$ coefficient of $x^9$ in $(1-x)^{-3}=\binom{22}{20}-\binom{11}{9}=176$
The asnwer is given as $121$. What's my mistake?
EDIT (after seeing @lulu's comment):
Finding coefficient of $x^{20}$ in $$\begin{align}
&\left(x^0+x^1+...+x^{20}\right)^2\left(x^0+x^2+...+x^{20}\right)\\
=&\left(\frac{1-x^{21}}{1-x}\right)^2\left(\frac{1-x^{22}}{1-x^2}\right)\\
=&\left(1-x^{21})^2(1-x)^{-2}(1-x^{22})(1-x^2\right)^{-1}
\end{align}$$
i.e.finding coefficient of $x^{20}$ in $$(1-x)^{-2}(1-x^2)^{-1}$$
Not able to proceed next.
| Found the mistake. Instead of $\left(x^0+x^1+\dots+x^{10}\right)$, it should have been $\left(x^0+x^2+\dots+x^{20}\right)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to deduce this factorization of $x^5+x+1$ by looking at $\int\frac{3x^4+2x^3-2x+1}{x^5+x+1}dx$? The question is:
$$\int\frac{3x^4+2x^3-2x+1}{x^5+x+1}dx$$
I tried a lot but couldn't solve it so I looked at the solution which is:
$$x^5+x+1=(x^2+x+1)(x^3-x^2+1)$$ and we can write $$3x^4+2x^3-2x+1=(x^3-x^2+1)+(3x^2-2x)(x^2+x+1)$$
which effectively reduces the integral to very simple ones.
My question is how they deduced the factorization of the denominator. After looking at the solution I think that if we put $x=1,x=\omega$ and $x=\omega^2$ we can deduce this but this was not immediately obvious to me. Is there some sort of hint you can get by looking at the integrand or is it simply a matter of less experience?
Any help would be appreciated.
| Just it's better to see that any polynomial $x^{3k-1}+x^{3n-2}+1$ has a factor $x^2+x+1$ for any naturals $k$ and $n$.
For example, your reasoning with $\omega\neq1$ and $\omega^3=1$ helps to understand it.
In our case we can get this factoring so:
$$x^5+x+1=x^5-x^2+x^2+x+1=(x^2+x+1)(x^3-x^2+1).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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For $n\ge 6$, can we partition the set $\{1 , 4 , 9 , ...,n^2\}$ into two subsets whose sums are equal or differ by one?
For $n\ge 6$, can we partition the set $\{1 , 4 , 9 , ...,n^2\}$ into two subsets such that the sums of the elements in the two subsets are equal or differ by one?
For example : for $n = 10$, we can form the subsets
$S_1 = \{100 , 64 , 25 , 4\}$ and $S2 = \{1 , 9 , 16, 36, 49, 81\}$.
$S_1$ adds up to $193$ and $S_2$ adds up to $192$.
Can we also identify the elements that we can assign to individual subsets that satisfies this property?
| The difference of sums is of the from $S:=\sum_{k=1}^n s_kk^2$ where each $s_k=\pm1$. Our task is to find $s_k$ such that the sum is $0$ or $1$.
Observe that $$\tag1a^2-(a+1)^2-(a+2)^2+(a+3)^2=4.$$
Therefore, we can pick four consecutive signs such that they contribute either $+4$ or $-4$.
Hence with $8$ consecutive signs, we can achieve zero contribution.
Here are the smallest non-negative sums we can achieve for some small $n$ with the eight distinct residues $\bmod 8$:
$$ \begin{align}S_0&=0&=0\\
S_1&=1^2&=1\\
S_6 &= 1^2-2^2{+3^2-4^2-5^2+6^2}&=1\\
S_7 &= 1^2+2^2-3^2{+4^2-5^2-6^2+7^2}&=0\\
S_{10}&=-1^2+2^2-3^2-4^2{+5^2-6^2-7^2+8^2}-9^2+10^2&=1\\
S_{11}&=-1^2+2^2-3^2-4^2-5^2+6^2+7^2+8^2-9^2+10^2-11^2&=0\\
S_{12}&=-1^2-2^2-3^2-4^2-5^2-6^2-7^2-8^2+9^2+10^2-11^2+12^2&=0\\
S_{13}&=-1^2-2^2-3^2-4^2-5^2-6^2-7^2+8^2+9^2-10^2+11^2+12^2-13^2&=1
\end{align}$$
We conclude that we can reach sum $=0$ or $=1$ at least when $n$ si one of $0,1,6,7,10,11,12,13$ plus a multiple of $8$. In particular, this covers all $n\ge 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Cauchy-Schwarz Inequality problems Let $a,$ $b,$ $c,$ $d,$ $e,$ $f$ be nonnegative real numbers.
(a) Prove that
$$(a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) \ge (ace + bdf)^4.$$
(b) Prove that
$$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2) \ge (ace + bdf)^2.$$
I'm not sure how I should start approaching both problems. I believe I should use Cauchy-Schwarz, but I'm not sure.
Any help would be appreciated! Thanks in advance.
| For (a),
Just use Holder's Inequality.
For (b),
Expand $(a^2 + b^2) (c^2 + d^2) (e^2 + f^2) $. (It is equivalent to the LHS of my answer)
Expand $(ace + bdf)^2$. (It is equivalent to the RHS of my answer)
The problem is just proving
$$
a^2 c^2 e^2 + a^2 c^2 f^2 + a^2 d^2 e^2 + a^2 d^2 f^2 + b^2 c^2 e^2 + b^2 c^2 f^2 + b^2 d^2 e^2 + b^2 d^2 f^2 \ge a^2 c^2 e^2 + b^2 d^2 f^2 + 2abcdef
$$ or
$$
a^2 c^2 f^2 + a^2 d^2 e^2 + a^2 d^2 f^2 + b^2 c^2 e^2 + b^2 c^2 f^2 + b^2 d^2 e^2 \ge 2abcdef \quad \textrm{By cancelling 2 terms on the left and right side.}
$$
Now, that is just
$$
(acf - bde)^2 + a^2 d^2 e^2 + a^2 d^2 f^2 + b^2 c^2 e^2 + b^2 c^2 f^2 \ge 0
$$
We have proved by this that the inequality is valid for any real numbers, not only non-negative ones.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a^2+b^2-ab=c^2$ for positive $a$, $b$, $c$, then show that $(a-c)(b-c)\leq0$
Let $a$, $b$, $c$ be positive numbers. If $a^2+b^2-ab=c^2$. Show that
$$(a-c)(b-c)\leq0$$
I have managed to get the equation to $(a-b)^2=c^2-ab$, but I haven't been able to make any progress.
Can someone help me?
| Equation, $(a^2-ab+b^2)=(c)^2$ has solution shown below:
$a=p^2-q^2$
$b=2pq-q^2$
$c=p^2-pq+q^2$
Where, $p>q>0$
We need:
$w=(a-c)(b-c)\leq0$
Or, $w=(c-a)(c-b)\leq0$
$(c-b)=(p-q)(p-2q)$
$(c-a)=-q(p-2q)$
Hence,
$w=(c-a)(c-b)=(-)(p-q)(q)(p-2q)^2$
Since, $p>q>0$:
$(p-2q)^2$ & $(p-q)$ & $(q)$ is a positive quantity and the expression 'w' is negative,
The expression 'w',
$(a-c)(b-c)\leq0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve differential equation: $y' = \frac{y^2}{x^3} + 2\frac{y}{x} - x$ I need to solve this differential equation: $y' = \frac{y^2}{x^3} + 2\frac{y}{x} - x$.
My attempt
I found that $y = x^2$ is a solution. Then I tried to put $y = x^2f(x)$, and solved this way:
$$2xf(x) + x^2f'(x) = xf^2(x) + 2xf(x) - x \implies x^2f'(x) = xf^2(x) - x \implies$$
$$xf'(x) = f^2(x) - 1 \implies \frac{df(x)}{f^2(x) - 1} = \frac{dx}{x} \implies$$
$$\frac{1}{2}\ln\left|\frac{f(x) - 1}{f(x) + 1}\right| = \ln|x| + C_* \implies \frac{f(x) - 1}{f(x) + 1} = Cx^2 \implies$$
$$f(x) = \frac{1 + Cx^2}{1 - Cx^2}$$
And we lost a solution $f(x) = -1$. So finally we have
$$y = x^2\frac{1 + Cx^2}{1 - Cx^2}, y = -x^2$$
Now I have 3 questions:
$\quad 1)$ Is my solution correct? I'm not sure that all solutions were found.
$\quad 2)$ When can we use particular solution to find all other solutions? I mean doing something like $y = g(x)h(x)$, where $g(x)$ is a particular solution.
$\quad 3)$ Is there an easier method to solve this equation?
| 1.) It is correct, if a little unusual.
2.a) The usual way for a Riccati equation with a particular solution is to set $y(x)=y_p(x)+\frac1{u(x)}$ which should result in a linear first order DE for $u$.
2.b) It is typical for Riccati equation that the solution family is similar to a circle, that is, that the real line for the parameter is closed at $\infty$ where you get the missing solution. In the present case replace $C=1/B$ to get the alternative solution formula
$$
y(x)=x^2\,\frac{B+x^2}{B-x^2},
$$
where now the parameter $B=0$ results in $y(x)=-x^2$.
3.) The alternative is to parametrize solutions as $y(x)=-x^3\frac{u'(x)}{u(x)}$ so that
$$
y'(x)=-x^3\frac{u''(x)}{u(x)}-3x^2\frac{u'(x)}{u(x)}+x^3\frac{u'(x)^2}{u(x)^2}
=x^3\frac{u'(x)^2}{u(x)^2}-2x^2\frac{u'(x)}{u(x)}-x
$$
which cancels and reduces to
$$
0=x^2u''(x)+xu'(x)-u(x)
$$
which is an Euler-Cauchy equation with basis solutions $x$ and $x^{-1}$,
$$
y(x)=-x^3\,\frac{A-Bx^{-2}}{Ax+Bx^{-1}}=-x^2\frac{Ax^2-B}{Ax^2+B}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the roots of $x^3 - 6x = 4$ This exercise is from the book Complex Analysis by Joseph Bak, and it says:
"Find the three roots of $x^{3}-6x=4$ by finding the three real-valued possibilities for $\sqrt[3]{2+2i}+\sqrt[3]{2-2i}$".
I know that these numbers were found by Cardan's method, but I don't understand why they give these numbers, because I found three real roots by common methods.
Pd: the three real roots are $-2$, $1-\sqrt{3}$ and $1+\sqrt{3}$.
| Let $x=\sqrt[3]{2+2i}+\sqrt[3]{2-2i}$. Then,
$x^3=2+2i+2-2i+3(\sqrt[3]{2+2i})^2(\sqrt[3]{2-2i})+3(\sqrt[3]{2+2i})(\sqrt[3]{2-2i})^2$
$=4+3(\sqrt[3]{2+2i}+\sqrt[3]{2-2i})(\sqrt[3]{2+2i})(\sqrt[3]{2-2i})$
$=4+3x\sqrt[3]{(2+2i)(2-2i)}$
$=4+3x\sqrt[3]{8}$
$=4+6x$
$\therefore x^3-6x=4$
Therefore, solving for $x=\sqrt[3]{2+2i}+\sqrt[3]{2-2i}$ can help you find the solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Area of triangle under tan conditions. For an acute triangle $ABC$, the following conditions hold.
$$\frac{1}{\tan A} + \frac{1}{\tan B} + \frac{1}{\tan C } =2$$
$$ a^2 + b^2 + c^2 =50 $$
Compute the area of such a triangle.
I used a tan addition law and get,
$$\tan A + \tan B + \tan C = \tan A \tan B \tan C $$.
And let $ \tan A \tan B \tan C = k $, so $\tan A ,
\tan B , \tan C $ are roots of the following cubic equation.
$x^3 - k x^2 +2k x - k $.
I am stuck right here.
| We have $$2=\sum_{cyc}\frac{ab\cos\gamma}{ab\sin\gamma}=\frac{\sum\limits_{cyc}(a^2+b^2-c^2)}{4S}=\frac{50}{4S},$$ which gives $$S=\frac{25}{4}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing $\lim_{x\to-5}\frac{x^2+2x-15}{|x+5|}$ The problem is: $$\lim_{x\to-5}\frac{x^2+2x-15}{|x+5|}$$
I factored the numerator to get: $\frac{(x-3)(x+5)}{|x+5|}$
How do i solve the rest?
| After the correction in the limit $x\rightarrow -5$.
Recall that $|x+5|=\begin{cases}x+5 \space\space \text{if $x\geq-5$} \\-(x+5)\space\space \text{if $x<-5$}\end{cases}.$ Then
*
*$\lim_{x\rightarrow (-5)^{-}}\frac{(x-3)(x+5)}{|x+5|}=\lim_{x\rightarrow (-5)^{-}}\frac{(x-3)(x+5)}{-(x+5)}=\lim_{x\rightarrow (-5)^{-}}-(x-3)=-(-5-3)=8$ since we approach from $x<-5$
*$\lim_{x\rightarrow (-5)^{+}}\frac{(x-3)(x+5)}{|x+5|}=\lim_{x\rightarrow (-5)^{+}}\frac{(x-3)(x+5)}{(x+5)}=\lim_{x\rightarrow (-5)^{+}}(x-3)=-5-3=-8$ since we approach from $x>-5.$
Thus the limit doesn't exist.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Real roots of $x^7+5x^5+x^3−3x^2+3x−7=0$? The number of real solutions of the equation,
$$x^7+5x^5+x^3−3x^2+3x−7=0$$
is
$$(A) 5 \quad (B) 7 \quad (C) 3 \quad (D) 1.$$
Using Descartes rule we may have maximum no. of positive real roots is $3$ and negative real root is $0.$ So there can be either $3$ real roots or $1$ real root but how to conclude what will be the no. of real roots exactly. Can you please help me?
| If the Descartes rule does not help there is always the following (or similar) way: $$x^7+5x^5+x^3−3x^2+3x−7=(x-1)(x^6+x^5+6x^4+6x^3+7x^2+4x+7).$$
But $$x^6+x^5+6x^4+6x^3+7x^2+4x+7=$$
$$=\left(x^3+\frac{1}{2}x^2-1\right)^2+\frac{1}{4}(23x^4+40x^3+36x^2+16x+12)=$$
$$=\left(x^3+\frac{1}{2}x^2-1\right)^2+\frac{1}{4}(23x^4+40x^3+18x^2+18x^2+16x+12)>0,$$
which says that our equation has an unique real root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$ ; $\frac{1}{a} + \frac{1}{c} + \frac{1}{a+y}$ ; $\frac{1}{a} + \frac{1}{x} + \frac{1}{y} = 0$.
If $a \neq 0$ , $b \neq 0$ , $c \neq 0$ and if :- $\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$ ; $\frac{1}{a} + \frac{1}{c} + \frac{1}{a+y}=0$ ; $\frac{1}{a} + \frac{1}{x} + \frac{1}{y} = 0$ , find $(a+b+c)$ .
What I Tried :- No information is given about $x$ and $y$ . So I thought of putting $x = y = 1$ , and this silly thing came out in the end .
Now, as $x = y = 1$ , I have $a = \frac{1}{-2}$ from the $3$rd equation .
So from the $1$st equation I get :- $$\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$$
$$ \rightarrow -2 + \frac{1}{b} + 2 = 0$$
$$ \rightarrow \frac{1}{b} = 0$$
This definitely looks absurd (also it's given that $b \neq 0$), so I guess putting $x = y = 1$ was a big mistake .
I don't have any other cool ideas for now as I see that doing it algebraically is going to include a lot of simplification and stuffs, and since there are $5$ variable there must be some shortcut of this .
Can anyone help?
| For those who love tedious algebra, from
$$\begin{cases}
\frac1a+\frac1b+\frac1{a+x}=0\\
\frac1a+\frac1c+\frac1{a+y}=0\\
\frac1a+\frac1x+\frac1y=0
\end{cases}$$
We have:
$$a = \frac {-1}{\frac1x+\frac1y} = -\frac{xy}{x+y}$$
$$b = \frac {-1}{\frac 1a+ \frac 1{a+x}}=-\frac {a(a+x)}{2a+x}$$
$$c = \frac {-1}{\frac 1a+ \frac 1{a+y}}=-\frac {a(a+y)}{2a+y}$$
$$\begin{align}a+b+c &= a\left(1-\frac{a+x}{2a+x}-\frac {a+y}{2a+y}\right)
\\&=a\left(1-\frac{x-\frac{xy}{x+y}}{x-\frac{2xy}{x+y}}-\frac {y-\frac{xy}{x+y}}{y-\frac{2xy}{x+y}}\right)
\\&=a\left(1-\frac{x^2}{x^2-xy}-\frac {y^2}{y^2-xy}\right)
\\&=a\left(1-\frac x{x-y}-\frac y{y-x}\right)
\\&=a\left(1-\frac {x-y}{x-y}\right)\\&=0
\end{align}$$
The question remains, to find a more elegant solution.
| {
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Another upper bound for the Stirling numbers of the first kind It is shown in this question that
$${n \brack n-k}\leq\frac{n^{2k}}{2^kk!}.$$
But a sharper bound seems to be $${n \brack n-k}\leq\frac{n^{k}}{2^k}{n-1 \choose k}.$$
I don't see how to derive this inequality. Any idea?
Hereafter is some numerical evidence: this is a representation of the natural logarithm of $f(n,k)$ as a function of $k$ in the range $1\le k \le n-1$, for $n=30$. The red dots are for $f(n,k)={n \brack n-k}$, the black dots for $f(n,k)=\frac{n^{2k}}{2^kk!}$ and the blue dots for $f(n,k)=\frac{n^{k}}{2^k}{n-1 \choose k}$.
| I think this proof by induction works:
\begin{align}
{n \brack n-k}
&= (n-1){n-1 \brack n-k}+{n-1\brack n-k-1}
\\&\le (n-1)\frac{(n-1)^{k-1}}{2^{k-1}}\binom{n-2}{k-1}+\frac{(n-1)^k}{2^k}\binom{n-2}{k}
\\&=\frac{(n-1)^k}{2^k}\binom{n-1}{k}\left[\frac{2k}{n-1}+\frac{n-k+1}{n-1}\right]
\\&=\frac{n^k}{2^k}\binom{n-1}{k}\cdot {\frac{(n-1)^k}{\color{blue}{n^k}}\cdot\frac{n+k-1}{n-1}}
\\&\le\frac{n^k}{2^k}\binom{n-1}{k}\cdot {\frac{(n-1)^k}{\color{blue}{(n-1)^k+k(n-1)^{k-1}}}\cdot\frac{n+k-1}{n-1}}
\\&=\frac{n^k}{2^k}\binom{n-1}{k}
\end{align}
To prove that second inequality, expand $n^k=((n-1)+1)^k$ with the binomial theorem, and only keep the first two terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Where did I go wrong in my proof that for all $n \in \mathbb{Z}^+$, $\sqrt{2} < a_n$ with $(a_n)$ being a particular recursive sequence? I am in Introduction to Abstract Math, and I have taken Calculus 1, Linear Algebra, and Discrete Math. I got stuck with an apparent contradiction in my proof and wanted to know where I messed up:
Consider the sequence $(a_n)$ defined recursively by:
$$a_1 = 3, \:\:\:\:\:\:\: a_{n+1} = \frac{a_n}{2}+\frac{1}{a_n} \:\:\:\:\:\:\: (n \geq 1).$$
Prove that $\sqrt{2}$ is a lower bound for $(a_n)$.
This can be rewritten as: Prove for all $n \in \mathbb{Z}^+$, $\sqrt{2} < a_n$, which we can prove with strong induction.
Base Step
Check for $n=1$.
$$a_1 = 3 > \sqrt{2}$$
Check for $n=2$.
$$a_2 = \frac{a_1}{2} + \frac{1}{a_1} = \frac{3}{2} + \frac{1}{3} = \frac{11}{6} > \sqrt{2}$$.
Therefore, the statement holds true for $n=1, n=2$.
Inductive Step
Assume that $a_n > \sqrt{2}$ for $1 \leq n \leq k$ with $k \in \mathbb{Z}^+$. Prove $a_{k+1} > \sqrt{2}$.
$$a_k = \frac{a_{k-1}}{2} + \frac{1}{a_{k-1}} > \sqrt{2}$$
$$a_{k+1} = \frac{a_k}{2} + \frac{1}{a_k}$$
$$a_{k+1} = \frac{\frac{a_{k-1}}{2} + \frac{1}{a_{k-1}}}{2} + \frac{1}{\frac{a_{k-1}}{2} + \frac{1}{a_{k-1}}}$$
$$a_{k+1} = \frac{2a_{k-1}}{a_{k-1}^2 + 2}+\frac{a_{k-1}}{4}+\frac{1}{2a_{k-1}}$$
$$a_{k+1} = \frac{2a_{k-1}}{a_{k-1}^2 + 2}+\frac{1}{2}(\frac{a_{k-1}}{2}+\frac{1}{a_{k-1}})$$
$$a_{k+1} = \frac{2a_{k-1}}{a_{k-1}^2 + 2}+\frac{1}{2}(a_k)$$
$$a_{k+1} = (\frac{2a_{k-1}}{a_{k-1}^2 + 2} \times \frac{1}{a_k})a_k+\frac{1}{2}(a_k)$$
$$a_{k+1} = a_k (\frac{2a_{k-1}}{a_{k-1}^2 + 2} \times (\frac{2}{a_{k-1}}+a_{k-1}) +\frac{1}{2})$$
Here's where I got stuck. I plugged it into Wolfram Alpha to do the algebra for that step and see if there was equivalent forms, and I saw that what was in the parenthesis was equal to $\frac{5}{2}$, which gives
$$a_{k+1} = \frac{5a_{k}}{2}$$
which doesn't make sense as the sequence would be increasing way too fast.
Where did I mess up in my proof?
| As another approach, let $f(x)=\frac x2+\frac 1x$ for $x>0$. It is easy to confirm that $$f(x)=\sqrt 2\iff x=\sqrt 2$$
As $f(x)$ is clearly continuous for all $x>0$ we see that $x>\sqrt 2$ must then imply that $f(x)>\sqrt 2$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculate $\lim_{(x,y)\to (0,0)}\frac{\arctan(x^2+y^4) }{ \sqrt{x^2+y^2+1} - 1}$ How can I calculate the limit of the following?
$$ \lim_{(x,y)\to (0,0)}\frac{\arctan(x^2+y^4) }{ \sqrt{x^2+y^2+1} - 1} $$
I've tried using the sandwich rule, but after more than 2 hours, with no success.
Would my appreciate help.
Thank you!
| We have that
$$\frac{\arctan(x^2+y^4) }{ \sqrt{x^2+y^2+1} - 1}=\frac{\arctan(x^2+y^4) }{ x^2+y^4}\cdot\frac{x^2+y^4 }{ \sqrt{x^2+y^2+1} - 1}$$
with
$$\frac{\arctan(x^2+y^4) }{ x^2+y^4}\to 1$$
and using polar coordinates we obtain
$$\frac{x^2+y^4 }{ \sqrt{x^2+y^2+1} - 1}=r^2\frac{\cos^2\theta+r^2\sin^4\theta }{ \sqrt{r^2+1} - 1}=r^2\frac{\cos^2\theta+r^2\sin^4\theta }{ \sqrt{r^2+1} - 1}\frac{\sqrt{r^2+1} + 1 }{ \sqrt{r^2+1} + 1}=$$
$$= r^2\frac{\cos^2\theta+r^2\sin^4\theta }{r^2}(\sqrt{r^2+1} + 1)=(\cos^2\theta+r^2\sin^4\theta)(\sqrt{r^2+1} + 1)\to 2\cos^2 \theta$$
which is dependent by $\theta$.
To see that without polar coordinates let consider
*
*$x=y=t \to 0$ then
$$\frac{x^2+y^4 }{ \sqrt{x^2+y^2+1} - 1}=\frac{t^2+t^4 }{ \sqrt{2t^2+1} - 1}=\frac{t^2+t^4 }{ \sqrt{2t^2+1} - 1}\cdot \frac{\sqrt{2t^2+1} + 1}{ \sqrt{2t^2+1} + 1}=$$
$$=\frac{t^2(1+t^2)(\sqrt{2t^2+1} + 1)}{ 2t^2}=\frac12(1+t^2)(\sqrt{2t^2+1}+1)\to 1$$
*
*$x=t \to 0$ and $y=0$ then
$$\frac{x^2+y^4 }{ \sqrt{x^2+y^2+1} - 1}=\frac{t^2 }{ \sqrt{t^2+1} - 1}=\frac{t^2 }{ \sqrt{t^2+1} - 1}\cdot \frac{\sqrt{t^2+1} + 1}{ \sqrt{t^2+1} + 1}=$$
$$=\frac{t^2(\sqrt{t^2+1} + 1)}{ t^2}=\sqrt{t^2+1}+1\to 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Find value of $\sin x-\frac{1}{\cot x}$ If $\sin x+\frac{1}{\cot x}=3$, calculate the value of $\sin x-\frac{1}{\cot x}$
Please kindly help me
Let $\sin x -\frac{1}{\cot x}=t$
Then, $$\sin x= \frac{3+t}{2}, \cot x= \frac{2}{3-t}$$
By using $$1+\cot ^2x= \frac{1}{\sin^2 x}$$
Then, the equation $$t^4-18t^2+48t+81=0$$
| Let $\tan{x}=y$.
Thus, $$\sin{x}+y=3,$$ which gives
$$\sin^2x=(3-y)^2$$ or
$$\frac{y^2}{1+y^2}=(3-y)^2$$ or
$$y^4-6y^3+9y^2-6y+9=0$$ or for any real $k$
$$(y^2-3y+k)^2-(2ky^2-6(k-1)y+k^2-9)=0.$$
Now, we'll choose a value of $k$, for which $k>0$ and $$2ky^2-6(k-1)y+k^2-9=(ay+b)^2,$$
for which we need $$9(k-1)^2-2k(k^2-9)=0$$ or
$$2k^3-9k^2-9=0,$$ which by the Cardano's formula gives:
$$k=\frac{3+3\sqrt[3]3+\sqrt[3]9}{2}$$ and we obtain:
$$\left(y^2-3y+\frac{3+3\sqrt[3]3+\sqrt[3]9}{2}\right)^2-(3+3\sqrt[3]3+\sqrt[3]9)\left(y-\frac{3+3\sqrt[3]3-\sqrt[3]9}{2}\right)^2=0,$$
which gives two quadratic equations.
One of them has no real roots.
The second gives two real roots:
$$\frac{\sqrt[3]3}{2}\left(\sqrt[3]9+\sqrt{1+\sqrt[3]3+\sqrt[3]9}-\sqrt{2\sqrt{2(\sqrt[3]3-1)}-(\sqrt[3]3-1)^2}\right)$$ and
$$\frac{\sqrt[3]3}{2}\left(\sqrt[3]9+\sqrt{1+\sqrt[3]3+\sqrt[3]9}+\sqrt{2\sqrt{2(\sqrt[3]3-1)}-(\sqrt[3]3-1)^2}\right)$$ and we obtain:
$$\sin{x}-\tan{x}=3-2y=\sqrt[3]3\left(-\sqrt{1+\sqrt[3]3+\sqrt[3]9}+\sqrt{2\sqrt{2(\sqrt[3]3-1)}-(\sqrt[3]3-1)^2}\right)$$ or $$\sin{x}-\tan{x}=-\sqrt[3]3\left(\sqrt{1+\sqrt[3]3+\sqrt[3]9}+\sqrt{2\sqrt{2(\sqrt[3]3-1)}-(\sqrt[3]3-1)^2}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3836076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Showing the pdfs of two normal distributions multiplied together is still the pdf of a normal distribution Given $p_1(x) = \cfrac{1}{\sqrt{2\pi}\sigma_1}e^{-x^2/2\sigma_1^2}$ and $p_2(x) = \cfrac{1}{\sqrt{2\pi}\sigma_2}e^{-x^2/2\sigma_2^2}$
I want to show that $p_1(x)p_2(x)$ is normal.
My thought
$$
\begin{align}
p_1(x)p_2(x)
&= \cfrac{1}{\sqrt{2\pi}\sigma_1}e^{-x^2/2\sigma_1^2} \cdot
\cfrac{1}{\sqrt{2\pi}\sigma_2}e^{-x^2/2\sigma_2^2} \\
&= \cfrac{1}{2\pi\sigma_1\sigma_2}
e^{-x^2/2\sigma_1^2}\cdot e^{-x^2/2\sigma_2^2} \\
&= \cfrac{1}{2\pi\sigma_1\sigma_2}
e^{-(x^2\sigma_1^2+x^2\sigma_2^2)/(2\sigma_1^2\sigma_2^2)} \\
&= \cfrac{1}{2\pi\sigma_1\sigma_2}
e^{-\frac{x^2}{2}\cdot(\sigma_1^2+\sigma_2^2)/(\sigma_1^2\sigma_2^2)} \\
\end{align}
$$
I'm given that the variance of these two functions mutiplied together is $\sigma_1\sigma_2/(\sigma_1^2\sigma_2^2) = \sigma^2$ and has mean 0.
So
$$
\begin{align}
\cfrac{1}{2\pi\sigma_1\sigma_2}
e^{-\frac{x^2}{2}\cdot(\sigma_1^2+\sigma_2^2)/(\sigma_1^2\sigma_2^2)}
=
\cfrac{1}{2\pi\sigma_1\sigma_2}
e^{-x^2/2\sigma^2}
\end{align}
$$
But I'm not sure how to show
$$
\cfrac{1}{2\pi\sigma_1\sigma_2}
=
\cfrac{1}{\sqrt{2\pi}\sigma}
$$
| $$
\frac{x^2}{\sigma_1^2} + \frac{x^2}{\sigma_2^2} = \frac{x^2}{\left( \dfrac{\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2} \right)} = \frac{x^2}{\text{a positive number}}
$$
It is not generally true that
$$ \require{cancel}
\xcancel{\sigma_1 \sigma_2 = \frac{\sigma_1\sigma_2}{\sqrt{\sigma_2^2 + \sigma_2^2}},}
$$
as may be seen by looking at any case in which $\sigma_2^2 + \sigma_2^2 \ne 1.$
What you get is a Gaussian function, but not a probability density function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3836236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
How to solve this absolute value inequality? I am new to absolute value inequalities.I was looking trough a book and I found this inequality,
I know a little bit about absolute value inequalities.
The inequality is given below:
$$
\left| \frac{n+1}{2n+3} - \frac{1}{2} \right| > \frac{1}{12}, \qquad n \in \mathbb{Z}
$$
| As it is an absolute value
Case 1)
$$\frac{n+1}{2n+3} - \frac{1}{2} > \frac{1}{12} $$
Case 2)
$$\frac{n+1}{2n+3} - \frac{1}{2} < \frac{-1}{12}$$
Now I can solve it a little for you , as in Case 1
$$\frac{n+1}{2n+3} - \frac{1}{2} - \frac{1}{12}>0 $$
$$\frac{n+1}{2n+3} - \frac{7}{12} > 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How to group people so everyone meets? I have sort of a stupid maths problem which I was thinking about recently, given the UK's covid rules about only being able to meet with 6 people max.
If I have a group of $N$ friends (e.g. 15) and everyone wants to hang out with everyone else, what's the fewest number of meet-ups we would need to accomplish this (with max 6 people in each)?
This seems similar to the social golfer problem. Is it possible to come up with a general solution?
(obviously not planning to go and do this in person, but I was thinking about this and couldn't come up with a solution)
| Denote $M+1$ the max number of friends per group, $N+1$ the number of people, and $S_{k}$ the minimum number of meetings required for $k$ people. There are ${N+1\choose 2}$ pairs of people which need to meet. We eliminate at most $M+1 \choose 2$ of these pairs per meeting. Hence $S_{N+1} \geq \lceil {N+1\choose 2}{M+1 \choose 2}^{-1} \rceil$.
Applying the identity
\begin{equation}
n = \lceil \frac{n}{m} \rceil + \cdots + \lceil \frac{n-m+1}{m} \rceil
\end{equation}
for all integers $n$ and positive integers $m$, counting with induction gives
\begin{align}
S_{N+1} &\leq S_{N} + \lceil \frac{N}{M}\rceil \\
&\leq \cdots \leq \lceil\frac{N}{M} \rceil + \lceil\frac{N-1}{M} \rceil + \cdots + \lceil \frac{M}{M} \rceil \\
&=\left(\lceil\frac{N}{M}\rceil + \cdots + \lceil \frac{N-M+1}{M} \rceil\right) +\left(\lceil\frac{N-M}{M}\rceil+\cdots+\lceil\frac{N-2M+1}{M}\rceil \right) + \cdots \\
&\leq N + (N-M) + (N-2M) + \cdots + (N-\left(\lfloor N/M \rfloor -1\right)M) \\
&=N\lfloor \frac{N}{M} \rfloor- M\lfloor \frac{N}{M} \rfloor \left(\lfloor \frac{N}{M} \rfloor - 1 \right)2^{-1} \\
&=\lfloor \frac{N}{M} \rfloor\left(N - \frac{M}{2} \left(\lfloor\frac{N}{M}\rfloor -1 \right) \right) \\
&\leq \frac{N}{M}\left(\frac{N}{2} + M\right) \\
&=\frac{N(N+2M)}{2M}
\end{align}
where the first $\leq$ followed by induction, the second by appending the missing terms to the final grouped summation. Lastly
\begin{equation}
\lceil \frac{N(N+1)}{M(M+1)} \rceil \leq S_{N+1} \leq \lfloor \frac{N(N+2M)}{2M} \rfloor
\end{equation}
The upper bound is rather large. It can be improved somewhat by assuming person $k+1$ has already met people from an earlier person's turn and doing the same computation.
$\textbf{Edit}$: A strategy like this is person $k+1$'s first meeting always includes $k$ and preferrably people $k$ has not met with numbers less than $k-1$. Then $k$ has always already met at least $M-1$ of the remaining people on their turn.
Similarly by induction
\begin{equation}
S_{N+1} \leq \lceil \frac{N}{M} \rceil + \lceil \frac{N-M}{M} \rceil + \lceil \frac{N-M-1}{M} \rceil + \cdots + \lceil \frac{1}{M} \rceil \\
\end{equation}
which is the previous calculation with $N$ replaced by $N-M$ plus $\lceil \frac{N}{M} \rceil + M-1$. So
\begin{equation}
S_{N+1} \leq \lfloor \frac{(N-M)(N+M)}{2M} + \frac{N}{M} + M \rfloor = \lfloor \frac{(N+1)^2+M^2-1}{2M} \rfloor
\end{equation}
$N = 14, M = 5$ gives $7 \leq S_{15} \leq 24$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Show that the solutions to the equation $ax^2 + 2bx + c =0$ are given by $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-ac}{a^2}}$
Show that the solutions to the equation $ax^2 + 2bx + c =0$ are given by $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-ac}{a^2}}$
Hint: Start by dividing the whole equation by $a$
At first I have tried solving the equation without using the hint provided in my exercise and directly applying completing square, I get $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-c}{a}}$. So if I am to use the hint, I obtain the appropriate answer. But I wonder if I am asked the same question in my exam where the hint will not be provided then how am I supposed to answer.
I would like to know how one should approach this kind of question and how do I realise when to divide the whole equation with in this case $a$ or is there any other ways so that I can avoid dividing the whole equation by $a$.Thanks in advance for any help you are able to provide!
EDIT: Here's my steps. Please see where have I done wrong.
\begin{align}
ax^2+2bx+c&=0 \\
a\left[\left(x+\frac{b}{a}\right)^2-\frac{b^2}{a^2}\right] + c&=0 \\
a\left(x+\frac{b}{a}\right)^2-\frac{b^2}{a} + c&=0 \\
\left(x+\frac{b}{a}\right)^2&=\left(\frac{b^2}{a}-c\right)\left(\frac{1}{a}\right)\\
\left(x+\frac{b}{a}\right)^2&=\frac{a(b^2-c)}{a^2}\\
\left(x+\frac{b}{a}\right)^2&=\frac{b^2-c}{a} \\
x+\frac{b}{a}&=\pm\sqrt{\frac{b^2-c}{a}} \\
\implies x&=-\frac{b}{a}\pm\sqrt{\frac{b^2-c}{a}}\\
\end{align}
| As JCAA said, you should assume $a\ne 0$, otherwise, it is not a quadratic equation, but a linear equation. Also, division by zero is undefined. We just could divide both sides by $a$ because of that.
$ax^2+2bx+c=0 \implies x^2+2\dfrac{b}{a}x+\dfrac{a}{c}=0 \implies x^2+2\dfrac{b}{a}x=-\dfrac{c}{a}$
Completing the square,
$x^2+2\dfrac{b}{a}x + \color{red}{\dfrac{b^2}{a^2}} =-\dfrac{c}{a}+ \color{red}{\dfrac{b^2}{a^2}}\\\\ \left(x+\dfrac{b}{a}\right)^2 = \dfrac{b^2-ac}{a^2} \\\boxed{x=-\dfrac{b}{a}\pm \sqrt{\dfrac{b^2-ac}{a^2}}} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the area between $r=1$ and $r=3\cos\theta$
Find the area between $r=1$ and $r=3\cos\theta$.
I squared both sides to get $r^2 = 1$, then did $r^2(\cos^2 \theta + \sin^2 \theta) = (r \cos \theta)^2 + (r \sin \theta)^2$$ = x^2+y^2 = 1$ to get $x^2+y^2=1$.
For $r = 3 \cos \theta$, I multiplied by $r$ on both sides to get $r^2 = 3r \cos \theta$, then substituted $x = r \cos \theta$ to get $x^2+y^2 =3x$. However, I don't know if it is easier to do it this way.
If not, how can I find this area?
| The area under a polar function $r(\theta)$ is:
$$\frac{1}{2} \int_a^b \big(r(\theta) \big)^2 \ d \theta$$
Here $r(\theta) = 3 \cos \theta - 1$. Now find the limits of integration, and use the identity $\cos^2 \theta = \frac{1}{2} (\cos 2 \theta+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$
$\cos^6x + \sin^6x = -3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = (-3/2)(2\cos^2x\sin^2x)$
$\cos^6x + \sin^6x = (-3/2)(\sin^22x)$
How can I get it into $ 1 - (3/4)\sin^2(2x)$?
| You have $$(\cos^2(x)+\sin^2(x))^3=\cos^6(x)+\sin^6(x)+3\cos^4(x)\sin^2(x)+3\sin^4(x)\cos^2(x)$$
$$=\cos^6(x)+\sin^6(x)+3\sin^2(x)\cos^2(x)(\cos^2(x)+\sin^2(x))$$
$$=\cos^6(x)+\sin^6(x)+\frac{3}{4}\cdot 4\sin^2(x)\cos^2(x)$$
$$=\cos^6(x)+\sin^6(x)+\frac{3}{4}(2\sin(x)\cos(x))^2$$
$$=\cos^6(x)+\sin^6(x)+\frac{3}{4}\sin^2(2x)$$
using the identity $\cos^2(x)+\sin^2(x)=1$ and the double angle formula $\sin(2x)=2\sin(x)\cos(x)$. Now the LHS is also equal to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 6
} |
PRMO level question of functions Consider the functions $f(x)$ and $g(x)$ which are defined as $f(x)=(x+1)(x^2+1)(x^4+1)\ldots\left({x^2}^{2007}+1\right)$ and $g(x)\left({x^2}^{2008}-1\right)= f(x)-1$.
Find $g(2)$
This is a PRMO level question of functions and I tried it with substituting values also but to no avail and the solution Of this question is also not available thought answer is given as $g(x)=2$.
| We can write
$$
g\left( x \right) = \frac{{\left( {x + 1} \right)\left( {x^2 + 1} \right) \ldots \left( {x^{2^{2007} } + 1} \right) - 1}}
{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {x^2 + 1} \right) \ldots \left( {x^{2^{2007} } + 1} \right)}}$$
and then get
$$
g\left( x \right) = \frac{1}
{{x - 1}} - \frac{1}
{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {x^2 + 1} \right) \ldots \left( {x^{2^{2007} } + 1} \right)}}$$
So $g(2) = 1-\epsilon$
where $\epsilon=\frac{1}
{{\left( {2 + 1} \right)\left( {2^2 + 1} \right) \ldots \left( {2^{2^{2007} } + 1} \right)}}=\frac{1}{4^{2008}-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
find the largest integer $m$ such that $2^m$ divides $3^{2n+2}-8n-9$
find the largest integer $m$ such that $2^m$ divides $\space 3^{2n+2}-8n-9$ when $n$ is a natural number.
If the answer was known it will be easy induction.
I started out like this :
$\space 3^{2n+2}-8n-9=9(3^{2n}-1)-8n=9\underbrace{(3^n-1)(3^n+1)}-8n$
Now we have $\frac{3^n-1}{3-1}$ is some integer (sum of GP),or
$ 2|\space 3^n-1$
also we have $3^n+1$ is even ,or
$2|3^n+1....(3)$
From this we conclude $4|(3^n-1)(3^n+1) ...(1)$
Let n be even then $3^n-1=3^{2m}-1=(3^m-1)(3^m+1)$,
by $(1)$ :
$4|(3^m+1)(3^m-1)$ meaning $4|3^n-1...........(2)$
combining $(2),(3)$ we have $8|3^{2n+2}-8n-9$
Similarly i was able to work out the same when $n=2m+1$ by noting that $3^n+1=3^{2m+1}+1$ is divisible by $4$.
I got the largest integer as $3$.
But i am wrong as the MCQ did not have the option $m=3$
how do i proceed.
Note: I have not learnt about fermat's little theorem
Also i am looking for Hints rather than complete solutions .use of >! may help
| Hint: For $n=1$, it is clear which is the largest power of $2$. Now consider $(8+1)^{n+1} - [(n+1)\cdot8+1]$ and use binomial expansion to conclude it works for all larger $n$.
| {
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$\frac{x-c}{x-y} |\frac{f(x) - f(c)}{x-c} - f'(c)| + \frac{c-y}{x-y}|\frac{f(y) - f(c)}{y-c} - f'(c)| < (\frac{x-c}{x-y} + \frac{c-y}{x-y}) \epsilon$? After substituting the green expression, how do you deduce the last two inequalities below? I'm guessing Triangle Inequality? But after you substitute the green expression, you have three terms, whilst Triangle Inequality has just two variables.
Since $f(x)−f(y)=f(x)\color{red}{−f(c)+f(c)}−f(y)$, a simple
calculation shows that
$\color{green}{\dfrac{f(x) - f(y)}{x-y} = \dfrac{x-c}{x-y}\cdot\dfrac{f(x) - f(c)}{x-c} +\dfrac{c-y}{x-y}\cdot \dfrac{f(y) - f(c)}{y-c}}$
Since both $\dfrac{x-c}{x-y}$ and $\dfrac{c-y}{x-y}$ are positive and sum to 1, it follows that
$\begin{align}
& \left|\color{green}{\dfrac {f(x)-f(y)}{x-y}}-f'(c)\right| \\
& \le (\frac{x-c}{x-y})\left|\frac{f(x) - f(c)}{x-c} - f'(c) \right| +(\frac{c-y}{x-y})\left|\frac{f(y) - f(c)}{y-c} - f'(c)\right| \\
& < (\frac{x-c}{x-y} + \frac{c-y}{x-y})\epsilon \\
& \qquad = \epsilon \end{align}$
| Supposing $x<c<y$, we justify it by
$$1=\frac{x-y}{x-y}=\frac{x-c}{x-y}+\frac{c-y}{x-y}$$
and with the calculations (using the triangle inequality in the "$\leq$" step)
\begin{align*}
\left|\frac{f(x)-f(y)}{x-y}-f'(c)\right| &=\left|\frac{f(x)-f(c)}{x-y}+\frac{f(c)-f(y)}{x-y}-\left(\frac{x-c}{x-y}+\frac{c-y}{x-y}\right)f'(c)\right|\\
&\leq\left|\frac{f(x)-f(c)}{x-y}-\frac{x-c}{x-y}f'(c)\right|+\left|\frac{f(c)-f(y)}{x-y}-\frac{c-y}{x-y}f'(c)\right|\\
&=\left(\frac{x-c}{x-y}\right)\left|\frac{f(x)-f(c)}{x-c}-f'(c)\right|+\left(\frac{c-y}{x-y}\right)\left|\frac{f(c)-f(y)}{c-y}-f'(c)\right|
\end{align*}
Addendum:
I had missed you asking about the second inequality. Here it goes:
By the fact that $f:I=[a,b]\to \mathbb{R}$ is differentiable at $c\in I$, we can find, for any arbitrarily fixed $\epsilon > 0$
a $\delta:=\delta(\epsilon)>0$ such for all $x\in I$ with $|x-c|<\delta$, we have $$\left|\frac{f(x)-f(c)}{x-c}-f'(c)\right|<\epsilon.$$
Start from the last expression above and assume $|x-c|<\delta$ and $|c-y|<\delta$. We get
$$\left(\frac{x-c}{x-y}\right)\left|\frac{f(x)-f(c)}{x-c}-f'(c)\right|+\left(\frac{c-y}{x-y}\right)\left|\frac{f(c)-f(y)}{c-y}-f'(c)\right| \\<
\left(\frac{x-c}{x-y}\right)\epsilon + \left(\frac{c-y}{x-y}\right)\epsilon =\left(\frac{x-c}{x-y}+\frac{c-y}{x-y}\right)\epsilon = \epsilon
$$
| {
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Find x so that $\left(\frac{3}{2}+ \sum_{i=1}^{x} 3^{i}\right)^2$ Find $x$ so that $$\left(\frac{3}{2}+ \sum_{i=1}^{x} 3^{i}\right)^2 = \frac{3^{22}}{4}$$
I've tried with simpler values for $x$ such as $0, 1$ and $2$. But I can't seem to find any pattern I can take advantage of. How do I solve it? Where would I learn how to solve things like these?
| $\dfrac{3^{22}}{4} = \left(\dfrac{3^{11}}{2}\right)^2$
So we have,
$$\frac{3}{2} + \sum_{i=1}^x 3^{i} = \frac{3^{11}}{2} \Rightarrow \sum_{i=1}^x 3^{i} = \frac{3^{11}-3}{2} \Rightarrow 3\left(\frac{3^\color{red}x-1}{2}\right) = 3\left(\frac{3^\color{red}{10}-1}{2}\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove that $\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$
prove that $$\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$$ for positives $a,b,c$
Attempt: By C-S; $$\left(\sum_{cyc}\frac{a}{b^2+c^2} \right) \left(\sum_{cyc} a(b^2+c^2) \right)\ge {(a+b+c)}^2$$ .
or as inequality is homogenous we take $a+b+c=1$.
or we have to prove (i am skipping the steps as it is just algebra) :
$$ 5(ab+bc+ca-abc)\ge 4(1+ab+bc+ca)(ab+bc+ca-3abc)$$
But i am not able to prove this by expanding.
How do i proceed?
Other methods are welcome!
| Let $p=a+b+c=1, \; q=ab+bc+ca, \; r=abc.$ We need to prove
$$5(ab+bc+ca-abc) \geqslant 4(1+ab+bc+ca)(ab+bc+ca-3abc),$$
equivalent to
$$5(q-r) \geqslant 4(1+q)(q-3r),$$
or
$$(12q+7)r \geqslant q(4q-1).$$
If $4q-1 < 0,$ then
$$(12q+7)r > 0 > q(4q-1).$$
If $4q-1 \geqslant 0,$ from Schur inequality
$$(a+b+c)^3+9abc \geqslant 4(a+b+c)(ab+bc+ca),$$
we get
$$r \geqslant \frac{p(4q-p^2)}{9} = \frac{4q-1}{9}.$$
It's remain to prove that
$$(12q+7) \cdot \frac{4q-1}{9} \geqslant q(4q-1),$$
or
$$\frac{(3q+7)(4q-1)}{9} \geqslant 0.$$
Which is true. The proof is completed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the smooth inverse of a function Problem: The mapping $\phi: S^2 \longrightarrow S^2 $ by $$\phi(x,y,z)=(x\cos z+y\sin z,x\sin z-y \cos z,z)$$ is a diffeomorphism. Where $S^2$ is a unit sphere in $\mathbb{R}^3$.
I've already shown that $\phi$ is smooth and bijective. The only thing I can't find is $\phi^{-1}$ that is smooth.
Any help would be much appreciated!
| To find $\phi^{-1}$ explicitly, let's define
$$ \begin{align*} a &= x \cos z + y \sin z \\
b &= x \sin z - y \cos z \\
c &= z \end{align*} $$
so that $\phi(x,y,z) = (a,b,c)$ and $\phi^{-1}(a,b,c) = (x,y,z)$. It just remains to write $x,y,z$ in terms of $a,b,c$.
Getting rid of $z$ is obvious:
$$ \begin{align*} a &= x \cos c + y \sin c \\
b &= x \sin c - y \cos c \end{align*} $$
To eliminate $x$, multiply the equations so both include the term $(x \cos c \sin c)$, then subtract:
$$ \begin{align*} a \sin c &= x \cos c \sin c + y \sin^2 c \\
b \cos c &= x \cos c \sin c - y \cos^2 c \\
a \sin c - b \cos c &= y (\sin^2 c + \cos^2 c) = y \end{align*}$$
Similarly eliminating $y$,
$$ \begin{align*} a \cos c &= x \cos^2 c + y \sin c \cos c \\
b \sin c &= x \sin^2 c - y \sin c \cos c \\
a \cos c + b \cos c &= x(\cos^2 c + \sin^2 c) = x \end{align*} $$
So finally,
$$ \phi^{-1}(a,b,c) = (a \cos c + b \sin c, a \sin c - b \cos c, c) $$
Look familiar? It happens that $\phi^{-1} = \phi$. This is because at each fixed value of $z=c$, the relationship between $(x,y)$ and $(a,b)$ describes a reflection of the $\mathbb{R}^2$ plane.
| {
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Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ We are given this limit to evaluate:
$$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this expression by the conjugate of the denominator.
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \right) = \lim _{x\to 2}\left(\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}\right) $$
But it still gives the indeterminate form $\frac{0}{0}$ .
But multiplying the expression by the conjugate of the demoninator and numerator we get
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \cdot \dfrac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\right) $$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{1} \cdot \dfrac{1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{2-x}{2-x} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
Now we can evaluate the limit:
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-2}+1}{\sqrt{6-2}+2}\right) = \dfrac{1}{2}$$
Taking this example, I would like to understand why rationalization was used. What did it change in the expression so the evaluation was possible? Especially, why multiplying by the numerator's and denominator's conjugate?
I am still new to limits and Calculus, so anything concerning concepts I'm missing is appreciated. I still couldn't understand how a limit supposedly tending to $\frac{0}{0}$ went to be $\frac{1}{2}$, I really want to understand it.
Thanks in advance for you answer.
| Option:
$z:=\sqrt{3-x}$; and consider $\lim z \rightarrow 1;$
$\dfrac{\sqrt{3+z^2}-2}{z-1}=$
$\dfrac{z^2-1}{(z-1)(\sqrt{3+z^2}+2)}=$
$\dfrac{z+1}{\sqrt{3+z^2}+2}.$
Take the limit.
| {
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Let $a_1,a_2,...,a_n>0$ be such that $\sum_{k=1}^n a_k=\frac{1}{2}n(n+1)$
Let $a_1,a_2,...,a_n>0$ be such that $\sum_{k=1}^n a_k=\frac{1}{2}n(n+1)$, then least value of
$$\sum_{k=1}^n\frac{(k^2-1)a_k+k^2+2k}{a_k^2+a_k+1}\text{ is,}$$
What I tried:
$$\sum_{k=1}^{n}k=\frac{1}{2}n(n+1)\implies k=a_k\tag{$\because\sum_{k=1}^n a_k=\frac{1}{2}n(n+1)$}$$
So replacing $a_k$ with $k$,
$$\begin{aligned}\require{cancel}
\sum_{k=1}^n\frac{(k^2-1)a_k+k^2+2k}{a_k^2+a_k+1}&=\sum_{k=1}^n\frac{(k^2-1)k+k^2+2k}{k^2+k+1}\\
&=\sum_{k=1}^{n}\frac{k^3-k+k^2+2k}{k^2+k+1}\\
&=\sum_{k=1}^{n}\frac{k^3+k+k^2}{k^2+k+1}\\
&=\sum_{k=1}^{n}\frac{k\color{red}{\bcancel{(k^2+k+1)}}}{\color{red}{\bcancel{(k^2+k+1)}}}\\
&=\sum_{k=1}^{n}k=\frac{1}{2}n(n+1)
\end{aligned}$$
Is this a correct method to solve this problem? because something feels off with this as the question asks for the least value but I get a direct result.
Please provide a correct method to solve if this is wrong
| You can use the fact that
$$ a_n=\sum_{k=1}^n a_k-\sum_{k=1}^{n-1} a_k=\frac{1}{2}n(n+1)-\frac{1}{2}(n-1)n=n. $$
The rest is correct.
| {
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Do the series converge? Let $g \colon \mathbb{N} \rightarrow \{0,1\}$ such that $g(n)=0$, if $n \equiv 0$ or $n \equiv 1 ~(\bmod~4)$; or $g(n)=1$, if $n \equiv 2$ ou $n \equiv 3 ~(\bmod~4)$.
Do $\sum_{n=1}^\infty \frac{(-1)^{g(n)}}{\sqrt{n}}$ converge?
The series is something like this
*
*If $n=1$, then $\frac{(-1)^{g(n)}}{\sqrt{n}} = \frac{(-1)^{0}}{\sqrt{1}} = 1$, partial sum of 1
*If $n=2$, then $\frac{(-1)^{g(n)}}{\sqrt{n}} = \frac{(-1)^{1}}{\sqrt{2}} \approx -0.707$, partial sum of $\approx$ 0.293
*If $n=3$, then $\frac{(-1)^{g(n)}}{\sqrt{n}} = \frac{(-1)^{1}}{\sqrt{3}} \approx -0.577$, partial sum of $\approx$ -0.284
*If $n=4$, then $\frac{(-1)^{g(n)}}{\sqrt{n}} = \frac{(-1)^{0}}{\sqrt{4}} = 0.5$, partial sum of $\approx$ 0.216
*$\cdot\cdot\cdot$ and goes on
What I thought of doing?
*
*Break the series (call it $a_n$) into two ($b_n$ and $c_n$), where $b_n$ is the positive terms and make the negative terms = $0$; and $c_n$ is only the negative terms, and the terms which are indices of positive = $0$. So we have an equality of $a_n = b_n - c_n$. If I show that either $b_n$ or $c_n$ diverges then $a_n$ diverges?
*I know, by using Alternate Series Test, that $\sum _{n=1}^{\infty }\frac{(-1)^n}{\sqrt{n}}$ converges, but does this also means $\sum_{n=1}^\infty \frac{(-1)^{g(n)}}{\sqrt{n}}$ converges? It is not properly a reordering, so I have my doubts.
I am not sure of any of the ways, so thats why I am here. Thank you.
| We have
$$\sum_{n=1}^\infty \frac{(-1)^{g(n)}}{\sqrt{n}}=\frac{1}{1}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\cdots$$
Now, note that for for $n\equiv 2\ (\text{mod}\ 4)$ we have the terms
$$\cdots-\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\frac{1}{\sqrt{n+3}}-\cdots$$
$$\cdots-\frac{\sqrt{n}+\sqrt{n+1}}{\sqrt{n(n+1)}}+\frac{\sqrt{n+2}+\sqrt{n+3}}{\sqrt{(n+2)(n+3)}}$$
That is, the sum can be rewritten as
$$\sum_{n=1}^\infty \frac{(-1)^{g(n)}}{\sqrt{n}}=1+\sum_{n=1}^\infty (-1)^n\frac{\sqrt{2n}+\sqrt{2n+1}}{\sqrt{2n(2n+1)}}$$
But we see that the term in this sum is decreasing. We conclude by the alternating series test that the sum converges
REQUESTED EDIT: Note that
$$0\leq \left(\frac{\sqrt{2n}+\sqrt{2n+1}}{\sqrt{2n(2n+1)}}\right)^2=\frac{4n+1+2\sqrt{2n(2n+1)}}{2n(2n+1)}$$
$$\leq\frac{5n+2\sqrt{2n(2n+1)}}{2n(2n+1)}=\frac{5}{4n+2}+\frac{2}{\sqrt{2n(2n+1)}}$$
$$<\frac{5}{4n}+\frac{2}{\sqrt{2n(2n-n)}}=\frac{5}{4n}+\frac{2}{n\sqrt{2}}$$
$$<\frac{2}{n}+\frac{2}{n}=\frac{4}{n}\to 0$$
SECOND REQUESTED EDIT: Note that
$$1=\frac{1}{1}$$
$$n=1:\ (-1)^1\frac{\sqrt{2\cdot 1}+\sqrt{2\cdot 1+1}}{\sqrt{2\cdot 1(2\cdot 1+1)}} =-\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}\sqrt{3}}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}$$
$$n=2:\ (-1)^2\frac{\sqrt{2\cdot 2}+\sqrt{2\cdot 2+1}}{\sqrt{2\cdot 2(2\cdot 2+1)}} =\frac{\sqrt{4}+\sqrt{5}}{\sqrt{4}\sqrt{5}}=\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}$$
$$n=3:\ (-1)^3\frac{\sqrt{2\cdot 3}+\sqrt{2\cdot 3+1}}{\sqrt{2\cdot 3(2\cdot 3+1)}} =-\frac{\sqrt{6}+\sqrt{7}}{\sqrt{6}\sqrt{7}}=-\frac{1}{\sqrt{6}}-\frac{1}{\sqrt{7}}$$
$$\vdots$$
The left hand side is the terms we found while the right hand side are the original terms.
| {
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"timestamp": "2023-03-29T00:00:00",
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Polynomial with root $α = \sqrt{2}+\sqrt{5}$ and using it to simplify $α^6$ Find a polynomial $\space P(X) \in \mathbb{Q}[X]\space$ of degree 4 such that
$$\alpha = \sqrt{2} + \sqrt{5}$$
Is a root of $P$.
Using this polynomial, find numbers $\space a, b, c, d \space$ such that
$$\alpha^{6} = a + b\alpha + c\alpha^{2} + d\alpha^{3}$$
What have I tried so far?
I know obviously that for $\alpha$ to be a root of $P$, then $(x-\alpha)$ must be part of the polynomial. Hence, $(x-\sqrt{2} - \sqrt{5})$ will be a factor of the polynomial. Where I’m getting stuck is what to do next in order to find the other factors of the polynomial such that I get values $a, b, c$ and $d$ that satisfy the equation with $\alpha^{6}$.
Any help would be greatly appreciated!
| We can use the following identity.
$$(a+b+c)\prod_{cyc}(a+b-c)=\sum_{cyc}(2a^2b^2-a^4).$$
We obtain:
$$2\alpha^2\cdot2+2\alpha^2\cdot5+2\cdot2\cdot5-\alpha^4-2^2-5^2=0$$ or
$$\alpha^4-14\alpha^2+9=0.$$
From here $$\alpha^6=14\alpha^4-9\alpha^2=14(14\alpha^2-9)-9\alpha^2=187\alpha^2-126,$$ which gives $$(a,b,c,d)=(-126,0,187,0).$$
| {
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Solution of first order inexact differential equation I have following differential equation before me:
$(2y sin(x)+3y^4 sin(x)cos(x))dx-(4y^3 cos^2(x)+cos(x))dy=0$
It is an inexact differential equation. Its Integrating factor comes out to be $cos(x)$.
New $M= 2ysin(x)cos(x)+3y^4sin(x)cos^2(x)$
New $N= -(4y^3cos^3(x)+cos^2(x))$
When I integrate $M$ treating $y$ as a constant, there are more than one way of doing this.
The term $2sin(x)cos(x)$ in $M$ can be integrated to give either $sin^2(x)$ or $-cos(2x)/2$ which differ by a constant. Also in $N$,$cos^2(x)$ could be written either as $1-sin^2(x)$ or $(1+cos(2x))/2$ giving the terms which do not contain $x$ as either $1$ or $1/2$.
All these integrands yield different solutions. I am confused as to which one of these is the correct solution? Are all of them correct?
| $$-\dfrac {\cos (2x) }2=\dfrac 12 (2\cos^2 x-1)=\dfrac 12(1-2\sin^2 x)$$
All the constants are absorbed by the constant on the right side of the solution:
$$F(x,y)=C$$
$$(2y \sin(x)+3y^4 \sin(x)\cos(x))dx-(4y^3 \cos^2(x)+\cos(x))dy=0$$
$$y d(\cos^2x)+y^4 d(\cos^3(x))+ \cos^3(x)dy^4+cos^2(x)dy=0$$
$$ d(y\cos^2x)+d(y^4\cos^3(x))=0$$
After integration:
$$\boxed {y\cos^2x+y^4\cos^3(x)=K}$$
The constant on the right side absorbs all the constants so all the solutions are equivalent and correct.
I am not sure to clearly understand the problem but let's integrate the way you did:
$$M= 2y\sin(x)\cos(x)+3y^4\sin(x)\cos^2(x)$$
We have that
$$\partial_x F=M$$
$$F= \int Mdx$$
$$F=\int 2y\sin(x)\cos(x)+3y^4\sin(x)\cos^2(x)dx$$
Don't forget the constant of integration:
$$F=y\sin^2(x)-y^4\cos^3(x)+\color{red}{g(y)}$$
$$F=-y\cos^2(x)-y^4\cos^3(x)+\color{red}{g(y)+y}$$
$$F=-y\cos^2(x)-y^4\cos^3(x)+\color{red}{h(y)}$$
Do the same for $N$:
$$F =\int -(4y^3\cos^3(x)+\cos^2(x))dy$$
$$F = -y^4\cos^3(x)-y\cos^2(x) +\color {green}{r(x)}$$
From these two equations about $F$ we deduce that $h(y)=r(x)=c$. So that the solution is now:
$$F(x,y)=K$$
$$y^4\cos^3(x)+y\cos^2(x) =\color {green}{C}$$
It ends with the same answer as we find before.
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Prove Segner's Recurrence Relation $C_{n+1} = \sum\limits_{i=0}^n C_i C_{n-i}$ on Catalan Numbers $C_n = \frac{1}{n+1} \binom{2n}{n}$
Prove Segner's Recurrence Relation $C_{n+1} = \sum\limits_{i=0}^n C_i C_{n-i}$ on Catalan Numbers $C_n = \frac{1}{n+1} \binom{2n}{n}$
Plugging in the Catalan Equation, we want to prove:
\begin{align*}
\frac{1}{n+2} \binom{2(n+1)}{n+1} &= \sum\limits_{i=0}^n \frac{1}{i+1} \binom{2i}{i} \frac{1}{n-i+1} \binom{2(n-i)}{n-i} \\
\end{align*}
Expanding the binomial coefficients to factorials (again this is a formula we want to prove, not a result we have demonstrated):
\begin{align*}
\frac{1}{n+2} \frac{(2n+2)!}{(n+1)!(n+1)!} &= \sum\limits_{i=0}^n \frac{1}{i+1} \frac{(2i)!}{i!i!} \frac{1}{n-i+1} \frac{(2n-2i)!}{(n-i)!(n-i)!} \\
\end{align*}
How would one go from here?
| I think the best way to solve Segner's recurrence relation is to use generator functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3871430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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System of equations (Problem $50$ from $101$ algebra by Titu)
Let $a$ and $b$ be given real numbers.
Solve the system of equations
$$\begin{aligned} \frac{x-y \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}} &=a \\ \frac{y-x \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}} &=b \end{aligned}$$
for real $x$ and $y$.
Solution -
Let $u=x+y$ and $v=x-y .$ Then
$$
0<x^{2}-y^{2}=u v<1, x=\frac{u+v}{2}, \text { and } y=\frac{u-v}{2}
$$
Adding the two equations and subtracting the two equations in the original system yields the new system
$$
\begin{aligned}
u-u \sqrt{u v} &=(a+b) \sqrt{1-u v} \\
v+v \sqrt{u v} &=(a-b) \sqrt{1-u v}
\end{aligned}
$$
Multiplying the above two equations yields
$$
u v(1-u v)=\left(a^{2}-b^{2}\right)(1-u v)
$$
hence $u v=a^{2}-b^{2} .$ It follows that
$$
u=\frac{(a+b) \sqrt{1-a^{2}+b^{2}}}{1-\sqrt{a^{2}-b^{2}}} \text { and } v=\frac{(a-b) \sqrt{1-a^{2}+b^{2}}}{1+\sqrt{a^{2}-b^{2}}}
$$
I did not get how they found values of $u$ and $v$ from
$u v=a^{2}-b^{2} .$ I mean obviously we can substitute the value in one of the equations and we will get some quadratic and we can find solution from there but the quadratic that I am getting is very large to handle, so is there some obvious step that directly lead solutions from $u v=a^{2}-b^{2}?$
Thank you
| $$u-u\sqrt{uv} = (a+b)\sqrt{1-uv}$$
$$u = (a+b)\dfrac{\sqrt{1-uv}}{1-\sqrt{uv}}$$
Now substitute $uv=a^2-b^2$
| {
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"source": "stackexchange",
"question_score": "2",
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Probability when choosing beads with repetitions There are 20 different beads, 4 of which are yellow. Choosing 30 beads with repetitions, I need to calculate:
*
*The probability that exactly 5 of the chosen beads are yellow
*The probability that the first and last beads are yellow and exactly 5 yellow beads were chosen (including the first and last beads)
The first one I calculated saying there are $4^5$ ways to chose 5 yellow beads and $16^{25}$ ways to choose the rest hence $$P_1 = \frac{4^5\cdot 16^{25}}{20^{30}}=\frac{4^{55}}{4^{30}\cdot5^{30}}=(0.8)^{25}\cdot(0.2)^{5}$$ As for the second probability I first tried choosing $\binom{5}{2}$ yellow beads for first and last and then multiply by 2 as they can change places and again multiply by the permutation of the remaining beads $28!$ hence $$P_2 = P_1\cdot\frac{\binom{5}{2}\cdot2\cdot28!}{30!}$$ but this is incorrect because there are repetitions. So the second attempt was choosing first and last yellow bead which can be done in $5^2$ and then $$P_2 = P_1\cdot\frac{5^2\cdot28!}{30!}$$ but this seems wrong too. My intuition says that the probability between 1 and 2 remains the same but I haven't found a way to prove (or disprove) it
| Because we are asked about probability, and not about the number of combinations, and because we are choosing with repetitions - we can ignore the identity of the individual beads and instead treat each choice as a $\frac{4}{20} = 20\%$ probability to choose yellow and $80\%$ probability to choose non-yellow.
For any given choice of the 5 places where yellow beads are chosen, the probability to choose it is $$
0.2^5\cdot 0.8^{25}
= \frac{2^5\cdot 8^{25}}{10^5\cdot 10^{25}}
= \frac{2^5\cdot 2^{75}}{10^5\cdot 10^{25}}
= \frac{2^{80}}{10^{30}}
$$
Since each such case is disjoint to the others, we can calculate how many ways there are to choose these 5 places and multiply: $$\boxed{
\frac{2^{80}}{10^{30}} \cdot \binom{30}{5}
= \frac{2^{80}}{10^{30}} \cdot \frac{30!}{5!\cdot 25!}
\approx 17.2279\%
}$$
For the second, the first and last places always have yellow beads so we only choose 3 places out of 28:
$$\boxed{
\frac{2^{80}}{10^{30}} \cdot \binom{28}{3}
= \frac{2^{80}}{10^{30}} \cdot \frac{28!}{3!\cdot 25!}
\approx 0.396\%
}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Construct an isomorphism between $V$ and $\Bbb F^2$ and justify Consider the subspace of $M_2(\Bbb F):V = \biggl\{\begin{pmatrix}a+b&a\\0&b\end{pmatrix}\Biggm\vert a,b\in \Bbb F\biggr\}$
Construct an isomorphism between $V$ and $\Bbb F^2$ and justify
| Consider $\phi:V\to F^2$ defined as $\phi\begin{pmatrix}a+b&a\\0&b\end{pmatrix}=(a,b)$.
Then
$\phi\Big(\begin{pmatrix}a+b&a\\0&b\end{pmatrix}+\begin{pmatrix}c+d&c\\0&d\end{pmatrix}\Big)=\phi\begin{pmatrix}a+b+c+d&a+c\\0&b+d\end{pmatrix}=(a+c,b+d)=(a,b)+(c,d)=\phi\begin{pmatrix}a+b&a\\0&b\end{pmatrix}+\phi\begin{pmatrix}c+d&c\\0&d\end{pmatrix}$
$\phi\Big(\lambda\begin{pmatrix}a+b&a\\0&b\end{pmatrix}\Big)=\phi\begin{pmatrix}\lambda a+\lambda b&\lambda a\\0&\lambda b\end{pmatrix}=(\lambda a,\lambda b)=\lambda(a,c)=\lambda\phi\begin{pmatrix}a+b&a\\0&b\end{pmatrix}$
So $\phi$ is linear.
If $\phi\begin{pmatrix}a+b&a\\0&b\end{pmatrix}=\phi\begin{pmatrix}c+d&c\\0&d\end{pmatrix}$ then $(a,b)=(c,d)$, so $\begin{pmatrix}a+b&a\\0&b\end{pmatrix}=\begin{pmatrix}c+d&c\\0&d\end{pmatrix}$, so $\phi$ is injective.
Take some $(r,s)\in F^2$, then we can consider $\begin{pmatrix}r+s&r\\0&s\end{pmatrix}$ and thus $\phi\begin{pmatrix}r+s&r\\0&s\end{pmatrix}=(r,s)$, so $\phi$ is surjective.
We conclude $\phi$ is an isomorphism, so $V\simeq F^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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cyclic rational inequalities $\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$ when $a+b+c=1$ I've been practicing for high school olympiads and I see a lot of problems set up like this:
let $a,b,c>0$ and $a+b+c=1$. Show that
$$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$$
Any problem that involves cyclic inequalities like these always stump me. I know I'm supposed to use Cauchy-Schwarz or AM-GM at some point, but I can never get to a place where this might be useful. My first instinct is to get common denominators and hope stuff simplifies, but I can never get farther than that. For example, in this problem I did the following:
$$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}$$
$$=\frac{(a^2+3)(b^2+3)+(b^2+3)(c^2+3)+(c^2+3)(a^2+3)}{(a^2+3)(b^2+3)(c^2+3)}$$
$$=\frac{a^2b^2+b^2c^2+c^2a^2+6(a^2+b^2+c^2)+27}{(a^2+3)(b^2+3)(c^2+3)}$$
but this is where I get stuck. I've tried using Cauchy-Schwarz on parts of this fraction to simplify it, but I can never get anything to work. How could you prove this inequality, and what are the important things to look out for in problems of this nature
| We need to prove $\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$ for $a, b, c \gt 0, a + b + c = 1$
Using tangent line method,
We consider the equation of the tangent line to $f(x) = \frac{1}{3+x^2}$ at $x = \frac{1}{3}$. Point is $(\frac{1}{3}, \frac{9}{28})$
$f'(x) = -\frac{2x}{(3+x^2)^2} = -\frac{27}{392}$
So equation of tangent line $y = -\frac{27}{392} x+ c$
Given the point on the line, $y = -\frac{27}{392} x + \frac{135}{392}$
We claim that $f(x) = \frac{1}{3+x^2} \leq -\frac{27}{392} x + \frac{135}{392}$ ...(i)
it is equivalent of saying $\frac{(135-27x)(3+x^2)}{392} \geq 1$ for $0 \lt x \leq 1$
which is true and equality occurs for $x = \frac{1}{3}$.
Now we know at $x = \frac{1}{3}, f(x) = \frac{9}{28}$
So, $\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$
EDIT: you could do it from (i) as follows too
$f(a) = \frac{1}{3+a^2} \leq -\frac{27}{392} a + \frac{135}{392}$ (same for $b$ and $c$)
$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3} \leq -\frac{27}{392} (a + b + c) + \frac{3 \times 135}{392} \leq \frac{27}{28} \, ($as $\, a + b + c = 1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Find $\frac{\mathrm{d} }{\mathrm{d} x}x\sin \left ( \sqrt{3x^{2}+5} \right )$ without using the chain rule. $$\frac{\mathrm{d} }{\mathrm{d} x}x\sin \left ( \sqrt{3x^{2}+5} \right )$$
I can't for the life of me differentiate this function while only using Trig Identities, Basic differentiation rules, and Limits (no L'Hopital, either.)
| $$f'=\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x)^2+5}}{t}$$ separate it into two limitation by $\pm(x)\sin\sqrt{3(x+t)^2+5}$
this means
$$f'=\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x)^2+5}}{t}=\\
=\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}\pm(x)\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x)^2+5}}{t}\\
=\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x+t)^2+5}}{t}+\lim_{t \to 0}\frac{+x\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x)^2+5}}{t}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3880196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
What I Tried: Here is a picture :-
I know the centroid divides each of the medians in the ratio $2:1$ . So $AD = 3\sqrt{3}$ , $BE = 3\sqrt{2}$ , $CF = 3$ .
From this site :- https://mathworld.wolfram.com/TriangleMedian.html, I find that the area of the triangle will be :- $$\frac{4}{3}\sqrt{s_m(s_m - m_1)(s_m - m_2)(s_m - m_3)}$$
Where $m_1,m_2,m_3$ are the medians of the triangle and $s_m = \frac{m_1 + m_2 + m_3}{2}$ .
After putting the respective values for the medians I get that $[\Delta ABC]$ is :-
$$\frac{4}{3}\sqrt{\Bigg(\frac{3(\sqrt{3} + \sqrt{2} + 1)}{2}\Bigg)\Bigg(\frac{3(\sqrt{2} + 1 - \sqrt{3})}{2}\Bigg)\Bigg(\frac{3(\sqrt{3} + 1 - \sqrt{2})}{2}\Bigg)\Bigg(\frac{3(\sqrt{3} + \sqrt{2} - 1)}{2}\Bigg)}$$
$$\rightarrow \frac{4}{3}\sqrt{\frac{81(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3})(\sqrt{3} + 1 - \sqrt{2})(\sqrt{3} + \sqrt{2} - 1)}{16}}$$
I am almost to the answer (assuming I made no mistake), but I think this simplification is getting complicated. How do I proceed next?
Can anyone help me?
| (This is not likely to be what you're looking for.)
I think in this problem you can use a simpler solution.
Construct point $H$ outside $\overline{AC}$ such that $AGCH$ forms a Parallelogram. We have
*
*$\overline{AH}=\overline{GC}=2$
*$\overline{AG}=2\sqrt3$
*$\overline{GE}=\overline{EH}\Longrightarrow \overline{GH}=\overline{GB}=2\sqrt2$
Since $\overline{AG}^2=\overline{AH}^2+\overline{GH}^2$, we know that $\angle AHG=90^{\circ}$.
Note that $\triangle AGE=\frac{1}2\triangle AGH=\frac{1}2\cdot\frac{1}2\cdot2\cdot2\sqrt2=\sqrt2$.
Therefore $\triangle ABC=6\triangle AGE=6\sqrt2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$.
Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$.
If $ab \mid (a+b)^²$, then $ab\mid a^2+2ab+b^2 \Longrightarrow ab\mid a^2, ab\mid 2ab$ and $ab\mid b^2$ right?
So since $(a-b)^2 = a^2-2ab+b^2$ from the assumption we have that $ab \mid a^2$ and $ab \mid b^2$. Now only remains to show that $ab \mid -2ab$ which is clearly true.
Is this valid? I'm not sure about the implication that $ab$ would divide all the terms in $a^2+2ab+b^2$.
| As others have said $x \mid (y+z) \nRightarrow (x \mid y) \land (x \mid z)$
What you can say is $ab \mid (a+b)^2 \iff (a+b)^2 = kab, k \in \mathbb{Z}$
This implies $a^2 + b^2 +2ab = kab \implies a^2 + b^2 = (k-2)ab \implies a^2 + b^2 - 2ab = (k-4)ab \implies (a-b)^2 = (k-4)ab \iff ab \mid (a-b)^2$, as required.
| {
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"timestamp": "2023-03-29T00:00:00",
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If the Symmedian/Lemoine point of triangle $ABC$ lies on the Altitude from vertex $C$, show that either $BC = AC$ or angle $C = 90 ^\circ$. If the Symmedian/Lemoine point of triangle $ABC$ lies on the Altitude from vertex $C$, show that either $BC = AC$ or angle $C = 90^\circ$.
I have proved it going the other way. When given that $C$ is $90^\circ$, I can show that the Lemoine point lies on the altitude, but I cannot figure out how to go backward or incorporate the $BC = AC$ part.
I have tried using the median from vertex $C$, which I know is an isogonal conjugate of the symmedian from $C$. But then from there I don't know where to go.
Thanks for your help.
| The coordinates of the symmedian point $X_6$ and
the foot of the altitude $H_c$
in terms of vertices $A(A_x,A_y),\ B(B_x,B_y),\ C(C_x,C_y)$ and side lengths $a,b,c$
can be found as
\begin{align}
X_6&= \frac{a^2\,A+b^2\,B+c^2\,C}{a^2+b^2+c^2}
\tag{1}\label{1}
,\\
H_c&=
\tfrac12\,(A+B)+\frac{a^2-b^2}{2c^2}\,(A-B)
\tag{2}\label{2}
.
\end{align}
Condition $X_6\in CH_c$ is equivalent to
\begin{align}
\operatorname{Im}
\left(
\frac{H_c-C}{X_6-C}
\right)
&=0
\tag{3}\label{3}
.
\end{align}
Let $C=(0,0)$, then the condition \eqref{3} simplifies to
\begin{align}
(b^2-a^2)(a^2+b^2-c^2)(A_x B_y-A_y B_x)
&=0
\tag{4}\label{4}
,
\end{align}
which holds either if $a=b$,
or $c^2=a^2+b^2$.
The third option, $A_x B_y=A_y B_x$,
corresponds to the degenerate case,
when all the vertices of $\triangle ABC$
are collinear.
| {
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How to write a polynomial function that has the roots $-2$ and $\sqrt7$? I need to write a polynomial function with integer coefficients that has the roots $-2$ and $\sqrt7$. I'm able to do this correctly when I'm given roots like $-3+i$ & $-3-i$, in which I set the roots equal to zero and then multiply them by one another. However, when I try this with $-2$ and $\sqrt7$ and multiply $x+2$ by $x-\sqrt7$, I get $x^2+2x-\sqrt7 x-2\sqrt7$. I don't know where to go from here, and I don't think that this is the correct next step. What do I do next?
| $x = \sqrt 7 \implies$
$x^2 = 7\implies$
$x^2 - 7 =0$. So $x^2 -7$ has $\sqrt 7$ as a root.
And $x=-20\implies x+2 = 0$ so $x+2$ has $-2$ as a root.
If $x^2-7=0$ and $x+2=0$ then $(x^2-7)(x+2)=0$ and $(x^2-7)(x+2)=x^3+2x^2-7x-14$ has $\sqrt 7$ and $-2$ as roots.
(It also has $-\sqrt 7$ as a root.)
=====
By the way. There is no 2nd degree polynomial with two roots $-2$ and $\sqrt 7$.
A second degree polynomial has at most two roots and if the two roots are $r_1$ and $r_2$ the polynomial is $(x-r_1)(x-r_2)= x^2 -(r_1+r_2)x + r_1r_2$ and if $r_1$ or $r_2$ are irrational we have no reason to assume $r_1+r_2$ or $r_1r_2$ are integers.
Is if $r_1,r_2 = -2,\sqrt 7$ they won't be. But to counter $\sqrt 7$ as a root we can add a third root of $-\sqrt 7$ to get a third degree polynomial:
$(x-\sqrt 7)(x+\sqrt 7)(x+2) = x^3+2x^2-7x-14$.
| {
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Integrals identity By numerical integration I found the identity
$$ \int_{-\infty}^{0}\frac{db}{\sqrt{R_4(b,W)}} = 2\int_{(W+1)^2}^{\infty}\frac{db}{\sqrt{R_4(b,W)}} $$
where $R_4(b,W)= b \ (b-4) \ (b-(W-1)^2) \ (b-(W+1)^2) $ and $W>3$.
Now I would like to prove that formally but I can't find the right way. I started by substituting
$$ b \rightarrow t=-b $$ and
$$ t \rightarrow x=t+(W+1)^2 $$
to have the same integration range on both sides, but $R_4(b,W)$ becomes quite strange.
| Let the roots of $R_4$ be $a=(W+1)^2,b=(W-1)^2,c=4,d=0$. Byrd and Friedman 251.00 and 258.00 resolves both sides in terms of elliptic integrals as follows:
$$gF(\varphi,m)=2gF(\psi,m)$$
where $g=\frac2{\sqrt{(a-c)(b-d)}}$, $m=k^2=\frac{(b-c)(a-d)}{(a-c)(b-d)}=\frac{(W-3)(W+1)^3}{(W+3)(W-1)^3}$, $\sin^2\varphi=\frac{a-c}{a-d}=1-\frac4{(W+1)^2}$ (hence $\cos\varphi=\frac2{W+1}$) and $\sin^2\psi=\frac{b-d}{a-d}=\frac{(W-1)^2}{(W+1)^2}$ (hence $\cos\psi=\frac{2\sqrt W}{W+1}$). Thus it remains to show $F(\varphi,m)=2F(\psi,m)$. Applying the $F$ addition formula to the RHS (see DLMF 19.11.14) gives
$$\sin\varphi=\frac{2\sin\psi\cos\psi\sqrt{1-m\sin^2\psi}}{1-m\sin^4\psi}=\frac{\sqrt{(W-1)(W+3)}}{W+1}$$
$$\sin^2\varphi=\frac{(W-1)(W+3)}{(W+1)^2}=1-\frac4{(W+1)^2}$$
which matches the expression we obtained for $\sin^2\varphi$ earlier. So the equality is true.
| {
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Show that three numbers form an arithmetic progression
The numbers $a,b$ and $c$ form an arithmetic progression. Show that the numbers $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ also form an arithmetic progression.
We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{ } \forall \text{ }n\ge2$). I am stuck here and I would be very grateful if you could give me a hint.
| Since $2b=a+c$
$$a^2+ab+b^2+b^2+bc+c^2=a^2+b(a+c)+2b^2+c^2=a^2+\frac{(a+c)^2}{2}+2\times \frac{(a+c)^2}{4}+c^2=a^2+(a+c)^2+c^2=2(a^2+ac+c^2)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$ $$\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$$
$$x=?$$
I solved this but the equation $ (2x + 1) (3x ^ 4-x ^ 3 + 2x ^ 2-2x + 1) = 0 $ is formed I answer $ x =- \frac {1} {2} $ I know there is, but I couldn't do the next expression. I need help with that, or someone will solve it in a better way. I'd be happy with that.
| Just to give a different approach, note that any solution to $\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$ must lie in $(-1,0)\cup(1,\infty)$ because the left hand side is negative on $(-\infty,-1)\cup(0,1)$ whereas the right hand side, being a square root, is always non-negative, and the expression is undefined at $x=0$ and $x=\pm1$. Now if $x\lt0$, then $3x^4-x^3+2x^2-2x+1$ is a sum of positive terms (i.e., $x^3$ and $2x$ are negative, so $-x^3$ and $-2x$ are positive), while if $x\gt1$, then
$$3x^4-x^3+2x^2-2x+1\gt3x^4-x^4+2x^2-2x^2+1=2x^4+1\gt0$$
so $3x^4-x^3+2x^2-2x+1$ has no roots in the feasible solution set for $x$. (In fact, as user's answer shows, it has no roots in $(0,1)$ either.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3895007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find a continuous path between matrices?
I have matrix $A=\begin{pmatrix} 0&-1&0\\1&0&0\\0&0&1\\\end{pmatrix}$ and matrix $B=\begin{pmatrix} 1&0&0\\0&\frac{1}{2}&-\frac{\sqrt{3}}{2}\\0&\frac{\sqrt{3}}{2}&\frac{1}{2}\\\end{pmatrix}$ and I want to find a continuous path between them. How do I do this?
I know that matrices in $SO(n)$ should be path connected but how do you actually construct such a path?
I'm not sure if just taking $\gamma(t)=A(t)+B(1-t)$ is going to work.
| You can check that any two dimensional rotation matrix can be written as $$R_\theta = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix} = e^{\theta J} \quad \text{where} \quad J = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}.$$
Using this fact, we have $$R_{\pi/2} = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} = e^{(\pi/2)J} \quad \text{and} \quad R_{\pi/3} = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} = e^{(\pi/3)J}.$$
Block-diagonal matrices $M = \begin{bmatrix}M_1 & 0 \\ 0 & M_2\end{bmatrix}$ satisfy $e^M = \begin{bmatrix}e^{M_1} & \\ & e^{M_2}\end{bmatrix}$.
Therefore, $$A = \begin{bmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} = e^{C} \quad \text{where} \quad C = \begin{bmatrix} \tfrac{\pi}{2}J & \\ & 0 \end{bmatrix} = \dfrac{\pi}{2}\begin{bmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix},$$ and $$B = \begin{bmatrix}1 & 0 & 0 \\ 0 & \tfrac{1}{2} & -\tfrac{\sqrt{3}}{2} \\ 0 & \tfrac{\sqrt{3}}{2} & \tfrac{1}{2}\end{bmatrix} = e^{D} \quad \text{where} \quad D = \begin{bmatrix} 0 & \\ & \tfrac{\pi}{3}J \end{bmatrix} = \dfrac{\pi}{3}\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0\end{bmatrix}.$$
Finally, you can check that for any anti-symmetric real matrix $H$, $e^H$ is in $SO(n)$. Can you use these facts to connect $A = e^C$ to $B = e^D$ by a continuous path in $SO(3)$? You have the right idea to use something like $t \cdot (\text{start point}) + (1-t) \cdot (\text{end point})$, but you need to tie in the matrix exponential somehow to guarantee the path stays in $SO(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3895837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Integrating partial Weierstrass products of $\sin(x)$ divided by $\sin(x)$ Recall the Weierstrass factorization of $\sin(z)$, valid for any complex $z$:
$$
\sin(z) = z \prod_{k=1}^{\infty}\left(1-\frac{z^2}{k^2\pi^2}\right)
$$Apropos of nothing, I thought, 'what if I integrated these partial products divided by sine over integer multiples of $\pi$?' The results Mathematica gave were somewhat striking:
\begin{align}
\int_{-\pi}^{\pi} \frac{x\left(1-\frac{x^2}{\pi^2}\right)}{\sin(x)}\,dx &= \frac{-21 \zeta(3)}{\pi}\\
\int_{-2\pi}^{2\pi} \frac{x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)}{\sin(x)}\,dx &= \frac{5 \left(35 \pi ^2 \zeta (3)-93 \zeta (5)\right)}{2 \pi ^3}\\
\int_{-3\pi}^{3\pi} \frac{x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)}{\sin(x)}\,dx &= \frac{7 \left(392 \pi ^4 \zeta (3)-2170 \pi ^2 \zeta (5)+1905 \zeta (7)\right)}{8 \pi ^5}\\
\int_{-4\pi}^{4\pi} \frac{x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\left(1-\frac{x^2}{16\pi^2}\right)}{\sin(x)}\,dx &= \frac{7 \left(3044 \pi ^6 \zeta (3)-24831 \pi ^4 \zeta (5)+51435 \pi ^2 \zeta (7)-22995 \zeta
(9)\right)}{16 \pi ^7}\\ &
\end{align}
Here as usual, $\zeta(s)$ is the Riemann zeta function, the analytic continuation of $\sum_{n=1}^{\infty} n^{-s}$; the odd positive-integer values are presently unknown. The results had a similar form with different integer coefficients if the range of integration was restricted to $(-\pi,\pi)$. I am seeking an explanation for these identities, as well as an explanation for how they were derived. I don't think computing antiderivatives is the best way to go; perhaps there is a transform that makes these integrals more manageable?
| Use integration by parts and induction in $k$ to find $$\int_{-\pi}^\pi x^{k}e^{-inx}dx$$
then
$$\int_{-\pi}^{\pi} \frac{x^{k+1}(1-x^2/\pi^2)}{\sin(x)}d=\lim_{r\to 1^-} \int_{-\pi}^\pi \frac{x^{k+1}(1-x^2/\pi^2) 2i e^{-ix}}{1-re^{-2ix}}dx$$ $$=\lim_{r\to 1^-} \int_{-\pi}^\pi \sum_{n=0}^\infty 2i x^{k+1}(1-x^2/\pi^2) r^n e^{-(2n+1)ix}dx$$ $$= 2i\lim_{r\to 1^-} \sum_{n=0}^\infty r^n \int_{-\pi}^\pi x^{k+1}(1-x^2/\pi^2) e^{-(2n+1)ix}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3896662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Factorial simplification help; how did wolfram get this? in this problem n is assumed to be odd:
Here's what I am trying to simplify:
$$ \frac {(n-2)!( \frac {n-1}{2})!} { (n-1)! (\frac {n-3} {2}) !}
$$
Wolfram is telling me that this should simplify to $$\frac {1}{2} $$ but I am very confused about how it gets there. I understand that the two factorial terms combine, but then I have $$ (n-2) (\frac {n-1} {2}) $$ as my final simplification and I really have no idea how to move from that to $$\frac{1}{2}$$ Any help would be appreciated.
| First, note that
$$\frac{(n-2)!}{(n-1)!} = \frac{(n-2)(n-3)(n-4) \cdots (2)(1)}{(n-1)(n-2)(n-3)(n-4) \cdots (2)(1)} = \frac{1}{n-1}$$
Similarly, notice that
$$\dfrac{ \left( \dfrac{n-1}{2} \right)! }{\left( \dfrac{n-3}{2} \right)!} = \dfrac{ \left( \dfrac{n-1}{2} \right)\left( \dfrac{n-3}{2} \right)\left( \dfrac{n-5}{2} \right)\cdots(2)(1) }{\left( \dfrac{n-3}{2} \right)\left( \dfrac{n-5}{2} \right)\cdots(2)(1) } = \frac{n-1}{2}$$
Multiply the two together for the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How I can solve this limit only with algebra I tried to resolve this using properties of limits, properties of logarithms and some substitutions, but i can't figure whats is the right procedure for this.
$$\lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 + x}{1 - x}}$$
first all I use the logarithms properties and rewrite like this:
$$\lim_{x\to 0} \log({\frac{1 + x}{1 - x}})^\frac{1}{2x}$$
and tried to make to this expression goes to $$\lim_{x\to 0} \log e^\frac{1}{2}$$
and then limit will be 1/2
I can't reach to this because the limit goes to 0 instead of ∞
| \begin{align*}
L &= \lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 + x}{1 - x}} \\
&= \lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 +(-x+x)+ x}{1 - x}} \\
&= \lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 -x +2x}{1 - x}} \\
&= \lim_{x\to 0} \frac{1}{x} \log{\sqrt{1 + \frac{2x}{1 - x}}} \\
&= \lim_{x\to 0} \frac{1}{x} \log{\left(1 + \frac{2x}{1 - x} \right)^{1/2}} \\
&= \lim_{x\to 0} \log{\left(1 + \frac{2x}{1 - x} \right)^{1/2x}} \\
&= \log \lim_{x\to 0} \left(1 + \frac{2x}{1 - x} \right)^{1/2x} \text{.}
\end{align*}
Let $x = \frac{1}{2u+1}$, so that $u = \frac{1-x}{2x}$. Then as $x \rightarrow 0^+$, $u \rightarrow \infty$ and as $x \rightarrow 0^-$, $u \rightarrow -\infty$. We investigate both. First,
\begin{align*}
M^+ &= \log \lim_{u \rightarrow \infty} \left( 1 + \frac{1}{u}\right)^{\frac{1}{2} + u} \\
&= \log \lim_{u \rightarrow \infty} \left( \left(1 + \frac{1}{u}\right)^{1/2} \left(1 + \frac{1}{u}\right)^{u} \right) \\
&= \log \left( 1 \cdot \mathrm{e} \right) \\
&= 1 \text{.}
\end{align*}
Then,
\begin{align*}
M^- &= \log \lim_{u \rightarrow -\infty} \left( 1 + \frac{1}{u}\right)^{\frac{1}{2} + u} \\
&= \log \lim_{u \rightarrow -\infty} \left( \left(1 + \frac{1}{u}\right)^{1/2} \left(1 + \frac{1}{u}\right)^{u} \right) \\
&= \log \lim_{u \rightarrow -\infty} \left( 1 + \frac{1}{u}\right)^{u} \\
&= \log \lim_{v \rightarrow \infty} \left( 1 + \frac{1}{-v}\right)^{-v} & & \text{u = -v} \\
&= \log \lim_{v \rightarrow \infty} \left(\left( 1 + \frac{-1}{v}\right)^{v}\right)^{-1} \\
&= \log \left(\lim_{v \rightarrow \infty} \left( 1 + \frac{-1}{v}\right)^{v}\right)^{-1} \\
&= -\log \lim_{v \rightarrow \infty} \left( 1 + \frac{-1}{v}\right)^{v} \\
&= -\log \mathrm{e}^{-1} \\
&= --1 \\
&= 1 \text{.}
\end{align*}
Since $M^- = M^+ = 1$, $L$ exists and $L = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Why is the gcd of $61+35\sqrt{3} $ and $170+32\sqrt{3}$ is $19 + 11\sqrt{3}$? As a completion or correction to the solution of my question here:
Find gcd of $a = 170 + 32\sqrt{3}$ and $b = 61 + 35\sqrt{3}.$ Then find $f,g \in \mathbb{Z}[\sqrt{3}]$ such that $af + bg = d$ using norm function.
I was told by my professor that: the gcd of $61+35\sqrt{3} $ and $170+32\sqrt{3}$ is $19 + 11\sqrt{3}$ and that we should not use the norm function and that we should use the ordinary procedure for calculating the gcd to get it. And my professor added that you will even get that $$19 + 11\sqrt{3} = (61 + 35\sqrt3) (718 - 413\sqrt{413}) +(170+32\sqrt{3})(-3 + \sqrt{3}),$$ As I mentioned in the above post I do not know how to complete after this step:
$$\frac{170 + 32 \sqrt{3}}{61 + 35 \sqrt{3}} = \frac{3505}{23} - \frac{1999 \sqrt{3}}{23},$$ specifically could anyone show me what should be the remainder in my previous line please?
| Division with remainder for number of these forms should be done as follows:
$$\frac{170 + 32 \sqrt{3}}{61 + 35 \sqrt{3}} = \frac{3505}{23} - \frac{1999 \sqrt{3}}{23}$$
This would be the fraction. Now to get the integral quotient, take the closest integer for the coefficients of the rational and irrational par
$$\frac{3505}{23} =152 + \frac{9}{23} \\
\frac{1999}{23} = 86 + \frac{21}{23} = 87 -\frac{2}{23}$$
Therefore, the integral quotient is
$$152 - 87 \sqrt{3}$$
and so
$$170 + 32 \sqrt{3} = (61 + 35 \sqrt{3})(152- 87 \sqrt{3})+ 33 + 19 \sqrt{3}$$
So $33 + 19 \sqrt{3}$ is the remainder.
Now divide $61 + 35 \sqrt{3}$ by $33 + 19 \sqrt{3}$. We have
$$\frac{61 + 35 \sqrt{3}}{33 + 19 \sqrt{3}} = 3 - \frac{2}{3}\sqrt{3}$$. Therefore the integral quotient is
$3 - \sqrt{3}$. The remainder is
$$(33 + 19 \sqrt{3}) \cdot\frac{\sqrt{3}}{3}=19 + 11 \sqrt{3}$$
Notice that the last remainder $19 + 11 \sqrt{3}$ divides the last divider $33 + 19 \sqrt{3}$. Therefore, we stop here, the $\gcd$ is $19 + 11 \sqrt{3}$.
Note: the division process is very similar to the usual division for integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
find all ordered pairs of natural numbers (x,y,z) that satisfies $x^{x^y} \cdot y^{y^z} \cdot z^{z^x} = 1990^{1990} \cdot xyz (x\lt y \lt z)$ find all ordered pairs of natural numbers (x,y,z) that satisfies $x^{x^y} \cdot y^{y^z} \cdot z^{z^x} = 1990^{1990} \cdot xyz (x\lt y \lt z)$
My approach was
$$x^{3(x^x-1)} \lt x^{x^y-1} \cdot y^{y^z-1} \cdot z^{z^x-1} = 1990^{1990}$$
and if x is 5 -> $5^{3(5^5-1)}=5^{9372} \gt 1990^{1990}$
so $x\in \{1,2,3,4\}$,and I tried to check in cases when x=1, x=2,x=3,x=4 but I couldn't proceed.
| First note the prime factorization of $1990 = 2 \times 5 \times 199$.
Hence at least one of $x,y,z$ must be divisible by $199$, from the equation $x^{x^y-1} \cdot y^{y^z-1} \cdot z^{z^x-1} = 1990^{1990}$.
Ignore the order for now and suppose $199 \mid x$. Then $x^{x^y-1} \ge 199^{199^y-1}$ and $z^{z^x-1} \ge z^{z^{199}-1}$.
For $y \ge 2$, $199^{199^y-1} \ge 199^{199^2-1} \gg 1990^{1990}$ (by observing logarithms.) Hence $y=1$.
Similarly we see that for $z \ge 2$, $z^{z^{199}-1} \ge 2^{2^{199}-1} \gg 1990^{1990}$. Hence $z = 1$ as well.
This reduces the equation to
$$x^{x-1} = 1990^{1990}$$
which has no solutions, since $2,5,199 \mid x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3907064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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