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Domain of $y=\arcsin\left({2x\sqrt{1-x^{2}}}\right)$ How do you find the domain of the function $y=\arcsin\left({2x\sqrt{1-x^{2}}}\right)$
I know that the domain of $\arcsin$ function is $[-1,1]$
So, $-1\le{2x\sqrt{1-x^{2}}}\le1$ probably?
or maybe $0\le{2x\sqrt{1-x^{2}}}\le1$ , since $\sqrt{1-x^{2}}\ge0$ ?
EDIT: So many people have answered that the domain would be $[-1,1]$ but my book says that its $[\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}].$
Can anyone explain how are those restrictions made in the given formulas?
| A proof without calculus or trigonometry.
$$\begin{aligned}
(1-2x^2)^2&\geq0
\\
(1-4x^2+4x^4)&\geq0
\\
x^2-x^4&\leq\frac{1}{4}
\\
0\leq \sqrt{x^2({1-x^2})}&\leq\frac{1}{2}
\\
0\leq |x\sqrt{1-x^2}|&\leq\frac{1}{2}
\\
-1\leq 2x\sqrt{1-x^2}&\leq1
\end{aligned}$$
Then, since $\arcsin:\,[-1,1]\mapsto\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$, the domain is $[-1,1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Applicability of L'Hôpital rule on infinite sum. $$\lim_{n \to \infty} \left( \frac{n}{n^2+1} + \frac{n}{n^2+2} + \frac{n}{n^2+3} + \space ... \space + \frac{n}{n^2+n}\right) $$
As $n$ is not $\infty$ but tends to $\infty$ I can split the limit of sums into sum of limits. i.e.
$$\lim_{n \to \infty} \frac{n}{n^2+1} +\lim_{n \to \infty} \frac{n}{n^2+2} +\lim_{n \to \infty} \frac{n}{n^2+3} + \space ... \space +\lim_{n \to \infty} \frac{n}{n^2+n} $$
Applying L'Hôpital rule.
$$\lim_{n \to \infty} \frac{1}{2n} +\lim_{n \to \infty} \frac{1}{2n} +\lim_{n \to \infty} \frac{1}{2n} + \space ... \space +\lim_{n \to \infty} \frac{1}{2n} $$
$$= \lim_{n \to \infty} \left( \frac{1}{2n} + \frac{1}{2n} + \frac{1}{2n} + \space ... \space + \frac{1}{2n} \right) = \lim_{n \to \infty}
\frac{n}{2n} $$
$$ =\lim_{n \to \infty } \frac{1}{2} = \frac{1}{2}$$
Whereas applying Sandwich theorem with $g(x) \leq f(x) \leq h(x)$, where
$$g(x) = \frac{n}{n^2+n} + \frac{n}{n^2+n} + \frac{n}{n^2+n} + \space ... \space + \frac{n}{n^2+n} = \frac{n^2}{n^2+n}$$
and
$$h(x) = \frac{n}{n^2+1} + \frac{n}{n^2+1} + \frac{n}{n^2+1} + \space ... \space + \frac{n}{n^2+1} = \frac{n^2}{n^2+1 }$$
Yields
$$\lim_{n \to \infty} g(x) = \lim_{n \to \infty} h(x) = 1
\implies \lim_{n \to \infty} f(x) = 1
$$
Sandwich theorem is very intuitive to discard hence I suppose there is some issue with the application of L'Hôpital's rule.
Is there any special condition involved with converting limit of sums to sum of limits ?
| Clearly the problem is way before L'Hopital's, splitting the limit into a sum of limits is not allowed (precisely because the number of terms tends to infinity, even though it is not infinite). For instance, consider:
$$ 1 = \lim_{n \to \infty} \left( n \frac{1}{n} \right)= \lim_{n \to \infty} \underbrace{\left( \frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n} \right)}_{n \ \text{times}} =\lim_{n \to \infty} \frac{1}{n}+\lim_{n \to \infty} \frac{1}{n} + \dots + \lim_{n \to \infty} \frac{1}{n} = 0 \neq 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Looking for a proof of an interesting identity Working on a problem I have encountered an interesting identity:
$$
\sum_{k=0}^\infty \left(\frac{x}{2}\right)^{n+2k}\binom{n+2k}{k}
=\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{x}\right)^n,
$$
where $n$ is a non-negative integer number and $x$ is a real number with absolute value less than 1 (probably a similar expression is valid for arbitrary complex numbers $|z|<1$).
Is there any simple proof of this identity?
| Using
$$\binom{n}{k}=\frac{1}{2 \pi i}\oint_C\frac{(1+z)^{n}}{z^{k+1}}dz$$ we get (integration contour is the unit cicrle)
$$
2\pi iS_n=\oint dz \sum_{k=0}^{\infty}\frac{(1+z)^{n+2k}x^{n+2k}}{z^{k+1}2^{n+2k}}=\oint dz \frac{(1+z)^n x^n}{z2^n}\sum_{k=0}^{\infty}\frac{(1+z)^{2k}x^{2k}}{2^{2k}z^k}=\\
4\frac{x^n}{2^n}\oint dz \underbrace{\frac{(1+z)^n}{4z-(1+z)^2x^2}}_{f(z)}
$$
for $|x|<1$ only we have just one pole of $f(z)$ inside the unit circle namely $z_0(x)=\frac2{x^2}-\frac{2\sqrt{1-x^2}}{x^2}-1$ , so
$$
S_n=4\frac{x^n}{2^n}\text{res}(f(z),z=z_0(x))=4\frac{x^n}{2^n}\left[ \frac{1}{4 \sqrt{1-x^2}}\left(2\frac{1-\sqrt{1-x^2}}{ x^2}\right)^n\right]
$$
or
$$
S_n=\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{ x}\right)^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2567158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the spectral decomposition of $A$ $$
A= \begin{pmatrix} -3 & 4\\ 4 & 3
\end{pmatrix}
$$
So i am assuming that i must find the evalues and evectors of this matrix first, and that is exactly what i did.
The evalues are $5$ and $-5$, and the evectors are $(2,1)^T$ and $(1,-2)^T$
Now the spectral decomposition of $A$ is equal to $(Q^{-1})^\ast$ (diagonal matrix with corresponding eigenvalues) * Q
$Q$ is given by [evector1/||evector1|| , evector2/||evector2||]
and for Q i got the matrix
$$
Q= \begin{pmatrix} 2/\sqrt{5} &1/\sqrt{5} \\ 1/\sqrt{5} & -2/\sqrt{5}
\end{pmatrix}
$$
the inverse of Q is the matrix...
$$
\begin{pmatrix} 2 \sqrt{5}/5 & \sqrt{5}/5 \\ \sqrt{5}/5 & -2 \sqrt{5}/5
\end{pmatrix}
$$
and the diagonal matrix with corresponding evalues is
$$
A= \begin{pmatrix} 5 & 0\\ 0 & -5
\end{pmatrix}
$$
so now i found the spectral decomposition of $A$, but i really need someone to check my work.
Did i take the proper steps to get the right answer, did i make a mistake somewhere?
| The needed computation is
$$\mathsf{A} = \mathsf{Q\Lambda}\mathsf{Q}^{-1}$$
Where $\Lambda$ is the eigenvalues matrix. And your eigenvalues are correct.
Hence you have to compute
$$\mathsf{AQ} = \mathsf{Q\Lambda}$$
Which gives you the solutions
$$a = 2c ~~~~~~~~~~~ d = -\frac{b}{2}$$
You can then choose easy values like $c = b = 1$ to get
$$Q = \begin{pmatrix} 2 & 1 \\ 1 & -\frac{1}{2} \end{pmatrix}$$
And easily
$$\mathsf{Q}^{-1} = \frac{1}{\text{det}\ \mathsf{Q}} \begin{pmatrix} -\frac{1}{2} & -1 \\ -1 & 2 \end{pmatrix}$$
Which you can compute alone.
| {
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"source": "stackexchange",
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Steiner inellipse Hello it's related to my answer for Prove the inequality $\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq 3\frac{a+b+c}{ax+by+cz}$
My answer fails but I don't know why ... So I was thinking a generalization of the following formula:
$$\frac{IA^2}{CA\cdot AB}+\frac{IB^2}{BC\cdot AB}+\frac{IC^2}{CA\cdot BC}=1$$
I know that it's related to the Steiner inellipse and we have for a triangle ABC and the ellipse of foci $P$ and $Q$:
$$\frac{PA\cdot QA}{BA\cdot CA}+
\frac{PB\cdot QB}{CB\cdot AB}+
\frac{PC\cdot QC}{BC\cdot AC}=1$$
But in my proof I have also use the following formula:
\begin{align}
\frac{1}{IA^2}+\frac{1}{IB^2}+\frac{1}{IC^2} &=
\frac{1}{r^2}-\frac{1}{2rR} \\
IA^2+IB^2+IC^2 &= s^2+r^2+8rR \\
CA\cdot AB+BC\cdot AB+CA\cdot BC &= s^2+(4R+r)r \\
\frac{1}{CA\cdot AB}+\frac{1}{BC\cdot AB}+\frac{1}{CA\cdot BC} &=
\frac{1}{2rR}
\end{align}
So what's the new expression of:
\begin{align}
\frac{1}{BA\cdot CA}+\frac{1}{CB\cdot AB}+\frac{1}{BC\cdot AC} &= ? \\
\frac{1}{PA\cdot QA}+\frac{1}{PB\cdot QB}+\frac{1}{PC\cdot QC} &= ? \\
PA\cdot QA+PB\cdot QB+PC\cdot QC &= ? \\
BA\cdot CA+CB\cdot AB+BC\cdot AC &= ?
\end{align}
In function of the parameters of the inellipse and the triangle $ABC$ like the area and the side of the triangle or the semi major semi minor axes of the ellipse?
Edit: I have a good news
The centroid $M$ of the triangle $ABC$ correspond to the centre of the inellipse and we have the following relation for $P$ any interior point related to the triangle $ABC$:
$$PA^2+PB^2+PC^2=MA^2+MB^2+MC^2+3MP^2$$
Thanks a lot.
| Sums $$CA\cdot AB+BC\cdot AB+CA\cdot BC = s^2+(4R+r)r$$ and
$$\frac{1}{CA\cdot AB}+\frac{1}{BC\cdot AB}+\frac{1}{CA\cdot BC}=\frac{1}{2rR}$$
are not related with Steiner inellipse.
Two remaining sums are based on the products of distances from the foci $P$ and $Q$ of the Steiner inellipse of the triangle $\Delta ABC$ to its vertices, which can be found as follows. Let the vertices $A$, $B$, and $C$ and the foci $P$ and $Q$ have the complex coordinates $z_A$, $z_B$, $z_C$, $z_P$, and $z_Q$, respectively. Accoriding to Steiner’s Theorem [MP, Th. 2.1], $z_P$ and $z_Q$ are given by the equality,
$$g\pm \sqrt{g^2-\frac f3},$$
where $g=\frac 13\left(z_A+z_B+z_C\right)$ is the centroid and $f=z_Az_B+ z_Bz_C+ z_Az_C$.
Then, for instance,
$$|PA\cdot QA|=|z_P-z_A||z_Q-z_A|=\left|(g –z_A)^2- g^2+\frac f3\right|=
\left|2gz_A+z_A^2+\frac f3\right|.$$
I don’t know whether we can further simplify the expressions for two remaining sums.
References
[MP] D. Minda, S. Phelps Triangles, ellipses, and cubic polynomials, American Mathematical Monthly, 115 (8) (2008), 679-689.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $3^n - 4(2^n) + (-1)^n + 6 \equiv 0 \mod 24 $ Is it possible to prove that $3^n - 4(2^n) + (-1)^n + 6 \equiv 0 \mod 24 $ for $n \geq 1 $ . I know that it is true because $ \frac{3^n - 4(2^n) + (-1)^n + 6}{24}$ represents the number of ways to uniquely $4$-colour an n-cycle , excluding permutations of colours.
| Using weak induction,
if $f(n)=3^n - 4(2^n) + (-1)^n + 6,$
$$f(m+2)-f(m)=3^m(3^2-1)-4\cdot2^m(2^2-1)$$ which is clearly divisible by $3\cdot8$ for $m\ge1$
So, $24\mid f(m)\iff24\mid f(m+2)$
Now establish the base cases $f(1),f(2)$
| {
"language": "en",
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In the triangle $ABC$ $R = \frac56 BH = \frac52OH$. Find the angles $ACB$ or $BAC$
In the triangle $ABC$, the height $BH$ is drawn, the point $O$ is the center of the circle circumscribed about it, the length of its radius $R$. Find the smallest of the angles $ACB$ and $BAC$, expressed in radians, if it is known that $R = \frac56 BH = \frac52OH$
My work so far:
1) In triangle $BOH$ $BO=R, BH=\frac65R, OH=\frac25R$. Then I can to find $\angle BOH, \angle BHO$ and $\angle OBH$
2) I proved that $\angle ABH= \angle OBC=90^{\circ}-\alpha$, where $\alpha=\angle A$
| The hint.
In $\Delta HOB$ we know that $OH=\frac{2}{5}R$, $BH=\frac{6}{5}R$ and $BO=R$.
Thus, by law of cosines we obtain $$\cos\cos\measuredangle HBO HBO=\frac{1+\frac{36}{25}-\frac{4}{25}}{2\cdot\frac{6}{5}},$$
which gives
$$\cos\measuredangle HBO=\frac{19}{20}.$$
In another hand, $\cos\measuredangle HBO=|\alpha-\gamma|$, which gives
$$\cos(\alpha-\gamma)=\frac{19}{20}.$$
Now, $$BH=c\sin\alpha=2R\sin\alpha\sin\gamma.$$
Thus, $$R=\frac{5}{6}\cdot2R\sin\alpha\sin\gamma$$ or
$$\sin\alpha\sin\gamma=\frac{3}{5}.$$
I hope the rest is smooth because
$$\cos(\alpha+\gamma)=\frac{19}{20}-2\cdot\frac{3}{5}=-\frac{1}{4}.$$
I got the following value.
$$\frac{\pi}{2}-\frac{\arccos\frac{19}{20}+\arccos\frac{1}{4}}{2}.$$
| {
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"url": "https://math.stackexchange.com/questions/2574230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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find the $\sum\limits_{k=0}^n \frac{(-1)^k}{k! (2k+1)} \frac{1}{(n-k)!}$ I'm stuck on computing the sum of
\begin{align*}
\sum\limits_{k=0}^n \frac{(-1)^k}{k! (2k+1)} \frac{1}{(n-k)!}
\end{align*}
I tried some manipulations which include
\begin{align*}
\frac{1}{n!} \binom{n}{k} = \frac{1}{k! (n-k)!}
\end{align*}
but still that $2k+1$ at the denominator complicates things. By the way, wolframalpha says that
\begin{align*}
\sum\limits_{k=0}^n \frac{(-1)^k}{k! (2k+1)} \frac{1}{(n-k)!} = \frac{\sqrt{\pi}}{2(n+\frac{1}{2})!}
\end{align*}
for $n\geq 1$.
Can anyone help me?
| Start with the binomial theorem:
$$\sum_{k=0}^n \binom{n}{k}x^n=(x+1)^n$$
Substitute $x=y^2$:
$$\sum_{k=0}^n \binom{n}{k}y^{2k}=(y^2+1)^n$$
Integrate both sides:
$$\sum_{k=0}^n \binom{n}{k}\frac{y^{2k+1}}{2k+1}=\int_0^y(t^2+1)^ndt$$
Divide across by $n!$:
$$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{y^{2k+1}}{2k+1}=\frac{1}{n!}\int_0^y(t^2+1)^ndt$$
Let $y=i$:
$$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{i(-1)^k}{2k+1}=\frac{1}{n!}\int_0^i(t^2+1)^ndt$$
$$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{i(-1)^k}{2k+1}=\frac{1}{n!}\int_0^1 i(1-t^2)^ndt$$
$$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{1}{2 n!}\int_0^1 t^{-1/2}(1-t)^ndt$$
Use Euler's Beta Function:
$$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{1}{2 n!}\frac{\Gamma(n+1)\Gamma(1/2)}{\Gamma(n+3/2)}$$
$$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{1}{2 n!}\frac{n!2^{n+1}}{(2n+1)!!}$$
$$\color{green}{\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{2^{n}}{(2n+1)!!}}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Evaluating $\big(\cot \frac{\pi}{18}-3\cot \frac{\pi}{6}\big)\cdot \big(\csc \frac{\pi}{9}+2\cot \frac{\pi}{9}\big)$
Finding value of $\displaystyle \bigg(\cot \frac{\pi}{18}-3\cot \frac{\pi}{6}\bigg)\cdot \bigg(\csc \frac{\pi}{9}+2\cot \frac{\pi}{9}\bigg)$
Try: $$\cot \frac{\pi}{18}\csc \frac{\pi}{9}-3\sqrt{3}\csc \frac{\pi}{9}+2\cot \frac{\pi}{18}\cot\frac{\pi}{9}-6\sqrt{3}\cot \frac{\pi}{9}$$
could some help me how can i simplify it,thanks
| $$\cot x-3\cot3x=\dfrac{\cos x\sin3x-3(\cos3x\sin x)}{\sin x\sin3x}$$
Again,
\begin{align}
2\cos x\sin3x-3(2\cos3x\sin x)
&=\sin4x+\sin2x-3(\sin4x-\sin2x)\\[4px]
&=4\sin2x-2\sin4x \\[4px]
&=4\sin2x(1-\cos2x)\\[4px]
&=4\sin2x(2\sin^2x)
\end{align}
Finally,
\begin{align}
\csc2x+2\cot2x
&=\dfrac{1+2\cos2x}{\sin2x}\\[4px]
&=\dfrac{1+2(1-2\sin^2x)}{\sin2x}\\[4px]
&=\dfrac{\sin3x}{\sin x\sin2x}
\end{align}
for $\sin x\ne0$ using $\sin3x$ formula.
Can you identify $x$ here?
| {
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Relation between integral, gamma function, elliptic integral, and AGM The integral $\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}}$ is equal to $\displaystyle \frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi }}$.
It is calculated or verified with a computer algebra system that $\displaystyle \frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi }} = K\left(\frac{1}{2}\right)$ , where $K(m)$ is the complete elliptic integral of the first kind. This is in relation to what is called the elliptic integral singular value.
It is also known or verified that
$\displaystyle K\left(\frac{1}{2}\right) =\displaystyle \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{1-\frac{\sin ^2(t)}{2}}} \, dt= \frac{1}{2} \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{\sin (t) \cos (t)}} \, dt$.
Can one prove directly or analytically that
$\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}} =\frac{1}{2} \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{\sin (t) \cos (t)}} \, dt =\displaystyle \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{1-\frac{\sin ^2(t)}{2}}} \, dt = K\left(\frac{1}{2}\right) $ ?
| Let $t=\frac{1}{{1+x^4}}$ and then
$$ dx=-\frac14(1-t)^{-3/4}t^{5/4} $$
So
\begin{eqnarray}
&&\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}}\\
&=&\frac14\int\limits_0^{1}(1-t)^{-3/4}t^{1/4}\mathrm dx\\
&=&\frac14B(\frac14,\frac54)\\
&=&\sqrt{\frac{\pi}{2}}\frac{\Gamma(\frac54)}{\Gamma(\frac34)}\\
&=&\frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi}}.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2579911",
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
Convergence/absolute convergence of $\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin \frac{1}{2n+1}\right)$
Does the following sum converge? Does it converge absolutely?
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}\right)$$
I promise this is the last one for today:
Using Simpson's rules:
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}\right) = \sum_{n=1}^\infty 2\cos\frac{4n+1}{4n² + 2n}\sin\frac{1}{8n² + 4n}$$
Now, $$\left|2\cos\frac{4n+1}{4n² + 2n}\sin\frac{1}{8n² + 4n}\right| \leq \frac{2}{8n² + 4n}$$
hence by the comparison test, the series converges absolutely, and hence it also converges. Is this correct?
| $$|\sin(\frac{1}{2n})-\sin(\frac{1}{2n+1})| = |cos(\xi)(\frac{1}{2n}-\frac{1}{2n+1})|$$
The above is followed by mean value theorem, then
$$|\sin(\frac{1}{2n})-\sin(\frac{1}{2n+1})| \leq \frac{1}{2n}-\frac{1}{2n+1}$$
So the series is absolutely convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
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$z ≤ x + y$ implies $z/(1 + z) ≤ x/(1 + x) + y/(1 + y)$
Suppose $x,y,z $ be nonnegative reals. Show that $z ≤ x + y\implies z/(1 + z) ≤ x/(1 + x) + y/(1 + y)$.
My Proof: If $z=0$, then we are done. So, suppose $z>0$. Since $z ≤ x + y$, and $x$ and $y$ nonnegative, $z ≤ x + y+2xy+xyz$, which leads to $z(1+x+y+xy)\le (x+y+2xy)(1+z)$ and since $z\ne 0$, we get $z/(1 + z) ≤x/(1 + x) + y/(1 + y)$.
Is my proof correct?
| Yes your proof is correct. I have provided another proof as follows. Notice that $f(t) = \dfrac{t}{1+t}$ is an increasing function on its domain. Therefore if $x$, $y$, and $z$ are non-negative real numbers and $z$ is less than or equal to $x+y$ then f(z) is less than or equal to f(x+y). That is $$\dfrac{z}{1+z} <= \dfrac{x+y}{1+x+y} <= \dfrac{x}{1+x} + \dfrac{y}{1+y}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Constant of integration change So, sometimes the constant of integration changes, and it confuses me a bit when and why it does. So for example, we have a simple antiderivative such as $$\int \frac{1}{x} dx $$ and we know that the result is $$\log|x| + C$$ and the domain is $$x\in\mathbb R \backslash \{0\} $$ If we want to show all the solutions, we need to do something like $$\begin{cases}
\log x+C_1 & x>0\\
\log(-x)+C_2 & x<0
\end{cases}$$
Do we need to do change the constant every time there is a gap in the domain or is it just when the expression changes? For example, $$ \int \frac {x^5} {x^2-1} dx$$ which is $$ \frac {1} {2} \log |x^2-1| + \frac {x^4} {4} + \frac {x^2}{2} + C$$ and the domain is $$x \in \mathbb R \backslash \{-1,1\}$$ When we want to write all the solutions, is it something like $$ \begin{cases} \ \frac {1} {2} \log (x^2-1) + \frac {x^4} {4} + \frac {x^2}{2} + C_1 & x>1\\ \frac {1} {2}\log (-x^2+1) + \frac {x^4} {4} + \frac {x^2}{2} + C_2 & -1<x<1\\ \frac {1} {2} \log (x^2-1) + \frac {x^4} {4} + \frac {x^2}{2} + C_3 & x<-1 \end{cases}$$ or is it that since the the first and last expression are the same they only have one constant associated? Meaning, the solutions are actually $$ \begin{cases} \ \frac {1} {2} \log (x^2-1) + \frac {x^4} {4} + \frac {x^2}{2} + C_1 & x \in \mathbb R \backslash [-1, 1]\\ \frac {1} {2}\log (-x^2+1) + \frac {x^4} {4} + \frac {x^2}{2} + C_2 & -1<x<1\\ \end{cases}$$
| Here is a 1-line compactification of Βασίλης Μάρκος' answer $\ddot\smile$
It can be shown that on every sub-interval of the domain of the original function, any two anti-derivatives differ by an additive constant, but constants for different sub-intervals can very well be different.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
Simplify this equation if $x$ is negative I am trying to simplify $7 \cdot\sqrt{5x}\cdot\sqrt{180x^5}$, given that $x$ is negative. The answer is $-210x^3$, but I am getting $210x^3$. Below is my reasoning:
For a $ k > 0 $, let $ k = - x $. Then, $7\cdot\sqrt{5x}\cdot\sqrt{180x^5}$ = $7\cdot5\cdot6\cdot i\sqrt{k} \cdot i\sqrt{k^5}$ = $210\cdot(-1)\cdot k^3$ = $ 210 \cdot (-1) \cdot k^3 = 210\cdot (-1) \cdot (-x)^3 = 210x^3$.
What am I doing wrong?
Any help is greatly appreciated
| $$\text{$7 \cdot\sqrt{5x}\cdot\sqrt{180x^5} \ $ where $ \ x<0$}$$
Lets let $x=-1$ to see what we should expect.
\begin{align}
\left. \left( 7 \cdot\sqrt{5x}\cdot\sqrt{180x^5}\right)\right|_{x=-1}
&= 7 \cdot\sqrt{-5}\cdot\sqrt{-180} \\
&= 7 \cdot i \cdot\sqrt{5}\cdot i \cdot \sqrt{180} \\
&= -7 \cdot \sqrt{900} \\
&= -210
\end{align}
Now Lets do it with $x=-n$ where $n > 0$.
\begin{align}
\left. \left( 7 \cdot\sqrt{5x}\cdot\sqrt{180x^5}\right)\right|_{x=-n}
&= 7 \cdot\sqrt{-5n}\cdot\sqrt{-180n^5} \\
&= 7 \cdot i \cdot\sqrt{5n}\cdot i \cdot \sqrt{180n^5} \\
&= -7 \cdot \sqrt{900n^6} \\
&= -210n^3 \\
&= -210(-x)^3 \\
&= 210x^3
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find possible values of angle C In an acute triangle $ABC$, $O$ is the circumcenter, $H$ is the orthocenter and $G$ is the
centroid. Let $OD$ be perpendicular to $BC$ and $HE$ be perpendicular to $CA$, with $D$ on $BC$ and $E$
on $CA$. Let $F$ be the midpoint of $AB$. Suppose the areas of triangles $ODC, HEA$ and $GFB$ are
equal. Find all the possible values of angle $C$.
My approach :
Let $R$ be the circumradius of $△ABC$ and $∆$ its area. We have $OD = R \cos A$ and
$DC =\frac{a}{2}$, so
$$[ODC] = \frac{1}{2}· OD · DC$$ $$=\frac{1}{2}· R \cos A · R \sin A $$ $$=\frac{1}{2}. R^2.\sin A \cos A .$$
Again $HE = 2R \cos C \cos A$ and $EA = c \cos A$. Hence
$$[HEA] = \frac{1}{2}· HE · EA$$
$$=\frac{1}{2}· 2R \cos C \cos A · c \cos A$$
$$=2R^2.\sin C .\cos C.{\cos}^2A .$$
Further
$$[GFB] = \frac{∆}{6}=\frac{1}{6}· 2R^2.\sin A.\sin B.\sin C$$ $$=\frac{1}{3}.R^2.\sin A.\sin B.\sin C$$
What to do next? Any help would be greatly appreciated.
| From $[ODC]=[HEA]$ one gets
$$
\tag{1}
\sin C\cos C={1\over4}{\sin A\over \cos A}.
$$
From $[ODC]=[GFB]$, taking into account that $\sin B=\sin(A+C)=\sin A\cos C+\cos A\sin C$, one gets
$$
\sin A \sin C\cos C+\cos A \sin^2C={3\over2}\cos A,
$$
that is, using $(1)$:
$$
{1\over4}{\sin^2 A\over \cos A}+\cos A \sin^2C={3\over2}\cos A,
$$
whence:
$$
\tag{2}
\sin^2 C={7\cos^2 A-1\over4\cos^2A}.
$$
Squaring $(1)$ and inserting there $(2)$ one finally gets
$$
20\cos^4A-9\cos A+1=0,
\quad\hbox{that is:}\quad
\cos^2A={1\over4}
\ \ \hbox{or}\ \
\cos^2A={1\over5}.
$$
From $(2)$ one then obtains
$$
\sin^2 C={3\over4}\ \hbox{and}\ C=60°,
\ \ \hbox{or}\ \
\sin^2C={1\over2}\ \hbox{and}\ C=45°.
$$
The first case obviously leads to an equilateral triangle, while the second case is not trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find minimum value of $\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$ If $a,b,c$ are sides of triangle Find Minimum value of
$$S=\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$$
My Try:
Let $$P=\sqrt{a}+\sqrt{b}+\sqrt{c}$$
we have $$S=\sum \frac{1}{\frac{\sqrt{b}}{\sqrt{a}}+\frac{\sqrt{c}}{\sqrt{a}}-1}$$
$$S=\sum \frac{1}{\frac{P}{\sqrt{a}}-2}$$
Let $x=\frac{P}{\sqrt{a}}$, $y=\frac{P}{\sqrt{b}}$,$z=\frac{P}{\sqrt{c}}$
Then we have $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$$
By $AM \ge HM$
$$\frac{x+y+z}{3} \ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$
Hence
$$x+y+z \ge 9$$
Any way to proceed further?
| When $a = b = c$, $S = 3$. Next it will be proved that $S \geqslant 3$ for all possible $a, b, c$.
Denote $\displaystyle u = \frac{\sqrt{a}}{\sqrt{a} + \sqrt{b} + \sqrt{c}}$, $\displaystyle v = \frac{\sqrt{b}}{\sqrt{a} + \sqrt{b} + \sqrt{c}}$, $\displaystyle w = \frac{\sqrt{c}}{\sqrt{a} + \sqrt{b} + \sqrt{c}}$, then $\sum u = 1$. Since$$
\sqrt{a} < \sqrt{b + c} < \sqrt{b} + \sqrt{c},
$$
then $\displaystyle 0 < u < \frac{1}{2}$. Analogously, $\displaystyle 0 < v, w < \frac{1}{2}$. It suffices to prove$$
S = \sum \frac{u}{v + w - u} = \sum \frac{u}{1 - 2u} \geqslant 3.
$$
Define $\displaystyle f(x) = \frac{x}{1 - 2x} \ (0 < x < \frac{1}{2})$. Because $\displaystyle f''(x) = \frac{4}{(1 - 2x)^3} > 0$, by Jensen's inequality,$$
S = \sum f(u) \geqslant 3 f\left(\frac{1}{3} \sum u\right) = 3f\left(\frac{1}{3}\right) = 3.
$$
Therefore the minimum of $S$ is $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the radius of curvature at the origin
Find the radius of curvature at the origin for the curve $3x^2+4y^2=2x$.
Below is my attempt:
We know that the radius of curvature $\rho=\dfrac{{(1+y_1^2)}^{\frac{3}{2}}}{y_2}$ .
Computing $y_1$ at $(0,0)$ : Differentiating we get $3x+4y\frac{dy}{dx}=1\implies \dfrac{dy}{dx}=\dfrac{1-3x}{4y}.$
How can I evaluate $\dfrac{dy}{dx}$ at the point $(0,0)$?
Please help.
| If we consider the differential $(6x-2)dx+8ydy$, then at $(0,0)$, we see $(6x-2) \neq 0$. By the implicit function theorem, we may write $x$ as a function of $y$.
Complete the square:
$$3x^2 -2x +4y^2=0$$
$$3(x-\frac{1}{3})^2+4y^2=\frac{1}{3}$$
From here we get:
$$x(y)=\frac{1}{3}+\sqrt{\frac{1}{9}-\frac{4}{3}y^2}$$
Let $r:(-\epsilon,\epsilon) \rightarrow \mathbb{R}^{2}$ be given by
$$r(y) =(x(y),y)$$
The curvature is given by the general formula:
$$\kappa=\frac{|x'y''-x''y'|}{(x'^{2}+y'^{2})^{\frac{3}{2}}}$$
Using our parameterization, we have:
$$\kappa(y)=\frac{|\frac{d^2x}{dy^2}|}{(1+(\frac{dx}{dy})^2)^\frac{3}{2}}$$
Now $\frac{dx}{dy}=0$ when $y=0$. If we compute $\frac{d^2
x}{dy^2}$ we have:
$$\frac{d^2
x}{dy^2}=y(junk)+\frac{4}{3}\bigg(\frac{1}{9}-(\frac{4}{3}y^2)\bigg)^\frac{-1}{2}$$
Evaluating at $y=0$ we find,
$$\kappa(0)=4$$.
Thus $\rho= \frac{1}{\kappa}=\frac{1}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Probability of a letter coming from a city I got this question:
A letter is known to have come either from $\text{TATANAGAR}$ or from $\text{CALCUTTA}$. On the envelope just two consecutive Letters $\text{TA}$ are visible. What is the probability that the letters came from $\text{TATANAGAR}$?
My attempt:
Total number of cases $=3$, as there are three such pairs.
Probability $={{\text{Favourable}}\over{\text{Total}}}=\frac23$
However, the answer is given to be $7\over11$.
How is this possible? Please help.
| $$T\equiv \text{comes from Tatanagar}\quad \text{and}\quad C\equiv \text{Comes from Calcutta}\rightarrow \left\{ \begin{array}{lcc}
p(T)=\dfrac{1}{2} \\
\\ p(C)=\dfrac{1}{2}
\end{array}
\right.$$
$$\text{Possible choices of two consecutive letters:}\left\{ \begin{array}{lcc}
\text{In TATANAGAR}: 8 \\
\\ \text{In CALCUTTA}: 7
\end{array}
\right.$$
$$\text{Event} D\equiv \text{The chosen couple of letters is TA}\rightarrow \left\{ \begin{array}{lcc}
p(D|T)=\dfrac{2}{8} \\
\\ p(D|C)=\dfrac{1}{7}
\end{array}
\right.$$
$$p(T|D)=\dfrac{p(T\cap D)}{p(D)}=\dfrac{\frac{1}{2}\cdot \frac{2}{8}}{\frac{1}{2}\cdot \frac{2}{8}+\frac{1}{2}\cdot \frac{1}{7}}=\dfrac{\frac{1}{8}}{\frac{7+4}{56}}=\dfrac{7}{11}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\int (2x+3) \sqrt {3x+1} dx$ Evaluate $\int (2x+3) \sqrt {3x+1} dx$
My Attempt:
Let $u=\sqrt {3x+1}$
$$\dfrac {du}{dx}= \dfrac {d(3x+1)^\dfrac {1}{2}}{dx}$$
$$\dfrac {du}{dx}=\dfrac {3}{2\sqrt {3x+1}}$$
$$du=\dfrac {3}{2\sqrt {3x+1}} dx$$
| Let us just split $2x+3 = \frac23 (3x+1)+\frac73$ and simplify to give us: $$I = \int (2x+3)\sqrt{3x+1}\, dx = \frac23 \int (3x+1)^{\frac32}\, dx+ \frac73 \int \sqrt {3x+1}\, dx$$ which can be easily solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
$P(X \geq 2Y) = ?$ where $P(n) = 2^{-n} , n =1,2,3,...$. Let $X,Y$ are independent identically distributed random variables.Then $P(X \geq 2Y) = ?$ where $P(n) = 2^{-n} , n =1,2,3,...$.
So it is the discrete case, So in order to calculate this -
$P(X \geq 2Y) = \sum_{y=1}^{\infty} \sum_{x = 2y}^{\infty} 2^{-x-y} = \sum_{y=1}^{\infty}2^{-y} (\frac{2^{-2y}}{1 - 2^{-2}}) = \sum_{y=1}^{\infty} \frac{4}{3}. 2^{-3y} =\frac{4}{3}. \frac{2^{-3}}{1 - 2^{-3}} = \frac{4}{21}.$
Is the above correct?
| Above Answer is not Correct. It should be
$P(X \geq 2Y) = \sum_{y=1}^{\infty} \sum_{x = 2y}^{\infty} 2^{-x-y}
= \sum_{y=1}^{\infty}2^{-y} (\frac{2^{-2y}}{1 - 2^{-1}})
= \sum_{y=1}^{\infty} \frac{2}{1}. 2^{-3y} =\frac{2}{1}. \frac{2^{-3}}{1 - 2^{-3}}
= \frac{2}{7} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $ a^4b^4+ a^4c^4+b^4c^4\le3$
Let $a,b,c>0$ and $a^3+b^3+c^3=3$. Prove that
$$ a^4b^4+ a^4c^4+b^4c^4\le3$$
My attempts:
1) $$ a^4b^4+ a^4c^4+b^4c^4\le \frac{(a^4+b^4)^2}{4}+\frac{(c^4+b^4)^2}{4}+\frac{(a^4+c^4)^2}{4}$$
2) $$a^4b^4+ a^4c^4+b^4c^4\le3=a^3+b^3+c^3$$
| By AM-GM and Schur we obtain
$$\sum_{cyc}a^4b^4=\frac{1}{3}\sum_{cyc}(3ab)a^3b^3\leq\frac{1}{3}\sum_{cyc}(1+a^3+b^3)a^3b^3=$$
$$=\frac{1}{3}\sum_{cyc}(a^3b^3+a^6b^3+a^6c^3)=\frac{1}{9}\sum_{cyc}a^3\sum_{cyc}a^3b^3+\frac{1}{3}\sum_{cyc}(a^6b^3+a^6c^3)=$$
$$=\frac{1}{9}\sum_{cyc}(4a^6b^3+4a^6c^3+a^3b^3c^3)=$$
$$=\frac{1}{9}\left(\sum_{cyc}(a^9+3a^6b^3+3a^6c^3+2a^2b^2c^2)-\sum_{cyc}(a^9-a^6b^3-a^6c^3+a^3b^3c^3)\right)\leq$$
$$\leq\frac{1}{9}\sum_{cyc}(a^9+3a^6b^3+3a^6c^3+2a^2b^2c^2)=\frac{1}{9}(a^3+b^3+c^3)^3=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing the sum $\sum_\limits{n=2}^\infty \left(\frac{1}{(n-1)!}-\frac{1}{n!}\right)\frac{1}{n+1}$ I have come across an infinite series, but I have no clue on how to compute its sum.
$$\sum_{n=2}^\infty \left(\frac{1}{(n-1)!}-\frac{1}{n!}\right)\frac{1}{n+1}$$
It should have something to do with the Taylor expansion of $e^x$, but I could not figure out how to do this.
| \begin{align*}
\sum_{n=2}^\infty \left (\frac{1}{(n-1)!}-\frac{1}{n!}\right)\frac{1}{(n+1)}
&=\sum_{n=2}^\infty \left (\frac{n}{(n+1)!}-\frac{1}{(n+1)!}\right)\\
&=\sum_{n=2}^\infty \left (\frac{n+1-1}{(n+1)!}-\frac{1}{(n+1)!}\right)\\
&=\sum_{n=2}^\infty \frac{1}{n!}-2\sum_{n=2}^\infty \frac{1}{(n+1)!}\\
&=-2+\sum_{n=0}^\infty \frac{1}{n!}-2\sum_{n=3}^\infty \frac{1}{n!}\\
&=-2+e-2(e- \frac 5 2 )=-e+3
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove: For $\triangle ABC$, if $\sin^2A + \sin^2B = 5\sin^2C$, then $\sin C \leq \frac{3}{5}$. We have a triangle $ABC$. It is given that $\sin^2A + \sin^2B = 5\sin^2C$. Prove that $\sin C \leq \frac{3}{5}$.
Let's say that $BC = a$, $AC=b$, $AB=c$.
According to the sine law,
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,$$ then
$\sin A = \frac{a}{2R}\implies \sin^2A = \frac{a^2}{4R^2}$
$\sin B = \frac{b}{2R}\implies \sin^2B = \frac{b^2}{4R^2}$
$\sin C = \frac{c}{2R}\implies \sin^2C = \frac{c^2}{4R^2}$
Then we get:
$$\frac{a^2}{4R^2} + \frac{b^2}{4R^2} = 5\frac{c^2}{4R^2}.$$
Since $4R^2 > 0$, we get that $a^2 + b^2 = 5c^2$
Guys, is that correct? Even if it is, do you have any ideas what shall I do next?
| Since $A + B + C = \pi$, we have $\sin (A + B) = \sin C$.
Also notice that
$$\cos^2 A + \cos^2 B = 1 - \sin^2A + 1 - \sin^2B = 2 - 5\sin^2(A + B)$$
Proceeding by CSB, we get:
\begin{align}
\sin(A + B) &= \sin A \cos B + \sin B \cos A \\
&\le \sqrt{\sin^2 A + \sin ^2 B} \sqrt{\cos^2A + \cos^2B} \\
&= \sqrt{5} \sin(A + B) \sqrt{2 - 5\sin^2(A + B)}
\end{align}
Therefore $$1 \le 5(2 - 5\sin^2(A + B)) = 10 - 25\sin^2(A + B)$$ which yields $$\sin C = \sin(A + B) \le \frac35$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that $(-1)^n\sum_{k=0}^{n}{{{n+k}\choose{n}}2^k}+1=2^{n+1}\sum_{k=0}^{n}{{{n+k}\choose{n}}(-1)^k}$ Define:
$$A_n:=\sum_{k=0}^{n}{{n+k}\choose{n}} 2^k,\quad{B_n}:=\sum_{k=0}^{n}{{n+k}\choose{n}}(-1)^k$$
I've found that (based on values for small $n$) this identity seems to be true:
$${\left(-1\right)}^nA_n+1=2^{n+1}B_n$$
However, I'm stuck on trying to find a proof. Any ideas? Thanks!
| Using formal power series we have the Iverson bracket
$$[[0\le k\le n]] = [z^n] z^k \frac{1}{1-z}.$$
We then get for $A_n$
$$\sum_{k\ge 0} [z^n] z^k \frac{1}{1-z} {n+k\choose k} 2^k
= [z^n] \frac{1}{1-z}
\sum_{k\ge 0} z^k {n+k\choose n} 2^k
\\ = [z^n] \frac{1}{1-z} \frac{1}{(1-2z)^{n+1}}.$$
This yields for $1+(-1)^n A_n$
$$1 + (-1)^n [z^n] \frac{1}{1-z} \frac{1}{(1-2z)^{n+1}}
= 1 + [z^n] \frac{1}{1+z} \frac{1}{(1+2z)^{n+1}}$$
This is
$$1 + \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{1+z} \frac{1}{(1+2z)^{n+1}}.$$
Now the residue at infinity is zero by inspection and we get
the closed form (residues sum to zero)
$$1 - \mathrm{Res}_{z=-1} \frac{1}{z^{n+1}}
\frac{1}{1+z} \frac{1}{(1+2z)^{n+1}}
- \mathrm{Res}_{z=-1/2} \frac{1}{z^{n+1}}
\frac{1}{1+z} \frac{1}{(1+2z)^{n+1}}
\\ = - \mathrm{Res}_{z=-1/2} \frac{1}{z^{n+1}}
\frac{1}{1+z} \frac{1}{(1+2z)^{n+1}}
\\ = - \frac{1}{2^{n+1}}\mathrm{Res}_{z=-1/2} \frac{1}{z^{n+1}}
\frac{1}{1+z} \frac{1}{(z+1/2)^{n+1}}
.$$
We evidently require (Leibniz rule)
$$\frac{1}{n!} \left(\frac{1}{z^{n+1}} \frac{1}{1+z} \right)^{(n)}
\\ = \frac{1}{n!}
\sum_{q=0}^n {n\choose q} \frac{(-1)^q (n+q)!}{n!}
\frac{1}{z^{n+1+q}}
\frac{(n-q)! (-1)^{n-q}}{(1+z)^{n-q+1}}
\\ = (-1)^n \sum_{q=0}^n {n+q\choose q}
\frac{1}{z^{n+1+q}}
\frac{1}{(1+z)^{n-q+1}}.$$
Evaluate at $z=-1/2$ to get
$$(-1)^n \sum_{q=0}^n {n+q\choose q}
(-2)^{n+1+q} 2^{n-q+1}
= 2^{2n+2} \sum_{q=0}^n {n+q\choose q} (-1)^{q+1}.$$
Restoring the multiplier in front now yields
$$- \frac{1}{2^{n+1}} 2^{2n+2}
\sum_{q=0}^n {n+q\choose q} (-1)^{q+1}
= 2^{n+1} \sum_{q=0}^n {n+q\choose q} (-1)^{q}.$$
This is $2^{n+1} B_n$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to find the limit:$\lim_{n\to \infty}\left(\sum_{k=n+1}^{2n}\left(2(2k)^{\frac{1}{2k}}-k^{\frac{1}{k}}\right)-n\right)$ How to find the limit:$$\lim_{n\to \infty}\left(\sum_{k=n+1}^{2n}\left(2(2k)^{\frac{1}{2k}}-k^{\frac{1}{k}}\right)-n\right)$$
I can't think of any way of this problem
Can someone to evaluate this?
Thank you.
| For small $x>0$, we have $1+x\leq\exp(x)\leq 1+x+x^{2}$, then for large $k$,
\begin{align*}
1+\dfrac{\log 2}{k}-\left(\dfrac{\log k}{k}\right)^{2}\leq 2(2k)^{1/2k}-k^{1/k}\leq 1+\dfrac{\log 2}{k}+\dfrac{(\log 2k)^{2}}{2k^{2}},
\end{align*}
and for large $n$,
\begin{align*}
\left(\sum_{k=n+1}^{2n}2(2k)^{1/2k}-k^{1/k}\right)-n\leq\log 2\sum_{k=n+1}^{2n}\dfrac{1}{k}+\sum_{k=n+1}^{2n}\dfrac{(\log 2k)^{2}}{2k^{2}},
\end{align*}
and
\begin{align*}
\left(\sum_{k=n+1}^{2n}2(2k)^{1/2k}-k^{1/k}\right)-n\geq\log 2\sum_{k=n+1}^{2n}\dfrac{1}{k}-\sum_{k=n+1}^{2n}\left(\dfrac{\log k}{k}\right)^{2},
\end{align*}
and note that
\begin{align*}
\sum_{k=1}^{\infty}\dfrac{(\log 2k)^{2}}{2k^{2}}&<\infty,\\
\sum_{k=1}^{\infty}\left(\dfrac{\log k}{k}\right)^{2}&<\infty,
\end{align*}
and we treat
\begin{align*}
\sum_{k=n+1}^{2n}\dfrac{1}{k}
\end{align*}
as
\begin{align*}
\int_{n+1}^{2n}\dfrac{1}{t}dt=\log 2n-\log(n+1)=\log\left(\dfrac{2n}{n+1}\right)
\end{align*}
when $n\rightarrow\infty$. So the limit is $(\log 2)^{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $\frac{1}{|x+1|} < \frac{1}{2x}$ Solving $\frac{1}{|x+1|} < \frac{1}{2x}$
I'm having trouble with this inequality. If it was $\frac{1}{|x+1|} < \frac{1}{2}$, then:
If $x+1>0, x\neq0$, then
$\frac{1}{(x+1)} < \frac{1}{2} \Rightarrow x+1 > 2 \Rightarrow x>1$
If $x+1<0$, then
$\frac{1}{-(x+1)} < \frac{1}{2} \Rightarrow -(x+1) > 2 \Rightarrow x+1<-2 \Rightarrow x<-3$
So the solution is $ x \in (-\infty,-3) \cup (1,\infty)$
But when solving $\frac{1}{|x+1|} < \frac{1}{2x}$,
If $x+1>0, x\neq0$, then
$\frac{1}{x+1} < \frac{1}{2x} \Rightarrow x+1 > 2x \Rightarrow x<1 $
If $x+1<0$, then
$\frac{1}{-(x+1)} < \frac{1}{2x} \Rightarrow -(x+1) > 2x \Rightarrow x+1 < -2x \Rightarrow x<-\frac{1}{3}$
But the solution should be $x \in (0,1)$.
I can see that there can't be negative values of $x$ in the inequality, because the left side would be positive and it can't be less than a negative number. But shouldn't this appear on my calculations?
| So from first case where $x\in(-1,\infty)$, you get the permissible via of $x$ to be $$(-1,\infty)\cap(-\infty,1)=(-1,1)$$
In the second case where $x\in(-\infty,-1)$ you get the permissible via of $x$ to be $$(\infty,-1)\cap\left(-\infty,-\frac13\right)=(-\infty,-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $f(2^{2017})$ The function $f(x)$ has only positive $f(x)$. It is known that $f(1)+f(2)=10$, and $f(a+b)=f(a)+f(b) + 2\sqrt{f(a)\cdot f(b)}$. How can I find $f(2^{2017})$?
The second part of the equality resembles $(\sqrt{f(a)}+\sqrt{f(b)})^2$, but I still have no idea what to do with $2^{2017}$.
| $f(2a) = f(a+a) = f(a) + f(a) + 2\sqrt{f(a)f(a)} = 4f(a)$
By induction $f(2^k)=4f(2^{k-1}) = 4^2f(2^{k-1}) = 4^kf(1)$.
$f(1) + f(2) = f(1) + 4f(1) = 5f(1) = 10$.
So $f(1) = 2$.
So $f(2^k) = 4^kf(1) = 2*4^k$
$f(2^{2017} = 2*4^{2017}=2^{4035}$
====
What what be interesting is what other values are.
$f(3) = f(1) + f(2) + 2\sqrt{f(1)f(2)} = 2 + 8 + 2\sqrt {16} = 18$
$..... = 3^2*2?????$
Does $f(k) = 2k^2$?????
$f(1) = 2*1^2$ and $f(2) = 8 = 2*2^2$ and
if $f(k) = 2k^2$ then $f(1+k) = f(1) + f(k) + 2\sqrt{f(1)f(k)}$
$= 2 + 2k^2 + 2\sqrt{2*2k^2} = 2+2k^2 + 4k = 2(k^2 + 2k + 1) = 2(k+1)^2$.
So that holds for integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this a combinatorial identity: $ \sum_{k=1}^{n+1}\binom{n+1}{k} \sum_{i=0}^{k-1}\binom{n}{i} = 2^{2n} $? $$
\sum_{k=1}^{n+1}\left(\binom{n+1}{k} \sum_{i=0}^{k-1}\binom{n}{i}\right) = 2^{2n}
$$
This is my first question, please feel free to correct/guide me. While solving a probability problem from a text book l reduced the problem to the above LHS. I couldn't reduce it any further. I tried a few values of n and it holds. I gave a half hearted attempt at induction before I gave up. Does this hold? Is there a combinatorial proof to it(assuming it holds) i.e count something one way and count the same thing other way and then equate them. Is there a name to it? Most importantly how to Google such questions?
To provide further context, the problem is as follows:-
Alice and Bob have a total of $2n+1$ fair coins. Bob tosses $n+1$ coins while Alice tosses $n$ coins. Tosses are independent. What is the probability that Bob tossed more heads than Alice? It is from a standard textbook "Introduction to Probability" by Dmitri and N John.
| The combinatorial interpretation provided by Lord Shark is really nice and elementary, but there also is a brute-force way to proving such identity.
$$ \sum_{k=1}^{n+1}\left[\binom{n+1}{k}\sum_{i=0}^{k-1}\binom{n}{i}\right]=\sum_{a=0}^{n}\left[\binom{n+1}{a+1}\sum_{b=0}^{a}\binom{n}{b}\right]=\sum_{0\leq b\leq a\leq n}\binom{n+1}{a+1}\binom{n}{b}. $$
Let $T_a=\sum_{b=0}^{a}\binom{n}{b}$. From $(1+x)^n = \sum_{b\geq 0}\binom{n}{b}x^b$ we get
$$ \frac{(1+x)^n}{1-x} = T_0+T_1 x+ \ldots + T_{n-1} x^{n-1} + 2^n x^n + 2^n x^{n+1} + 2^{n} x^{n+2}+\ldots $$
hence
$$ \frac{(1+x)^n-2^n x^{n+1}}{1-x} = \sum_{a=0}^{n} T_a x^a $$
$$ \frac{\left(1+\frac{1}{x}\right)^n-\frac{2^n}{ x^{n+1}}}{x-1} = \sum_{a=0}^{n} T_a x^{-(a+1)} $$
$$ (1+x)^{n+1}\frac{\left(1+\frac{1}{x}\right)^n-\frac{2^n}{ x^{n+1}}}{x-1} = \sum_{a=0}^{n} T_a x^{-(a+1)}(1+x)^{n+1} $$
and
$$ \sum_{a=0}^{n}\binom{n+1}{a+1}T_a =\operatorname*{Res}_{x=0}\frac{(x+1)^{n+1}}{x}\cdot\frac{x(x+1)^n-2^n}{x^{n+1}(x-1)}.$$
On the other hand,
$$ \operatorname*{Res}_{x=0}\frac{2^n(x+1)^{n+1}}{x^{n+2}(x-1)}=-\operatorname*{Res}_{x=1}\frac{2^n(x+1)^{n+1}}{x^{n+2}(x-1)}=-2^{2n+1}$$
and
$$\begin{eqnarray*} \operatorname*{Res}_{x=0}\frac{(x+1)^{2n+1}}{x^{n+1}(x-1)}&=&-\operatorname*{Res}_{x=0}\frac{(x+1)^{2n+1}}{x^{n+1}}-\operatorname*{Res}_{x=0}\frac{(x+1)^{2n+1}}{x^{n}}-\ldots-\operatorname*{Res}_{x=0}\frac{(x+1)^{2n+1}}{x}\\&=&-\binom{2n+1}{n}-\binom{2n+1}{n-1}-\ldots-\binom{2n+1}{0}=-2^{2n}\end{eqnarray*}$$
so
$$\sum_{a=0}^{n}\binom{n+1}{a+1}T_a = 2^{2n+1}-2^{2n} = \color{red}{2^{2n}}$$
as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Factoring the polynomial $3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$ I'm trying to factor the following polynomial:
$$3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$$
What I've done:
$$3(x^2 -1)^3 + 7(x^2-1)^2 + 4(x^2 -1)$$
Then I set $p=x^2 -1$ so the polynomial is:
$$3p^3 + 7p^2 + 4p$$
Therefore: $$p(3p^2 + 7p + 4)$$
I apply Cross Multiplication Method: $$p(p+3)(p+4)$$
I substitute $p$ with $x^2-1$:
$$(x^2-1)(x^2-1+3)(x^2-1+4)$$
$$(x-1)(x+1)(x^2-2)(x^2-3)$$
I don't know if I've done something wrong or if I have to proceed further and how. The result has to be: $x^2(3x^2+1)(x+1)(x-1)$. Can you help me? Thanks.
| Taking $ (x^2 - 1) $ as common, you get:
$$ (x^2 - 1) (3(x^2 - 1)^2 + 7(x^2 - 1) + 4) $$
$$ = (x+1)(x-1)( 3 (x^2-1)^2 + 3(x^2 - 1) + 4(x^2 - 1) + 4 )$$
$$ = (x+1)(x-1)( 3(x^2 - 1) + 4 ) ( x^2 - 1 + 1 ) $$
$$ = (x+1)(x-1)( 3x^2 - 3 + 4)(x^2) $$
$$ = x^2(x+1)(x-1)(3x^2 + 1) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
} |
Derivative of $e^{2x^4-x^2-1}$ with limit definition of derivative Let $f:\mathbb{R}\to\mathbb{R}$ be defined as
$f(x) = e^{2x^4-x^2-1}$.
I have to find the derivative using the defintion:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
My approach:
$$
\begin{align}
&\lim_{h \to 0} \frac{\exp({2(x+h)^4-(x+h)^2-1})-\exp({2x^4-x^2-1})}{h}\\
&=\lim_{h \to 0} \frac{\exp(2 h^4 + 8 h^3 x + h^2 (12 x^2 - 1) + h (8 x^3 - 2 x) + 2 x^4 - x^2 - 1) - e^{2 x^4 - x^2 - 1}}{h}
\end{align}$$
How can I remove $h $from denominator?
| Hint
Factor the term $e^{2 x^4 - x^2 - 1}$ in each of the two terms of the difference. Then you get the product
$$\frac{f(x+h)-f(x)}{h}=e^{2 x^4 - x^2 - 1} \times \frac{e^{hA(x,h)}-1}{h}$$
Where $A(x,h)=h(8x^2 -x) +h^2 B(x,h)$ with $A,B$ polynomials.
Then as $e^{y} -1 \approx y$ around $0$, you get
$$e^{2 x^4 - x^2 - 1} \times \frac{e^{hA(x,h)}-1}{h} \approx e^{2 x^4 - x^2 - 1} \left((8x^2-x) +hB(x,h)\right)$$
which converges to $(8x^2-x)e^{2 x^4 - x^2 - 1}$ as $h \to 0$. Proving that $f^\prime(x)=(8x^2-x)e^{2 x^4 - x^2 - 1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Intersection of two cylinders What is the easiest way to find the area of the surface created when the cylinder $$x^2+z^2=1\text { intersects the cylinder }y^2+z^2=1.$$
I have used double integrals and also line integrals to find the area of the surface created when the cylinder $x^2+z^2=1$ intersects $ y^2+z^2=1.$
The line integral is much easier than the double integral. Is there a way to find this surface area without using integrals?
| \begin{align}
R &= \{ 0<x<1,-x<y<x,z=\sqrt{1-x^2} \} \\
S &= 8\iint_R dA \\
z &= \sqrt{1-x^2} \\
z_x &= -\frac{x}{\sqrt{1-x^2}} \\
z_y &= 0 \\
dA &= \sqrt{1+z_x^2+z_y^2} \, dx \, dy \\
&= \frac{1}{\sqrt{1-x^2}} \, dx \, dy \\
S &= 8\int_{0}^{1} \int_{-x}^{x} \frac{1}{\sqrt{1-x^2}} \, dy \, dx \\
S &= 8\int_{0}^{1} \frac{2x}{\sqrt{1-x^2}} \, dx \\
&= 16
\end{align}
See also ancient Chinese work, Nine Chapters《九章算術》commented by Liu Hui(劉徽) here.
Addendum
In the figure above, the intersections between two cylinders (yellow and cyan) are two ellipses, namely $$x^2=y^2=1-z^2$$
Consider the purple region $T=\{ 0<y<x<1,x=\sqrt{1-z^2} \}$ which is bounded by a semi-circle in $xz$-plane and a semi-ellipse namely
$$(x,y,z)=(\sin \theta,\sin \theta,\cos \theta)$$
Now unwrapping the cylinder,
$$s=\theta \implies y=\sin s$$
therefore we get a sine curve (green).
The area of purple region $T$ is
$$\int_0^\pi \sin \theta \, d\theta=2$$
which is $\dfrac{1}{8}$ of the total surface area.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find the limits $\lim_{x\to \infty} \lfloor f(x) \rfloor=? $ let $f(x)=\dfrac{4x\sin^2x-\sin 2x}{x^2-x\sin 2x}$ the fine the limits :
$$\lim_{x\to \infty} \lfloor f(x) \rfloor=? $$
$$f(x)=\dfrac{4x\sin^2x-\sin 2x}{x^2-x\sin 2x}=\dfrac{2\sin x(2x\sin x-\cos x)}{x(x-\sin2x)}$$
what do i do ?
| A more heuristic argument is as follows: The key observation is that $\sin u$ is always between $-1$ and $1$, no matter what $u$ is. Consider the impact of this on the numerator. We have
\begin{align}
\text{numerator} & = 4x \times \text{(something between $-1$ and $1$)} - \text{(something between $-1$ and $1$)} \\
& = \text{(something between $-4|x|$ and $4|x|$)} - \text{(something between $-1$ and $1$)} \\
& = \text{(something between $-4|x|-1$ and $4|x|+1$)}
\end{align}
Now, consider the impact of that observation on the denominator. We have
\begin{align}
\text{denominator} & = x^2 - x \times \text{(something between $-1$ and $1$)} \\
& = x^2 - \text{(something between $-|x|$ and $|x|$)} \\
& = \text{(something between $x^2-|x|$ and $x^2+|x|$)}
\end{align}
So our fraction consists of something which is linear in $x$, divided by something that is quadratic in $x$. Therefore, the limit must be $0$, as $x$ increases without bound (in either direction).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the greatest possible value for the difference in ratings. Seven experts evaluate the picture. Each of them makes an assessment - an integer number of points from 0 to 10 inclusive. It is known that all experts have put different points. According to the old rating system, the rating of the picture is the arithmetic mean of all expert points. According to the new rating system, the rating of the picture is estimated as follows: the smallest and largest points are discarded and the arithmetic mean of the remaining points is calculated. Find the greatest possible value for the difference in ratings calculated from the old and new assessment systems.
I find seven points 0, 1, 2, 3, 4, 5, 10 for which this difference is $\frac{4}{7}$ but I can't prove that it is maximum value.
| Let the seven ratings be $a,b,c,d,e,f,g$ where, WLOG, $a<b<c<d<e<f<g$.
The old system would give $\frac{a+b+c+d+e+f+g}{7}$, whereas the new one would give $\frac{b+c+d+e+f}{5}$
We would like to maximize the absolute difference between the two, which is given by
\begin{align}
\ & |\frac{a+b+c+d+e+f+g}{7}-\frac{b+c+d+e+f}{5}| \\
\ = & |\frac{a+g}{7}-\frac{2(b+c+d+e+f)}{35}| \\
\end{align}
WLOG, let's say we make $\frac{a+g}{7}$ as large as possible, and $\frac{2(b+c+d+e+f)}{35}$ as small as possible.
Then we would want $g=10$. And whatever value that $a$ takes, the smallest possible value for $\frac{2(b+c+d+e+f)}{35}$ would occur when $b=a+1, c=a+2, d=a+3, e=a+4, f=a+5$. And now it is just a matter of directly computing the cases for when $a=0,1,2,3,4$.
It turns out that,
when a=0, the difference is $\frac47$
when a=1, the difference is $\frac37$
when a=2, the difference is $\frac27$
when a=3, the difference is $\frac17$
when a=4, the difference is $0$.
So the answer is indeed $\frac47$
Of course, a more elegant way of doing this would be to say that $$\frac{a+(a+1)+(a+2)+(a+3)+(a+4)+(a+5)+10}{7}=\frac{6a+25}{7},$$ and that
$$\frac{(a+1)+(a+2)+(a+3)+(a+4)+(a+5)}{5}=\frac{5a+15}{5},$$ where $\frac{6a+25}{7} - \frac{5a+15}{5} = \frac{4-a}{7}$ so that the smallest value of $a$, namely $a=0$, gives the largest difference.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the integral $\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx.$ The question is $$\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx.$$
I have tried to multiply both numerator and denominator by $1-\sqrt{x}$ but can't proceed any further, help!
| I think its useful to learn some standard substitution you can use for such kind of problem.
for this case instead of using $x = \cos^2\theta$ I'm gonna try using $x=\cos^22\theta$ $$x=\cos^22\theta \Rightarrow dx=\left(-4\cos 2\theta \sin 2\theta \right)d\theta$$
so we have $$\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx$$
$$-4\int\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\cos 2\theta \left(\sin 2\theta \right),d\theta$$
$$-4\int \frac{\ \sin \theta }{\cos \theta }\cos 2\theta \left(\sin 2\theta \right)$$
$$-8\int (\sin^2 \theta\cos2\theta)d\theta $$
$$-8\int \frac{\ 1-\cos 2\theta }{2}\times \ (\cos 2\theta)d\theta $$
$$-4\int \cos 2\theta \ -\frac{\left[\cos 4\theta +1\right]}{2}d\theta $$
$$-2\sin 2\theta +\frac{\sin 4\theta }{2}+(2\theta) + C$$
we want to write this in terms of $x$ so $\ \theta =\frac{\cos ^{-1}\left(\sqrt{\ x}\right)}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2605444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $|ax^2+bx+c|\le \frac12$ for all $|x|\le1$, then $|ax^2+bx+c|\le x^2-\frac12$ for all $|x|\ge1$
Prove that if $|ax^2+bx+c|\le \frac12$ for all $|x|\le1$ then $|ax^2+bx+c|\le x^2-\frac12$ for all $|x|\ge1$.
My attempts:
Let $f(x)=ax^2+bx+c$
I know that
1) if $f(a)<0$ and $f(b)>0$ then exist $x_0\in[a;b]$ then $f(x_0)=0$
2) $|ax^2+bx+c|\le \frac12 \Leftrightarrow -\frac12<ax^2+bx+c\le \frac12$
| Since $$f(-1)=a-b+c,\quad f(0)=c,\quad f(1)=a+b+c$$ we can write
$$a=\frac{f(1)+f(-1)-2f(0)}{2},\quad b=\frac{f(1)-f(-1)}{2},\quad c=f(0)$$
Suppose here that there exists a real number $p$ such that
$$|p|\ge 1\qquad\text{and}\qquad |ap^2+bp+c|\gt p^2-\frac 12$$
Then,
$$\begin{align}p^2-\frac 12&\lt\left|\frac{f(1)+f(-1)-2f(0)}{2}p^2+\frac{f(1)-f(-1)}{2}p+f(0)\right|\\\\&=\left|f(1)\cdot\frac{p^2+p}{2}+f(-1)\cdot\frac{p^2-p}{2}+f(0)(-p^2+1)\right|\\\\&\le \left|f(1)\right|\left|\frac{p^2+p}{2}\right|+\left|f(-1)\right|\left|\frac{p^2-p}{2}\right|+\left|f(0)\right||p^2-1|\\\\&\le \frac 12\left|\frac{p^2+p}{2}\right|+\frac 12\left|\frac{p^2-p}{2}\right|+\frac 12|p^2-1|\\\\&=\frac 14\left|p(p+1)\right|+\frac 14\left|p(p-1)\right|+\frac 12|(p-1)(p+1)|\\\\&=\frac 14p(p+1)+\frac 14p(p-1)+\frac 12(p-1)(p+1)\\\\&=p^2-\frac 12\end{align}$$
which is impossible.
| {
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"url": "https://math.stackexchange.com/questions/2606384",
"timestamp": "2023-03-29T00:00:00",
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Solve $ \int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx$ I came across this question in my textbook and have been trying to solve it for a while but I seem to have made a mistake somewhere.
$$ \int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx$$ and here is what I did. First I simplified the equation as $$ \int_\frac{-π}{3}^{\frac{π}{3}}(1-\tan^2(x))dx=x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}(\tan^2(x))dx$$
Then I simplified $\tan^2(x)\equiv\frac{\sin^2(x)}{\cos^2(x)}, \sin^2(x)\equiv1-\cos^2(x)$ so it becomes, $\tan^2(x)\equiv\frac{1-\cos^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}-1$ making the overall integral
$$\int_\frac{-π}{3}^{\frac{π}{3}}(1-\tan^2(x))dx=x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}(\frac{1}{\cos^2(x)}-1)dx$$
$$=x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}\frac{1}{\cos^2(x)}dx-\int_\frac{-π}{3}^{\frac{π}{3}}1dx=x|_{\frac{-π}{3}}^{\frac{π}{3}}-x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}\frac{1}{\cos^2(x)}dx$$
I know that $\int\frac{1}{\cos^2(x)}dx=\tan(x)+c$ but that's off by heart and not because I can work it out. Since $x|_{\frac{-π}{3}}^{\frac{π}{3}}-x|_{\frac{-π}{3}}^{\frac{π}{3}}=0$, the final equation becomes
$$\int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx=\tan(x)|_{\frac{-π}{3}}^{\frac{π}{3}}=2 \sqrt3$$
Is what I did correct because I feel like I've made a mistake somewhere but can't find it. Also why does $\int\frac{1}{\cos^2(x)}dx=\tan(x)+c$.
EDIT - Made an error in $\int\frac{1}{\cos^2(x)}dx=\tan^2(x)+c$, it's actually $\int\frac{1}{\cos^2(x)}dx=\tan(x)+c$.
| $\begin{align} J=\int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx=2\int_0^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx\end{align}$
Observe that for $x\in [0;\frac{\pi}{3}]$,
$\begin{align} \frac{\cos^2(x)-\sin^2(x)}{\cos^2 x}&=\frac{\cos^2 x(1-\tan^2 x)}{\cos^2 x}\\
&=\frac{1-\tan^2 x}{1+\tan^2 x}\times \frac{1}{\cos^2x}
\end{align}$
Perform the change of variable $y=\tan x$,
$\begin{align}J&=2\int_{0}^{\sqrt{3}}\frac{1-y^2}{1+y^2}\,dy\\
&=2\int_{0}^{\sqrt{3}} \left(\frac{2}{1+y^2}-1\right)\,dy\\
&=2\Big[2\arctan y-y\Big]_{0}^{\sqrt{3}}\\
&=2\Big[2\times \frac{\pi}{3}-\sqrt{3}\Big]\\
&=\boxed{\frac{4\pi}{3}-2\sqrt{3}}
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Arc length of curve of intersection between cylinder and sphere
Given the sphere $x^2+y^2+z^2 = \frac{1}{8}$ and the cylinder $8x^2+10z^2=1$, find the arc length of the curve of intersection between the two.
I tried parametrizing the cylinder (the task specifies this as a hint). My attempt:
$$x(t) = \frac{1}{\sqrt{8}} \sin(t)$$
$$z(t) = \frac{1}{\sqrt{10}} \cos(t)$$
Plugging this into $x^2+y^2+z^2 = \frac{1}{8}$, I solve for $y$ to get
$$y = \sqrt{\frac{\cos(2t)+1}{4\sqrt{5}}}$$
I then tried integrating $|x(t), y(t), z(t)|$ from $0$ to $2\pi$
with no luck. I suspect my parametrization is wrong as my expression for $y$ looks rather ugly. Any ideas?
| From $8x^2 + 10z^2 = 1$,you get $z^2 = \frac{1}{10}.(1-8x^2)$. Substitute this in the other equation $ x^2+y^2+z^2 = \frac{1}{8}$ you get
$$x^2 + 5y^2 = \frac{1}{8}$$
This is the curve of intersection, now parameterize this ellipse with
$x = \frac{1}{\sqrt{8}} sint$
and
$ y = \frac{1}{\sqrt{40}} cost$
$ z = \frac{1}{\sqrt{10}} cost$
Now arc length $ L= \int_0^{2\pi} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 +(\frac{dz}{dt})^2} dt$
$$ = \int_0^{2\pi} \sqrt{\frac{1}{8} cos^2t + \frac{1}{40} sin^2t+\frac{1}{10} sin^2t}dt$$
$$ = \int_0^{2\pi} \sqrt{\frac{1}{40}}.\sqrt{ 5cos^2t + sin^2t +4sin^2t} dt$$
$$=\int_0^{2\pi} \sqrt{\frac{5}{40}} dt$$
$$L= \frac{1}{2\sqrt{2}}\int_0^{2\pi} dt\tag 1$$
$$ L = \frac{2\pi}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculating the integral $\int \sqrt{1+\sin x}\, dx$. I want to calculate the integral $\int \sqrt{1+\sin x}\, dx$.
I have done the following:
\begin{equation*}\int \sqrt{1+\sin x}\, dx=\int \sqrt{\frac{(1+\sin x)(1-\sin x)}{1-\sin x}}\, dx=\int \sqrt{\frac{1-\sin^2 x}{1-\sin x}}\, dx=\int \sqrt{\frac{\cos^2x}{1-\sin x}}\, dx=\int \frac{\cos x}{\sqrt{1-\sin x}}\, dx\end{equation*}
We substitute $$u=\sqrt{1-\sin x} \Rightarrow du=\frac{1}{2\sqrt{1-\sin x}}\cdot (1-\sin x)'\, dx \Rightarrow du=-\frac{\cos x}{2\sqrt{1-\sin x}}\, dx \\ \Rightarrow -2\, du=\frac{\cos x}{\sqrt{1-\sin x}}\, dx $$
We get the following:
\begin{equation*}\int \frac{\cos x}{\sqrt{1-\sin x}}\, dx=\int(-2)\, du=-2\cdot \int 1\, du=-2u+c\end{equation*}
Therefore \begin{equation*}\int \frac{\cos x}{\sqrt{1-\sin x}}\, dx=-2\sqrt{1-\sin x}+c\end{equation*}
In Wolfram the answer is a different one. What have I done wrong?
| As pointed out by other answers, you need to take signs into consideration. Indeed, starting from your computation we know that
$$ \int \sqrt{1+\sin x} \, dx = \int \frac{\left|\cos x\right|}{\sqrt{1-\sin x}} \, dx $$
Now let $I$ be an interval on which $\cos x$ has the constant sign $\epsilon \in \{1, -1\}$. That is, assume that $\left| \cos x \right| = \epsilon \cos x$ for all $x \in I$. Then
\begin{align*}
\text{on } I \ : \qquad
\int \sqrt{1+\sin x} \, dx
&= \epsilon \int \frac{\cos x}{\sqrt{1-\sin x}} \, dx \\
&= -2\epsilon \sqrt{1-\sin x} + C \\
&= - \frac{2\cos x}{\sqrt{1+\sin x}} + C
\end{align*}
In the last line, we utilized the equality $\cos x = \epsilon \left|\cos x\right| = \epsilon \sqrt{1-\sin^2 x}$.
Notice that maximal choices of $I$ are of the form $I_k := [(k-\frac{1}{2})\pi, (k+\frac{1}{2})\pi]$. So if you want a solution which works on a larger interval, you have to stitch solutions on $I_k$ for different $k$'s together in continuous way. This causes values of $C$ change for different intervals $I_k$. But from the periodicity, it is not terribly hard to describe a global solution and indeed it can be written as
$$
\int \sqrt{1+\sin x} \, dx
= - \frac{2\cos x}{\sqrt{1+\sin x}} + 2\sqrt{2} \left( \left\lceil \frac{x+\frac{\pi}{2}}{2\pi} \right\rceil+ \left\lfloor \frac{x+\frac{\pi}{2}}{2\pi} \right\rfloor \right) + C
$$
The extra term of floor/ceiling function is introduces to compensate jumps of $y=-2\frac{\cos x}{\sqrt{1+\sin x}}$:
$\hspace{2em}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Convergence of the sequence $ \sqrt {n-2\sqrt n} - \sqrt n $ Here's my attempt at proving it:
Given the sequence $$ a_n =\left( \sqrt {n-2\sqrt n} - \sqrt n\right)_{n\geq1} $$
To get rid of the square root in the numerator:
\begin{align}
\frac {\sqrt {n-2\sqrt n} - \sqrt n} 1 \cdot \frac {\sqrt {n-2\sqrt n} + \sqrt n}{\sqrt {n-2\sqrt n} + \sqrt n} &= \frac { {n-2\sqrt n} - \ n}{\sqrt {n-2\sqrt n} + \sqrt n} = \frac { {-2\sqrt n}}{\sqrt {n-2\sqrt n} + \sqrt n} \\&= \frac { {-2}}{\frac {\sqrt {n-2\sqrt n}} {\sqrt n} + 1}
\end{align}
By using the limit laws it should converge against:
$$
\frac { \lim_{x \to \infty} -2 } { \lim_{x \to \infty} \frac {\sqrt {n-2\sqrt n}}{\sqrt n} ~~+~~\lim_{x \to \infty} 1}
$$
So now we have to figure out what $\frac {\sqrt {n-2\sqrt n}}{\sqrt n}$ converges against:
$$
\frac {\sqrt {n-2\sqrt n}}{\sqrt n} \leftrightarrow \frac { {n-2\sqrt n}}{ n} = \frac {1-\frac{2\sqrt n}{n}}{1}
$$
${\frac{2\sqrt n}{n}}$ converges to $0$ since:
$$
2\sqrt n = \sqrt n + \sqrt n \leq \sqrt n ~\cdot ~ \sqrt n = n
$$
Therefore $~\lim_{n\to \infty} a_n = -1$
Is this correct and sufficient enough?
| It's perfect except for the justification that $2\sqrt {n}/n $ converges to $0 $. What you have written only proves that it is bounded above by $1 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Using elementary row operations to solve intersection of two planes The question I am struggling with is the following:
Solve the following using elementary row operations, and interpret each system of equations geometrically:
\begin{align*}
x - 3y + 2z &= 8\\
3x - 9y + 2z &= 4
\end{align*}
The answer given in the book is $x = -2+3t, y = t, z = 5$, and the planes meet in a line.
I put this in matrix form, so
$$\left(\begin{array}{c c c|c}
1 & -3 & 2 & 8\\
3 & -9 & 2 & 4
\end{array}\right)
$$
Then I subtracted R1 from R2
$$\left(\begin{array}{c c c|c}
1 & -3 & 2 & 8\\
2 & -6 & 0 & -4
\end{array}\right)
$$
Then R1 - $\frac{1}{2}$R2
$$\left(\begin{array}{c c c|c}
0 & 0 & 2 & 10\\
-2 & -6 & 0 & -4
\end{array}\right)
$$
So I got that $2z =10$, so $z = 5$, but I am stumped on how to continue. How can I get $x$ and $y$ and interpret the results in terms of plane intersection?
| $2z = 10$, so $z = 5$.
$2x - 6y = -4 \Rightarrow2x = -4 + 6y \Rightarrow x = -2 + 3y$. Let $y = t$, we get $x = -2 + 3t$, $y = t$, $z = 5$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Probability problem involving geometric distribution and expected value We're given the following problem:
"An experiment is conducted until it results in success: the first step has probability $\frac{1}{2}$ to be successful, the second step (only conducted if the first step had no success) has probability $\frac{1}{3}$ to be successful, the third step (only conducted if the first two steps had no success) has probability $\frac{1}{4}$ to be successful : if none of the steps were successful, we repeat the experiment until success is achieved. Assuming the first step has cost $2$, and the second step as well as the third step have cost $1$, what is the overall cost until success ? "
Here's my approach. Let $X$ be the random variable the counts the number of steps required to achieve success. Thus giving me: $$P(X=1) = \frac{1}{2} \\ P(X=2)=\frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \\ P(X=3) = \frac{1}{2} \cdot \frac{5}{6} \cdot \frac{1}{4} = \frac{5}{48} $$
Now, to compute the expected cost, I did the following: $$E(X) = 2 \cdot 1 \cdot P(X=1) + 1 \cdot 2 \cdot P(X=2) + 1 \cdot 3 \ P(X=3)$$ (where the first coeffeicient of each term of the sum is the cost of each step). This gave me $\frac{79}{48}$ , which is far from the correct answer.
So, I do not understand what I did wrong (I guess there must be a problem with my reasoning), and I do not know how to find the correct result (which is $\frac{34}{9}$).
| Let $\mu_{0}$ denote the expectation of the cost of the steps yet
to be done if no steps have been made.
Let $\mu_{1}$ denote the expectation of the cost of the steps yet
to be done if step 1 has been made without success.
Let $\mu_{2}$ denote the expectation of the cost of the steps yet
to be done if step 1 and step 2 have been made without success.
Then we have the following equalities.
$\mu_{0}=\frac{1}{2}2+\frac{1}{2}\left[2+\mu_{1}\right]=2+\frac{1}{2}\mu_{1}$
$\mu_{1}=\frac{1}{3}1+\frac{2}{3}\left[1+\mu_{2}\right]=1+\frac{2}{3}\mu_{2}$
$\mu_{2}=\frac{1}{4}1+\frac{3}{4}\left[1+\mu_{0}\right]=1+\frac{3}{4}\mu_{0}$
Then:$$\mu_0=2+\frac12\left[1+\frac23\left[1+\frac34\mu_0\right]\right]$$
The solution of this equality is: $$\mu_0=\frac{34}9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving that the sequence $\{\frac{3n+5}{2n+6}\}$ is Cauchy. I'm not quite sure how to tackle these kinds of questions in general, but I tried something that I thought could be right. Hoping to be steered in the right direction here!
Let $\{\frac{3n+5}{2n+6}\}$ be a sequence of real numbers. Prove that this sequence is Cauchy.
Proof:
We want to establish that $\forall_{\epsilon>0}\exists_{{n_0}\in{\mathbb{N}}}\forall_{n,m\geq n_0}\big(|f(n)-f(m)|\big)<\epsilon.$
Suppose $n>m$ without loss of generality. We then know that $\frac{3n+5}{2n+6}>\frac{3m+5}{2m+6}$ and thus that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}>0$ such that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=|\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}|=|f(n)-f(m)|.$
Let us work out the original sequence:
$\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=\frac{(3n+5)(2m+6)-(3m+5)(2n+6)}{(2n+6)(2m+6)} = \frac{8(n-m)}{(n2+6)(2m+6)}<\frac{8(n-m)}{nm}= 8(\frac{1}{n}- \frac{1}{m}).$
We know that $\frac{1}{n}<\frac{1}{m}$ as $n>m$ and that $\frac{1}{n}\leq\frac{1}{n_0}, \frac{1}{m}\leq\frac{1}{n_0}$ for $n,m\geq n_0$.
This means that $\frac{1}{n}-\frac{1}{m}\leq \frac{1}{n_0}- \frac{1}{m}\leq\frac{1}{n_0}$, and thus $8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.
Let $\epsilon=\frac{8}{n_0}$, as it only depends on $n_0$ it can become arbitrarily small.
Then the following inequality holds: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.
So: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<\epsilon$, and thus the sequence is Cauchy.$\tag*{$\Box$}$
| As you were told in the comments, you must take some $\varepsilon>0$ and then prove that $\left|\frac{3n+5}{2m+6}-\frac{3n+5}{2n+6}\right|<\varepsilon$ if $m$ and $n$ are large enough.
Note that$$(\forall n\in\mathbb{N}):\frac{3n+5}{2n+6}=\frac{3n+9}{2n+6}-\frac4{2n+6}=\frac32-\frac2{n+3}.$$So, take $p\in\mathbb N$ such that $\frac2{p+3}<\frac\varepsilon2$. Then$$m,n\geqslant p\implies\left|\frac{3m+5}{2m+6}-\frac{3n+5}{2n+6}\right|=\left|\left(\frac32-\frac2{m+3}\right)-\left(\frac32-\frac2{n+3}\right)\right|<\varepsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Show that $2n\choose n$ is divisible by $2n-1$ I have found many questions asking to prove $2n\choose n$ is divisible $2$, but I also observed by trying the first few "$n$" that $2n\choose n$ is divisible by $2n-1$.
It sure seems true when $2n-1$ is prime, but is it true in general for all $n$?
| Note that $\dbinom{2n}{n}$ is an integer. Note also that $1$ divides $\binom 21 $ and assume $n>1$ hereafter.
$\begin{align}
\dbinom{2n}{n} &= \dfrac{2n!}{n!\cdot n!} \\ &= \dfrac{2n\cdot (2n-1)\cdots (n+1)}{n\cdot (n-1)\cdots 1}\\
&= \dfrac{(2n)(2n-1)}{n\cdot n}\dbinom{2(n-1)}{n-1}\\
&= 2\dfrac{2n-1}{n}\dbinom{2(n-1)}{n-1}\\
\end{align}$
Now $\gcd(2n-1,n)=1$ so $n$ must divide $2\dbinom{2(n-1)}{n-1}$, that is, $2\dbinom{2(n-1)}{n-1}= kn$ for some integer $k$.
Then $\dbinom{2n}{n} = k(2n-1)$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
} |
Compute the determinant. If $a$, $b$, and $c$ are roots of unity. Then what is the value of
\begin{vmatrix}
e^a & e^{2a} & (e^{3a}-1) \\
e^b & e^{2b} & (e^{3b}-1) \\
e^c & e^{2c} & (e^{3c}-1)
\end{vmatrix}
I tried expanding it but the expression becomes unmanageable, is there some kind of simplification I can do?
| This is
$$\det\pmatrix{x&x^2&x^3-1\\y&y^2&y^3-1\\z&z^2&z^3-1}
=\det\pmatrix{x&x^2&x^3\\y&y^2&y^3\\z&z^2&z^3}
-\det\pmatrix{x&x^2&1\\y&y^2&1\\z&z^2&1}
$$
for $x=e^a$ etc. Both of these are essentially Vandermonde determinants.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\arg(\frac{z_1}{z_2})$ of complex equation If $z_1,z_2$ are the roots of the equation $az^2 + bz + c = 0$, with $a, b, c > 0$; $2b^2 > 4ac > b^2$; $z_1\in$ third quadrant; $z_2 \in$ second quadrant in the argand's plane then, show that $$\arg\left(\frac{z_1}{z_2}\right) = 2\cos^{-1}\left(\frac{b^2}{4ac}\right)^{1/2}$$
| $z_2 = \overline{z_1}\,$ since the quadratic has real coefficients, so $\,\arg\left(\dfrac{z_1}{z_2}\right)=\arg(z_1)-\arg(z_2)=2 \arg(z_1)\,$.
Since $\,a,b,c \gt 0\,$ the roots are in the left half-plane $\,\operatorname{Re}(z_1)=\operatorname{Re}(z_2) = -\,\dfrac{b}{2a} \lt 0\,$, and given the condition that "$z_1\in$ third quadrant" $\,\operatorname{Im}(z_1) \lt 0\,$, so $\,z_1\,$ is the root with negative imaginary part:
$$z_1 = \dfrac{-b - i \sqrt{4ac - b^2}}{2a} = \sqrt{\dfrac{c}{a}}\left(-\dfrac{b}{2\sqrt{ac}} - i \sqrt{1 - \dfrac{b^2}{4ac}}\right)$$
The latter can be written as $\,z_1=\sqrt{\dfrac{c}{a}}\big(\cos(\varphi)+ i \sin(\varphi)\big)\,$ where $\,\varphi=\arg(z_1)\,$ and:
$$
\begin{cases}
\begin{align}
\cos(\varphi) &= -\dfrac{b}{2\sqrt{ac}} \\[5px]
\sin(\varphi) &= -\sqrt{1 - \dfrac{b^2}{4ac}}
\end{align}
\end{cases}
$$
Therefore $\,\varphi=2k\pi \pm \arccos\left(-\dfrac{b}{2\sqrt{ac}}\right)\,$, and the $\,\sin(\varphi) \lt 0\,$ condition trims the solution set down to $\,\varphi=2k\pi \color{red}{-} \arccos\left(-\dfrac{b}{2\sqrt{ac}}\right) = (2k-1)\pi + \arccos\left(\dfrac{b}{2\sqrt{ac}}\right)\,$, so in the end:
$$\,\arg\left(\dfrac{z_1}{z_2}\right)= 2 \varphi \;\;\equiv\;\; 2 \arccos\left(\dfrac{b}{2\sqrt{ac}}\right) = 2 \arccos\left(\sqrt{\dfrac{b^2}{4ac}}\right) \pmod{2 \pi} \,$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Number of answers : $f(x)=f^{-1}(x)$
let $f(x)= 1+\sqrt{x+k+1}-\sqrt{x+k} \ \ k \in \mathbb{R}$
Number of answers :
$$f(x)=f^{-1}(x) \ \ \ \ :f^{-1}(f(x))=x$$
MY Try :
$$y=1+\sqrt{x+k+1}-\sqrt{x+k} \\( y-1)^2=x+k+1-x-k-2\sqrt{(x+k+1)(x+k)}\\(y-1)^2+k-1=-2\sqrt{(x+k+1)(x+k)}\\ ((y-1)^2+k-1)^2=4(x^2+x(2k+1)+k^2+k)$$
now what do i do ?
| Hint:
Point of intersection of $f(x)$ and $f^{-1}(x)$ while same as that of $f(x)$ and the line $y=x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Value of $\sec^2 a+2\sec^2 b$
If $a,b$ are $2$ real number such that $2\sin a \sin b +3\cos b+6\cos a\sin b=7,$ Then $\sec^2 a+2\sec^2 b$ is
Try: i am trying to sve it using cauchy schwarz inequality
$$\bigg[(\sin b)(2\sin a+6\cos a)+(\cos b)(3)\bigg]^2\leq (\sin^2b+\cos^2b)\bigg((2\sin a+6\cos a)^2+3^2\bigg)$$
Could some help me to solve it , thanks
| Hint:
$2\sin a+6\cos a=\sqrt{40}\sin (a+x)$ where $x=\sin^{-1}\frac{6}{\sqrt{40}}$.
\begin{align*}
2\sin a \sin b +3\cos b+6\cos a\sin b&=\sqrt{40}\sin (a+x)\sin b+3\cos b\\
&=\sqrt{40\sin^2(a+x)+9}\cdot \sin(b+y)
\end{align*}
where $\sqrt{40}\sin(a+x)=\sqrt{40\sin^2(a+x)+9}\cdot\cos y$ and $3=\sqrt{40\sin^2(a+x)+9}\cdot\sin y$.
Since the maximum value of $\sqrt{40\sin^2(a+x)+9}\cdot\sin(b+y)$ is $7$, we need $\sqrt{40\sin^2(a+x)+9}=7$ and $\sin (b+y)=1$.
The answer should be $12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How to prove this definite integral does not depend on the parameter? I am working on some development formulas for surfaces and as a byproduct of abstract theory i get that:
$$\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{1+\sin^2\theta}{(\cos^4\theta+(\gamma\cos^2\theta-\sin\theta)^2)^\frac{3}{4}}d\theta$$
is independent on the parameter $\gamma\in\mathbb{R}$. I thought that there was something wrong with my calculations but actually turns out that using Mathematica that the value is somewhat near $5,24412$ independently on the $\gamma$ I plug in the calculation of the integral. Is there any way to verify that actually this is a constant by direct computations, complex analysis, or at least is this kind of integrals studied?
Edit:obviously differentiating in the integral does not help much
| Put
\begin{equation*}
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{1+\sin^2\theta}{(\cos^4\theta +(\gamma\cos^2\theta-\sin \theta)^2)^{\frac{3}{4}}}\, d\theta
\end{equation*}
If $x = \dfrac{\sin\theta}{\cos^2\theta}$, $\, y = \gamma-x$ and $y = \sqrt{z}$ then
\begin{equation*}
dx = \dfrac{\cos^2\theta+2\sin^2\theta}{\cos^3\theta}\, d\theta
\end{equation*}
and
\begin{gather*}
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\cos^2\theta+2\sin^2\theta}{\cos^3\theta\left(1 +\left(\gamma-\frac{\sin \theta}{\cos^2\theta}\right)^2\right)^{\frac{3}{4}}}\, d\theta = \int_{-\infty}^{\infty}\dfrac{1}{\left(1 +\left(\gamma- x\right)^2\right)^{\frac{3}{4}}}\, dx = \int_{-\infty}^{\infty}\dfrac{1}{\left(1 +y^2\right)^{\frac{3}{4}}}\, dy = \\[2ex]
\int_{0}^{\infty}\dfrac{z^{\frac{1}{2}-1}}{(1+z)^{\frac{1}{2}+\frac{1}{4}}}\, dz = {\rm B}\left(\frac{1}{4},\frac{1}{2}\right) \approx 5.244115109
\end{gather*}
where ${\rm B}$ is the Beta function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How many numbers from $1$ to $99999$ have the sum of the digits $= 15$? The problem:
How many numbers from $1$ to $99999$ have the sum of the digits $= 15$?
I thought of using the bitstring method and $x_1 + x_2 + x_3 + x_4 + x_5$ will be the boxes therefor we have $4$ zeros and $15$ balls. I'd say the answer would have to be $\binom{19}{15}$ is it right?
| Your answer is incorrect since you have not considered the restriction that a digit in the decimal system cannot exceed $9$.
We want to find the number of positive integers between $1$ and $99~999$ inclusive that have digit sum $15$. Since $0$ does not have digit sum $15$, we get the same answer by considering nonnegative numbers less than or equal to $99~999$ with digit sum $15$.
A nonnegative number with fewer than five digits such as $437$ can be viewed as a string of five digits by appending leading zeros. In this case, $437$ can be represented as the string $00437$. Hence, we can view the problem as finding the number of five-digit decimal strings with digit sum $15$. Hence, we seek the number of solutions in the nonnegative integers of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 15 \tag{1}$$
subject to the restrictions that $x_j \leq 9$ for $1 \leq j \leq 5$.
As you determined, the number of solutions of equation 1 is
$$\binom{15 + 5 - 1}{5 - 1} = \binom{19}{4} = \binom{19}{15}$$
From these, we must subtract those solutions in which one or more of the variables exceeds $9$. Since $2 \cdot 10 = 20 > 15$, at most one of the variables can exceed $9$.
Suppose $x_1 > 9$. Then $x_1' = x_1 - 10$ is a nonnegative integer. Substituting $x_1' + 10$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 10 + x_2 + x_3 + x_4 + x_5 & = 15\\
x_1' + x_2 + x_3 + x_4 + x_5 & = 5 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{5 + 5 - 1}{5 - 1} = \binom{9}{4}$$
solution. By symmetry, there are $\binom{9}{4}$ solutions that violate the restrictions for each of the five variables that could exceed $9$. Hence, the number of positive integers less than or equal to $99~999$ with digit sum $15$ is
$$\binom{19}{4} - \binom{5}{1}\binom{9}{4}$$
| {
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"source": "stackexchange",
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Compute $\iint_D(x^2-y^2)e^{2xy}dxdy$. Compute $$\iint_D(x^2-y^2)e^{2xy}dxdy,$$ where $D=\{(x,y):x^2+y^2\leq 1, \ -x\leq y\leq x, \ x\geq 0\}.$
The area is a circlesector disk with radius $1$ in the first and fourth quadrant. Going over to polar coordinates I get
$$\left\{
\begin{array}{rcr}
x & = & r\cos{\theta} \\
y & = & r\sin{\theta} \\
\end{array}, \ \ \implies E:\left\{
\begin{array}{rcr}
0 \leq r\leq 1 \\
-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} \\
\end{array}
\right.
\right.$$
and $$J(r,\theta)=\frac{d(x,y)}{d(r,\theta)}=r.$$
So $$\iint_D(x^2-y^2)e^{2xy}r \ dxdy=\iint_Er^3(\cos^2{\theta}-\sin{2\theta})e^{r^22\cos{\theta}\sin{\theta}}drd\theta= \\ =\iint_Er^3\cos{2\theta} e^{r^2\sin{2\theta}}drd\theta = 2\int_0^{4/\pi}\cos{2\theta}\cdot\left(\int_0^1 r^3e^{r^2\sin{2\theta}}dr\right)d\theta.$$
I have no idea how to compute the inner integral. I seem to get quite complicated integrals everytime I do this.
| Call $\alpha = \sin(2\theta)$ for simplicity.
$$\int_0^1 r^3 e^{\alpha r^2}\ dr = \int_0^1 \frac{d}{d\alpha} r e^{\alpha r^2} = \frac{d}{d\alpha} \int_0^1 r e^{\alpha r^2}\ dr$$
The latter can easily be done by parts once to get
$$\int r e^{\alpha r}\ dr = \frac{e^{a r} (a r-1)}{a^2}$$
Hence with the extrema it becomes
$$\frac{e^a (a-1)+1}{a^2}$$
Hence
$$\frac{d}{d\alpha} \left(\frac{e^a (a-1)+1}{a^2}\right) = \frac{\left(a^2-2 a+2\right) e^a-2}{a^3}$$
Getting back $\alpha = \sin(2\theta)$ and thou hast
$$\frac{\left(\sin^2(2\theta)-2 \sin(2\theta)+2\right) e^{\sin(2\theta)}-2}{\sin^3(2\theta)}$$
| {
"language": "en",
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"source": "stackexchange",
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How to solve the equation $x^2+2=4\sqrt{x^3+1}$? From the Leningrad Mathematical Olympiad, 1975:
Solve $x^2+2=4\sqrt{x^3+1}$.
In answer sheet only written $x=4+2\sqrt{3}\pm \sqrt{34+20\sqrt{3}}$.
How to solve this?
| HINT.-The given answer is root of the quadratic equation
$$x^2-2(4+2\sqrt3)x+c=0$$ where $c$ is certain constant. It follows
$$x=4+2\sqrt3\pm\sqrt{{(4+2\sqrt3)^2-c}}=4+2\sqrt3\pm\sqrt{34+20\sqrt3}$$
You get so the value of $c$ from which you can verify the solution.
| {
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} |
Inequality Proof $\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac{3\sqrt{3}}{2}$ Let $a,b,c\in \mathbb{R}^+$, and $a^2+b^2+c^2=1$, show that:
$$ \frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac{3\sqrt{3}}{2}$$
| The function $f(x)=\frac{1}{x(1-x^2)}$ takes its minimum at $x=\frac{1}{\sqrt 3}$ on $(0,1)$.
Thus
$$\begin{eqnarray*}\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2} & = & a^2 f(a) + b^2 f(b) + c^2 f(c)\\
& \ge & (a^2+b^2+c^2)f\left(\frac{1}{\sqrt 3}\right)= \frac{3\sqrt 3}{2}.\end{eqnarray*}$$
| {
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"url": "https://math.stackexchange.com/questions/2650458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How many bit strings?
How many bit strings of length $8$ have either exactly two $1$-bit among the first $4$ bits or exactly two $1$-bit among the last $4$ bits?
My solution:
A bit only contains $0$ and $1$, so $2$ different numbers, i.e., $0$ and $1$. For the first part we have $2^6=64$ ways. Similar for the other way. Hence there exists $2^4=16$ bit strings. Is my answer true?
Update: I mean $2^6+2^6-2^4=112$ bit strings
| Let $A$ be the set of bit strings with exactly two $1$-bit among the first $4$ bits, and $B$ be the set of bit strings with exactly two $1$-bit among the last $4$ bits.
\begin{align}
\#A &= \binom{4}{2} 2^4 = 6\cdot2^4 \\
\#B &= 2^4 \binom{4}{2} = 6\cdot2^4 \\
\#A\cap B &= \binom{4}{2}^2 = 6^2 \\
\#A\cup B &= \#A + \#B - \# A \cap B \\
&= 6 (2^4 \cdot 2 - 6) \\
&= 6 \cdot 26 = 156
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2650571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is this sum integer? $b! \pi+1/(b+1)+1/(b+2)(b+1)+...$, where $b \neq 0$. Is $S$ an integer ?
$S= \: b! \:\pi+\frac{1}{b+1}+\frac{1}{(b+2)(b+1)} + \frac{1}{(b+3)(b+2)(b+1)} + ...$
$b \neq 0$.
Also, from here Is this sum rational or not? $1/(q+1)+1/(q+2)(q+1)...$ where $q$ is an integer
$\frac{1}{b+1}+\frac{1}{(b+2)(b+1)} + \frac{1}{(b+3)(b+2)(b+1)} + ...$ is irrational and between $(0,1)$.
| Hint:
$$S_b=\frac{1}{b+1}+\frac{1}{(b+2)(b+1)} + \frac{1}{(b+3)(b+2)(b+1)} + ...+\: b! \:\pi=\dfrac{1}{b+1}(1+\frac{1}{b+2}+\frac{1}{(b+3)(b+2)} + \frac{1}{(b+4)(b+3)(b+2)} + ...+\: (b+1)! \:\pi)=\dfrac{S_{b+1}+1}{b+1}\to\\S_b=\dfrac{S_{b+1}+1}{b+1}$$can you finish now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Algorithm complexity using iteration method I want to find the complexity of
$$T(n) = T\left(\frac{n}{2}\right) + n \left(\sin\left(n-\frac{π}{2}\right) +2\right)$$ by iteration method.
Assume $T(1) = 1$.
\begin{align*}
T(n) &= T\left(\frac{n}{2}\right) + n \left(\sin\left(n-\frac{π}{2}\right) +2\right)\\
&= T\left(\frac{n}{2^2}\right) + \frac{n}{2} \left(\sin\left(\frac{n}{2}-\frac{π}{2}\right) +2\right) + n \left(\sin\left(n-\frac{π}{2}\right) +2\right)\\
&= T\left(\frac{n}{2^3}\right) + \frac{n}{2^2} \left(\sin\left(\frac{n}{2^2}-\frac{π}{2}\right) +2\right) + \frac{n}{2} \left(\sin\left(\frac{n}{2}-\frac{π}{2}\right) +2\right)\\
&\mathrel{\phantom{=}}{}+ n \left(\sin\left(n-\frac{π}{2}\right) +2\right)\\
&=\cdots\\
&= T\left(\frac{n}{2^k}\right)+ n \sum_{i = 0}^{k-1} \frac{1}{2^i} \sin\left(\frac{n}{2^i} - \frac{π}{2}\right) + 2n \sum_{i = 0}^{k-1} \frac{1}{2^i}\\
&= T\left(\frac{n}{2^k}\right)+ n \sum_{i = 0}^{k-1} \frac{1}{2^i} \sin\left(\frac{n}{2^i} - \frac{π}{2}\right) + 2n \frac{1-(\frac{1}{2})^{k}}{1-\frac{1}{2}}.
\end{align*}
Now, how to I can simplify the second summation?
Thanks.
| The term to be simplified is a combination of $\sum 2^k\cos 2^k$ and $\sum 2^k\sin 2^k$ or collectively $\sum 2^ke^{i2^k}$.
As far as I know, there is no closed form expression for such a "super-geometric" summation.
Anyway, using that $|e^{it}|\le1$, the absolute value of the sum is bounded by $2n$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Reduced row echelon with imaginary numbers Working on the following problem:
Let $v = \begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix}
w = \begin{bmatrix} -3 \\ i \\ 8 \\ \end{bmatrix}
y = \begin{bmatrix} h \\ -5i \\ -3 \\ \end{bmatrix}$
For what values of $h$ is the vector $y$ in the plane generated by the vectors $v$ and $w$?
My work so far: I know that for $y$ to be in the plane generated by $v$ and $w$, $y$ has to be a linear combination of $v$ and $w$. This means that I need to find the solution of the augmented matrix:
$$
\begin{bmatrix}
1 & -3 & h \\
0 & i & -5i \\
-2 & 8 & -3 \\
\end{bmatrix}
$$
I assume that I need to find the reduced echelon form of this matrix. However, I don't know how to deal with the imaginary numbers when doing the row reduction. Could anyone give me some pointers?
| \begin{align}
\begin{bmatrix}
1 & -3 & h \\
0 & i & -5i \\
-2 & 8 & -3
\end{bmatrix}\xrightarrow{R_3\leftarrow R_3+2R_1}
\begin{bmatrix}
1 & -3 & h \\
0 & i & -5i \\
0 & 2 & -3+2h
\end{bmatrix}\xrightarrow{R_3\leftarrow R_3+2iR_2}
\begin{bmatrix}
1 & -3 & h \\
0 & i & -5i \\
0 & 0 & 7+2h
\end{bmatrix}
\end{align}
so the condition is $\;h=-\dfrac72$.
If you really need the reduced row echelon form, just proceed:
$$
\begin{bmatrix}
1 & -3 & h \\
0 & i & -5i \\
0 & 0 & 7+2h
\end{bmatrix}\xrightarrow{R_2\leftarrow -iR_2}
\begin{bmatrix}
1 & -3 & h \\
0 & 1 & -5 \\
0 & 0 & 7+2h
\end{bmatrix}\xrightarrow{R_1\leftarrow R_1+3R_2}
\begin{bmatrix}
1 & 0 & h-15 \\
0 & 1 & -5 \\
0 & 0 & 7+2h
\end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Finding value of Quadratic If the quadratic equations $3x^2+ax+1=0$ and $2x^2+bx+1=0$ have a common root, then the value of $5ab-2a^2-3b^2$ has to be find.
I tried by eliminating the terms but ended with $(2a-b)x=1$. Can you please suggest how to proceed further?
| The common root, $r$, is also root of
\begin{align*}
3x^2+ax+1-(2x^2+bx+1)&=0\\
x^2+(a-b)x&=0\\
x(x+a-b)&=0
\end{align*}
The root cannot be $zero$, so we get $r=b-a$ as the common root.
Also, the common root is a root of
\begin{align*}
2(3x^2+ax+1)-3(2x^2+bx+1)&=0\\
(2a-3b)x-1&=0\\
x&=\frac1{2a-3b}\implies r=\frac1{2a-3b}
\end{align*}
Now,
\begin{align*}
5ab-2a^2-3b^2&=(b-a)(2a-3b)\\
&=r\left(\frac1r\right)\\
&=1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve recurrence relation. I've got an recurrent function to solve.
$T_1 = 1$
$T_n = 2T_{\frac n2} + \frac{1}{2}n\log n$
I've got a tip to this excercise to determine additional variable as $k = 2^n$, where $ k = 1, 2, 3, ...$
But after some calulations I'm wondering if $k=2^n$ can i say that $2^{k-1} = n - 1$ ?
I recieve following equation:
$T_{2^k} = 2T_{2^{k-1}} + 2^{k-1}k*\log2$ , am i correct? What can i do next?
| We can prove by induction that
$$T_n = 2^kT_{\frac{n}{2^k}} + \frac{k}{2}n\log(n)-\sum_{i=0}^{k-1}\frac{in}{2}$$
This clearly holds for the base case $k = 1$. So assuming it holds for some $k$, we find:
\begin{align}
T_n &= 2^kT_{\frac{n}{2^k}}+\frac{k}{2}n\log(n) - \sum_{i=0}^{k-1}\frac{in}{2}\\
&= 2^k\left(2T_{\frac{n}{2^{k+1}}} + \frac{1}{2}\frac{n}{2^k}\log\left(\frac{n}{2^k}\right)\right)+\frac{k}{2}n\log(n) - \sum_{i=0}^{k-1}\frac{in}{2}\\
&=2^{k+1}T_{\frac{n}{2^{k+1}}} + \frac{n}{2}\log(n) - \frac{n}{2}\log(2^k)+\frac{k}{2}n\log(n) - \sum_{i=0}^{k-1}\frac{in}{2}\\
&= 2^{k+1}T_{\frac{n}{2^{k+1}}}+\frac{k+1}{2}n\log(n)-\sum_{i=0}^{k}\frac{in}{2}
\end{align}
So we have the claim. The sum has an easy closed form, so we may rewrite $T_n$ as:
$$T_n = 2^kT_{\frac{n}{2^k}}+\frac{k}{2}n\log(n)-\frac{k(k-1)}{4}n$$
Letting $k= \log(n)$, we have:
\begin{align}
T_n&=nT_1+\frac{n}{2}\log^2(n)-\frac{n}{4}\log(n)(\log(n) - 1)\\
&=n + \frac{n}{4}\log^2(n)+\frac{n}{4}\log(n)
\end{align}
| {
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>Finding range of $f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$
Finding range of $$f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$$
Try: put $\sin x=t$ and $-1\leq t\leq 1$
So $$y=\frac{t^2+4t+5}{2t^2+8t+8}$$
$$2yt^2+8yt+8y=t^2+4t+5$$
$$(2y-1)t^2+4(2y-1)t+(8y-5)=0$$
For real roots $D\geq 0$
So $$16(2y-1)^2-4(2y-1)(8y-5)\geq 0$$
$$4(2y-1)^2-(2y-1)(8y-5)\geq 0$$
$y\geq 0.5$
Could some help me where I have wrong, thanks
| Hint: Use
$$\sin^2 x +4\sin x+5 = (\sin x +2)^2 +1$$
and
$$2\sin^2 x +8 \sin x +8 = 2\left(\sin x + 2 \right)^2.$$
Also, break the fraction into two pieces
$$\dfrac{(\sin x +2)^2 +1}{2\left(\sin x + 2 \right)^2}=\dfrac{1}{2}+\dfrac{1}{2}\dfrac{1}{\left(\sin x + 2 \right)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Alternative way to calculate the sequence Given $a>0$ that satisfies $4a^2+\sqrt 2 a-\sqrt 2=0$.
Calculate $S=\frac{a+1}{\sqrt{a^4+a+1}-a^2}$.
Attempt:
There is only one number $a>0$ that satisfies $4a^2+\sqrt 2 a-\sqrt 2=0$, that is
$a=\frac{-\sqrt{2}+\sqrt{\Delta }}{2\times 4}=\frac{-\sqrt{2}+\sqrt{2+16\sqrt{2}}}{8}$
However obviously if you replace it directly to calculate $S$, it would be extremely time-consuming.
Is there another way? I think the first equation can be changed (without needing to solve it) to calculate $S$, as when I use brute force (with calculator), I found out that $S\approx 1.414213562\approx\sqrt 2$.
| First, we have
$$\begin{align}S&=\frac{a+1}{\sqrt{a^4+a+1}-a^2}\\\\&=\frac{a+1}{\sqrt{a^4+a+1}-a^2}\cdot\frac{\sqrt{a^4+a+1}+a^2}{\sqrt{a^4+a+1}+a^2}\\\\&=\frac{(a+1)(\sqrt{a^4+a+1}+a^2)}{a+1}\\\\&=\sqrt{a^4+a+1}+a^2\tag1\end{align}$$
We have
$$4a^2+\sqrt 2 a-\sqrt 2=0\implies (4a^2)^2=(\sqrt 2-\sqrt 2\ a)^2\implies a^4=\frac{a^2-2a+1}{8}$$
from which
$$a^4+a+1=\frac{a^2-2a+1}{8}+a+1=\frac{(a+3)^2}{8}\tag2$$
follows.
Also, we have
$\sqrt 2\ (4a^2+\sqrt 2 a-\sqrt 2)=0\implies 2\sqrt 2\ a^2+a=1\tag3$
From $(1)(2)(3)$,
$$S=\frac{a+3}{2\sqrt 2}+a^2=\frac{2\sqrt 2a^2+a+3}{2\sqrt 2}=\frac{1+3}{2\sqrt 2}=\color{red}{\sqrt 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2669096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Roots of $x^3+5x-18=0$ using Cardano's method Given that $x^3+5x-18=0$. We have to solve it using Cardano's method.
Using trial $x=2$ will be a root. Dividing the equation by $x-2$ we shall get the other quadratic equation and solving that one, we shall obtain all the roots.
But when I am trying to solve the equation using Cardan\rq s method, the calculation is becoming very difficult. I don\rq t know why. Please help.
Here is how did I proceed.
Let $x=u+v$. Then $x^3=u^3+v^3+3uvx$ i.e. $x^3-3uvx-(u^3+v^3)=0$. So $-3uv=5$ and $u^3+v^3=18$. Clearly $u^3, v^3$ are the roots of
\begin{align}
&t^2-(u^3+v^3)t+(uv)^3=0\\
\Rightarrow &t^2-18t-\frac{125}{27}=0\\
\Rightarrow &27t^2-(27\times 18)t-125=0
\end{align}
and from here when we are getting the roots of $t$, they are very complicated. Hence I do not know how to simplify them so that $x=2$ finally be achieved along with the other two roots.
Please help me
| By your work we obtain:
$$x=\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}+\sqrt[3]{9-\sqrt{81+\frac{125}{27}}}=$$
$$=
\sqrt[3]{9+\frac{34}{3}\sqrt{\frac{2}{3}}}+\sqrt[3]{9+\frac{34}{3}\sqrt{\frac{2}{3}}}=\frac{1}{3}\left(\sqrt[3]{243+102\sqrt6}+\sqrt[3]{243-102\sqrt6}\right)=$$
$$=\frac{1}{3}\left(\sqrt[3]{(3+2\sqrt6)^3}+\sqrt[3]{(3-2\sqrt6)^3}\right)=\frac{1}{3}(3+2\sqrt6+3-2\sqrt6)=2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing $x^4-x^3+x^2-x+1>\frac{1}{2}$ for all $x \in \mathbb R$
Show that
$$x^4-x^3+x^2-x+1>\frac{1}{2}. \quad \forall x \in \mathbb{R}$$
Let $x \in \mathbb{R}$,
\begin{align*}
&\mathrel{\phantom{=}}x^4-x^3+x^2-x+1-\frac{1}{2}=x^4-x^3+x^2-x+\dfrac{1}{2}\\
&=x^2(x^2-x)+(x^2-x)+\dfrac{1}{2}=(x^2-x)(x^2+1)+\dfrac{1}{2}\\
&=x(x-1)(x^2+1)+\dfrac{1}{2}.
\end{align*}
Is there any way to solve this question?
| We have to prove $$x^4-x^3+x^2-x+0.5>0\forall x\in\mathbb{R}$$
So let $$f(x)=\frac{1}{2}\bigg[2x^4-2x^3+2x^2-2x+1\bigg]$$
$$f(x)=\frac{1}{2}\bigg[x^4+(x^2-x)^2+(x-1)^2\bigg]>0\forall x\in\mathbb{R}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Evaluate the limit $\lim_{n\to \infty}\sqrt{n^2 +2} - \sqrt{n^2 +1}$ I know that
$$\lim_{n\to \infty}(\sqrt{n^2+2} - \sqrt{n^2+1})=0.$$
But how can I prove this?
I only know that $(n^2+2)^{0.5} - \sqrt{n^2}$ is smaller than $\sqrt{n^2+2} - \sqrt{n^2}$ = $\sqrt{n^2+2} - n$.
Edit:
Thank Y'all for the nice and fast answers!
| Note that
$$( \sqrt{n^2+2}-\sqrt{n^2+1})(\sqrt{n^2+2}+\sqrt{n^2+1})=1$$
Thus
$$\sqrt{n^2+2}+\sqrt{n^2+1} \to \infty \implies \sqrt{n^2+2}-\sqrt{n^2+1}\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Adding Absolute value to a complex number: $ z+| z|=2+8i$ I would like to know my error in this problem.
Find the complex number such that:
$$ z+|z|=2+8i$$
So far, I have:
$$
\begin{split}
a+bi+\sqrt{a^2+b^2} &= 2 + 8i\\
a^2-b^2+a^2+b^2&=4-64\\
2a^2 -b^2 + b^2&=-60\\
a^2&=-30
\end{split}
$$
But I should end up with
$$a^2=-15$$
No matter how hard I try, I can't seem to find what I did wrong. Any suggestions?
| $a+bi+\sqrt{a^2+b^2} = 2 + 8i$ so
$a + \sqrt{a^2 + b^2} = 2$ and $b = 8$.
So $a + \sqrt{a^2 + 64} = 2$
So $\sqrt{a^2 + 64} = 2- a$
$a^2 + 64 = 4 -4a + a^2$
$4a = -60$
$a = -15$.
$z = -15 + 8i$.
....
To do what you were attempting
You have to realize that the $Re(z) = a + \sqrt{a^2 + b^2}$ and $Im(z) = b$. I think somehow you were thinking there were threee parts $Re(z)=a$ and $Im(z) = b$ and some $Weird(z) = \sqrt{a^2 + b^2}$ and that $z\overline z = Re^2(z) - Im^2(z) + Weird^2(z)$. That simply isn't true....
$(a+bi+\sqrt{a^2+b^2})(a - bi +\sqrt{a^2 + b^2}) = (2 + 8i)(2-8i)$
$(a + \sqrt{a^2 + b^2})^2 - b^2 = 4 - 64$
$2a^2 + b^2 + 2a\sqrt{a^2 + b^2} -b^2 = -60$
$a^2 + a \sqrt{a^2 + b^2} = -30$
which is a pain to solve but can be done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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RREF Practice Check I am going over the following exercise
Find a condition on $a,b,c$ so that $(a,b,c)$ in $\mathbb{R}^{3}$
belongs to the space spanned by $u = (2,1,0)$, $v=(1,-1,2)$, and $w =
(0,3,-4)$.
I write out the span of $u,v,w$ and set it equal to $a,b,c$. Then I reduce that linear system to
The last row tells me that one of my vectors was dependent on another. So I must have $\frac{1}{2}c = -\frac{2}{3}(b - \frac{1}{2}a)$. Then I solve for $c_2$ to get $c_2 = 2c_3 - \frac{2}{3}(b - \frac{1}{2}a)$. Now $c_1 = \frac{1}{2}a - c_3 + \frac{1}{3}(b - \frac{1}{2}a)$.
So, I think this is right, but how can I check?
| You row reduce the matrix
\begin{align}
\begin{bmatrix}
2 & 1 & 0 & a \\
1 & -1 & 3 & b \\
0 & 2 & -4 & c
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1/2 & 0 & a/2 \\
0 & -3/2 & 3 & b-a/2 \\
0 & 2 & -4 & c
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 1/2 & 0 & a/2 \\
0 & 1 & -2 & -\frac{2}{3}(b-a/2) \\
0 & 2 & -4 & c
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 1/2 & 0 & a/2 \\
0 & 1 & -2 & -\frac{2}{3}(b-a/2) \\
0 & 0 & 0 & c+\frac{4}{3}(b-a/2)
\end{bmatrix}
\end{align}
Thus the condition is
$$
c+\frac{4}{3}\left(b-\frac{a}{2}\right)=0
$$
that can be rewritten
$$
c=\frac{2}{3}a-\frac{4}{3}b
$$
No more steps are required.
Once you have a vector $(a,b,c)$ satisfying the condition, it's easy to find how you can express it in terms of the given vectors (actually of the first two, which form a basis of the span of $u,v,w$). Just compute the RREF:
$$
\begin{bmatrix}
1 & 1/2 & 0 & a/2 \\
0 & 1 & -2 & -\frac{2}{3}(b-a/2) \\
0 & 0 & 0 & 0
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 1 & a/2+\frac{1}{3}(b-a/2) \\
0 & 1 & -2 & -\frac{2}{3}(b-a/2) \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
Thus you see that a vector satisfying the condition above can be written as
$$
\left(\frac{a}{3}+\frac{b}{3}\right)u+
\left(\frac{a}{3}-\frac{2b}{3}\right)v=\frac{a+b}{3}u+\frac{a-2b}{3}v
$$
You also see that $w=u-2v$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum to $n$ terms of the series: $1^2.1+2^2.3+3^2.5+....$ Find the sum to $n$ terms of the series:
$$1^2.1+2^2.3+3^2.5+.....$$
My Attempt:
Here, $n^{th}$ term of $1,2,3,....=n$
$n^{th}$ term of $1^2,2^2,3^2,....=n^2$
Also, $n^{th}$ term of $1,3,5,....=2n-1$
Hence, $n^{th}$ term of the given series is $t_n=n^2(2n-1)$
| As , $$T_n=n^2(2n-1)$$ Hence , $$S_n=\sum T_n$$ Or , $$S_n=\sum_{k=1}^{n}2k^3-k^2$$ $$S_n=2\sum_{k=1}^{n}k^3-\sum_{k=1}^{n}k^2$$ Hence ,
$$S_n=\frac{(n(n+1))^2}{2}-\frac{n(n+1)(2n+1)}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2680816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove that $3^{2n-1} + 2^{n+1}$ is always a multiple of $7$. I'm trying to prove the following statement:
$P(n) = 3^{2n-1} + 2^{n+1}$ is always a multiple of $7$ $\forall n\geq1$.
I want to use induction, so the base case is $P(1) = 7$ so that's okay.
Now I need to prove that if $P(n)$ is true then $P(n+1)$ is true.
So there exists a $d \in \mathbb{N}$ such that
$$ 3^{2n-1} + 2^{n+1} = 7d $$
From this I need to say that there exists a $k \in \mathbb{N}$ such that:
$$ 3^{2n+1} + 2^{n+2} = 7k $$
With a little algebraic manipulation, I have managed to say:
$$ 2 \cdot 3^{2n+1} + 9 \cdot 2^{n+2} = 7\cdot(18d) $$
But now I am stuck. How should I keep going?
| Alternatively, suppose $3^{2n-1}+2^{n+1}= 7x$.
Then, $x=\frac17\left(3^{2n-1}+2^{n+1}\right)=\frac{1}{21}\left(3^{2n}+6(2^{n})\right)=\frac{1}{21}\left(9^{n}+6(2^{n})\right)\equiv 0 \pmod3$
Now, $9\equiv 2 \pmod7\Rightarrow9^n\equiv 2^n \pmod3$, $6\equiv -1 \pmod7$ and hence, $6\cdot2^n\equiv (-1)\cdot2^n\pmod7$
Therefore, $9^{n}+6(2^{n})\equiv 0 \pmod7$, because $9^{n}\equiv 2^n \pmod7$ and $6\cdot2^{n}\equiv -2^n \pmod7$
Therefore, since $9^{n}+6(2^{n})\equiv 0 \pmod3$, $9^{n}+6(2^{n})\equiv 0 \pmod{21}$, hence $\frac{1}{21}\left(9^{n}+6(2^{n})\right)=\frac{1}{7}\left(3^{2n-1}+2^{n+1}\right)$ is an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Given that $P(x)$ is a polynomial such that $P(x^2+1) = x^4+5x^2+3$, what is $P(x^2-1)$? How would I go about solving this? I can't find a clear relation between $x^2+1$ and $x^4+5x^2+3$ to solve $P(x^2-1)$.
| One more alternative is to start by deducing that $P(x)$ is a quadratic based on the fact that its degree has to be $2$ in order to output a quartic when the argument is a quadratic expression. So let $P(x) = ax^2 + bx + c$.
When $x^2 + 1 = 0, x = \pm i$, so we get From $P(0) = 1 - 5 + 3 = -1$, and also $P(0) = c$ we can immediately get $c=-1$.
When $x=0, x^2 + 1 = 1$ so we have that $P(1) = 3$ and $P(1) = a+b-1$, so $a+b = 4$.
When $x = 1, x^2 + 1 = 2$, so we have that $P(2) = 9$ and $P(2) = 4a + 2b - 1$, so $4a + 2b = 10$.
Solving the latter two for $a$ and $b$, we get $a=1, b=3$.
Hence $P(x) = x^2 + 3x - 1$ and $P(x^2-1)$ can be worked out.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
A quicker way to do a Lagrange multiplier problem I was working on the problem: minimize $x + 4z$ subject to $x^2 + y^2 + z^2 \le 2 $. I have it solved, I want a faster method for use in standardized exams.
My work:
I tackled this using Lagrange Multipliers, considering the interior by looking for points where all individual partial derivatives of $x+ 4z$ are zero, (of which there are none). Then considering the boundary
$$ (x + 4z) - \lambda ( x^2 + y^2 +z^2 - 2) $$
From here I differentiated w.r.t x,y,z, $\lambda$ and set equal to 0 to yield
$$ 1 - 2\lambda x = 0 \rightarrow 1 = 2\lambda x $$
$$ - 2 \lambda y = 0 \rightarrow 0 = 2 \lambda y \rightarrow y=0$$
$$ 4 - 2 \lambda z = 0 \rightarrow 4 = 2\lambda z$$
$$ - (x^2 + y^2 +z^2 -2 ) = 0 \rightarrow x^2 +z^2 = 2$$
Looking at equations 1, 3 we have
$$ \frac{1}{2} = \lambda x, 2 = \lambda z $$ And therefore
$$ \frac{1}{4} + 4 = \lambda^2 (x^2 +z^2 ) = 2 \lambda ^2 $$
$$ \frac{17}{8} = \lambda ^2 $$
And thus $$ \lambda = \pm \sqrt{ \frac{17}{8} } $$
$x = \frac{1}{2\lambda}, z = \frac{2}{\lambda} $
Yields
$$ x + 4z = \frac{1}{2\lambda} + 4 \frac{2}{\lambda} = \frac{17}{2 \lambda}
= \pm \sqrt{17} \sqrt{2} = \pm \sqrt{34}$$
Clearly $-\sqrt{34}$ is smaller, so we opt for that as our solution.
Now while this works, and makes sense, its not satisfactory as it TAKES SO LONG. And on a Math GRE where the expectation is to do this under 30 seconds a problem, I was hoping there was a faster method. Any suggestions? [Also open to ways to speed up the process, since even the same method with a different angle might be superior]
| By Cauchy-Schwarz,
$$|x + 4z| \le \sqrt{1^2 + 4^2} \sqrt{x^2+z^2} \le \sqrt{34}.$$
Then think about when Cauchy-Schwarz attains equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Calculete $a$ and $b$ in a limit Calculate $a$ and $b$ in
$$\lim_{x \to 1} \frac{ax^2+(3a+1)x+3}{bx^2+(2-b)x-2} = \frac{3}{2}$$
I tried this
$$\lim_{x \to 1} \frac{(ax+1)(x+3)}{(bx+2)(x-1)} = \frac{3}{2}$$
but I could not see the next step
I tried to look but it did not help. Solve for $a$ and $b$ in a limit
and Find A and B in this limit
| Your first step
$$\lim_{x \to 1} \frac{ax^2+(3a+1)x+3}{bx^2+(2-b)x-2} =\lim_{x \to 1} \frac{(ax+1)(x+3)}{(bx+2)(x-1)}$$
is correct, now observe that you need to cancel out the term $(x-1)$ in order to have a finite limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why: $z^{-1}=\cos (-\phi)+i\sin(-\phi)$ Let's suppose that we have a complex number with
$$r=1 \implies z=\cos\phi+i\sin\phi$$
Then why is $$z^{-1} =\cos (-\phi)+i\sin (-\phi)=\cos\phi-i\sin\phi$$
| As your complex number as $r=1$, you can express it like $z=e^{i\theta}$, where $\theta$ is the argument. Then,
$$z^{-1}=\frac{1}{z}=\frac{1}{e^{i\theta}}=e^{-i\theta}$$
Now, using the trigonometric form of complex numbers,
$$e^{-i\theta}=\cos(-\theta)+i\sin(-\theta)=\cos(\theta)-i\sin(\theta),$$
where we used that $\cos(\theta)=\cos(-\theta)$ and $\sin(\theta)=-\sin(-\theta)$ for all $\theta$.
Edit:
The power series of the exponential, the sine and the cosine are
$$e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$$
$$\sin x=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\dfrac{x^9}{9!}+...$$
$$\cos x=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!}x^{2n}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+...$$
Now, substituting $i\theta$ instead of $x$ in the power series of $e^x$, we have
\begin{align*}
e^{i\theta} & =1+i\theta+\dfrac{(i\theta)^2}{2!}+\dfrac{(i\theta)^3}{3!}+\dfrac{(i\theta)^4}{4!}+...=1+i\theta+\dfrac{i^2\theta^2}{2!}+\dfrac{i^3\theta^3}{3!}+\dfrac{i^4\theta^4}{4!}+... \\ \\
& =1+i\theta-\dfrac{\theta^2}{2!}-\dfrac{i\theta^3}{3!}+\dfrac{\theta^4}{4!}+...=\left(1-\dfrac{\theta^2}{2!}+\dfrac{\theta^4}{4!}-...\right)+i\left(\theta-\dfrac{i\theta^3}{3!}+\dfrac{\theta^5}{5!}-...\right) \\ \\
& =\cos\theta +i\sin\theta
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2686442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to deal with $(-1)^{k-1}$ It's a problem on mathematical induction.
$$1^2-2^2+3^2-.....+(-1)^{n-1}n^2=(-1)^{n-1}\frac{n.(n+1)}{2}$$
I have proved it for values of $n=1,2$.
Now I assume for $n=k$
$$P(k):1^2-2^2+3^2-.....+(-1)^{k-1}k^2=(-1)^{k-1}\frac{k.(k+1)}{2}$$.
$$P(k+1):1^2-2^2+3^2-.....+(-1)^{k-1}k^2+(k+1)^2=(-1)^{k-1}\frac{k.(k+1)}{2}+(k+1)^2\\=\frac{(k+1)}{2} [(-1)^{k-1}.k+2k+2]$$
I need suggestion to deal with the $(-1)^{k-1}$ so that I can prove the whole. Any help is appreciated.
| you have to prove that $$(-1)^k\frac{k(k+1)}{2}+(-1)^k(k+1)^2=(-1)^k\frac{(k+1)(k+2)}{2}$$ or
$$(-1)^{k-1}\frac{k(k+1)}{2}=(-1)^k\frac{(k+1)(k+2)}{2}-(-1)^k(k+1)^2$$
the right-hand side is given by
$$(-1)^k(k+1)\left(\frac{(k+2)}{2}-k-1\right)$$ and this is equal to
$$(-1)^k(k+1)\left(\frac{k+2-2k-2}{2}\right)$$ Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2688026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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} |
Double Antiderivation problem I have to find $f(x)$ given $f''(x)$ and certain initial conditions.
$$f"(x) = 8x^3 + 5$$ and $f(1) = 0$ and $f'(1) = 8$
$$f'(x) = 8 \cdot \frac{x^4}{4} + 5x + C = 2x^4 + 5x + C$$
Since $f'(1) = 8 \Rightarrow 2 + 5 + C = 8$ so $C = 1$
$$ f'(x) = 2x^4 + 5x + 1$$
$$f(x) = 2 \cdot \frac{x^5}{5} + 5 \cdot \frac{x^2}{2} + X = \frac{2}{5} x^5 + \frac{5}{2} x^2 + X + D$$
Since $f(1) = 0$ then:
$$\frac{2}{5} + \frac{5}{2} + 1 + D = 0$$
$$\frac{29}{10} + 1 + D = 0$$
So $D = \dfrac{-39}{10}$
So $$f(x) = \frac{2}{5} \cdot x^5 + \frac{5}{2} \cdot x^2 + x - \frac{39}{10}$$
Does that look right?
| Let check
$$f(x) = \frac{2}{5} \cdot x^5 + \frac{5}{2} \cdot x^2 + x - \frac{39}{10}\implies f'(x)= 2x^4+5x+1\implies f''(x)=8x^3+5$$
$$f(1)=\frac{2}{5}+ \frac{5}{2} + 1 - \frac{39}{10}=\frac{4+25+10-39}{10}=0$$
$$f'(1)=2+5+1=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2688163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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When point is inside of circle Let we have points $A_1,A_2,A_3,A_4$.In time $t$ point $A_i$ has coordinates $(x_i,y_i) + (v_{xi},v_{yi}) * t$, all parametrs are given. Describe algorithm to find all $t$ when point $A_4$ inside circle circumscribed around $A_1,A_2,A_3$ , or to find first moment when this happens.
| Let $A_i(X_i,Y_i)$ where
$$X_i=x_i+v_{xi}t\quad\text{and}\quad Y_i=y_i+v_{yi}t$$
for $i=1,2,3$ and $4$.
According to MathWorld, the center of the circle passing through three points $(X_1,Y_1),(X_2,Y_2)$ and $(X_3,Y_3)$ is given by
$$\left(-\frac{b}{2a},-\frac{c}{2a}\right)$$
and its radius is given by
$$\sqrt{\frac{b^2+c^2}{4a^2}-\frac da}$$
where
$$a=\begin{vmatrix}
X_1 & Y_1 & 1 \\
X_2 & Y_2 & 1 \\
X_3 & Y_3 & 1 \\
\end{vmatrix},\ b=-\begin{vmatrix}
X_1^2+Y_1^2 & Y_1 & 1 \\
X_2^2+Y_2^2 & Y_2 & 1 \\
X_3^2+Y_3^2 & Y_3 & 1 \\
\end{vmatrix},$$
$$c=\begin{vmatrix}
X_1^2+Y_1^2 & X_1 & 1 \\
X_2^2+Y_2^2 & X_2 & 1 \\
X_3^2+Y_3^2 & X_3 & 1 \\
\end{vmatrix},\ d=-\begin{vmatrix}
X_1^2+Y_1^2 & X_1 & Y_1 \\
X_2^2+Y_2^2 & X_2 & Y_2 \\
X_3^2+Y_3^2 & X_3 & Y_3 \\
\end{vmatrix}
$$
Therefore, under the conditions
$$a\not=0\quad\text{and}\quad \frac{b^2+c^2}{4a^2}-\frac da\gt 0$$
which are needed in order for such a circle to exist, we have
$$\begin{align}&\text{$A_4$ is inside the circle}
\\\\&\iff\sqrt{\left(X_4-\left(-\frac{b}{2a}\right)\right)^2+\left(Y_4-\left(-\frac{c}{2a}\right)\right)^2}\le \sqrt{\frac{b^2+c^2}{4a^2}-\frac da}
\\\\&\iff \left(X_4+\frac{b}{2a}\right)^2+\left(Y_4+\frac{c}{2a}\right)^2\le \frac{b^2+c^2}{4a^2}-\frac da
\\\\&\iff X_4^2+\frac{bX_4}{a}+Y_4^2+\frac{cY_4}{a}+\frac da\le 0
\\\\&\iff a^2X_4^2+abX_4+a^2Y_4^2+acY_4+ad\le 0\end{align}$$
The LHS of the last inequality is a sixth degree polynomial on $t$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Number of divisors of the number $2079000$ which are even and divisible by $15$
Find the number of divisors of $2079000$ which are even and divisible by $15$?
My Attempt: Since they are divisible by $15$ and are even, $2$ and $5$ have to included from the numbers prime factors.
$2079000 = 2^3 \cdot 3^3 \cdot 5^3 \cdot 7 \cdot 11$
Therefore, the number of divisors should be
$2 \cdot 2 \cdot (3+1) \cdot (1+1) \cdot (1+1)$
But however this answer is wrong.
Any help would be appreciated.
| Your factorization of $2079000$ is incorrect.
\begin{align*}
20790000 & = 2079 \cdot 1000\\
& = 2079 \cdot 10^3\\
& = 2079 \cdot 2^3 \cdot 5^3\\
& = 3 \cdot 693 \cdot 2^3 \cdot 5^3\\
& = 3 \cdot 3 \cdot 231 \cdot 2^3 \cdot 5^3\\
& = 3 \cdot 3 \cdot 3 \cdot 77 \cdot 2^3 \cdot 5^3\\
& = 2^3 \cdot 3^3 \cdot 5^3 \cdot 7 \cdot 11
\end{align*}
If a divisor of $2079000$ is a multiple of $2$ and $15 = 3 \cdot 5$, it must be a multiple of $2 \cdot 15 = 30$ since $2$ and $15$ are relatively prime. If a divisor of $2079000$ is a multiple of $30 = 2 \cdot 3 \cdot 5$, then $\frac{1}{30}$ of it must be a factor of
$$\frac{2079000}{30} = \frac{2^3 \cdot 3^3 \cdot 5^3 \cdot 7 \cdot 11}{2 \cdot 3 \cdot 5} = 2^2 \cdot 3^2 \cdot 5^2 \cdot 7 \cdot 11$$
Hence, the number of such divisors is
$$(2 + 1)(2 + 1)(2 + 1)(1 + 1)(1 + 1) = 3 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 108$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2690113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Fraction and simplification
solve: $\frac{1}{x(x-1)} + \frac{1}{x} = \frac{1}{x-1}$
What are the possible answers ?
(A) -1 (B) Infinitely Many Solutions (C) No solution (D) 0
The answer from where i've referred this is (B), but when i simplify it I get (D)
My solution:
$$\frac{1}{x(x-1)} + \frac{1}{x} = \frac{1}{x-1}$$
$$ \frac{x +x(x-1)}{x(x-1)\cdot x} = \frac{1}{x-1} \text{ (took l.c.m on l.h.s)}$$
$$ \frac{x + (x^2 -x)}{(x^2 - x)\cdot x}= \frac{1}{x-1}$$
$$\frac{x^2}{x^3 - x^2} = \frac{1}{x-1}$$
$$ x^2(x-1) = x^3 - x^2$$
$$ x^3 - x^2 = x^3 - x^2$$
Have I simplified it correctly?
| You can make it much simpler.
First you have to set the domain of validity: you must have $x\ne 0,1$.
Next, on this domain, remove the denominators multiplying both sides by the l.c.m. of the denominators, and simplify; you get:
$$1+(x-1)=x\iff x=x.$$
Hence any number $x\ne 0,1$ is a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $x^2 - 37y^2 =2$ does not have integer solutions We need to prove that $x^2 - 37y^2 =2$ does not have integer solutions.
I have two angles I thought about approaching it from:
*
*Since 37 is prime, I can show that for $x$ not divisible by $37$, we have $x^{36} ≡ 1mod(37)$ but I don't see how that's useful
*I could manipulate the equation and make it to: $x^2 - 4 = 37y^2 - 2$
$\implies (x-2)(x+2) = 37y^2 - 2$
Then if the RHS is even, then $y^2$ is even $\implies$ $y^2$ ends with $0, 4,$ or $6$ $\implies$ $37y^2$ ends with $0, 8,$ or $2$
$\implies 37y^2 -2$ ends with $0, 6,$ or $8$
But then I reach a dead end here too
Any suggestions or ideas?
| $x^2 - 37y^2 = 2$ is even. So as $odd \pm even = odd$, $even + even = even$, $odd + odd = even$ we can see that either $x^2$ and $37y^2$ are either both even or both odd and we can pursue that and get a contradiction.
But now would be a nice time to point out that for all integer $m^2 \not \equiv 2 \mod 4$ and $m^2 \not \equiv 3 \mod 4$ and that either $m^2 \equiv 0 \mod 4$ or $m^2 \equiv 1 \mod 4$.
Prook: Let $m = 2k + i$ where $i= 0,1$. Then $m^2 \equiv 4k^2 + 4ki +i^2 \equiv i^2 \mod 4$. And $i^2$ is either $0$ or $1$.
So $x^2 \equiv \{0,1\} \mod 4$ and $37y^2 \equiv y^2 \equiv \{0,1\} \mod 4$.
So $x^2 - 37y^2 \equiv \{0,1\} - \{0,1\} \equiv \{0-0,0-1,1-0,1-1\} \equiv \{0,3,1,0\} \mod 4$. And $x^2 - 37y^2 \equiv 2 \mod 4$ is just about the only possibility that can never happen.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that $ \int_{0}^{c} \frac{\sin(\frac{x}{2})x~dx}{\sqrt{\cos(x) - \cos(c)}} = \sqrt{2} \pi \ln(\sec(\frac{c}{2}))$ As the title says, I want to find a way calculate the following integral$ \int_{0}^{c} \frac{\sin(x/2)x~dx}{\sqrt{\cos(x) - \cos(c)}}$, which I know is equal to $\sqrt{2} \pi \ln(\sec(\frac{c}{2}))$.
At first glance, I thought this would not be very difficult to prove (and maybe it isn't), but after some straighforward manipulations, I was unable to make the $\pi$ factor appear.
Obs: $ 0<c < \pi$.
| We can transform
\begin{align}
I(c)&=\int_{0}^{c} \frac{\sin(x/2)}{\sqrt{\cos(x) - \cos(c)}}x\,dx\\
&=\frac{1}{\sqrt{2}}\int_{0}^{c} \frac{\sin(x/2)}{\sqrt{\cos^2(x/2) - \cos^2(c/2)}}x\,dx\\
&=2\sqrt{2}\int_0^{c/2}\frac{\sin y}{\sqrt{\cos^2y-\cos^2(c/2)}}y\,dy
\end{align}
Now, denoting $C=\cos (c/2)$ and enforcing the substitution $\cos y=Cu$, it comes
\begin{equation}
I(c)=2\sqrt{2}\int_1^{1/C}\frac{\arccos(Cu)}{\sqrt{u^2-1}}\,du
\end{equation}
By differentiation of this expression with respect to $C$, noticing that $\arccos(1)=0$, we obtain
\begin{align}
\frac{dI(c)}{dC}&=2\sqrt{2}\left[-\frac{1}{C^2}\frac{\arccos(1)}{\sqrt{\tfrac{1}{C^2}-1}}-\int_1^{1/C}\frac{u}{\sqrt{u^2-1}\sqrt{1-C^2u^2}}\,du\right]\\
&=\frac{-2\sqrt{2}}{\sqrt{1-C^2}}\int_0^{\frac{\sqrt{1-C^2}}{C}}\frac{dw}{\sqrt{1-\left( \frac{Cw}{\sqrt{1-C^2}} \right)^2}}\\
&=\frac{-2\sqrt{2}}{\sqrt{1-C^2}}\left[\frac{\sqrt{1-C^2}}{C}\arcsin\left(\frac{Cw}{\sqrt{1-C^2}}\right)\right]_{w=0}^{w=\frac{\sqrt{1-C^2}}{C}}\\
&=-\sqrt{2}\frac{\pi}{C}
\end{align}
(we have used the substitution $u^2=1+w^2$). Then,
\begin{equation}
I(c)=-\pi\sqrt{2}\ln\left( \frac{C}{A} \right)
\end{equation}
where $A$ is a constant. For $c=0$, we have $C=1 $ and $I(0)=0$, thus $A=1$.
Finally,
\begin{equation}
I(c)=\pi\sqrt{2}\ln\left(\sec\left( \frac{c}{2} \right)\right)
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
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prove that $\{a_n\}_{n=1}^{\infty}$ is bounded and monotone increasing. I am trying to prove the sequence $\{a_n\}_{n=1}^{\infty}$, which is defined by $a_{n+1} = \frac {2(a_n+1)}{a_n+2}$, and $a_1=1$.
(1) prove that it is monotone increasing. ($a_{n+1} \ge a_n$)
proof by induction.
$P(1)$ : $a_2= \frac {2(1+1)}{2+1} = \frac 43 \ge a_1 = 1$
Suppose $P(k)$ holds true : $a_k \le a_{k+1}$.
$P(k+1)$ : $a_{k+2}= \frac {2(a_k+1)}{a_k+2}$. Then, how can I proceed from here??
(2) prove that it is bounded ($a^2_n<2$)
$P(1): a^2_1=1<2 $
Suppose $P(k)$ holds true: $a^2_k<2$
$P(k+1): a^2_{k+1}= (\frac {2(a_k+1)}{a_k+2})^2= \frac {2a_k^2+4a_k+2}{a^2_k+4a_k+4}$ I also don't know how to proceed from here.
Thank you in advance !!
| I'll play around
and see what happens.
$a_{n+1} = \frac {2(a_n+1)}{a_n+2},
a_1=1
$
$\begin{array}\\
a_{n+1}^2-2
&= \dfrac {4(a_n+1)^2}{(a_n+2)^2}-2\\
&= \dfrac {4a_n^2+8a_n+4-2(a_n^2+4a_n+4)}{(a_n+2)^2}\\
&= \dfrac {4a_n^2+8a_n+4-2a_n^2-8a_n-8}{(a_n+2)^2}\\
&= \dfrac {2a_n^2-4}{(a_n+2)^2}\\
&= 2\dfrac {a_n^2-2}{(a_n+2)^2}\\
\end{array}
$
So $a_{n+1}^2-2$
has the same sign as
$a_n^2-2$.
Since $a_1 = 1$,
all $a_n$ satisfy
$a_n^2 < 2$.
$\begin{array}\\
a_{n+1} -a_n
&= \dfrac {2(a_n+1)}{a_n+2}-a_n\\
&= \dfrac {2a_n+2-a_n^2-2a_n}{a_n+2}\\
&= \dfrac {2-a_n^2}{a_n+2}\\
&> 0
\qquad\text{since } a_n^2 < 2\\
\end{array}
$
Note that if
$a_1^2 > 2$
then
all $a_n^2 > 2$
and the $a_n$
are decreasing.
In either case,
the $a_n$ are bounded and
monotonic,
so they approach a limit.
If $L$ is this limit,
since
$a_{n+1} -a_n
= \dfrac {2-a_n^2}{a_n+2}
$,
$\dfrac {2-a_n^2}{a_n+2} \to 0$
so
$a_n^2 \to 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all positives integers $n$ such that $n^3+1$ is a perfect square A solution as follows:
$n^3+1=x^2$
$n^3=x^2-1$
$n^3=(x-1)(x+1)$
$x-1=(x+1)^2~~or~~x+1=(x-1)^2$
$x^2+x+2=0~~or~~x^2-3x=0$
$x(x-3)=0$
$x=0~~or~~x=3~~\Longrightarrow~~n=2$
Does it cover all possible solutions? How to prove that 2 is the only which solves the problem.
| Hint: see that $m^2=n^3+1$ gives $(m-1)(m+1)=n^3$. What factors can $m-1$ and $m+1$ have in common? How can their product be a perfect cube?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Inequality $\frac{a^3+b^3+c^3}{3}\geq\sqrt{\frac{a^4+b^4+c^4}{3}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b)(a+c)(b+c)=8$. Prove that:
$$\frac{a^3+b^3+c^3}{3}\geq\sqrt{\frac{a^4+b^4+c^4}{3}}$$
Some attempts:
*
*From the condition follows $a^3+b^3+c^3 = (a+b+c)^3 -24$
*It is known (see here)
$$\frac{a+b+c}{3}\geq\sqrt[27]{\frac{a^3+b^3+c^3}{3}}$$
*Setting $2x=a+b$, $2y = b+c$, $2z = a+c$, we can express $a =x+z-y$ etc. The condition then becomes $xyz = 1$ which can be parametrized with free variables $0\leq q \leq 2 \pi /3 $ and arbitrary $r$ by
$$
x = \exp(r \cos q) \; ; \; y = \exp(r \cos (q + 2 \pi /3)) \; ; \; z = \exp(r \cos (q - 2 \pi /3))
$$
Using that, the condition can be removed and then calculus may be used.
*The question may be interpreted geometrically. Expressions such as $a^3+b^3+c^3 = $const. and $a^4+b^4+c^4 =$ const. can be interpreted as hypersurfaces of what has been called an N(3)-dimensional ball in p-norm, see here. A nice visualization is given in here. Then properties such as extrema, convexity etc. of these surfaces can be used.
I couldn't put the pieces together.
| I repeat the above hints:
Setting $2x=a+b$, $2y = b+c$, $2z = a+c$, we can express $a =x+z-y$ etc. Further, we have $a^3+b^3+c^3 = (a+b+c)^3 -24 = (x+y+z)^3 -24$.
The condition then becomes $xyz = 1$ which can be parametrized with free variables $0\leq q \leq 2 \pi /3 $ and arbitrary $r$ by
$$
x = \exp(r \cos q) \; ; \; y = \exp(r \cos (q + 2 \pi /3)) \; ; \; z = \exp(r \cos (q - 2 \pi /3))
$$
Using that, the condition can be removed and then calculus may be used.
So we have to show
$$f = ((x+z+y)^3 -24)^2- 3 \sum_{cyc}(x+z-y)^4 \geq 0$$
From the parametrization, we see that $f$ is periodic in $q$ with period length $2 \pi/3$. The maxima are at $q_+ = n 2 \pi/3$ and the minima are at $q_- = \pi/3 + n 2 \pi/3$. Hence, it is enough to investigate $f$ at $q = \pi/3$. This gives
$$f(r,q) \ge f(r,q = \pi/3) = \\
(\exp(-3r)(2\exp(3r/2) + 1)^3 - 24)^2 - 6\exp(-4r) - 3\exp(-4r)(2\exp(3r/2) - 1)^4$$
It is now a matter of calculus to show that the minimum of $f(r,q = \pi/3)$ occurs at $r=0$, giving $f(r,q) \ge f(r=0,q = \pi/3) = 0 \geq 0$ which proves the claim. $\qquad \Box$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find the value of $\frac{\tan A}{\tan B}$, given $\frac{\sin A}{\sin B}=5$
If $\displaystyle \frac{\sin A}{\sin B}=5$, then find the value of $\displaystyle \frac{\tan A}{\tan B}$
Try using the Componendo and Dividendo formula:
$$\frac{\sin A+\sin B}{\sin A-\sin B}=\frac{3}{2}$$
$$\frac{\tan(A+B)/2}{\tan(A-B)/2}=\frac{3}{2}$$
Can someone help me find: $$\frac{\tan A}{\tan B}$$
Thanks
| suppose $\dfrac{SinA}{SinB}=\dfrac{x}{y}=5$
so from right tringle we know $cosA=\sqrt{1-x^{2}}$ and so $$tanA=\dfrac{x}{\sqrt{1-x^{2}}}.$$ the same for y so :
$$\dfrac{tanA}{tanB}=\dfrac{\frac{x}{\sqrt{1-x^{2}}}}{\frac{y}{\sqrt{1-y^{2}}}}$$
and we have $x=5y$ so $x^{2}=25y^{2}$ , $1-x^{2}=1-25y^{2}$
$$\dfrac{tanA}{tanB}=\dfrac{\frac{5y}{\sqrt{1-25y^{2}}}}{\frac{y}{\sqrt{1-y^{2}}}}=\dfrac{5y\sqrt{1-y^{2}}}{y\sqrt{1-25y^{2}}}$$
now you need to know what is $y$!
| {
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"timestamp": "2023-03-29T00:00:00",
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Is $\frac{cos^4(x)}{4}-\frac{cos^2(x)}{2}+C$ a correct evalution of $\int sin^3(x)cos(x)\ dx$ The answer on Khan Academy states that the integral evaluates to $1 \over 4$$\sin^4(x)+C$
However, I performed a u-substitution that I cannot find a mistake in (maybe I am blind).
Here's the working:
$I=\int \sin^3(x)\cos(x)dx$
let $\ u=\cos(x)$
$du=-\sin(x)dx$
$dx= \frac{-1}{\sin(x)}\ du$
$I=\int \sin^3(x)u\cdot-\frac{1}{\sin(x)}dx$
$I=\int -\sin^2(x)\cdot u\ du$
$I=\int -(1-\cos^2(x))\ u\ du$
$I=\int (\cos^2(x)-1)\cdot u\ du$
$I=\int u(u^2-1)\ du$
$I=\int u^3-u\ du$
$I=\frac{u^4}{4}-\frac{u^2}{2}$
$I=\frac{\cos^4(x)}{4}-\frac{\cos^2(x)}{2}+C$
| Your solution is correct (with minor error at the end) and it is equivalent to the one given in Khan Academy:
$$\begin{align}I&=\frac{\cos^4(x)}{4}\overbrace{\require{cancel}\cancel{+}}^{-}\frac{\cos^2(x)}{2}+C=I= \\
&=\frac{(1-\sin^2 x)^2}{4}-\frac{\cos^2(x)}{2}+C=\\
&=\frac14-\frac{\sin^2 x}{2}+\frac{\sin^4 x}{4}-\frac{\cos^2 x}{2}+C= \\
&=\frac{\sin^4 x}{4}+C+\frac14-\frac12=\frac{\sin^4 x}{4}+A.\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/2704467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
My attempt:
$|x^2-3|=(x-3)^2$
So $-(x^2-3)=(x-3)^2$ or $(x^2-3)=(x-3)^2$
If $-(x^2-3)=(x-3)^2=x^2+9-6x$
So no solutions in $\mathbb R$
And if $(x^2-3)=(x-3)^2$
So $x^2-3=x^2+9-6x$
Now, can I delete $x^2$ with $x^2$ ? Like this
$x^2-x^2-3-9+6x=0$
$6x=12$
$x=2$
But $f(2)$ isn’t equal to $0$?
| I was taught to always find the domain of possible solutions first. We have
\begin{cases}
x-3 \ge 0, \\
x^2-3 \ge0
\end{cases} or \begin{cases}
x \ge 3, \\
-\sqrt{3} \le x \le \sqrt{3}
\end{cases}
which has no solutions
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A Problem on Theory Of Equations Let $f(x) = x^2 + x$, for all real $x$. There exist positive integers $m$ and $n$, and distinct nonzero real
numbers $y$ and $z$, such that $f(y) = f(z) = m + \sqrt{n}$ and $f(1/y) + f(1/z) = 1/10$ . Compute $100m + n$.
| Hint. Let $a=m+\sqrt{n}>0$. Note that $y$ and $z$ are the two solutions of the quadratic equation $x^2+x-a=0$. Therefore $y+z=-1$ and $yz=-a$. Hence
$$\frac{1}{y}+\frac{1}{z}=\frac{y+z}{yz}=\frac{-1}{-a}=\frac{1}{a}.$$
Moreover
$$\frac{1}{y^2}+\frac{1}{z^2}=\frac{y^2+z^2}{y^2z^2}=\frac{(y+z)^2-2yz}{(yz)^2}=\frac{(-1)^2-2(-a)}{(-a)^2}=\frac{1}{a^2}+\frac{2}{a}.$$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Expressing $\tan 20°$ in terms of $\tan 35°$ If $\tan 35^\circ = a$, we are required to express $\left(\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ\tan 125^\circ}\right)$ in terms of $a$.
Here's one way to solve this: $$\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ\tan 125^\circ} = \tan (145^\circ - 125^\circ) = \tan 20^\circ = \tan (90^\circ - 70^\circ) = \cot 70^\circ = \frac{1}{\tan 70^\circ} = \frac{1}{\tan (2 \times 35^\circ)} = \frac{1 - \tan^2 35^\circ}{2\tan 35^\circ} = \frac{1 - a^2}{2a}$$
However, I tried to solve it using another method as described below, and faced a problem:
$$\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ\tan 125^\circ} = \tan 20^\circ = \tan (35^\circ - 15^\circ) = \frac{\tan 35^\circ - \tan15^\circ}{1 + \tan 35^\circ\tan 15^\circ} = \frac{a - (2 - \sqrt3)}{1 + a(2 - \sqrt3)} = \frac{a - 2 + \sqrt3}{1 + 2a - \sqrt3a}$$
I tried to simplify it to get $\frac{1 - a^2}{2a}$, but I couldn't. So my question is, is there any way to show that $\frac{a - 2 + \sqrt3}{1 + 2a - \sqrt3a}$ is equal to $\frac{1 - a^2}{2a}$? If not, why are we getting two different answers?
| $\tan145^\circ=\tan(180^\circ-35^\circ) =-\tan35^\circ $ and $\tan125^\circ=\tan(90^\circ+35^\circ) =-\frac{1}{\tan35^\circ} $
For your two answers, have you find their values with a calculator?
Actually, you have proven an equality in $a$
| {
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What values of $a$ make matrix $A$ diagonalisable? I have the following question in an assignment paper.
Let $$A=\begin{bmatrix} 0 & a & 0\\ 1 & 0 & a\\ a & 1 & 0\end{bmatrix}$$ For what values of $a$ is $A$ diagonalisable?
Simply put, I don't know how to do it. In the $2 \times 2$ case we were asked, I completed the square of the characteristic polynomial and found that in all but $1$ choice of the unknown entry you got distinct eigenvalues and, therefore, distinct eigenvectors. At which point I just had to consider the one case for which I had eigenvalue of algebraic multiplicity $2$ and show that the geometric multiplicity of the eigenvector associated with it was $1$, I was done.
Any tips would be hugely appreciated, I've said it an assignment so reservation on full solution I understand but some hints would be amazing. Thank you.
| Taking $a^3 = \frac{32}{27} \; , $ this includes complex $a:$
$$
\frac{1}{864a}
\left(
\begin{array}{rrr}
40 & 36 a^2 & 24 a \\
-100 & -90 a^2 & 156 a \\
-9 a^2 & -24 a & 32
\end{array}
\right)
\left(
\begin{array}{rrr}
0 & a & 0 \\
1 & 0 & a \\
a & 1 & 0
\end{array}
\right)
\left(
\begin{array}{rrr}
6 a & -12 a & 54 a^2 \\
9 a^2 & 9 a^2 & -60 \\
10 & 4 & 0
\end{array}
\right) =
\left(
\begin{array}{ccc}
\frac{3}{2} a^2 & 0 & 0 \\
0 & - \frac{3}{4} a^2 & 1 \\
0 & 0 & - \frac{3}{4} a^2
\end{array}
\right)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero) The problem in question is as follows:
18b Suppose that $b^2 -4c \lt 0$. Show that there are no numbers $x$ that satisfy $x^2 + bx + c = 0$; in fact, $x^2 + bx + c \gt 0$ for all $x$. Hint: complete the square.
Trying to apply the hint, I began by constructing $b^2 - 4c < 0 \therefore (b-\frac{2c}{b})^2 - \frac{4c^2}{b^2} \lt 0$, but manipulating this ultimately just leads you to $b^2 \lt 4c$ which you didn't need to complete the square to get anyway.
The only other idea I had was that one could construct the quadratic equation beginning from the assumption that $x^2 + bx + c = 0$ and then go for proof by contradiction e.g.
$x^2 + bx + c =0$
$x^2 + bx = -c$
$x^2 + bx + (\frac{b}{2})^2 = -c + (\frac{b}{2})^2$
$(x + \frac{b}{2})^2 = \frac{b^2 - 4c}{4}$
$\therefore$ Given that for all real values of $x$ and $b$, $(x + \frac{b}{2})^2 \gt 0$, by transitivity of equality, $\frac{b^2 - 4c}{4} \gt 0$
$\therefore 4(\frac{b^2 - 4c}{4}) \gt 4(0)$
$\therefore b^2 - 4c \gt 0$ for all x such that $x^2 + bx + c = 0$
But that still leaves the statement "in fact, $x^2 + bx + c \gt 0$ for all $x$" unproven, unless it's supposed to obviously follow, in which case I'm not seeing how.
| What you are missing is that$$x^2+bx+c=\left(x+\frac b2\right)^2-\frac{b^2-4c}4.$$
| {
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"url": "https://math.stackexchange.com/questions/2706487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solve the differential equation 2y'=yx/(x^2 + 1) - 2x/y I have to solve the equation:
$$ 2y'= {\frac{xy}{x^2+1}} - {\frac{2x}{y}} $$
I know the first step is to divide by y, which gives the following equation:
$$ {\frac{2y'}{y}} ={\frac{x}{x^2+1}} - {\frac{2x}{y^2}}$$
According to my notes I get that I should make a substitution:
$$ z = {\frac{1}{y^2}} $$
and the derivative of z:$$ z' = {\frac{y'}{y^3}}$$
But I don't know how to proceed after this... Any help is appreciated!
| Hint
$$2y'= {\frac{xy}{x^2+1}} - {\frac{2x}{y}}$$
You can also multiply by y the Bernouilli 's equation
$$2y'y= {\frac{xy^2}{x^2+1}} - 2x$$
Observe that $(y^2)'=2y'y$ substitute $z=y^2$
$$z'= {\frac{xz}{x^2+1}} - 2x$$
Now it's a linear first ode
But you can do it your way
$${\frac{2y'}{y}} ={\frac{x}{x^2+1}} - {\frac{2x}{y^2}}$$
$${\frac{2y'}{y^2}} ={\frac{x}{y(x^2+1)}} - {\frac{2x}{y^3}}$$
Substitute $z=1/y$
$$-2z' =\frac{zx}{(x^2+1)} - 2xz^3$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What's the algebraic trick to evaluate $\lim_{x\rightarrow \infty} \frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$? $$\lim_{x \rightarrow \infty} \frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$$
I got the first half:
$$\frac{x\sqrt{x}}{\sqrt{x^{3}-1}+x}=\frac{x\sqrt{x}}{\sqrt{x^{3}(1-\frac{1}{x^3})}+x}=\frac{1}{\sqrt{1-\frac{1}{x^3}}+\frac{1}{x^2}}$$
which evaluates to$\frac{1}{1+0}$.
For the second term $\frac{\sqrt[3]{x+1}}{{\sqrt{x^{3}-1}+x}}$ I can't get the manipulation right. Help is much apreciated!
| Note that
$$\frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}=\frac{\sqrt{x^3}}{\sqrt{x^{3}}}\frac{1+\sqrt[6]{\frac{(x+1)^2}{x^9}}}{\sqrt{1-1/x^3}+1/\sqrt x}\to \frac{1+\sqrt{0}}{\sqrt{1-0}+0}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$x_{0}= \cos \frac{2\pi }{21}+ \cos \frac{8\pi }{21}+ \cos\frac{10\pi }{21}$ Prove that $x_{0}= \cos \frac{2\pi }{21}+ \cos \frac{8\pi }{21}+ \cos\frac{10\pi }{21}$ is a solution of the equation
$$4x^{3}+ 2x^{2}- 7x- 5= 0$$
My try:
If $x_{0}$, $x_{1}$, $x_{2}$ be the solutions of the equation then
$$\left\{\begin{matrix}
x_{0}+ x_{1}+ x_{2} = -\frac{1}{2} \\
x_{0}x_{1}+ x_{1}x_{2}+ x_{2}x_{0}= -\frac{7}{4} \\
x_{0}x_{1}x_{2} = -\frac{5}{4}
\end{matrix}\right.$$
I can' t continue! Help me!
| HINT:
The polynomial $4x^{3}+ 2x^{2}- 7x- 5$ factors as $(x + 1) (4 x^2 - 2 x - 5)$.
So we need to prove that $x_{0}= \cos \frac{2\pi }{21}+ \cos \frac{8\pi }{21}+ \cos\frac{10\pi }{21}$ is a root of $(4 x^2 - 2 x - 5)$. What is the other root? It is $x_1=\cos \frac{4\pi }{21}+ \cos \frac{16\pi }{21}+ \cos\frac{20\pi }{21}$.
How does this work? Note that we are dealing with conjugate elements $\cos\frac{2 k \pi}{21}$, where $k \in (\mathbb{Z}/21)^{\times}/{\pm 1}$. The Galois group of the extension generated by them is isomorphic to $G\colon =(\mathbb{Z}/21)^{\times}/{\pm 1}$, a cyclic group of order $6$ (note that $(\mathbb{Z}/21)^{\times}$ itself is not cyclic). The transformation $\rho_{a}$ maps $\cos \frac{2 k \pi}{21}$ to $\cos \frac{2 a k \pi}{21}$.
Now $G$ has a unique subgroup of order $3$ with representatives ${1,4,5}$ (the squares of elements of $G$). ${2,8,10}$ is the coset of $2$. Now it's not that hard to see that $x_0$, $x_1$ are conjugate algebraic numbers.
It's not that hard to check that $x_0+x_1=-\frac{1}{2}$, using the cyclotomic polynomial $\Phi_{21}(x)$ to produce an equation of degree $6$ with roots $\cos\frac{2 k \pi}{21}$, $k \in {1,2,4,5,8,10}$. It seems like an interesting exercise to check that $x_0 \cdot x_1=-\frac{5}{4}$.
ADDED: One can check that the sum $\sum_{a \in \mathbb{Z}/21)^{\times}} e^{\frac{2 a\pi }{21}}$ is just $2(x_0+x_1)$. But the sum of the roots of the cyclotomic polynomial $\Phi_{21}(z)$ is $-1$. Now for $x_0 \cdot x_1$ one can do calculations by hand using formulas for $\cos \alpha \cdot \cos \beta$.
Comment: The method in the other answer of @Will Jagy: works fine when we need to check that the expression is a root of a given polynomial, it's all automatic.
However, when one has a number like $\sum_{a \in H} e^{{2 a \pi}{n}}$ for a subgroup $H$ of $(\mathbb{Z}/n)^{\times}$, one finds right away the conjugates ( in fact one can do this for any elements of a cyclotomic field), then, using sufficient precision, select the distinct ones, and finds the minimal polynomial(now preferably to work with algebraic integers).
| {
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"timestamp": "2023-03-29T00:00:00",
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Volume of Ellipsoid using Triple Integrals Given the general equation of the ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} =1$, I am supposed to use a 3D Jacobian to prove that the volume of the ellipsoid is $\frac{4}{3}\pi abc$
I decided to consider the first octant where $0\le x\le a, 0\le y \le b, 0 \le z \le c$
I then obtained $8\iiint _E dV$ where $E = \{(x, y, z): 0\le x \le a, 0\le y \le b\sqrt{1-\frac{x^2}{a^2}}, 0\le z \le c\sqrt{1-\frac{x^2}{a^2} - \frac{y^2}{b^2}} \}$
I understood that a 3D Jacobian requires 3 variables, $x$, $y$ and $z$, but in this case I noticed that I can simple reduce the triple integral into a double integral:
$$8 \int_0^a \int_0^{b\sqrt{1-\frac{x^2}{a^2}}} c\sqrt{1-\frac{x^2}{a^2} - \frac{y^2}{b^2}} dydx$$ which I am not sure what substitution I should do in order to solve this, any advise on this matter is much appreciated!
| HINT
Let use spherical coordinates with
*
*$x=ra\sin\phi\cos\theta$
*$y=rb\sin\phi\sin\theta$
*$z=rc\cos\phi$
and with the limits
*
*$0\le \theta \le \frac{\pi}2$
*$0\le r \le 1$
*$0\le \phi \le \frac{\pi}2$
Remember also that in this case
$$dx\,dy\,dz=r^2abc\sin \phi \,d\phi \,d\theta \,dr$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2711676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Can $7n + 13$ ever equal a square? If not, why not? Can it be proved? And if it can be proved that it does equal a square (which I doubt), what is the smallest value for which this occurs?
| \begin{align}
1^2 &\equiv 1\pmod 7 \\
2^2 &\equiv 4\pmod 7 \\
3^2 &\equiv 2\pmod 7 \\
4^2 &\equiv 2\pmod 7 \\
5^2 &\equiv 4\pmod 7 \\
6^2 &\equiv 1\pmod 7 \\
7^2 &\equiv 0\pmod 7 \\
8^2 &\equiv 1\pmod 7 \\
9^2 &\equiv 4\pmod 7 \\
10^2 &\equiv 2\pmod 7 \\
11^2 &\equiv 2\pmod7\\
12^2 &\equiv 4\pmod7 \\
& \space\space\space\space\space\space\space\vdots
\end{align}
It repeats, so a perfect square can only be 0,1,2,4 mod 7
$7n+13 \equiv 6 \mod 7$ which is clearly not a perfect square.
| {
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"url": "https://math.stackexchange.com/questions/2712121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\frac{a}{a+ b}+ \frac{b}{b+ c}+ \frac{c}{2c+ 5a}\geq \frac{38}{39}$ $$a, b, c \in \left [ 1, 4 \right ] \text{ }
\frac{a}{a+ b}+ \frac{b}{b+ c}+ \frac{c}{2c+ 5a}\geq \frac{38}{39}$$
This is an [old problem of mine in AoPS]
(https://artofproblemsolving.com/community/u372289h1606772p10020940).
First solution
$$ab\geq 1\Leftrightarrow \frac{1}{1+ a}+ \frac{1}{1+ b}- \frac{2}{1+ \sqrt{ab}}= \frac{\left ( \sqrt{ab}- 1 \right )\left ( \sqrt{a}- \sqrt{b} \right )^{2}}{\left ( 1+ a \right )\left ( 1+ b \right )\left ( 1+ \sqrt{ab} \right )}\geq 0$$
then
$$\frac{1}{1+ a}+ \frac{1}{1+ b}\geq \frac{2}{1+ \sqrt{ab}}$$
Thus, we have
$$P= \frac{1}{1+ \frac{b}{a}}+ \frac{1}{1+ \frac{c}{b}}+ \frac{c}{2c+ 5a}\geq \frac{2}{1+ \sqrt{\frac{c}{a}}}+ \frac{\frac{c}{a}}{2\frac{c}{a}+ 5}\geq \frac{38}{39}$$
which is true by $\frac{c}{a}\leq 4$
How about another solution? I hope to see more. Thanks!
| You want to minimize the smooth function $F(a,b,c) = {\frac {a}{a+b}}+{\frac {b}{b+c}}+{\frac {c}{2\,c+5\,a}}$ over the cube $1 \le a,b,c \le 4$. The candidates for minimizer are critical points in the interior, points on a face where the gradient is orthogonal to the face, points on an edge where the gradient is orthogonal to the edge, and vertices. It's a bit tedious, but quite routine. The minimum value turns out to be $38/39$, achieved at $(a,b,c) = (1,2,4)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simon's Favorite Factoring Trick Problem Suppose that $x,y,z$ are positive integers satisfying $x \le y \le z$, and such that the product of all three numbers is twice their sum. What is the sum of all possible values of $z$?
Since this is for only positive integers, and there are sums and products involved, I think that this can be approached using Simon's Favorite Factoring Trick. I am not sure how though. Help is greatly appreciated
For those who do not know what Simon's Favorite Factoring Trick is, it is a method of factoring by grouping. For example, say that we want to factor $xy+x+y+1$. We can factor this as so:
$$xy+x+y+1$$
$$x(y+1)+y+1$$
$$x(y+1)+1(y+1)$$
$$(x+1)(y+1)$$
| Suppose $x,y,z$ are positive integers, with $x \le y \le z$, such that $xyz=2(x+y+z)$.
\begin{align*}
\text{Then}\;\;&xyz=2(x+y+z)\\[4pt]
\implies\;&xyz \le 2(3z)\\[4pt]
\implies\;&xy \le 6\\[4pt]
\implies\;&x^2 \le 6\\[4pt]
\implies\;&x \le 2\\[4pt]
\end{align*}
Consider two cases . . .
Case $(1)$:$\;x=1$.
\begin{align*}
\text{Then}\;\;&xyz=2(x+y+z)\\[4pt]
\iff\;&yz =2(1+y+z)\\[4pt]
\iff\;&yz - 2y -2z - 2 = 0\\[4pt]
\iff\;&(y-2)(z-2)=6
\end{align*}
which leads to a small number of possibilities for $y,z$, left for you to complete.
Case $(2)$:$\;x=2$.
Since $xy\le 6$, and $x \le y$, we get $2\le y\le 3$.
Using $x=2$, for each of $y=2,y=3$, solve the equation $xyz=2(x+y+z)$ for $z$.
| {
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"url": "https://math.stackexchange.com/questions/2713827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Derivative of the function $\sin^{-1}\Big(\frac{2^{x+1}}{1+4^x}\Big)$
Find y' if $y=\sin^{-1}\Big(\frac{2^{x+1}}{1+4^x}\Big)$
In my reference $y'$ is given as $\frac{2^{x+1}\log2}{(1+4^x)}$. But is it a complete solution ?
Attempt 1
Let $2^x=\tan\alpha$
$$
\begin{align}
y=\sin^{-1}\Big(\frac{2\tan\alpha}{1+\tan^2\alpha}\Big)=\sin^{-1}(\sin2\alpha)&\implies \sin y=\sin2\alpha=\sin\big(2\tan^{-1}2^x\big)\\
&\implies y=n\pi+(-1)^n(2\tan^{-1}2^x)
\end{align}
$$
$$
\begin{align}
y'&=\pm\frac{2.2^x.\log2}{1+4^x}=\pm\frac{2^{x+1}.\log2}{1+4^x}\\
&=\color{blue}{\begin{cases}
\frac{2^{x+1}.\log2}{1+4^x}\text{ if }-n\pi-\frac{\pi}{2}\leq2\tan^{-1}2^x\leq -n\pi+\frac{\pi}{2}\\
-\frac{2^{x+1}.\log2}{1+4^x}\text{ if }n\pi-\frac{\pi}{2}\leq2\tan^{-1}2^x\leq n\pi+\frac{\pi}{2}
\end{cases}}
\end{align}
$$
Attempt 2
$$
\begin{align}
y'&=\frac{1}{\sqrt{1-\frac{(2^{x+1})^2}{(1+4^x)^2}}}.\frac{d}{dx}\frac{2^{x+1}}{1+4^x}\\
&=\frac{1+4^x}{\sqrt{1+4^{2x}+2.4^x-4^x.4}}.\frac{(1+4^x)\frac{d}{dx}2^{x+1}-2^{x+1}\frac{d}{dx}(1+4^x)}{(1+4^x)^2}\\
&=\frac{(1+4^x).2^{x+1}.\log2-2^{x+1}.4^x.\log2.2}{\sqrt{1+4^{2x}-2.4^x}.(1+4^x)}\\&=\frac{2^{x+1}\log2\big[1+4^x-2.4^x\big]}{\sqrt{(1-4^x)^2}.(1+4^x)}\\&=\frac{2^{x+1}\log2\big[1-4^x\big]}{|{(1-4^x)}|.(1+4^x)}\\
&=\color{blue}{\begin{cases}\frac{2^{x+1}\log2}{(1+4^x)}\text{ if }1>4^x>0\\
-\frac{2^{x+1}\log2}{(1+4^x)}\text{ if }1<4^x
\end{cases}}
\end{align}
$$
In both my attempts i am getting both +ve and -ve solutions. Is it the right way to find the derivative?
And how do I connect the domain for each cases in attempt 1 and attempt 2 ?
| $y = arcsin(\frac{2^{x+1}}{1+4^x})\\\implies y = arcsin (\frac{2.2^x}{1+(2^x)^2})$
consider $y = arcsin(\frac{2x}{1+x^2})$
here let $x = tan(\theta) \implies \theta = arctan(x)$
$y= arcsin(\frac{2tan(\theta)}{1+ tan^2(\theta)}) = arcsin(sin(2\theta)) = 2\theta = 2arctan(x)$
so in your given question, $y = arcsin(\frac{2^{x+1}}{1+4^x}) = 2arctan(2^x) $
differentiate wrt x ;
$y' = \frac{2^{x+1}ln(2)}{1+4^x}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve recurrence $T(n) = T(\frac34n) + \sqrt n$ How can I solve the following recurrence relation?
$T(n) = T(\frac34n) + \sqrt n$
I have attempted to solve it with the substitution method.
I believe that the general pattern is $T(n) = T((\frac34)^kn) + n^{\frac1{2^k}}$,
but I cannot think of what method I need to follow to reach the closed form.
| An elementary way which doesn't use the Master Theorem could be:
$\begin{align*}
T(n) = T\left(\frac{3}{4} n\right) + \sqrt{n} \iff T(4n^2) = T(3n^2) + 2n
\end{align*}$
Thus, let us assume that $T(n) = C n^{\alpha}$ for some $C, \alpha$, is solution for the previous relation.
Then: $C(4^{\alpha} - 3^{\alpha})n^{2\alpha} = 2n$.
Then: $C(4^{\alpha} - 3^{\alpha})n^{2\alpha - 1} = 2$.
Now, this is true for all $n \in \mathbb{N}$, so $\alpha = \dfrac{1}{2}$ in order for $n^{2\alpha - 1} = 1$.
Then: $C = \dfrac{2}{\sqrt{4} - \sqrt{3}}$.
Now: $T(n) = \dfrac{2}{\sqrt{4} - \sqrt{3}} \sqrt{n}$.
Let us verify if $T$ is indeed a solution of your relation.
$\begin{align*}
T\left(\dfrac{3}{4} n\right) + \sqrt{n} & = C\sqrt{\dfrac{3}{4} n} + \sqrt{n} \\
& = \left[\dfrac{C}{2}\sqrt{3} + 1\right] \sqrt{n} \\
& = \left[\dfrac{1}{\sqrt{\frac{4}{3}} - 1} + 1\right] \sqrt{n} \\
& = \dfrac{2}{\sqrt{4} - \sqrt{3}} \sqrt{n} \\
& = C \sqrt{n} \\
& = T(n)
\end{align*}$
Which is coherent with the Master Theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the sum of all of the odd divisors of $6300$? What is the sum of all of the odd divisors of $6300$?
Hello! I am a middle school student, so simply-worded answers would be super helpful.
To solve this, I tried to find the sum of the odd divisors of a few smaller numbers, like 10. I know $10 = 2 * 5$, so I thought that, for the number to be odd, I'd have to exclude 2. Therefore, including 1, the sum would be $1 + 5 = 6$. This was correct, but I think that's only because the number is so small. When I tried it again with 18, which is $3^{2} * 2$, I got a different answer from the correct sum. How should I start this problem?
| $$6300 = 2^2\cdot3^2\cdot 5^2\cdot 7$$
The set of odd factors of $6300$ is equal to the set of factors of $3^2 \cdot 5^2 \cdot 7$.
The sum of factors of $3^2 \cdot 5^2 \cdot 7$ is
\begin{align}
\sum_{a=0}^2 \sum_{b=0}^2 \sum_{c=0}^1 3^a\cdot 5^b \cdot 7^c &= \sum_{a=0}^2 3^a\sum_{b=0}^2 5^b\sum_{c=0}^1 7^c \\
&=(1+3+3^2)(1+5+5^2)(1+7)
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2715635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving limits for fractions using epsilon-delta definition Using the $\epsilon - \delta $ definition of the limit, prove that: $$\lim_{x\to 0} \frac{(2x+1)(x-2)}{3x+1} = -2$$
I firstly notice that my delta can never be greater than $\frac{1}{3}$ because there is a discontinuity at $x=-\frac{1}{3}$.
I applied the standard steps as follows:
$\vert \frac{(2x+1)(x-2)}{3x+1} +2 \vert = \vert\frac{2x+3}{3x+1}\vert \vert x\vert$
Right now I need to restrict $x$ to some number, but I am not sure which value should I choose in order to easily bound my fraction, any help on choosing the correct delta is appreciated!
| Let $x > -\dfrac13$ and $|x| < \delta$, then $2x+3, 3x+1 > 0$, where $\delta > 0$ is to be determined.
$$\frac{2x+3}{3x+1} \le \frac{2\delta+3}{\underbrace{1-3\delta}_{\mbox{take $\delta < \frac13$}}}$$
Take $\delta < \dfrac13$ so that the denominator is positive. Observe that when $|x| < \delta$, the fraction is positive, so the absolute sign can be omitted.
$$0<\frac{1}{1-3\delta} < 2 \iff 1-3\delta > \frac12 \iff \delta < \frac16 \implies \delta < \frac13$$
When $|x| < \delta < \dfrac16$, $2x + 3 < 2\delta + 3 < \dfrac{10}{3}$, so $\dfrac{2x+3}{3x+1} < \dfrac{20}{3}$. If you want to cancel this factor in the final inequality, multiply $\epsilon$ with its inverse while defining $\delta$, i.e. set $\delta = \min\{\dfrac{3}{20} \epsilon, \dfrac16\}$. When $|x| < \delta$,
$$\left\vert \frac{(2x+1)(x-2)}{3x+1} +2 \right\vert = \frac{2x+3}{3x+1} \: \vert x\vert = \frac{20}{3} \cdot \dfrac{3}{20} \epsilon = \epsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof by induction with factorials I need help with proving this:
$$\sum_{i=1}^n \frac{i-1}{i!}=\frac{n!-1}{n!}$$
My induction hypothesis is:
$$\sum_{i=1}^{n+1} = \sum_{i=1}^n \frac{i-1}{i!}+\frac{(n+1)!-1}{(n+1)!}=\frac{(n+1)!-1}{(n+1)}$$
I tried a few things and landed here:
$$\frac{(n+1)n!-1+n}{(n+1)n!}=\frac{(n+1)n!-1}{(n+1)n!}$$
there is one $n$ too much in my last equation and I don't know how to get rid of it.
Thanks for your help.
| "My induction hypothethesis is
$\sum_{i=1}^{n+1} = \sum_{i=1}^n \frac{i-1}{i!}+\frac{(n+1)!-1}{(n+1)!}=\frac{(n+1)!-1}{(n+1)}$"
WHY?!?!?!?!?!?!?!?
$\sum_{i=1}^{n+1}\frac {i -1}{i} = \sum_{i=1}^n \frac {i-1}{i} + \frac {n+ 1 - 1}{(n+1)!}$ and $\frac {n+1 - 1}{(n+1)!} \ne \frac{(n+1)!-1}{(n+1)!}$ And setting $n\to n+1$ will give you $\frac {n!-1}{n!} \to \frac {(n+1)! - 1}{(n+1)!}$ and not $\frac{(n+1)! - 1}{n+1}$.
Surely you induction hypothethesis should have been
$\sum_{i=1}^{n+1}\frac {i -1}{i} = \sum_{i=1}^n \frac {i-1}{i} + \frac {n+ 1 - 1}{(n+1)!}= \frac {(n+1)! -1}{(n+1)!}$
Which is a matter of proving
$\frac {n! - 1}{n!} + \frac {n}{(n+1)!} = \frac {(n+1)! -1}{(n+1)!}$
which should be very easy to prove:
$\frac {n! - 1}{n!} + \frac {n}{(n+1)!} =
$\frac {(n! - 1)(n+1)}{n!(n+1} + \frac {n}{(n+1)!} =$
$\frac {(n+1)! - (n+1)}{(n+1)!} + \frac n{(n+1)!} =$
$\frac {(n+1)! - n - 1 +n}{(n+1)!} = \frac {(n+1)! - 1}{(n+1)!}$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/2717394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
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