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Find the limit: limit x tends to zero [arcsin(x)-arctan(x)]/(x^3) I'm having difficulty in finding the following limit.
$$\lim_{x\to 0}\frac{\arcsin(x)-\arctan(x)}{x^3}$$
I tried manipulating the given limit in standard limit(s) but I got nowhere. I tried L'Hôpital's rule and then realized it would be very lengthy and daunting so I left it. The answer is 1/2.
Thanks for any help in advance.
|
Without using L'Hospital's and Series Expansion,
Using this and Proof of $\arctan(x) = \arcsin(x/\sqrt{1+x^2})$,
$$\arcsin x-\arctan x=\arcsin x-\arcsin\frac x{\sqrt{1+x^2}}=\arcsin\left(x\cdot\frac1{\sqrt{1+x^2}}-\sqrt{1-x^2}\cdot\frac x{\sqrt{1+x^2}} \right)$$
$$=\arcsin\left(x\cdot\frac{(1-\sqrt{1-x^2})}{\sqrt{1+x^2}}\right)$$
$$=\arcsin\left(x\cdot\frac{1-(1-x^2)}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}\right)=\arcsin\left(\frac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}\right)$$
$$\implies\lim_{x\to 0}\frac{\arcsin(x)-\arctan(x)}{x^3}$$
$$=\lim_{x\to 0}\frac{\arcsin\left(\dfrac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}\right)}{\dfrac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}}\cdot\frac1{\lim_{x\to 0}\sqrt{1+x^2}(1+\sqrt{1-x^2})}$$
Set $\displaystyle\arcsin\left(\dfrac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}\right)=\theta\implies\dfrac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}=\sin\theta$ in the first limit to find the limit to be $1$
The second limit is easy enough
|
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|
Finding the Asymptotic Curves of a Given Surface I have to find the asymptotic curves of the surface given by $$z = a \left( \frac{x}{y} + \frac{y}{x} \right),$$ for constant $a \neq 0$.
I guess that what was meant by that statement is that surface $S$ can be locally parametrized by $$X(u,v) = \left( u, v, a \left( \frac{u}{v} + \frac{u}{v} \right) \right).$$ Do you think that my parametrization is correct (meaning that I read the description of the surface correctly), and do you know of a more convenient parametrization?
Assuming that parametrization, I derived the following ($E$, $F$, $G$, are the coefficients of the first fundamental form; $e$, $f$, $g$ are coefficients of the second fundamental form; $N$ is the normal vector to surface $S$ at a point; these quantities are all functions of local coordinates $(u,v)$):
$$E = 1 + a^2 \left( \frac{1}{v} - \frac{v}{u^2} \right)^2,$$
$$F = -\frac{a^2 (u^2 - v^2)^2}{u^3 v^3},$$
$$G = 1 + a^2 \left( \frac{1}{u} - \frac{u}{v^2} \right)^2.$$
$$N = \frac{1}{\sqrt{E G - F^2}} \left( a \left( \frac{v}{u^2}-\frac{1}{v} \right), a \left( \frac{u}{v^2}-\frac{1}{u} \right), 1 \right).$$
$$X_{u,u} = \left( 0,0, \frac{2 a v}{u^3} \right), X_{u,v} = \left( 0,0, -a \left( \frac{1}{u^2} + \frac{1}{v^2} \right) \right), X_{v,v} = \left( 0, 0, \frac{2 a u}{v^3} \right).$$
$$e = \frac{2 a v}{u^3 \sqrt{E G - F^2}},$$
$$f = - \frac{a (\frac{1}{u^2} + \frac{1}{v^2})}{\sqrt{E G - F^2}},$$
$$g = \frac{2 a u}{v3 \sqrt{E G - F^2}}.$$
Thus, the Gaussian curvature (from these calculations) is:
$$K = -\frac{a^2 u^4 v^4 (u^2 - v^2)^2}{(u^4 v^4 +
a^2 (u^2 - v^2)^2 (u^2 + v^2))^2}.$$
And the mean curvature would be:
$$H = \frac{a u^3 v^3 (u^4 + v^4)}{(u^4 v^4 + a^2 (u^2 - v^2)^2 (u^2 + v^2))^{3/2
}}.$$
So, the principal curvatures are:
$$k_{\pm} = H \pm \sqrt{H^2 - K} = a u^2 v^2 \frac{u v (u^4 + v^4) \pm \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}}{(u^4 v^4 + a^2 (u^2 - v^2)^2 (u^2 + v^2))^{3/2}}.$$
In order to find the asymptotic curves, but trying to avoid the differential equation, I was hoping to find the angles $\theta (u,v)$ such that the normal curvature would always be $0$. In other words I was trying:
$0 = k_n = k_{+} \cos{(\theta)}^2 + k_{-} \sin{(\theta)}^2$, and solving for $\theta$.
Assuming sufficient niceness, this calculate would result in: $$(u v (u^4 + v^4) + \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}) \cos{(\theta)}^2 + (u v (u^4 + v^4) - \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}) \sin{(\theta)}^2$$
First of all, is this approach (solving for $\theta$ rather than solving the differential equation) valid? If it is, after I find that angle $\theta$, determined by location $(u,v)$ on $S$, what more work do I have to do? How do I find the equations for the asymptotic curves based on this angle?
If this whole method was for naught, how does one solve the differential equation. in this case, of: $$e (u')^2 + 2f u' v' + g (v')^2 = 2a v^4 (u')^2 - 2a u^3 v^3 \left( \frac{1}{u^2} + \frac{1}{v^2} \right)u' v' + 2a u^4 (v')^2 = 0?$$ (Again, assuming sufficient niceness.)
(See: https://math.stackexchange.com/questions/762195/differential-equation-for-the-asymptotic-directions-of-a-given-surface)
Thank you!
|
I don't have much time for elaborating now, but you definitely can do this yourself (feel free to add your own answer).
Just observe that substitutions $x = r \cos \theta$ and $y = r \sin \theta$ to $z = a \left( x/y + y/x \right)$ turn this equation to $z = \tfrac{a}{\cos \theta \cdot \sin \theta}$ which clearly does not depend on $r$. This means that for each $\theta$ the ray $\theta = const$, $r>0$ lies on the surface. Now you can use Corollary 18.6. from Chapter 18 of A.Gray, E.Abbena and S.Salamon "Modern Differential Geometry of Curves and Surfaces with Mathematica" stating that "A straight line that is contained in a regular surface is necessarily an asymptotic curve".
Alternatively you may try to use the differential equation for asymptotic curves from here.
|
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|
Proof that $n$ is prime when $1+x+x^2+...+x^{n-1}$ = ${1-x^n}\over{1-x}$ is prime. Given $x$ and $n$ are positive integers and $1+x+x^2+...+x^{n-1}$ is a prime number.
Then prove that $n$ is a prime number.
Can the formula $1+x+x^2+...+x^{n-1} = \dfrac{1-x^n}{1-x}$ be somehow used?
|
An equivalent statement is that if $n = km$ then $\frac {1-x^n} {1-x}$ is also composite. See how you can use the formula $1+x+x^2+...+x^{n-1} = \frac {1-x^n} {1-x}$ to factorize $\frac {1-x^{km}} {1-x}$.
|
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|
Prove $\cos 3\theta = 4 \cos^3\theta − 3 \cos \theta$ $\cos 3θ = 4 \cos^3 θ − 3 \cos θ$
Here's my attempt. Is it correct? Thanks!
$\cos(3θ)$
$= \cos(2θ + θ)$
$= \cos(2θ)\cos(θ) - \sin(2θ)\sinθ$
$= (2\cos^2θ - 1)\cosθ - (2\sinθ\cosθ)\sinθ$
$= 2\cos^3θ - \cosθ - 2\sin^2θ\cosθ$
$= 2\cos^3θ - \cosθ - 2(1 - \cos^2θ)\cosθ$
$= 2\cos^3θ - \cosθ - (2\cosθ - 2\cos^3θ)$
$= 4\cos^3θ - 3\cosθ $
|
Almost good. You made a mistake on the second last line: the $2$ should have been distributed to both the $1$ and the $\cos^2\theta$
Instead of $2\cos^3\theta-\cos\theta-(2\cos\theta-\cos ^3\theta)$, it should be $2\cos^3\theta-\cos\theta-(2\cos\theta-2\cos^3\theta)$
|
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"language": "en",
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|
Find roots of $3^x+x^3=17$ Find $x$ in the following equation:
$$3^x+x^3=17$$
|
Note that if $y\gt 0$we have $3^y\gt 1$ so that if $x\ge 0$ $$3^{x+y}+(x+y)^3=3^x\cdot 3^y +x^3+(3x^2y+3xy^2+y^3)\gt 3^x+x^3$$
And it is very easy to show that for $x\le 0$ we have $3^x+x^3\le 1+0\lt 17$
So the function is increasing for positive $x$ (without calculus) and there is at most a single solution.
|
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|
Non-homogenous System where did I go wrong? Solve the system $\vec{x^{'}}=\begin{pmatrix}2 & -5\\1 & -2 \end{pmatrix}\vec{x}+ \begin{pmatrix} -\cos t\\ \sin t \end{pmatrix}$
The Eigenvalues are $(2-\lambda)(-2-\lambda)+5=0 \implies \lambda=\pm i$
The Eigevalue we choose is $\lambda=i$ and get a fundamental matrix of $$ \phi(t)=\begin{pmatrix} 5\cos t & 5\sin t\\ 2\cos t + \sin t & 2\sin t - \cos t \end{pmatrix}$$
We find the inverse by \begin{gather*}
\phi^{-1}(t)=\dfrac{1}{5\cos t(2\sin t-\cos t)- 5\sin t(2 \cos t+ \sin t)} \begin{pmatrix} 2\sin t - \cos t & -5\sin t\\ -2 \cos t - \sin t & -5\cos t \end{pmatrix}=\\[8pt]\dfrac{1}{5}\begin{pmatrix} \cos t-2\sin t & 5\sin t\\ 2 \cos t + \sin t & 5\cos t \end{pmatrix}
\end{gather*}
We multiply by $g(t)$ and have \begin{gather*}\dfrac{1}{5}\begin{pmatrix} \cos t-2\sin t & 5\sin t\\ 2 \cos t + \sin t & 5\cos t \end{pmatrix}\begin{pmatrix} -\cos t\\ \sin t \end{pmatrix}=\begin{pmatrix} -\cos^2 t+2\sin t\cos t +5 \sin^2 t\\ -2\cos^2 t -2\cos t \sin t +5\cos t \sin t \end{pmatrix}=\\[8pt]\begin{pmatrix} 5 \sin^2 t-\cos^2 t +2 \sin t \cos t\\ -2\cos^2 t +3\sin t \cos t \end{pmatrix}\end{gather*}
Integrating all of this yields $$\begin{pmatrix} \frac{5}{2}t - \frac{5}{4}\sin 2t - \frac{1}{2}t -\frac{1}{4}\sin 2t + \sin^2 t\\[8pt] -t-\frac{1}{2}\sin 2t+\frac{3}{2}\sin^2 t\end{pmatrix}=\begin{pmatrix} 2t-3\sin 2t +\sin^2 t\\ -t-\frac{1}{2}\sin 2t+\frac{3}{2}\sin^2 t \end{pmatrix}$$
Multiply by $\phi(t)$ and we have \begin{gather*}\begin{pmatrix} 5\cos t & 5\sin t\\ 2\cos t + \sin t & 2\sin t - \cos t \end{pmatrix}\begin{pmatrix} 2t-3\sin 2t +\sin^2 t\\ -t-\frac{1}{2}\sin 2t+\frac{3}{2}\sin^2 t \end{pmatrix}=\end{gather*}
$$\begin{pmatrix} 10t\cos t -15 \sin 2t \cos t +5 \sin^2 t \cos t-5t\sin t -\frac{5}{2}\sin 2t \sin t+\frac{15}{2}\sin^3 t \\ 4t \cos t +2t \sin t -6 \sin 2t \cos t -3 \sin t \sin 2t+2\cos t \sin^2 t + \sin^3 t -2t\sin t +2t\cos t -\sin 2t \sin t+\frac{1}{2}\sin 2t \cos t +3\sin^3 t-\frac{3}{2}\sin^2 t \cos t \end{pmatrix}
\dfrac{1}{5}$$
The book gets $x_h+\begin{pmatrix}2\\1 \end{pmatrix} t \cos t -\begin{pmatrix}1\\
0 \end{pmatrix} t \sin t$
So how did they simplify all that?
|
Note: the author used $\lambda = i$ and $v_1 = (5, 2-i)$ for the calculations.
You have a sign issue:
$$\phi^{-1}(t)=\dfrac{1}{5}\begin{pmatrix} \cos t-2\sin t & 5\sin t\\ 2 \cos t + \sin t & -5\cos t \end{pmatrix}$$
Look at the sign of entry $\phi^{-1}(t)_{22}$, which should be a negative.
Multiplying by $g(t)$, we get:
$$\left(
\begin{array}{c}
\sin ^2(t)-\frac{1}{5} \cos (t) (\cos (t)-2 \sin (t)) \\
-\cos (t) \sin (t)-\frac{1}{5} \cos (t) (2 \cos (t)+\sin (t)) \\
\end{array}
\right)$$
Integrating, we get:
$$\left(
\begin{array}{c}
\frac{2 t}{5}-\frac{1}{10} \cos (2 t)-\frac{3}{10} \sin (2 t) \\
\frac{3 \cos ^2(t)}{5}-\frac{1}{10} \sin (2 t)-\frac{t}{5} \\
\end{array}
\right)$$
Expanding the trig terms, we get:
$$\left(
\begin{array}{c}
-\frac{1}{10} \cos ^2(t)-\frac{3}{5} \sin (t) \cos (t)+\frac{\sin ^2(t)}{10}+\frac{2 t}{5} \\
\frac{3 \cos ^2(t)}{10}-\frac{1}{5} \sin (t) \cos (t)-\frac{t}{5}-\frac{3 \sin ^2(t)}{10}+\frac{3}{10} \\
\end{array}
\right)$$
Multiplying by $\phi(t)$, we get:
$$\left(
\begin{array}{c}
\left(2 t-\frac{1}{2}\right) \cos (t)-t \sin (t) \\
\frac{1}{10} (2 (5 t-4) \cos (t)+\sin (t)) \\
\end{array}
\right)$$
Of course the $\cos t$ and $\sin t$ terms get absorbed into the constants and you are left with the author's result.
|
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|
Calculation of $\int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$ Calculation of $\displaystyle \int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$
$\bf{My\; Try::}$ Let $x=\tan \psi\;,$ Then $\displaystyle dx = \sec^2 \psi$
So Integral convert into $\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi-1+\sec \psi}{\tan \psi+1+\sec \psi}d\psi = \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi+\sec \psi - 1}{\tan \psi+\sec \psi+1}d\psi$
Now Multiply both Numerator and Denominator by $\left(\tan \psi+\sec \psi-1\right)$
So $\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi+\sec \psi-1}{\tan \psi+\sec \psi+1} \times \frac{\tan \psi+\sec \psi-1}{\tan \psi+\sec \psi-1}d\psi = \int_{0}^{\frac{\pi}{4}}\frac{\left(\tan \psi+\sec \psi-1\right)^2}{\left(\tan \psi+\sec \psi\right)^2-1}d\psi$
Now I did not understand how can i solve after that
Help required
Thanks.
|
Letting $x=\tan \theta$ gives
$$
\begin{aligned}
\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}} &=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1+\sec \theta} \\
&=\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta+1} \\
&=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}+2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}+2 \cos ^2 \frac{\theta}{2}} \\
&=\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2} }=\frac{\sin \theta}{1+\cos \theta}
\end{aligned}
$$
Then let $c=\cos \theta,$ we have$$
I= -\int_0^{\frac{1}{\sqrt 2} } \frac{d c}{(1+c) c^2}= \int_0^{\frac{1}{\sqrt{2}}}\left(\frac{1}{c}-\frac{1}{c^2}-\frac{1}{1+c}\right) d c= \sqrt{2}-1+\ln [2(\sqrt{2}-1)]
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How Do I Integrate? $\int \frac{-2x^{2}+6x+8}{x^{2}(x+2)}$ How do I integrate this one? $$\int \frac{-2x^{2}+6x+8}{x^{2}(x+2)}\,dx$$
Is my answer correct: $$-3\ln\left \| x+2 \right \|+\ln\left \| x \right \|+\frac{4}{x}+C$$
|
Express in partial fractions:
$$\frac{8 + 6x - 2x^{2}}{x^{2}(x + 2)} = \frac{4}{x^{2}} + \frac{1}{x} - \frac{3}{x + 2}$$
Integrate term by term:
$$\int \frac{4}{x^{2}} + \frac{1}{x} - \frac{3}{x + 2} \ \mathrm{d}x = -\frac{4}{x} + \ln{|x|} - 3\ln{|x + 2|} + C$$
So you are missing a negative sign.
|
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|
What is the value of $\sin 47^{\circ}+\sin 61^{\circ}- \sin25^{\circ} -\sin11^{\circ}$? After simplification using sum to product transformation equations I keep ending up with
$$4\cos36^\circ\cdot\cos7^\circ\cdot\cos18^\circ$$
How do I simplify this to a single term?
|
First we rearrange
\begin{align}
X = \sin 47 + \sin 61 - \sin 25 - \sin 11 & = \sin 47 - \sin 25 + \sin 61- \sin 11
\end{align}
In general, we have $\sin a - \sin b = 2\cos(\frac{a+b}{2})\sin(\frac{a-b}{2})$, and therefore, $\sin 47 - \sin 25 = 2\cos 36 \sin 11$, and $\sin 61 - \sin 11 = 2\cos 36\sin 25$. Substituting to above, we obtain
\begin{align}
X = 2\cos 36(\sin 11 + \sin 25).
\end{align}
We now have in general $\sin a + \sin b = 2\cos(\frac{a-b}{2})\sin(\frac{a+b}{2})$, and thus
\begin{align}
X = 4\cos 36\sin 18 \cos 7.
\end{align}
On the other hand, $\cos 36 = \frac{1+\sqrt{5}}{4}$ and $\sin 18 = \frac{\sqrt{5}-1}{4}$, and therefore, in fact
$$X = \cos 7.$$
|
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|
If $x\equiv2\pmod{3}$ prove that $3|4x^2+2x+1$ I've tried many different things to get a factor of $k-2$ but keep failing.
If $x\equiv2\pmod{3}$ prove that $3 \mid 4x^2+2x+1$
|
Another way to prove it:
Since $x \equiv 2 \mod 3$, $\exists k \in \mathbb{Z}$ such that $x=2+3k$. Therefore:
$$4x^2+2x+1=4(2+3k)^2+2(2+3k)+1=21+54 k+36 k^2 = 3(12k^2+18k+7)$$
So clearly $3 \mid (4x^2+2x+1)$.
|
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|
Is this a good way to prove that $3^x+4^x =5^x $ has $x=2$ as the only real solution? Divide both sides of the equation $3^x+4^x=5^x$ by $5^x$.
$$ \Rightarrow \frac { 3^x }{ 5^x } +\frac { 4^x }{ 5^x } =\frac { 5^x }{ 5^x }$$
$$\tag1 \Rightarrow \left( \frac 3 5 \right)^x + \left( \frac { 4 }{ 5 } \right)^x =1$$
$\because \frac { 3 }{ 5 } \leqslant 1 \Rightarrow \sin \theta =\frac { 3 }{ 5 }$ would be valid.
We know, $\sin^2 \theta +\cos ^2 \theta =1$.
$\Leftrightarrow \cos^2 \theta =1-\sin ^2 \theta$.
$$ \Leftrightarrow \cos \theta =\sqrt { 1- \left( \frac { 3 }{ 5 } \right)^2 } =\frac { 4 }{ 5 }$$
$\therefore \sin \theta =\frac { 3 }{ 5 } \ \land \ \cos\theta =\frac { 4 }{ 5 }$.
The equation $(1)$ can be rewritten as
$$ \Rightarrow \left( \cos\theta \right)^x + \left( \sin \theta \right)^x =1.$$
We know that the above equation would hold true for only $ x=2$.
Hence Proved.
Thanks to lab bhattacharjee for reminding me to link the question details to other answers.
Please tell me if my answere is better and more reasonable than the other answers on the following pages:
Prove that $x = 2$ is the unique solution to $3^x + 4^x = 5^x$ where $x \in \mathbb{R}$
Proving that $ 2 $ is the only real solution of $ 3^x+4^x=5^x $
If it is not the best one, please give the link to the best solution.
Thanks a ton for your time!
Thanks to Hagen von Eitzen for reformatting the question details into a better format.
|
Let $f(x)=(\frac{3}{5})^x+(\frac{4}{5})^x$
This function is strictly decreasing , and as such it's graph will only intersect the line $y=1$ in only one point(suppose there are two points, due to the fact that it is strictly decreasing, these points must coincide).
|
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|
Solving limit without using L'Hopital $$\lim_{x\to 0} \frac{\ln(\cos x)}{\ln(\cos 3x)}$$
The answer is $1/9$, but I don't want to use L'Hopital because I'm not supposed to.
Any help would be greatly appreciated and thanks in advance.
|
Since $\ln(x)=\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}(x-1)^n}{n!}}}$ and $\cos x=\displaystyle{\sum_{n=0}^{\infty}{\frac{(-1)^nx^{2n}}{(2n)!}}}=1-\frac{x^2}{2}+\frac{x^4}{24}+\ldots$ we can estimate $\ln(\cos x)$ and $\ln(\cos 3x)$ as follows
\begin{align}
\ln(\cos x)&=-\frac{1}{2}x^2+O(x^4) &\text{and} & &\ln(\cos 3x)&=-\frac{1}{2}(3x)^2+O(x^4)=-\frac{9}{2}x^2+O(x^4) \\
\end{align}
where $O(x^4)$ means that all remaining terms are of order $4$ and greater. So
\begin{align}
\frac{\ln(\cos x)}{\ln(\cos 3x)}&=\frac{-\frac{1}{2}x^2+O(x^4)}{-\frac{9}{2}x^2+O(x^4)} \\
&= \frac{-\frac{1}{2}x^2[1+O(x^2)]}{-\frac{1}{2}x^2[9+O(x^2)]}\\
&= \frac{1+O(x^2)}{9+O(x^2)}\\
\end{align}
Hence $\displaystyle{\lim_{x\rightarrow0}{\frac{\ln(\cos x)}{\ln(\cos 3x)}}=\frac{1}{9}}$.
|
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|
Solve $\cos x+8\sin x-7=0$
Solve $\cos x+8\sin x-7=0$
My attempt:
\begin{align}
&8\sin x=7-\cos x\\
&\implies 8\cdot \left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)=7-\cos x\\
&\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=7-1+2\sin ^2\frac{x}{2}\\
&\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=6+2\sin^2 \frac{x}{2}\\
&\implies 8\sin \frac{x}{2}\cos \frac{x}{2}=3+\sin^2 \frac{x}{2}\\
&\implies 0=\sin^2 \frac{x}{2}-8\sin \frac{x}{2}\cos \frac{x}{2}+3\\
&\implies 0=\sin \frac{x}{2}\left(\sin \frac{x}{2}-8\cos \frac{x}{2}\right)+3
\end{align}
I'm not sure how to proceed from here (if this process is even right at all?) . Any help would be appreciated. Thanks.
|
Use the formula for linear combinations of sine and cosine:
$$A \cos x+B \sin x=C \sin (x+\phi) $$ where $C=\sqrt{A^2+B^2},\phi=\arg (A+Bi)$.
This makes transforms your equation to:
$$\sqrt{65} \sin(x+\tan^{-1} 8)=7 $$
which can be easily solved.
|
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|
Integral $\int_0^\infty \log(1+x^2)\frac{\cosh{\frac{\pi x}{2}}}{\sinh^2{\frac{\pi x}{2}}}\mathrm dx=2-\frac{4}{\pi}$ Hi I am trying to show$$
I:=\int_0^\infty \log(1+x^2)\frac{\cosh{\frac{\pi x}{2}}}{\sinh^2{\frac{\pi x}{2}}}\mathrm dx=2-\frac{4}{\pi}.
$$
Thank you.
What a desirable thing to want to prove! It is a work of art this one. I wish to prove this in as many ways as we can find.
Note I tried writing
$$
I=\int_0^\infty \log(1+x^2)\coth \frac{\pi x}{2} \sinh^{-2} \frac{\pi x}{2}\mathrm dx
$$
but this didn't help me much. We can also try introducing a parameter as follows
$$
I(\alpha)=\int_0^\infty \log(1+x^2)\frac{\cosh{\frac{\alpha \pi x}{2}}}{\sinh^2{\frac{\pi x}{2}}}\mathrm dx,
$$
But this is where I got stuck. How can we calculate I? Thanks.
|
As Lucian stated in the comments, integrating by parts shows that the integral is equivalent to showing that $$ \int_{0}^{\infty} \frac{x}{1+x^{2}} \frac{1}{\sinh \frac{\pi x}{2}} \ dx = \frac{\pi}{2}-1 .$$
Let $ \displaystyle f(z) = \frac{z}{1+z^{2}} \frac{1}{\sinh \frac{\pi z}{2}} $
and integrate around a rectangle with vertices at $\pm N$ and $\pm N+ i (2N+1)$ where $N$ is some positive integer.
As $N$ goes to infinity through the integers, the integral vanishes on the left and right sides of the rectangle and along the top of the rectangle.
In particular, the absolute value of the integral along the top of the rectangle is bounded by $$\frac{3N+1}{(2N+1)^{2}-1}\int_{-\infty}^{\infty} \frac{1}{\cosh \frac{\pi x}{2}} \ dx = \frac{6N+2}{(2N+1)^{2}-1} \to 0 \ \text{as} \ N \to \infty .$$
Then
$$\int_{-\infty}^{\infty} \frac{x}{1+x^{2}} \frac{1}{\sinh \frac{\pi x}{2}} \ dx = 2 \pi i \left(\text{Res}[f(z),i]+ \sum_{k=1}^{\infty} \text{Res}[f(z),2ki] \right)$$
where
$$ \text{Res}[f(z),i] = \lim_{z \to i} \frac{z}{z+i} \frac{1}{\sinh \frac{\pi x}{2}} =\frac{1}{2i}$$
and
$$ \text{Res}[f(z),2ki] = \lim_{z \to 2ki} \frac{z}{2z \sinh \frac{\pi z}{2}+(1+z^{2}) \frac{\pi}{2} \cosh \frac{\pi z}{2}} = \frac{4i}{\pi} \frac{(-1)^{k} k}{1-4k^{2}} . $$
And notice that
$$ \begin{align} \sum_{k=1}^{\infty} \frac{(-1)^{k} k}{1-4k^{2}} &= -\frac{1}{4} \sum_{k=1}^{\infty} \left( \frac{(-1)^{k}}{2k+1} + \frac{(-1)^{k}}{2k-1} \right) \\ &= -\frac{1}{4} \left(\arctan(1)-1 - \arctan(1) \right) \\ &= \frac{1}{4} . \end{align}$$
Therefore,
$$ \int_{-\infty}^{\infty} \frac{x}{1+x^{2}} \frac{1}{\sinh \frac{\pi x}{2}} \ dx = 2 \pi i \left(\frac{1}{2i} + \frac{i}{\pi} \right) = \pi - 2$$
which implies
$$ \int_{0}^{\infty} \frac{x}{1+x^{2}} \frac{1}{\sinh \frac{\pi x}{2}} \ dx = \frac{\pi}{2} -1 .$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Algebra Iranian Olympiad Problem If:
$x^2+y^2+z^2=2(xy+xz+zy)$
and $x,y,z \in R^+$
Prove:
$\frac{x+y+z}{3} \ge \sqrt[3]{2xyz}$
I tried my best to solve this thing but no use.
Hope you guys can help me.Thanks in advance.
|
You have:
\begin{align}
(x + y + z)^2
&= x^2 + y^2 + z^2 + 2 (x y + x z + y z) \\
&= 4 (x y + x z + y z) \\
&= 4 \cdot 3 \cdot \frac{x y + x z + y z}{3} \\
&\ge 12 \sqrt[3]{x^2 y^2 z^2} \\
x + y + z
&\ge 2 \sqrt[3]{3 x y z} \\
\frac{x + y + z}{3}
&\ge \sqrt[3]{\frac{8}{9} x y z}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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|
Simplifying $\frac1{1+x}+\frac2{1+x^2}+\frac4{1+x^4}+\frac8{1+x^8}+\frac{16}{x^{16}-1}$ We need to simplify $$\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{x^{16}-1}$$
The last denominator can be factored and we can get all the other denominators as factors of $x^{16}-1$. I tried handling the expressions in pairs,starting from the right.I also tried to take a common factor of two out of the numerators to help simplify,but that has yielded nothing.I then tried multiplying all the fractions to get $x^{16}-1$ in the denominator but that worsens things(I think so anyway).
So after doing the above things(and much more),I feel like I am running out of ideas.A really small hint will be appreciated.
|
$$\frac{16}{x^{16}-1}=\frac{8}{x^8-1}+\frac{-8}{x^8+1}$$
So the $4^{th}$ term of the original sum and the $2^{nd}$ part of the decomposition above are canceled. You are left with:
$$
\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{x^{8}-1}
$$
Continue similarly. In the end you will get $\frac{1}{x-1}$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
The 'sine and cosine theorem' - formulas for the sum and difference I've read somewhere that the sine and cosine functions can be fully described by this theorem:
*
*$\sin(0) = 0, \cos(0) = 1$
*$\sin(a-b) = \sin(a)\cos(b) - \sin(b)\cos(a)$
*$\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$
*There is na $r>0$ such that:
$$0<\sin(x)<x<\tan(x), x \in(0, r), \tan(x) = \frac{\sin(x)}{\cos(x)} $$
With this theorem, we can prove things like:
*
*$\sin^2(x) + \cos^2(x) = 1$ by doing $\cos(a-a) = \cos(a)\cos(a) + \sin(a)\sin(a)$
*$\sin(-x) = -\sin(x)$ by doing $\sin(0-x) = \sin(0)\cos(x) - \sin(x)\cos(0)$
*$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$ by doing $\cos(a-(-b)) = \cos(a)\cos(-b) + \sin(a)\sin(-b)$
*$\sin(a+b) = \sin(a)\cos(b)+\sin(b)\cos(a)$ by doing $\sin(a-(-b)) = \sin(a)\cos(-b) - \sin(-b)\cos(a)$
And other trigonometric identities that also follows from what I've already done.
The problem is that there are many definitions for the sine and cosine function. Let's begin by the classical definition:
Classical Definition
The sine function is defined as the ratio between the opposite side of the angle, and the hipotenuse of this right triangle.
The cosine function is defined as the ratio between the adjacent side of the angle, and the hipotenuse of this right triangle.
The tangent function is defined as the ratio between the sine function and the cosine function (with $\cos (x) \neq 0)$
The other trigonometric identities can be proven geometrically for an angle less than or equal $\frac{\pi}{2}rad$ because it's a right triangle. So, we can't prove $\sin(a-b)$ geometrically and then prove $\sin(a+b)$ analytically like I did, because we assumed a negative $b$, something that's not defined geometrically in the right triangle.
The identity $\sin^2(x) + \cos^2(x) = 1$ can be proven with a simple pythagorean theorem in a triangle with hypotenuse 1.
The formulas for the sum and difference of sines and cosines can be proven geometrically like in this images I found in this answer:
Unit circle definition
Imagine a circle centered at the origin of the cartesian plane, then:
The sine function for an $x \in R, x>0$ can be defined as the $y$ position of point of the circle where the angle stops if we travel anti-clockwise inside the circle's line.
The cosine for an $x \in R, x>0$ function can be defined as the $x$ position of point of the circle where the angle stops if we travel anti-clockwise inside the circle's line.
We can make the same definition for negative angles, so for an $x \in R, x<0$ the same holds, but we're know travelling clockwise.
The tangent function is defined as the ratio between the sine function and the cosine function (with $\cos (x) \neq 0)$
Then, we can define these functions for all real numbers, since when we travel $2\pi$ we get back to the initial point. So we defined sine and cosine as periodic functions.
Open question: How do I prove, with the unit circle periodic definition and without being circular, the $\sin(a-b)$ and $\cos(a-b)$ formulas? (same for $\sin(a+b)$ and $\cos(a+b)$.
Taylor Series definition
$$\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$$
$$\cos x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n}$$
$$\tan x = \sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n (1-4^n)}{(2n)!} x^{2n-1}$$
Here, in the same question, there's an analytic proof of the trigonometric identites, for these sums.
What's the best definition for calculus?
Well, in calculus we use trigonometric functions a LOT: in integral substitutions, in series, taylor series (like the ones I showed now), derivatives, convergence tests (like the Euler onde in the basel problem) and other things...
All the definitions I see, are kinda circular or not rigorous enough to make me feel good taking some derivatives or integral substitution, because I always care about the domain of these things. So I want to define it very nicely and be able to use all the trigonometric identities.
I've seen many geometrical proofs of $\sin(a-b)$, $\sin(a+b)$, $\cos(a-b)$, $\cos(a+b)$ using a right triangle and then suddenly the person starts using this formula for all real numbers. I need a complete definition of the trigonometric functions that works periodically and for all reals. The taylor series definition seems good but they're generated using trigonometric identities that are not yet proven (assuming this definition).
ps: I know that I used some primitive words in some definitions, like 'travel' so I let them in emphasis, but I hope you guys understand. And sorry by the long post, but I needed to do it, because I've never seen a complete definition in any book. Thanks.
|
I created this shortcut proof for the cos addition formula, it uses the cosine and sine formula
The proof I know usually involves complex triangle shapes, so I'll be working with a basic triangle here
$a+b+c = \pi$, and $c = \pi -(a+b)$
$$ C^2 = A^2+B^2-2AB\cos{c}$$
$$ C^2 = A^2+B^2-2AB\cos{(\pi-(a+b))}$$
$$\frac{C^2}{A^2} = \frac{A^2}{A^2}+\frac{B^2}{A^2}-2\frac{AB}{A^2}\cos{(\pi-(a+b))}$$
$${\frac{C}{A}}^2 = 1+{\frac{B}{A}}^2-2\frac{B}{A}\cos{(\pi-(a+b))}$$
$$\frac{\sin{a}}{A} = \frac{\sin{b}}{B} = \frac{\sin{c}}{C} $$
$$\frac{\sin^2{c}}{\sin^2{a}} = 1+\frac{\sin^2{b}}{\sin^2{a}}-2\frac{\sin{b}}{\sin{a}}\cos{(\pi-(a+b))}$$
$$\sin^2{c} = \sin^2{a} \cdot ( 1+\frac{\sin^2{b}}{\sin^2{a}}-2\frac{\sin{b}}{\sin{a}}\cos{(\pi-(a+b))} )$$
$$1-\sin^2{c} = 1-\sin^2{a} \cdot ( 1+\frac{\sin^2{b}}{\sin^2{a}}-2\frac{\sin{b}}{\sin{a}}\cos{(\pi-(a+b))} )$$
$$\cos^2{c} = 1-\sin^2{a} \cdot ( 1+\frac{\sin^2{b}}{\sin^2{a}}-2\frac{\sin{b}}{\sin{a}}\cos{(\pi-(a+b))} )$$
$$\cos^2{c} = 1-\sin^2{a}-\sin^2{b}+2\sin{a}\sin{b}\cos{c}$$
$$\cos^2{c} - 2\sin{a}\sin{b}\cos{c}+\sin^2{a}+\sin^2{b}-1 = 0$$
$$\cos{c} = \frac{ 2\sin{a}\sin{b} \pm \sqrt{ (2\sin{a}\sin{b})^2-4(\sin^2{a}+\sin^2{b}-1) } }{2}$$
$$\cos{c} = \frac{ 2\sin{a}\sin{b} \pm \sqrt{ 4\sin^2{a}\sin^2{b}-4\sin^2{a}-4\sin^2{b}+4 } }{2}$$
$$\cos{c} = \frac{ 2\sin{a}\sin{b} \pm 2\sqrt{ \sin^2{a}\sin^2{b}-\sin^2{a}-\sin^2{b}+1 } }{2}$$
$$\cos{c} = \frac{ 2\sin{a}\sin{b} \pm 2\sqrt{ (1-\sin^2{a})(1-\sin^2{b}) } }{2}$$
$$\cos{c} = \frac{ 2\sin{a}\sin{b} \pm 2\sqrt{ \cos^2{a}\cos^2{b} } }{2}$$
$$\cos{c} = \sin{a}\sin{b} \pm \cos{a}\cos{b}$$
$$\cos{(\pi-(a+b))} = \sin{a}\sin{b} - \cos{a}\cos{b}$$
$$-\cos{(a+b)} = \sin{a}\sin{b} - \cos{a}\cos{b}$$
$$\cos{(a+b)} = \cos{a}\cos{b}- \sin{a}\sin{b}$$
|
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|
To find relatively prime ordered pairs of positive integers $(a,b)$ such that $ \dfrac ab +\dfrac {14b}{9a}$ is an integer How many ordered pairs $(a,b)$ of positive integers are there such that g.c.d.$(a,b)=1$ , and
$ \dfrac ab +\dfrac {14b}{9a}$ is an integer ?
|
$$\frac ab+\frac{14b}{9a}=\frac{9a^2+14b^2}{9ab}$$
If this expression is an integer, since $a$ and $b$ are coprime, $a$ divides $14$ and $b$ divides $9$.
If $b=1$, we get
$$a+\frac{14}{9a}$$
which is not an integer.
If $b=9$,
$$\frac{a}9+\frac{14}a=\frac{a^2+126}{9a}$$
which is neither an integer, since $a$ can't be a multiple of $9$.
If $b=3$,
$$\frac a3+\frac{14}{3a}=\frac{a^2+14}{3a}$$
which is an integer for every possible value of $a$ (namely, $1$, $2$, $7$ and $14$).
So there is $4$ ordered pairs of positive integers.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Sequence $x_{n+1}=\sqrt{x_n+a(a+1)}$ $a\gt0$, and $x_0=0$, $x_{n+1}=\sqrt{x_n+a(a+1)}, n=0,1,2,\dotsc$.
compute
$$\lim_{n\to\infty}\big(a+1\big)^{2n}\big(a+1-x_n\big)$$
The problem is difficult, I have no idea.
Thank you!
|
From the recursion given, we get
$$
x_{n+1}^2=x_n+a(a+1)\tag{1}
$$
Subtracting $(1)$ from $(a+1)^2$ gives
$$
(a+1)^2-x_{n+1}^2=(a+1)-x_n\tag{2}
$$
which is equivalent to
$$
\frac{(a+1)-x_{n+1}}{(a+1)-x_n}
=\frac1{(a+1)+x_{n+1}}\tag{3}
$$
which inductively implies
$$
(a+1)-x_n\le\frac{(a+1)-x_0}{(a+1)^n}\tag{4}
$$
Multiplying $(3)$ by $2(a+1)$ gives
$$
\begin{align}
2(a+1)\frac{(a+1)-x_{n+1}}{(a+1)-x_n}
&=\frac{2(a+1)}{(a+1)+x_{n+1}}\\
&=1+\frac{(a+1)-x_{n+1}}{(a+1)+x_{n+1}}\tag{5}
\end{align}
$$
which inductively implies
$$
2^n(a+1)^n((a+1)-x_n)=((a+1)-x_0)\prod_{k=1}^n\left(1+\frac{(a+1)-x_k}{(a+1)+x_k}\right)\tag{6}
$$
Inequality $(4)$ guarantees that the product in $(6)$ converges. Thus, there is a $0\lt C_a\lt\infty$ so that
$$
\lim_{n\to\infty}2^n(a+1)^n((a+1)-x_n)=C_a\tag{7}
$$
Therefore, if $a\gt1$, then the given limit is $\infty$, and if $a\lt1$, then the given limit is $0$. If $a=1$, then the given limit is $C_1$.
Computation of $C_1$
When $a=1$, recursion $(1)$ becomes
$$
x_{n+1}^2=x_n+2\tag{8}
$$
which is satified by
$$
x_n=2\cos(2^{-n}t_0)\tag{9}
$$
with $t_0=\frac\pi2$. Therefore,
$$
\begin{align}
C_1
&=\lim_{n\to\infty}4^n(2-2\cos(2^{-n}t_0))\\
&=t_0^2\\
&=\frac{\pi^2}{4}\tag{10}
\end{align}
$$
As has been mentioned, this is related to this answer.
|
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|
Evaluating $\int (\tan^3x+\tan^4x) dx $ using substitution solving $$\int (\tan^3x+\tan^4x) dx $$ using substitution $$t = \tan x$$
My approach has led me to $ \int (1+t)t\sin^2xdt$ which has an $x$ too much and isn't easily solvable for me. If I remove the $x$ I get $\sin^2(\arctan(t))$ and that's not too nice to work with...
|
unfortunately $\sec$ is not used in my country
If you feel unconfortable with $\sec $ you may proceed from your last integral
\begin{equation*}
I=\int (1+t)t\sin ^{2}\left( \arctan t\right) \,dt
\end{equation*}
by using the following trigonometric identity
\begin{equation*}
\sin ^{2}x=\frac{\tan ^{2}x}{1+\tan ^{2}x},\qquad x=\arctan t,
\end{equation*}
which you can derive from the fundamental identity $\sin^2 x+\cos^2 x=1$ to obtain
\begin{equation*}
\sin ^{2}\left( \arctan t\right) =\frac{t^{2}}{1+t^{2}}.
\end{equation*}
Consequently
\begin{equation*}
I=\int (1+t)t\frac{t^{2}}{1+t^{2}}\,dt=\int \frac{t^{3}+t^{4}}{1+t^{2}}\,dt,
\end{equation*}
which is integrable by partial fractions. By long division we compute
\begin{equation*}
\frac{t^{4}+t^{3}}{1+t^{2}}=t^{2}+t-1+\frac{-t+1}{1+t^{2}}.
\end{equation*}
So
\begin{equation*}
I=\int \left( t^{2}+t-1\right) dt+\int \frac{-t+1}{1+t^{2}}dt.
\end{equation*}
Since
\begin{eqnarray*}
\int \frac{-t+1}{1+t^{2}}dt &=&-\frac{1}{2}\int \frac{2t}{1+t^{2}}dt+\int
\frac{1}{1+t^{2}}dt \\
&=&-\frac{1}{2}\ln \left( 1+t^{2}\right) +\arctan t+C,
\end{eqnarray*}
we thus have
\begin{eqnarray*}
I &=&\frac{1}{3}t^{3}+\frac{1}{2}t^{2}-t-\frac{1}{2}\ln \left(
1+t^{2}\right) +\arctan t+C \\
&=&\frac{1}{3}\tan ^{3}x+\frac{1}{2}\tan ^{2}x-\tan x-\frac{1}{2}\ln \left(
1+\tan ^{2}x\right) +x+C.
\end{eqnarray*}
|
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|
6 keys and a door( probabilities) We have a key ring with 6 keys, but only one key opens the door. We test the keys to the door with no preference until we find the right key. Now we have two different cases:
1) At the first case when a key is not the right one we discard it and try the next one until we find the right one.
2) At the second case we have amnesia so when we try a key and its a wrong one we put it back in the key ring and we start again from the beginning to find the right key.
For both cases find the probabilities to:
How many tries we will need in average, and the probability to find the right key exactly at the 4th try.
|
For 1), let X be the number of tries. Then E(X) is the expected number of tries. Recall that $E(X)=\sum_{x=1}^{x=6}xP(X=x)$.
$E(X)=1(\frac{1}{6})+2(\frac{5}{6})(\frac{1}{5})+3(\frac{5}{6})(\frac{4}{5})(\frac{1}{4})+4(\frac{5}{6})(\frac{4}{5})(\frac{3}{4})(\frac{1}{3})+5(\frac{5}{6})(\frac{4}{5})(\frac{3}{4})(\frac{2}{3})(\frac{1}{2})+6(\frac{5}{6})(\frac{4}{5})(\frac{3}{4})(\frac{2}{3})(\frac{1}{2}(\frac{1}{1})=\frac{7}{2}$
For 2) if you pick the correct key, it's a success and if you don't it's a failure. Let p be the probability of a success. Then $p=\frac{1}{6}$. So, you should notice that $X is Geometric(p)=Geometric(\frac{1}{6})$.
Now, you know that for a Geometric Distribution, $E(X) = \frac{1}{p}=6$
Finally, $P(X=4) =(1-p)^3p$
|
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|
How find the maximum value of $|bc|$ Question:
Given complex numbers $a,b,c$, we have that $|az^2 + bz +c| \leq 1$ holds true for any complex number $z, |z| \leq 1$. Find the maximum value of $|bc|$
It is said this is answer is $$|bc|\le \dfrac{3\sqrt{3}}{16}$$
My idea: let $z=1$,then
$$|a+b+c|\le 1$$
let $z=-1$,then $$|a-b+c|\le 1$$
let $z=0$, then
$$|c|\le 1$$
let $|\theta|=1$,then we have
$$|a(z/\theta)^2+b(z/\theta)+c|\le 1\Longrightarrow |az^2+b\theta z+c\theta^2|\le 1$$
and only this can't solve this problem,Thank you
|
You can assume that $a,b,c$ are reals and $b,c\ge0$. For $c$ this is obvious. For $b$ you can get this substituting $\theta z$ for $z$, as you wrote in the question. Finally, you can replace the polynomial by $\frac{(az^2+bz+c)+(\bar{a}z^2+bz+c)}2$.
By the maximum principle it suffices to verify the condition on the circle.
At the point $e^{it}$ we have
$$
1 \ge |ae^{2it}+be^{it}+c|^2 = |a+be^{-it}+ce^{-2it}|^2 \\
= \big(a+b\cos t+c\cos 2t\big)^2 + \big(b\sin t+c\sin 2t\big)^2 \\
= \big(a+b\cos t+c\cos 2t\big)^2 + \big(b\sin t-c\sin 2t\big)^2
+ 4bc \sin t \sin 2t \\
\ge 4bc \sin t \sin 2t.
$$
The maximum of $\sin t \sin 2t$ is $\frac{16}{3\sqrt3}$ (at $t=\arcsin\frac{\sqrt2}{\sqrt3}$ where $\cos t=\frac1{\sqrt3}$, $\sin t=\frac{\sqrt2}{\sqrt3}$, $\cos 2t=-\frac13$ and $\sin 2t=\frac{2\sqrt2}{3}$). So, at this point we get the bound $bc\le \frac{3\sqrt3}{16}$.
Of course this is only a bound...
To preserve equality in the extreme case, we have to set $b=\frac{\sqrt3}{2\sqrt2}$,
$c=\frac{3}{4\sqrt2}$, $a=-\frac1{4\sqrt2}$.
Then
$$
|ae^{2it}+be^{it}+c|^2 = |ae^{it}+b+ce^{-it}|^2 \\
= \big((a+c)\cos t+b\big)^2 + \big((a-c)\sin t\big)^2 \\
= (a-c)^2+b^2 +2(a+c)b\cos t+4ac\cos^2t \\
=
\frac78+\frac{\sqrt3}4\cos t -\frac38\cos^2 t
= 1 - \frac38\left(\cos t-\frac1{\sqrt3}\right)^2 \le 1.
$$
Hence, $bc=\frac{3\sqrt3}{16}$ is possible.
|
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|
Fourier Series Fourier Transform Method I understand $f$ is even about $\pi$ but i'm struggling conceptually with the part I have underlined.
|
For completeness, I will work through the entire problem so it can benefit others who view the post. To find, the Fourier coefficients we use
\begin{align}
a_0 &= \frac{1}{2\pi}\int_0^{2\pi}\frac{(\pi - \theta)^2}{4}d\theta\\
&= \frac{\pi^2}{12}\\
a_n &= \frac{1}{\pi}\int_0^{2\pi}\frac{(\pi - \theta)^2}{4}\cos(n\theta)d\theta\tag{1}\\
&= \frac{1}{n^2}
\end{align}
where the solution to $(1)$ comes about from integration by parts twice. Thus,
$$
f(\theta) = \frac{\pi^2}{12} + \sum_{n=1}^{\infty}\frac{\cos(n\theta)}{n^2}
$$
Now, we can use the exponential form of the Fourier series, $\sum_{n=-\infty}^{\infty}c_ne^{in\theta}$.
\begin{align}
c_n &= \frac{1}{2\pi}\int_0^{2\pi}\frac{(\pi - \theta)^2}{4}e^{-in\theta}d\theta\\
&= \cdots\\
&= \frac{1}{4\pi ni}\Bigl[\frac{-2\pi}{in} + \frac{e^{-in\theta}}{n^2}\Bigr|_0^{2\pi}\\
&= \frac{1}{2n^2}
\end{align}
That was for $c_n$ when $n\neq 0$ since we would have $\frac{1}{2\cdot 0}$ if zero was included. When $n = 0$, $e^{-in\theta} = 1$ so the integral is
$$
c_0 = \frac{1}{2\pi}\int_0^{2\pi}\frac{(\pi - \theta)^2}{4}d\theta = \frac{\pi^2}{12},
$$
Now, we have that
$$
f(\theta) = \frac{(\pi - \theta)^2}{4} = \underbrace{\frac{\pi^2}{12}}_{c_0\text{ term}} + \underbrace{\sum_{n\neq 0}\frac{e^{in\theta}}{2n^2}}_{\sum_{n =1}^{\infty}\frac{e^{in\theta}}{2n^2}+\sum_{n =1}^{\infty}\frac{e^{-in\theta}}{2n^2}} = \frac{\pi^2}{12} + \sum_{n=1}^{\infty}\frac{\cos(n\theta)}{n^2} \tag{2}
$$
So we now have $\sum_{n =1}^{\infty}\frac{e^{in\theta}}{2n^2}+\sum_{n =1}^{\infty}\frac{e^{-in\theta}}{2n^2}$. Since $\cos$ is even, $\cos(x) = \cos(-x)$ so $e^{in\theta} + e^{-in\theta}=2\cos(n\theta)$ which leads to the desired cosine series.
In the exponential Fourier series, the $c_0$ term is accounted for by $\frac{\pi^2}{12}$ so the exponential series is for $n\in\mathbb{Z}\setminus\{0\}$. If we let $\theta = 0$ in $(2)$, we have
$$
\frac{\pi^2}{4} = \frac{\pi^2}{12} + \underbrace{\sum_{n\neq 0}\frac{1}{2n^2}}_{2\sum_{n=1}^{\infty}\frac{1}{2n^2}} = \frac{\pi^2}{12} + \sum_{n=1}^{\infty}\frac{1}{n^2}
$$
Taking the RHS and the last equality, we have
\begin{align}
\frac{\pi^2}{4} &= \frac{\pi^2}{12} + \sum_{n=1}^{\infty}\frac{1}{n^2}\\
\sum_{n=1}^{\infty}\frac{1}{n^2} &= \frac{\pi^2}{6}
\end{align}
|
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|
How prove $\frac{a}{11a+9b+c}+\frac{b}{11b+9c+a}+\frac{c}{11c+9a+b}\le\frac{1}{7}$ Qustion:
$a,b,c\ge 0$,show that
$$\dfrac{a}{11a+9b+c}+\dfrac{b}{11b+9c+a}+\dfrac{c}{11c+9a+b}\le\dfrac{1}{7}\tag{1}$$
I found this method can't work,
$$x=11a+9b+c,y=11b+9c+a,z=11c+9a+b$$
$$\Longrightarrow a=\dfrac{8x-7y+5z}{126},b=\dfrac{5x+8y-7z}{126},c=\dfrac{5y-7x+8z}{126}$$
so
$$LHS=\dfrac{1}{126}\left(\dfrac{8x-7y+5z}{x}+\dfrac{5x+8y-7z}{y}+\dfrac{5y-7x+8z}{z}\right)$$
so
$$LHS=\dfrac{1}{126}\left(24-7\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)+5\left(\dfrac{z}{x}+\dfrac{x}{y}+\dfrac{y}{z}\right)\right)$$
then I can't use AM-GM inequality to Continue,
By the way
I use AM-GM inequality can solve this famous
Crux problem (2009)
let $x,y,z\ge 0$,and $a,b,c>0$, if $b^2\le ca ,c^2\le ab$,then we can use this methods to solve this Crux problem
$$\dfrac{x}{ax+by+cz}+\dfrac{y}{ay+bz+cz}+\dfrac{z}{az+bx+cy}\le\dfrac{3}{a+b+c}$$
so how can we prove this (1) inequality? and I Think this inequality is very sharp,maybe there are other methods.
Thank you
|
Let $a=\min\{a,b,c\}$,$b=a+u$ and $c=a+v$.
Hence, we need to prove that
$$49(u^2-uv+v^2)a+(9u+v)(u-3v)^2\geq0$$
Done!
|
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|
Points on $(x^2 + y^2)^2 = 2x^2 - 2y^2$ with slope of $1$ Let the curve in the plane defined by the equation: $(x^2 + y^2)^2 = 2x^2 - 2y^2$
How can i graph the curve in the plane and determine the points of the curve where $\frac{dy}{dx} = 1$.
My work:
First i found the roots of this equation with a change of variable $z = y^2$ and get:
and then i tried to graph the point $ x - y$ and $x + y $ but i stuck i can't graph this and find the point where the derivative is 1. Some help please.
|
I would try to solve this as an implicit derivative:
$$\frac{d}{dx}(x^2 + y^2)^2 = \frac{d}{dx}(2x^2 - 2y^2)$$
$$2(x^2+y^2)(2x+2y*y') =4x-4y*y'$$
$$2 x^3+2x^2 y* y'+2 y^3 y'+2 x y^2 = 4x-4y*y'$$
$$2x^2y*y' + 2y^3y' + 4y*y' = 4x-2xy^2-2x^2$$
$$y'(2x^2y + 2y^3+4y) =4x-2xy^2-2x^2$$
$$y' = \frac{4x-2xy^2-2x^2}{2x^2y + 2y^3+4y} = 1$$
$$\implies 4x-2xy^2-2x^2 = 2x^2y + 2y^3+4y$$
|
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|
When a quadratic involving three primes is a perfect square How do we find all primes $p,q,r$ such that $p^2+q^2+rpq$ is a perfect square ?
with $r=7$ and $p=q$ we have the expression a perfect square
|
Generally before searching primes need to understand what formula solutions of this equation.
Generally speaking, this equation has a lot of formulas for the solution. Because it is symmetrical.
Write the formula can someone come in handy. the equation:
$Y^2+aXY+X^2=Z^2$
Has a solution:
$X=as^2-2ps$
$Y=p^2-s^2$
$Z=p^2-aps+s^2$
more:
$X=(4a+3a^2)s^2-2(2+a)ps-p^2$
$Y=(a^3-8a-8)s^2+2(a^2-2)ps+ap^2$
$Z=(2a^3+a^2-8a-8)s^2+2(a^2-2)ps-p^2$
more:
$X=(a+4)p^2-2ps$
$Y=3p^2-4ps+s^2$
$Z=(2a+5)p^2-(a+4)ps+s^2$
more:
$X=8s^2-4ps$
$Y=p^2-(4-2a)ps+a(a-4)s^2$
$Z=-p^2+4ps+(a^2-8)s^2$
For the particular case: $Y^2+XY+X^2=Z^2$ You can draw more formulas.
$X=3s^2+2ps$
$Y=p^2+2ps$
$Z=p^2+3ps+3s^2$
more:
$X=3s^2+2ps-p^2$
$Y=p^2+2ps-3s^2$
$Z=p^2+3s^2$
In the equation: $X^2+aXY+bY^2=Z^2$ there is always a solution and one of them is quite simple.
$X=s^2-bp^2$
$Y=ap^2+2ps$
$Z=bp^2+aps+s^2$
$p,s$ - integers asked us.
|
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|
If $4xy+3=c^2+3d^2$, is $xy$ necessarily a square? I have a polynomial which, simplified, ends up in the form
$$4xy+3 = c^2+3d^2.$$
Evidently $4xy+3$ is of the form $a^2+3b^2$, in light of the equality. But does
$$
c^2 + 3d^2 = 4xy + 3 = xy(2)^2 + 3(1)^2
$$
necessarily force $xy$ to be a square? I just can't see how to prove it.
Thanks.
|
For the equation: $4xy+3=c^2+3d^2$
Can be written like such a simple solution:
$x=2a^2+6t^2-6at-2a+3t-1$
$y=2a^2+6t^2-6at-a+3t-1$
$c=2a^2+6t^2-6at-2$
$d=2a^2+6t^2-6at-2a+4t-1$
$a,t$ - what some integers any sign.
|
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|
Solving a differential equation using the laplace transform involving convolution The problem is the following
The thing that puzzles me here is the integral on the right hand side, so:
How to take the laplace transform on the right hand side?
Any help to get me going would be greatly appreciated.
|
The convolution theorem is
\begin{align}
F(s)*G(s) = \int_{0}^{t} g(t-\tau) f(\tau) d\tau
\end{align}
such that
\begin{align}
L\{ f*g\} = L\{ \int_{0}^{t} g(t-\tau) f(\tau) d\tau \} = F(s)G(s).
\end{align}
The differential equation then becomes
\begin{align}
L\{ y^{''} + 3 y^{'} + 2y \} = L\{ H(t-\pi) \sin(2t) \} + \delta(s) f(s) - \delta(s-2) f(s).
\end{align}
Completing the transform the equation becomes
\begin{align}
(s+1)(s+2) y(s) - (s+1) y(0) - y^{'}(0) = \frac{2 e^{-\pi s}}{s^{2} + 4} + \delta(s) f(s) - \delta(s-2) f(s).
\end{align}
Since $y^{'}(0) = 0$ and $y(0) = 1$, and $f(t+2) = f(t)$ then
\begin{align}
y(s) &= \frac{1}{s+2} + \frac{1}{(s+1)(s+2)} \left[ \frac{2 e^{-\pi s}}{s^{2} + 4} + \delta(s) f(s) - \delta(s-2) f(s) \right] \\
&= \frac{1}{s+2} + \frac{1}{(s+1)(s+2)} \left[ \frac{2 e^{-\pi s}}{s^{2} + 4} + f(0) - f(2) \right] \\
&= \frac{1}{s+2} + \frac{1}{(s+1)(s+2)} \left[ \frac{2 e^{-\pi s}}{s^{2} + 4} \right] \\
\end{align}
Inversion leads to
\begin{align}
y(t) &= e^{-2t} + L^{-1}\{ \frac{1}{(s+1)(s+2)} \cdot \frac{2 e^{-\pi s}}{s^{2} + 4} \} \\
&= e^{-2t} + \frac{1}{20} H(t-\pi) \left[ 8 e^{\pi - t} - 5 e^{2(\pi -t)} - \sin(2t) - 3 \cos(2t) \right].
\end{align}
note: the convolution theorem is in most standard tables such as this one
http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
|
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|
Values of $m$ for which $y^2 + 2xy + 2x -my -3$ can be factorised
For what values of $m$, will the expression $y^2 + 2xy + 2x -my -3$ be capable of resolution into two rational linear factor?
This is how I did it:
$$y^2 + 2xy + 2x -my -3 = y^2+(2x-m)y+2x-3$$
This can always be factorized if $b^2-4ac>0$, so if $4ac$ will be negative ($\forall x\in(-\infty,3/2)$), then $b^2-4ac > 0, . $ We need to only worry about $x>\frac32$.
I tried using the quadratic formula next, but couldn't get any further .
|
Hint: The factorization can be taken to have shape $(y+ax+b)(y+cx+d)$. Since there is no $x^2$ term, we have $a=0$ or $c=0$. We can take $c=0$. The term in $xy$ is $2xy$, so $a=2$. Continue. We are pretty close to the end.
Added: Your discriminant approach will also work. The discriminant is
$(2x-m)^2-4(2x-3)$. This expands to $4x^2-4x(m+2)+m^2+12$.
This discriminant must be the square of something linear in $x$. So the discriminant of the polynomial $4x^2-4x(m+2)+m^2+12$ must be $0$. We get
$16(m+2)^2-16(m^2+12)=0$. Solve for $m$.
|
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|
Reducing Function to be Expressed in Terms of 1 Trigonometric Function I am looking for a way to express $\tan(x) + \sec(x)$ as a function expressed in terms of a single trigonometric function. So far I have it down to:
$$
\frac{\sqrt{1 - \cos^2 x} + 1}{\cos(x)}
$$
Is there a more clean way to define this. Basically I have:
$$
y = \frac{\sqrt{1 - \cos^2 x} + 1}{\cos(x)}
$$
and I need to express $x(y)$.
|
$\tan x + \sec x = \dfrac{\sin x + 1}{\cos x} = \dfrac{(\sin(\frac{x}{2}) + \cos(\frac{x}{2}))^2}{(\cos(\frac{x}{2}) - \sin(\frac{x}{2}))(\cos(\frac{x}{2}) + \sin(\frac{x}{2}))} = \dfrac{\sin(\frac{x}{2}) + \cos(\frac{x}{2})}{\cos(\frac{x}{2}) - \sin(\frac{x}{2})} = \dfrac{\tan(\frac{x}{2}) + 1}{1 - \tan(\frac{x}{2})}$.
Let $t = \tan(\frac{x}{2})$, then:
$f(x) = f(t) = \dfrac{t + 1}{1 - t}$
|
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|
Formula for the number of integer solutions of an equation (using generating functions) Let $a_n$ be the number of integer solutions of $$i+3j+3k=n$$ where $i \geq 0, j \geq 2, k \geq 3$.
I want to use the generating function of $(a_n)_{n \in \mathbb N}$ to get a formula for $a_n$.
We just introduced generating functions, so I'm fairly new to this stuff and hope you can help me solve this problem.
I began by interpreting the problem as a power series. Without the restrictions for $i, j, k$ I get
$$(1+x+x^2+x^3+\dots)(1+x^3+x^6+x^9+\dots)^2.$$
Accounting for $i \geq 0, j \geq 2, k \geq 3$ should get me something like
$$(1+x+x^2+x^3+\dots)(x^6+x^9+x^{12}+\dots)(x^9++x^{12}+x^{15}+\dots).$$
Is that correct?
Now I tried to simplify this like this:
$$1+x+x^2+x^3+\dots = \frac{1}{1-x}$$
$$x^6+x^9+x^{12}+\dots
=(1+x^3+x^6+x^9+x^{12}+\dots)-(1+x^3)
=\frac{1}{1-x^3}-(1+x^3)
=\frac{x^6}{1-x^3}$$
and likewise
$$x^9++x^{12}+x^{15}+\dots = \frac{x^9}{1-x^3}.$$
So I have
$$(1+x+x^2+x^3+\dots)(x^6+x^9+x^{12}+\dots)(x^9++x^{12}+x^{15}+\dots)
=\frac{x^{15}}{(1-x)(1-x^3)^2}.$$
I hope I didn't make a mistake already. In any case, here I'm stuck. I have to find a formula for the $n$-th coefficient in this power series, but I don't know how to do it.
[edit]
I'm still quite confused, but I'll try to go over it again one by one. Here is again the complete (and hopefully correct) partial fraction decomposition and my try to expand it:
$$\frac{1}{(1 - x)^3 (x^2 + x + 1)^2} \\
=\frac{1}{27} \frac{7x + 8}{x^2 + x + 1}
+ \frac{1}{9} \frac{2x+1}{(x^2 + x + 1)^2}
+ \frac{1}{27} \frac{7}{1-x}
- \frac{1}{9} \frac{2}{(1-x)^2}
+ \frac{1}{9} \frac{1}{(1-x)^3} \\
=\frac{1}{27} \frac{(7x + 8)(1-x)}{(1-x^3)}
+ \frac{1}{9} \frac{(2x+1)(1-x)^2}{(1-x^3)^2}
+ \frac{1}{27} \frac{7}{1-x}
- \frac{1}{9} \frac{2}{(1-x)^2}
+ \frac{1}{9} \frac{-1}{(1-x)^3} \\
=\frac{ (7x + 8)(1-x)}{27} \sum x^{3n}
+ \frac{ (2x+1)(1-x)^2}{9} \sum (n+1) x^{3n}
+ \frac{7}{27} \sum x^n
- \frac{2}{9} \sum (n+1) x^n
+ \frac{1}{9} \sum x^{3n}.$$
Still to do: Multiply by $x^{15}$ and find the coefficient of $x^k$.
|
OK, use generating functions. Note that $1 - z^3 = (1 - \omega z) (1 - \omega^2 z) (1 - z)$, where $\omega = -\frac{1}{2} + \mathrm{i} \frac{\sqrt{3}}{2}$ and $\omega^3 = 1$:
\begin{align}
[z^n] z^{15} (1 + z &+ z^2 + \ldots) (1 + z^3 + z^6 + \ldots)^2 \\
&= [z^{n - 15}] \frac{1}{(1 - z) (1 - z^3)^2} \\
&= [z^{n - 15}] \frac{1}{(1 - z)^3 (1 - \omega z)^2 (1 - \omega^2 z)^2}
\end{align}
This can be split into partial fractions, but the coefficients are very ugly. They can be simplified by $\omega^2 + \omega + 1 = 0$ and $\omega^3 = 1$. Use of a computer algebra system is mandatory.
Once you've got the partial fractions, you can use:
$$
\binom{-m}{k} = (-1)^k \binom{k + m - 1}{m - 1}
$$
The imaginary components cancel out, and you get real coefficients. One way to express them is to use Euler's formula:
$$
\omega^n
= \exp( \mathrm{i} \frac{2 \pi n}{3} )
= \cos \frac{2 \pi n}{3} + \mathrm{i} \sin \frac{2 \pi n}{3}
$$
|
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|
What is wrong with this limit $\lim_{x \rightarrow \infty} \frac{2x^4-1}{-4x^5+x^2}$? The arrow indicates an elimination due to numbers becoming negligible at high values of x.
The correct answer is 0, but I got infinity and I'm not sure where my reasoning went wrong.
$$\lim_{x \rightarrow \infty} \frac{2x^4-1}{-4x^5+x^2}=\lim_{x \rightarrow \infty} \frac{x^4(2-\frac{1}{x^4})}{x^5(-4+\frac{x^2}{x^5})}=\lim_{x \rightarrow \infty} \frac{x^4(2)}{x^5(\frac{1}{x^3})}=\lim_{x \rightarrow \infty} \frac{2x^4}{x^2}=\lim_{x \rightarrow \infty} 2x^2 = \infty$$
Thanks!
|
Your mistake is not that you did not evaluate the numerator and denominator at the same time or in the same way. Your mistake is strictly a computational one: $$\lim_{x \rightarrow \infty} \frac{x^4(2-\frac{1}{x^4})}{x^5(-4+\frac{x^2}{x^5})}=\lim_{x \rightarrow \infty} \frac{x^4(2)}{x^5(\frac{1}{x^3})}$$ is incorrect. The RHS must instead read $$\lim_{x \rightarrow \infty} \frac{x^4(2)}{x^5(-4 + \frac{1}{x^3})},$$ and this is the source of the error. This is because as $x \to \infty$, $x^5(-4 + x^{-3})$ asymptotically behaves as $-4x^5$, not $x^2$, and it is this faster growth of the denominator compared to the numerator that makes the limit zero.
|
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|
Finding integral solutions to the equation $x^4-ax^3-bx^2-cx-d=0$ How many integral solutions exist for the equation:
$$x^4-ax^3-bx^2-cx-d=0\qquad a\ge b\ge c\ge d\qquad a,b,c,d\in\Bbb{N}$$
I have no idea where to begin even.Please help.
|
Let's find integer root $x$.
As $d\ne 0$, then $x\ne 0$.
Denote $y=1/x$.
Then equation for $y$:
$$
ay+by^2+cy^3+dy^4=1.\tag{1}
$$
A. If $x=-n$, where $n\in\mathbb{N}$, then $y=-\dfrac{1}{n}$, and $(1) \implies$
$$
\dfrac{-a}{n}+\dfrac{b}{n^2}+\dfrac{-c}{n^3}+\dfrac{d}{n^4}=1.\tag{2}
$$
$$
\dfrac{-an+b}{n^2}+\dfrac{-cn+d}{n^4}=1.
$$
LHS is non-positive, since $a\ge b$, $c\ge d$. So, there are no solutions in the form $x=-n, n\in \mathbb{N}$.
B. If $x=n$, where $n\in\mathbb{N}$, then $y=\dfrac{1}{n}$, and $(1) \implies$
$$
\dfrac{a}{n}+\dfrac{b}{n^2}+\dfrac{c}{n^3}+\dfrac{d}{n^4}=1.\tag{3}
$$
$(3)$ $\implies$ $n\ne 1$, and
$$
\dfrac{a}{n}<1,\tag{4'}
$$
$$
1\le\dfrac{a}{n}+\dfrac{a}{n^2}+\dfrac{a}{n^3}+\dfrac{a}{n^4}< \sum_{j=1}^{\infty}\dfrac{a}{n^j}=\dfrac{a}{n-1},\tag{4''}
$$
$(4'), (4'') \implies$
$$
\dfrac{n}{a}>1,\tag{5'}
$$
$$
\dfrac{n-1}{a}<1.\tag{5''}
$$
There are no such $n\in \mathbb{N}$ to fit both $(5'), (5'')$ (remember, that $a\in \mathbb{N}$): if $n=a$, then $\dfrac{n}{a}=1$.
Answer: $0$ integer solutions.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Finding generators for an ideal of $\Bbb{Z}[x]$ We know that $\Bbb{Z}$ is Noetherian. Hence, we can conclude that $\Bbb{Z}[x]$ is Noetherian, too. Consider the ideal generated by $\langle 2x^2+2,3x^3+3,5x^5+5,…,px^p+p,…\rangle$ for all prime natural numbers $p$. How can I determine a finite number of elements which generate this ideal? Clearly the Euclidean algorithm is not valid in $\Bbb{Z}[x]$.
Thank you
|
$$ (1 - 11 x - x^2 + 5 x^3 - 5 x^4) (2 x^2 + 2) + 4 (1 + x) (3 x^3 + 3) -
2 (1 - x) (5 x^5 + 5) = 4$$
$$-112 (1 - x^3 - x^5) (5 x^5 + 5) - 80 (x + x^3) (7 x^7 + 7) + \\
187 (1 - x^3 + x^6 - x^9) (3 x^3 + 3) + 51 x (11 x^{11} + 11) = x+1$$
Conversely,
$2x^2+2 = 2(x+1)(x-1)+4$, $px^p+p = p(x+1)(1-x+\dots + x^{p-1})$ for $p$ odd.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Calculation of integral $\int\exp \left(-\alpha \sin^2 \left(\frac{x}{2} \right) \right) dx$ Given $\alpha$ is a constant. How to calculate the following integral?
\begin{equation}
\int \exp \bigg(-\alpha \sin^2 \bigg(\frac{x}{2} \bigg) \bigg) dx
\end{equation}
Thanks for your answer.
|
Let $u=\dfrac{x}{2}$ ,
Then $x=2u$
$dx=2~du$
$\therefore\int e^{-\alpha\sin^2\frac{x}{2}}~dx$
$=2\int e^{-\alpha\sin^2u}~du$
$=2\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\alpha^n\sin^{2n}u}{n!}du$
$=2\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\alpha^n\sin^{2n}u}{n!}\right)du$
For $n$ is any natural number,
$\int\sin^{2n}u~du=\dfrac{(2n)!u}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}u\cos u}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts.
$\therefore2\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\alpha^n\sin^{2n}u}{n!}\right)du$
$=2u+2\sum\limits_{n=1}^\infty\dfrac{(-1)^n\alpha^n(2n)!u}{4^n(n!)^3}-2\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n\alpha^n(2n)!((k-1)!)^2\sin^{2k-1}u\cos u}{4^{n-k+1}(n!)^3(2k-1)!}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\alpha^n(2n)!2u}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n\alpha^n(2n)!((k-1)!)^2\sin^{2k-1}u\cos u}{2^{2n-2k+1}(n!)^3(2k-1)!}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\alpha^n(2n)!x}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n\alpha^n(2n)!((k-1)!)^2\sin^{2k-1}\dfrac{x}{2}\cos\dfrac{x}{2}}{2^{2n-2k+1}(n!)^3(2k-1)!}+C$
Or you can express in terms of the incomplete bessel K function:
$\int e^{-\alpha\sin^2\frac{x}{2}}~dx$
$=\int e^\frac{\alpha(\cos x-1)}{2}~dx$
$=e^{-\frac{\alpha}{2}}\int e^\frac{\alpha\cos x}{2}~dx$
$=e^{-\frac{\alpha}{2}}\int e^\frac{\alpha(e^{ix}+e^{-ix})}{4}~dx$
$=e^{-\frac{\alpha}{2}}\int e^{\frac{\alpha e^{-ix}}{4}+\frac{\alpha}{4e^{-ix}}}~dx$
Let $u=e^{-ix}$ ,
Then $x=i\ln u$
$dx=\dfrac{i}{u}du$
$\therefore e^{-\frac{\alpha}{2}}\int e^{\frac{\alpha e^{-ix}}{4}+\frac{\alpha}{4e^{-ix}}}~dx$
$=ie^{-\frac{\alpha}{2}}\int\dfrac{e^{\frac{\alpha u}{4}+\frac{\alpha}{4u}}}{u}du$
$=ie^{-\frac{\alpha}{2}}\int_0^u\dfrac{e^{\frac{\alpha t}{4}+\frac{\alpha}{4t}}}{t}dt+C$
$=ie^{-\frac{\alpha}{2}}\int_0^1\dfrac{e^{\frac{\alpha ut}{4}+\frac{\alpha}{4ut}}}{ut}d(ut)+C$
$=ie^{-\frac{\alpha}{2}}\int_0^1\dfrac{e^{\frac{\alpha ut}{4}+\frac{\alpha}{4ut}}}{t}dt+C$
$=ie^{-\frac{\alpha}{2}}\int_\infty^1te^{\frac{\alpha u}{4t}+\frac{\alpha t}{4u}}~d\left(\dfrac{1}{t}\right)+C$
$=ie^{-\frac{\alpha}{2}}\int_1^\infty\dfrac{e^{\frac{\alpha t}{4u}+\frac{\alpha u}{4t}}}{t}dt+C$
$=ie^{-\frac{\alpha}{2}}K_0\left(-\dfrac{\alpha}{4u},-\dfrac{\alpha u}{4}\right)+C$ (according to http://artax.karlin.mff.cuni.cz/r-help/library/DistributionUtils/html/incompleteBesselK.html)
$=ie^{-\frac{\alpha}{2}}K_0\left(-\dfrac{\alpha e^{ix}}{4},-\dfrac{\alpha e^{-ix}}{4}\right)+C$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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|
How to solve $P=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^2}\right)\left(1+\frac{1}{3^3}\right)\ldots \infty$ How do I find the following product
$$P=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^2}\right)\left(1+\frac{1}{3^3}\right)\ldots \infty$$
|
This is not an answer but I hope others will help to conclude something.
First of all, take a logarithm from both sides:
$$
\log P=\log\prod_n\left(1+\frac{1}{3^n}\right)=\sum_n\log\left(1+\frac{1}{3^n}\right)
$$
Now using Taylor series, you get:
$$
\log(1+\frac{1}{3^m})=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \frac{1}{3^{mn}}.
$$
Therefore we have:
$$
\log P=\log\prod_n\left(1+\frac{1}{3^n}\right)=\sum_{m=1}^\infty\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \frac{1}{3^{mn}}\\
=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sum_{m=1}^\infty\frac{1}{3^{mn}}\\
=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \frac{1}{3^{n}-1}.
$$
The preceding series is convergent. Now this is a Lambert series with $a_n=\frac{(-1)^{n+1}}{n}$ and $q=\frac{1}{3}$. For the moment, I am not sure if there is a closed form in terms of some special function or not.
Update 1:
$$
\sum_{n=1}^\infty \frac{1}{3^{n}-1}=\frac{\psi_{\frac{1}{3}}(1)+\log\frac 23}{\log \frac 13}
$$
where $\psi_{q}(z)$ is q-Polygamma function.
|
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"url": "https://math.stackexchange.com/questions/804006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
}
|
How prove $\left(\frac{b+c}{a}+2\right)^2+\left(\frac{c}{b}+2\right)^2+\left(\frac{c}{a+b}-1\right)^2\ge 5$ Let $a,b,c\in R$ and $ab\neq 0,a+b\neq 0$. Find the minimum of:
$$\left(\dfrac{b+c}{a}+2\right)^2+\left(\dfrac{c}{b}+2\right)^2+\left(\dfrac{c}{a+b}-1\right)^2\ge 5$$
if and only if $$a=b=1,c=-2$$
My idea: Since
$$\left(\dfrac{b+c+2a}{a}\right)^2+\left(\dfrac{c+2b}{b}\right)^2+\left(\dfrac{c-a-b}{a+b}\right)^2$$
let
$$x=\dfrac{b+c+2a}{a},y=\dfrac{c+2b}{b},z=\dfrac{c-a-b}{a+b}$$
then I can't work.
Thank you.
|
Continuing from Calvin Lin' answer, the minimum value will be otained when all partial derivatives will be zero. The only partial derivative which is easy to write is with respect to $c$ which only appears in numerator. Almost as Calvin Lin wrote, this derivative cancels when $$c=-\frac{b (a+b) (2 a+b)}{a^2+a b+b^2}$$ Back to the expression, this leads to a constant value of 5.
The expansion is not so tedious since, replacing $c$ by its expression, we can find $$\frac{b+c}{a}+2=-\frac{(a+b) (a+2 b)}{a^2+a b+b^2}$$ $$\frac{c}{b}+2=\frac{b (b-a)}{a^2+a b+b^2}$$ $$\frac{c}{a+b}-1=\frac{a (2 a+b)}{a^2+a b+b^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/807404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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|
How to prove: prove $\frac{1+\tan^2\theta}{1+\cot^2\theta} = \tan^2\theta$ I need to prove that $\frac{1+\tan^2\theta}{1+\cot^2\theta}= \tan^2\theta.$
I know that $1+\tan^2\theta=\sec^2\theta$ and that $1+\cot^2\theta=\csc^2\theta$, making it now $$\frac{\sec^2\theta}{\csc^2\theta,}$$ but I don't know how to get it down to $ \tan^2\theta.$
HELP!!!!
I also need help proving that $\tan\theta + \cot\theta = \sec\theta\cdot\csc\theta.$
|
Look at $\sin^2 \theta + \cos^2 \theta = 1$. Dividing by $\cos^2 \theta$ we get $\tan^2 \theta + 1 = \sec^2 \theta$. Try to find another relation by dividing by $\sin^2 \theta $ and see what appears.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/810453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Series Question: $\sum_{n=1}^{\infty}\frac{n+1}{2^nn^2}$ How to compute the following series:
$$\sum_{n=1}^{\infty}\frac{n+1}{2^nn^2}$$
I tried
$$\frac{n+1}{2^nn^2}=\frac{1}{2^nn}+\frac{1}{2^nn^2}$$
The idea is using Riemann zeta function
$$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$
but the term $2^n$ makes complicated. I know that
$$\sum_{n=1}^{\infty}\frac{1}{2^n}=1$$
using geometric series but I don't know how to use those series to answer the question. Any help would be appreciated. Thanks in advance.
|
Consider Maclaurin series of natural logarithm
$$
\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}.
$$
Taking $x=\dfrac12$ yields
\begin{align}
\ln\left(1-\frac12\right)&=-\sum_{n=1}^\infty\frac{1}{2^n\ n}\\
\ln2&=\sum_{n=1}^\infty\frac{1}{2^n\ n}.
\end{align}
Now, dividing the Maclaurin series of natural logarithm by $x$ yields
\begin{align}
\frac{\ln(1-x)}{x}&=-\sum_{n=1}^\infty\frac{x^{n-1}}{n},
\end{align}
then integrating both sides and taking the limit of integration $0<x<\dfrac12$. We obtain
\begin{align}
\int_0^{\Large\frac12}\frac{\ln(1-x)}{x}\ dx&=-\int_0^{\Large\frac12}\sum_{n=1}^\infty\frac{x^{n-1}}{n}\ dx\\
&=-\sum_{n=1}^\infty\int_0^{\Large\frac12}\frac{x^{n-1}}{n}\ dx\\
&=-\left.\sum_{n=1}^\infty\frac{x^{n}}{n^2}\right|_{x=0}^{\Large\frac12}\\
-\int_0^{\Large\frac12}\frac{\ln(1-x)}{x}\ dx&=\sum_{n=1}^\infty\frac{1}{2^n\ n^2}.
\end{align}
The integral in the LHS is $\text{Li}_2\left(\dfrac12\right)=\dfrac{\pi^2}{12}-\dfrac{\ln^22}{2}$, but since you are not familiar with dilogarithm function, we will evaluate the LHS integral using IBP. Taking $u=\ln(1-x)$ and $dv=\dfrac1x\ dx$, we obtain
\begin{align}
\int_0^{\Large\frac12}\frac{\ln(1-x)}{x}\ dx&=\left.\ln(1-x)\ln x\right|_0^{\large\frac12}+\int_0^{\Large\frac12}\frac{\ln x}{1-x}\ dx\\
&=\ln^22-\lim_{x\to0}\ln(1-x)\ln x-\int_1^{\Large\frac12}\frac{\ln (1-x)}{x}\ dx\ ;\\&\color{red}{\Rightarrow\text{let}\quad x=1-x}\\
\int_0^{\Large\frac12}\frac{\ln(1-x)}{x}\ dx+\int_1^{\Large\frac12}\frac{\ln (1-x)}{x}\ dx&=\ln^22-0\\
-\left.\sum_{n=1}^\infty\frac{x^{n}}{n^2}\right|_{x=0}^{\Large\frac12}-\left.\sum_{n=1}^\infty\frac{x^{n}}{n^2}\right|_{x=1}^{\Large\frac12}&=\ln^22\\
\sum_{n=1}^\infty\frac{1}{2^n\ n^2}+\sum_{n=1}^\infty\frac{1}{2^n\ n^2}-\sum_{n=1}^\infty\frac{1}{n^2}&=-\ln^22\\
2\sum_{n=1}^\infty\frac{1}{2^n\ n^2}-\frac{\pi^2}{6}&=-\ln^22\\
\sum_{n=1}^\infty\frac{1}{2^n\ n^2}&=\frac{\pi^2}{12}-\frac{\ln^22}{2}.
\end{align}
Thus,
\begin{align}
\sum_{n=1}^\infty\frac{n+1}{2^n\ n^2}&=\sum_{n=1}^\infty\left(\frac{1}{2^n\ n}+\frac{1}{2^n\ n^2}\right)\\
&=\large\color{blue}{\ln2+\frac{\pi^2}{12}-\frac{\ln^22}{2}}.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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|
Prove the inequality $\frac{1}{1999}<\frac{1\cdot3\cdot5\cdot7\cdots1997}{2\cdot4\cdot6\cdot8\cdots1998}<\frac{1}{44}$ prove this inequality.
$\dfrac{1}{1999}<\dfrac{1\cdot3\cdot5\cdot7\cdots1997}{2\cdot4\cdot6\cdot8\cdots1998}<\dfrac{1}{44}$
I have tried to convert this series in factorial form. I am not getting what to do with this type of numbers $44$ and $1999$.
|
Just to expand on the comment I made, let $X = \dfrac{1\cdot3\cdot5\cdot7\cdots1997}{2\cdot4\cdot6\cdot8\cdots1998}$. Then the right inequality is equivalent to showing
$$ \frac1{X^2} > 44^2 \iff \frac{2^2}{1 \cdot 3} \cdot \frac{4^2}{3 \cdot 5}\cdot \frac{6^2}{5 \cdot 7} \cdots \frac{1998^2}{1997 \cdot 1999}> \frac{44^2}{1999}$$
$$\iff \prod_{k=1}^{999} {\frac{(2k)^2}{(2k)^2-1}} > \frac{1936}{1999}$$
As all the factors on the left are greater than $1$ the inequality is evident.
For the left inequality, as the first comment mentions, simply multiply by $1999$ to get
$$\iff 1 < 1 \cdot \frac32 \cdot \frac54 \cdots\frac{1999}{1998}$$
which is obvious.
|
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 4,
"answer_id": 3
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|
$4 \sin 72^\circ \sin 36^\circ = \sqrt 5$ How do I establish this and similar values of trigonometric functions?
$$
4 \sin 72^\circ \sin 36^\circ = \sqrt 5
$$
|
A different approach
$ a = n\pi/5$
$5a = n\pi$
$3a = n\pi-2a$
$\sin 3a = \sin{(n\pi-2a)}$
$\sin 3a = \sin2a$
Expand using identities
Remove a sin
Square both sides
You get a quadratic in $\sin^2$
Let $\sin^2 a = x$
So
$16x^2 -20 a +5 = 0$
So
$\sin^2 72 \cdot\sin^2 36 = 5/16$
Hence
$4\sin72\sin36 = \sqrt5$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/816195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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|
Solve a system of linear equations $\newcommand{\Sp}{\phantom{0}}$There is a system of linear equations:
\begin{alignat*}{4}
&x - &&y - 2&&z = &&1, \\
2&x + 3&&y - &&z =-&&2.
\end{alignat*}
I create the matrix of the system:
$$
\left[\begin{array}{rrr|r}
1 & -1 & -2 & 1 \\
2 & 3 & -1 & -2
\end{array}\right]
$$
then with GEM,
$$
\left[\begin{array}{rrr|r}
1 & -1 & -2 & 1 \\
0 & 5 & 3 & -4
\end{array}\right]
$$
I don't know how to proceed after that?
I have found the correction of this exercise but I still don't understand the way to solve it. Can someone help me please?
|
This is the problem of finding the complete solution to $Ax=b$. To learn how to do this, you may want to watch this video or Gilbert Strang's full video lecture or notes on the topic.
The idea is that we need to find
(a) the null space $N$ of $ \left[\begin{array}{rrr|r}
1 & -1 & -2 \\
2 & 3 & -1
\end{array}\right]$ and
(b) a single solution $c$ to $ \left[\begin{array}{rrr}
1 & -1 & -2 \\
2 & 3 & -1
\end{array}\right]c =
\left[\begin{array}{r} 1 \\-2
\end{array}\right] $.
Then all the solutions are exactly $x = N + c$.
Row reduction is the first step regardless of what we do next.
(a) We have $\left[\begin{array}{rrr}
1 & -1 & -2 \\
0 & 5 & 3
\end{array}\right] \left[\begin{array}{r} 7/5 \\ -3/5 \\ 1 \end{array}\right] = \left[\begin{array}{r} 0 \\ 0 \end{array}\right]$. Thus the null-space is $N = \left[\begin{array}{r} 7/5 \\ -3/5 \\ 1 \end{array}\right]a$.
(b) Notice that $ \left[\begin{array}{rrr}
1 & -1 & -2 \\
0 & 5 & 3
\end{array}\right] \left[\begin{array}{r} 1/5 \\ -4/5 \\ 0 \end{array}\right] = \left[\begin{array}{r} 1 \\ -4 \end{array}\right] $.
Hence the solutions are exactly all $x$ such that
$$ x = \left[\begin{array}{r} 7/5 \\ -3/5 \\ 1 \end{array}\right]a + \left[\begin{array}{r} 1/5 \\ -4/5 \\ 0 \end{array}\right].$$
Another way of writing this is
\begin{align*}
x &= \frac{7}{5}a + \frac{1}{5} \\
y &= -\frac{3}{5}a - \frac{4}{5} \\
z &= a.
\end{align*}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/817183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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|
Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$ Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$
My approach :
I used $\sin A +\sin B = 2\sin(A+B)/2\times\cos(A-B)/2 $
$\Rightarrow \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7} = 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(8\pi/7) $
$= 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(\pi + \pi/7) $
$= 2\sin(3\pi)/7\times\cos(\pi/7)-\sin(\pi/7) $
I am not getting any clue how to proceed further or whether it is correct or not. Please help thanks..
|
The idea in this answer is: if you can introduce a trigonometric sum with arguments in arithmetic progression, then you have a nice formula that does all the job (see here for the statement and two proofs).
Let
$$S=\sin \frac{2\pi}{7}+\sin \frac{4\pi}{7}+\sin \frac{8\pi}{7}$$
Then
$$S^2=\sin^2 \frac{2\pi}{7}+\sin^2 \frac{4\pi}{7}+\sin^2 \frac{8\pi}{7}\\
+2\sin \frac{2\pi}{7}\sin \frac{4\pi}{7}+2\sin \frac{2\pi}{7}\sin \frac{8\pi}{7}+2\sin \frac{4\pi}{7}\sin \frac{8\pi}{7}$$
$$S^2=\frac{1-\cos \frac{4\pi}{7}}{2}+\frac{1-\cos \frac{8\pi}{7}}{2}+\frac{1-\cos \frac{16\pi}{7}}{2}+\cos \frac{2\pi}{7}-\cos \frac{6\pi}{7}\\
+\cos \frac{6\pi}{7}-\cos \frac{10\pi}{7}+\cos \frac{4\pi}{7}-\cos \frac{12\pi}{7}$$
$$S^2=\frac32+\cos \frac{2\pi}{7}+\frac12\cos \frac{4\pi}{7}-\frac12\cos \frac{8\pi}{7}-\cos \frac{10\pi}{7}-\cos \frac{12\pi}{7}-\frac12\cos \frac{16\pi}{7}$$
$$S^2=\frac32+\cos \frac{2\pi}{7}-\frac12\cos \frac{3\pi}{7}+\frac12\cos \frac{\pi}{7}+\cos \frac{3\pi}{7}-\cos \frac{2\pi}{7}-\frac12\cos \frac{2\pi}{7}$$
$$S^2=\frac32+\frac12\cos \frac{\pi}{7}-\frac12\cos \frac{2\pi}{7}+\frac12\cos \frac{3\pi}{7}$$
$$S^2=\frac32-\frac12\left(\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7}\right)$$
And using the following formula from this answer, with $n=3, a=b=\frac{2\pi}{7}$:
$$\sum_{k=0}^{n-1} \cos (a+kb) = \frac{\sin \frac{nb}{2}}{\sin \frac{b}{2}} \cos \left( a+ (n-1)\frac{b}{2}\right)$$
$$\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7}=\frac{\sin \frac{3\pi}{7}\cos \frac{4\pi}{7}}{\sin \frac{\pi}{7}}=\frac{\sin\frac{7\pi}{7}-\sin\frac{\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac12$$
Hence
$$S^2=\frac32+\frac14=\frac74$$
And we know that $S>0$ because the only negative term is $\sin\frac{8\pi}{7}=-\sin\frac{\pi}{7}$, and $0<\sin\frac{\pi}{7}<\sin\frac{2\pi}{7}$. So we have finally
$$\sin \frac{2\pi}{7}+\sin \frac{4\pi}{7}+\sin \frac{8\pi}{7}=\frac{\sqrt7}{2}$$
|
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}
|
How to integrate $\int \frac{1}{\sin^4x + \cos^4 x} \,dx$?
How to integrate
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx$$
I tried the following approach:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{\sin^4x + (1-\sin^2x)^2} \,dx = \int \frac{1}{\sin^4x + 1- 2\sin^2x + \sin^4x} \,dx \\
= \frac{1}{2}\int \frac{1}{\sin^4x - \sin^2x + \frac{1}{2}} \,dx = \frac{1}{2}\int \frac{1}{(\sin^2x - \frac{1}{2})^2 + \frac{1}{4}} \,dx$$
The substitution $t = \tan\frac{x}{2}$ yields 4th degree polynomials and a $\sin$ substitution would produce polynomials and expressions with square roots while Wolfram Alpha's solution doesn't look that complicated. Another approach:
$\sin^4x + \cos^4 x = (\sin^2 x + \cos^2x)(\sin^2 x + \cos^2 x) - 2\sin^2 x\cos^2 x = 1 - 2\sin^2 x\cos^2 x = (1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)$
and then I tried substituting: $t = \sin x \cos x$ and got
$$\int\frac{t\,dt}{2(1-2t^2)\sqrt{1-4t^2}}$$
Another way would maybe be to make two integrals:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{(1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)} \,dx = \\ \frac{1}{2}\int \frac{1}{1-\sqrt2\sin x \cos x} \,dx + \frac{1}{2}\int\frac{1}{1+\sqrt2\sin x \cos x} \,dx$$
... and again I tried $t = \tan\frac{x}{2}$ (4th degree polynomial) and $t=\sqrt2 \sin x \cos x$ and I get $\frac{\sqrt 2}{2} \int \frac{\,dt}{(1-t)\sqrt{1-2t^2}}$ for the first one.
Any hints?
|
Convert the exponential powers to multiple angles. From deMoivre's theorem, with $n\in\mathbb{N}$:
$$
\begin{align}
\left( e^{i \theta} \right)^{n} &= e^{i n\theta} \\
\left( \cos \theta + i \sin \theta \right)^{n} &= \cos n\theta + i \sin n\theta
\end{align}
$$
These intermediate formulas may help:
$$
\begin{align}
\cos 2\theta &= \cos^{2} \theta - \sin^{2} \theta \\
\sin 2\theta &= 2 \cos \theta \sin 2 \theta \\
\end{align}
$$
Reduce the denominator
$$
\sin ^4(x)+\cos ^4(x) = \frac{1}{4} (\cos (4 x)+3)
$$
The primitive is
$$
\int \frac{1}{\cos^{4}x + \sin^{4}x} \, dx =
\int \frac{1}{\cos (4 x)+3} \, dx =
\left(4 \sqrt{2}\right)^{-1}\arctan \left(\frac{\tan (2 x)}{\sqrt{2}}\right)
$$
Here is a look at the integrand:
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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"answer_id": 1
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|
Trouble with factoring polynomial fractions and understanding wolfram alpha I hope the community can excuse me if I'm making excessive posts, I have a calculus quiz tomorrow and I want to be as prepared as possible.
I'm going through a review problem on wolfram alpha and I have trouble understanding a step. The problem begins like this:
$$\frac{2}{3}x^{\frac{-1}{3}}(x-5) + x^{\frac{2}{3}} = 0$$
Wolfram alpha combines and rewrites as follows:
$$\frac{2(x - 5)}{3\sqrt[3]{x}} + x^{\frac{2}{3}} = 0$$
So far so good. But now wolfram alpha subtracts $x^\frac{2}{3}$ from both sides...
$$\frac{2x^{\frac{2}{3}}}{3} – \frac{10}{3(\sqrt[3]{x})} = -x^{\frac{2}{3}}$$
And that's my problem, right there—why is there a three under the $\frac{2x^{\frac{2}{3}}}{3}$? If wolfram alpha multiplied out 2(x - 5) then wouldn't the resulting numbers share a denominator of $3(\sqrt[3]{x})$?
|
Let's write radicals in fractional notation:
$$\frac{2(x - 5)}{3\sqrt[3]{x}} + x^{2/3} = 0\iff \frac{2(x - 5)}{3x^{1/3}} + x^{2/3}=0.$$
Now, it will become obvious to you: $$\require{cancel}\eqalign{\frac{2(x - 5)}{3x^{1/3}} + x^{2/3}
&=\frac{2x - 10}{3x^{1/3}} + x^{2/3}\\ \ \\
&=\frac{2x}{3x^{1/3}} - \frac{10}{3x^{1/3}} + x^{2/3}\\ \ \\
&=\frac{2x^{1/3+2/3}}{3x^{1/3}} - \frac{10}{3x^{1/3}} + x^{2/3}\\ \ \\
&=\frac{2x^{1/3}\cdot x^{2/3}}{3x^{1/3}} - \frac{10}{3x^{1/3}} + x^{2/3}\\ \ \\
&=\frac{2\cancel{x^{1/3}}\cdot x^{2/3}}{3\cancel{x^{1/3}}} - \frac{10}{3x^{1/3}} + x^{2/3}\\ \ \\
&=\frac{{2}x^{2/3}}{3} - \frac{10}{3x^{1/3}} + x^{2/3}.
}$$
|
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"url": "https://math.stackexchange.com/questions/820916",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluation of tricky integral I want to evaluate the integral
$$\int _ {b} ^ {\infty} \mathrm{d} x \, \frac{e ^ {x ^ {2} / s} (b^2 + 3 x ^ 2) ^ {2}}{x (x^2 + b^2)}$$,
where $b$ and $s$ are positive real numbers. I thought of writing $x$ as $x = - i y$. Then $y$ would be a pure imaginary number and the integral would become
$$\int _ {i b} ^ {i \infty} \mathrm{d} y \, \frac{e ^ {- y ^ {2} / s} (b^2 - 3 y ^ 2) ^ {2}}{y (- y^2 + b^2)}$$.
I tried then to do the integral by contour integration and consider the integral
$$\int \mathrm{d} z \, \frac{e ^ {- z ^ {2} / s} (b^2 - 3 z ^ 2) ^ {2} \log(z - i b)}{z (- z^2 + b^2)}$$,
with a branch cut from $ib$ to $i \infty$. Am I on the right track? Could someone please help me?
|
As proposed the integral has a issue with large values, ie $x \rightarrow \infty$, since the integral
is unbounded. This can be remedied by replacing $s$ with $- 1/s$. In this view consider the integral
\begin{align}
I = \int_{b}^{\infty} e^{-s x^{2}} \ \frac{(a x^{2} + b^{2})^{2}}{x ( x^{2} + b^{2}) } \ dx.
\end{align}
It is readily seen that
\begin{align}
\frac{(a x^{2} + b^{2})^{2}}{x ( x^{2} + b^{2}) } &= \frac{ [a (x^{2}+b^{2}) + (1-a) b^{2}]^{2}}
{x(x^{2}+b^{2})} \\
&= \frac{a^{2} (x^{2}+b^{2})^{2} + 2 a (1-a) b^{2} (x^{2}+b^{2}) + (1-a)^{2}b^{4}}{x(x^{2}+b^{2})} \\
&= a^{2} \ \frac{x^{2}+b^{2}}{x} + 2 a(1-a) b^{2} \ \frac{1}{x} + \frac{(1-a)^{2} b^{4}}{x(x^{2}
+ b^{2})} \\
&= a^{2} x + \frac{a(2-a)b^{2}}{x} + \frac{(1-a)^{2} b^{4}}{x(x^{2}+ b^{2})}
\end{align}
for which the integral becomes
\begin{align}
I = a^{2} \int_{b}^{\infty} e^{-s x^{2}} x \ dx + a(2-a)b^{2} \int_{b}^{\infty} e^{-s x^{2}} \
\frac{dx}{x} + (1-a)^{2} \int_{b}^{\infty} \frac{e^{-s x^{2}} \ dx}{x(x^{2} + b^{2})}.
\end{align}
Let $x = \sqrt{t/s}$ in each of the integrals to obtain
\begin{align}
I = \frac{a^{2}}{2s} \ e^{- s b^{2}} + \frac{a(2-a)b^{2}}{2} \int_{sb^{2}}^{\infty} \frac{e^{-t} \ dt}
{t} + \frac{s (1-a)^{2}}{2} \int_{sb^{2}}^{\infty} \frac{e^{-t} \ dt}{t(t+sb^{2})}.
\end{align}
The last integral can be reduced in the following. Using
\begin{align}
\frac{1}{t(t+sb^{2})} = \frac{1}{s b^{2}} \left( \frac{1}{t} - \frac{1}{t+s b^{2}} \right)
\end{align}
then
\begin{align}
\int_{sb^{2}}^{\infty} \frac{e^{-t} \ dt}{t(t+sb^{2})} &= \frac{1}{s b^{2}} \int_{sb^{2}}^{\infty}
e^{-t} \left( \frac{1}{t} - \frac{1}{t+s b^{2}} \right) dt \\
&= \frac{1}{s b^{2}} \int_{sb^{2}}^{\infty} \frac{e^{-t} \ dt}{t} - \frac{1}{s b^{2}} \int_{sb^{2}
}^{\infty} \frac{e^{-t} \ dt}{t+s b^{2}} \\
&= \frac{1}{s b^{2}} \int_{sb^{2}}^{\infty} \frac{e^{-t} \ dt}{t} - \frac{e^{s b^{2}}}{s b^{2}}
\int_{2 s b^{2}}^{\infty} \frac{ e^{-t} \ dt}{t}.
\end{align}
Using this reduction in the integral $I$ leads to
\begin{align}
I = \frac{a^{2}}{2s} \ e^{- s b^{2}} + \frac{a(2-a)b^{4}+(1-a)^{2}}{2 b^{2}} \int_{sb^{2}}^{\infty}
\frac{e^{-t} \ dt}{t} - \frac{(1-a)^{2}}{2 b^{2}} \ e^{s b^{2}} \int_{2 s b^{2}}^{\infty}
\frac{e^{-t} \ dt}{t}.
\end{align}
The exponential integral $E_{1}(z)$ is given by
\begin{align}
E_{1}(z) = \int_{z}^{\infty} \frac{e^{-t}}{t} \ dt
\end{align}
and can be used to reduce the integral in question to the value
\begin{align}
I = \frac{a^{2}}{2s} \ e^{- s b^{2}} + \frac{a(2-a)b^{4}+(1-a)^{2}}{2 b^{2}} E_{1}(s b^{2})
- \frac{(1-a)^{2}}{2 b^{2}} \ e^{s b^{2}} E_{1}(2 s b^{2}).
\end{align}
Hence
\begin{align*}
& \int_{b}^{\infty} e^{-s x^{2}} \ \frac{(a x^{2} + b^{2})^{2}}{x ( x^{2} + b^{2}) } \ dx \\
& \hspace{5mm} = \frac{a^{2}}{2s} \ e^{- s b^{2}} + \frac{a(2-a)b^{4}+(1-a)^{2}}{2 b^{2}} E_{1}(s b^{2})
- \frac{(1-a)^{2}}{2 b^{2}} \ e^{s b^{2}} E_{1}(2 s b^{2}).
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to calculate the sum of the infinite series Please help me with the sum of the infinite series:
$$
\Large \frac{1+\frac{\pi^4}{5!}+\frac{\pi^8}{9!}+\frac{\pi^{12}}{13!}+...}
{\frac{1}{3!}+\frac{\pi^4}{7!}+\frac{\pi^8}{11!}+\frac{\pi^{12}}{15!}+...}
$$
|
Evaluating the numerator first.
Consider the series
$$\sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}=1+\frac{x^4}{4!}+\frac{x^8}{8!}+\cdots$$
Also,
$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots$$
$$\cosh x=\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\frac{x^8}{8!}+\cdots$$
Add the above two and you obtain:
$$\sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}=\frac{1}{2}(\cos x+\cosh x)\,\,\,\,(*)$$
Integrate both the sides from the limits $0$ to $\pi$ to get:
$$\sum_{n=0}^{\infty} \frac{\pi^{4n+1}}{(4n+1)!}=\frac{\sinh(\pi)}{2} \Rightarrow \sum_{n=0}^{\infty} \frac{\pi^{4n}}{(4n+1)!}=\frac{\sinh(\pi)}{2\pi}\,\,\,\,\,\,(1)$$
For denominator, integrate both the sides of $(*)$ wrt $x$ thrice and substitute $x=\pi$ to obtain:
$$\sum_{n=0}^{\infty} \frac{\pi^{4n+3}}{(4n+3)!}=\frac{1}{2}\sinh(\pi)\Rightarrow \sum_{n=0}^{\infty} \frac{\pi^{4n}}{(4n+3)!}=\frac{\sinh(\pi)}{2\pi^3}\,\,\,\,(2)$$
Divide $(1)$ and $(2)$ to obtain:
$$\boxed{\pi^2}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/821628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Integrate : $\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}dx$
$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$
My approach :
Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then
$$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\tan^2x)}{x^2\tan x -x\tan^2x+x-\tan x}dx$$
But I am not getting any relation between numerator and denominator so that I will take any substitution and solve further please suggest whether it is correct and how to proceed in this. Thanks.
|
$\bf{Another \; Solution::}$ Let $\displaystyle I = \int\frac{x^2}{(x\sin x+\cos x)\cdot (x\cos x-\sin x)}dx$
Now $\displaystyle I = \int\frac{2x^2}{(x^2\sin 2x+2x\cos^2 x-2x\sin^2 x-\sin 2x)}dx$
So $\displaystyle I = \int\frac{2x^2}{(x^2-1)\sin 2x+2x\cos 2x}dx$
Now $\displaystyle (x^2-1)\sin 2x+2x\cos 2x = \sqrt{(x^2-1)^2+(2x)^2}\cdot \left\{\left(\frac{x^2-1}{x^2+1}\right)\sin 2x+\left(\frac{2x}{x^2+1}\right)\cos 2x\right\}$
Now We can write $\displaystyle (x^2-1)\sin 2x+2x\cos 2x = (x^2+1)\cdot \sin (2x+\alpha)$
Where $\displaystyle \displaystyle \left(\frac{x^2-1}{x^2+1}\right) = \cos \alpha$
and $\displaystyle \left(\frac{2x}{x^2+1}\right)=\sin \alpha$
and so $\displaystyle\tan \alpha = \left(\frac{2x}{x^2-1}\right)\Rightarrow \alpha = \tan^{-1}\left(\frac{2x}{x^2-1}\right)$
So Integral $\displaystyle I = \int \csc(2x+\alpha)\cdot \left(\frac{2x^2}{x^2+1}\right)dx$
Now let $\displaystyle (2x+\alpha) = t\Rightarrow \left\{2x+\tan^{-1}\left(\frac{2x}{x^2-1}\right)\right\} = t\;,$Then $\displaystyle \left(\frac{2x^2}{x^2+1}\right)dx=dt$
So Integral $\displaystyle I = \int \csc tdt = \ln \left|\tan \frac{t}{2}\right|+\mathcal{C} = \ln \left|\tan \left(x+\frac{\alpha}{2}\right)\right|+\mathcal{C} $
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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|
Limit $\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}$ $$\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}=\lim_{x\rightarrow\infty}1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}=\lim_{x\rightarrow\infty}1=1\neq2$$ as Wolfram Alpha state. Where I miss something?
|
For convention, let $\lim$ denote $\lim_{x\rightarrow \infty}$. I avoid big-O notation and power series here and perhaps it is clearer this way?
Now, $$\lim [1-x+\sqrt{2+2x+x^2}] \\
= \lim [1 + \frac{2+2x}{x + \sqrt{2+2x+x^2}}] \\
= \lim [1 + \frac{2+2x}{x(1+\sqrt{2/x^2+2/x+1})}] \\
= \lim [1 + \frac{2/x + 2}{1+\sqrt{2/x^2+2/x+1}}] \\
= 1 + \frac{2}{1+\sqrt{1}} = 2.$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Math Competition Question Hong Kong I was (as again) doing practise papers, when i saw this question:
Triangle ABC is divided by segments BD, DF and FE into four triangles. E and D lie on CA and F lies on BC. These four small triangles have equal areas. If BF=2DE, find the ratio of AC:BC.
I have completely no idea on how to solve it. Anyone help me?
|
Since $\triangle CEF$ and $\triangle EDF$ have equal area and equal height, you get $CE=ED$. Without loss of generality, we can assume this length to be equal to $1$. I will also assume the height of these two triangles to be equal to $2$ so that all four areas should be equal to $1$, but read below for comments on this.
The whole triangle $\triangle ABC$ has a base of length $AC=b$ and a height $h$. The resulting area should be $\frac12bh=4$ while the area of $\triangle ABD$ should be $\frac12(b-2)h=1$. You can combine these two to get $b=4(b-2)$, so $b=\frac83$ and threfore $h=3$.
So now you can assume some coordinates:
$$
C=\begin{pmatrix}0\\0\end{pmatrix}\quad
E=\begin{pmatrix}1\\0\end{pmatrix}\quad
D=\begin{pmatrix}2\\0\end{pmatrix}\\
A=\begin{pmatrix}\tfrac83\\0\end{pmatrix}\quad
F=\begin{pmatrix}2x\\2\end{pmatrix}\quad
B=\begin{pmatrix}3x\\3\end{pmatrix}
$$
From this you get
\begin{align*}
\lVert B-F\rVert &= 2\lVert D-E\rVert \\
\lVert B-F\rVert^2 &= 4\lVert D-E\rVert^2 \\
x^2+1 &= 4 \\
x^2 &= 3 \\
x_{1,2} &= \pm\sqrt3
\end{align*}
So there are two solutions, but both lead to the same ratio
$$\frac{\lVert A-C\rVert}{\lVert B-C\rVert}=\frac49$$
You could have choosen a different height for point $F$ up front, which would have led to different solutions for $x$. But the final ratio would nevertheless have been the same. If you wanted to, you could formulate the generic solution like this:
$$
C=\begin{pmatrix}0\\0\end{pmatrix}\quad
E=\begin{pmatrix}1\\0\end{pmatrix}\quad
D=\begin{pmatrix}2\\0\end{pmatrix}\\
A=\begin{pmatrix}\tfrac83\\0\end{pmatrix}\quad
F=\begin{pmatrix}2x\\2y\end{pmatrix}\quad
B=\begin{pmatrix}3x\\3y\end{pmatrix}
$$
\begin{align*}
\lVert B-F\rVert &= 2\lVert D-E\rVert \\
\lVert B-F\rVert^2 &= 4\lVert D-E\rVert^2 \\
x^2+y^2 &= 4
\end{align*}
$$\frac{\lVert A-C\rVert}{\lVert B-C\rVert}=
\frac{\tfrac83}{3\sqrt{x^2+y^2}} =
\frac{8}{3\cdot 3\cdot \sqrt4} =
\frac{4}{9}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Expand $(x-7)^2 + x^2 = (x+2)^2$ algebraically $(x-7)^2 + x^2 = (x+2)^2$
$(x-7)(x-7) + x^2 = (x+2)(x+2)$
$x^2 -7x -7x + 49 + x^2 = x^2 + 4x + 4$
$x^2 + 18x - 45 = x^2 + x^2$
From that point on, everything I do is incorrect. I don't know what to do with the three $x^2$.
|
You don't need to expand the squares. Use $a^2-b^2 = (a+b)(a-b)$ to get
$$
x^2=(x+2)^2-(x-7)^2=(x+2+x-7)(x+2-x+7)=9(2x-5)=18x-45
$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Integration by parts - $\int \ln (2x+1) \text{dx}$ Use integration by parts to find
$\int \ln (2x+1) \text{dx}$.
So far I have:
$$x\ln(2x+1)-\int\dfrac{2x}{2x+1}dx+c$$
Using integration by substitution to find the integral
$$u=2x+1\Rightarrow\text{du}=2\text{dx}$$
$$\int\dfrac{2x}{2x+1}\cdot\dfrac{1}{2}\text{du}=\int xu^{-1}$$
$$=\int \left(\dfrac{u}{2}-\dfrac{1}{2}\right)u^{-1}\text{du}=\int\left[\dfrac{1}{2}-\dfrac{1}{2}u^{-1}\right]\text{du}$$
$$=\dfrac{1}{2}x-\dfrac{1}{2} \ln \left|2x+1\right|$$
Looking at the answer in the back, this is wrong.
The answer is $x \ln(2x+1)-x+\dfrac{1}{2}\ln(2x+1)+c$.
What have I done wrong?
|
$$
\begin{aligned}
\int \ln (2 x+1) d x &=\frac{1}{2} \int \ln (2 x+1) d(2 x+1) \\
&=\frac{(2 x+1) \ln (2 x+1)}{2}-\frac{1}{2} \int(2 x+1) \frac{2 d x}{2 x+1} \\
&=\frac{(2 x+1) \ln (2 x+1)}{2}-x+C
\end{aligned}
$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Find a closed form for $\sum_{k=0}^{n} k^3$ Find a closed form for $\sum_{k=0}^{n} k^3$.
I would appreciate ideas for approaching questions like this in general as well.
Thanks.
|
First, we know that: $$\sum_{i \mathop = 1}^n i = \frac {n \left({n + 1}\right)} 2.$$ Thus: $$\left({\sum_{i \mathop = 1}^n i}\right)^2 = \frac{n^2 \left({n + 1}\right)^2} 4.$$ Next we use induction on $n$ to show that $$\sum_{i \mathop = 1}^n i^3 = \frac{n^2 \left({n + 1}\right)^2} 4.$$
The base case holds since $$1^3 = \frac{1 \left({1 + 1}\right)^2} 4.$$
Now we need to show that if it holds for $n$, then it holds for $n+1$. $$\eqalign{
\sum_{i \mathop = 1}^n i^3 + \left({n + 1}\right)^3 &= \sum_{i \mathop = 1}^n i^3 + \left({n + 1}\right)^3 \\
&= \frac{n^2 \left({n + 1}\right)^2} 4 + \left({n + 1}\right)^3 \\
&= \frac{n^4 + 2 n^3 + n^2} 4 + \frac {4 n^3 + 12 n^2 + 12 n + 4} 4 \\
&= \frac{n^4 + 6 n^3 + 13 n^2 + 12 n + 4} 4 \\
&= \frac{\left({n + 1}\right)^2 \left({n + 2}\right)^2} 4.
}$$
So by induction we have that: $$\boxed{\ \displaystyle\sum_{i \mathop = 1}^n i^3 = \frac{n^2 \left({n + 1}\right)^2} 4. \ }$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How prove $\frac{1-xy}{\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{(1-x)^2+(1-y)^2}}\le\frac{\sqrt{5}-1}{4}$ Question:
let $x,y\in [0,1]$, show that
$$\dfrac{1-xy}{\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{(1-x)^2+(1-y)^2}}\le\dfrac{\sqrt{5}-1}{4}$$
Thank you (I think this inequality can use Geometric interpretation)
my idea:
$$\Longleftrightarrow 4(1-xy)\le (\sqrt{5}-1)[\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{(1-x)^2+(1-y)^2}]$$
$$\Longleftrightarrow (\sqrt{5}-1)[\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{(1-x)^2+(1-y)^2}]+4xy\ge 4$$
then I can't prove it.
Thank you
|
The in equality becomes $\sqrt{1+a^2}+\sqrt{1+b^2}+\sqrt{(1-a)^2+(1-b)^2}\geq (1+\sqrt{5})(1-ab)$
We'll prove that
$\sqrt{(1-a)^2+(1-b)^2}+\sqrt{5}ab\geq \sqrt{1+(1-a-b)^2}$ (1)
$\sqrt{1+a^2}+\sqrt{1+b^2}+ab\geq 1+\sqrt{1+(a+b)^2}$ (2)
To prove (1) you just need to square it
To prove (2), set :
A=$1+\sqrt{1+(a+b)^2}+\sqrt{1+a^2}+\sqrt{1+b^2}$
B=$\sqrt{1+(a+b)^2}+\sqrt{(1+a^2)(1+b^2)}$
We have :
$1+\sqrt{1+(a+b)^2}-\sqrt{1+a^2}-\sqrt{1+b^2}
=\frac{2ab+2[\sqrt{1+(a+b)^2}-\sqrt{(1+a^2)(1+b^2)}]}{A}
=\frac{2ab+\frac{2ab(2-ab)}{B}}{A}\le \frac{2ab+\frac{4ab}{B}}{A}$
So we only need to prove :
$\frac{1}{A}\geq\frac{2}{B}+1$
This is right because $A\geq4$ and $B\geq2$
From (1) and (2), the inequation become :
$\sqrt{1+(a+b)^2}+\sqrt{1+(1-a-b)^2} \geq\sqrt{5}$
This is right due to minkowski inequality
$\sqrt{1+(a+b)^2}+\sqrt{1+(1-a-b)^2} \geq\sqrt{(1+1)^2+(a+b+1-a-b)^2}=\sqrt{5}$
|
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|
Prove the sum $\sum_{n=1}^\infty \frac{\arctan{n}}{n}$ diverges. I must prove, that sum diverges, but...
$$\sum_{n=1}^\infty \frac{\arctan{n}}{n}$$
$$\lim_{n \to \infty} \frac{\arctan{n}}{n} = \frac{\pi/2}{\infty} = 0$$
$$\lim_{n \to \infty} \frac{ \sqrt[n]{\arctan{n}} }{ \sqrt[n]{n} } = \frac{1}{1} = 1$$
Cauchy's convergence test undefined.
There is a $E_0\gt0$:
$$\left|\frac{\arctan{n+1}}{n+1} + \frac{\arctan{n+2}}{n+2} + ... + \frac{\arctan{n+p}}{n+p}\right| \ge \frac{\pi}{4}\left|\frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{n+p}\right| \ge \frac{\pi}{4} \frac{p}{n+p} (Let\, p = n) \ge \frac{\pi}{4} \frac{\bcancel{n}}{2\bcancel{n}} = \frac{\pi}{8} (\sim0.4) \ge E_0 = \frac{1}{8} \gt 0;$$
Now am I correct?
|
At the end, you have to take the limit as $n$ and $p$ go to infinity, or, if you prefer, the inequality $|S_{n+p}-S_n|\lt \varepsilon$ has to take place for any $n\geqslant N(\varepsilon)$ and $p\geqslant 0$.
Instead, use the inequality $\arctan n\geqslant \pi /4$ for $n\geqslant 1$.
|
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|
Intersection of ellipse and hyperbola at a right angle Need to show that two functions intersect at a right angle.
Show that the ellipse
$$ \frac{x^2}{a^2} +\frac{y^2}{b^2} = 1
$$
and the hyperbola
$$
\frac{x^2}{α^2} −\frac{y^2}{β^2} = 1
$$
will intersect at a right angle if
$$α^2 ≤ a^2 \quad \text{and}\quad a^2 − b^2 = α^2 + β^2$$
Not sure how to tackle this question, graphing didn't help.
|
Note that if $a^2=b^2+\alpha^2+\beta^2$ holds, then $a^2\ge \alpha^2$ holds since $$a^2=(b^2+\beta^2)+\alpha^2\ge 0+\alpha^2=\alpha^2.$$
There are four intersection points $(x,y)$ where
$$x^2=\frac{a^2\alpha^2 (b^2+\beta^2)}{\alpha^2 b^2+a^2\beta^2},\qquad y^2=\frac{\beta^2b^2(a^2-\alpha^2)}{\alpha^2b^2+a^2\beta^2}=\frac{\beta^2b^2(b^2+\beta^2)}{\alpha^2b^2+a^2\beta^2}\tag1$$
Now
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\implies \frac{yy_{\text{e}}'}{b^2}=-\frac{x}{a^2}$$
$$\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1\implies \frac{yy_{\text{h}}'}{\beta^2}=\frac{x}{\alpha^2}$$
giving
$$\frac{yy_{\text{e}}'}{b^2}\cdot \frac{yy_{\text{h}}'}{\beta^2}=-\frac{x^2}{a^2\alpha^2}\tag2$$
From $(1)(2)$, we have
$$y_{\text{e}}'y_{\text{h}}'=-\frac{x^2}{a^2\alpha^2}\cdot\frac{b^2\beta^2}{y^2}=-\frac{b^2\beta^2}{a^2\alpha^2}\cdot\frac{a^2\alpha^2}{b^2\beta^2}=-1$$
Therefore, the two curves intersect at right angle.
|
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|
Proof by induction that $(1^3 + 2^3 + 3^3+\cdots+n^3) = (1 + 2 + \cdots + n)^2$ I'm sitting with the proof in front of me, but I do not understand it.
$$A = \{n \in Z^{++} \mid (1^3 + 2^3 + 3^3+\cdots+n^3) = (1 + 2 + \cdots + n)^2\}$$
The first step of proof by induction is simple enough,to prove that $1 \in A$
$1^3 = 1^2$
The next step is where I get tripped up. So I add $n + 1$ to the right hand side
$$(1 + 2 + \cdots + n + (n + 1))^2 = (1 + 2 + \cdots + n)^2 + 2(1 + 2 + \cdots + n)(n + 1) + (n + 1)^2$$
My algebra is failing me here, because I do not understand how the equation was expanded.
|
The last step uses the identity:
$$(a + b)^2 = a ^2 + 2ab + b^2$$
So:
$$(\underbrace{1 + 2 + \cdots + n}_a + \underbrace{(n + 1)}_b)^2 =\\= (1 + 2 + \cdots + n)^2 + 2(1 + 2 + \cdots + n)(n+1) + (n + 1)^2\tag{1}$$
By induction hypotesis we have that $(1 + 2 + \cdots + n) ^2 = (1^3 + 2^3 + \cdots + n^3)$. We also use another identity, namely $(1 + 2 + \cdots + n) = n(n+1)/2$.
Combining those two facts and substituting into $(1)$:
$$(1^3 + 2^3 + \cdots + n^3) + 2\times n(n+1)/2 \times (n + 1) + (n+1)^2 = \\(1^3 + 2^3 + \cdots + n^3) + n(n+1)^2 + (n+1)^2 =\\1^3 + 2^3 + \cdots + n^3 + (n + 1)^3$$
$\blacksquare$
|
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|
Partial fraction (doubt) I have this partial fraction
$$\displaystyle\frac{1}{(2+x)^2(4+x)^2}$$
I tried to resolve using this method:
$$\displaystyle\frac{A}{2+x}+\displaystyle\frac{B}{(2+x)^2}+\displaystyle\frac{C}{4+x}+\displaystyle\frac{D}{(4+x)^2}$$
$$1=A(2+x)(4+x)^2+B(4+x)^2+C(4+x)(2+x)^2+D(2+x)^2$$
When x=-2
$$1=B(4-2)^2$$
$$B=\displaystyle\frac{1}{4}$$
When x=-4
$$1=D(2-4)^2$$
$$D=\displaystyle\frac{1}{4}$$
When x=0
$$1=A(2)(16)+B(16)+C(4)(4)+D(4)$$
$$1=A(32)+B(16)+C(16)+D(4)$$
Replacing the values for B y D
$$1=A(32)+4+C(16)+1$$
$$1-4-1=A(32)+c(16)$$
$$-4=A(32)+C(16)$$
How I can get the values of $A$ and $D$?
|
Try another value of $x$ as well, say $x=2$, giving $1 = 144A + 9 + 64C + 4$, or $144A + 96C = -12$. Combined with the equation you already have, $32A+16C=-4$, solving gives $A=-\frac{1}{4}$, $C=\frac{1}{4}$.
|
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|
Factor the Expression completely$ (a+b)^2 - (a-b)^2$ I don't understand this question. The answer in the book is $4ab$, but how is that term a factor?
I was thinking along the line that this was a difference of squares example. $a^2-b^2 = (a+b)(a-b)$
My answer is $[(a+b)-(a-b)][(a+b)+(a-b)]$
What do I not understand?
|
Just See this,
$(a+b)^2 = a^2 + b^2 +2ab$,
$(a-b)^2 = a^2 + b^2 -2ab$
Subtract them,
You will get $4ab$
Ok I will show that , you have already done everything
$(a+b)^2 - (a-b)^2 = [(a+b)-(a-b)][(a+b)+(a-b)]$
$\hspace{2cm} = [a+b-a+b][a+b+a-b]$
$\hspace{2cm} = [a-a+b+b][a+a+b-b]$
$\hspace{2cm} = [2b][2a]$
$\hspace{2cm} = 4ab$
|
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|
Solve trigonometric equation $\sin14x - \sin12x + 8\sin x - \cos13x= 4$ I am trying to solve the trigonometric equation
$$ \sin14x - \sin12x + 8\sin x - \cos13x= 4 $$
The exact task is to find the number of real solutions for this equation on the range $[0, 2\pi]$. Thanks.
|
We can use $\sin{x}-\sin{y}=2\sin{\frac{x-y}{2}}\cos{\frac{x+y}{2}}$.
So, $$(\sin{14x}−\sin{12x})+8\sin{x}−\cos13x=4 \Longleftrightarrow 2\sin{x}\cos{13x}−\cos13x+8\sin{x}-4=0$$ $$\Longleftrightarrow 2\cos{13x}(\sin{x}-\frac{1}{2})+8(\sin{x}-\frac{1}{2})=0 \Longleftrightarrow (\sin{x}-\frac{1}{2})(2\cos{13x}+8)=0$$
So, we have $\sin{x}=\frac{1}{2}$ and $\cos{13x}=-4$(which can't be achieved), now we take just $\sin{x}=\frac{1}{2}$, and we get $x_1=\frac{\pi}{6}$, $x_2=\frac{5\pi}{6}$ on the given interval.
|
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|
Prove that diophantine equation has only two solutions. I am looking at the following exercise:
$$\text{Prove that the diophantine equation } x^4-2y^2=1 \text{ has only two solutions.}$$
That's what I thought:
We could set $x^2=k$,then we would have $k^2-2y^2=1 \Rightarrow 2y^2=k^2-1 \Rightarrow 2 \mid k^2-1$
Can I use this fact or do I have to do it in an other way?
EDIT: Or could we do it maybe like that:
$k^2-2y^2=1 \Rightarrow (k-\sqrt{2}y)(k+\sqrt{2}y)=1 \Rightarrow k-\sqrt{2}y= \pm1 \text{ and } k+\sqrt{2}y=\pm 1 \Rightarrow k-\sqrt{2}y=1 \text{ and } k+\sqrt{2}y=1 \text{ OR } k-\sqrt{2}y=-1 \text{ and } k+\sqrt{2}y=-1$
The second case is rejected,because then we get $k<0$
$$k-\sqrt{2}y=1 \text{ and } k+\sqrt{2}y=1 \Rightarrow k=1 \text{ and } y=0 \text{ OR } k=0 \text{ and } y=\frac{1}{\sqrt{2}} \notin \mathbb{Z}$$
So, $k=1 \Rightarrow x^2=1 \Rightarrow x=\pm 1 \text{ and } y=0$
So,the only two solutions are $(1,0) \text{ and } (-1,0)$
|
You are doing it right. $x^4 - 2 y^2 = 1 \Leftrightarrow x^4 -1 = 2 y^2 \Leftrightarrow (x-1 ) ( x+ 1 ) ( x^2 +1 ) = 2 y^2$.
Now prove that $x$ is odd, then $x = 2 k +1 $, and show that $k ( k+1) ( 2 k^2 + 2 k + 1)$ is a square, and since each factor is relatively prime, $k$ and $k+1$ are squares, conclude that $k=0$ and $x=1$ and $y=0$.
|
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|
How to sum $\frac{1}{9} + \frac{1}{18}+\frac{1}{30}+\frac{1}{45} + ......$ How to sum this series :
$\frac{1}{9} + \frac{1}{18}+\frac{1}{30}+\frac{1}{45} + \frac{1}{65}......$
I am not getting any clue only a hint will be suffice please help. thanks..
|
Observe
\begin{align*}
\frac{1}{9}+\frac{1}{18}+\frac{1}{30}+\frac{1}{45}+\ldots&=\frac{1}{3+3\cdot2}+\frac{1}{3+3\cdot2+3\cdot3}+\frac{1}{3+3\cdot2+3\cdot3+3\cdot4}+\ldots\\
&=\sum_{k=2}^{\infty}{\frac{1}{3\sum_{j=1}^{k}{j}}}\\
&=\sum_{k=2}^{\infty}{\frac{1}{3\left[\frac{k(k+1)}{2}\right]}}\\
&=\sum_{k=2}^{\infty}{\frac{2}{3\left[k(k+1)\right]}}\\
&=\frac{2}{3}\sum_{k=2}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)\\
\end{align*}
Last one is a telescopic series, note we have
$$\frac{2}{3}\sum_{k=2}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)=\frac{2}{3}\left(\frac{1}{2}\right)=\frac{1}{3}$$
|
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|
$a^{12} \equiv 1 \pmod{35}$,knowing that $(a,35)=1$ Prove that $\forall a \text{ with } (a,35)=1:$
$$a^{12} \equiv 1 \pmod{35}$$
$$35 \mid a^{12}-1 \Leftrightarrow 5 \cdot 7 \mid a^{12}-1 \overset{(5,7=1)}{ \Leftrightarrow} 5 \mid a^{12}-1 \text{ and } 7 \mid a^{12}-1$$
Therefore, $\displaystyle{ a^{12} \equiv 1 \pmod{35} \Leftrightarrow a^{12} \equiv 1 \pmod 5, a^{12} \equiv 1 \pmod 7}$
$$(a,35=1) \Rightarrow (a, 5 \cdot 7)=1 \overset{(5,7)=1}{\Rightarrow } (a,5)=1 \text{ and } (a,7)=1$$
According to Fermat's theorem:
$$a^4 \equiv 1 \pmod 5$$
$$a^{12} \equiv (a^4)^3 \equiv 1 \pmod 5$$
Also:
$$a^6 \equiv 1 \pmod 7$$
$$a^{12} \equiv (a^6)^2 \equiv 1 \pmod 7$$
So,we conclude that:
$$a^{12} \equiv 1 \pmod{35}$$
Could you tell me if it is right?
|
Hint : For $(a,5)=1$ only possibilities are : $a=5k+1;5k+2;5k+3;5k+4$
$(5k+1)^{12}\equiv 1\mod5$ case should be clear...
$(5k+2)^{12}\equiv 1\mod 5$ should be clear if you know what $2^{12}$ is...
$(5k+3)^{12}\equiv 1\mod 5$ should be clear if you know what $3^{12}$ is...
$(5k+4)^{12}\equiv 1\mod 5$ should be clear if you know what $4^{12}$ is...
You may repeat this kind of strategy for case of $7$ and conclude what you wanted to...
|
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|
Trigonometric equation with complex numbers Let $x$, $y$, and $z$ be real numbers such that $\cos x+\cos y+\cos z=\sin x+\sin y+\sin z=0$.
Prove that $\cos 2x+\cos 2y+\cos 2z=\sin 2x+\sin 2y+\sin 2z=0$.
Starting with the given equation, I got that $i\sin x+i\sin y+i\sin z=0$.
Adding this to the other part of the given equation, I then got $\cos x+\cos y+\cos z+i\sin x+i\sin y+i\sin z=0$, which can also be written as $(\cos x+i\sin x)+(\cos y+i\sin y)+(\cos z+i\sin z)=0$.
Here, I let $a=e^{ix}$, $b=e^{iy}$, and $c=e^{iz}$, which, after substituting in to the above equation gives $a+b+c=0$.
What we want to prove can be written as $a^2+b^2+c^2$, but I am not quite sure how to find this result from what I have.
Some help would be very appreciated. Thanks!
|
Multiply the equations by $\sin(x+y+z)$ and $\cos(x+y+z)$ and subtract them to get
$$\sin(x+y+z)\cos x -\cos(x+y+z)\sin x +\\
\sin(x+y+z)\cos y -\cos(x+y+z)\sin y +\\
\sin(x+y+z)\cos z -\cos(x+y+z)\sin z\\
=\sin(y+z)+\sin(x+z)+\sin(x+y)=0$$
In a similar way we get
$$\cos(y+z)+\cos(x+z)+\cos(x+y)=0$$
Now if we square both equations and subtract them we have
$$\cos 2x+ \cos 2y+ \cos 2z +2\cos(y+z)+2\cos(x+z)+2\cos(x+y)=0$$ and if we multiply the two equations together we have
$$\sin 2x+ \sin 2y+ \sin 2z +2\sin(y+z)+2\sin(x+z)+2\sin(x+y)=0$$
|
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|
Simplest form of $h'(y)$ given $h(y)= (1-3y^2)^5 \cdot ( y^2 + 2)^6$ Find $h'(y)$ in the simplest form if the $$h(y)= (1-3y^2)^5 \cdot ( y^2 + 2)^6$$
My answer was: $$-30y(1-3y^2)^4 \cdot (y^2+2)^6 + 12y(y^2+2)^5 \cdot (1-3y^2)^5$$
But according to wolfram alpha the answer was $$ -30y(1-3y^2)^4 \cdot (y^2+2)^6 $$ only. which makes it weird that they did not differentiate the second term my answer was different.
I posted it again because I did not get an answer.
|
You differentiated correctly.
Since you have been asked to write the derivative in its simplest form:
you can factor out common factors in each term of your sum:
$$-30y(1-3y^2)^4 \cdot (y^2+2)^6 + 12y(y^2+2)^5(1-3y^2)^5$$ $$ = 6y(1-3y^2)^4(y^2 +2)^5\Big((-5)(y^2+2) + 2(1-3y^2)\Big)$$ $$ = 6y(1-3y^2)^4(y^2 +2)^5\Big(-5y^2-10 +2 -6y^2\Big) $$ $$=6y(1-3y^2)^4(y^2 +2)^5(-11y^2 - 8)$$
|
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|
Locus of vertex of triangle moving inside circle A right triangle with sides $3,4$ and $5$ lies inside the circle $2x^2+2y^2=25$. The triangle is moved inside the circle in such a way that its hypotenuse always forms a chord of the circle. The locus of the vertex opposite to the hypotenuse is ____ ?
A) $2x^2+2y^2=1$
B) $x^2+y^2=1$
C) $x^2+y^2=2$
D) $2x^2+2y^2=5$
From the centre of circle distance to chord is $\displaystyle\frac{5}{2}$ units. From the vertex of triangle, distance to hypotenuse is $2.4$ units. However I am unable to sum up these two parts to obtain the final answer.
|
Some of the results are summarized into the following figure.
$θ = \angle DAX - \angle CAX = sin^{-1} \dfrac {2.5}{2.5√2} – sin^{-1} \dfrac {2.4}{4} = 8.130xxxxxx^0$
In ⊿CAD, $r^2$, the square of the required radius = … by cosine law … = 0.5
Thus, the required equation of the circle is $2x^2 + 2y^2 = 1$
Note:- The strange thing is that such a complicated problem is classified as an MC question. There must be some other easier method in finding the solution.
|
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|
Finding $ \lim_{x \to - \infty} \left( \frac{1 - x^{3}}{x^{2} + 7 x} \right)^{5} $. I have to evaluate the following limit:
$$
\lim_{x \to - \infty} \left( \frac{1 - x^{3}}{x^{2} + 7 x} \right)^{5}.
$$
What I did was to divide both the numerator and the denominator of the expression inside the parentheses by $ \dfrac{1}{x^{3}} $ to get
$$
\left[
\frac{\left( \dfrac{1}{x^{3}} - 1 \right)}{\left( \dfrac{1}{x} + \dfrac{7}{x^{2}} \right)}
\right]^{5}.
$$
As $ x \to - \infty $, each term in the expression above goes to $ 0 $ except for the $ -1 $.
Question: Why is the final limit $ \infty $?
|
Consider dividing top and bottom of $\dfrac{1-x^3}{x^2+7x}$ by $x^2$. We get
$$\frac{\frac{1}{x^2}-x}{1+\frac{7}{x}}.$$
Now let $x$ get large negative. It is clear that the bottom approaches $1$, and that the top blows up.
|
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|
Find all integer solution Find all integer solutions such that $$a+1|2a^2+9$$
Solution. I could solve this by writing $$\frac{2a^2+9}{a+1}=2a-2+\frac{11}{a+1}.$$ So, the only integer solution for the last equation are $a=10, a=-12.$
But, i want to get a solution using divisibility properties.
|
You pretty much have the right idea. Observe that $a+1 | 2(a^2-1)$. Moreover we can write
$$2a^2+9=2(a^2-1)+11.$$
Therefore for $a+1$ to divide $2a^2+9$, it has to divide $11$ (because $11$ is the difference of two expression both of which are divisible by $a+1$).
Thus the only values of $a+1$ are all divisors of $11$, namely $\pm 1, \pm11$. Thus $4$ solutions in all.
|
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|
Is $\mathbb{Q}\left( \sqrt[3]{2}, \frac{-1 + i\sqrt{3}}{2}\right):\mathbb{Q}$ a simple extension? Is the extension $$\mathbb{Q}\left( \sqrt[3]{2}, \frac{-1 + i\sqrt{3}}{2}\right):\mathbb{Q}$$ simple? If so find the minimal polynomial and the basis for the extension.
|
This extension can be written as $\mathbb{Q}(\sqrt[3]{2}, \zeta_3)$, where $\zeta_3 = e^{2\pi i /3}$. This is a degree 6 extension.
$$p(x) = x^6+3 x^5+6 x^4+3 x^3+9 x+9$$ is the minimal polynomial for $\sqrt[3]{2}+ \zeta_3$. Since this polynomial is degree 6, its roots form number fields of degree 6 over $\mathbb{Q}$ contained in $\mathbb{Q}(\sqrt[3]{2}, \zeta_3)$.
But this actually implies that $\mathbb{Q}(\sqrt[3]{2}, \zeta_3) = \mathbb{Q}(\sqrt[3]{2}+ \zeta_3)$. This gives us our simple extension.
As Rene Schipperus notes, this holds true for any finite extension of a field of characteristic $0$. If we let $\theta = \sqrt[3]{2}+ \zeta_3$ then $1,\theta, \theta^2,\ldots, \theta^5$ should form a basis for the extension.
If we sub in for $\theta$ in the minimal polynomial, we get
$$p(\theta) = 63 + 45 \sqrt[3]{2} + 36 (\sqrt[3]{2})^2 + 63\zeta^2 +
45 \sqrt[3]{2}\zeta^2 \\~~~~~~~~~~+ 36 (\sqrt[3]{2})^2\zeta^2 +
63 \zeta + 45 \sqrt[3]{2}\zeta +
36 (\sqrt[3]{2})^2 \zeta$$
Note that we can group terms as follows:
$$63(1 + \zeta + \zeta^2)\cdot 45\sqrt[3]{2}(1 + \zeta + \zeta^2)\cdot 36(\sqrt[3]{2})^2(1 + \zeta + \zeta^2)$$
Note that since $1 + \zeta + \zeta^2 = 0$ (Look up cyclotomic polynomials), we have that the above product is $0$. Hence $p(\theta) = 0$.
|
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Math probability combination explanation A group of four components is known to contain two defectives. An inspector tests the components one at a time until the two defectives are located. once she locates the two defectives, she stops testing, but the second defective is tested to ensure accuracy. let Y denote the number of the test on which the second defective is found. Find the probability distribution for Y.
I know the answers are
P[2] = 2/4 1/3 = 1/6
P[3] = 2c1(2/4 *2/3) * 1/2 = 1/3
P[4] = 3c2*(2/4 * 1/3 * 2/2) *1/1 = 1/2
But could someone please explain where those fractions came from?
|
The numbers are small, so we do not need general theory to do the calculation.
The random variable $Y$ can take on values $2$, $3$, or $4$.
$\Pr(Y=2)$: We need to have the first defective, and the second. The probability the first is defective is $\frac{2}{4}$. Given the first was defective, there are $3$ items left, of which $1$ is defective. So the probability the second is defective given the first was is $\frac{1}{3}$, and therefore $\Pr(Y=2)=\frac{2}{4}\cdot\frac{1}{3}$.
$\Pr(Y=3)$: This can happen in $2$ ways: (i) BGB (bad, good, bad) or GBB. By the same reasoning as above, the probability of BG is $\frac{2}{4}\cdot\frac{2}{3}$. If we started with BG, the probability the next is B is $\frac{1}{2}$. So the probability of BGB is $\frac{2}{4}\cdot\frac{2}{3}\cdot\frac{1}{2}$. Similar reasoning shows the probability of GBB is $\frac{2}{4}\cdot\frac{2}{3}\cdot\frac{1}{2}$, the same. So we need to multiply $\frac{2}{4}\cdot\frac{2}{3}\cdot\frac{1}{2}$ by $2$.
$\Pr(Y=4)$: This happens with the patterns GGBB, GBGB, and BGGB. It turns out that each has probability $\frac{2}{4}\cdot \frac{1}{3}$, giving probability $3\cdot\frac{2}{4}\cdot\frac{1}{3}$.
Remark: The probabilities must add up to $1$, so when we have found two of them, the other can be found by arithmetic. In this case, the probabilities simplify to $\frac{1}{6}$, $\frac{1}{3}$, and $\frac{1}{2}$.
There is a nice formula for the general case, where we have $b$ bad and $g$ good, and want to find the probability that the $m$-th bad is found on the $n$-th trial. However, since we only had three cases, direct calculation is easier.
|
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|
expected value and variance of the difference of number of people in a row. I need to calculate the expected value and the variance of the following variable:
$n$ people sit in a row, among them person 'a' and person 'b'.
Define $X$ to be the amount of people between 'a' and 'b'. Calculate $E(X)$ and $Var(X)$.
I have tried to calculate $P(X=k)$, and all I managed to find is $P(X=0)= \frac{n}{2}$ which I dont know if is true. And besides that, I guess that calculating the expectation and the variance directly will be easier than finding the whole distribution, but I simply don't know how to do that.
|
$P\left(X=k\right) = \frac{\mbox{#positions of A and B with $k$ seats between them}}{\mbox{#total positions for A and B}}$.
When A is to the left of B, there are $n-k-1$ possible positions they can have with $k$ seats between them. When A is to the right of B, this is a mirror image so there is a further $n-k-1$ positions. Total such positions is then $2\left(n-k-1\right)$.
Total number of positions for A and B is $n\left(n-1\right)$ because there are $n$ choices for A's seat and then $n-1$ choices for B's seat.
So, $P\left(X=k\right) = \frac{2\left(n-k-1\right)}{n\left(n-1\right)}$.
Then,
\begin{eqnarray*}
E\left[X\right] &=& \sum_{k=0}^{n-2}{kP\left(X=k\right)} \\
&& \\
&=& \sum_{k=0}^{n-2}{\frac{2k\left(n-k-1\right)}{n\left(n-1\right)}} \\
&& \\
&=& \frac{2}{n\left(n-1\right)} \left[\left(n-1\right) \sum_{k=0}^{n-2}{k} - \sum_{k=0}^{n-2}{k^2} \right] \\
&& \\
&=& \frac{2}{n\left(n-1\right)} \left[\left(n-1\right) \frac{\left(n-2\right)\left(n-1\right)}{2} - \frac{\left(n-2\right)\left(n-1\right)\left(2n-3\right)}{6} \right] \\
&& \\
&=& \frac{1}{3n} \left[3\left(n-2\right)\left(n-1\right) - \left(n-2\right)\left(2n-3\right) \right] \\
&& \\
&=& \frac{1}{3n} \left[3n^2 - 9n + 6 - \left(2n^2 - 7n + 6\right) \right] \\
&& \\
&=& \frac{1}{3n} \left[n^2 - 2n \right] \\
&& \\
&=& \frac{n-2}{3}.
\end{eqnarray*}
And
\begin{eqnarray*}
E\left[X^2\right] &=& \sum_{k=0}^{n-2}{k^2 P\left(X=k\right)} \\
&& \\
&=& \sum_{k=0}^{n-2}{\frac{2k^2 \left(n-k-1\right)}{n\left(n-1\right)}} \\
&& \\
&=& \frac{2}{n\left(n-1\right)} \left[\left(n-1\right) \sum_{k=0}^{n-2}{k^2} - \sum_{k=0}^{n-2}{k^3} \right] \\
&& \\
&=& \frac{2}{n\left(n-1\right)} \left[\left(n-1\right) \frac{\left(n-2\right)\left(n-1\right)\left(2n-3\right)}{6} - \frac{\left(n-2\right)^2\left(n-1\right)^2}{4} \right] \\
&& \\
&=& \frac{\left(n-2\right)\left(n-1\right)}{6n} \left[2\left(2n-3\right) - 3\left(n-2\right) \right] \\
&& \\
&=& \frac{\left(n-2\right)\left(n-1\right)}{6}.
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
Var\left(X\right) &=& E\left[X^2\right] - E\left[X\right]^2 \\
&& \\
&=& \frac{\left(n-2\right)\left(n-1\right)}{6} - \frac{\left(n-2\right)^2}{9} \\
&& \\
&=& \frac{n-2}{18} \left(3n - 3 - \left(2n - 4\right) \right) \\
&& \\
&=& \frac{\left(n-2\right) \left(n+1\right)}{18}.
\end{eqnarray*}
Identities used to evaluate the sums above.
|
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|
L'Hôpital's Rule and Infinite Limits I was wondering if anyone could help me with computing a limit using L'Hôpital's Rule.
Using L'Hôpital Rule for the following limit, I get the following result:
\begin{equation}
\lim_{x \to 0} \frac{e^x-1-x^2}{x^4+x^3+x^2} \therefore \lim_{x \to 0} \frac{e^x-1-x^2}{x^4+x^3+x^2} \stackrel{L'H}{=} \lim_{x \to 0} \frac{e^x-2x}{4x^3+3x^2+2x}
\end{equation}
(1/0) which means that the limit is either $+\infty$, $-\infty$ or it does not exist.
If I want to find out the answer, I should take the side limits - but here lies my question: the side mimits of what? The original function or the function after L'Hôpital rule? Or any of the two?
Should I do this
\begin{equation}
\lim_{x \to 0^-} \frac{e^x-2x}{4x^3+3x^2+2x} = -\infty
\end{equation}
\begin{equation}
\lim_{x \to 0^+} \frac{e^x-2x}{4x^3+3x^2+2x} = +\infty
\end{equation}
Or this?
\begin{equation}
\lim_{x \to 0^-} \frac{e^x-1-x^2}{x^4+x^3+x^2} = -\infty
\end{equation}
\begin{equation}
\lim_{x \to 0^+} \frac{e^x-1-x^2}{x^4+x^3+x^2} = +\infty
\end{equation}
Coincidentally (or not) both limits (from the right of both equations and from the left of both equations) give the same answer. Therefore, the limit does not exist.
For the other function below, the same thing happens:
\begin{equation}
\lim_{x \to 0} \frac{sin(x)}{x^4+x^3}
\end{equation}
So what I dind't find out is which function I should use, or if it really doens't matter.
Thank you
|
The right way to proceed is the following from what you have found
$$\lim_{x \to 0^+} \frac{e^x-2x}{4x^3+3x^2+2x} = \infty \implies \lim_{x \to 0^+} \frac{e^x-1-x^2}{x^4+x^3+x^2} = \infty$$
$$\lim_{x \to 0^-} \frac{e^x-2x}{4x^3+3x^2+2x} = -\infty \implies \lim_{x \to 0^-} \frac{e^x-1-x^2}{x^4+x^3+x^2} = -\infty$$
where the LHS are evaluated by l'Hospital's rule and the RHS are derived form this result.
Therefore in this case we conclude that limit doesn't exist.
For the other one we can also proceed as follows
$$\frac{\sin(x)}{x^4+x^3}=\frac{\sin(x)}{x}\frac{x}{x^4+x^3}=\frac{\sin(x)}{x}\frac{1}{x^3+x^2} \to 1 \cdot \infty=\infty$$
|
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|
Evaluation of $ \int \tan x\cdot \sqrt{1+\sin x}dx$ Calculation of $\displaystyle \int \tan x\cdot \sqrt{1+\sin x}dx$
$\bf{My\; Try::}$ Let $\displaystyle (1+\sin x)= t^2\;,$ Then $\displaystyle \cos xdx = 2tdt\Rightarrow dx = \frac{2t}{\sqrt{2-t^2}}dt$
So Integral is $\displaystyle = \displaystyle 2\int \frac{t^2}{\sqrt{2-t^2}} \frac{(t^2-1)}{\sqrt{2-t^2}}dt = 2\int\frac{t^4-t^2}{2-t^2}dt $
Now How Can I solve after that
Help me
Thanks
|
I think you made some mistakes is the substitution,
$$t^2=\sin x +1$$ so
$$2tdt=\cos x dx$$
now $$\frac{\tan x}{\cos x}= \frac{\sin x}{\cos^2 x}= \frac{\sin x}{1-\sin^2 x}=
\frac{t^2-1}{1-(t^2-1)^2 }=\frac{t^2-1}{2t^2-t^4 }$$
So $$\int \tan x \sqrt{\sin x +1} dx= \int \frac{\tan x}{\cos x} \sqrt{\sin x +1} \cos x dx=\int \frac{t^2-1}{2t^2-t^4 } 2t^2dt=
\int 2\frac{t^2-1}{2-t^2 } dt$$
To integrate
$$\int \frac{t^2-1}{t^2-2}dt=\int 1+ \frac{1}{t^2-2}dt =
\int 1+ \frac{1}{2\sqrt{2}} \left( \frac{1}{t-\sqrt{2}} -\frac{1}{t-\sqrt{2}} \right)dt $$
Now we can integrate getting some logarithms.
|
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|
How to calculate $\int\frac{1}{x + 1 + \sqrt{x^2 + 4x + 5}}\ dx$? How to calculate $$\int\frac{1}{x + 1 + \sqrt{x^2 + 4x + 5}}dx?$$ I really don't know how to attack this integral. I tried $u=x^2 + 4x + 5$ but failed miserably. Help please.
|
\begin{align}
\int\frac1{x+1+\sqrt{x^2+4x+5}}\ dx&=\int\frac1{x+1+\sqrt{(x+2)^2+1}}\ dx\\
&\stackrel{\color{red}{[1]}}=\int\frac{\sec^2y}{\sec y+\tan y-1}\ dy\\
&\stackrel{\color{red}{[2]}}=\int\frac{\sec y}{\sin y-\cos y+1}\ dy\\
&\stackrel{\color{red}{[3]}}=\int\frac{1+t^2}{t(1+t)(1-t^2)}\ dt\\
&\stackrel{\color{red}{[4]}}=\int\left[\frac1{t}-\frac1{2(t+1)}-\frac1{2(t-1)}-\frac{1}{(t+1)^2}\right]\ dt.
\end{align}
The rest is yours.
Notes :
$\color{red}{[1]}\;\;\;$Put $x+2=\tan y\;\Rightarrow\;dx=\sec^2y\ dy$.
$\color{red}{[2]}\;\;\;$Multiply by $\dfrac{\cos y}{\cos y}$.
$\color{red}{[3]}\;\;\;$Use Weierstrass substitution, $\tan\frac{y}{2}=t$.
$\color{red}{[4]}\;\;\;$Use partial fractions decomposition.
|
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|
How to calculate the integral $\int_{-1}^{1}\frac{dz}{\sqrt[3]{(1-z)(1+z)^2}}$? The integral is $I=\displaystyle\int_{-1}^{1}\dfrac{dz}{\sqrt[3]{(1-z)(1+z)^2}}$. I used Mathematica to calculate, the result was $\dfrac{2\pi}{\sqrt{3}}$, I think it may help.
|
Rearrange the integral
$$\int \frac{1}{1+z}\sqrt[3]{\frac{1+z}{1-z}}dz$$
Now set $$t^3=\frac{1+z}{1-z}$$
so
$$z=\frac{t^3-1}{t^3+1}=1-\frac{2}{t^3+1}$$
$$dz=\frac{6t^2}{(t^3+1)^2}dt$$
And the integral becomes
$$\int \frac{1}{1+z}\sqrt[3]{\frac{1+z}{1-z}}dz=\int \frac{t^3+1}{2t^3}t\frac{6t^2}{(t^3+1)^2}dt=\int \frac{3}{t^3+1}dt
$$
Taking the limits into account,
$$\int_{-1}^1 \frac{1}{1+z}\sqrt[3]{\frac{1+z}{1-z}}dz=\int_0^{\infty} \frac{3}{t^3+1}dt
$$
Note that
$$\frac{3}{t^3+1}=\frac{1}{t+1}-\frac{t-2}{t^2-t+1}=
\frac{1}{t+1}-\frac{1}{2}\frac{2t-1}{t^2-t+1}+\frac{1}{2}\frac{3}{t^2-t+1}$$
Now
$$\int_0^{\infty} \frac{1}{t+1}-\frac{1}{2}\frac{2t-1}{t^2-t+1}dt=\ln\left(\frac{t+1}{\sqrt{t^2-1+1}}\right) |_0^{\infty}=0$$
(how do you make the long line for the integral ?)
That leaves $$\int_0^{\infty} \frac{1}{2}\frac{3}{t^2-t+1}dt
=\frac{3}{2} \int_0^{\infty} \frac{1}{(t-\frac{1}{2})^2+\frac{3}{4}}dt
=\sqrt{3} \arctan \frac{2}{\sqrt{3}} (t-\frac{1}{2})|_0^{\infty}=
\sqrt{3}(\arctan (\infty)-\arctan (-\frac{1}{\sqrt3})=\sqrt{3}(\frac{\pi}{2}+\frac{\pi}{6})=\frac{2\pi}{\sqrt{3}}$$
|
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|
Find the minimum of $\displaystyle \frac{1}{\sin^2(\angle A)} + \frac{1}{\sin^2(\angle B)} + \frac{1}{\sin^2(\angle C)}$ Is it possible to find the minimum value of $E$ where
$$E = \frac{1}{\sin^2(\angle A)} + \frac{1}{\sin^2(\angle B)} + \frac{1}{\sin^2(\angle C)}$$for any $\triangle ABC$.
I've got the feeling that $\min(E) = 4$ and that the critical value occurs when $ABC$ is equilateral.
|
fix $A$, and now we have:
$$\frac{1}{\sin^2(B)}+\frac{1}{\sin^2(C)}\geq \frac{2}{\sin^2\left(\frac{B+C}{2}\right)}$$
becaues $\dfrac{1}{\sin^2(x)}$ is a convex function over $[0,\pi]$. thus when we fix an angle like $A$ it is better that two other angles be equal. Now the function $\dfrac{1}{\sin^2(A)}+\dfrac{1}{\sin^2(B)}+\dfrac{1}{\sin^2(C)}$ is defined over
$$X=\{(A,B,C)| A,B,C \geq 0 , A+B+C=\pi\}$$
that is a compact subest of $\mathbb{R}^3$. and this continous function will get its minimum over its domain. now if minimum takes in a place like $(A,B,C)$ then if we have $B\neq C$ then by the above we can find another minimum that is less than this minimum. thus $B=C$ and $A=B$ too. thus $A=B=C=\dfrac{\pi}{3}$ and minimum is $4$.
|
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|
Find the exact value of $\sin (\theta)$ and $\cos (\theta)$ when $\tan (\theta)=\frac{12}{5}$ So I've been asked to find $\sin(\theta)$ and $\cos(\theta)$ when $\tan(\theta)=\cfrac{12}{5}$; my question is if $\tan (\theta)=\cfrac{\sin (\theta) }{\cos (\theta)}$ does this mean that because $\tan (\theta)=\cfrac{12}{5}$ then $\sin (\theta) =12$ and $\cos(\theta)=5$?
It doesn't seem to be the case in this example, because $\sin (\theta)\ne 12 $ and $\cos (\theta)\ne 12 $.
Can someone tell me where the error is in my thinking?
|
The error is that $\tan\theta$ merely gives the ratio of $\sin\theta$ and $\cos\theta$. If it is given as a fraction, that does not mean that the numerator must be $\sin\theta$ and the denominator must be $\cos\theta$. It is necessary to use trigonometric identities to solve for $\sin\theta$ and $\cos\theta$.
We have
$$1+\tan^2\theta=\sec^2\theta$$
so
$$\cos\theta=\sqrt[]{\frac{1}{1+\tan^2\theta}}=\sqrt[]{\frac{1}{1+(\frac{12}{5})^2}}=\sqrt[]{\frac{25}{169}}=\pm\frac{5}{13}$$
Then by the trigonometric identity $\sin^2\theta+\cos^2\theta=1$ we have
$$\sin\theta=\sqrt[]{1-\cos^2\theta}=\sqrt[]{1-\frac{25}{169}}=\sqrt[]{\frac{144}{169}}=\pm\frac{12}{13}$$
Now since $\tan\theta>0$, $\theta$ is in either the first or third quadrant, so $\sin\theta$ and $\cos\theta$ have the same sign. So
$$(\sin\theta,\cos\theta)=(\frac{12}{13},\frac{5}{13}) \text{ or } (\sin\theta,\cos\theta)=(-\frac{12}{13},-\frac{5}{13})$$
Verifying our answer,
$$\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\pm\frac{12}{13}}{\pm\frac{5}{13}}=\frac{12}{5}$$
as the problem stated.
|
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|
How to solve this linear equation? which has an x on each side I have made this equation.
$5x + 8 = 10x + \dfrac{3}{6}$
And I have achieved this result:
$x = 9$
Is my result correct?
I have already posted two other questions related to this topic, I'm a programmer and am learning Math out of my interest, this is not homework.
My steps taken:
10x - 5x = 48 - 3
5x = 45
5x/5 = 45/5
x = 9
This is my steps which led to a wrong answer.
|
$5x + 8 = 10x + 3/6$
Let's gather all the x's on one side and the other stuff on the other side as well as reduce the fraction to lowest terms first:
$10x-5x = 8-\frac{1}{2}$
$5x = 8-.5$
$5x = 7.5$
$x = 1.5$ or $x=\frac{3}{2}$ or $x=1+\frac{1}{2}$
Now, to compute each side as a verfication:
$LHS= 5x+8 = 5*\frac{3}{2}+8 = 7.5+8 = 15.5$
$RHS = 10x+\frac{3}{6} = 10*\frac{3}{2}+\frac{1}{2} = 15+0.5 = 15.5$
|
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|
The sum of three consecutive cubes numbers produces 9 multiple I want to prove that $n^3 + (n+1)^3 + (n+2)^3$ is always a $9$ multiple
I used induction by the way.
I reach this expression: $(n+1)^3 + (n+2)^3 + (n+3)^3$
But is a lot of time to calculate each three terms, so could you help me to achieve the induction formula
Thanks in advance
|
As an alternative to induction, we take any $3$ consecutive cubes as follows:
$$(n-1)^3 + n^3 + (n+1)^3$$
$$= 3n^3 + 6n$$
$$=3n(n^2 +2)$$
Notice that
$$\begin{align}n(n^2 + 2) &\equiv n(n^2-1)\pmod 3 \\&\equiv (n-1)(n)(n+1)\pmod 3 \end{align}$$
Since either one of $(n-1),n$ or $(n+1)$ must be divisible by $3$, it follows that$3|n(n^2+2)$. This implies that $3\cdot3=9$ divides $3n(n^2 +2)$.
|
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|
Evaluation of $\int\sqrt[4]{\tan x}dx$ Evaluation of $\displaystyle \int\sqrt[4]{\tan x}dx$
$\bf{My\; Try}::$ Let $\tan x = t^4\;\;,$ Then $\sec^2 xdx = 4t^3dt$. So $\displaystyle dx = \frac{4t^3}{1+t^8}dt$
So Integral Convert into $\displaystyle 4\int\frac{t^4}{1+t^8}dt = 2\int \frac{(t^4+1)+(t^4-1)}{t^8+1}dt$
So Integral is $\displaystyle 2\int\frac{t^4+1}{t^8+1}dt+2\int\frac{t^4-1}{t^4+1}dt$
Now How can I solve after that
Help me
Thanks
|
Transforming the integral into an integral of rational function by letting $y^{4}=\tan x$, then
$$
I=4 \int \frac{y^{4}}{1+y^{8}} d y
$$
By my post,
$$ \displaystyle \int \frac{x^{4}}{1+x^{8}} d x =\frac{1}{4 \sqrt{2}}\left[\frac{1}{2 \sqrt{2+\sqrt{2}}} \ln \left|\frac{x^{2}-\sqrt{2+\sqrt{2}} x+1}{x^{2}+\sqrt{2+\sqrt{2}} x+1}\right|+\frac{1}{\sqrt{2-\sqrt{2}}} \tan^{-1}\left(\frac{x^{2}-1}{\sqrt{2-\sqrt{2}} x}\right)\\ \quad - \frac{1}{2 \sqrt{2-\sqrt{2}}} \ln \left|\frac{x^{2}-\sqrt{2-\sqrt{2}} x+1}{x^{2}+\sqrt{2-\sqrt{2} x}+1}\right|-\frac{1}{\sqrt{2+\sqrt{2}}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{2+\sqrt{2}} x}\right)\right]+C$$
Now we can conclude that
$$ \displaystyle \int \sqrt[4]{\tan x} d x =\frac{1}{ \sqrt{2}}\left[\frac{1}{2 \sqrt{2+\sqrt{2}}} \ln \left|\frac{\sqrt{\tan x} -\sqrt{2+\sqrt{2}} \sqrt[4]{\tan x} +1}{\sqrt{\tan x} +\sqrt{2+\sqrt{2}} \sqrt[4]{\tan x} +1}\right|+\frac{1}{\sqrt{2-\sqrt{2}}} \tan^{-1}\left(\frac{\sqrt{\tan x}-1}{\sqrt{2-\sqrt{2}} \sqrt[4]{\tan x}}\right)\\ - \frac{1}{2 \sqrt{2-\sqrt{2}}} \ln \left|\frac{\sqrt{\tan x}-\sqrt{2-\sqrt{2}} \sqrt[4]{\tan x} +1}{\sqrt{\tan x} +\sqrt{2-\sqrt{2} \sqrt[4]{\tan x}}+1}\right|-\frac{1}{\sqrt{2+\sqrt{2}}} \tan ^{-1}\left(\frac{\sqrt{\tan x}-1}{\sqrt{2+\sqrt{2}} \sqrt[4]{\tan x}}\right)\right]+C$$
:|D Wish you enjoy the solution!
|
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"url": "https://math.stackexchange.com/questions/860315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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|
Reliability Probability problem What is the Probability that at least one close path is formed from $A$ to $B$ where each switch has a probability of close $p$ and each switch is mutually independent of each other?
Proposed Solution
Let event $A$ be such that
$$\begin{align}
A &= \{\text{Current Flows From A to B}\} \\
&= \{ S_0^c S_1^c S_2^c S_3^c, S_0^o S_1^o S_2^c S_3^c , S_0^c S_1^o S_2^c S_3^c, S_0^o S_1^c S_2^c S_3^c, S_0^c S_1^c S_2^o S_3^c \} \\
\end{align}$$
where $S_0^c$ or $S_0^o$ shows the switch $S_0$ Close or Open respectively. Now the probability of $A$ is
$$\begin{align}
p[A] &= p^4 + p^2(1-p)^2 + p^3(1-p) + p^3(1-p) + p^3(1-p) \\
&= p^2(1+pq)\\
\end{align}$$
where $q = 1-p$.
|
Notice that
$$\begin{align}
p[A] &= p^4 + p^2(1-p)^2 + p^3(1-p) + p^3(1-p) + p^3(1-p) \\
&= p^4+3p^3(1-p)+p^2(1-p)^2\\
&= -p^4+p^3+p^2
\end{align}$$
where $q = 1-p$ and $-p^4+p^3+p^2=p^2(1+pq)$ so as others nocited: your calculation for the event probability is correct.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/860522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Convergence of an improper integral $\int_0^\infty \frac {1-e^{-x}} {\sqrt x^3}dx$ I need to find if
$$\int_0^\infty \frac {e^{-x}-1} {\sqrt x^3}dx$$
converges.
Let $$f(x) = \int_0^\infty \frac {1- e^{-x}} {\sqrt x^3}dx \ge 0$$
$$\int_0^\infty \frac {1-e^{-x}} {\sqrt x^3}dx = \int_0^1 \frac {1- e^{-x}} {\sqrt x^3}dx + \int_1^\infty \frac {1-e^{-x}} {\sqrt x^3}dx$$
The integral
$$\int_1^\infty \frac {1-e^{-x}} {\sqrt x^3}dx$$
converges by applying the comparison test:
$$\lim_{x \to \infty}\frac {\frac {1-e^{-x}} {\sqrt {x^3}}} {\frac 1 {\sqrt {x^3}}} = \lim_{x \to \infty}1- e^{-x}=1$$
so $f(x)$ converges iff $$\int_1^\infty \frac 1 {x^{\frac 3 2}}dx$$
converges.
the problem i have is with $\int_0^1 f(x)dx$
i tried to use the same comparison test but got limit 0, so divergence of
$$\int_0^1 \frac 1 {x^{\frac 3 2}}dx$$
doesn't indicate that $\int_0^1 f(x)dx$ diverges.
how do i handle $\int_0^1 f(x)dx$ ?
|
Given
$$
\int_0^\infty \frac{ \exp(-x) - 1 }{ \sqrt{x}^3 } dx.
$$
Use substitution $x = y^2$, then we obtain
$$
\begin{eqnarray}
\int_0^\infty \frac{ \exp(-x) - 1 }{ \sqrt{x}^3 } dx &=& 2 \int_0^\infty \frac{ \exp(-y^2) - 1 }{ y^2 } dy\\
&=& \color{red}{\left[ - 2 \frac{ \exp(-y^2) - 1 }{y} \right]_0^\infty}
- \color{blue}{\int_0^\infty \exp(-y^2) dy}\\
&=& \underbrace{\color{red}{- 2 \lim_{y \rightarrow 0} \frac{ \exp(-y^2) - 1 }{y}}}_{\textit{use l'Hôpital}} - \color{blue}{2 \sqrt{\pi}}\\
&=& \color{red}{4 \lim_{y \rightarrow 0} \exp(-y^2)} - \color{blue}{2 \sqrt{\pi}}\\
&=& \color{blue}{- 2 \sqrt{\pi}}.
\end{eqnarray}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/863694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Series Convergence/Divergence $\frac{n^n}{(n+1)^{n+1}}$ Trying to establish whether $\sum x_n$ for $x_n := \frac{n^n}{(n+1)^{n+1}}$ converges or diverges. Here's what I've done so far:
1) n-th term: $x_n < \frac{n^n}{n^{n+1}} = \frac{1}{n}$, so $\lim(x_n) = 0$; n-th term test inconclusive.
2) quotient test: $$
\begin{eqnarray*}
\lim \frac{x_{n+1}}{x_n} &=&\lim \frac{(n+1)^{n+1}/(n+2)^{n+2}}{n^n/(n+1)^{n+1}} \\
&=&\lim \frac{(n+1)^{2n+2}}{(n+2)^{n+2}n^n}\\
&=&\lim \left(\frac{n+1}{n+2}\right)^{n+2}\left(\frac{n+1}{n}\right)^{n}\\
&=& \lim \left(\frac{n+1}{n+2} \frac{n+1}{n}\right)^{n}\\
&=& \lim \left(\frac{n^2+2n+1}{n^2+2n}\right)^{n}\\
&=& \lim \left(1+ \frac{1}{2n+n^2}\right)^{n}
\end{eqnarray*}
$$
I am not sure where to go from here. This looks a lot like the classic e limit $\lim (1+1/n)^n$, but it is sufficiently different that I cannot see how to transform it.
3) Root Test:
$$
\begin{eqnarray*}
\lim x_n^{1/n} &=& \frac{n}{(n+1)^{1+1/n}} \\
&=& \lim \frac{n}{n+1} \frac{1}{(n+1)^{1/n}}
\end{eqnarray*}
$$
And again stuck here.
Thanks for your help,
Best wishes,
Leon
|
It's not hard to see that $(1+\frac{1}{n})^{-n}> \frac{1}{e}$, so $(1+\frac{1}{n})^{-n}\frac{1}{n+1}>\frac{1}{e}\frac{1}{n+1}$. But $\sum \frac{1}{e}\frac{1}{n+1}$ diverges so the original series diverges too.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/864278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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|
What is the closed form for $\sum_{n=1}^\infty \frac1n - \frac1{n+1/p}$? A while ago, I started to look at expressions of the following form:
$$
S_p:=\sum_{n=1}^\infty \frac1n - \frac1{n+1/p},
$$
where $p$ is prime, because otherwise things get too complicated for me at the moment.
What I found so far is the following:
$$
S_p= p -\log(2p) - \frac12 \pi \cot\left(\frac\pi p\right) + T_p,
$$
where $T_p$ contains $\frac{p-1}2$ terms $t_{p,k}$ of the form:
\begin{cases}
\pm 2 \sin\left(\frac{k\pi}{2p}\right) \log\left(\sin\left(\frac{k\pi}{2p}\right)\right) \\
\pm 2 \cos\left(\frac{k\pi}{2p}\right) \log\left(\sin\left(\frac{k\pi}{2p}\right)\right)\\
\pm 2 \sin\left(\frac{k\pi}{2p}\right) \log\left(\cos\left(\frac{k\pi}{2p}\right)\right)
\\
\pm 2 \cos\left(\frac{k\pi}{2p}\right) \log\left(\cos\left(\frac{k\pi}{2p}\right)\right)
\end{cases}
An example can be found here.
How does the closed form for $\sum_{n=1}^\infty \frac1n - \frac1{a+n}$ look like? (EDIT: where $a=1/p$)
|
First, for any $z \in \mathbb{C}$ with $|z| < 1$, we have following expansion
$$-\log(1-z) = \sum_{n=1}^\infty \frac{z^n}{n}$$
Let $\displaystyle\;\omega = e^{\frac{i2\pi}{p}}\;$ be the primitive $p$-root of unity, we know for any integer $n$,
$$\frac{1}{p}\sum_{\ell=0}^{p-1} (\omega^\ell)^n = \delta_p(n)
\stackrel{def}{=}
\begin{cases}1,& n \equiv 0\pmod p\\0,& \text{otherwise}\end{cases}$$
This implies for any integer $k \in \{\;1,2,\ldots,p\;\}$, we have
$$-\frac{1}{p}\sum_{\ell=0}^{p-1}\omega^{-k\ell}\log(1-z\omega^\ell)
= \frac{1}{p}\sum_{\ell=0}^{p-1}\sum_{n=1}^\infty \frac{z^n}{n}( \omega^\ell )^{n-k}
= \sum_{n=1}^\infty \delta_p(n-k)\frac{z^n}{n}
= \frac{1}{p}\sum_{n=0}^\infty \frac{z^{np+k}}{n+\frac{k}{p}}
$$
This leads to
$$\sum_{n=0}^\infty\left(\frac{1}{n+1} - \frac{1}{n+\frac{k}{p}}\right)z^{np}
= - \sum_{\ell=0}^{p-1}\left(z^{-p} - z^{-k}\omega^{-k\ell}\right)\log(1-z\omega^\ell)
\tag{*1}$$
Since the sequences $\displaystyle\;\frac{1}{n+1} - \frac{1}{n+\frac{k}{p}} \sim O\left(\frac{1}{n^2}\right)\;$, the series on the LHS converges absolutely as $z \to 1^{-}$.
The RHS is a finite sum, the only term that may cause trouble is the term for $\ell = 0$.
It is clear the logarithm singularity from the $\log(1-z)$ get killed by the
$z^{-p} - z^{-k}$ factor as $z \to 1^{-}$. This allow us to take the limit $z \to 1^{-}$ in
$(*1)$ and get
$$\mathcal{S}_{k/p} \stackrel{def}{=}
\sum_{n=1}^\infty\left(\frac{1}{n} - \frac{1}{n+\frac{k}{p}}\right)
= \frac{p}{k} + \sum_{n=0}^\infty\left(\frac{1}{n+1} - \frac{1}{n+\frac{k}{p}}\right)\\
= \frac{p}{k} - \sum_{\ell=1}^{p-1}\left(1 - \omega^{-kl}\right)\log(1-\omega^\ell)
$$
Using the identities
$$\sum_{\ell=1}^{p-1}\log(1-\omega^\ell) = \log p
\quad\text{ and }\quad
\sum_{\ell=1}^{p-1}\omega^\ell = -1$$
The sum reduces to
$$\begin{align}
&\frac{p}{k} - \log(2p) + \sum_{l=1}^{p-1}\omega^{-k\ell}\log\left(\frac{1-\omega^\ell}{2}\right)\\
= & \frac{p}{k} - \log(2p) + \sum_{l=1}^{p-1}
\left( \cos\left(\frac{2\pi k\ell}{p}\right) - i\sin\left(\frac{2\pi k\ell}{p}\right)
\right)\left(\log\sin\left(\frac{\ell\pi}{p}\right) + i\pi\left(\frac{\ell}{p}-\frac12\right)\right)
\end{align}
$$
Using another set of trigonometric identities
$$\sum_{\ell=1}^{p-1}\frac{\ell}{p}\sin\left(\frac{2\pi k\ell}{p}\right) = -\frac12\cot\left(\frac{k\pi}{p}\right)
\quad\text{ and }\quad
\sum_{\ell=1}^{p-1}\sin\left(\frac{2\pi k\ell}{p}\right) = 0
$$
We finally get
$$
\bbox[4pt,border:1px solid blue]{
\mathcal{S}_{k/p} =
\frac{p}{k} - \log(2p) -\frac{\pi}{2}\cot\left(\frac{k\pi}{p}\right) +
\sum_{l=1}^{p-1}
\cos\left(\frac{2\pi k\ell}{p}\right) \log\sin\left(\frac{\ell\pi}{p}\right)
}
\tag{*2}
$$
On the wiki page of digamma function, there is a formula for digamma function at $\frac{k}{p}$.
$$\psi\left(\frac{k}{p}\right) = -\gamma - \log(2p) - \frac{\pi}{2}\cot\left(\frac{k\pi}{p}\right) + 2\sum_{\ell=1}^{\lfloor\frac{p-1}{2}\rfloor} \cos\left(\frac{2\pi k\ell}{p}\right)\log\sin\left(\frac{\ell\pi}{p}\right)$$
Compare this with what we have in $(*2)$, we can simplify our sum as
$$\mathcal{S}_{k/p} = \frac{p}{k}+\psi\left(\frac{k}{p}\right) + \gamma
= \psi\left(\frac{k}{p}+1\right) + \gamma$$
Replace $\frac{k}{p}$ by $a$, this matches the result derived in another answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/864609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Diagonalizable Matrix real values Let $A=\begin{pmatrix}
a & -b \\
b & a \\
\end{pmatrix}$. For which real values of $a$ and $b$ is $A$ a diagonalizable matrix?
|
Although David's answer already provides a sufficient algebraic explanation, I'd like to give a bit of a geometric one. Consider where this matrix sends the standard orthonormal basis
$$ \begin{pmatrix} a &- b\\ b & a \end{pmatrix}\begin{pmatrix} 1\\ 0\end{pmatrix} = \begin{pmatrix} a\\ b \end{pmatrix} =v_1, \begin{pmatrix} a &- b\\ b & a \end{pmatrix}\begin{pmatrix} 0\\ 1\end{pmatrix} = \begin{pmatrix} -b\\ a \end{pmatrix} =v_2 $$
Note that $v_1, v_2$ are orthogonal with lengths $\sqrt{a^2 + b^2}$. If you look at some cases youu'll probably notice that $v_2$ is always 90 degrees counterclockwise from $v_1$. This, is because if we write $a/b = \tan(\theta)$, and thus have $\sin(\theta) = b/\sqrt{a^2 + b^2}$, and $cos(\theta) = a/\sqrt{a^2 + b^2}$ then we can rewrite
$$ \begin{pmatrix} a &- b\\ b & a \end{pmatrix} = \sqrt{a^2 + b^2}\begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos \theta \end{pmatrix} $$
Which shows that our matrix is the composition of a scaling and a rotation. A rotation has real eigenvectors only if it is by $2\pi, \pi$ (half rotation or full/no rotation) in which case it already diagonal. The scaling of course, will not affect our analysis of diagonalizability.
As a final note, there is a beautiful reason why we should care about matrices of this form. They form a subring of all two by two matrices (that is, they are closed under addition, subtraction and multiplication). And in fact this subring is isomorphic to the complex numbers by the map $$a+bi \mapsto \begin{pmatrix} a &- b\\ b & a \end{pmatrix}$$
And even moreso, the map $\mathbb{R}^2 \to \mathbb{R}^2$ given by these matrices, is the same as the map of multiplication by $a+bi$ from $\mathbb{C} \to \mathbb{C}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/866514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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|
Is $2013^{2014}+2014^{2015}+2015^{2013}+1$ a prime? (usage of a computer not allowed)
Prove or disprove: $$2013^{2014}+2014^{2015}+2015^{2013}+1$$ is a prime number, without using a computer.
I tried to transform the expression $n^{n+1}+(n+1)^{n+2}+(n+2)^{n}+1$, but couldn't reach useful conclusions.
|
For the original version, which asked abuot
$$2014^{2015} + 2015^{2016} + 2016^{2014} + 1$$
then no: This is an even number (sum of two evens and two odds), so not prime. It's also divisible by $3$.
For the new question: The number is divisible by $7$, so it is not prime. This can be computed by hand by reducing everything modulo $7$ and using, e.g. successive squaring to compute the residue on the left side. For example,
\begin{align*}
2013 \equiv 4 &\pmod 7 \\
2013^2 \equiv 2 &\pmod 7 \\
2013^4 \equiv 4 &\pmod 7 \\
2013^8 \equiv 2 &\pmod 7
\end{align*}
and so on; it keeps oscillating. Now after writing $2014$ as a sum of powers of $2$,
\begin{align*}
2013^{2014} &\equiv 4^{2014} \\
&\equiv 4^{1024} \cdot 4^{512} \cdot 4^{256} \cdot 4^{128} \cdot 4^{64} \cdot 4^{16} \cdot 4^8 \cdot 4^4 \cdot 4^2 \\
&\equiv 4\cdot2\cdot4\cdot2\cdot4\cdot4\cdot2\cdot 4 \cdot 2 \\
&\equiv 16384 \equiv 4
\end{align*}
modulo $7$. Hence, the first term is $4$ modulo $7$; similar computations can be done for the other terms, and summing indeed gives $0$.
Alternatively, for a vastly faster computation, Fermat's little theorem implies that $4^6 \equiv 1 \pmod 7$. Writing $2014 = 6 \cdot 335 + 4$, we have
$$2013^{2014} \equiv \Big(4^{6}\Big)^{335} \cdot 4^4 \equiv 4 \pmod 7$$
It's worth mentioning that even without any special form, this is really, really unlikely to be prime. As a consequence of the prime number theorem, if you choose a number on this scale ($\sim 2^{15000}$), then the probability that it's prime is about $1$ in $15000$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/866619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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|
Blocking lines of length $5$ in a $7 \times 8$ matrix; how can we know the solutions have a specific form? A friend shared with me the following puzzle he encountered in a Chinese math competition:
In a $7 \times 8$ matrix, we place tokens so that any straight line of length $5$ (horizontal, vertical, or on either diagonal) intersects at least one token. What is the smallest number of tokens required?
I'm not asking for a solution to this problem. I know the solution, I want to know the following:
Q: I found, by computation, the possible solutions all have a particular form (I list them below). Can we prove this? (In a way other than "I tried all the other possibilities".)
To illustrate the problem,
*
*This arrangement doesn't work
$$\begin{array}{|ccccccc|}
\hline
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \bullet & \bullet & \cdot & \cdot \\
\bullet & \bullet & \bullet & \bullet & \cdot & \bullet & \bullet & \bullet \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\hline
\end{array}$$
since e.g. there is the line
$$\begin{array}{|ccccccc|}
\hline
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\cdot & \cdot & \square & \cdot & \bullet & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \square & \bullet & \bullet & \cdot & \cdot \\
\bullet & \bullet & \bullet & \bullet & \square & \bullet & \bullet & \bullet \\
\cdot & \cdot & \cdot & \cdot & \bullet & \square & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \square & \cdot \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\hline
\end{array}.$$
*This arrangement works
$$\begin{array}{|ccccccc|}
\hline
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\hline
\end{array}$$
and uses $14$ tokens (which is not the minimum).
The solutions are (spoiler alert):
It requires $11$ tokens, realized by these matrices: $\begin{array}{cc} \begin{array}{|ccccccc|} \hline \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot \\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot \\ \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot \\ \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\ \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet \\ \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot \\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot \\ \hline \end{array} & \begin{array}{|ccccccc|} \hline \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot \\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot \\ \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot \\ \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet \\ \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\ \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot \\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot \\ \hline \end{array} \\ \begin{array}{|ccccccc|} \hline \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot \\ \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\ \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet \\ \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot \\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot \\ \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot \\ \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\ \hline \end{array} & \begin{array}{|ccccccc|} \hline \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet \\ \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\ \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot \\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot \\ \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot \\ \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet \\ \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\ \hline \end{array} \\ \begin{array}{|ccccccc|} \hline \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot \\ \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot \\ \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet \\ \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\ \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot \\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot \\ \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot \\ \hline \end{array} & \begin{array}{|ccccccc|} \hline \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot \\ \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot \\ \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\ \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet \\ \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot \\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot \\ \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot \\ \hline \end{array} \\ \begin{array}{|ccccccc|} \hline \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\ \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot \\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot \\ \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot \\ \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet \\ \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\ \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot \\ \hline \end{array} & \begin{array}{|ccccccc|} \hline \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\ \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet \\ \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot \\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot \\ \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot \\ \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\ \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet \\ \hline \end{array} \\ \end{array}$
|
Part 1.
Prove that it is impossible to place $10$ tokens.
Let's try to build with $10$ tokens.
First, note that in each column/row must be at least $1$ token.
Lets call column with token in $1$st cell as column $a$; and row with token in $2$nd cell as column $b$.
Column $a$ and column $b$ must have at least $2$ tokens.
Column $a$ must have cells $1,6$ with tokens; column $b$ must have cells $2,7$ with tokens. (otherwise we'll need to have third column with multiple tokens: more then $10$ tokens at total).
If we want to use only $10$ tokens, then columns $a$ and $b$ need to be placed as $4$th and $5$th columns (otherwise red $5$-rows need to have at least $1$ token, that derives other column $a'$ or $b'$).
Then we have to place $6$ tokens in two $3\times 3$ green squares:
Each $3\times 3$ square must have $3$ tokens in different rows/columns. There are only $6$ such placements for each square. No one of them has tokens on "main" diagonals (see violet diagonals for the left square):
$10$ tokens are too few for such constructing.
Examples with $11$ tokens are shown by author of question.
Part 2. (thanks to Gerry Myerson for helpful comments)
Prove that there are only eight placements of $11$ tokens (all they are shown in the spoiler (see question text).
If we'll use $11$ tokens, then there are $3$ ways:
*
*to use columns with $1,1,1,1,1,1,1,4$ tokens;
*to use columns with $1,1,1,1,1,1,2,3$ tokens;
*to use columns with $1,1,1,1,1,2,2,2$ tokens.
First two cases require to place "rich" columns on the center of $7\times 8$ table (as above), and we'll need to place $7$ tokens in wide green area (as in part $1$). It is impossible (some of four "main" diagonals will be empty).
Third case: there can be
$\qquad$- $3$ "long" columns (like $a$, $b$: with distance between tokens $=5$);
$\qquad$- or $2$ "long" and one "short" column (with distance $\le 4$ between tokens).
If one of columns is "short", then $2$ "long" columns must be on the center (as above), and some of four "main" diagonals will be empty again.
So, we'll consider third case with $3$ "long" columns. Then third case requires to place one of $2$-token columns on the center ($4$th or $5$th column). WLOG, let it is column $b$ as $4$th column. Other possibilities are symmetric to this one (left-right mirroring, up-down mirroring, and both).
There are only $9$ such variants (where we need to place $5$ tokens in the green area):
these $3$ cases are impossible:
$\quad$
$\quad$
these $3$ cases are impossible too:
$\quad$
$\quad$
this one is impossible (at least $3$rd or $5$th column must have two tokens):
and these $2$ cases are possible:
$\qquad$
These cases generate only $2$ possible placements (with column $b$ on the $4$th place).
Applying symmetries, we'll have all $8$ possible placements (shown by author).
|
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"url": "https://math.stackexchange.com/questions/867462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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|
Multidimensional integral involving delta functions The question is to compute the following multidimensional integral:
\begin{equation}
\omega^{(T)}({\bf c}) := \int\limits_{{\mathbb R}^{2 T}}
\delta\left( c_{1,1} - \sum\limits_{j=1}^T x_{1,j}^2 \right)
\delta\left( c_{2,2} - \sum\limits_{j=1}^T x_{2,j}^2 \right)
\delta\left( c_{1,2} - \sum\limits_{j=1}^T x_{1,j} x_{2,j} \right)
\prod\limits_{j=1}^T d x_{1,j} d x_{2,j}
\end{equation}
This is the ``density of states'' of the estimator of covariances in a random matrix ensemble.
Using the definition of delta function and elementary integration I have checked that :
\begin{eqnarray}
\omega^{(1)}({\bf c}) &=& \delta\left(\det(c)\right) \\
\omega^{(2)}({\bf c}) &=& \pi (\det(c))^{-1/2} \\
\omega^{(3)}({\bf c}) &=& \pi^2 1_{\det(c) >0} \\
\omega^{(4)}({\bf c}) &=& \pi^3 \left(\det(c)\right)^{1/2}
\end{eqnarray}
where
\begin{equation}
{\bf c} :=
\left(\begin{array}{cc} c_{1,1} & c_{1,2} \\ c_{1,2} & c_{2,2}
\end{array}
\right)
\end{equation}
The question is what is the result for generic values of $T$. I suspect that the result depends only on the determinant of the matrix ${\bf c}$.
|
Below is the result for $T=2$. We have:
\begin{eqnarray}
\int\limits_{{\mathcal R}^4}
\delta\left(c_{1,1} - x_{1,1}^2 - x_{1,2}^2\right)
\delta\left(c_{2,2} - x_{2,1}^2 - x_{2,2}^2\right)
\delta\left(c_{1,2} - x_{1,1}x_{2,1} - x_{1,2}x_{2,2}\right) d x_{1,1} d x_{1,2} d x_{2,1} d x_{2,2} = \\
\int\limits_{{\mathcal R}^3}
\delta\left(c_{1,1} x_{2,1}^2 - \left({c_{1,2} - x_{1,2} x_{2 2}}\right)^2 - x_{1,2}^2 x_{2,1}^2\right)
\delta\left(c_{2,2} - x_{2,1}^2 - x_{2,2}^2\right) {x_{2,1}}d x_{1,2} d x_{2,1} d x_{2,2} = \\
\frac{1}{2} \int\limits_{{\mathcal R}^2}
\delta\left((c_{1,1}-x_{1,2}^2)(c_{2,2}-x_{2,2}^2) - (c_{1,2}-x_{1,2} x_{2,})^2\right) d x_{1,2} d x_{2,2} = \\
\frac{1}{2} \frac{1}{\det(c)} \int\limits_{{\mathcal R}^2}
\delta\left(1 - \vec{x}^T c^{-1} \vec{x}\right) d^2 x = \\
\frac{\pi}{2} \frac{1}{\sqrt{\det(c)}}
\end{eqnarray}
In the second line we have integrated over $x_{1,1}$ using the definition of the delta function.In the third line we integrated over $x_{2,1}$.In the fourth line we took out the determinant from the inside of the delta function and we wrote the remaining terms as a quadratic form.Finally, we diagonalized the quadratic form and computed the integral by going to radial coordinates.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/868206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How find the value of the $x+y$ Question:
let $x,y\in \Bbb R $, and such
$$\begin{cases}
3x^3+4y^3=7\\
4x^4+3y^4=16
\end{cases}$$
Find the $x+y$
This problem is from china some BBS
My idea: since
$$(3x^3+4y^3)(4x^4+3y^4)=12(x^7+y^7)+x^3y^3(9y+16x)=112$$
$$(3x^3+4y^3)^2+(4x^4+3y^4)^2=9(x^6+y^8)+16(y^6+x^8)+24x^3y^3(1+xy)=305$$
then I can't Continue
|
Assuming the question is typed correctly as shown, there are two unique real solutions. Let $$\begin{align*} u(z) &= -983749-111132z^3+786432z^4+71442z^6-196608z^8-20412z^9+18571z^{12}, \\ v(z) &= -178112-351232z^3+186624z^4+301056z^6-34992z^8-114688z^9+18571z^{12}. \end{align*}$$ These polynomials have exactly two distinct real roots; let $r(u,+)$, $r(u,-)$ be the positive and negative real roots of $u$, and $r(v,+)$, $r(v,-)$ be the positive and negative real roots of $v$, respectively. Then $$(x,y) \in \{(r(u,-),r(v,+)), (r(u,+),r(v,-))\}$$ are the desired solutions. The sum $x+y$ can then be expressed by the solution to a third polynomial $$f(z) = 819447-537600z-8998752z^3+3291428z^3+22132992z^4-17875200z^5+3163146z^6+1042512z^8-437500z^9+18571z^{12},$$ for which there are again two real roots, both positive. All of these polynomials are irreducible. So I highly doubt that this is a problem that can be reasonably solved by hand.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/868635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Area of a triangle whose each side is less than 2 and greater than1. What is the area of a triangle if each of its sides is greater than 1 and less than 2?
My Try:Let a,b,c be the sides of triangle,then
Area$=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{\frac{a+b+c}{2}\frac{b+c-a}{2}\frac{a+c-b}{2}\frac{a+b-c}{2}}=\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$
$1< a,b,c<2$
So $3< a+b+c<6$ and $0< a+b-c,a+c-b,b+c-a<3$
Therefore $0< Area<\frac{9}{4}\sqrt{2}$
Am I correct? Please also tell me some other method to solve this.
|
You can get arbitrarily small area, since you can get arbitrarily close to the degenerate triangle $a=b=1$, $c=2$, which has zero area.
The maximal possible area is obtained when forming equilateral triangle with $a=b=c=2$. (I.e., the area is $\sqrt3$.)
One of possible arguments is the following:
If two sides are shorter than $2$ units, you can increase the area by leaving the third side the same as it was and changing the height of the triangle by some small $\varepsilon>0$. (Which also increases the length of the two sides, but if the change is small enough, you still can keep those lengths $\le2$.)
So it suffices to look at isosceles triangles with $b=c=2$ and $1\le a \le 2$. This means that we want to maximize the expression $\frac{a\sqrt{4-\frac{a^2}4}}2=\frac{\sqrt{a^2\left(4-\frac{a^2}4\right)}}2$. Since the function $t\mapsto t\left(4-\frac{t}4\right)$ is increasing for $t\in[1,4]$, the maximum is attained for $a=2$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/868706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Evaluate $\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$ $$\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$
My approaches so far has been using substitution with $\tan x = t$ and $\tan \frac x2 = t$ but the calculations has been harder than I think they should.
I've also tried using ordinary polynom integration to simplify the integral but I'm having problems with factorizing the denominator.
|
Choose $t=\tan x$ and then $dt=(1+\tan^2x)dx$.
So your integral is:
$$\int \frac{t}{t^3+3 t^2+2 t+6} dt=\int \frac{t}{(t+3)(t^2+2)} dt.$$
We have:
$$\frac{1}{(t+3)(t^2+2)}=\frac{At+B}{t^2+2}+\frac{C}{t+3}=\frac{(A+C)t^2+(3A+B)t+(3B+2C)}{(t+3)(t^2+2)}.$$
We easily get: $A=-C=-\dfrac{1}{11}$ and $B=\dfrac{3}{11}$.
Therefore:
$$\begin{equation}\begin{split}\int \frac{t}{t^3+3 t^2+2 t+6} dt&=\dfrac{1}{11}\int t\left(\frac{3-t}{t^2+2}+\frac{1}{t+3} \right)dt,\\&=\dfrac{1}{11}\int \frac{3t-t^2}{t^2+2}+\frac{t}{t+3}dt\\&=\dfrac{1}{11}\int \frac{3t-t^2}{t^2+2}dt+\dfrac{1}{11}\int\frac{t}{t+3}dt\\&=-\dfrac{1}{11}\int \frac{-3t+t^2+2-2}{t^2+2}dt+\dfrac{1}{11}\int\frac{t+3-3}{t+3}dt\\&=\cdots\end{split}\end{equation}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/869142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Calculating the area For the two graphs $ \frac{x^3+2x^2-8x+6}{x+4} $ and $ \frac{x^3+x^2-10x+9}{x+4} $, calculate the area which is confined by them;
Attempt to solve: Limits of the integral are $1$ and $-3$, so I took the definite integral of the diffrence between $ \frac{x^3+2x^2-8x+6}{x+4} $ and $ \frac{x^3+x^2-10x+9}{x+4} $ since the latter is a graph of a lower-degree polynomial than the first one, but as a result I got a strange negative area result, which is quite funny since the offical answer is $12 -5 \ln5$...
What am I doing wrong?
Note that in this exercise you need to somehow imagine the two graphs by yourself without the use of any comupters programs or whatever. Hence it is not as trivial to evalutae the position of each graph to the other...
|
For $-3<x<1$:
$$\frac{x^3+2x^2-8x+6}{x+4} < \frac{x^3+x^2-10x+9}{x+4}$$
So you have to calculate the integral of the difference $\frac{x^3+x^2-10x+9}{x+4}-\frac{x^3+2x^2-8x+6}{x+4}$ to find the area.
EDIT:
$$\frac{x^3+x^2-10x+9}{x+4}-\frac{x^3+2x^2-8x+6}{x+4}=\frac{-x^2-2x+3}{x+4}=-\frac{x^2+4x-2x}{x+4}+\frac{3}{x+4}=-\frac{x(x+4)}{x+4}+\frac{2x}{x+4}+\frac{3}{x+4}=-x+2\frac{x+4-4}{x+4}+\frac{3}{x+4}=-x+2-\frac{5}{x+4}$$
$$\int_{-3}^1 -x+2-\frac{5}{x+4} dx=[-\frac{x^2}{2}+2x-5\ln{|x+4|}]_{-3}^1=-\frac{1}{2}+2-5\ln{5}+\frac{9}{2}+6+5 \ln{1}=12-5\ln{5}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/871646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
find a polynomial whose roots are inverse of squares of roots of $x^3+px+q$ Question is :
Given a polynomial $f(x)=x^3+px+q\in \mathbb{Q}[x]$ find a polynomial whose roots are inverse of sqares of roots of $f(x)$
Supposing $a,b,c$ as roots of $f(x)$ we have :
*
*$a+b+c=0$
*$ab+bc+ca=p$
*$abc=-q$
Now i need to know what
*
*$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}$
*$\dfrac{1}{a^2}\cdot\dfrac{1}{b^2}+\dfrac{1}{b^2}\cdot\dfrac{1}{c^2}+\dfrac{1}{c^2}\cdot\dfrac{1}{a^2}$
*$\dfrac{1}{(abc)^2}$
All i have to do is use $(a+b+c)^2$ formula and others and conclude what those sums,products are.. I am fairly comfortable with that...
But then, this question was from a Galois theory course.. So, i some how guess there is a better way to do this...
Can some one suggest something..
|
This has been answered by Mr.Michael But then it is not complete answer..
So, Just for my convenience I would write it explicitly...
Given a polynomial $f(x)\in \mathbb{Q}[x]$ with roots $a,b,c$ we need to find a poly. whose roots are $\dfrac{1}{a^2},\dfrac{1}{b^2},\dfrac{1}{c^2}$
Given that $f(x)=x^3+px+q$ with roots $a,b,c$ . So, we have :
*
*$a+b+c=0$
*$ab+bc+ca=p$
*$abc=-q$
With this we would find polynomial with roots $-a,-b,-c$. For that we need to know :
*
*$(-a)+(-b)+(-c)=-(a+b+c)=0$
*$(-a)(-b)+(-b)(-c)+(-c)(-a)=ab+bc+ca=p$
*$(-a)(-b)(-c)=-abc=q$
So, polynomial with roots $-a,-b,-c$ is $g(x)=x^3+px-q$
So, polynomial with roots $a,b,c,-a,-b,-c$ is :
$$(x^3+px+q)(x^3+px-q)=x^6+2px^4+p^2x^2-q^2$$
Now we see that $(x-a)(x-b)(x-c)=x^3+px+q$ and $(x+a)(x+b)(x+c)=x^3+px-q$
So, we have $$(x-a)(x-b)(x-c)(x+a)(x+b)(x+c)$$
$$=(x^3+px+q)(x^3+px-q)=x^6+2px^4+p^2x^2-q^2$$
i.e., $$(x-a)(x-b)(x-c)(x+a)(x+b)(x+c)=(x-a)(x+a)(x-b)(x+b)(x-c)(x+c)$$
$$=(x^2-a^2)(x^2-b^2)(x^2-c^2)=x^6+2px^4+p^2x^2-q^2$$
i.e., $$(x^2-a^2)(x^2-b^2)(x^2-c^2)=x^6+2px^4+p^2x^2-q^2$$
So, to get polynomial with roots $a^2,b^2,c^2$ we replace $x^2$ with $y$
polynomial with roots $a^2,b^2,c^2$ is $y^3+2py^2+p^2y-q^2$
Polynomial with roots $\dfrac{1}{a^2},\dfrac{1}{b^2},\dfrac{1}{c^2}$ is $-q^2x^3+p^2y^2+2py+1$
So, the polynomial that i need is $g(x)=-q^2x^3+p^2y^2+2py+1$
|
{
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"url": "https://math.stackexchange.com/questions/872264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Show that this expression is a perfect square? Show that this expression is a perfect square?
$(b^2 + 3a^2 )^2 - 4 ab*(2b^2 - ab - 6a^2)$
|
Observe that $4ab(2b^2-6a^2)=8ab(b^2-3a^2)$
Using $(A+B)^2=(A-B)^2+4AB$
$$(b^2+3a^2)^2=(b^2-3a^2)^2+4(b^2)(3a^2)$$
$$\implies(b^2+3a^2)^2-8ab(b^2-3a^2)+4a^2b^2$$
$$=(b^2-3a^2)^2+12a^2b^2-8ab(b^2-3a^2)+4a^2b^2$$
$$=(\underbrace{b^2-3a^2}_p)^2 -2(\underbrace{b^2-3a^2}_p)(\underbrace{4ab}_q)+(\underbrace{4ab}_q)^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/872331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Which contour is best for $\int_0^\infty\frac{1}{x^2 + x + 1}dx$ The following is a complex analysis problem. Does anyone have any idea what contour would be good to use?
$$\int_0^\infty\frac{1}{x^2 + x + 1}dx$$
Its roots on the bottom are are $\frac{-1 \pm i\sqrt{3}}{2}$.
|
Let $ \displaystyle f(z) = \frac{\log z}{z^{2}+z+1}$ and integrate around a keyhole contour where the branch cut for $\log z$ is placed on the positive real axis.
As the radius of the little circle goes to $0$ and the radius of the big circle goes to $\infty$, $ \int f(z) \ dz$ will vanish along both circles. You can use the ML inequality to show this.
So integrating counterclockwise around the contour,
$$ \int_{0}^{\infty} \frac{\log x}{x^{2}+x+1} \ dx + \int_{\infty}^{0} \frac{\log x + 2 \pi i }{x^{2}+x+1} \ dx = 2 \pi i \Big(\text{Res} [f(z), e^{2 \pi i /3}] + \text{Res}[f(z), e^{4 \pi i /3}] \Big)$$
where
$$\text{Res} [f(z), e^{2 \pi i /3}] = \lim_{z \to e^{2 \pi i /3}} \frac{\log z}{2z+1} = \frac{2 \pi i /3}{2e^{2 \pi i /3}+1} = \frac{2 \pi}{3 \sqrt{3}} $$
and
$$ \text{Res} [f(z), e^{4 \pi i /3}] = \lim_{z \to e^{4 \pi i /3}} \frac{\log z}{2z+1} = \frac{4 \pi i /3}{2e^{4 \pi i /3}+1} = - \frac{4 \pi}{3 \sqrt{3}}. $$
Therefore,
$$ -2 \pi i \int_{0}^{\infty} \frac{1}{x^{2}+x+1} \ dx = 2 \pi i \left(- \frac{2 \pi}{3 \sqrt{3}} \right) $$
which implies
$$ \int_{0}^{\infty} \frac{1}{x^{2}+x+1} \ dx = \frac{2 \pi }{3 \sqrt{3}} .$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/872742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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|
Functional equation with cyclic function. Find all functions $f:\mathbb R \to \mathbb R$ that satisfy:
$$ f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x.$$
Some progress: I plugged-in $\dfrac{x-1}{x}$ and $\dfrac{1}{1-x}$, got a system of equations and solving I got $f(x) = \dfrac{x^3-4x^2-3x-3}{4x^2-4x}$. But after testing it back looks like it's not sufficient.
|
Set $T(x) = \frac{x}{x-1}$. Then the composition $T^3(x) = x$. We have $f(Tx) = a x + b f(x)$ for $ a= \frac{7}{3}$ and $b = -\frac{1}{3}$, so
\begin{align*}
f(x)
&= f(T^3(x)) \\
&= a T^2(x) + b f(T^2 x) \\
&= a T^2(x) + a b T(x) + b^2 f(Tx) \\
&= a T^2(x) + a b T(x) +a b^2 x + b^3 f(x).
\end{align*}
That relation gives $f$ as a rational function of $x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/875030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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|
Evaluation of $\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx$
Compute the indefinite integral
$$
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx
$$
My Attempt:
$$
\begin{align}
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx &= \int\frac{\cos 2x}{\sin^2 x\sqrt{\cos 2x}}\sin xdx\\
&= \int\frac{2\cos^2 x-1}{(1-\cos^2 x)\sqrt{2\cos^2 x-1} }\sin x \,dx
\end{align}
$$
Let $\cos x = t$, so that $\sin x\,dx = -dt$. This changes the integral to
$$
\begin{align}
\int\frac{(2t^2-1)}{(t^2-1)\sqrt{2t^2-1}}\,dt &= \int\frac{(2t^2-2)+1}{(t^2-1)\sqrt{2t^2-1}}\,dt\\
&= 2\int\frac{dt}{\sqrt{2t^2-1}}+\int \frac{dt}{(t^2-1)\sqrt{2t^2-1}}
\end{align}
$$
How can I solve the integral from here?
|
\begin{align}
\int\frac{\sqrt{\cos 2x}}{\sin x}\ dx&=\int\frac{\sqrt{\cos^2x-\sin^2x}}{\sin x}\ dx\\
&\stackrel{\color{red}{[1]}}=\int\frac{\sqrt{t^4-6t^2+1}}{t^3+t}\ dt\\
&\stackrel{\color{red}{[2]}}=\frac12\int\frac{\sqrt{u^2-6u+1}}{u^2+u}\ du\\
&\stackrel{\color{red}{[3]}}=\int\frac{(y^2-6y+1)^2}{(y-1)(y-3)(y+1)(y^2+2t-7)}\ dy\\
&\stackrel{\color{red}{[4]}}=\int\left[\frac1{y-1}+\frac1{y-3}-\frac1{y+1}-\frac{16}{y^2+2y-7}\right]\ dt\\
&=\int\left[\frac1{y-1}+\frac1{y-3}-\frac1{y+1}-\frac{16}{(y+1)^2-8}\right]\ dt
\end{align}
The rest is yours.
Notes :
$\color{red}{[1]}\;\;\;$Use Weierstrass substitution, $\tan\left(\dfrac{x}{2}\right)=t$.
$\color{red}{[2]}\;\;\;$Use substitution $u=t^2$.
$\color{red}{[3]}\;\;\;$Use Euler substitution, $y-u=\sqrt{u^2-6u+1}\;\color{blue}{\Rightarrow}\;y=\dfrac{u^2-1}{2u-6}$.
$\color{red}{[4]}\;\;\;$Use partial fractions decomposition.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/876175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 3
}
|
If $(a, b,c)$ is a Pythagorean triple with $b, c$ consecutive integers then $c \mid a^b – 1$, proof/disproof? If $(a, b,c)$ is a Pythagorean triple with $b, c$ consecutive integers then $c \mid a^b – 1$, proof/disproof?
Here are some examples:
$(3, 4, 5)$ is a Primitive Pythagorean Triple (PPT), $3^2 + 4^2 = 5^2$, where $4$ and $5$ are consecutive integers.
$(3^4 – 1)/5
= 80/5
= 16$
$(5, 12, 13)$ is a PPT, $5^2 + 12^2 = 13^2$, where $12$ and $13$ are consecutive integers.
$(5^{12} – 1)/13
= 244140624/13
= 18780048$
$(7, 24, 25)$ is a PPT, $7^2 + 24^2 = 25^2$, where $24$ and $25$ are consecutive integers.
$(7^{24} – 1)/25
= 191581231380566414400/25
= 7663249255222656576$
|
If $b=c-1$ is divisible by $4$, this is true.
$$a^2+(c-1)^2=c^2 \iff a^2=2c-1.$$
Write $b=4k$, and the above yields: $$a^b = (a^2)^{2k} =(2c-1)^{2k} \equiv 1 (\mod c).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/876344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
}
|
Simplifying $\displaystyle\sum_{k=0}^{20}(k+4)\binom{23-k}{3}$ In trying to simplify my answer to a problem posted recently, I am trying to show that
$\displaystyle\sum_{k=0}^{20}(k+4)\binom{23-k}{3}=8\binom{24}{4}$.
I know that $\displaystyle\sum_{k=0}^{20}\binom{23-k}{3}=\binom{24}{4}$, but how can I simplify $\displaystyle\sum_{k=0}^{20}k\binom{23-k}{3}$?
|
Maple gives
$$\sum_{k=1}^n k\binom{m+n-k}{m}
=\frac{(m+n)(m+n+1)}{(m+1)(m+2)}\binom{m+n-1}{m}\ ,$$
which can be checked by induction if you wish. So your sum becomes
$$4\binom{24}{4}+\frac{23\times24}{4\times5}\binom{22}{3}
=4\binom{24}{4}
+\frac{23\times24}{4\times5}\frac{4\times20}{23\times24}\binom{24}{4}
=8\binom{24}{4}\ .$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/882127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How does $x^3 - \sin^3 x$ become $x^3 + \frac{1}{4}\sin{3x}-\frac{3}{4}\sin x$? I was going through answers on this question and came across this answer and I was wondering how the user arrived at the first line where they state:
$$f(x) \equiv x^3 - \sin^3 x = x^3 + {1 \over 4} \,\sin {3x} - {3 \over 4}\,\sin x$$
How does $x^3 - \sin^3 x$ become $x^3 + \frac{1}{4}\sin{3x}-\frac{3}{4}\sin x$?
Are they using some simple identity or is there some other observation happening?
Thanks!
|
HINT: $$\sin{3x}=3\sin {x} -4\sin^3 x \\ \sin {(A+B)}=\sin A \cos B + \cos A \sin B$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/882538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.