How should I show the following by using the signum function?

[Math100]

Homework Statement

Show that $\operatorname{sgn}(\sin\theta)sin^{2}\theta=\frac{8}{\pi}\sum_{k=1}^{\infty}\frac{sin((2k-1)\theta)}{(2k-1)(3+4k-4k^2)}$.

Relevant Equations

The signum function is defined as $\operatorname{sgn}(x)=1$ for $x>0$, $\operatorname{sgn}(x)=0$ for $x=0$ and $\operatorname{sgn}(x)=-1$ for $x<0$.

Proof:

Let $f(x)$ be a function of the real variable $x$ such that the integral $\int_{-\pi}^{\pi}f(x)dx$ exists and if the Fourier coefficients $(a_{n}, b_{n})$ are defined by $a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx dx, n=0, 1, ...,$ and $b_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin nx dx, n=1, 2, ...,$ then the Fourier series of $f(x)$ is the periodic function $F(x)=\frac{1}{2}a_{0}+\sum_{k=1}^{\infty}(a_{n}\cos nx+b_{n}\sin nx)$.
Consider the function $f(\theta)=\operatorname{sgn}(\sin\theta)sin^{2}\theta$.
Then $\int_{-\pi}^{\pi}\operatorname{sgn}(\sin\theta)sin^{2}\theta d\theta=[\frac{1}{2}(\theta-\sin\theta\cos\theta)\operatorname{sgn}(sin\theta)]_{-\pi}^{\pi}=0$ because $\operatorname{sgn}(sin\pi)=0$ and $\operatorname{sgn}(sin(-\pi))=0$.
Note that $a_{0}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)cos(0)d\theta=\frac{1}{\pi}\int_{-\pi}^{\pi}\operatorname{sgn}(sin\theta)sin^{2}\theta d\theta=\frac{1}{\pi}(0)=0$.
This gives $a_{k}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)cos k\theta d\theta=\frac{1}{\pi}\int_{-\pi}^{\pi}\operatorname{sgn}(sin\theta)sin^{2}\theta\cos k\theta d\theta=0$ for all $n$ because $f(\theta)=\operatorname{sgn}(sin\theta)sin^{2}\theta$ is an odd function and $\cos k\theta$ is an even function.
Hence, $F(\theta)=\frac{1}{2}a_{0}+\sum_{k=1}^{\infty}(a_{k}\cos k\theta+b_{k}\sin k\theta)=\sum_{k=1}^{\infty}b_{k}\sin k\theta$ where $b_{k}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\sin k\theta d\theta$ for $k=1, 2, ...$.

From here, how should I evaluate/simplify $b_{k}=\frac{1}{\pi}\int_{-\pi}^{\pi}\operatorname{sgn}(sin\theta)sin^{2}\theta\cdot\sin k\theta d\theta$ for $k=1, 2, ...$ in order to get the right hand side of $\frac{8}{\pi}\sum_{k=1}^{\infty}\frac{sin((2k-1)\theta)}{(2k-1)(3+4k-4k^2)}$? Where does $sin((2k-1)\theta)$ come from? Is it because of the odd function and so we're only considering the $2k-1$ terms? Or it might be the case that $\operatorname{sgn}(\sin\theta)=1$ from the signum function? Is everything correct in my work up to here?

 

[Mark44]

Notice that $\operatorname{sgn}(\sin(\theta))\sin^2(\theta) = -\sin^2(\theta)$ on the interval $[-\pi, 0]$ and $\sin^2(\theta)$ on the interval $[0, \pi]$. So you should be working with two integrals for $b_k$.
I'm not sure about the $\sin((2k - 1)\theta)$ part but they might be using the double angle identity for $\cos(2\theta)$ here. That is, $\cos(2\theta) = 1 - 2\sin^2(\theta) \Rightarrow \sin^2(\theta) = \frac{1 - \cos(2\theta)}2$.

 

[pasmith]

Note that \begin{split}<br /> \int_{-\pi}^\pi \operatorname{sgn}(\sin(\theta))\sin k\theta\,d\theta &amp;= <br /> \int_0^\pi \sin^2 \theta \sin k\theta\,d\theta - \int_{-\pi}^0 \sin^2\theta \sin k\theta\,d\theta \\<br /> &amp;= 2\int_0^\pi \sin^2 \theta \sin k\theta\,d\theta\end{split} by oddness of \sin. Now the idea is to use trig identities to express \sin^2\theta \sin k\theta as a linear combination of sines. Setting \sin^2 \theta = (1 - \cos 2\theta)/2 is the first step; then you can use <br /> \cos 2\theta \sin k\theta = \frac{\sin(k+2)\theta + \sin (k-2)\theta}{2}.

 

[Math100]

Note that \begin{split}<br /> \int_{-\pi}^\pi \operatorname{sgn}(\sin(\theta))\sin k\theta\,d\theta &amp;=<br /> \int_0^\pi \sin^2 \theta \sin k\theta\,d\theta - \int_{-\pi}^0 \sin^2\theta \sin k\theta\,d\theta \\<br /> &amp;= 2\int_0^\pi \sin^2 \theta \sin k\theta\,d\theta\end{split} by oddness of \sin. Now the idea is to use trig identities to express \sin^2\theta \sin k\theta as a linear combination of sines. Setting \sin^2 \theta = (1 - \cos 2\theta)/2 is the first step; then you can use <br /> \cos 2\theta \sin k\theta = \frac{\sin(k+2)\theta + \sin (k-2)\theta}{2}.

So $\sin^{2}\theta\sin k\theta=(\frac{1-cos(2\theta)}{2})\cdot\sin k\theta=\frac{1}{2}[sin k\theta-(\frac{sin(k+2)\theta+sin(k-2)\theta}{2})]=\frac{1}{2}(\frac{2sin k\theta-sin(k+2)\theta-sin(k-2)\theta}{2})=\frac{2sin k\theta-sin(k+2)\theta-sin(k-2)\theta}{4}$.
And I've got $2\int_{0}^{\pi}sin^{2}\theta\sin k\theta d\theta=2\int_{0}^{\pi}(\frac{2sin k\theta-sin(k+2)\theta-sin(k-2)\theta}{4})d\theta=\frac{1}{2}\int_{0}^{\pi}(2sin k\theta-sin(k+2)\theta-sin(k-2)\theta)d\theta=\int_{0}^{\pi}sin k\theta d\theta-\frac{1}{2}\int_{0}^{\pi}sin(k+2)\theta d\theta-\frac{1}{2}\int_{0}^{\pi}sin(k-2)\theta d\theta=[\theta\cdot\sin k\theta]_{0}^{\pi}-\frac{1}{2}[\frac{\theta^{2}\cdot\sin(k+2)}{2}]_{0}^{\pi}-\frac{1}{2}[\frac{\theta^{2}\cdot\sin(k-2)}{2}]_{0}^{\pi}=\pi\cdot\sin k\pi-\frac{1}{2}(\frac{\pi^{2}\cdot\sin(k+2)}{2})-\frac{1}{2}(\frac{\pi^{2}\cdot\sin(k-2)}{2})$.
Thus, $\int_{-\pi}^{\pi}\operatorname{sgn}(sin\theta)sin k\theta d\theta=-\frac{(sin(k+2)\cdot\pi+sin(k-2)\cdot\pi-4\sin k\theta)\pi}{4}$.

From here, what should I do in order to find/evaluate $b_{k}=\frac{1}{\pi}\int_{-\pi}^{\pi}\operatorname{sgn}(\sin\theta)\sin^{2}\theta\cdot\sin k\theta d\theta$ for $\sum_{k=1}^{\infty}b_{k}\sin k\theta$ for $k=1, 2, ...$?

 

[pasmith]

\sin k\theta means \sin(k\theta). Evaluate the integrals.

 

[Math100]

\sin k\theta means \sin(k\theta). Evaluate the integrals.

Observe that $b_{k}=\frac{1}{\pi}\int_{-\pi}^{\pi}\operatorname{sgn}(sin\theta)sin^{2}\theta\cdot\sin k\theta d\theta=\frac{1}{\pi}\int_{-\pi}^{\pi}\operatorname{sgn}(sin\theta)[(\frac{1-cos(2\theta)}{2})\cdot\sin k\theta]d\theta=\frac{1}{\pi}\int_{-\pi}^{\pi}\operatorname{sgn}(sin\theta)[\frac{1}{2}(sin k\theta-(\frac{sin(k+2)\theta+sin(k-2)\theta}{2}))]d\theta=\frac{1}{\pi}\int_{-\pi}^{\pi}\operatorname{sgn}(sin\theta)[\frac{1}{2}({2sin k\theta-sin(k+2)\theta-sin(k-2)\theta}{2})]d\theta=\frac{1}{\pi}\int_{-\pi}^{\pi}\operatorname{sgn}(sin\theta)[\frac{2sin k\theta-sin(k+2)\theta-sin(k-2)\theta}{4}]d\theta=\frac{1}{4\pi}\int_{-\pi}^{\pi}\operatorname{sgn}(sin\theta)\cdot(2sin k\theta-sin(k+2)\theta-sin(k-2)\theta)d\theta=\frac{1}{4\pi}[-\frac{\operatorname{sgn}(sin\theta)\cdot(\theta^{2}k\cos(2)\sin k+2\cos(\theta k))}{k}]_{-\pi}^{\pi}=\frac{1}{4\pi}[-\frac{2\cos(\pi k)}{k}+\frac{2\cos(-\pi k)}{k}]=0$.

So as a result, I've got $b_{k}=0$, which means that $\sum_{k=1}^{\infty}b_{k}\sin k\theta=0$ but that doesn't makes sense because I didn't get the desired result as $\frac{8}{\pi}\sum_{k=1}^{\infty}\frac{\sin((2k-1)\theta)}{(2k-1)(3+4k-4k^2)}$. What's wrong in here?

 

[pasmith]

The \operatorname{sgn}(\sin \theta) is the difficult bit, so you should get rid of that before doing any trig manipulations. So start with \begin{split}<br /> b_k &amp;= \frac{1}{\pi} \int_{-\pi}^\pi \operatorname{sgn}(\sin \theta) \sin^2 \theta \sin (k\theta)\,d\theta \\<br /> &amp;= \frac{2}{\pi} \int_0^\pi \sin^2 \theta \sin (k\theta)\,d\theta \qquad \mbox{(see my first post)} \\<br /> &amp;= \frac{1}{\pi} \int_0^\pi (1 - \cos (2\theta)) \sin (k\theta)\,d\theta \\<br /> &amp;= \frac{1}{\pi} \int_0^\pi \sin (k\theta) - \tfrac 12 \sin ((k-2)\theta) - \tfrac 12\sin((k+2)\theta) \,d\theta.\end{split}

 

[Math100]

The \operatorname{sgn}(\sin \theta) is the difficult bit, so you should get rid of that before doing any trig manipulations. So start with \begin{split}<br /> b_k &amp;= \frac{1}{\pi} \int_{-\pi}^\pi \operatorname{sgn}(\sin \theta) \sin^2 \theta \sin (k\theta)\,d\theta \\<br /> &amp;= \frac{2}{\pi} \int_0^\pi \sin^2 \theta \sin (k\theta)\,d\theta \qquad \mbox{(see my first post)} \\<br /> &amp;= \frac{1}{\pi} \int_0^\pi (1 - \cos (2\theta)) \sin (k\theta)\,d\theta \\<br /> &amp;= \frac{1}{\pi} \int_0^\pi \sin (k\theta) - \tfrac 12 \sin ((k-2)\theta) - \tfrac 12\sin((k+2)\theta) \,d\theta.\end{split}

So $b_{k}=\frac{1}{\pi}([\theta\cdot\sin k\theta]_{0}^{\pi}-\frac{1}{2}[\frac{\theta^{2}\cdot\sin(k-2)}{2}]_{0}^{\pi}-\frac{1}{2}[\frac{\theta^{2}\cdot\sin(k+2)}{2}]_{0}^{\pi})\implies b_{k}=\frac{1}{\pi}[\pi\cdot\sin k\pi-\frac{1}{2}(\frac{\pi^{2}\cdot\sin(k-2)}{2})-\frac{1}{2}(\frac{\pi^{2}\cdot\sin(k+2)}{2})]\implies b_{k}=\frac{1}{\pi}(\pi\cdot\sin k\pi-\frac{1}{4}(\pi^{2}\cdot\sin(k-2)+\pi^{2}\cdot\sin(k+2)))\implies b_{k}=\frac{1}{\pi}(\pi\cdot\sin k\pi-\frac{\pi^{2}}{4}(sin(k-2)+sin(k+2)))\implies b_{k}=\frac{1}{4\pi}(4\pi\cdot\sin k\pi-\pi^{2}(sin(k-2)+sin(k+2)))\implies b_{k}=\frac{1}{4}(4\sin k\pi-\pi(sin(k-2)+sin(k+2)))$.

Hence, $b_{k}=-\frac{\pi}{4}(sin(k-2)+sin(k+2))$ because $sin(k\pi)=0, \forall k\in\mathbb{Z}$.

 

[SammyS]

So $b_{k}=\frac{1}{\pi}([\theta\cdot\sin k\theta]_{0}^{\pi}-\frac{1}{2}[\frac{\theta^{2}\cdot\sin(k-2)}{2}]_{0}^{\pi}-\frac{1}{2}[\frac{\theta^{2}\cdot\sin(k+2)}{2}]_{0}^{\pi})\implies b_{k}=\frac{1}{\pi}[\pi\cdot\sin k\pi-\frac{1}{2}(\frac{\pi^{2}\cdot\sin(k-2)}{2})-\frac{1}{2}(\frac{\pi^{2}\cdot\sin(k+2)}{2})]\implies b_{k}=\frac{1}{\pi}(\pi\cdot\sin k\pi-\frac{1}{4}(\pi^{2}\cdot\sin(k-2)+\pi^{2}\cdot\sin(k+2)))\implies b_{k}=\frac{1}{\pi}(\pi\cdot\sin k\pi-\frac{\pi^{2}}{4}(sin(k-2)+sin(k+2)))\implies b_{k}=\frac{1}{4\pi}(4\pi\cdot\sin k\pi-\pi^{2}(sin(k-2)+sin(k+2)))\implies b_{k}=\frac{1}{4}(4\sin k\pi-\pi(sin(k-2)+sin(k+2)))$.

Hence, $b_{k}=-\frac{\pi}{4}(sin(k-2)+sin(k+2))$ because $sin(k\pi)=0, \forall k\in\mathbb{Z}$.

That integration doesn't look correct at all.

What is the antiderivative of $\sin x$ ?

Also:
Please break up those long lines of LATEX.

 

[Mark44]

Please break up those long lines of LATEX.

+1

 

[Math100]

That integration doesn't look correct at all.

What is the antiderivative of $\sin x$ ?

Also:
Please break up those long lines of LATEX.

Observe that $b_{k}=\frac{1}{\pi}([-\frac{\cos\theta k}{k}]_{0}^{\pi}-\frac{1}{2}[-\frac{\cos((k-2)\theta)}{k-2}]_{0}^{\pi}-\frac{1}{2}[-\frac{\cos((k+2)\theta)}{k+2}]_{0}^{\pi})\implies b_{k}=\frac{1}{\pi}[\frac{1-\cos\pi k}{k}-\frac{1}{2}(-\frac{\cos((k-2)\pi)}{k-2}+\frac{1}{k-2})-\frac{1}{2}(-\frac{\cos((k+2)\pi)}{k+2}+\frac{1}{k+2})]$.
So we get
$b_{k}=\frac{1}{\pi}[\frac{1-\cos\pi k}{k}-\frac{1}{2}(\frac{1-\cos((k-2)\pi)}{k-2})-\frac{1}{2}(\frac{1-\cos((k+2)\pi)}{k+2})]\implies b_{k}=\frac{1}{\pi}[\frac{1-\cos\pi k}{k}-\frac{1}{2}(\frac{1-\cos((k-2)\pi)}{k-2}+\frac{1-\cos((k+2)\pi)}{k+2})]\implies b_{k}=\frac{1}{\pi}[\frac{1-\cos\pi k}{k}-\frac{1}{2}(\frac{-2k(\cos(k\pi)-1)}{(k-2)(k+2)})]$.
Thus
$b_{k}=\frac{1}{\pi}(\frac{1-\cos\pi k}{k}+\frac{k(\cos(k\pi)-1)}{(k-2)(k+2)})\implies b_{k}=\frac{1}{\pi}(\frac{-(k-2)(k+2)(\cos(k\pi)-1)+k^{2}(\cos(k\pi)-1)}{k(k-2)(k+2)}\implies b_{k}=\frac{1}{\pi}[\frac{(\cos(k\pi)-1)(k^{2}-(k-2)(k+2))}{k(k-2)(k+2)}]\implies b_{k}=\frac{4}{\pi}(\frac{\cos(k\pi)-1}{k(k-2)(k+2)})$.

From here, how should I find/evaluate/simplify $\sum_{k=1}^{\infty}b_{k}\sin k\theta$ for $k=1, 2, ...$ in order to get $\frac{8}{\pi}\sum_{k=1}^{\infty}\frac{\sin((2k-1)\theta)}{(2k-1)(3+4k-4k^2)}$?

 

[pasmith]

What is \cos(k\pi) for integer k?

 

[Mark44]

Observe that
$b_{k}=\frac{1}{\pi}([-\frac{\cos\theta k}{k}]_{0}^{\pi}-\frac{1}{2}[-\frac{\cos((k-2)\theta)}{k-2}]_{0}^{\pi}-\frac{1}{2}[-\frac{\cos((k+2)\theta)}{k+2}]_{0}^{\pi})\implies b_{k}=\frac{1}{\pi}[\frac{1-\cos\pi k}{k}-\frac{1}{2}(-\frac{\cos((k-2)\pi)}{k-2}+\frac{1}{k-2})-\frac{1}{2}(-\frac{\cos((k+2)\pi)}{k+2}+\frac{1}{k+2})]$.

As already requested by @SammyS, please break up long lines of LaTeX so that a reader doesn't have to scroll off what's visible on the screen to see the far ends of them.

Like so:
$b_{k}=\frac{1}{\pi}([-\frac{\cos\theta k}{k}]_{0}^{\pi}-\frac{1}{2}[-\frac{\cos((k-2)\theta)}{k-2}]_{0}^{\pi}-\frac{1}{2}[-\frac{\cos((k+2)\theta)}{k+2}]_{0}^{\pi})$
$\implies b_{k}=\frac{1}{\pi}[\frac{1-\cos\pi k}{k}-\frac{1}{2}(-\frac{\cos((k-2)\pi)}{k-2}+\frac{1}{k-2})-\frac{1}{2}(-\frac{\cos((k+2)\pi)}{k+2}+\frac{1}{k+2})]$

 

[Math100]

What is \cos(k\pi) for integer k?

$\cos(k\pi)=-1$ for odd integers $k$. And this has reminded me of something where I finally got the desired result/answer.

 

[Math100]

Thank you, Pasmith, Sammy and Mark44.