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

  • #1
Math100
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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?
 
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  • #2
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##.
 
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  • #3
Note that [tex]\begin{split}
\int_{-\pi}^\pi \operatorname{sgn}(\sin(\theta))\sin k\theta\,d\theta &=
\int_0^\pi \sin^2 \theta \sin k\theta\,d\theta - \int_{-\pi}^0 \sin^2\theta \sin k\theta\,d\theta \\
&= 2\int_0^\pi \sin^2 \theta \sin k\theta\,d\theta\end{split}[/tex] by oddness of [itex]\sin [/itex]. Now the idea is to use trig identities to express [itex]\sin^2\theta \sin k\theta[/itex] as a linear combination of sines. Setting [itex]\sin^2 \theta = (1 - \cos 2\theta)/2[/itex] is the first step; then you can use [tex]
\cos 2\theta \sin k\theta = \frac{\sin(k+2)\theta + \sin (k-2)\theta}{2}.[/tex]
 
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  • #4
pasmith said:
Note that [tex]\begin{split}
\int_{-\pi}^\pi \operatorname{sgn}(\sin(\theta))\sin k\theta\,d\theta &=
\int_0^\pi \sin^2 \theta \sin k\theta\,d\theta - \int_{-\pi}^0 \sin^2\theta \sin k\theta\,d\theta \\
&= 2\int_0^\pi \sin^2 \theta \sin k\theta\,d\theta\end{split}[/tex] by oddness of [itex]\sin [/itex]. Now the idea is to use trig identities to express [itex]\sin^2\theta \sin k\theta[/itex] as a linear combination of sines. Setting [itex]\sin^2 \theta = (1 - \cos 2\theta)/2[/itex] is the first step; then you can use [tex]
\cos 2\theta \sin k\theta = \frac{\sin(k+2)\theta + \sin (k-2)\theta}{2}.[/tex]
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, ... ##?
 
  • #5
[itex]\sin k\theta[/itex] means [itex]\sin(k\theta)[/itex]. Evaluate the integrals.
 
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  • #6
pasmith said:
[itex]\sin k\theta[/itex] means [itex]\sin(k\theta)[/itex]. 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?
 
  • #7
The [itex]\operatorname{sgn}(\sin \theta)[/itex] is the difficult bit, so you should get rid of that before doing any trig manipulations. So start with [tex]\begin{split}
b_k &= \frac{1}{\pi} \int_{-\pi}^\pi \operatorname{sgn}(\sin \theta) \sin^2 \theta \sin (k\theta)\,d\theta \\
&= \frac{2}{\pi} \int_0^\pi \sin^2 \theta \sin (k\theta)\,d\theta \qquad \mbox{(see my first post)} \\
&= \frac{1}{\pi} \int_0^\pi (1 - \cos (2\theta)) \sin (k\theta)\,d\theta \\
&= \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}[/tex]
 
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  • #8
pasmith said:
The [itex]\operatorname{sgn}(\sin \theta)[/itex] is the difficult bit, so you should get rid of that before doing any trig manipulations. So start with [tex]\begin{split}
b_k &= \frac{1}{\pi} \int_{-\pi}^\pi \operatorname{sgn}(\sin \theta) \sin^2 \theta \sin (k\theta)\,d\theta \\
&= \frac{2}{\pi} \int_0^\pi \sin^2 \theta \sin (k\theta)\,d\theta \qquad \mbox{(see my first post)} \\
&= \frac{1}{\pi} \int_0^\pi (1 - \cos (2\theta)) \sin (k\theta)\,d\theta \\
&= \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}[/tex]
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} ##.
 
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  • #9
Math100 said:
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.
 
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  • #10
SammyS said:
Please break up those long lines of LATEX.
+1
 
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  • #11
SammyS said:
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)} ##?
 
  • #12
What is [itex]\cos(k\pi)[/itex] for integer [itex]k[/itex]?
 
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  • #13
Math100 said:
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})] ##
 
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  • #14
pasmith said:
What is [itex]\cos(k\pi)[/itex] for integer [itex]k[/itex]?
## \cos(k\pi)=-1 ## for odd integers ## k ##. And this has reminded me of something where I finally got the desired result/answer.
 
  • #15
Thank you, Pasmith, Sammy and Mark44.
 
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