- #1
Math100
- 794
- 221
- 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?
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?