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<math>=\frac{1}{2L}\int_{-L}^L f(x)(\cos(nx)-i\sin(nx))\,dx=</math> | <math>=\frac{1}{2L}\int_{-L}^L f(x)(\cos(nx)-i\sin(nx))\,dx=</math> | ||
| + | |||
| + | <math>=\frac{1}{2L}(\int_{-L}^L f(x)(\cos(nx)\,dx - | ||
| + | i\int_{-L}^L f(x)\sin(nx))\,dx)=</math> | ||
<math>=\frac{1}{2}(a_n-ib_n).</math> | <math>=\frac{1}{2}(a_n-ib_n).</math> | ||
Revision as of 07:05, 5 November 2010
Homework 11 collaboration area
Question: I'm having trouble getting HWK 11, Page 499, Problem 3 started.
Answer: You will need to use Euler's identity
$ e^{i\theta}=\cos\theta+i\sin\theta $
and separate the definitions of the complex coefficients into real and imaginary parts. For example,
$ c_n=\frac{1}{2L}\int_{-L}^L f(x)e^{-inx}\,dx= $
$ =\frac{1}{2L}\int_{-L}^L f(x)(\cos(-nx)+i\sin(-nx))\,dx= $
$ =\frac{1}{2L}\int_{-L}^L f(x)(\cos(nx)-i\sin(nx))\,dx= $
$ =\frac{1}{2L}(\int_{-L}^L f(x)(\cos(nx)\,dx - i\int_{-L}^L f(x)\sin(nx))\,dx)= $
$ =\frac{1}{2}(a_n-ib_n). $
Do the same thing for $ c_{-n} $ and combine.
