Fourier Example - Revision history

Cdxskier at 22:26, 25 January 2010

2010-01-25T22:26:31Z

← Older revision		Revision as of 15:26, 25 January 2010
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	==References:==		==References:==
	[[Fourier Series: Basic Results]]		[[Fourier Series: Basic Results]]
			==Readers==
			[[Lau, Chris\|Christopher Garrison Lau I]]

2010-01-21T21:18:53Z

← Older revision		Revision as of 14:18, 21 January 2010
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	==Solution Graph==		==Solution Graph==
	[[Image:Fourier.gif]]		[[Image:Fourier.gif]]


			==*BONUS* VIDEO! (For people completely lost on Fourier Series)==


			http://www.youtube.com/watch?v=nXEqrOt-nB8


	==References:==		==References:==
	[[Fourier Series: Basic Results]]		[[Fourier Series: Basic Results]]

2010-01-19T08:43:42Z

← Older revision		Revision as of 01:43, 19 January 2010
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	~~'''~~
	== Solution ==		== Solution ==
	~~'''~~


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	<math>f(x)=\frac{\pi}{2}+2(sin(x)+\frac{sin(3x)}{3}+\frac{sin(5x)}{5}+...)</math>		<math>f(x)=\frac{\pi}{2}+2(sin(x)+\frac{sin(3x)}{3}+\frac{sin(5x)}{5}+...)</math>


			==Solution Graph==
	[[Image:Fourier.gif]]		[[Image:Fourier.gif]]

	==References:==		==References:==
	[[Fourier Series: Basic Results]]		[[Fourier Series: Basic Results]]

2010-01-19T08:36:38Z

Solution

← Older revision		Revision as of 01:36, 19 January 2010
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	<math>f(x)=\frac{\pi}{2}+2(sin(x)+\frac{sin(3x)}{3}+\frac{sin(5x)}{5}+...)</math>		<math>f(x)=\frac{\pi}{2}+2(sin(x)+\frac{sin(3x)}{3}+\frac{sin(5x)}{5}+...)</math>

			[[Image:Fourier.gif]]

	==References:==		==References:==
	[[Fourier Series: Basic Results]]		[[Fourier Series: Basic Results]]

2010-01-19T08:31:06Z

Solution

← Older revision		Revision as of 01:31, 19 January 2010
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	<math>~~a_0~~=\frac{1}{2\pi}(\int_{-\pi}^00\ dx+\int_{0}^\pi\pi\ dx)=\frac{\pi}{2}</math>		<math>a_o=\frac{1}{2\pi}(\int_{-\pi}^00\ dx+\int_{0}^\pi\pi\ dx)=\frac{\pi}{2}</math>

2010-01-19T08:28:57Z

← Older revision		Revision as of 01:28, 19 January 2010
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	\pi,& 0 \le x \le \pi\\		\pi,& 0 \le x \le \pi\\
	\end{cases}</math>		\end{cases}</math>



	'''		'''
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	Here we have		Here we have




			<math>a_0=\frac{1}{2\pi}(\int_{-\pi}^00\ dx+\int_{0}^\pi\pi\ dx)=\frac{\pi}{2}</math>



			<math>a_n=\int_{0}^\pi\pi cos(nx)\ dx=0, n\ge1,</math>


			and


	:<math>b_n = \int_{0}^\pi \pi\sin(nx)\, dx = \frac{1}{n}(1-cos(x\pi))=\frac{1}{n}(1-(-1)^n)</math>		:<math>b_n = \int_{0}^\pi \pi\sin(nx)\, dx = \frac{1}{n}(1-cos(x\pi))=\frac{1}{n}(1-(-1)^n)</math>


			We obtain <math>b_{2n}</math> = 0 and

	~~We obtain b_2n = 0 and~~

			<math>b_{2n+1}=\frac{2}{2n+1}</math>

	~~<math>b_{2n+1}=\frac{2}{2n+1}</math>~~

	Therefore, the Fourier series of f(x) is		Therefore, the Fourier series of f(x) is

2010-01-19T08:01:43Z

2010-01-19T07:55:06Z

← Older revision		Revision as of 00:55, 19 January 2010
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	:<math>f(x) = \begin{cases}0,& -\pi<x<0\\		:<math>f(x) = \begin{cases}0,& -\pi\le x<0\\
	\pi,& 0<x<\pi\\		\pi,& 0 \le x \le \pi\\
	\end{cases}</math>		\end{cases}</math>

	:<math>b_n = ~~\frac{1}{2\pi}~~\int_{0}^\pi \pi\sin(nx)\, dx, = \frac{1}{n}(1-cos(x\pi))=\frac{1}{n}(1-(-1)^n)</math>		'''
			== Solution ==
			'''

			Here we have

			:<math>b_n = \int_{0}^\pi \pi\sin(nx)\, dx = \frac{1}{n}(1-cos(x\pi))=\frac{1}{n}(1-(-1)^n)</math>


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	<math>f(x)=\frac{\pi}{2}+2(sin(x)+\frac{sin(3x)}{3}+\frac{sin(5x)}{5}+...)</math>		<math>f(x)=\frac{\pi}{2}+2(sin(x)+\frac{sin(3x)}{3}+\frac{sin(5x)}{5}+...)</math>

			==References:==
			[[Fourier Series: Basic Results]]

2010-01-19T07:41:53Z

@@ Line 1: / Line 1: @@
 Find the Fourier Series of the function:
 :<math>f(x) = \begin{cases}0,& -\pi<x<0\\
@@ Line 5: / Line 6: @@
 \end{cases}</math>
-:<math>b_n = \frac{1}{2}{\pi}\int_{0}^\pi  \pi\sin(nx)\, dx, = \frac{1}{n}(1-cos(x\pi))=\frac{1}{n}(1-(-1)^n)</math>
+:<math>b_n = \frac{1}{2\pi}\int_{0}^\pi  \pi\sin(nx)\, dx, = \frac{1}{n}(1-cos(x\pi))=\frac{1}{n}(1-(-1)^n)</math>

2010-01-19T07:28:49Z

@@ Line 1: / Line 1: @@
 Find the Fourier Series of the function:
-<br />
+:<math>f(x) = \begin{cases}0,& -\pi<x<0\\
-<center><math>\f(x)=begin{cases}
+:<math>b_n = \frac{1}{2}{\pi}\int_{0}^\pi  \pi\sin(nx)\, dx, = \frac{1}{n}(1-cos(x\pi))=\frac{1}{n}(1-(-1)^n)</math>