Serie de Fourier para una serie de funciones Herramientas de

Serie de Fourier para una serie de funciones Herramientas de

Postby Guest » Mon May 13, 2019 5:56 pm

From equations 1) and 2) use the appropriate values of [tex]\theta[/tex] generalmente [tex]( 0 ,\frac{\pi }{2},\pi )[/tex] to get to the indicated equation.

1)


[tex]f(\theta )=e^{b\theta }(-\pi <\theta <\pi ) \quad | \quad\frac{sinh(b\pi )}{\pi }\sum_{-\infty }^{\infty }\frac{(-1)^{n}}{b-in}e^{in\theta }[/tex]


2)


[tex]f(\theta )=e^{b\theta }(0<\theta <2\pi )\quad | \quad\frac{e^{2\pi b}-1}{2\pi }\sum_{-\infty }^{\infty }\frac{e^{in\theta }}{b-in}[/tex]



Indicated equation: [TEX=null]\sum_{1 }^{\infty }\frac{(-1)^{n}}{n^{2}+b^{2}}=\frac{\pi }{2b}csch(b\pi )-\frac{1}{2b^{2}}[/TEX]


Well I start with equation 1):


[tex]e^{b\theta }=\frac{sinh(b\pi )}{\pi }\sum_{-\infty }^{\infty }\frac{(-1)^{n}}{b-in}e^{in\theta }[/tex]


If [tex]\theta =0[/tex]


[tex]e^{b(0)}=\frac{sinh(b\pi )}{\pi }\sum_{-\infty }^{\infty }\frac{(-1)^{n}}{b-in}e^{in(0) }[/tex]


[tex]1=\frac{sinh(b\pi )}{\pi }\sum_{-\infty }^{\infty }\frac{(-1)^{n}}{b-in}[/tex]


Using the conjugate of the complex:


[tex]1=\frac{sinh(b\pi )}{\pi }\sum_{-\infty }^{\infty }\frac{(-1)^{n}}{b-in}\frac{b+in}{b+in}[/tex]


[tex]1=\frac{sinh(b\pi )}{\pi }\sum_{-\infty }^{\infty }(-1)^{n}\frac{b+in}{b^{2}+n^{2}}[/tex]


[tex]\pi csch(b\pi )=\sum_{-\infty }^{\infty }(-1)^{n}\frac{b+in}{b^{2}+n^{2}}[/tex]


Up to now I have only arrived, I do not know if it is for equation 1 to arrive at the result or for equation 2, another doubt that I have is that the summations of equations 1) and 2) tend to [tex](-\infty ,\infty )[/tex] and the result of the summation tends to [tex](1,\infty )[/tex], as I do so that the sum of [tex](-\infty ,\infty )[/tex] tend a [tex](1,\infty )[/tex] ,my idea was to see if the functions were even or odd but since both are exponential they are not odd or even, they can help me with my problem in advance thanks.
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