The Fourier series of the function f(x)=cosx is given by the Euler formula: cosx=2eix+e−ix​=21​e−ix+21​eix So if f(x)=co

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The Fourier series of the function f(x)=cosx is given by the Euler formula: cosx=2eix+e−ix​=21​e−ix+21​eix So if f(x)=co

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The Fourier Series Of The Function F X Cosx Is Given By The Euler Formula Cosx 2eix E Ix 21 E Ix 21 Eix So If F X Co 1
The Fourier Series Of The Function F X Cosx Is Given By The Euler Formula Cosx 2eix E Ix 21 E Ix 21 Eix So If F X Co 1 (55.88 KiB) Viewed 52 times
The Fourier Series Of The Function F X Cosx Is Given By The Euler Formula Cosx 2eix E Ix 21 E Ix 21 Eix So If F X Co 2
The Fourier Series Of The Function F X Cosx Is Given By The Euler Formula Cosx 2eix E Ix 21 E Ix 21 Eix So If F X Co 2 (64.52 KiB) Viewed 52 times
The Fourier Series Of The Function F X Cosx Is Given By The Euler Formula Cosx 2eix E Ix 21 E Ix 21 Eix So If F X Co 3
The Fourier Series Of The Function F X Cosx Is Given By The Euler Formula Cosx 2eix E Ix 21 E Ix 21 Eix So If F X Co 3 (44.71 KiB) Viewed 52 times
The Fourier Series Of The Function F X Cosx Is Given By The Euler Formula Cosx 2eix E Ix 21 E Ix 21 Eix So If F X Co 4
The Fourier Series Of The Function F X Cosx Is Given By The Euler Formula Cosx 2eix E Ix 21 E Ix 21 Eix So If F X Co 4 (67.68 KiB) Viewed 52 times
The Fourier series of the function f(x)=cosx is given by the Euler formula: cosx=2eix+e−ix​=21​e−ix+21​eix So if f(x)=cosx the Fourier coefficients of f are f^​(n)=⎩⎨⎧​1/21/20​ if n=1 if n=−1 otherwise ​ Find the Fourier series of the function g(x)=sinx. (Your solution should resemble what's given above for the cosine function.)
Suppose f and g are two integrable functions on the interval (−π,π], and extended to be 2π-periodic on the real ner Then the convolution of f and g is the function defined for all x by: f∗g(x)=2π1​∫−ππ​f(x−y)g(y)dy Example Let f be any function, and let g be defined as follows. Select a number 0<h<π. Then let g(x)={2π/h0​ if −h/2≤x≤h/2 otherwise ​ The function g is a tall narrow rectangle (if h is relatively small):
The function g is a tall narrow rectangle (if h is relatively small): Now the convolution of f and g is f∗g(x)=2π1​∫−ππ​f(x−y)g(y)dy=2π1​∫−h/2h/2​f(x−y)(2π/h)dy=h1​∫−h/2h/2​f(x−y)dy
Now change variables: Let u=x−y. So du=−dy or −du=dy. If y=−h/2 then u=x−y=x−(−h/2)=x+h/2 If y=h/2 then u=x−y=x−h/2. This gives f∗g(x)=2π1​∫−h/2h/2​f(x−y)(1/h)dy=−h1​∫x+h/2x−h/2​f(u)du=h1​∫x−h/2x+h/2​f(u)du This is the average of the function f over the interval from x−h/2 to x+h/2. If x is the current time, this gives the average of over the interval of time of duration h, centered at x.
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