Distance of functions and fourier coefficientsby sridhar10chitta Tags: coefficients, distance, fourier, functions 

#1
Nov2107, 10:34 AM

P: 27

There are two functions f(t) and g(t); t is the independent variable.
The distance between the two functions will be given by [1/2pi integral{f(t)g(t)}^2 dt]^1/2 between pi and +pi. Apparently, this distance also is the fourier coefficient of each term in the fourier expansion of a periodic function f(t) such that it is closest to f(t). Why is this so ? why is not the distance given by f(t)g(t) simply ? i.e 1/sqrt(2pi) integral{f(t)g(t)}dt between pi and +pi. 



#2
Nov2107, 01:56 PM

Math
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PF Gold
P: 38,877

In [itex]L_2[/itex], the space of square integrable functions, it can be show that the finite Fourier series is essentially the "least squares approximation" to the function using that distance (with the square root). You CAN define "distance from f to g" to be [itex]\int f(t) g(t)dt[/itex] where the integral is taken over whatever interval you are interested in. That is the "L_{1} norm. You can also define "distance from f to g" to be maxf(t)g(t) where the "max" is the largest value that takes on over whatever interval you are interested in. That's called the "uniform norm" because convergence defined in that norm is equivalent to uniform convergence. We use the "root square" distance formula in Fourier Analysis precisely because the finite Fourier series is the "least squares approximation" with that distance formula. 


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