- #1
drawar
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Homework Statement
Let [itex]t_j=j/100, a_j=j, b_j=-j[/itex], for j=0,1,...,99. Define [tex]f(t)=\sum\limits_{k=0}^{99} (a_k\cos(2\pi kt)+b_k\sin(2\pi kt))[/tex]
Determine the values of [itex]c_l, d_m[/itex] for l= 0,...5, m=1,...,4, so that [tex]P(t)=\frac{c_0}{2}+\sum\limits_{k=1}^4 (c_k\cos(2\pi kt)+d_k\sin(2\pi kt))+c_5\cos(10\pi kt)[/tex]
is the least squares approximation to the data point [itex](t_j,f(t_j))[/itex] for j=0,...,99.
Homework Equations
[tex] c_k=\frac{1}{50}\sum\limits_{j=0}^{99} f(t_j)\cos(2\pi kt_j)[/tex]
[tex] d_k=\frac{1}{50}\sum\limits_{j=0}^{99} f(t_j)\sin(2\pi kt_j)[/tex]
The Attempt at a Solution
It's clear that I have to evaluate [itex]f(t_j)[/itex] first, but I don't know how to. I've tried simplifying the expression for [itex]f(t_j)[/itex] a bit but this is all I can get
[tex]f(t_j)=100\sum\limits_{k=1}^{49}\cos(2\pi kt_j)+50\cos(2\pi50t_j)+\sum\limits_{k=1}^{49}(100-2k)\sin(2\pi kt_j))[/tex]
since [itex]\cos(2\pi(100-k)t_j)=\cos(2\pi kt_j), \sin(2\pi(100-k)t_j)=-\sin(2\pi kt_j), \sin(2\pi 50t_j)=0.[/itex]
I would be much appreciated if someone could help me evaluate this summation, thanks!