Trigonometric interpolation polynomial

[drawar]

Homework Statement

Let $t_j=j/100, a_j=j, b_j=-j$, for j=0,1,...,99. Define $$f(t)=\sum\limits_{k=0}^{99} (a_k\cos(2\pi kt)+b_k\sin(2\pi kt))$$
Determine the values of $c_l, d_m$ for l= 0,...5, m=1,...,4, so that $$P(t)=\frac{c_0}{2}+\sum\limits_{k=1}^4 (c_k\cos(2\pi kt)+d_k\sin(2\pi kt))+c_5\cos(10\pi kt)$$
is the least squares approximation to the data point $(t_j,f(t_j))$ for j=0,...,99.

Homework Equations

$$c_k=\frac{1}{50}\sum\limits_{j=0}^{99} f(t_j)\cos(2\pi kt_j)$$
$$d_k=\frac{1}{50}\sum\limits_{j=0}^{99} f(t_j)\sin(2\pi kt_j)$$

The Attempt at a Solution

It's clear that I have to evaluate $f(t_j)$ first, but I don't know how to. I've tried simplifying the expression for $f(t_j)$ a bit but this is all I can get
$$f(t_j)=100\sum\limits_{k=1}^{49}\cos(2\pi kt_j)+50\cos(2\pi50t_j)+\sum\limits_{k=1}^{49}(100-2k)\sin(2\pi kt_j))$$
since $\cos(2\pi(100-k)t_j)=\cos(2\pi kt_j), \sin(2\pi(100-k)t_j)=-\sin(2\pi kt_j), \sin(2\pi 50t_j)=0.$
I would be much appreciated if someone could help me evaluate this summation, thanks!

 

[vela]

You have
$$f(t) = \sum_{k=0}^{99} [k \cos(2\pi k t) - k \sin(2\pi k t)].$$ You could try using the fact that
$$k \cos(2\pi k t) - k \sin(2\pi k t) = \frac{1}{2\pi}\frac{\partial}{\partial t} [\sin(2\pi k t) + \cos (2\pi k t)].$$ You can evaluate the sums of sin and cos by considering the geometric series $(e^{i2\pi t})^k$.

 

[drawar]

Thanks for the idea, but I can't seem to arrive at anything useful. Here's what I have done
$$f(t) = \sum_{k=0}^{99} [k \cos(2\pi k t) - k \sin(2\pi k t)]=\frac{1}{2\pi}\frac{\partial}{\partial t}\sum_{k=0}^{99} [\sin(2\pi k t) + \cos (2\pi k t)].$$
$$\sum_{k=0}^{99}\sin(2\pi k t)=\mathrm{Im}\sum_{k=0}^{99} (e^{i2\pi t})^k=\ldots=\frac{\sin(100\pi t)\sin(99\pi t)}{\sin (\pi t)}$$
Similarly, $$\sum_{k=0}^{99}\sin(2\pi k t)=1+\frac{\cos(100\pi t)\sin(99\pi t)}{\sin (\pi t)}$$
I'm not going to differentiate this whole thing wrt t, am I?

 

[vela]

Yeah, that does look like it's going to be tedious, doesn't it? Perhaps you can differentiate a little earlier when you have the results of the sums in terms of complex exponentials still.