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 July 23rd, 2018, 04:13 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 200 Thanks: 2 Need some insights for approximating curves Data Point: (-3, 1), (-1, 0), (0, 5), (2, 0), (4, 1) Using the following basis to approxmiate the above points.. The five basis: $\displaystyle 1, \sin(x \frac{\pi}{2}), \sin(x \frac{\pi}{4}), \cos(x \frac{\pi}{2}), \cos(x \frac{\pi}{6})$ $\displaystyle \begin{bmatrix} 1 & 1 & -\frac{1}{\sqrt{2}} & 0 & 0 \\ 1 & -1 & -\frac{1}{\sqrt{2}} & 0 & \frac{\sqrt{3}}{2} \\ 1 & 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & -1 & 0.5 \\ 1 & 0 & 0 & 1 & -0.5 \end{bmatrix} \begin{bmatrix} t_4 \\ t_3 \\ t_2 \\ t_1 \\ t_0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 5 \\ 0 \\ 1 \end{bmatrix}$ Would anyone please share any insights as to how the above big matrix was formed? Last edited by skipjack; July 24th, 2018 at 01:54 AM.
 July 23rd, 2018, 07:52 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,200 Thanks: 1155 let that matrix be $A$ let $b$ be a vector of the 5 basis functions as a function of $x$ let $d = (d_1,d_2)$ be the vector of ordered pairs $A_{i,j} = b_j(d_{i,1})$ Thanks from topsquark and zollen
 July 24th, 2018, 04:25 PM #3 Senior Member   Joined: Jan 2017 From: Toronto Posts: 200 Thanks: 2 How the basis were chosen? I sensed there were hidden rules for choosing these 5 basis. I did some experiments with the following basis of my choosing. $\displaystyle 1, \sin(x \frac{\pi}{2}), \cos(x \frac{\pi}{4}), \sin(x \frac{\pi}{6}), \cos(x \frac{\pi}{8})$ It did not yield the correct result of t0, t1, t2, t3 and t4. Last edited by skipjack; July 24th, 2018 at 06:35 PM.
 July 24th, 2018, 04:32 PM #4 Senior Member   Joined: Jan 2017 From: Toronto Posts: 200 Thanks: 2 I think it is based on Fourier series....
 July 24th, 2018, 05:11 PM #5 Senior Member   Joined: Jan 2017 From: Toronto Posts: 200 Thanks: 2 Here is the formula: $\displaystyle y(x) = a_0 + a_1 \sin(x \frac{\pi}{2}) + a_2 \sin(x \frac{\pi}{4}) + a_k \sin(x \frac{\pi}{2k}) + b_1 \cos(x \frac{\pi}{2}) + b_2 \cos(x \frac{\pi}{4}) +\, ... + b_{k-1} \cos( x \frac{\pi}{2(k - 1)})$ Last edited by skipjack; July 24th, 2018 at 06:34 PM.
 July 25th, 2018, 06:14 AM #6 Senior Member   Joined: Jan 2017 From: Toronto Posts: 200 Thanks: 2 But.. what term should I pick for the matrix of nxn sizze??
July 25th, 2018, 04:57 PM   #7
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Quote:
 Originally Posted by zollen Here is the formula: $\displaystyle y(x) = a_0 + a_1 \sin(x \frac{\pi}{2}) + a_2 \sin(x \frac{\pi}{4}) + a_k \sin(x \frac{\pi}{2k}) + b_1 \cos(x \frac{\pi}{2}) + b_2 \cos(x \frac{\pi}{4}) +\, ... + b_{k-1} \cos( x \frac{\pi}{2(k - 1)})$
Would anyone able to tell me what formula is that? I thought it was fourier but it wasn't....

July 25th, 2018, 06:14 PM   #8
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Quote:
 Originally Posted by zollen Would anyone able to tell me what formula is that? I thought it was fourier but it wasn't....
it looks like maybe it's a truncated combo of sine and cosine series but I suspect it's just some sinusoidal basis they threw together for this problem.

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