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July 23rd, 2018, 03:13 PM   #1
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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 12:54 AM.
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July 23rd, 2018, 06:52 PM   #2
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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
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July 24th, 2018, 03:25 PM   #3
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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 05:35 PM.
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July 24th, 2018, 03:32 PM   #4
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I think it is based on Fourier series....
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July 24th, 2018, 04:11 PM   #5
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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 05:34 PM.
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July 25th, 2018, 05:14 AM   #6
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But.. what term should I pick for the matrix of nxn sizze??
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July 25th, 2018, 03:57 PM   #7
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Quote:
Originally Posted by zollen View Post
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....
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July 25th, 2018, 05:14 PM   #8
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Quote:
Originally Posted by zollen View Post
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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