- #1
NeoDevin
- 334
- 2
Let's say I have some data, as a function of a variable x. I want to fit this to the real part of the function
\frac{A}{1-ix}\left(1+\sum_{n=1}^\infty\frac{c_n}{(1-ix)^n}\right)
by numerically fitting the first N of the c_n's (A is fixed). I tried something like
However, this way does not work. For the case N=5, I put in the summation explicitly, calling my parameters c1, c2, etc, and it produces different (better) results than the above code. This is not feasible if I want the first 100 or 1000 coefficients, so any help with the above method would be appreciated. Thanks in advance.
\frac{A}{1-ix}\left(1+\sum_{n=1}^\infty\frac{c_n}{(1-ix)^n}\right)
by numerically fitting the first N of the c_n's (A is fixed). I tried something like
Code:
A = 1; N = 100;
fit = FindFit[data, Re[A(1+Sum[c[k]/(1-i x)^k,{k,1,N}])], Table[c[k], {k,1,N}], x];
Plot[Re[A(1+Sum[c[k]/(1-i x)^k,{k,1,N}])]/.fit, {x,0,10}]However, this way does not work. For the case N=5, I put in the summation explicitly, calling my parameters c1, c2, etc, and it produces different (better) results than the above code. This is not feasible if I want the first 100 or 1000 coefficients, so any help with the above method would be appreciated. Thanks in advance.
