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I don't have an analytical formula for the function f[x,t] but I have a list of values (table) obtained by applying the inverse of Laplace Transform to the function f[x,s].

How can I apply the Taylor expansion to the following function?

g[x,t]= (-1+Sqrt[1+2* k* Interpolation[f[x,t]]])/2*k, 

where k is a constant.

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    $\begingroup$ It does not sound like a very good idea to me. Interpolation is polynomial based, so if you work directly with the InterpolatingFunction, you can go only up to a certain order (typically 3). You'd be better off estimating the derivatives directly from the discrete data rather than going through interpolation. What is your actual goal here? If you just want a polynomial approximation to a function then something akin to fitting may be better. $\endgroup$ – Szabolcs Oct 29 '16 at 9:10
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I think you can do the series first.

Let's suppose that you have the function expr1={Cos[x]*Sin[t]}

expr2 = Normal[Series[expr1, {x, 0, 5}, {t, 0, 5}]];
A = Table[expr2, {x, 0, 3, 0.5}, {t, 1, 5, 0.5}];
B = Sqrt[1 + 2*k*Interpolation[A]]/2 k - 1

with the result: enter image description here

or just as you wish

Table[Evaluate[{{x, t}, expr1}], {x, 0, 3, 1.}, {t, 0, 3, 1.}];
expr3 = Sqrt[1 + 2*k*%]/2 k - 1;
Series[expr3, {x, 0, 5}, {t, 0, 5}]
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  • $\begingroup$ Thank you, but I don't have a formula for f[x,t] to make the series, it is a table of values! and I tried: exper=Series[ F[x,t],{x,o,5}] where F[x,t] is the table of f[x,t] and it gives the same List of f[x,t] $\endgroup$ – Essam Oct 31 '16 at 15:02
  • $\begingroup$ @Essam Try FindFit $\endgroup$ – tablecircle Oct 31 '16 at 16:32
  • $\begingroup$ then...maybe use the fitting to do the series? $\endgroup$ – tablecircle Oct 31 '16 at 16:35
  • $\begingroup$ Before interpolation I got this list: {{0., 0.0647956}, {0.5, 0.034067}, {1., 0.0173953}, {1.5, 0.00873915}, {2., 0.00432485}, {2.5, 0.00210854}, {3., 0.00101385} and after interpolation I got the following list {{0., 0.064587}, {0.488088, 0.0340092}, {0.954451, 0.0173802}, {1.40175, .00873534}}, and when I plotted them on one graph I got an identical graph (curves). How can I show the very slight difference between these two lists on one graph? $\endgroup$ – Essam Nov 1 '16 at 1:45
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    $\begingroup$ @essam If this answer helps you, you can vote them up by clicking the gray triangles or accept it by clicking the checkmark sign. $\endgroup$ – xzczd Dec 23 '16 at 5:53

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