How could I use the Wynn-epsilon alghoritm in Matlab to accelerate the convergence of a Maclaurin series?
I want to extimate the first derivative of $f(x)$, so
$$f'(x)= \sum_{k=0}^\infty {ka_kx^{k-1}}$$
where $a_k$ is the coefficient
$$a_k=\frac{f^{(k)}(0)}{k!}$$
The epsilon alghoritm is like
$\epsilon_{-1}^{(l)}=0 \\\epsilon_{0}^{(l)}=S_{l}(x) \ \ \ \ l=0,...,L \\\epsilon_{k+1}^{(l)}=\epsilon_{k-1}^{(l+1)}+\left[\epsilon_{k}^{(l+1)}-\epsilon_{k}^{(l-1)}\right]^{-1} \ \ \ \ ,k=0,...,L; \ \ \ l=0,...,L-k-1.$
Where $S_{l}(x)$ is the succession of partial sums, i.e., in our case
$$S_{l}(x)=\sum_{k=1}^l {ka_kx^{k-1}}$$
Definitively we got a tableau from $\epsilon$ values like $$ \begin{matrix} \epsilon_0^{0} & \epsilon_1^{0} & \epsilon_2 ^{0} & \cdots & \epsilon_{L-1}^{0} & \epsilon_{L}^{0}\\ \epsilon_{0}^{1} & \epsilon_{1}^{1} & \epsilon_{2}^{1} &\cdots& \epsilon_{L-1}^{1}\\ \epsilon_{0}^{2} &\cdots & \\ \vdots \\ \ & \epsilon_{1}^{(L-1)}\\ \epsilon_{0}^{L}\end{matrix} $$
where each element of $\mathbf{even}$ columns extimates the limit. I tried to solve it, but i got many problems. I know it could be a not immediate question but i would really appreciate your help. I need this to solve Blasius and Falkner Scan problem. If you need any other information please ask me. Thanks in advance for any incoming comments!
EDIT $\\$
for Antonio Vargas $\\$
$f(x)$ is the function i get from the solution of Falkner Scan equation for boundary layer: $$f'''(x)+\beta_0 f(x)f''(x)+\beta (1-f'(x)^2)=0$$
The coefficients of the series are known since they are found by recurrence by applying the differential operator to the equation $$\frac{d^k}{dx^k}$$
You can see the results at page 4 and 5 of the pdf here. If you could look at it and tell me where does the $\delta_{k0}$ in the eq. (3) come from i would appreciate your help. For the Matlab problem of convergence I should probably post in the other section as the previous user suggested, maybe I have solved, but I have to finish and be sure before posting solution. Thanks for interesting and, if you want, for help.