## an issue with solving an IVP by Taylor Series

Okay so suppose I have the Initial Value Problem:

$$\left. \begin{array}{l} \frac {dy} {dx} = f(x,y) \\ y( x_{0} ) = y_{0} \end{array} \right\} \mbox{IVP}$$

NB. I am considering only real functions of real variables.

If $$f(x,y)$$ is analytic at x0 and y0 then that means that we can construct its Taylor Series centered around the point (x0,y0) and that the Taylor Series will have a positive radius of convergence, and also that within this radius of convergence the function $$f(x,y)$$ will equal its Taylor Series.

$$f(x,y)$$ being analytic at x0 and y0 also means that the IVP has a unique solution in a neighbourhood of the point x0, as follows from The Existence and Uniqueness Theorem (Picard–Lindelöf). Let y(x) be the function that satisfies the IVP in a neighbourhood of x0.

The idea behind the Taylor Series method is to use the differential equation and initial condition to find the Taylor coefficients of y(x):

$$\frac {dy} {dx} = f(x,y) \; \; \rightarrow \; \; y'(x_0)=f(x_0,y(x_0))=f(x_0,y_0)$$

$$\frac {d^2y} {dx^2} = \frac {d} {dx} f(x,y)= \frac {\partial f} {\partial x} \frac {dx} {dx} \; + \; \frac {\partial f} {\partial y} \frac {dy} {dx} \\ \indent \rightarrow \; \; y''(x_0)= \frac {\partial f} {\partial x}|_{(x_0,y(x_0))} \; + \; \frac {\partial f} {\partial y}|_{(x_0,y(x_0))}\cdot f(x_0,y_0)$$

etc.

Obviously we can do this because if f is analytic, all the partial derivatives of f exist at (x0,y0), and by the relationship given by the ODE (which we know y(x) satisfies at least in a neighbourhood of x0), all the derivatives of y exist at x0, that is,
$$y(x) \in C^\infty _x (x_0)$$

Okay, so we can construct the Taylor Series of y(x) at x0:
$$\sum \frac{y^{(n)}(x_0)}{n!}x^n$$

But!
1)How do we know that this series has a non-zero radius of convergence?
2)And secondly, if it does have a positive radius of convergence, how do we know it is equal to the solution of the IVP within its radius of convergence? That is, obviously within its radius of convergence, the series represents a certain function which is analytic at x0, but how do we know that this function is indeed the function y(x) which is what we called the solution to the IVP? I mean, maybe the solution to the IVP is not analytic at x0 (even though it is infinitely derivable at x0) and thus its Taylor series (the one we constructed) does not represent it in any neighbourhood of x0. After all, the existence and uniqueness theorem does not require or state that the solution be analytic at all.
Um, if you can affirm that the Taylor Series we have constructed is indeed a solution to the IVP within its radius of convergence, then by the uniqueness of the solution, the function it represents must be the solution of the IVP, y(x). But can this be affirmed?

If anyone has any ideas please help, I'm just not happy with this method coz yes, I can calculate a number of terms of the Taylor Series to approximate the solution of the IVP, but I really have no assurance that the series I am constructing is indeed solution of the IVP (not to mention whether it even does converge at all for any x near x0).
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 Sometimes I feel I speak in Martian, especially when my posts are long ones:P Did anyone actually get was I was on about? Just curious, as sometimes I myself don't lol :P
 Yay I think I might be able to (partially) answer my own question xD -gingerly tries to demostrate- Let $$T_{y,x_0} (x) = \sum_{n=0}^\infty \frac{y^{(n)}(x_0)}{n!}x^n$$ be the Taylor series of y(x) that I've constructed using the initial condition and the relationship given by the o.d.e. If it has a positive radius of convergence R>0 (unfortunately I'll only be able to know this if I can deduce an expression for the general term of the series), then within its radius of convergence, it converges to an analytic (at x0) function, lets say u(x): [tex] u(x) = T_{y,x_0} (x) = \sum_{n=0}^\infty \frac{y^{(n)}(x_0)}{n!}x^n = T_{u,x_0} (x) = \sum_ {n=0}^\infty \frac{u^{(n)}(x_0)}{n!}x^n \ \ \ \ \ \ \ \ \ \mbox{for} \ \ |x-x_0|

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## an issue with solving an IVP by Taylor Series

Wikipedia says if the a function is analytic then it is locally given by a convergent power series.

http://en.wikipedia.org/wiki/Analytic_function

 Quote by John Creighto Wikipedia says if the a function is analytic then it is locally given by a convergent power series. http://en.wikipedia.org/wiki/Analytic_function
Fanku, John :)
I've always gone by that definition of analicity (for real functions anyhow, which are what concern me). That is why I equated f(x,u) to its power series (Taylor series) under the assumption that it was analytic xD.

 Tags convergence, existence, ivp, taylor, uniqueness