[tex]

\left. \begin{array}{l}

\frac {dy} {dx} = f(x,y) \\

y( x_{0} ) = y_{0}

\end{array} \right\} \mbox{IVP}

[/tex]

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

If [tex]f(x,y) [/tex] is analytic at

*x*and

_{0}*y*then that means that we can construct its Taylor Series centered around the point

_{0}*(x*and that the Taylor Series will have a positive radius of convergence, and also that within this radius of convergence the function [tex]f(x,y) [/tex] will equal its Taylor Series.

_{0},y_{0})[tex]f(x,y) [/tex] being analytic at

*x*and

_{0}*y*also means that the IVP has a unique solution in a neighbourhood of the point

_{0}*x*, as follows from The Existence and Uniqueness Theorem (Picard–Lindelöf). Let

_{0}*y(x)*be the function that satisfies the IVP in a neighbourhood of

*x*.

_{0}The idea behind the Taylor Series method is to use the differential equation and initial condition to find the Taylor coefficients of

*y(x)*:

[tex]

\frac {dy} {dx} = f(x,y) \; \; \rightarrow \; \;

y'(x_0)=f(x_0,y(x_0))=f(x_0,y_0)

[/tex]

[tex]

\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)

[/tex]

etc.

Obviously we can do this because if

*f*is analytic, all the partial derivatives of

*f*exist at

*(x*, and by the relationship given by the ODE (which we know

_{0},y_{0})*y(x)*satisfies at least in a neighbourhood of

*x*), all the derivatives of

_{0}*y*exist at

*x*, that is,

_{0}[tex]

y(x) \in C^\infty _x (x_0)

[/tex]

Okay, so we can construct the Taylor Series of

*y(x)*at

*x*:

_{0}[tex]

\sum \frac{y^{(n)}(x_0)}{n!}x^n

[/tex]

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

*x*, but how do we know that this function is indeed the function

_{0}*y(x)*which is what we called the solution to the IVP? I mean, maybe the solution to the IVP is not analytic at

*x*(even though it is infinitely derivable at

_{0}*x*) and thus its Taylor series (the one we constructed) does not represent it in any neighbourhood of

_{0}*x*. After all, the existence and uniqueness theorem does not require or state that the solution be analytic at all.

_{0}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

*x*).

_{0}