Consider the(adsbygoogle = window.adsbygoogle || []).push({}); IVP:

[tex]

\left. \begin{array}{l}

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

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

\end{array} \right\} \mbox{ze IVP :p}

[/tex]

Hypothesis:

[tex] f(x,y)\subset C^\infty_{x,y}(D)\; \; / \; \;(x_0,y_0)\in D [/tex]

[Note that this condition automatically satisfies the hypotheses of theExistence and Uniqueness Theorem(ie, [tex] f \in C_{x,y}(D)[/tex] and [tex]f \in L_y(D)[/tex] ([tex] L [/tex]=Lipschitzian)), hence we know that a unique solution to the IVP does indeed exist at least in a certain interval centred around [tex] x_0, \; x \in |x-x_0| \leq h [/tex]].

So then, let [tex]y(x)[/tex] be the solution to the IVP (at least in that interval [tex]x \in |x-x_0| \leq h [/tex]). Then we know that in that interval the ode is satisfied, and so,

[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]

[tex]\frac {d^3y} {dx^3} = \frac {d} {dx} (\frac {d^2y} {dx^2}) = \frac {\partial ^2f} {\partial x^2} \frac {dx} {dx} \; + \; \frac {\partial ^2f} {\partial y^2} \frac {dy} {dx} \; + \; \frac {\partial^2 f} {\partial x \partial y } \frac {dy} {dx} \; + \; \frac {\partial ^2f} {\partial y^2}(\frac {dy} {dx})^2 \; + \; \frac {\partial f} {\partial y} \frac {d^2y} {dx^2}\\ = \frac {\partial ^2f} {\partial x^2} \; + \; f\cdot \frac {\partial ^2f} {\partial y^2} \; + \; \frac {\partial f} {\partial y} (\frac {\partial f} {\partial x} \; + \; f\cdot \frac {\partial f} {\partial y}) \; + \; f\cdot \frac {\partial^2 f} {\partial x \partial y } \; + \; f^2 \frac {\partial ^2f} {\partial y^2} \\ \;

\indent \rightarrow \; \; [/tex] get [tex]y'''(x_0)[/tex] -wipes brow-

[Note that for the derivatives ofto exist we need the derivatives of all orders of [tex]f [/tex] to exist, at least in the small region around [tex](x_0, y_0)[/tex], which justifies the hypothesis of the method].y

So! We can then construct the Taylor series:

[tex]y(x_0)+y'(x_0)\cdot (x-x_0) + y''(x_0)\cdot \frac {(x-x_0)^2}{2!} + . . . [/tex]

BUT!!

1) How do you know that the Taylor series converges around? (Do all Taylor series converge in some interval aroundx_{0}? :P )x_{0}

2) Okay so assuming the Taylor series converges in some interval around, how do we know that it is equal to the function [tex]y(x)[/tex] solution of the IVP? That [tex]y(x) \in C^\infty _x [/tex] does NOT mean that [tex]y(x)[/tex] is analytic! :>x_{0}

PALEEZE HELP!! 0:

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Taylor method for IVPs

Loading...

Similar Threads - Taylor method IVPs | Date |
---|---|

Taylor expansion with multi variables | Jan 8, 2016 |

Stuck on Taylor series DE | Oct 30, 2013 |

Taylor series and the forward finite difference method | Sep 4, 2011 |

Multi-Variable Second Order Taylor Series Expansion, Ignoring SOME second order terms | Jul 18, 2011 |

Taylor Expansion of Gradient | May 27, 2011 |

**Physics Forums - The Fusion of Science and Community**