University of Florida/Egm4313/s12.team11.gooding/R4

Problem 4.4 Parts 1,2

Part 1

Problem Statement

Find n sufficiently high so that $y_n(x_1),y{\text{'}}_n(x_1)$ do not differ from the numerical solution by more than $10^{-5}$ at $x_1=0.9$

Solution

Using a program in MATLAB that iteratively added terms onto the taylor series of $log(1+x)$, terms were added until the error between the exact answer and the series was less than $10^{-5}$.

File:Taylor series 441a.jpg

It was found after trial and error that $n=39$ for the error to be of a magnitude of $10^{-5}$. This error found was 9.7422e-005

Similarly, for $y{\text{'}}_n(x_1)$.

File:Taylor series441b.jpg

It was found after trial and error that $n=74$ for the error to be of a magnitude of $10^{-5}$. This error found was 9.3967e-005

Part 2

Problem Statement

Develop $log(1+x)$ in Taylor series about $\hat{x}=1$ for $n=4,7,11$ and plot these truncated series vs. the exact function.
What is now the domain of convergence by observation?

Solution

A MATLAB program was created, which calculated the Taylor series of each n value, along with the exact function, then plotted these together to show the comparison of all the series.
Below is the Taylor series for $n=7$ expanded at $\hat{x}=1$.
$\frac{x-1}{2\ln \left(10\right)}-\frac{{(x-1)}^2}{8\ln \left(10\right)}+\frac{{(x-1)}^3}{24\ln \left(10\right)}-\frac{{(x-1)}^4}{64\ln \left(10\right)}+\frac{{(x-1)}^5}{160\ln \left(10\right)}-\frac{{(x-1)}^6}{384\ln \left(10\right)}+\frac{\ln \left(2\right)}{\ln \left(10\right)}$

File:Taylor series 442 code.jpg
File:Taylor series 442 graph.jpg
It can be seen by observation that the domain of convergence has shifted to the right one unit.

--egm4313.s12.team11.gooding (talk) 03:48, 14 March 2012 (UTC)