User:Egm4313.s12.team11.arrieta/Report 4

Part 1
Solved by Francisco Arrieta

Problem Statement

 * Develop $$ \log(x+1) \!$$ in Taylor Series about $$ x=0 \!$$ to reproduce the figure on page 7-25

Given

 * $$ f(x)= \sum_{n=0}^\infty \frac{f^{(n)}(\hat x)}{n!}(x- \hat x)^n \!$$

Solution
For n=4:
 * $$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)} \!$$

For n=7:
 * $$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)} \!$$

For n=11:
 * $$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})} \!$$

For n=16:
 * $$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})}\!$$
 * $$ -\frac{x^{12}}{\log(10^{12})}+\frac{x^{13}}{\log(10^{13})}-\frac{x^{14}}{\log(10^{14})}+\frac{x^{15}}{\log(10^{15})}-\frac{x^{16}}{\log(10^{16})} \!$$



Using MATLAB

--Egm4313.s12.team11.arrieta (talk) 00:44, 14 March 2012 (UTC)