User:Egm4313.s12.team8.koester/R4.4.2

Problem Statement
Develop $$ \displaystyle log(1+x)$$ in a Taylor Series about â=1.

Compute for n= 4, 7, 11.

Plot these three functions along with the real solution.

Find the domain of convergence.

Solution
Taylor Series expansions follow the form:
 * $$ \ \sum_{n = 0}^{\infty} \frac { f^n(a) (x-a)^n }{n!}\ $$

For our problem this translates to:
 * $$ \ \sum_{n = 1}^{\infty} (-1)^{n+1}\frac { (x-1)^n }{n}\ $$

We start the summation at 1 because log(0) is not defined.

For n=4 terms this expands to:
 * $$ (x-1)- \frac {(x-1)^2}{2} + \frac {(x-1)^3}{3} - \frac {(x-1)^4}{4} $$

For n=7 terms this expands to:
 * $$ (x-1)- \frac {(x-1)^2}{2} + \frac {(x-1)^3}{3} - \frac {(x-1)^4}{4} + \frac {(x-1)^5}{5} - \frac {(x-1)^6}{6} + \frac {(x-1)^7}{7}  $$

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

Graph
Now these 3 functions will be plotted against the actual solution:

It can then be found from observation of the graph that the domain of convergence is (0,2).