# Taylor/Maclaurin polynomials

• Mar 5th 2008, 08:35 AM
larson
Taylor/Maclaurin polynomials
Hey guys, I just wanted to make sure that I did this question correctly:

The fourth-order Taylor polynomial for $\displaystyle f(x) = \frac {1}{x}$ based at $\displaystyle x = 1$ is $\displaystyle P4(x) = 1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + (x - 1)^4$. Find the fourth-order Maclaurin polynomial for $\displaystyle g(x) = \frac {1}{(1 + x)}$

So all I did was place (1 + x) wherever I see an x up in the original function, which would mean...

$\displaystyle 1 - ((1 + x)-1) + ((1 + x)-1)^2 - ((1 + x) - 1)^3 + ((1 + x) -1)^4$

which would then mean the final answer is:
$\displaystyle 1 - x + x^2 - x^3 + x^4$ right?
• Mar 5th 2008, 07:43 PM
TheEmptySet
Quote:

Originally Posted by larson
Hey guys, I just wanted to make sure that I did this question correctly:

The fourth-order Taylor polynomial for $\displaystyle f(x) = \frac {1}{x}$ based at $\displaystyle x = 1$ is $\displaystyle P4(x) = 1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + (x - 1)^4$. Find the fourth-order Maclaurin polynomial for $\displaystyle g(x) = \frac {1}{(1 + x)}$

So all I did was place (1 + x) wherever I see an x up in the original function, which would mean...

$\displaystyle 1 - ((1 + x)-1) + ((1 + x)-1)^2 - ((1 + x) - 1)^3 + ((1 + x) -1)^4$

which would then mean the final answer is:
$\displaystyle 1 - x + x^2 - x^3 + x^4$ right?

You could have also rewritten

$\displaystyle \frac{1}{1+x}=\frac{1}{1-(-x)}$ This is the def of a geometric series with r =-x so you would get

$\displaystyle \sum_{n=0}^\infty(-x)^n=\sum_{n=0}^\infty(-1)^nx^n$

if you write out the first few terms you get...

$\displaystyle 1-x+x^2-x^3+x^4$