# Thread: Taylor Series Expansion Around 0.25

1. ## Taylor Series Expansion Around 0.25

Hi, I am having to expand the taylor series 1/(1-x) around the point 0.25 to 3 degrees.

I am okay expanding it around x=0 and x=.5 but for this question I am having difficulties due to the point being 0.25.

Currently I have gotten 4/3 for the first one and (16/9x - 4/9) for the 1st order.

The second order I got is (32/27)x^2 - (16/27)x + (2/27).

Due to all these extra terms being generated, I am unable to cancel out terms to make it into a geometric series. Any help would be appreciated.

2. Originally Posted by todo
Hi, I am having to expand the taylor series 1/(1-x) around the point 0.25 to 3 degrees.

I am okay expanding it around x=0 and x=.5 but for this question I am having difficulties due to the point being 0.25.

Currently I have gotten 4/3 for the first one and (16/9x - 4/9) for the 1st order.

The second order I got is (32/27)x^2 - (16/27)x + (2/27).

Due to all these extra terms being generated, I am unable to cancel out terms to make it into a geometric series. Any help would be appreciated.
Taylor expansion about $x_0$:

$\displaymstyle f(x)=\sum_{n=0}^{\infty} \frac{1}{n!}(x-x_0)^nf^{(n)}(x_0)$

$\displaymstyle f(x)=\frac{1}{0.75}+(x-0.25)\frac{1}{0.75^2}+\frac{(x-0.25)^2}{2}\frac{2}{0.75^3}+ ...$

which simplifies to:

$f(x)=\frac{28}{27}+\frac{16}{27} x+\frac{64}{27} x^2+O(x-0.25)^3$

CB

3. If you can do it for $x_0= 0.5$, why would $x_0= 0.25$ be any different?

Captain Black gives the formula for a general Taylor's series. Since you specifically mention "geometric series", you can also do it this way-

Write $\frac{1}{1- x}$ as $\frac{1}{1-x+ .25- .25}= \frac{1}{1-.25- (x- .25)}= \frac{1}{.75- (x- .25)}$.

Now divide both numerator and denominator by .75: $\frac{1}{1- x}= \frac{\frac{4}{3}}{1- \frac{x- .25}{.75}}$

Now, you should be able to recognise that as $\frac{A}{1- r}$ with A= 4/3 and $r= \frac{x- .25}{.75}$ and you know that that is the sum of the geometric series $\sum_{n=0}^\infty Ar^n$.