# Thread: Taylor Series

1. ## Taylor Series

1. Find the first five terms in the Taylor series about x = 0 for $\displaystyle f(x) = \frac{1}{1-2x}$ and find the interval of convergence for the series.

2. Then, using partial fractions, find the first five terms in the Taylor series about x = 0 for $\displaystyle g(x) = \frac{1}{(1-2x)(1-x)}.$

Originally I took the first four derivatives of f(x), plugged in x=0 for them, then used those 5 results (four derivatives + original) to get the first five terms. However, my teacher told me that was incorrect, so I instead just plugged in x=0, x=1, up to x=4 into f(x) to get 1, -1, -1/3, -1/5, -1/7. Is that right?

(I've got issues with later parts too but I want to clear up the first part.)

2. Originally Posted by Lord Darkin
1. Find the first five terms in the Taylor series about x = 0 for $\displaystyle f(x) = \frac{1}{1-2x}$ and find the interval of convergence for the series.

2. Then, using partial fractions, find the first five terms in the Taylor series about x = 0 for $\displaystyle g(x) = \frac{1}{(1-2x)(1-x)}.$

Originally I took the first four derivatives of f(x), plugged in x=0 for them, then used those 5 results (four derivatives + original) to get the first five terms. However, my teacher told me that was incorrect, so I instead just plugged in x=0, x=1, up to x=4 into f(x) to get 1, -1, -1/3, -1/5, -1/7. Is that right?

(I've got issues with later parts too but I want to clear up the first part.)

If your teacher told you that was wrong perhaps she/he meant that you didn't derivate or substitute or something properly, because taking the first four derivatives and plugging in x = 0 is what must be done.

The series' first 5 elements are $\displaystyle f(0)+\frac{f'(0)x}{1!}+\frac{f''(0)x^2}{2!}+\frac{ f'''(0)x^3}{3!}+\frac{f^{(iv)}(0)x^4}{4!}$ $\displaystyle =1+2x+4x^2+8x^3+16x^4$ , which is exactly the same as plugging in $\displaystyle 2x$ instead of $\displaystyle x$ in the well-known Taylor series $\displaystyle \frac{1}{1-x}=1+x+x^2+x^3+x^4+\ldots$ , which is true for $\displaystyle |x|<1$ . In your case it'd be for $\displaystyle |2x|<1\Longleftrightarrow |x|<\frac{1}{2}$

Tonio

3. Ah, I understand part 1 now. It was a little misunderstanding between us.

My first four terms come out to be:

1, 2, -4, 12, 48

Now for the other part you said ... are you saying the interval of convergence is from -1/2 to 1/2? The way I was taught to do it was to put the series in series/sigma notation, such as this:

$\displaystyle \sum (2x)^n$

Then do a ratio test or limit comparision test to get 2 possibilities for the endpoints of x, then plug the 2 endpoints back in the original series and determine whether it convergers/diverges to determine if I need a bracket or a paranthesis (includes the endpoint or does not include the endpoint).

If I plug in -1/2 and 1/2 in for the series, I get:

$\displaystyle \sum (-1)^n$

and

$\displaystyle \sum (1)^n$

Which ultimately causes both series to diverge, so my final interval of convergence is:

(-1/2,1/2)

With paranthesis to indicate that the halves are NOT INCLUDED.

4. Okay, I understand the first part. But I"m not sure how to use partial fractions for the last part. I know how to aply them to get an integrand, but not a derivative. And it's hard to manually find the derivatives because taking derivatives of g(x) results in messy equations due to the chain rule.

5. Anyone?

6. Partial fractions:
$\displaystyle \frac{1}{(1- 2x)(1- x)}= \frac{A}{1- 2x}+ \frac{B}{1- x}$

Multiply on both sides by (1- 2x)(1- x) to get
1= A(1- x)+ B(1- 2x) for all x.

Taking x= 1 and x= 1/2 gives you A and B very easily.

Now use the result before to find the Taylor series for this function.