# Finding Taylor series for f

• Nov 17th 2011, 10:40 PM
Intrusion
Finding Taylor series for f
Hello there, I was having some trouble with this problem:

Find the Taylor series for f centered at 4 if:

$$f^n(0) = \frac{(-1)^nn!}{3^n(n+1)}$$

What is the radius of convergence of the Taylor series?

Help appreciates.
• Nov 17th 2011, 11:37 PM
alexmahone
Re: Finding Taylor series for f
Quote:

Originally Posted by Intrusion
Hello there, I was having some trouble with this problem:

Find the Taylor series for f centered at 4 if:

$$f^n(0) = \frac{(-1)^nn!}{3^n(n+1)}$$

What is the radius of convergence of the Taylor series?

Help appreciates.

To find the Taylor series for f centered at 4, you need the values of $f^n(4)$.
• Nov 18th 2011, 03:44 AM
chisigma
Re: Finding Taylor series for f
Quote:

Originally Posted by Intrusion
Hello there, I was having some trouble with this problem:

Find the Taylor series for f centered at 4 if:

$$f^n(0) = \frac{(-1)^nn!}{3^n(n+1)}$$

What is the radius of convergence of the Taylor series?

Help appreciates.

The Taylor expansion of f(*) around z=0 is...

$f(z)= \sum_{n=0}^{\infty} \frac{(-1)^{n}\ z^{n}}{3^{n}\ (n+1)}$ (1)

... and the series (1) has radius of convergence 3 because the singularity of the function in z=-3. Therefore the series of f(*) centered in z=4 has radius of convergence 7...

Kind regards

$\chi$ $\sigma$
• Nov 18th 2011, 09:38 PM
FernandoRevilla
Re: Finding Taylor series for f
The problem is ill defined, we have no sufficient information about $f$ . For example

$f(x)=\begin{Bmatrix} \displaystyle \sum_{n=0}^{\infty}\dfrac{(-1)^n}{3^n(n+1)}x^n& \mbox{if}& |x|<3\\x-4 & \mbox{if}& |x|\geq 3\end{matrix}\quad g(x)=\begin{Bmatrix} \displaystyle \sum_{n=0}^{\infty}\dfrac{(-1)^n}{3^n(n+1)}x^n& \mbox{if}& |x|<3\\|x-4| & \mbox{if}& |x|\geq 3\end{matrix}$

We have $f^{(n)}(0)=g^{(n)}(0)=\dfrac{(-1)^nn!}{3^n(n+1)}$ . However: a) The Taylor series of $f$ centered at $4$ has radius of convergence $+\infty$ . b) The Taylor series of $g$ centered at $4$ does not exist .
• Nov 19th 2011, 01:07 AM
chisigma
Re: Finding Taylor series for f
Quote:

Originally Posted by chisigma
The Taylor expansion of f(*) around z=0 is...

$f(z)= \sum_{n=0}^{\infty} \frac{(-1)^{n}\ z^{n}}{3^{n}\ (n+1)}$ (1)

... and the series (1) has radius of convergence 3 because the singularity of the function in z=-3. Therefore the series of f(*) centered in z=4 has radius of convergence 7...

May be that a further explanation from me is necessary... if we consider the series expansion...

$g(x)= \sum_{n=0}^{\infty} \frac{(-1)^{n}\ x^{n}}{3^{n}}= \frac{1}{1+\frac{x}{3}}$ (1)

...it is not too difficult to obtain...

$f(x)= \sum_{n=0}^{\infty} \frac{(-1)^{n}\ x^{n}}{3^{n}\ (n+1)}= \frac{1}{x}\ \int_{0}^{x} g(\xi)\ \d \xi = \frac{3\ \ln (1+\frac{x}{3})}{x}$ (2)

Now the f(*) defined in (2) is analytic for $x>-3$ and that means that its series expansion centered in $x_{0}>-3$ has radius of convergence $x_{0}+3$ ...

Kind regards

$\chi$ $\sigma$
• Nov 19th 2011, 01:48 AM
FernandoRevilla
Re: Finding Taylor series for f
Quote:

Originally Posted by chisigma
May be that a further explanation from me is necessary...

Well, independently of your explanation I would insist that the only solution to the problem is: The radius of convergence of the Taylor series for $f$ centered at $x_0=4$ (if exists) can be any $\rho\in[0,+\infty]$ depending on the expression of $f^{(n)}(4)$ .