# A problem!

• June 25th 2010, 01:54 PM
Also sprach Zarathustra
A problem!
$f(x)=x-\frac{3}{2}x^2+\frac{11}{6}x^3-\frac{50}{24}x^4+\frac{274}{120}x^5+...$
This is Taylor series of function $f(x)=\frac{ln(1+x)}{1+x}$

My question is:

$f(x)=x-\frac{3}{2}x^2+\frac{11}{6}x^3-\frac{50}{24}x^4+\frac{274}{120}x^5+...=\Sigma^\in fty_{n=0}(-1)^{n+1}a_n x^n$

What is form of $a_n$?

Thank you very much!
• June 25th 2010, 04:37 PM
Ackbeet
Well, the denominators look something like a factorial (except for that pesky 3 in the squared term; perhaps you could massage it to look better). Maybe you could set up a recurrence relation to figure out the numerators?
• June 25th 2010, 04:49 PM
Also sprach Zarathustra
Ohhh... it's not 2/3 it's 3/2 :(

Sorry!!!

(I'm going to change it in my original post)

Thank you Mr. Keister!
• June 25th 2010, 05:11 PM
Ackbeet
Aha. So it is a nice little factorial in the denominator. Well, the recurrence relation idea sounds nice, but I'm not sure how it would work in principle. Why don't you show all the details of the computations for those first few terms? Maybe something'll pop out.
• June 25th 2010, 05:15 PM
Also sprach Zarathustra
Quote:

Originally Posted by Ackbeet
Aha. So it is a nice little factorial in the denominator. Well, the recurrence relation idea sounds nice, but I'm not sure how it would work in principle. Why don't you show all the details of the computations for those first few terms? Maybe something'll pop out.

Here it is: :) Function calculator

Moreover... look here...

1 3 11 50 - OEIS Search Results
• June 25th 2010, 05:25 PM
Ackbeet
No, that's not what I'm interested in. I'm interested in seeing the step-by-step calculations for computing each coefficient. I'm curious to see if a pattern pops out by examining that process.
• June 25th 2010, 05:39 PM
Also sprach Zarathustra
Unfortunately, I can't do so, it is very complicated, but the thing I can do is to post the original question asked.

Q.
Find Taylor polynomial of $f(x)=\frac{ln(1+x)}{1+x}$ around $x=0$, find formula for $f^{(n)}(x)$.
Find the radius of convergence of infinite Taylor polynomial(power series).
• June 25th 2010, 07:32 PM
simplependulum
Hi , it is $H_n = \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n}$
• June 25th 2010, 07:41 PM
Also sprach Zarathustra
Make sense!

How you get this result please?
• June 25th 2010, 08:05 PM
Also sprach Zarathustra
So, the radius of convergence $R$ of $\Sigma^{\infty}_{n=0}(-1)^nH_nx^n$ is R=0?

Thanks!
• June 25th 2010, 08:39 PM
simplependulum
No , the radius is 1 , i obtain the general term by expanding $\ln(1+x)$ and $\frac{1}{1+x}$ followed by coefficients' consideration .
• June 26th 2010, 12:11 AM
chisigma
The general formula for the product of two series is...

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

... and in Your case is...

$\displaystyle \ln (1+x)= \sum_{n=1}^{\infty} (-1)^{n+1}\ \frac{x^{n}}{n}$

$\displaystyle \frac{1}{1+x} = \sum_{k=0}^{\infty} (-1)^{k}\ x^{k}$ (2)

Kind regards

$\chi$ $\sigma$
• June 26th 2010, 08:56 AM
Also sprach Zarathustra
I have difficulties in finding radius of convergence! Why it equals to 1?

In other words, WHY $R=\lim_{n\to\infty}\frac{1}{\sqrt[n]{H_n}} = 1$?
• June 26th 2010, 09:01 AM
Also sprach Zarathustra
Maybe it from the "sandwich" rule?
$1
$\sqrt[n]{1}<\sqrt[n]H_n<\sqrt[n]{n}$

so when $n\to\infty$

$H_n \to 1$
• June 26th 2010, 09:42 AM
theodds
If $\lim_{n \to \infty} \frac{a_{n + 1}}{a_n}$ exists, then $\lim_{n \to \infty} \sqrt[n]{a_n}$ does as well, and they are equal. Since the first expression exists, you can use it to find the radius of convergence, which will work out to be 1.

If you think to calculate $a_n - a_{n - 1} = \frac{1}{n}$ from the terms you listed out then it becomes pretty easy to figure out the form of the $a_n$, though this isn't exactly a rigorous solution.