Taylor Polynomials and Series..

**Vedicmaths** · June 25th 2008, 04:24 PM

Hello!

I need help figuring out the Taylor series..I keep getting undetermined form in these functions..Please help!

Q1) Suppose f(x) = cos(2x) is replaced by its Taylor Polynomial P2(X) about the point X(0) = pi/2 on the interval 0<= x <= pi. Estimate the error?

Q2) Derive the Taylor Series for f(x) = ln(1+x) with X(0) = 0 and determine for which value of X the series converges.

Thanks!

**Mathstud28** · June 25th 2008, 04:54 PM

Originally Posted by Vedicmaths

Hello!

I need help figuring out the Taylor series..I keep getting undetermined form in these functions..Please help!

Q1) Suppose f(x) = cos(2x) is replaced by its Taylor Polynomial P2(X) about the point X(0) = pi/2 on the interval 0<= x <= pi. Estimate the error?

Q2) Derive the Taylor Series for f(x) = ln(1+x) with X(0) = 0 and determine for which value of X the series converges.

Thanks!

For the first one this is an alternating series, so use the appropriate error test.

For the second one

Consider that

$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n\quad\forall{x}\in(-1,1)$

and that

$\ln(1+x)=\int\frac{dx}{1+x}$

**badgerigar** · June 25th 2008, 04:56 PM

Edit: sorry, ignore this, didn't see mathstud's post

Q1) Suppose f(x) = cos(2x) is replaced by its Taylor Polynomial P2(X) about the point X(0) = pi/2 on the interval 0<= x <= pi. Estimate the error?

The formula you are probably using is
$error \leq M\frac{r^{n+1}}{(n+1)!}$
where M is the maximum value of $f^{(n+1)}$ on the interval and r is the maximum distance from the point around which the taylor series is expanded and a point in the interval.

Q2) Derive the Taylor Series for f(x) = ln(1+x) with X(0) = 0 and determine for which value of X the series converges.

If this doesn't make sense I don't know how you are supposed to do this, but I am sure there is a way.

$\log(1+x) = \int\frac{1}{x+1}dx$
$=\int\frac{1}{1--x}$
using geometric series, with |-x|<1
$=\int \sum_{n=0}^{\infty}(-x)^n$
$=\sum_{n=0}^{\infty}\int (-x)^n$
$=\sum_{n=0}^{\infty}\frac{1}{n+1}(-x)^{n+1}$
$=\sum_{n=1}^{\infty}\frac{1}{n}(-x)^{n}$
$=\sum_{n=1}^{\infty}\frac{(-1)^n}{n}x^{n}$

The series converges whenever |x|<1 because 1 is the distance to the nearest value of x that produces a singularity in C.

**Mathstud28** · June 25th 2008, 04:59 PM

Originally Posted by badgerigar

If this doesn't make sense I don't know how you are supposed to do this, but I am sure there is a way.

$\log(1+x) = \int\frac{1}{x+1}dx$
$=\int\frac{1}{1--x}$
using geometric series, with |-x|<1
$=\int \sum_{n=0}^{\infty}(-x)^n$
$=\sum_{n=0}^{\infty}\int (-x)^n$
$=\sum_{n=0}^{\infty}\frac{1}{n+1}(-x)^{n+1}$
$=\sum_{n=1}^{\infty}\frac{1}{n}(-x)^{n}$
$=\sum_{n=1}^{\infty}\frac{(-1)^n}{n}x^{n}$

The series converges whenever |x|<1 because 1 is the distance to the nearest value of x that produces a singularity in C.

Uhm, the other way would be using the definition of the Maclaurin series

$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)x^n}{n!}$ and notcing the pattern

And the interval of convergence is $(-1,1)$

By the root test.

**badgerigar** · June 25th 2008, 05:15 PM

Uhm, the other way would be using the definition of the Maclaurin series

and notcing the pattern

oh. good point

.

I just looked at the wikipedia entry for the root test and it looked useful too
Thanks

**Mathstud28** · June 25th 2008, 05:29 PM

Originally Posted by badgerigar

oh. good point

.

I just looked at the wikipedia entry for the root test and it looked useful too
Thanks

How is it that you are working with power series but are ignorant of the Root test (not at all meant to be hostile, I am just curious)

**badgerigar** · June 25th 2008, 05:43 PM

How is it that you are working with power series but are ignorant of the Root test (not at all meant to be hostile, I am just curious)

I dunno. I'm a second year student at Melbourne uni and I don't recall having seen it. The most advanced stuff I have done in power series was in a semester of complex analysis, although Taylor series have been a major part of every single subject I have done since starting uni except my first year programming subjects. I also learn stuff as I read on this forum, which gives me a more patchy education on certain topics, but that doesn't really apply to power series because I have been doing that as part of my course.

**Mathstud28** · June 25th 2008, 05:45 PM

Originally Posted by badgerigar

I dunno. I'm a second year student at Melbourne uni and I don't recall having seen it. The most advanced stuff I have done in power series was in a semester of complex analysis, although Taylor series have been a major part of every single subject I have done since starting uni except my first year programming subjects. I also learn stuff as I read on this forum, which gives me a more patchy education on certain topics, but that doesn't really apply to power series because I have been doing that as part of my course.

Hmm, that is interesting. You have taken complex analysis, with all its Laurent series and such I am sure then that you are well aquainted with poewr series, but usually the Root test which is an integral part of series convergence/divergence you skipped. It's not a bad thing I suppose, just peculiar

. You should probably learn it

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