# Thread: Taylor Polynomials and Series..

1. ## Taylor Polynomials and Series..

Hello!

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!

2. Originally Posted by Vedicmaths
Hello!

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

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

and that

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

3. 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
$\displaystyle error \leq M\frac{r^{n+1}}{(n+1)!}$
where M is the maximum value of $\displaystyle f^{(n+1)}$ 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.

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

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

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.

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

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

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

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

By the root test.

5. 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

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)

7. 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.