maclaurin series

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Apr 13th 2013, 03:40 AM
iragequit

maclaurin series

given f(x)=x^2/(1-x^2) find the value of f^(2012)(0)
where f^(2012)(0) is the 2012'th derivative of f(x) evaluated at a=0

answer is 2012! , have no idea how they did this...

also, another quesiton --

what is the coefficient of x^2012 in the maclaurin series of the function f(x) = sin(-x) ?
Apr 13th 2013, 04:00 AM
Prove It

Re: maclaurin series

For starters, you should know that a MacLaurin series for a function f(x) looks like $\displaystyle \begin{align*} f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(0)}{n!}\, x^n \end{align*}$ .

Now if $\displaystyle f(x) = \frac{x^2}{1 - x^2} = x^2 \left( \frac{1}{1 - x^2} \right)$ we can recognise that $\displaystyle \frac{1}{1 - x^2}$ is the closed form of the geometric series $\sum_{m = 0}^{\infty} \left( x^2 \right) ^m$ with $\displaystyle \left| x^2 \right| < 1 \implies |x| < 1$ , so that means

$\displaystyle \begin{align*} f(x) &= \frac{x^2}{1 - x^2} \\ &= x^2 \sum_{m = 0}^{\infty} \left( x^2 \right) ^m \\ &= \sum_{m = 0}^{\infty} x^{2m + 2} \\ &= 0 + 0x + 1x^2 + 0x^3 + 1x^4 + 0x^5 + 1x^6 + \dots \end{align*}$

which is a MacLaurin Series. It can be seen that the coefficient of every even-powered term (except the constant term) is 1, so that means the coefficient of the 2012th term is 1. Remember that every coefficient in the MacLaurin Series is of the form $\displaystyle \frac{f^{(n)}(0)}{n!}$ , this means for the 2012th term we have

$\displaystyle \begin{align*} \frac{f^{(2012)}(0)}{2012!} &= 1 \\ f^{(2012)}(0) &= 2012! \end{align*}$
Apr 13th 2013, 04:07 AM
Prove It

Re: maclaurin series

As for your second question, note that if $\displaystyle f(x) = \sin{(-x)}$ then

$\displaystyle \begin{align*} f(x) &= \sin{(-x)} \\ \\ f'(x) &= -\cos{(-x)} \\ \\ f''(x) &= -\sin{(-x)} \\ \\ f'''(x) &= \cos{(-x)} \\ \\ f^{(iv)}(x) &= \sin{(-x)} \end{align*}$

So every fourth derivative is the same as the starting function, and since 2012 is a multiple of 4, that means $\displaystyle f^{(2012)}(x) = \sin{(-x)} \implies f^{(2012)}(0) = 0$ .

So the coefficient of $\displaystyle x^{2012}$ is $\displaystyle \frac{f^{(2012)}(0)}{2012!} = 0$ .
Apr 13th 2013, 04:42 AM
iragequit

Re: maclaurin series

Quote:

Originally Posted by Prove It

For starters, you should know that a MacLaurin series for a function f(x) looks like $\displaystyle \begin{align*} f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(0)}{n!}\, x^n \end{align*}$ .

Now if $\displaystyle f(x) = \frac{x^2}{1 - x^2} = x^2 \left( \frac{1}{1 - x^2} \right)$ we can recognise that $\displaystyle \frac{1}{1 - x^2}$ is the closed form of the geometric series $\sum_{m = 0}^{\infty} \left( x^2 \right) ^m$ with $\displaystyle \left| x^2 \right| < 1 \implies |x| < 1$ , so that means

$\displaystyle \begin{align*} f(x) &= \frac{x^2}{1 - x^2} \\ &= x^2 \sum_{m = 0}^{\infty} \left( x^2 \right) ^m \\ &= \sum_{m = 0}^{\infty} x^{2m + 2} \\ &= 0 + 0x + 1x^2 + 0x^3 + 1x^4 + 0x^5 + 1x^6 + \dots \end{align*}$

which is a MacLaurin Series. It can be seen that the coefficient of every even-powered term (except the constant term) is 1, so that means the coefficient of the 2012th term is 1. Remember that every coefficient in the MacLaurin Series is of the form $\displaystyle \frac{f^{(n)}(0)}{n!}$ , this means for the 2012th term we have

$\displaystyle \begin{align*} \frac{f^{(2012)}(0)}{2012!} &= 1 \\ f^{(2012)}(0) &= 2012! \end{align*}$

Thanks a lot. You're actually a life saver. I just have one more question. I had this one question on my quiz a few months ago, and I still don't understand it:

Find the value of c for which the integral 0 to infinity of { x / (x^2+1) - c / (3x+1) }dx converges, and evaluate the integral for this c.

I just need help with finding the value of c. I vaguely remember it being c=3, but I don't know how. I know how to do the integral, so if you could just help me understand why c=3... thanks.

I know you combine the fractions and you get { (3-c)x^2 + x - c } / (x^2+1)(3x+1) and you set c =3 to make the x^2 vanish, but why does x^2 have to vanish for it to converge?
Apr 13th 2013, 04:49 AM
iragequit

Re: maclaurin series

Actually I have two more questions... I got these two wrong on my midterm, and I don't understand how to do it to this day.

Why is sum n=0 to infinity of { n*cos(n) / 2^n } only conditionally convergent (I think that was the answer?)? I know you use the ratio test and you eventually get |cos(n+1)/2cos(n)| < 1, but what do you do after?

and also...

Suppose sum n=1 to infinity of a_n diverges. Which of the following statements are always true?
i) sum n=1 to inf |a_n| diverges
ii) sum n=1 to infinity (-1)^n *a_n diverges
iii) sum n=1 to infinity (a_n)^2 diverges

I chose i and iii and it's wrong. Not sure what the answer is.
Apr 15th 2013, 07:30 PM
iragequit

Re: maclaurin series

pls respond
Apr 15th 2013, 07:45 PM
Prove It

Re: maclaurin series

For starters, you did not get any responses because your new questions were not posted in a new thread.

For the first, have you tried evaluating the integral?

I don't see why you would use the ratio test (which would show absolute convergence) when you are wanting to show conditional convergence. A series is conditionally convergent if it can be shown that it converges although the series of absolute values does not!

iii) can not possibly be true. Take for example the harmonic series, it is divergent, but if you square each term it's a convergent p-series. So only i) is correct.