# Thread: finding coefficient of nth term

1. ## finding coefficient of nth term

Dear sir,
i would be glad if someone can help me in the below problem
thanks

Find the coefficient of x to the power of n in the expansion of
cos 2x.

2. The expansion about what value of x?

The coefficient of $x^n$ in the Taylor series expansion of f(x) about x= a is $\frac{f^{(n)}(a)}{n!}$ where $f^{(n)}(a)$ is the nth derivative of f at x= a. Can you find the nth derivative of cos(2x)?

3. Thanks for the reply but the problem is that i got something times 2x to the power of 2n instead of what i am requiring to find is x to the power of n. I need some help starting from here.

thanks you

4. Originally Posted by kingman
Thanks for the reply but the problem is that i got something times 2x to the power of 2n instead of what i am requiring to find is x to the power of n. I need some help starting from here.

thanks you

5. ## here is my answer

here i wonder how get coefficient of x to the power of n but not 2n

6. Dear Sir,
I would be most grateful if someone can help me in the posted question
thanks

7. Originally Posted by kingman
Dear Sir,
I would be most grateful if someone can help me in the posted question
thanks

8. Originally Posted by kingman
Dear Sir,
I would be most grateful if someone can help me in the posted question
thanks
People have. You have refused to do what you have been asked. You say you can get this "to the power of n but not 2n" but your original problem says nothing about a power of 2n. Do you mean you could find the coefficient of x^n if the problem were to expand cos(x) rather than cos(2x)?
(Around, I assume, x= 0. You still haven't answered the very first question asked of you.)

If so, please show us what you would get if this were cos(x) rather than cos(2x).

I have got my working inserted but need to find coefficient of x to the power of n

10. Hello, kingman!

$\text{Find the coefficient of }x^n\text{ in the expansion of }\cos 2x$

$\displaystyle \cos2x \;=\;1 - \frac{(2x)}{2!} + \frac{(2x)^2}{4!} - \frac{(2x)^6}{6!} + \frac{(2x)^8}{8!} - \hdots$

. . . . . $\displaystyle =\;1 - \frac{2^2}{2!}x^2 + \frac{2^4}{4!}x^4 - \frac{2^6}{6!}x^6 + \frac{2^8}{8!}x^8 - \hdots$

$\,n$ is an even nonnegative integer.
. . . The coefficient of $\,x^n$ is: . $(\text{-}1)^{\frac{n}{2}}\dfrac{2^n}{n!}$

$\displaystyle \cos2x \;=\;1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \frac{(2x)^8}{8!} - \hdots$

My physics teacher once showed us how to do it (we are not supposed to know this however) but I tend to forget. It was nice seeing it again.