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.
The expansion about what value of x?
The coefficient of $\displaystyle x^n$ in the Taylor series expansion of f(x) about x= a is $\displaystyle \frac{f^{(n)}(a)}{n!}$ where $\displaystyle f^{(n)}(a)$ is the nth derivative of f at x= a. Can you find the nth derivative of cos(2x)?
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).
Hello, kingman!
$\displaystyle \text{Find the coefficient of }x^n\text{ in the expansion of }\cos 2x$
$\displaystyle \displaystyle \cos2x \;=\;1 - \frac{(2x)}{2!} + \frac{(2x)^2}{4!} - \frac{(2x)^6}{6!} + \frac{(2x)^8}{8!} - \hdots $
. . . . . $\displaystyle \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 $
$\displaystyle \,n$ is an even nonnegative integer.
. . . The coefficient of $\displaystyle \,x^n$ is: .$\displaystyle (\text{-}1)^{\frac{n}{2}}\dfrac{2^n}{n!} $
Don't you mean instead:
$\displaystyle \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.