Question

how to find the Taylor series for
f(x)=cos(2x^4)?
Please help! The answer should be in summation form Sigma from k=0 to infinity

Answer 1

Compute the series of \(\cos(y)\). Then choose \(y=2x^4\) and see what to coeffients become. It's a simple plugin method.

Answer 2

http://www.webassign.net/userimages/knownseries.jpg?db=v4net&id=154152
So i use the cos(x) taylor series but how do you find the summation of the series cos(2x^4)?
I only know how to find the first few terms of the sequence.

Answer 3

You can use the general term: \(y^k\) becomes \((2x^4)^k\). This will be the new general term.

Answer 4

without forgetting the \((-1)^{\dots}\frac1{(2k)!}\) factor of course

Answer 5

ah lol. I was overthinking it. Thanks!
But could you also explain how to find taylor series of f(x)=sin^2(5x)? It says to use the sin^2(theta) = 1/2(1-cos(2theta)) identity.

Answer 6

if you forget the infinite feature of the Taylor expansion, adding 1 or 1/2 or 'x' to a polynomial is simple. Actually, it's just as simple as that. Example:
\[e^x - 1 = (1+\frac{x}{1!} + \frac{x^2}{2!} +\dotsb) - 1\\
\quad = \frac{x}{1!} + \frac{x^2}{2!} +\dotsb
\]
and
\[
\frac12 e^x = \frac12 + \frac{x}{2\times1!} + \frac{x^2}{2\times2!} +\dotsb
\]

Answer 7

so for the next you just need to find the taylor series of cos(2x) by the same substitution method and then plug the series in 1/2*(1-series) and find the new coefficients.