Question

maclaurian series of f(x) =x^2 cos(x)^2

Answer 1

The Maclaurin series for \(\cos x\) is \[\sum_{n=1}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}\]Replace \(x\) with \(x^2\), and you will get the Maclaurin series for \(\cos x^2\).
Then multiply by \(x^2\), so that you will get the Maclaurin series for \(x^2\cos x^2\).

Answer 2

Sorry, n = 1 must be n = 0 instead.

Answer 3

cos(x)^2 = (cosx)^2

Answer 4

The idea is that you need to find the series expansion for cosx, then do
( series of cosx) * (series of cosx) to get the series of (cosx)^2,

then do x^2 * (series of (cosx)^2)

Of course, the process is extremely tedious

Answer 5

But isn't \(\cos x^2\) and \(\cos^2 x=(\cos x)^2\) completely different?

Answer 6

here is the series for cos(x)^2 done by wolfram alpha

Answer 7

when you write cos(x)^2, it is actually (cosx)^2, not cos(x^2).

Answer 8

(cosx)^2 = cos^2(x) = cos(x)^2

there are many ways to write it.

Answer 9

Oh, I see. :)

Answer 10

so answer is