Question

maclaurin series cos(x^2) - 1

Answer 1

start with \[\cos y = \sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}y^{2n}\] or
\[\cos y = 1-\frac{y^2}{2!}+\frac{y^4}{4!}-\frac{y^6}{y!}+\cdots\]

Answer 2

The McLaurin serie of the function cos x is :
\[\Large \cos x= 1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots+(-1)^p\frac{x^{2p}}{(2p)!}+\cdots\]
So :
\[\Large \cos x^2-1=-\frac{x^4}{2!}+\frac{x^8}{4!}+\cdots+(-1)^p\frac{x^{4p}}{(2p)!}+\cdots \]

Answer 3

I need it in summation notation?

Answer 4

What do I do with the remaining - 1?

Answer 5

OK !
IT IS LIKE THIS :
\[\Large \cos x^2-1=\sum_{k=1}^{+\infty}(-1)^k\frac{x^{4k}}{(2k)!}\]

Answer 6

Does it matter is k starts at 0 or 1?

Answer 7

It is matter, because we have "-1" so it begins from k=1

Answer 8

Ok thank you so much! :)

Answer 9

You are welcome !

Answer 10

So it's for sure that and not