Question

MacLaurin series cos(x^2)

Answer 1

Supposed to find the first 4 non-zero terms of the series above

Answer 2

I get:
f(0)=1
f'(0)=0
f''(0)=0
f'''(0)=0

Answer 3

wondering if its ever gonna not be zero besides the first term...

Answer 4

f(x) = cos(x)^2
f'(x) = -2 cos(x) sin(x)
f''(x) = 2 sin^2(x) - 2 cos^2(x)
f'''(x) = 8 cos(x) sin(x)
f''''(x) = -8 sin(x) + 8 cos(x)
f'''''(x) = -32 cos(x) sin(x)
f''''''(x) = 32 sin^2(x) - 32 cos^2(x)

f(0) = 1
f'(0) = 0
f''(0) = -2
f'''(0) = 0
f''''(0) = 8
f'''''(0) = 0
f''''''(0) = -32

Is this of any help?

Answer 5

I think remnant is true

Answer 6

Not (cos(x))^2 but cos(x^2)

Answer 7

please read question properly: cos(x^2)

Answer 8

So for the first four series the answer is 1

Answer 9

i got:\[f^{(4)}(0)=-8\] so i guess it does become non zero

Answer 10

im sorry but it looks like everyone else has got this so i will go elsewhere

Answer 11

\[\Large{ \cos(x^2) = \sum_{k=0}^\infty \frac{(-1)^k (x^2)^{2 k}}{(2 k)!}\\=1-\frac{x^4}{2}+\frac{x^8}{24}-\frac{x^{12}}{720}+\cdots}\]

Answer 12

just replace x from the maclaurin series for cos x as x^2, wallah you are done, no differentiation needed