Let's see if I remember any of Cal II
I believe that since we're expanding about 0 that we have a Maclaurin series. The expansion of this is given by
Sometimes, the most tedious thing about this activity (as you have probably discovered) is taking the derivatives. Let's do that for the first few terms and see if we can arrive at a "pattern".
f(0) = 1 - cos(0^2) =1- cos(0) = 1- 1 = 0
f'(x) = 2xsin(x^2) f'(0) = 0
f"(x) = 2sin(x^2) + 4x^2cos(x^2) f"'(0) = 0
f'''(x) = -8x^3sin(x^2) + 12xcos(x^2) f'''(0) = 0
f''''(x) = -48x^2sin(x^2) - 16^4cos(x^2) + 12cos(x^2) f''''(0) = 12
At this point, we can probably stop.
So far, the only term is (12)x^4 / 4! = (x^4) /2
Taking successive derivatives (very tedious, indeed) until we reached f8(x) would yield "0' terms
From an online app, the next non-zero terms occur at f8(x), f12(x), and f16(x) - as we might now expect.
So it appears that we have this .....
[(-1) n+1( x 4n )] / [(2n)! ] for n= 1, 2, 3.......
And the series would be written as
f(x) = 1 + (n=0, ∞)∑ [(-1) n+1( x 4n )] / [(2n)! ]
Note that we need the (1) because of the "-1" that the series would generate by starting at n=0. In effect, this "erases" that term. We could start the series at n=1, but then it wouldn't be in Maclaurin form.
Sorry...I don't know (yet) how to put the "n=0" and the infinity sign "under" and "over" the summa symbol, but I think you get the idea!!
I think that's it !!
