Taylor Series for 1-cos(x^2) around a=0

Are you allowed to quote the Maclaurin series for cos(x) ? If yes then a simple substitution will do.

$$\cos x = 1 - \frac{x^{2}}{2!}+\frac{x^{4}}{4!} - \dots$$ ,

so, replacing x by x^2,

$$\cos x^{2}= 1 - \frac{x^{4}}{2!}+\frac{x^{8}}{4!}- \dots$$ .

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 .....