+0

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

0
354
2

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

Guest May 4, 2014

#2
+889
+5

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,

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

Bertie  May 4, 2014
Sort:

#1
+80983
+3

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 !!

CPhill  May 4, 2014
#2
+889
+5

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,

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

Bertie  May 4, 2014

### 5 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details