What is power series for sin(x^2)

Question

austinl · Answer

@zepdrix 
Are you familiar with a way to work to it? Without just giving the answer?

zepdrix · Answer

I don't really remember power series that well. I would have to work it out the long way. :D lol

I have to go to school though :( I don't have time to help right now :c

austinl · Answer

I have always just had access to tables....

amistre64 · Answer

its the power series for sin, but using an x^2 instead of x

amistre64 · Answer

$$sin(u)=\sum\frac{(-1)^n}{(2n+1)!}u^{2n+1}$$
$$sin(x^2)=\sum\frac{(-1)^n}{(2n+1)!}x^{2(2n+1)}$$

austinl · Answer

That is what I get as well @amistre64

amistre64 · Answer

was the intent to construct it as a taylor series?  derivatives and such?

amistre64 · Answer

spose we have a function that is a constant, plus some powers of (x-c)

f = c0 + powers
f' = c1 + powers'
f'' = 2c2 + powers''
f''' = 3.2c3 + powers'''
f'''' = 4.3.2c4 + powers''''

when the (x-c) parts are set to zero, we have a way to define the coefficients that are popping out

anonymous · Answer

yeah thanks alot, tha helpped !

amistre64 · Answer

f = c0 
f' = c1
f'' = 2c2
f''' = 3.2c3
f'''' = 4.3.2c4

f = c0 
f' = c1
f''/2 = c2
f'''/3! = c3
f''''/4! = c4

$$c_n=f^{(n)}\frac1{n!}$$