Maclaurin series help

1. Apr 26, 2005

vabamyyr

i have trouble expandind f(x)= (sin2x)^2 into Maclaurin series
for sin(x) Maclaurin series is

$$\sum^{\infty}_{n=0} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$

probably the key is to change (sin2x)^2 into new shape. I found that

(sin2x)^2=2sin(2x^2), but that coefficent 2 is bothering me, what to do?

2. Apr 26, 2005

Galileo

That relation does not hold (the left side is always positive, the right side isn't).

But you can use another identity:

$$\sin^2 x=\frac{1}{2}(1-\cos 2x)$$

3. Apr 26, 2005

vabamyyr

$$\sin^2 2x=\frac{1}{2}(1-\cos 4x)=0.5-0.5cos4x$$

and now if i apply for cosx maclaurin series expansion considering the function f(x)= -0,5cos4x i get the right answer but im puzzled, where does the coefficent 0,5 go????? i dont have to count that??

Last edited: Apr 26, 2005
4. Apr 26, 2005

mathman

The constant term of the series for sin2(2x) is 0.
The 0.5 cancels the -0.5 from the expansion of -0.5*cos(4x).

Last edited: Apr 26, 2005
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