I want to get the first four terms of the expansion of this expression
(x - x^3/3! + x^5/5! - x^7/7!)^2
I already got the answer:
x^2 - x^4/3 + 2x^6/45 - x^8/315
However, I believe I went the long manual route. What's the shortcut to this kind of expansion?
The book gives
x^2 - 2^3/4! x^4 + 2^5/6! x^6 - 2^7/8! x^8
That works out to the same answer, but it looks like they used a shortcut to get that more easily.
Expand the polynomial as follows, but retain the lowest order terms by collecting the coefficients of products that produce x2,x4, x6 and x8, since they have to be all even powers.
The first four terms are:
a2x2 +
(2ab)x4 +
(b2+2ac)x6 +
(2ad+2bd)x8
By substituting
a=+1
b=-(1/3!)
c=+(1/5!)
d=-(1/7!)
you should get the book results quickly.
Note: The expression had been abbreviated as:
(ax+bx3+c5+d7)2
where
a=+1
b=-(1/3!)
c=+(1/5!)
d=-(1/7!)
I looked at it again, it looks like that the book result uses a specific property of the series, namely
sin(x)=(x - x^3/3! + x^5/5! - x^7/7!)
so, instead of expanding algebraically, we can proceed to use the trigonometric identity of sin2(x), where
sin2(x)
= (1/2)(cos(x-x)-cos(x+x))
= (1/2)(1-cos(2x))
= (1/2)(1 - (1-(2x)2/2! + (2x)4/4! - (2x)6/6! + (2x)8/8! -...)
= what your book shows.
impressive!
Thanks, it was just a fluke!
Thanks! You guys are awesome. I would tip for this service if I could.
The book's question was to get the first four terms of sin^2(x). Makes perfect sense.