Question

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.

Answer 1

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.

Answer 2

Note: The expression had been abbreviated as:

(ax+bx³+c⁵+d⁷)²
where
a=+1
b=-(1/3!)
c=+(1/5!)
d=-(1/7!)

Answer 3

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.

Answer 4

impressive!

Answer 5

Thanks, it was just a fluke!

Answer 6

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.

Answer 7

Tips are welcome, to your favourite charity!