Question

How do you use the binomial series to expand `y=f(x)` as a power function?

Answer 1

`(1+x)^{p}=1+px+(p(p-1))/(2!)x^2+(p(p-1)(p-2))/(3!)x^3+(p(p-1)(p-2)(p-3))/(4!)x^4+\cdots`,

In general, this converges for `|x|<1` (though it converges for all `x` if `p` is a non-negative integer).

Explanation:

Perhaps your question is meant to say: how do you use the binomial series to expand `(1+x)^p` as a power series?

If so, the answer is:

`(1+x)^{p}=1+px+(p(p-1))/(2!)x^2+(p(p-1)(p-2))/(3!)x^3+(p(p-1)(p-2)(p-3))/(4!)x^4+\cdots`

In general, this converges for `|x|<1` (though it converges for all `x` if `p` is a non-negative integer).

The expansion of `f(x)=(1+x)^{p}` as a power series can be computed from the Taylor (Maclaurin) series formula:

`f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+\cdots`

Give it a try!

Proving that the Taylor series equals `(1+x)^{p}` for `|x|<1` is harder, and I won't go into it here.