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

Jul 9, 2015

(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 ${\left(1 + x\right)}^{p}$ as a power series?

(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 $f \left(x\right) = {\left(1 + x\right)}^{p}$ 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 ${\left(1 + x\right)}^{p}$ for $| x | < 1$ is harder, and I won't go into it here.