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

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Answer 1 · 2015-07-09T03:23:19

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

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

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.

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