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

Question

Paisley Johnson · Accepted Answer

#(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.

Luke Alvarez · Answer

undefined To expand $ y = f(x) $ as a power function using the binomial series, you can follow these steps:

1. Write the function $ f(x) $ in the form $ f(x) = (1 + g(x))^n $, where $ g(x) $ is a function such that $ g(0) = 0 $ and $ n $ is a constant.

2. Apply the binomial theorem, which states that $ (1 + g(x))^n = \sum_{k=0}^{\infty} \binom{n}{k} g(x)^k $, where $ \binom{n}{k} $ represents the binomial coefficient.

3. Expand the series using the binomial coefficients and simplify.

4. If necessary, express the result as a power function of $ x $.