P4.4 - Binomial expansion

Syllabus
2019
Topic
P4.4
Level
A2

Expand a rational power and keep its valid range

The binomial series expands a rational power around x=0x=0. Put the expression into the standard form (1+u)n(1+u)^n first; the coefficients come from the descending products of nn, while the size of uu controls where a non-terminating series is valid.

(1+u)n=1+nu+n(n1)2!u2+n(n1)(n2)3!u3+(1+u)^n=1+nu+\frac{n(n-1)}{2!}u^2+\frac{n(n-1)(n-2)}{3!}u^3+\cdots

Rewrite (ax+b)n=bn(1+abx)n(ax+b)^n=b^n\left(1+\frac{a}{b}x\right)^n. For a negative or non-integer rational nn, require abx<1\left|\frac{a}{b}x\right|<1, so x<b/a|x|<\left|b/a\right|. A non-negative integer power terminates and does not need this convergence restriction.

For (4+5x)1/2(4+5x)^{1/2}, take out 41/2=24^{1/2}=2 and use u=5x/4u=5x/4: 2(1+12u18u2+116u3+)=2+54x2564x2+125512x3+.2\left(1+\frac12u-\frac18u^2+\frac1{16}u^3+\cdots\right)=2+\frac54x-\frac{25}{64}x^2+\frac{125}{512}x^3+\cdots. This series is valid for 5x/4<1|5x/4|<1, or x<4/5|x|<4/5.

For a rational function, decompose first and expand each simple fraction. For example, 212x+34+x=2(12x)1+34(1+x/4)1\frac{2}{1-2x}+\frac{3}{4+x}=2(1-2x)^{-1}+\frac34(1+x/4)^{-1} =114+6116x+51564x2+.=\frac{11}{4}+\frac{61}{16}x+\frac{515}{64}x^2+\cdots. The two component ranges are x<1/2|x|<1/2 and x<4|x|<4, so the combined expansion requires the narrower condition x<1/2|x|<1/2.

Keep the extracted factor bnb^n outside the series, substitute the whole expression for uu into every power, and state the range in terms of xx. A finite truncation is an approximation within that range; it is not an identity equal to the original function for every xx.