# Binomial series a. Find the first four nonzero terms of the binomial series centered at 0 for the given function. b. Use the first four terms of the series to approximate the given quantity. f(x)=(1+x)^{frac{2}{3}}, approximate (1.02)^{frac{2}{3}}

Question
Series
Binomial series
a. Find the first four nonzero terms of the binomial series centered at 0 for the given function.
b. Use the first four terms of the series to approximate the given quantity.
$$f(x)=(1+x)^{\frac{2}{3}}$$, approximate $$(1.02)^{\frac{2}{3}}$$

2020-12-02
Given function is $$f(x)=(1+x)^{\frac{2}{3}}$$
For function $$f(x)=(1+x)^n$$ binomial series centered at 0 is given as:
$$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+...$$
We have, $$f(x)=(1+x)^{\frac{2}{3}}$$
Here, $$n=\frac{2}{3}$$
Putting $$n=\frac{2}{3}$$ in the series, we get
$$(1+x)^{\frac{2}{3}}=1+\frac{2}{3}x+\frac{\frac{2}{3}(\frac{2}{3}-1)}{2}x^2+\frac{\frac{2}{3}(\frac{2}{3}-1)(\frac{2}{3}-2)}{6}x^3+...$$
$$=1+\frac{2}{3}x+\frac{\frac{2}{3}\cdot\frac{-1}{3}}{2}x^2+\frac{\frac{2}{3}\cdot\frac{-1}{3}\cdot\frac{-4}{3}}{6}x^3+...$$
$$=1+\frac{2}{3}x-\frac{1}{9}x^2+\frac{4}{81}x^3+...$$
First four non-zero terms are:
$$(1+x)^{\frac{2}{3}}=1+\frac{2}{3}x-\frac{1}{9}x^2+\frac{4}{81}x^3$$
Now, to approximate $$(1.02)^\frac{2}{3}$$
$$(1.02)^{\frac{2}{3}}=(1+0.02)^{\frac{2}{3}}$$
Put x=0.02 in the series we obtained above.
$$(1+0.02)^{\frac{2}{3}}=1+\frac{2}{3}\cdot(0.02)-\frac{1}{9}\cdot(0.02)^2+\frac{4}{81}\cdot(0.02)^3$$
$$=1+\frac{0.04}{3}-\frac{0.0004}{9}+\frac{4}{81}\cdot(0.000008)$$
$$=1+0.0133-0.000044+0.000000392$$
$$=1.0128$$

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