Using linear approximation edit

The function

${\displaystyle f(x)=(1+x)^{\alpha }}$ 

is a smooth function for x near 0. Thus, standard linear approximation tools from calculus apply: one has

${\displaystyle f'(x)=\alpha (1+x)^{\alpha -1}}$ 

and so

${\displaystyle f'(0)=\alpha .}$ 

Thus

${\displaystyle f(x)\approx f(0)+f'(0)(x-0)=1+\alpha x.}$ 

By Taylor's theorem, the error in this approximation is equal to ${\textstyle {\frac {\alpha (\alpha -1)x^{2}}{2}}\cdot (1+\zeta )^{\alpha -2}}$  for some value of ${\displaystyle \zeta }$  that lies between 0 and x. For example, if ${\displaystyle x<0}$  and ${\displaystyle \alpha \geq 2}$ , the error is at most ${\textstyle {\frac {\alpha (\alpha -1)x^{2}}{2}}}$ . In little o notation, one can say that the error is ${\displaystyle o(|x|)}$ , meaning that ${\textstyle \lim _{x\to 0}{\frac {\textrm {error}}{|x|}}=0}$ .

Using Taylor series edit

The function

${\displaystyle f(x)=(1+x)^{\alpha }}$ 

where ${\displaystyle x}$  and ${\displaystyle \alpha }$  may be real or complex can be expressed as a Taylor series about the point zero.

${\displaystyle {\begin{aligned}f(x)&=\sum _{n=0}^{\infty }{\frac {f^{(n)}(0)}{n!}}x^{n}\\f(x)&=f(0)+f'(0)x+{\frac {1}{2}}f''(0)x^{2}+{\frac {1}{6}}f'''(0)x^{3}+{\frac {1}{24}}f^{(4)}(0)x^{4}+\cdots \\(1+x)^{\alpha }&=1+\alpha x+{\frac {1}{2}}\alpha (\alpha -1)x^{2}+{\frac {1}{6}}\alpha (\alpha -1)(\alpha -2)x^{3}+{\frac {1}{24}}\alpha (\alpha -1)(\alpha -2)(\alpha -3)x^{4}+\cdots \end{aligned}}}$ 

If ${\displaystyle |x|<1}$  and ${\displaystyle |\alpha x|\ll 1}$ , then the terms in the series become progressively smaller and it can be truncated to

${\displaystyle (1+x)^{\alpha }\approx 1+\alpha x.}$ 

This result from the binomial approximation can always be improved by keeping additional terms from the Taylor series above. This is especially important when ${\displaystyle |\alpha x|}$  starts to approach one, or when evaluating a more complex expression where the first two terms in the Taylor series cancel (see example).

Sometimes it is wrongly claimed that ${\displaystyle |x|\ll 1}$  is a sufficient condition for the binomial approximation. A simple counterexample is to let ${\displaystyle x=10^{-6}}$  and ${\displaystyle \alpha =10^{7}}$ . In this case ${\displaystyle (1+x)^{\alpha }>22,000}$  but the binomial approximation yields ${\displaystyle 1+\alpha x=11}$ . For small ${\displaystyle |x|}$  but large ${\displaystyle |\alpha x|}$ , a better approximation is:

${\displaystyle (1+x)^{\alpha }\approx e^{\alpha x}.}$ 

The binomial approximation for the square root, ${\displaystyle {\sqrt {1+x}}\approx 1+x/2}$ , can be applied for the following expression,

${\displaystyle {\frac {1}{\sqrt {a+b}}}-{\frac {1}{\sqrt {a-b}}}}$ 

where ${\displaystyle a}$  and ${\displaystyle b}$  are real but ${\displaystyle a\gg b}$ .

The mathematical form for the binomial approximation can be recovered by factoring out the large term ${\displaystyle a}$  and recalling that a square root is the same as a power of one half.

${\displaystyle {\begin{aligned}{\frac {1}{\sqrt {a+b}}}-{\frac {1}{\sqrt {a-b}}}&={\frac {1}{\sqrt {a}}}\left(\left(1+{\frac {b}{a}}\right)^{-1/2}-\left(1-{\frac {b}{a}}\right)^{-1/2}\right)\\&\approx {\frac {1}{\sqrt {a}}}\left(\left(1+\left(-{\frac {1}{2}}\right){\frac {b}{a}}\right)-\left(1-\left(-{\frac {1}{2}}\right){\frac {b}{a}}\right)\right)\\&\approx {\frac {1}{\sqrt {a}}}\left(1-{\frac {b}{2a}}-1-{\frac {b}{2a}}\right)\\&\approx -{\frac {b}{a{\sqrt {a}}}}\end{aligned}}}$ 

Evidently the expression is linear in ${\displaystyle b}$  when ${\displaystyle a\gg b}$  which is otherwise not obvious from the original expression.

While the binomial approximation is linear, it can be generalized to keep the quadratic term in the Taylor series:

${\displaystyle (1+x)^{\alpha }\approx 1+\alpha x+(\alpha /2)(\alpha -1)x^{2}}$ 

Applied to the square root, it results in:

${\displaystyle {\sqrt {1+x}}\approx 1+x/2-x^{2}/8.}$ 

Quadratic example edit

Consider the expression:

${\displaystyle (1+\epsilon )^{n}-(1-\epsilon )^{-n}}$ 

where ${\displaystyle |\epsilon |<1}$  and ${\displaystyle |n\epsilon |\ll 1}$ . If only the linear term from the binomial approximation is kept ${\displaystyle (1+x)^{\alpha }\approx 1+\alpha x}$  then the expression unhelpfully simplifies to zero

${\displaystyle {\begin{aligned}(1+\epsilon )^{n}-(1-\epsilon )^{-n}&\approx (1+n\epsilon )-(1-(-n)\epsilon )\\&\approx (1+n\epsilon )-(1+n\epsilon )\\&\approx 0.\end{aligned}}}$ 

While the expression is small, it is not exactly zero. So now, keeping the quadratic term:

${\displaystyle {\begin{aligned}(1+\epsilon )^{n}-(1-\epsilon )^{-n}&\approx \left(1+n\epsilon +{\frac {1}{2}}n(n-1)\epsilon ^{2}\right)-\left(1+(-n)(-\epsilon )+{\frac {1}{2}}(-n)(-n-1)(-\epsilon )^{2}\right)\\&\approx \left(1+n\epsilon +{\frac {1}{2}}n(n-1)\epsilon ^{2}\right)-\left(1+n\epsilon +{\frac {1}{2}}n(n+1)\epsilon ^{2}\right)\\&\approx {\frac {1}{2}}n(n-1)\epsilon ^{2}-{\frac {1}{2}}n(n+1)\epsilon ^{2}\\&\approx {\frac {1}{2}}n\epsilon ^{2}((n-1)-(n+1))\\&\approx -n\epsilon ^{2}\end{aligned}}}$ 

This result is quadratic in ${\displaystyle \epsilon }$  which is why it did not appear when only the linear terms in ${\displaystyle \epsilon }$  were kept.