Binomial approximation

The binomial approximation is useful for approximately calculating powers of sums of a small number and 1. It states that if $|x|<1$ and $|\alpha x|$ ≪ $1$ where $x$ and $\alpha$ are real or complex numbers, then

(1 + x)^\alpha \approx 1 + \alpha x.

If either $x$ or $\alpha$ are complex then the absolute value denotes the modulus of the complex number.

The benefit of this approximation is that $\alpha$ is converted from a power to multiplicative factor. This can greatly simplify mathematical expressions (see example) and is a common tool in physics.^[1]

This approximation can be proven several ways including the binomial theorem and ignoring the terms beyond the first two.

By Bernoulli's inequality, the left-hand side of this relation is greater than or equal to the right-hand side whenever $x>-1$ and $\alpha \geq 1$ .

Derivation using linear approximation

The function

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

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

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

and so

f'(0) = \alpha.

Thus

f(x) \approx f(0) + f'(0)(x - 0) = 1 + \alpha x.

Derivation using Taylor Series

The function

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

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

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

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

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

Example simplification

Consider the following expression where $a$ and $b$ are real but $a$ ≫ $b$ .

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

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

{\begin{aligned}{\frac {1}{\sqrt {a+b}}}-{\frac {1}{\sqrt {a-b}}}&={\frac {1}{\sqrt {a}}}{\Big (}{\big (}1+{\frac {b}{a}}{\big )}^{-1/2}-{\big (}1-{\frac {b}{a}}{\big )}^{-1/2}{\Big )}\\&\approx {\frac {1}{\sqrt {a}}}{\Big (}{\big (}1+{\big (}-{\frac {1}{2}}{\big )}{\frac {b}{a}}{\big )}-{\big (}1-{\big (}-{\frac {1}{2}}{\big )}{\frac {b}{a}}{\big )}{\Big )}\\&\approx {\frac {1}{\sqrt {a}}}{\Big (}1-{\frac {b}{2a}}-1-{\frac {b}{2a}}{\Big )}\\&\approx -{\frac {b}{a{\sqrt {a}}}}\end{aligned}}

Evidently the expression is linear in $b$ when $a$ ≫ $b$ which is otherwise not obvious from the original expression.

Example keeping the quadratic term

Consider the expression:

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

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

{\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. It is possible to extract a nonzero approximate solution by keeping the quadratic term in the Taylor Series, i.e. $(1+x)^{\alpha }\approx 1+\alpha x+{\frac {1}{2}}\alpha (\alpha -1)x^{2}$ so now,

{\begin{aligned}(1+\epsilon )^{n}-(1-\epsilon )^{-n}&\approx (1+n\epsilon +{\frac {1}{2}}n(n-1)\epsilon ^{2})-(1+(-n)(-\epsilon )+{\frac {1}{2}}(-n)(-n-1)(-\epsilon )^{2})\\&\approx (1+n\epsilon +{\frac {1}{2}}n(n-1)\epsilon ^{2})-(1+n\epsilon +{\frac {1}{2}}n(n+1)\epsilon ^{2})\\&\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 $\epsilon$ which is why it did not appear when only the linear in terms in $\epsilon$ were kept.

References

↑ For example calculating the multipole expansion. Griffiths, D. (1999). Introduction to Electrodynamics (Third ed.). Pearson Education, Inc. pp. 146–148.

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[1] For example calculating the multipole expansion. Griffiths, D. (1999). Introduction to Electrodynamics (Third ed.). Pearson Education, Inc. pp. 146–148.