Binomial approximation

The binomial approximation is useful for approximately calculating powers of sums of 1 and a small number ''x''. It states that

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

It is valid when $, x, <1$ and $, \alpha x, \ll 1$ where $x$ and $\alpha$ may be real or complex numbers.

The benefit of this approximation is that $\alpha$ is converted from an exponent to a multiplicative factor. This can greatly simplify mathematical expressions (as in the example below) and is a common tool in physics.For example calculating the multipole expansion.

The approximation can be proven several ways, and is closely related to the binomial theorem. By Bernoulli's inequality, the left-hand side of the approximation is greater than or equal to the right-hand side whenever $x>-1$ and $\alpha \geq 1$.

Derivations

Using linear approximation

The function
:$f(x) = (1 + x)^$
is a smooth function for ''x'' near 0. Thus, standard linear approximation tools from calculus apply: one has
:$f'(x) = \alpha (1 + x)^$
and so
:$f'(0) = \alpha.$
Thus
:$f(x) \approx f(0) + f'(0)(x - 0) = 1 + \alpha x.$

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

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.

:$\begin f(x) &= \sum_^ \frac x^n\\ f(x) &= f(0) + f'(0) x + \frac f''(0) x^2 + \frac f(0) x^3 + \frac f^(0) x^4 + \cdots\\ (1+x)^ &= 1 + \alpha x + \frac \alpha (\alpha-1) x^2 + \frac \alpha (\alpha-1)(\alpha-2)x^3 + \frac \alpha (\alpha-1)(\alpha-2)(\alpha-3)x^4 + \cdots \end$

If $, x, < 1$ and $, \alpha x, \ll 1$, 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 $, x, \ll 1$ is a sufficient condition for the binomial approximation. A simple counterexample is to let $x=10^$ 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:

: $(1 + x)^\alpha \approx e^ .$

Example

The binomial approximation for the square root, $\sqrt \approx 1+x/2$, can be applied for the following expression,
:$\frac - \frac$
where $a$ and $b$ are real but $a \gg b$.

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.

:$\begin \frac - \frac &= \frac \left(\left(1+\frac\right)^ - \left(1-\frac\right)^\right)\\ &\approx\frac \left(\left(1+\left(-\frac\right)\frac\right) - \left(1-\left(-\frac\right)\frac\right)\right) \\ &\approx\frac \left(1-\frac - 1 -\frac\right) \\ &\approx -\frac \end$

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

Generalization

While the binomial approximation is linear, it can be generalized to keep the quadratic term in the Taylor series:
:$(1+x)^\alpha \approx 1 + \alpha x + (\alpha/2) (\alpha-1) x^2$

Applied to the square root, it results in:
:$\sqrt \approx 1 + x/2 - x^2 / 8.$

Quadratic example

Consider the expression:
:$(1 + \epsilon)^n - (1 - \epsilon)^$
where $, \epsilon, <1$ and $, n \epsilon, \ll 1$. 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
:$\begin (1 + \epsilon)^n - (1 - \epsilon)^ &\approx (1+ n \epsilon) - (1 - (-n) \epsilon)\\ &\approx (1+ n \epsilon) - (1 + n \epsilon)\\ &\approx 0 . \end$

While the expression is small, it is not exactly zero.
So now, keeping the quadratic term:
:$\begin (1+\epsilon)^n - (1 - \epsilon)^&\approx \left(1+ n \epsilon + \frac n (n-1) \epsilon^2\right) - \left(1 + (-n)(-\epsilon) + \frac (-n) (-n-1) (-\epsilon)^2\right)\\ &\approx \left(1+ n \epsilon + \frac n (n-1) \epsilon^2\right) - \left(1 + n \epsilon + \frac n (n+1) \epsilon^2\right)\\ &\approx \frac n (n-1) \epsilon^2 - \frac n (n+1) \epsilon^2\\ &\approx \frac n \epsilon^2 ((n-1) - (n+1)) \\ &\approx - n \epsilon^2 \end$

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

References

{{Reflist

Factorial and binomial topics
Approximations