Approximation of powers of some binomials
The binomial approximation is useful for approximately calculating powers of sums of 1 and a small number x. It states that
![{\displaystyle (1+x)^{\alpha }\approx 1+\alpha x.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c05e5fb59215d783e80ac0bcace42f1e91cf6083)
It is valid when
and
where
and
may be real or complex numbers.
The benefit of this approximation is that
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.[1]
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
and
.
Derivations
Using linear approximation
The function
![{\displaystyle f(x)=(1+x)^{\alpha }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a97d0a3c44850e129c1a5f4ced0eb7e347b70a6)
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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6e24f5625d18d4ba9d7bc4d7890f87952920527)
and so
![{\displaystyle f'(0)=\alpha .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63c5c5765b45086fe0f43e95510ce32d5a50ec82)
Thus
![{\displaystyle f(x)\approx f(0)+f'(0)(x-0)=1+\alpha x.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e39cbcb7d94f1e75f3f3eba6f88cce156610714f)
By Taylor's theorem, the error in this approximation is equal to
for some value of
that lies between 0 and x. For example, if
and
, the error is at most
. In little o notation, one can say that the error is
, meaning that
.
Using Taylor series
The function
![{\displaystyle f(x)=(1+x)^{\alpha }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a97d0a3c44850e129c1a5f4ced0eb7e347b70a6)
where
and
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ffe94aa526854742eff2675a5aee0440d4dbb035)
If
and
, then the terms in the series become progressively smaller and it can be truncated to
![{\displaystyle (1+x)^{\alpha }\approx 1+\alpha x.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c05e5fb59215d783e80ac0bcace42f1e91cf6083)
This result from the binomial approximation can always be improved by keeping additional terms from the Taylor series above. This is especially important when
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
is a sufficient condition for the binomial approximation. A simple counterexample is to let
and
. In this case
but the binomial approximation yields
. For small
but large
, a better approximation is:
![{\displaystyle (1+x)^{\alpha }\approx e^{\alpha x}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c609bdb5a7832b436400ae7bf8ce7235b78410dc)
Example
The binomial approximation for the square root,
, can be applied for the following expression,
![{\displaystyle {\frac {1}{\sqrt {a+b}}}-{\frac {1}{\sqrt {a-b}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de887b5faf32edd8cd644aca7886446854b0c0c8)
where
and
are real but
.
The mathematical form for the binomial approximation can be recovered by factoring out the large term
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee0a76a5bf01f9ecb0fe6a890bedd24ab1a9e3a5)
Evidently the expression is linear in
when
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:
![{\displaystyle (1+x)^{\alpha }\approx 1+\alpha x+(\alpha /2)(\alpha -1)x^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96d33624f395fb73f00b2c5c770c4072ceb7bddd)
Applied to the square root, it results in:
![{\displaystyle {\sqrt {1+x}}\approx 1+x/2-x^{2}/8.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7cc685cbde56f6a2dc5dafd7dce2eb1c62a18807)
Quadratic example
Consider the expression:
![{\displaystyle (1+\epsilon )^{n}-(1-\epsilon )^{-n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5d263fa54a60931aa6ea6b472d9a03241b77e64)
where
and
. If only the linear term from the binomial approximation is kept
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7cf665755ebc4609d5fe077bae3b28573fe6cdc5)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec4696b992bd3364f5d48f72ca2994308c2dc22a)
This result is quadratic in
which is why it did not appear when only the linear terms in
were kept.
References
- ^ For example calculating the multipole expansion. Griffiths, D. (1999). Introduction to Electrodynamics (Third ed.). Pearson Education, Inc. pp. 146–148.