Harman Patil (Editor)

Binomial approximation

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

Contents

( 1 + x ) α ≈ 1 + α x .

If either x or α are complex then the absolute value denotes the modulus of the complex number.

The benefit of this approximation is that α is converted from a power to multiplicative factor. This can greatly simplify mathematical expressions (see example) and is a common tool in physics.

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 α ≥ 1 .

Derivation using linear approximation

The function

f ( x ) = ( 1 + x ) α

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

f ′ ( x ) = α ( 1 + x ) α − 1

and so

f ′ ( 0 ) = α .

Thus

f ( x ) ≈ f ( 0 ) + f ′ ( 0 ) ( x − 0 ) = 1 + α x .

Derivation using Taylor Series

The function

f ( x ) = ( 1 + x ) α

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

f ( x ) = ∑ n = 0 ∞ f ( n ) ( 0 ) n ! x n f ( x ) = f ( 0 ) + f ′ ( 0 ) x + 1 2 f ″ ( 0 ) x 2 + 1 6 f ‴ ( 0 ) x 3 + 1 24 f ( 4 ) ( 0 ) x 4 + ⋯ ( 1 + x ) α = 1 + α x + 1 2 α ( α − 1 ) x 2 + 1 6 α ( α − 1 ) ( α − 2 ) x 3 + 1 24 α ( α − 1 ) ( α − 2 ) ( α − 3 ) x 4 + ⋯

If | x | < 1 and | α x | ≪ 1 , then the terms in the series become progressively smaller and it can be truncated to

( 1 + x ) α ≈ 1 + α 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 | α 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 α = 10 7 . In this case ( 1 + x ) α > 22 , 000 but the binomial approximation yields 1 + α x = 11 .

Example simplification

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

1 a + b − 1 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.

1 a + b − 1 a − b = 1 a ( ( 1 + b a ) − 1 / 2 − ( 1 − b a ) − 1 / 2 ) ≈ 1 a ( ( 1 + ( − 1 2 ) b a ) − ( 1 − ( − 1 2 ) b a ) ) ≈ 1 a ( 1 − b 2 a − 1 − b 2 a ) ≈ − b a a

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 + ϵ ) n − ( 1 − ϵ ) − n

where | ϵ | < 1 and | n ϵ | ≪ 1 . If only the linear term from the binomial approximation is kept ( 1 + x ) α ≈ 1 + α x then the expression unhelpfully simplifies to zero

( 1 + ϵ ) n − ( 1 − ϵ ) − n ≈ ( 1 + n ϵ ) − ( 1 − ( − n ) ϵ ) ≈ ( 1 + n ϵ ) − ( 1 + n ϵ ) ≈ 0 .

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 ) α ≈ 1 + α x + 1 2 α ( α − 1 ) x 2 so now,

( 1 + ϵ ) n − ( 1 − ϵ ) − n ≈ ( 1 + n ϵ + 1 2 n ( n − 1 ) ϵ 2 ) − ( 1 + ( − n ) ( − ϵ ) + 1 2 ( − n ) ( − n − 1 ) ( − ϵ ) 2 ) ≈ ( 1 + n ϵ + 1 2 n ( n − 1 ) ϵ 2 ) − ( 1 + n ϵ + 1 2 n ( n + 1 ) ϵ 2 ) ≈ 1 2 n ( n − 1 ) ϵ 2 − 1 2 n ( n + 1 ) ϵ 2 ≈ 1 2 n ϵ 2 ( ( n − 1 ) − ( n − 1 ) ) ≈ − n ϵ 2

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

References

Binomial approximation Wikipedia
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