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Derivations

Using linear approximation

The function

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

and so

f ′ ( 0 ) = α . {\displaystyle f'(0)=\alpha .}

Thus

f ( x ) ≈ f ( 0 ) + f ′ ( 0 ) ( x − 0 ) = 1 + α x . {\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 α ( α − 1 ) x 2 2 ⋅ ( 1 + ζ ) α − 2 {\textstyle {\frac {\alpha (\alpha -1)x^{2}}{2}}\cdot (1+\zeta )^{\alpha -2}} for some value of ζ {\displaystyle \zeta } that lies between 0 and x. For example, if x < 0 {\displaystyle x<0} and α ≥ 2 {\displaystyle \alpha \geq 2} , the error is at most α ( α − 1 ) x 2 2 {\textstyle {\frac {\alpha (\alpha -1)x^{2}}{2}}} . In little o notation, one can say that the error is o ( | x | ) {\displaystyle o(|x|)} , meaning that lim x → 0 error | x | = 0 {\textstyle \lim _{x\to 0}{\frac {\textrm {error}}{|x|}}=0} .

Using Taylor series

The function

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

where x {\displaystyle x} and α {\displaystyle \alpha } 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 + ⋯ {\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 | x | < 1 {\displaystyle |x|<1} and | α x | ≪ 1 {\displaystyle |\alpha x|\ll 1} , then the terms in the series become progressively smaller and it can be truncated to

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

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

Example

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

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

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

The mathematical form for the binomial approximation can be recovered by factoring out the large term a {\displaystyle 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 {\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 b {\displaystyle b} when a ≫ b {\displaystyle a\gg b} which is otherwise not obvious from the original expression.

Generalization

Further information: Binomial series

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

( 1 + x ) α ≈ 1 + α x + ( α / 2 ) ( α − 1 ) x 2 {\displaystyle (1+x)^{\alpha }\approx 1+\alpha x+(\alpha /2)(\alpha -1)x^{2}}

Applied to the square root, it results in:

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

Quadratic example

Consider the expression:

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

where | ϵ | < 1 {\displaystyle |\epsilon |<1} and | n ϵ | ≪ 1 {\displaystyle |n\epsilon |\ll 1} . If only the linear term from the binomial approximation is kept ( 1 + x ) α ≈ 1 + α x {\displaystyle (1+x)^{\alpha }\approx 1+\alpha 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. {\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:

( 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 {\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 ϵ {\displaystyle \epsilon } which is why it did not appear when only the linear terms in ϵ {\displaystyle \epsilon } were kept.

References

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