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
(
1
+
x
)
α
≈
1
+
α
x
.
{\displaystyle (1+x)^{\alpha }\approx 1+\alpha x.}
It is valid when
|
x
|
<
1
{\displaystyle |x|<1}
and
|
α
x
|
≪
1
{\displaystyle |\alpha x|\ll 1}
where
x
{\displaystyle x}
and
α
{\displaystyle \alpha }
may be real or complex numbers .
The benefit of this approximation is that
α
{\displaystyle \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.[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
x
>
−
1
{\displaystyle x>-1}
and
α
≥
1
{\displaystyle \alpha \geq 1}
.
Derivations [ edit ]
Using linear approximation [ edit ]
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 [ edit ]
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 [ edit ]
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 [ edit ]
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 [ edit ]
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 [ edit ]
^ For example calculating the multipole expansion . Griffiths, D. (1999). Introduction to Electrodynamics (Third ed.). Pearson Education, Inc. pp. 146–148.