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| 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. 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 .

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| 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. 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 . 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) = \alpha (1 + x)^{\alpha - 1} 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{\alpha(\alpha - 1) x^2}{2} \cdot (1 + \zeta)^{\alpha - 2} for some value of \zeta that lies between 0 and . For example, if x and \alpha \geq 2 , the error is at most \frac{\alpha(\alpha - 1) x^2}{2} . In little o notation, one can say that the error is o(|x|) , meaning that \lim_{x \to 0} \frac{\textrm{error}} = 0 . 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{align} 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{align} If |x| 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^{-6} 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^{\alpha x} . \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{align} 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 a quadratic approximation keeping the second 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{1+x} \approx 1 + x/2 - x^2 / 8. ===Quadratic example=== Consider the expression: (1 + \epsilon)^n - (1 - \epsilon)^{-n} where |\epsilon| 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{align} (1 + \epsilon)^n - (1 - \epsilon)^{-n} &\approx (1+ n \epsilon) - (1 - (-n) \epsilon)\\ &\approx (1+ n \epsilon) - (1 + n \epsilon)\\ &\approx 0 . \end{align} While the expression is small, it is not exactly zero. So now, keeping the quadratic term: \begin{align} (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{align} This result is quadratic in \epsilon which is why it did not appear when only the linear terms in \epsilon were kept. ==References==

Text derived from https://en.wikipedia.org/w/index.php?oldid=1346694792 under CC BY-SA 4.0; modifications have been made.