binomial approximation

The function

:$f(x)\; =\; (1\; +\; x)^\{\backslash alpha\}$

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

:$f\text{'}(x)\; =\; \backslash alpha\; (1\; +\; x)^\{\backslash alpha\; -\; 1\}$

and so

:$f\text{'}(0)\; =\; \backslash alpha.$

Thus

:$f(x)\; \backslash approx\; f(0)\; +\; f\text{'}(0)(x\; -\; 0)\; =\; 1\; +\; \backslash alpha\; x.$

By Taylor's theorem, the error in this approximation is equal to $\backslash frac\{\backslash alpha(\backslash alpha\; -\; 1)\; x^2\}\{2\}\; \backslash cdot\; (1\; +\; \backslash zeta)^\{\backslash alpha\; -\; 2\}$ for some value of $\backslash zeta$ that lies between 0 and {{mvar|x}}. For example, if and $\backslash alpha\; \backslash geq\; 2$, the error is at most $\backslash frac\{\backslash alpha(\backslash alpha\; -\; 1)\; x^2\}\{2\}$. In little o notation, one can say that the error is $o(|x|)$, meaning that $\backslash lim\_\{x\; \backslash to\; 0\}\; \backslash frac\{\backslash textrm\{error\}\}$

x

= 0.

The function

:$f(x)\; =\; (1+x)^\backslash alpha$

where $x$ and $\backslash alpha$ may be real or complex can be expressed as a Taylor series about the point zero.

:$\backslash begin\{align\}$

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 and $|\backslash alpha\; x|\; \backslash ll\; 1$, then the terms in the series become progressively smaller and it can be truncated to

:$(1+x)^\backslash alpha\; \backslash approx\; 1\; +\; \backslash alpha\; x\; .$

This result from the binomial approximation can always be improved by keeping additional terms from the Taylor series above. This is especially important when $|\backslash alpha\; x|$ 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 $|x|\; \backslash ll\; 1$ is a sufficient condition for the binomial approximation. A simple counterexample is to let $x=10^\{-6\}$ and $\backslash alpha=10^7$. In this case $(1+x)^\backslash alpha\; >\; 22,000$ but the binomial approximation yields $1\; +\; \backslash alpha\; x\; =\; 11$. For small $|x|$ but large $|\backslash alpha\; x|$, a better approximation is:

: $(1\; +\; x)^\backslash alpha\; \backslash approx\; e^\{\backslash alpha\; x\}\; .$

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

:$\backslash frac\{1\}\{\backslash sqrt\{a+b\}\}\; -\; \backslash frac\{1\}\{\backslash sqrt\{a-b\}\}$

where $a$ and $b$ are real but $a\; \backslash gg\; b$.

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.

:$\backslash begin\{align\}$

\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 $a\; \backslash gg\; b$ which is otherwise not obvious from the original expression.

{{further|Binomial series}}

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

:$(1+x)^\backslash alpha\; \backslash approx\; 1\; +\; \backslash alpha\; x\; +\; (\backslash alpha/2)\; (\backslash alpha-1)\; x^2$

Applied to the square root, it results in:

:$\backslash sqrt\{1+x\}\; \backslash approx\; 1\; +\; x/2\; -\; x^2\; /\; 8.$

Consider the expression:

:$(1\; +\; \backslash epsilon)^n\; -\; (1\; -\; \backslash epsilon)^\{-n\}$

where and $|n\; \backslash epsilon|\; \backslash ll\; 1$. If only the linear term from the binomial approximation is kept $(1+x)^\backslash alpha\; \backslash approx\; 1\; +\; \backslash alpha\; x$ then the expression unhelpfully simplifies to zero

:$\backslash begin\{align\}$

(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:

:$\backslash begin\{align\}$

(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 $\backslash epsilon$ which is why it did not appear when only the linear terms in $\backslash epsilon$ were kept.

{{Reflist}}

Category:Factorial and binomial topics

Category:Approximations