How to approximate $(x+1)^\alpha-x^\alpha$

Question

Let $\alpha$ be a positive real. Then how do you approximate $$ (x+1)^{\alpha}-x^\alpha $$ as $x\to \infty$?

For example, we know the answer if $\alpha=1/2$, which is $$ \sqrt{x+1}-\sqrt{x}=(\sqrt{x+1}-\sqrt{x})\cdot \frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}=\frac{1}{\sqrt{x+1}+\sqrt{x}} \sim \frac{1}{2\sqrt{x}}. $$ What about the other cases?

Answer 1

Applying the Generalized Binomial Theorem, we can write immediately for |x|>1$

$$\bbox[5px,border:2px solid #C0A000]{(1+x)^\alpha-x^\alpha=x^\alpha\,\sum_{k=1}^\infty \binom{\alpha}{k}\left(\frac{1}{x}\right)^k} \tag 1$$

where $\binom{\alpha}{k}$ is the Generalized Binomial Coefficient.

The first few terms of the expansion are

$$(1+x)^\alpha-x^\alpha=\alpha x^{\alpha-1}+\frac{\alpha(\alpha -1)}{2!}\,x^{\alpha -2}+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}\,x^{\alpha-3}+O\left(x^{\alpha-4}\right) $$

---

EXAMPLE:

For $\alpha =1/2$, $(1)$ yields

$$\bbox[5px,border:2px solid #C0A000]{(1+x)^{1/2}-x^{1/2}=x^{1/2}\,\sum_{k=1}^\infty \binom{1/2}{k}\left(\frac{1}{x}\right)^k} $$

and the first few terms are

$$(1+x)^{1/2}-x^{1/2}=\frac12 x^{-1/2}-\frac18 x^{-3/2}+\frac{1}{16}x^{-5/2}-\frac{5}{128}x^{-7/2}+O(x^{-9/2})$$

Answer 2

Hint: You may use $$ (1+x)^\alpha-x^\alpha= x^\alpha (( 1+ \frac{1}{x})^\alpha -1) $$ and do a Taylor expansion.

Answer 3

Given $g(x)=x^\alpha$, if $\alpha$ isn't too big, then $g'(x)=\alpha x^{\alpha-1}$ is good. If $\alpha<2$, then this approximation gets better as $x$ increases (if $\alpha=2$, the error stays at exactly $1$ no matter what $x$ is).

For higher $\alpha$, if you want this approach to still work, you need to take into account higher and higher derivatives of $g$. For instance, for $2<\alpha<3$, you need to use $g'(x)+\frac{g''(x)}{2}$ for the error to decrease as $x$ increases. If this looks familiar, that's because this is using the second degree Taylor approximation of $g$ at $x$.

In general, for positive $\alpha$, you need to use the $\lfloor \alpha\rfloor$ degree expansion for the error to decrease as $x$ increases, but more terms will of course increase accuracy.