Approximation of $x\log(x/a)$ for $x$ near a

Question

I'm trying to see where the approximation $$(x-a) + ((x-a)^2)/2a$$ of $x\log(x/a)$ comes from (for x near a). Might be missing something very trivial but I've already tried the usual expansions unsuccessfully. Thanks.

Answer 1

Take the Taylor series centered at $a$ for the function

$$f(x) = x \log \left(\frac x a\right) = x \log x - x \log a$$

Since $f(a) = 0$, the constant term is $0$. Next,

$$f'(a) = \log x + x \frac 1 x - \log a \Big|_{x = a} = 1$$

so the linear term is $x - a$. The quadratic term is then derived from

$$f''(a) = \frac 1 x \Big|_{x = a} = \frac 1 a$$

giving

$$\frac{f''(a)}{2!} (x - a)^2 = \frac 1 {2a} (x - a)^2$$

as desired.

Answer 2

We have

$$f(x) = x \log (x/a)$$

and we want to find an approximation of $f$ near $a$. However,

$$f(x) = a\cdot g(x/a)$$

where

$$g(x) = x \log (x).$$

Now, it is enough to find an approximation of $g$ near $a/a = 1$. That is,

$$g(x) = (x-1) + \frac{1}{2}(x-1)^2 + O\big((x-1)^3\big).$$

Now, going back we get

\begin{align} f(x) &= a \cdot g(x/a) \\ &= a \cdot \left((x/a-1) + \frac{1}{2}(x/a-1)^2 + O\big((x/a-1)^3\big)\right) \\ &= (x-a) + \frac{a^2}{2a}(x/a - 1)^2 + O\left(\frac{a^3}{a^2}(x/a-1)^3\right) \\ &= (x-a) + \frac{1}{2a}(x-a)^2 + O\left(\frac{1}{a^2}(x-a)^3\right) \end{align}

which is exactly as desired.

I hope this helps $\ddot\smile$