2
$\begingroup$

Let $f\colon X\to\mathbb{R}$ be a differentiable function.

What is interpretation of the following quantity:

$$h(x_{0}):=\frac{f'(x_{0})}{f(x_{0})}$$

where $x_{0}\in X$.


My own reaserch.

a)

We know that

$$f'(x_{0})=\lim_{h\to 0}\frac{f(x_{0}+h)-f(x_{0})}{h}$$

so

$$f'(x_{0})h\approx f(x_{0}+h)-f(x_{0})$$

If we take $h=1$ we get

$$f'(x_{0})\approx f(x_{0}+1)-f(x_{0})$$

so we have the following interpretation:

If we increase an argument of function $f$ by 1 unit form level $x_{0}$ then the value of function $f$ will change (approximately) by $f'(x_{0})$ units.

This interpretation is used in economics.

b)

We know that

$$h(x_{0})=\lim_{h\to0}\frac{f(x_{0}+h)-f(x_{0})}{hf(x_{0})}$$

so

$$h(x_{0})h\approx \frac{f(x_{0}+h)-f(x_{0})}{f(x_{0})}$$

and if we take $h=1$ we obtain

$$h(x_{0})\approx \frac{f(x_{0}+1)-f(x_{0})}{f(x_{0})}$$

How can we interpret this quantity?

$\endgroup$
1
  • $\begingroup$ It's the logarithmic derivative: d/dx(log[f(x)]) $\endgroup$ Jul 23, 2016 at 15:44

1 Answer 1

5
$\begingroup$

As you explained, we can write $$\frac{f'(x)}{f(x)}=\frac{df(x)}{dx}\frac{1}{f(x)}\approx \frac{(\Delta f/f)}{\Delta x}$$

where $\Delta x$ is a discrete change in $x$ and $\Delta f$ is the resulting change in $f$.

Thus the ratio is the proportional change in $f$ per unit change in $x$. For example, if $f$ gave the quantity demanded at each given price $x$, then the ratio would tell you the approximate proportional change in quantity demanded for each dollar rise in $x$.

In economics the ratio is known as a semi-elasticity: https://en.wikipedia.org/wiki/Elasticity_of_a_function#Semi-elasticity

Note also that the ratio is the derivative of $\ln f$, which is known as the logarithmic derivative: https://en.wikipedia.org/wiki/Logarithmic_derivative

$\endgroup$
2
  • $\begingroup$ So we can say: If we increase an argument of function $f$ by 1 unit form level $x_0$ then the value of function $f$ will change (approximately) by $h(x_0)$ percent? $\endgroup$
    – Leon
    Jul 23, 2016 at 14:44
  • 1
    $\begingroup$ Yes, that is the correct interpretation. $\endgroup$
    – smcc
    Jul 23, 2016 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.