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I've just learnt about price elasticity of demand. Since we're only doing 10th grade economics, our teacher tried to make it simple by using an example where the demand curve is linear (i.e. quantity of demand is in linear relationship with the price), and we were supposed to treat the price elasticity of demand $e_d$ as having a linear relationship with the price (but it is not).

However, as the teacher pointed out to a few curious students, $e_d$ is in fact not having such a linear relationship at all, as seen in its formula:

\begin{align} e_d = \frac{\text{variation in demand}}{\text{variation in price}} = \frac{\Delta Q}{\Delta P} \times \frac{P}{Q} \end{align} where $Q$ stands for quantity of demand and $P$ stands for price, and $\Delta Q$ and $\Delta P$ respectively stand for change in demand and price.

I then wanted to find a case where $e_d$ is actually linear. By assuming the demand $q$ for a certain price $p$ is given by a function $q = f(p)$, I could express the price elasticity at that price $p$ as

\begin{align} e_d = \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h} \times \frac{x}{f(x)} = \frac{f'(x)x}{f(x)} \end{align}

This where I got stuck. My limited knowledge in calculus and in mathematics in general does not give an explicit way to find a $f(x)$ that makes $e_d$ a linear equation, i.e. $\frac{f'(x) \cdot x}{f(x)} = ax + c$, where $a$ and $c$ are real numbers.

I might have mixed too much mathematics in economics here, but I am just curious to know if this could actually happen.

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    $\begingroup$ $f$ is not a polynomial since $\deg(f(x))=\deg(f'(x)*x)$, so the quotient can't be linear if that's the case. You can have functions like $f(x)=e^{ax}$ though such that $\frac{f'(x)x}{f(x)}=ax$ $\endgroup$
    – user304329
    Mar 13, 2017 at 20:01
  • $\begingroup$ @vrugtehagel could you explain to me in a little bit more detail $\endgroup$
    – Jingjie Yang
    Mar 13, 2017 at 20:02
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    $\begingroup$ I mean, if $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, then $n$ is its degree; since the degree decreases by $1$ when taking the derivative, $f'(x)x$ and $f(x)$ have the same degree and as such, one divided by the other cannot have degree 1 (which is the degree of $ax+c$). $\endgroup$
    – user304329
    Mar 13, 2017 at 20:04
  • $\begingroup$ @vrugtehagel ah I see. It would be great if you answer this question directly :) $\endgroup$
    – Jingjie Yang
    Mar 13, 2017 at 20:05

2 Answers 2

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Firstly, we know that $f$ cannot be a polynomial, that is, of the form

$$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$

since then $xf'(x)$ and $f(x)$ have the same degree, thus, their quotient can never be linear.

We can however use functions like (for any real $k\neq 0$)

$$f(x)=k\cdot x^c\cdot e^{ax}$$ to get $$\frac{f'(x)x}{f(x)}=ax+c$$

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If you rewrite the differential equation as $f'/f=a+c/x$ and use the fact that $f'/f=(\log f)'$, then integrating both sides we get $\log f = ax + c\log x +d$, and exponentiating both sides, $f=e^d x^c e^{ax}$. Rewriting $e^d=k$ (because $k$ is a more standard name for a multiplicative constant), we get $f=kx^c e^{ax}$.

This is where @vrugtehagel's answer comes from.

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