Elasticity of a Function as a Exponential Constant

Question

I am looking for an explanation as to why the the elasticity of a function of the form $\\f(x)=Cx^\alpha$, where $\\C\gt0$, is a constant $\alpha$.

Furthermore, if anyone also knows of any material that expounds on this, that would also be much appreciated.

Just have a look here : https://en.wikipedia.org/wiki/Elasticity_of_a_function — Claude Leibovici, Jun 13 '17 at 06:49
@ClaudeLeibovici But this article doesn't explain how to derive the form above from the more general concept of elasticity. — Ryan da Silva, Jun 13 '17 at 07:36
@callculus I am missing some kind of proof as to why this constant is the elasticity of the function. What was the line of reasoning of the mathematicians who figured this out? — Ryan da Silva, Jun 13 '17 at 20:06
@RyandaSilva :"I am missing some kind of proof as to why this constant is the elasticity of the function " This can be shown mathematically. I can show you if you want. The second question is more a historical one rather than a math question. — callculus42, Jun 13 '17 at 20:12

callculus42 · Accepted Answer · 2017-06-14T06:37:36.120

The elacsticity of a differentiable function $y=f(x)$ is

$$\large{\mathcal E_{y,x}}=\frac{dy}{dx}\cdot\frac{x}{y}=f^{'}(x)\cdot\frac{x}{y}$$

If $f(x)=C\cdot x^{\alpha}$ then the derivative w.r.t $x$ is $f^{'}(x)=\alpha\cdot C\cdot x^{\alpha-1}$

Therefore $\large{\mathcal E_{y,x}}=\alpha\cdot C\cdot x^{\alpha-1}\cdot\frac{x}{C\cdot x^{\alpha}}$

It is $x^{\alpha-1}=\frac{x^{\alpha}}{x}$.

$\large{\mathcal E_{y,x}}=\alpha\cdot C\cdot \frac{x^{\alpha}}{x}\cdot\frac{x}{C\cdot x^{\alpha}}$

Finally $x^{\alpha}, C$ and $x$ can be cancelled.

The elasticity is defined as the relative change of the dependent variable divided by the relative change of the independent variable.

$\large{\frac{\frac{\Delta y}{y}}{\frac{\Delta x}{x}}}\normalsize = \frac{\Delta y}{\Delta x}\cdot \frac{x}{y}$

$\Delta x$ is the distance between two x-values. This is the definition of the elasticity. If $\Delta x \to 0$ then $\frac{\Delta y}{\Delta x}$ becomes $\frac{dy}{dx}$. We want a constant elasticity $\alpha$. This condition we can regard and see what kind of function we will get. The equation is

$$\frac{dy}{dx}\cdot \frac{x}{y}=\alpha$$

This is a $\texttt{first order linear ordinary differential equation}$ which can be solved. Multiplying both sides by $dx$ and dividing both sides by $x$.

$$\frac1y \, dy=\alpha\cdot \frac1x \,dx $$

Integrating both sides

$\int \frac1y \, dy=\int \alpha\cdot \frac1x \,dx $

$\int \frac1y \, dy=\alpha\cdot\int \frac1x \,dx $

$\ln(y)=\alpha\cdot ln(x)+c$

Taking both sides as exponents of $e$

$$e^{\ln(y)}=e^{\alpha\cdot ln(x)+c}$$

$$y=e^{\alpha\cdot ln(x)}\cdot e^{c}$$

$$y=\left(e^{ ln(x)}\right)^{\alpha}\cdot e^{c}$$

$$y=x^{\alpha}\cdot e^{c}$$

With the substitution $e^c=C$ we get

$$y=C\cdot x^{\alpha}$$

Thanks. Do you think the mathematicians that discovered this followed the same line of reasoning? In other words, were they looking for the elasticity of the $f(x)=Cx^\alpha$ equation or were they looking for some equation that contain a term that encapsulated the elasticity of the function as whole? If the latter, could you also provide a proof that starts with the definition of elasticity and ends with the $f(x)=Cx^\alpha$ equation? Is that possible? — Ryan da Silva, Jun 14 '17 at 04:09
@RyandaSilva I´ve made an edit. From the definition of the elasticity you can derive the function with a constant elasticity. — callculus42, Jun 14 '17 at 06:09
I just have one more question. Let's say we wanted to derive something like the Cobb-Douglas function, $Y=cL^\alpha K^\beta$, from the definition of elasticity. Could we just do derivation you did above twice - once for $Y=cL^\alpha$ and once for $cK^\beta$ - and then substitute $cL^\alpha$ for the $c$ in $cK^\beta$? — Ryan da Silva, Jun 23 '17 at 04:41

Elasticity of a Function as a Exponential Constant

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