In my previous post, we saw that we can derive Cobb-Douglas production function from CES production function when the elasticity of substitution is equal to zero. In this blog, we will show that it is quite easy to obtain a Cobb-Douglas production function from CES production function and to check the results with Wolfram Alpha.

We can recall the expression of the CES function that we used (with constant returns to scale):

Y=[\alpha K^\rho+(1-\alpha) L^\rho]^{\frac{1}{\rho}} \quad (1)

Now, we use the exponential:

Y=\exp\Big\lbrace\frac{1}{\rho}\ln[\alpha K^\rho+(1-\alpha) L^\rho]\Big\rbrace\quad (2)

Consider the first-order Maclaurin expansion (Taylor expansion centered at zero) of the term inside the logarithm, with respect to ρ, as explain here:

\small \begin{align*} & a (K^{\rho}) +(1-a) (L^{\rho}) \\ & = a (K^{0}) +(1-a) (L^{0}) +a (K^{0})\rho\ln K \\ & +(1-a) (L^{0})\rho\ln L + O(\rho^2) \\ & =a+(1-a)+a\rho \ln K+(1-a)\rho \ln L + O(\rho^2) \\ & = 1 +\rho \big[\ln K^{a}L^{(1-a)}\big]+ O(\rho^2) \end{align*}

You can check that the symbolic value the derivative of the term inside the brackets here for the capital and here for the labor. In a previous post, I tried to give an intuitive explanation of the Big-O notation. The first-order Maclaurin expression can be checked by typing the following formula in Wolfram Alpha:

`Series[a K^ρ + (1 - a) L^ρ, {ρ, 0, 0}]`

Insert this in equation (1):

Y=\Big(1 +\rho \big[\ln K^{a}L^{(1-a)}\big]+ O(\rho^2)\Big)^{\frac{1}{\rho}}

Define r = 1/ρ and re-write,

Y=\Big(1 +\frac{ \big[\ln K^{a}L^{(1-a)}\big]}{r}+ O(r^{-2})\Big)^{r}

The Big-O terms that capture the remainder of the approximation will approach tends to 0 where *r* tends to infinity (I thank Mario Veruete for his crystal-clear explanation on this point!). Now it gives an expression whose limit at infinity will give us an exponential:

\lim_{\rho\rightarrow 0}Y = \lim_{r\rightarrow \infty}Y = \left(\exp\left\{ \ln K^{a}L^{(1-a)}\right\} \right)

\lim_{\rho\rightarrow 0}Y = K^{a}L^{(1-a)} \quad (3)

In Wolfram Alpha, you can enter the following following formula to check the results:

`Limit[(a K^ρ + (1 - a) L^ρ)^(1/ρ), ρ -> 0]`

This show that the CES production function gives the Cobb-Douglas production function when ρ tends to 0.

## 7 Comments

In Equation 2, I don’t think it’s necessary to use the exponential function trick. I don’t see any use in using this trick. I sketched out a proof without using $\exp$ and I think it worked. Could you tell me if this trick is essential? I couldn’t see it.

To simplify the limit calculation, you can use the exponential. Then, apply the L’Hopital’s rule.

Dear Professor:

This is an astonishing wonderful illustration of the CD derivation from CES! However, might it be due to my misconception that the fourth line of Maclaurin expansion may have an “(1-a)ρlnL” instead of “(1-a)ρln”? Thank you very much and please forgive my boldness.

You’re welcome. Can you expand a bit your question?

Dear Professor, my question is that I’m afraid the fourth line of the Maclaurin expansion lacks an “L” in the four term, i.e. “(1-a)ρlnL” instead of “(1-a)ρln”. This might be due to typo or my browser issues.

Many thanks for your very good point, it was a typo. It’s corrected now.

Kind regards,

Jamel

I would like to thank Jonathan Benchimol for spotting a typo in equation (2)!