# Utility Theory - Certainty equivalent approximation formula derivation

I have a question on an exercise from chapter 9 of D. Luenberger, Investment Science, International Edition, where I suspect there may be a typo.

Exercise 8 (Certainty approximation)

There is a useful approximation to the certainty equivalent that is easy to derive. A second-order expansion near $\bar x=E(x)$ gives $$U(x)\approx U(\bar x)+U^{'}(\bar x)(x-\bar x)+\frac12U^{''}(\bar x)(x-\bar x)^2$$

Hence, $$E[U(x)]\approx U(\bar x)+\frac 12 U^{''}(\bar x)var(x)$$ On the other hand, if we let c denote the certainty equivalent and assume it is close to $\bar x$, we can use the first-order expansion $$U(c)\approx U(\bar x)+U^{'}(\bar x)(c-\bar x)$$ Using these approximations, show that $$c \approx \bar x+{U^{''}(\bar x) \over U^{'}(\bar x)}var(x)$$

Now, I used general methods of algebra along with the fact that $E[U(x)] = U(c)$ to show directly that $$c \approx \bar x+\frac 12 ({U^{''}(\bar x) \over U^{'}(\bar x)})var(x)$$ as follows:

Take the third equation and transform it into $$c\approx \bar x + {U(c)-U(\bar x) \over U^{'}(\bar x)}$$ Now all I have to do is show that the numerator in the fraction part is $\frac 12 U^{''}(\bar x)var(x)$ which is done by putting $E[U(x)] = U(c)$ into the second formula and you can see the result is immediately there.

On top of this work, I wrote out an example of an investment with the log utility function and showed that my approximation for c worked whereas the book's formula without the "2" didn't.

However, I would like to post all this here just to verify that this is a typo from the book and not some misunderstanding on my part.

Thanks in advance for any feedback.