Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm currently working through a book on the Theory of Interest. Currently, I'm looking at continuous annuities, or annuities that are paid continuously. The book gives the following formula, the derivation of which makes perfect sense to me. Little delta is the force of interest and nu is the discount factor.

$$a_n = \int_0^n\nu^t$$

$$= \frac{1-\nu^n}{\delta}$$

This I follow so far. The author then states that the formula can be expressed entirely in terms of the force of interest, and gives the following formula but does not show its derivation.

$$a_n= \frac{1-e^{-n\delta}}{\delta}$$

How is this arrived at? I don't understand how the discount factor is equivalent to e to the "-n delta."

share|cite|improve this question
up vote 1 down vote accepted

Remember that $\ln(1+i) = \delta$ so that $1+i = e^{\delta}$. Thus, overall, we have

$$v^n = (1+i)^{-n} = \left(e^{\delta}\right)^{-n} = e^{-n\delta}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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