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If r per annum is the rate at which the principal A is compounded annually, then at the end of k years, the money due is
Q = A (1 + r)k
Suppose compounding is done continuously. i.e. at every instant the principal A is compounded at R per annum. Then,
Q = A eRk
The relationship between R and r is given by the following reasoning:
A (1 + r)k = A eRk
Example
If R = 5.25%, then ln(1 + r) = 5.25% or r = 5.39%
Suppose Rs.100 is being compounded annually at the rate of 10% per annum. What is the future value of Rs.100 at the end of the third year? What is the effective continuously compounded rate of interest? What is the future value of Rs.100 at the end of the third year, using this interest rate?
FV(Rs.100) = 100 x (1.10)3 = 133.1
If r = 0.1, then the continuously compounded rate of interest R is given by
R = ln(1 + 0.1) = 0.0953
FV(Rs.100) = 100 e0.0953 x 3 = 100 x 1.331 = 133.1
Write an octave program that will take a set of points {x k , f k } representing a function and compute the derivative at the same points x k using 1. 2-point forward dierence
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