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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
01010011 01100101 01101101 01110000 01100101 01110010 00100000 01000110 01101001 00100001
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Ashow that sec^2x+cosec^2x cannot be less than 4
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Use Newton's Method to find out an approximation to the solution to cos x = x which lies in the interval [0,2]. Determine the approximation to six decimal places. Solution
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