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The compound interest on a sum of money for 2years is832 and the simple interest on the same sum for same period rs.800.the difference between compound interestand thesimple  interest for 3years will be ?

I can calculate the answer in terms of the parameters of the problem but because there are more unknowns than independent equations, I can't provide a numerical solution. Also, because the equation for one of the parameters is implicit, I need to use an approximation to arrive at a solution.

Define the parameters P = initial principal, r = annual interest rate, y = number of years  and n = number of compounding periods per year. Then using the formula for compound interest we have

return after 2 years (y=2) for compounded interest R(2,n) = P(1 +r/n)^2n = 832. The interest is then R(2,n) - P = P[(1+r/n)^2n-1].

For the simple interest (y=2, n=1), we have for the interest R(2,1)-P = P[(1+r)^2-1] = 800. We can easily solve this for r in terms of P so let's assume we "know" both r and P and solve for n.

Rewrite the R(2,n) equation as (1+r/n)^2n = (832/P)-1 = C (constant), then taking the natural log

2n・ln(1+r/n) = lnC.

This is the implicit equation and numerical techniques are need to solve it, at least in the sense of getting an arbitrarily accurate answer. Graphing can be used, but since r/n is much less than 1, i.e., r/n << 1 where r = interest rate, say 4%  of r = 0.04 and n is > 1, say n = 4 (quarterly compounding) so that r/n ~ 0.01, we can use the Taylor series expansion of ln(1+x) for small x to get

2n・ln(1+r/n) ≈ 2n・(r/n - (r/n)^2 + ...) = lnC or

n = 2r^2/(2r-lnC).

A little messy but if you had values for any 2 of P, r or n, a good numerical solution for the 3 year simple interest could be obtained.

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randy patton


college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography


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