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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
If the p th , q th & r th term of an AP is x, y and z respectively, show that x(q-r) + y(r-p) + z(p-q) = 0 Ans: p th term ⇒ x = A + (p-1) D q th term ⇒ y = A + (
Decision-Making Under Conditions of Uncertainty With decision making under uncertainty, the decision maker is aware of different possible states of nature, but has insufficient
from 0->1: Int sqrt(1-x^2) Solution) I=∫sqrt(1-x 2 )dx = sqrt(1-x 2 )∫dx - ∫{(-2x)/2sqrt(1-x 2 )}∫dx ---->(INTEGRATION BY PARTS) = x√(1-x 2 ) - ∫-x 2 /√(1-x 2 ) Let
2*9
3x^2+19x-14=0
We will firstly notice the undamped case. The differential equation under this case is, mu'' + ku = F(t) It is just a non-homogeneous differential equation and we identify h
integrate x over x+1
100+5000
Let m be a positive integer with m>1. Find out whether or not the subsequent relation is an equivalent relation. R = {(a,b)|a ≡ b (mod m)} Ans: Relation R is illust
In this case we will require deriving a new formula for variation of parameters for systems. The derivation now will be much simpler than the when we first noticed variation of pa
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