Assume compounded and i 0 If 1 grows to K in x years at rat

Assume compounded and i > 0. If $1 grows to $K in x years at rate i per year, and $1 grows to $K in y years at rate 2i, prove that y > x/2. This result says that if we double the rate of interest, it takes more that half the amount of time to reach the same accumulation.

Hint: (1 + i)^x = ((1 + i)^2)^(x/2).

Solution

the compound intersent formula is :

A = P[1 + (r/n)]^(nt)

for case 1:

P = principal = $ 1

A = $ K

r = i % per year = i/100

t = time period = x years

n = number of times compounded = annual = 1

=> K = 1*(1 + i/100)^x

    K = [(100 + i)/100]^x -------------(1)

Similarly for the other condition

K = [(100 +2i)/100]^y ----------(2)

So from (1) and (2)

[(100 + i)/100]^x = [(100 + 2i)/100]^y

[(100 + i)]^x = [(100 + 2i)]^y

or we could write it as

[(1 + i)]^x = [(1 + 2i)]^y

now we could write the Left hand side as

[((1 + i)^2)^x/2]

[((1 + 2i + i^2)^x/2]

assuming i is very small

=> [((1 + 2i)^x/2]

=> [((1 + 2i)^x/2] = [(1 + 2i)]^y

now we have the same base on both the sides

=> the exponents will be equal

hence y = x/2

or y > x/2