# Thread: Comparison of simple and compounded interest function

1. ## Comparison of simple and compounded interest function

Hello everyone,

I had an assignment that one of my actuarial math teacher gave me and there is a question that need to be proved.

6. Assume that 0<i<1. Show that

a) (1+i)^t < 1+it for 0<t<1

b) (1+i)^t > 1+it for 1<t

HINT: if f(a)=0 and f'(x)>0 for a<x<b then f(b)>0

The problem must be solved with the HINT he gave us.

Thank for your help.

2. Originally Posted by solvj
Just to see where you're at, do you agree with following:

> 6. Assume that 0<i<1.

12% annual compounded monthly means i = .01,
and 1.01^12 is the future value of $1 after 1 year, and the effective annual rate becomes 12.6825% >a) (1+i)^t < 1+it for 0<t<1 If rate = 12% compounded annually, then i = .12, and above means the future value of$1 after a period of time less than 1 year.

> b) (1+i)^t > 1+it for 1<t

If rate = 12% compounded annually, then i = .12, and above means
the future value of \$1 after a period of time greater than 1 year, and
if the future value is to be after 3 years and 7 months, then the
value of t would be 3 + 7/12 = 43/12.

.

3. This can be an exemple but it doesn't prove that

a) (1+i)^t < 1+it for 0<t<1
or
b) (1+i)^t > 1+it for 1<t

I need a proof for "t" in the real number

Thank

4. What I said has NOTHING to do with a proof.
I asked if you agreed with (thus understood) what I said.
If not, not much use working on a proof.

Also, I'm a little perplexed with your subject title:
"Comparison of simple and compounded interest function".
There is nothing in your post that refers to simple interest.

5. Sorry for the misunderstood, but yes I agree with what you said.