Math Help - 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.

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.