# Problem 31 Solution

Let f(t) equal the money in the account at time t. Since the interest rate is 100% (compounded infinitely) f'(t) = f(t), in other words at any moment in time the money is doubling (which is not the case with simple interest). So for what function does f'(t)=f(t)? This answer is ex. So the value of the account after one year is f(1)=e1=e.

Below is another approach:

The value of \$1 invested for n years at interest rate i, compounded x times per year is (1+(i/x))nx.

In this case f(x) = the value of \$1 invested at 100% interest compounded n times per year = (1+(1/x))x, where x approaches infinity.

The following table shows values of f(x) for various values of x:

```     x              f(x)
-----------     ----------
1	2.00000000
4	2.44140625
12	2.61303529
52	2.69259695
365	2.71456748
1,000	2.71692393
10,000	2.71814593
100,000	2.71826824
1,000,000	2.71828047
10,000,000	2.71828169
100,000,000	2.71828180
```

As x approaches infinity it can be seen that f(x) approaches e =~ 2.71828182846.

Michael Shackleford, A.S.A.