Suppose you borrow 1 Rupee from Uncle Scrooge, at an interest rate compounded annually of 100%. At the end of the year you have to pay Scrooge Rs 2. Money costs Money.
Suppose Scrooge, the scrooge that he is, starts to wonder …. why only 2 Rupees ? After all, the money owing to him should be accumulating as long as it is in the possession of the borrower. Why should he wait for a year to benefit from this ?
He now insists you should compute interest every month, and not at the end of the year. Thus, by the end of the first month, the borrowed capital should have grown to Rs (1+1/12), at the end of 2 months – to
〖Rs (1+1/12)〗^2 …. and at the end of the year Scrooge’s loaned amount should have grown to
〖Rs (1+1/12)〗^12. This works out to 〖(13/12)〗^12= which is the same as Rs 2.613035. (Try it out in Excel). That’s better. Now coming to think of it, why should Scrooge remain contented with just monthly computations ?
What if we compute interest every day , the year being reckoned at 365 days ? We would see that the capital grows by the end of the year to 〖(1+1/365)〗^365 which works out to Rs 2 .714567 … even better !
Why not compute the money every second ? There are 365x24x60x60 = 31536000 seconds in the year !
We get Rs 〖(1+1/31536000)〗^31536000 , which works out to 2.718282….
What if we find a way of Scrooge’s money keeping on growing every nanosecond, every instant … all the
time ? We would have to calculate 〖(1+1/n)〗^n where n grows to infinity in the limit ! The amount levels off to this grand, magic constant named by Euler as ‘e’ , namely, 2.718281828………… . Whew !
The way Euler computed this number was by expanding 〖(1+1/n)〗^n using the binomial theorem and getting an infinite series. We can see that this expansion is
〖nC^0×1〗^n+nC^1×1^(n-1)×1/n+nC^2×1^(n-2)×1/n^2 +nC^3×1^(n-3)×1/n^3 +⋯ =
n!/n!0!×1+n!/(n-1!1!)×1×1/n+n!/(n-2!2!)×1×1/n^2 +n!/(n-3!3!)×1×1/n^3 …. =
1/0!+1/1!+(n(n-1))/2!×1/n^2 +(n(n-1)(n-2))/3!×1/n^3 …. = 1/0!+1/1!+((1-1/n))/2!+((1-1/n)(1-2/n))/3!….
In the limit as n →∞ , this becomes
1/0!+1/1!+1/2!+1/3!+1/4!+⋯
(don’t forget, 0!=1)
This expression enables the computation of e in a far more rapid and easier manner, and we find the series sum approaches 2.718281828... Try it out on EXCEL.