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First Calculation of E

Date: 12/18/97 at 13:50:49
From: Nick Vlassis
Subject: The first calculation of the value of e


I am a student doing a project on the discovery of e. I already know 
that D_x e^x = e^x, and that this is how e came to be found. However, 
I *don't* know (and really hope to learn how Euler first calculated 
the value of e starting from this equation. I am specifically 
interested in the *historical* facts, so please respond, if possible, 
with the exact method originally used by Euler.

Thanks in advance,

Nickolas Vlassis

Date: 12/20/97 at 08:46:15
From: Doctor Anthony
Subject: Re: The first calculation of the value of e

In the 1730's Euler investigated the result of compounding interest 
continuously when a sum of money, say, is invested at compound 

If interest is added once a year we have the usual formula for the 
amount, A,  with principal P, rate of interest r% per annum, and t the 
time in years:

   A = P(1 + r/100)^t

If interest were added twice a year, then we replace r by r/2 and we 
replace t by 2t. So formula becomes

   A = P(1 + r/(2x100))^(2t)   = amount after t years.

If 3 times a year then A at the end of t years would be:

   A = P(1 + r/(3x100))^(3t)

and if we added interest N times a year, then after t years the amount 
A would be

   A = P[1 + r/(Nx100)]^(Nt)

Now to simplify the working we put r/(100N) = 1/n  so  N = nr/100

and  A = P[1 + 1/n]^(nrt/100)

     A = P[(1 + 1/n)^n]^(rt/100)

We now let n -> infinity and we must see what happens to the 

    (1 + 1/n)^n  as n tends to infinity.

Expanding by the binomial theorem

(1 + 1/n)^n = 1 + n(1/n) + n(n-1)/2! (1/n)^2 + n(n-1)(n-2)/3! (1/n)^3 
+ ...

now take the n's in 1/n^2,  1/n^3 etc, in the denominators and 
distribute one n to each of the terms n, n-1, n-2, etc in the 
numerator, getting

  1 x (1-1/n) x (1-2/n) x .....  so we now have

(1 + 1/n)^n = 1 + 1 +  1(1-1/n)/2! + 1(1-1/n)(1-2/n)/3! + .....

Now let n -> infinity and the terms 1/n, 2/n etc all go to zero, 

(1 + 1/n)^n = 1 + 1 + 1/2! + 1/3! + 1/4! + .....

and this series converges to the value we now know as e.

If you consider e^x you get

(1 + 1/n)^(nx) and expand this by the binomial theorem you have

(1 + 1/n)^(nx) = 1 + (nx)(1/n) + nx(nx-1)/2! (1/n)^2 + ....

and carrying through the same process of putting the n's in the 
denominator into each term in the numerator as described above you 

  e^x = 1 + x + x^2/2! + x^3/3! + .... 

and differentiating this we get

 d(e^x)/dx = 0 + 1 + 2x/2! + 3x^2/3! + ...

           =     1 + x + x^2/2! + x^3/3! + ....

           =   e^x

Reverting to our original problem of compounding interest 
continuously, the formula for the amount becomes

    A = P.e^(rt/100)

You might like to compare the difference between this and compounding 

If P = 5000,   r = 8,   t = 12 years

Annual compounding gives  A = 5000(1.08)^12  = 12590.85

Continuous compounding gives  A = 5000.e^(96/100)  = 13058.48

The difference is not as great as might be expected.

-Doctor Anthony,  The Math Forum
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Associated Topics:
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