The Number E And The Exponential Function - Galileo

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The Number e and the Exponential Function

Michael Fowler

Disclaimer: these notes are not mathematically rigorous.  Instead, they present quick, and, I hope, plausible, derivations of the properties of e,  e x  and the natural logarithm.

The Limit lim n→∞ (1+ 1 n ) n =e

Consider the following series: (1+1), (1+ 1 2 ) 2 ,  (1+ 1 3 ) 3 , ... , (1+ 1 n ) n ,...  where n  runs through the positive integers. What happens as n  gets very large?

It’s easy to find out with a calculator using the function x^y.  The first three terms are 2, 2.25, 2.37.  You can use your calculator to confirm that for n  = 10, 100, 1000, 10,000, 100,000, 1,000,000 the values of (1+ 1 n ) n  are (rounding off) 2.59, 2.70, 2.717, 2.718, 2.71827, 2.718280.  These calculations strongly suggest that as n  goes up to infinity, (1+ 1 n ) n  goes to a definite limit.  It can be proved mathematically that (1+ 1 n ) n  does go to a limit, and this limiting value is called e.   The value of e is 2.7182818283… .

To try to get a bit more insight into (1+ 1 n ) n  for large n,  let us expand it using the binomial theorem. Recall that the binomial theorem gives all the terms in ( 1+x ) n , as follows:

(1+x) n =1+nx+ n(n−1) 2! x 2 + n(n−1)(n−2) 3! x 3 +...+ x n

To use this result to find (1+ 1 n ) n ,  we obviously need to put x=1/n,  giving:

(1+ 1 n ) n =1+n. 1 n + n(n−1) 2! ( 1 n ) 2 + n(n−1)(n−2) 3! ( 1 n ) 3 +...

We are particularly interested in what happens to this series when n  gets very large, because that’s when we are approaching e.   In that limit, n( n−1 )/ n 2  tends to 1, and so does n( n−1 )( n−2 )/ n 3 . .  So, for large enough n,  we can ignore the n  -dependence of these early terms in the series altogether! 

When we do that, the series becomes just:

1+1+ 1 2! + 1 3! + 1 4! +...

And, the larger we take n,  the more accurately the terms in the binomial series can be simplified in this way, so as n  goes to infinity this simple series represents the limiting value of (1+ 1 n ) n . Therefore, e  must be just the sum of this infinite series.

(Notice that we can see immediately from this series that e  is less than 3,  because 1/3! is less than 1/22, and 1/4! is less than 1/23, and so on, so the whole series adds up to less than  1 + 1 + ½  + 1/22  + 1/23 + 1/24 + … = 3.) 

The Exponential Function ex

Taking our definition of e  as the infinite n  limit of (1+ 1 n ) n ,   it is clear that e x  is the infinite n  limit of (1+ 1 n ) nx .   Let us write this another way: put y=nx,   so 1/n=x/y.    Therefore, e x  is the infinite y  limit of (1+ x y ) y .   The strategy at this point is to expand this using the binomial theorem, as above, and get a power series for e x .   

(Footnote: there is one tricky technical point.  The binomial expansion is only simple if the exponent is a whole number, and for general values of x, y=nx  won’t be.  But remember we are only interested in the limit of very large n,  so if x  is a rational number a/b,  where a  and b  are integers, for n  ny multiple of b, y  will be an integer, and pretty clearly the function (1+ x y ) y  is continuous in y,   so we don’t need to worry.  If x  is an irrational number, we can approximate it arbitrarily well by a sequence of rational numbers to get the same result.)

So, we need to do the binomial expansion of (1+ x y ) y  where y  is an integer — to make this clear, let us write y=m:   

(1+ x m ) m =1+m. x m + m(m−1) 2! ( x m ) 2 + m(m−1)(m−2) 3! ( x m ) 3 +...

Notice that this has exactly the same form as the binomial expansion of (1+ 1 n ) n  in the paragraph above, except that now a power of x  appears in each term.  Again, we are only interested in the limiting value as m  goes to infinity, and in this limit m( m−1 )/ m 2  goes to 1, as does m( m−1 )( m−2 )/ m 3 .   Thus, as we take m  to infinity, the m  dependence of each term disappears, leaving

e x = lim m→∞ (1+ x m ) m =1+x+ x 2 2! + x 3 3! +...

Differentiating ex

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

so when we differentiate e x ,  we just get e x  back. This means e x  is the solution to the equation dy dx =y, and also the equation d 2 y d x 2 =y, etc.  More generally, replacing x  by ax  in the series above gives

e ax =1+ax+ a 2 x 2 2! + a 3 x 3 3! +...

and now differentiating the series term by term we see d dx e ax =a e ax ,   d 2 d x 2 e ax = a 2 e ax , etc., so the function e ax  is the solution to differential equations of the form dy dx =ay,  or of the form d 2 y d x 2 = a 2 y  and so on.

Instead of differentiating term by term, we could have written

d dx e ax = lim Δx→0 e a( x+Δx ) − e ax Δx = lim Δx→0 e ax ( e aΔx −1 ) Δx =a e ax

where we have used ( e aΔx −1 )→aΔx  in the limit Δx→0.

The Natural Logarithm

We define the natural logarithm function lnx  as the inverse of the exponential function, by which we mean

y=lnx  if  x= e y .  

Notice that we’ve switched x  and y  from the paragraph above!  Differentiating the exponential function x= e y  in this switched notation,

dx dy = e y =x  so dy dx = 1 x .

That is to say,

d dx lnx= 1 x .

Therefore, lnx  can be written as an integral,

lnx= ∫ 1 x dz z .

You can check that this satisfies the differential equation by taking the upper limit of the integral to be x+Δx,  then x,  subtracting the second from the first, dividing by Δx,  and taking Δx  very small.  But why have I taken the lower limit of the integral to be 1?  In solving the differential equation in this way, I could have set the lower limit to be any constant and it would still be a solution — but it would not be the inverse function to e y  unless the integral has lower limit 1, since that gives for the value x=1  that y=lnx=0.   We need this to be true to be consistent with x= e y ,  since e 0 =1.  

Exercise: show from the integral form of lnx  that for small x, ln( 1+x )  is approximately equal to x.  Check with your calculator to see how accurate this is for x=0.1,  0.01.  

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