2017-06-20 13:25:45 UTC
Suppose you are Isaac Newton, and it is your Annus Mirabilis 1666. You've just invented calculus and power series.
You've discovered that the antiderivative of x^n is x^(n+1)/(n+1).
You know nothing of logarithms -- having never heard of Alphonse Antonio de Sarasa (Google him).
Something is bothering you terribly; what is the form of the integral of x^n when n=-1? The problem is, the denominator of x^(n+1)/(n+1) vanishes when n=-1, so what exactly does this mean?
You know that the definite integral of x^n from x=a to x=b is
= b^(n+1)/(n+1) - a^(n+1)/(n+1)
In particular, you know that the definite integral of x^n from x=1 to x=b is
You quickly grow tired of dealing with "n+1", so you replace "n+1" by "m", so the definite integral of x^(m-1) from x=1 to x=b is
You, being Isaac Newton, believe in the fundamental continuity of the universe; in fact, you believe in continuity so much that you cannot even comprehend the possibility of non-continuous functions.
So you investigate what happens as m "approaches" 0 (i.e., as n approaches -1).
But since we will be investigating small numbers, it will be more convenient to change variables once more so that m=1/p, and p will be an integer we will make very large. Yes, we have changed from integer m (and hence n) to non-integer m,n, but Newton did this all the time, as it is a necessary consequence of his belief that everything is continuous.
So your formula for the integral of x^n (= x^(1/p-1)) from 1 to b is now:
You (or actually, your assistant, Max Gnuplot) now plots these functions for p=1,2,4,8,16,32,64,128:
You can hardly believe your eyes: the curves are converging! As p grows and grows, the plots become more and more similar, which convinces Newton that he may be onto something.
But you're still suspicious, so you also check negative values of p:
Aha! you say. The positive p's converge from above, the negative p's converge from below, so there must be a specific curve in between the upper and lower bounds that is the answer to our question: what happens to x^(n+1)/(n+1) as n approaches -1?
Now that Newton knows that a function exists which answers his question, he wants to give it a name. Having just participated in a recent misadventure with an apple tree, now reduced to firewood, Newton decides to call the function "log(b)":
log(b) = lim (b^(1/p)-1)*p
[p->-oo also works.]
Newton, being Newton, wants to better understand this function, so he asks about its inverse function, which he temporarily calls "ilog".
If c=log(b), then ilog(c)=b, so after simple algebra,
ilog(c) = b = lim (1+c/p)^p
Now Newton immediately notices the connection between ilog(c) and continuously compounding interest; Newton didn't become warden of the Royal Mint for nothing! (Yes, Newton was "Mint Newton", not "Fig Newton".)
Given the relationship with compound interest, Newton decides to call this inverse function "exp(c)", as he fully EXPects to reap a reward of exp(c) from an interest rate of c. exp(1)=2.71828... is thus the "Mark of the Beast" that one might expect from a year numbered 1666.
exp(c) = b = lim (1+c/p)^p
Newton knows about power series (including power series inversion), so he starts laboriously computing the power series for exp(c):
exp(c) = 1 + c + c^2*approx(1/2) + c^3*approx(1/6) + c^4*approx(1/24) + ...
where "approx(x)" = very close approximation in *base p* (!) of x.
Furthermore, Newton notices that as p grows larger, this approximation becomes better and better.
Newton leaps (parabolically, as was his wont!) to the conclusion that
exp(c) = c^0/0! + c^1/1! + c^2/2! + c^3/3! + ...
Now, Newton was quite familiar with simplifying formulae like
x^2*(x-1)*(x-2)2/(x-1) = x^2*(x-2),
where a potential problem at x=1 disappears when taking the limit as x approaches 1, so he finally concludes that his new function
log(b) ~ (b^0-1)/0
so that his antiderivative formula becomes unexceptional.
In other words, the problem with using x^(n+1)/(n+1) as the antiderivative of x^n is the fact that it implicitly uses 0 as the lower bound of the integration:
integrate(x^n,x) = integrate(x^n,x,0,x)
What we have discovered in the discussion above is that the problem vanishes if we instead use *one* as the implicit lower bound of the integration:
integrate(x^n,x) = integrate(x^n,x,1,x) = (x^(n+1)-1)/(n+1)
So log(b) ~ (b^0-1)/0 now "makes sense".