Exponential

Let us define \(e(x)=\Sigma x^n/n!\). We have already seen that \(e(x)\) is defined everywhere. We also note that since \((x+y)^n=\Sigma_{i=0}^n n!/(k!n-k!)x^ky^{n-k}\) and so (x+y)^n/n!=Sigma_{i=0}^n (x^k/k!)(y^{n-k}/(n-k)!) then by the fact that it converges absolutely and by the permutation theorem, we have \(e(x+y)=e(x)e(y)\). In particular, we have: \(e(q)=e(1)^q\) for any rational \(q\). By definition, we extend this to all numbers, and putting \(e=e(1)\) we have \(e(x)=e^x\). This is the so-called “natural exponent” function.

We note that \((e(x+h)-e(x)/h)=e(x)[e(h)-1]/h=e(x)[1+h\Sigma h^n/(n-1)!]\). If \(|h|<1\), we have \(|\Sigma h^n/(n-1)!`|(e-1)n\), so for any \(t>1\), there is a number \(e^s=t\). If \(t1\), then we can write \(b^2=a\), \(b=1+c\). \(n/a^n=n/(b^2)^n=n/(b^n)^2=n/((1+c)^n)^2>n/(cn)^2=1/(c^2)n\) converges to \(0\). By sandwich, \(n/a^{n^2}\) also converges to \(0\). By continuity, \(n^k/(a^k)^{n^2}\) converges to \(0\) for any real \(a>1\). If \(c>1\), let \(a\) be such that \(a^k=c\), then \(n^k/c^{n^2}\) converges to \(0\) for any real \(c>1\). Therefore, the limit of c^{-1/x^2}/x^k is \(0\) at \(0\). Away from \(0\), \(e^{-1/x^2}\) has a derivative. For any \(n`th order derivative, it is :math:`e^{-1/x^2}/(a_kx^k+....+a_0)\). So we have that we can complete \(e^{-1/x^2}\) to \(0\) with \(0\), and it has all derivatives at \(0\) be \(0\).

For this reason, if we define \(f(x)=0\) if \(x\leq 0\), \(f(x)=e^{-1/x^2}\) if \(x>0\), we have \(f\) have all derivatives. If we define \(g(x)=f(x)f(1-x)\) then \(g\) has all derivatives, and \(g\) is 0 outside of \([0,1]\). \(h=\int_0^x g\) is bounded by \(\int_0^1 g\), and has all derivatives. If we assume the bound is \(M\), \(t=h/M\) is \(0\) for \(x1\) and increasing between \(0\) and \(1\). If we put \(s(x)=t(x/\epsilon)\), then we have \(s(x)=0\) if \(xepsilon\). If we take \(q(x)=s(x)s(1-x)\) then we have the function we have promised at the beginning: it is \(0\) for \(x<0\), it is \(1\) for \(\epsilon<x1\). Therefore, if \(q\) defines your position on a hundred-meter dash course, then at the beginning, you’re at the beginning. After a minute, you are back at the beginning – and for most of the intervening minute (all but \(\epsilon\) of it, say a microsecond) you are chatting with your friend at the other end. Note that \(q\) is a “nice” function: it has all derivatives, so at any given time, you will have a well-defined speed and acceleration.