Hurray for pi and especially e

Pi-day is around the corner, but what about e-day? Here in Europe we’re better off than in the US. Let me explain. e-Day is celebrated on February 7 in countries where the date format is 2/7. Over here we celebrate it on January 27, because that’s 27/1; and that’s one decimal more precise than for the US. So basically we win.

But then again it’s the e that counts. It’s a special number. You’ll encounter it everywhere, together with its exponential function e^x.
That function is even more special because e^x stays true to itself. Integrate it, differentiate it, do it again and again, and you’ll end up with e^x. It means that everywhere on the graph the change in function value is equal to the function value itself. Amazing. And the surface under any point of the graph is equal to the function value at that point. Bonus points.

You can show that the derivative of e^x is equal to itself in a nice and elegant way with the knowledge that if f(x)=x then f'(x)=1. Now if x=\ln(e^x) then by virtue of the chain rule the derivative is \frac{1}{e^x}\cdot \tfrac{d}{dx}e^x=1 so \tfrac{d}{dx}e^x has to be e^x.

The more formal way is to use the definition of the derivative

    \[f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\]

Let’s see what happens:

    \[\lim_{h\rightarrow 0}\frac{e^{x+h}-e^x}{h}=\lim_{h\rightarrow 0}\frac{e^xe^h-e^x}{h}\]

This is

    \[\lim_{h\rightarrow 0}\frac{e^x(e^h-1)}{h}=e^x\cdot\lim_{h\rightarrow 0}\frac{e^h-1}{h}\]

All we have to do now is look at

    \[\lim_{h\rightarrow 0}\frac{e^h-1}{h}\]

According to the graph below it should be 1.

Graph of \frac{e^h-1}{h} At h=0 there’s a hole at (0,1)

But let’s look at it a bit more precise. The definition of e is

    \[\lim_{h\rightarrow\infty}(1+\tfrac{1}{h})^h =\lim_{h\rightarrow 0}(1+h)^{\tfrac{1}{h}}\]

If we substitue this in the limit above, the limit for e is preserved within this limit:

    \[e^x\cdot\lim_{h\rightarrow 0}\frac{((1+h)^{\tfrac{1}{h}})^h-1}{h}\]

which of course is

    \[e^x\cdot\lim_{h\rightarrow 0}\frac{1+h-1}{h}=e^x\cdot 1 =e^x\]

So, if

    \[f(x)=e^x\]

then

    \[f'(x)=e^x\]

Those who forgot to celebrate e-day can always celebrate e^x-day in private.