Page:Calculus Made Easy.pdf/163

${\begin{aligned}&=1+x+{\frac {x^{2}-{\dfrac {x}{n}}}{2!}}+{\frac {x^{3}-{\dfrac {3x^{2}}{n}}+{\dfrac {2x}{n^{2}}}}{3!}}+{\text{etc}}.\end{aligned}}$

But, when $n$ is made indefinitely great, this simplifies down to the following:

$\epsilon ^{x}=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+{\frac {x^{4}}{4!}}+{\text{etc.}}\dots$

This series is called the exponential series.

The great reason why $\epsilon$ is regarded of importance is that $\epsilon ^{x}$ possesses a property, not possessed by any other function of $x$, that when you differentiate it its value remains unchanged; or, in other words, its differential coefficient is the same as itself. This can be instantly seen by differentiating it with respect to $x$, thus:

${\begin{aligned}{\frac {d(\epsilon ^{x})}{dx}}&=0+1+{\frac {2x}{1\cdot 2}}+{\frac {3x^{2}}{1\cdot 2\cdot 3}}+{\frac {4x^{3}}{1\cdot 2\cdot 3\cdot 4}}\\&{\phantom {=0+1+{\frac {2x}{1\cdot 2}}+{\frac {3x^{2}}{1\cdot 2\cdot 3}}\ }}+{\frac {5x^{4}}{1\cdot 2\cdot 3\cdot 4\cdot 5}}+{\text{etc}}.\\or&=1+x+{\frac {x^{2}}{1\cdot 2}}+{\frac {x^{3}}{1\cdot 2\cdot 3}}+{\frac {x^{4}}{1\cdot 2\cdot 3\cdot 4}}+{\text{etc}}.,\end{aligned}}$

which is exactly the same as the original series.

Now we might have gone to work the other way, and said: Go to; let us find a function of $x$, such that its differential coefficient is the same as itself. Or, is there any expression, involving only powers