Page:Calculus Made Easy.pdf/164

of $x$, which is unchanged by differentiation? Accordingly; let us assume as a general expression that

$y=A+Bx+Cx^{2}+Dx^{3}+Ex^{4}+{\text{etc}}$.,

(in which the coefficients $A$, $B$, $C$, etc. will have to be determined), and differentiate it.

${\dfrac {dy}{dx}}=B+2Cx+3Dx^{2}+4Ex^{3}+{\text{etc}}$.

Now, if this new expression is really to be the same as that from which it was derived, it is clear that $A$ must $=B$; that $C={\dfrac {B}{2}}={\dfrac {A}{1\cdot 2}}$; that $D={\dfrac {C}{3}}={\dfrac {A}{1\cdot 2\cdot 3}}$; that $E={\dfrac {D}{4}}={\dfrac {A}{1\cdot 2\cdot 3\cdot 4}}$, etc.

The law of change is therefore that

$y=A\left(1+{\dfrac {x}{1}}+{\dfrac {x^{2}}{1\cdot 2}}+{\dfrac {x^{3}}{1\cdot 2\cdot 3}}+{\dfrac {x^{4}}{1\cdot 2\cdot 3\cdot 4}}+{\text{etc}}.\right)$.

If, now, we take $A=1$ for the sake of further simplicity, we have

$y=1+{\dfrac {x}{1}}+{\dfrac {x^{2}}{1\cdot 2}}+{\dfrac {x^{3}}{1\cdot 2\cdot 3}}+{\dfrac {x^{4}}{1\cdot 2\cdot 3\cdot 4}}+{\text{etc}}$.

Differentiating it any number of times will give always the same series over again.

If, now, we take the particular case of $A=1$, and evaluate the series, we shall get simply

${\begin{aligned}{\text{when }}x&=1,\quad &y&=2.718281{\text{ etc.}};&{\text{that is, }}y&=\epsilon ;\\{\text{when }}x&=2,\quad &y&=(2.718281{\text{ etc.}})^{2};&{\text{that is, }}y&=\epsilon ^{2};\\{\text{when }}x&=3,\quad &y&=(2.718281{\text{ etc.}})^{3};&{\text{that is, }}y&=\epsilon ^{3};\end{aligned}}$