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144

Calculus Made Easy

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

${\displaystyle {\begin{aligned}{\text{when }}x&=1,\quad &y&=2.718281{\text{ etc.}};&{\text{that is, }}y&=\epsilon$

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