# Taylor expansion for the exponential series

Inside this question, a key derivation was made in which

$$x(a)=\exp\Big(a\frac{\partial}{\partial{t}}\Big)x(t)\Big|_{t=0}$$

I know that we can write $$x(a)$$ as a series by

$$x(a)=\sum_{n=0}^\infty\frac{x^{(n)}}{n!}a^n$$

But, I don't see how

$$x(a)=\sum_{n=0}^\infty\frac{x^{(n)}}{n!}a^n=\cdots=\exp\Big(a\frac{\partial}{\partial{t}}\Big)x(t)\Big|_{t=0}$$

How do you derive $$x(a)=\exp\Big(a\frac{\partial}{\partial{t}}\Big)x(t)\Big|_{t=0}$$?

Through the related question, I think I understand the derivation.

We are given that

$$x(a)=\sum_{n=0}^\infty\frac{x^{(n)}}{n!}a^n$$

We also know that

$$\text{exp}(a)=\sum_{n=0}^\infty\frac{a^n}{n!}$$

Therefore,

$$\text{exp}\Big(a\frac{\partial}{\partial{t}}\Big)=\sum_{n=0}^\infty\frac{a^n}{n!}\Big(\frac{\partial}{\partial{t}}\Big)^n$$

Hence if we evaluate this at $$x(t)$$ we have

$$\text{exp}\Big(a\frac{\partial}{\partial{t}}\Big)x(t)=\sum_{n=0}^\infty\frac{a^n}{n!}\Big(\frac{\partial{x(t)}}{\partial{t}}\Big)^n =\sum_{n=0}^\infty\frac{a^n}{n!}x^{n}(t)=\sum_{n=0}^\infty\frac{x^{n}(t)}{n!}a^n$$

It is necessary for $$t=0$$ in order to compute the Maclaurin expansion. If $$t\neq0$$, then we would have

$$x(a)=\sum_{n=0}^\infty\frac{x^{n}(t)}{n!}(a-t)^n$$

Therefore,

$$\text{exp}\Big(a\frac{\partial}{\partial{t}}\Big)x(t)\Big|_{t=0}=\sum_{n=0}^\infty\frac{x^{n}(t)}{n!}a^n=x(a)$$