2

For example:

$f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{1}{2!}f''(x_0)(x-x_0)^2+\dots=\exp\left((x-x_0)\frac{\mathrm{d}}{\mathrm{d}x}\right)f(x)\Big|_{x_0}$

I don't know how to write the $\Big|_{x_0}$ to the front … Is this thing an operator? How to write it down more elegantly?

LePtC
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    Not exactly because $\left((x-x_0)\frac{d}{dx}\right)^n \neq (x-x_0)^n\left(\frac{d}{dx}\right)^n$ – amcalde Oct 01 '14 at 13:37
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    The squared term in the exponential is not the squared term in the expansion though as you will have a derivative of $x-x_0$ – Paul Oct 01 '14 at 13:51
  • good point, any suggestion to improve it? – LePtC Oct 01 '14 at 14:13
  • I doubt there is a special, more elegant way to write $1+ab+\frac{a^2b^2}{2}+\frac{a^3b^3}{3!}+\cdots$ when $ab\neq ba$. – Jonas Meyer Oct 01 '14 at 17:22

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