# Translation operator

by matematikuvol
Tags: operator, translation
 P: 190 $$e^{\alpha\frac{d}{dx}}=1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^ 2}{dx^2}+...=\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\frac{d^n}{dx^n}$$ Why this is translational operator? ##e^{\alpha\frac{d}{dx}}f(x)=f(x+\alpha)##
 PF Patron HW Helper Sci Advisor Thanks P: 25,496 taylor expansion?
 PF Patron HW Helper P: 2,688 Consider alpha to be an infinitesimal translation. Expand $f(x+\alpha )$ for small $\alpha$ to first order. Do the same for the LHS of the equation and you should see that the equality is true for infinitesimal translations. We say that the operator $\frac{d}{dx}$ (technically $\frac{d}{idx}$) is the 'generator' of the translation. EDIT: Beaten to the punch by TT!
P: 190

## Translation operator

I have a problem with that. So
$$f(x+\alpha)=f(x)+\alpha f'(x)+...$$
My problem is that we have ##\frac{df}{dx}## and that isn't value in some fixed point ##x##. This is the value in some fixed point ##(\frac{df}{dx})_{x_0}##.
P: 8
 Quote by matematikuvol I have a problem with that. So $$f(x+\alpha)=f(x)+\alpha f'(x)+...$$ My problem is that we have ##\frac{df}{dx}## and that isn't value in some fixed point ##x##. This is the value in some fixed point ##(\frac{df}{dx})_{x_0}##.
I am not sure about your question. But that translation operator is a generic operator, which translate a function value at x to x+a.
 P: 181 In Taylor series x is fixed, while in ##\frac{df}{dx}## ##x## isn't fixed. Well you suppose that is.
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P: 25,496
 Quote by matematikuvol I have a problem with that. So $$f(x+\alpha)=f(x)+\alpha f'(x)+...$$ My problem is that we have ##\frac{df}{dx}## and that isn't value in some fixed point ##x##. This is the value in some fixed point ##(\frac{df}{dx})_{x_0}##.
yes, but x here is a constant, and only α is the variable

if you prefer, write f(xo + α) = f(xo) + α∂f/∂x|xo + …
P: 190
 Quote by tiny-tim yes, but x here is a constant, and only α is the variable if you prefer, write f(xo + α) = f(xo) + α∂f/∂x|xo + …
Ok but that is equal to
$$\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}(\frac{df}{dx})_{x_0}$$
and how to expand now
$$e^{\alpha\frac{d}{dx}}$$
 PF Patron HW Helper Sci Advisor Thanks P: 25,496 no, it's $\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\left(\frac{d^nf}{dx^n}\right)_{x _0}$
 P: 190 I made a mistake. But I'm asking when you get that ##(\frac{d^n}{dx^n})_{x_0}##? Please answer my question if you know. In $$e^{\alpha\frac{d}{dx}}=1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^ 2}{dx^2}+...$$ you never have ##x_0##.
 PF Patron HW Helper Sci Advisor Thanks P: 25,496 $$\left(e^{\alpha\frac{d}{dx}}(f(x))\right)_{x_o}$$ $$= \left(\left(1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^2}{dx^2}+.. .\right)(f(x))\right)_{x_o}$$ $$= \sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\left(\frac{d^nf}{dx^n}\right)_{x _0}$$

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