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Let $T_v$ be a translation by $v$ (i.e. $T_v(x,y)=(x+v_x,y+v_y)$).

Let $f:\mathbb{R}^2\to\mathbb{R}$ be a smooth function (i.e. have derivatives of all orders).

We define the following differential operator : $P_x(f)(v)=(\frac{d}{dt}(f(T_{(t,0)}v))(0)$ (evaluated at t=0).

I don't know how to derive this expression, can someone please help ? (The translations should be with a 1x2 vector and not 2x1, but I don't know this notation in tex).

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You've given a definition - what is there to derive? –  Gerry Myerson Apr 21 '12 at 7:48
    
@GerryMyerson I wish to get an expression like in Giuseppe answer –  Belgi Apr 21 '12 at 7:51
    
You're lucky that Giuseppe was able to read your mind and figure out what you wanted. –  Gerry Myerson Apr 21 '12 at 7:56
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up vote 1 down vote accepted

You have by your same definition $$P_x(f)(v)=\left.\tfrac{d}{dt}\right|_{t=0}f(T_{(t,0)}v)=\left.\tfrac{d}{dt}\right|_{t=0}f(v_x+t,v_y).$$ Applying the chain rule for the derivation you get $$(P_xf)(v)=((\partial_x f)(v),(\partial_y f)(v))\cdot(1,0)=\frac{\partial f}{\partial x}(v).$$ I hope it helps.

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Can you please elaborate about how you used ? thanks! –  Belgi Apr 21 '12 at 7:49
    
Thanks again, it really does help –  Belgi Apr 21 '12 at 8:03
    
Does this mean that if we were to evaluate at a different $t$ we would get the same result ? –  Belgi Apr 21 '12 at 8:20
    
No, $$\left.\tfrac{d}{dt}\right|_{t}f(T_{(t,0)}v)=((\partial_x f,\partial_y f)(T_{(t,0)}v))\cdot(1,0)=(\partial_xf)(T_{(t,0)}v)$$ –  Giuseppe Tortorella Apr 21 '12 at 8:27
    
hmm, where did I go wrong ? I did: $F(t)=f(x(t),y(t))$ where $x(t)=x+t,y(t)=y\implies x'(t)=1,y'(t)=0\implies{P_{x}(f)(v)=(\frac{d}{dt}(f(T_{\begin{pmatrix}t\\ 0 \end{pmatrix}}v))|_{t=0}=(\frac{\partial f}{\partial x}1+\frac{\partial f}{\partial x}0)(v)=\frac{\partial f}{\partial x}v}$. –  Belgi Apr 21 '12 at 8:32
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