Translation operator


by matematikuvol
Tags: operator, translation
matematikuvol
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#1
Feb21-13, 02:01 PM
P: 192
[tex]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}[/tex]
Why this is translational operator?
##e^{\alpha\frac{d}{dx}}f(x)=f(x+\alpha)##
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tiny-tim
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#2
Feb21-13, 02:29 PM
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taylor expansion?
G01
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#3
Feb21-13, 02:30 PM
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Consider alpha to be an infinitesimal translation. Expand [itex]f(x+\alpha )[/itex] for small [itex]\alpha[/itex] 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 [itex]\frac{d}{dx}[/itex] (technically [itex]\frac{d}{idx}[/itex]) is the 'generator' of the translation.

EDIT: Beaten to the punch by TT!

matematikuvol
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#4
Feb21-13, 04:19 PM
P: 192

Translation operator


I have a problem with that. So
[tex]f(x+\alpha)=f(x)+\alpha f'(x)+...[/tex]
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}##.
Questionasker
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#5
Feb21-13, 04:25 PM
P: 8
Quote Quote by matematikuvol View Post
I have a problem with that. So
[tex]f(x+\alpha)=f(x)+\alpha f'(x)+...[/tex]
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.
LagrangeEuler
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#6
Feb21-13, 04:35 PM
P: 276
In Taylor series x is fixed, while in ##\frac{df}{dx}## ##x## isn't fixed. Well you suppose that is.
tiny-tim
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#7
Feb21-13, 04:36 PM
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Quote Quote by matematikuvol View Post
I have a problem with that. So
[tex]f(x+\alpha)=f(x)+\alpha f'(x)+...[/tex]
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 + …
matematikuvol
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#8
Feb21-13, 05:06 PM
P: 192
Quote Quote by tiny-tim View Post
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
[tex]\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}(\frac{df}{dx})_{x_0}[/tex]
and how to expand now
[tex]e^{\alpha\frac{d}{dx}}[/tex]
tiny-tim
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#9
Feb21-13, 05:20 PM
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no, it's [itex]\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\left(\frac{d^nf}{dx^n}\right)_{x _0}[/itex]
matematikuvol
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#10
Feb22-13, 02:21 AM
P: 192
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
[tex]e^{\alpha\frac{d}{dx}}=1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^ 2}{dx^2}+...[/tex]
you never have ##x_0##.
tiny-tim
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#11
Feb23-13, 01:55 AM
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[tex]\left(e^{\alpha\frac{d}{dx}}(f(x))\right)_{x_o}[/tex]
[tex]= \left(\left(1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^2}{dx^2}+.. .\right)(f(x))\right)_{x_o}[/tex]
[tex]= \sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\left(\frac{d^nf}{dx^n}\right)_{x _0}[/tex]


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