A little help with this Taylor expansion please?

  • Thread starter jeebs
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  • #1
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I have this translation operator T(a) that acts on a function y(x) and causes the transformation T(a)y(x) = y(x+a).
I am supposed to be "expanding y(x+a) as a taylor series in a" to show that T(a)=eipa, where p is the operator p = -i.d/dx]
So, I've started out with the general equation for the Taylor expansion, to expand, say, f(x) about the point x=a:
[tex] f(x) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}f(x=a)}{n!} (x-a)^n[/tex]

I've used this plenty of times before but only when I have been expanding something like f(x) rather than, say, f(x+c), so I'm making mistakes and messing up when I try and work through this problem.
Anyway, back to the original thing I'm trying to expand out. What I've got so far is:

[tex] y(x+a) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}y(x+a = ?)}{n!}[(x+a)- ?]^n [/tex]

Where I've left the question marks is where I'm not certain what to use. My y(x) expansion is otherwise the same way as the general formula for f(x) just with x+a replacing f(x).
I know I'm supposed to end up with something of the same form as [tex] e^x = \sum^{\infty}_{0} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + ...[/tex] but I just can't seem to get there.
 
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  • #2
Dick
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The taylor expansion of y(x+a) is the sum over n of D_n(y(x))*a^n/n! where D_n(y(x)) denotes the nth derivative of y evaluated at x, isn't it? Compare that with the operator expansion of exp(ipa) applied to y(x).
 
  • #3
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Thanks. This does work apparently:

[tex] Y(x+a) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}Y(x)}{n!}a^n = \frac{1}{0!}a^0Y(x) + \frac{1}{1!}a^1\frac{d}{dx}Y(x) + \frac{1}{2!}a^2\frac{d^2}{dx^2}Y(x) + ... [/tex]
[tex] = Y(x) + a\frac{d}{dx}Y(x) + \frac{1}{2!}a^2\frac{d^2}{dx^2}Y(x) + ... [/tex]
[tex] = Y(x) + aipY(x) + \frac{1}{2}(aip)^2Y(x) + ... [/tex]
[tex] = Y(x) + aipY(x) - \frac{1}{2}a^2p^2Y(x) + ... [/tex]
[tex] = (1 + aip - \frac{1}{2}a^2p^2 + ...)Y(x) [/tex]
[tex] = e^{iap}Y(x) = T(a)Y(x) [/tex] where [tex] e^{iap} = 1 + aip - \frac{1}{2}a^2p^2 + ...[/tex] and [tex] p = -i(d/dx)[/tex]

However, I'm not sure why this is. I would not have been able to do this if you hadn't told me. Certainly this sort of thing would shaft me in an exam (which I have soon). How do we get from the basic Taylor series expansion for, say, y(x), to the one that you gave me for y(x+a)? If it's too much to be bothered typing out, a link to a webpage would be nice.
 
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  • #4
Dick
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Your original form is
[tex]f(x) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}f(a)}{n!} (x-a)^n[/tex]
Interchange a and x since you want the derivatives evaluated at x,
[tex]f(a) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}f(x)}{n!} (a-x)^n[/tex]
Now you don't want f(a) you want f(x+a), so put change a->x+a.
[tex]f(x+a) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}f(x)}{n!} (a)^n[/tex]
There's lot's of ways to write the taylor series.
 
  • #5
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Your original form is
[tex]f(x) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}f(a)}{n!} (x-a)^n[/tex]
Interchange a and x since you want the derivatives evaluated at x,
[tex]f(a) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}f(x)}{n!} (a-x)^n[/tex]
Now you don't want f(a) you want f(x+a), so put change a->x+a.
[tex]f(x+a) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}f(x)}{n!} (a)^n[/tex]
There's lot's of ways to write the taylor series.
ahh of course, thanks man... maths ey? it's always so simple in hindsight.
 

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