A little help with this Taylor expansion please?

[jeebs]

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)=e^ipa, 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:
f(x) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}f(x=a)}{n!} (x-a)^n

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:

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

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 e^x = \sum^{\infty}_{0} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + ... but I just can't seem to get there.

 

[Dick]

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).

 

[jeebs]

Thanks. This does work apparently:

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) + ...
= Y(x) + a\frac{d}{dx}Y(x) + \frac{1}{2!}a^2\frac{d^2}{dx^2}Y(x) + ...
= Y(x) + aipY(x) + \frac{1}{2}(aip)^2Y(x) + ...
= Y(x) + aipY(x) - \frac{1}{2}a^2p^2Y(x) + ...
= (1 + aip - \frac{1}{2}a^2p^2 + ...)Y(x)
= e^{iap}Y(x) = T(a)Y(x) where e^{iap} = 1 + aip - \frac{1}{2}a^2p^2 + ... and p = -i(d/dx)

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.

 

[Dick]

Your original form is
f(x) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}f(a)}{n!} (x-a)^n
Interchange a and x since you want the derivatives evaluated at x,
f(a) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}f(x)}{n!} (a-x)^n
Now you don't want f(a) you want f(x+a), so put change a->x+a.
f(x+a) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}f(x)}{n!} (a)^n
There's lot's of ways to write the taylor series.

 

[jeebs]

Your original form is
f(x) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}f(a)}{n!} (x-a)^n
Interchange a and x since you want the derivatives evaluated at x,
f(a) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}f(x)}{n!} (a-x)^n
Now you don't want f(a) you want f(x+a), so put change a->x+a.
f(x+a) = \sum^{\infty}_{0} \frac{\frac{d^n}{dx^n}f(x)}{n!} (a)^n
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.