A little help with this Taylor expansion please?

In summary: T(a) that acts on a function y(x) and causes the transformation T(a)y(x) = y(x+a) can be written in terms of the operator p = -i.d/dx]
  • #1
jeebs
325
4
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
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
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
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
Dick said:
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.
 

1. What is a Taylor expansion?

A Taylor expansion is a mathematical technique used to approximate a function with a polynomial. It is often used in calculus to simplify the calculation of derivatives and integrals.

2. When is a Taylor expansion used?

A Taylor expansion is commonly used when a function is difficult to evaluate directly, or when the function is defined only by a few of its derivatives. It is also used to approximate functions with a simpler form for easier calculations.

3. How is a Taylor expansion calculated?

A Taylor expansion is calculated by taking the derivatives of a function at a specific point, and using those derivatives to construct a polynomial. The polynomial is then used to approximate the function at that point and in its surrounding neighborhood.

4. What is the significance of the order of a Taylor expansion?

The order of a Taylor expansion refers to the highest degree of the polynomial used to approximate the function. The higher the order, the more accurate the approximation will be. However, using a higher order also requires more derivatives to be calculated.

5. What are some applications of Taylor expansions?

Taylor expansions have a wide range of applications in mathematics, physics, and engineering. They are commonly used in optimization problems, numerical analysis, and differential equations. They are also used in the development of computer algorithms and in the study of chaotic systems.

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