# (Physicist version of) Taylor expansions

• Jason Bennett
In summary: That makes a big difference. (Hint: think about what happens to ##\ln (1+\delta a)## as ##\delta a \to 0##.)
Jason Bennett
Homework Statement
in the context of garnering Lie algebras from Lie groups
Relevant Equations
see below
3) Taylor expansion question in the context of Lie algebra elements:

Consider some n-dimensional Lie group whose elements depend on a set of parameters $$\alpha =(\alpha_1 ... \alpha_n)$$ such that $$g(0) = e$$ with e as the identity, and that had a d-dimensional representation $$D(\alpha)=D(g( \alpha),$$ such that $$D(0)=\mathbb{1}_{d \times d}$$. Then in some small neighborhood of $$\mathbb{1}$$, we can expand $$D(\alpha)$$ as,
$$D(d\alpha) = \mathbb{1} + i d \alpha_i X^i,$$ where $$X^a = -i \frac{\partial}{\partial \alpha_i} D(\alpha)|_{i=0}$$

I have always had trouble with this from quantum mechanics class and on ward. For instance, this process seems identical to the following, from Lancaster and Blundell's QFT for the gifted amateur:

Using this terminology on the Lie case:

$$\begin{eqnarray} D(0+d\alpha) &=& D(0) + \frac{ \partial D(\alpha)}{\partial \alpha_i}d\alpha \\ &=& \mathbb{1} + (i) (-i) \frac{ \partial D(\alpha)}{\partial \alpha_i}d\alpha \\ &=& \mathbb{1} + (i) X^i d\alpha \end{eqnarray}$$

is this correct? Also, why is the "taking the derivative at $$\alpha=0$$ important? And can you please point me towards a place to learn these types of Taylor expansions?

Also having some trouble understanding the limit of N to infinity in eq. 9.13 of the included picture. In my mind the limit of $$(1+a)^x$$ as x goes to infinity, is infinity... Can someone help me grasp this limit in the case of going from infinitesimal variations with Taylor expansions, to finite variations?
[1]: https://i.stack.imgur.com/yAXum.png

Jason Bennett said:
Homework Statement: in the context of garnering Lie algebras from Lie groups
Homework Equations: see below

3) Taylor expansion question in the context of Lie algebra elements:

Consider some n-dimensional Lie group whose elements depend on a set of parameters $$\alpha =(\alpha_1 ... \alpha_n)$$ such that $$g(0) = e$$ with e as the identity, and that had a d-dimensional representation $$D(\alpha)=D(g( \alpha),$$ such that $$D(0)=\mathbb{1}_{d \times d}$$. Then in some small neighborhood of $$\mathbb{1}$$, we can expand $$D(\alpha)$$ as,
$$D(d\alpha) = \mathbb{1} + i d \alpha_i X^i,$$ where $$X^a = -i \frac{\partial}{\partial \alpha_i} D(\alpha)|_{i=0}$$

I have always had trouble with this from quantum mechanics class and on ward. For instance, this process seems identical to the following, from Lancaster and Blundell's QFT for the gifted amateur:

Using this terminology on the Lie case:

$$\begin{eqnarray} D(0+d\alpha) &=& D(0) + \frac{ \partial D(\alpha)}{\partial \alpha_i}d\alpha \\ &=& \mathbb{1} + (i) (-i) \frac{ \partial D(\alpha)}{\partial \alpha_i}d\alpha \\ &=& \mathbb{1} + (i) X^i d\alpha \end{eqnarray}$$

is this correct? Also, why is the "taking the derivative at $$\alpha=0$$ important? And can you please point me towards a place to learn these types of Taylor expansions?

Also having some trouble understanding the limit of N to infinity in eq. 9.13 of the included picture. In my mind the limit of $$(1+a)^x$$ as x goes to infinity, is infinity... Can someone help me grasp this limit in the case of going from infinitesimal variations with Taylor expansions, to finite variations?
[1]: https://i.stack.imgur.com/yAXum.png

Your question is not specific to Lie algebras/groups. You simply don't know Taylor series.

A Taylor series is a polynomial expansion of a function around a certain point:

##f(x) =\displaystyle \sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}##

##a## is the point around which you are expanding.

Since you wrote ##
D(0+d\alpha) = D(0) + \frac{\partial D(\alpha)}{\partial \alpha_i}d\alpha##, the point you are expanding around is ##\alpha = 0##, since the right hand side has the constant term ##D(0)##. That's why the derivative is at zero.

In equations 1,2,3, ##d\alpha## should be changed to ##d\alpha _i## and there should be a sum over i.

As for the limit of ##(1+a)^x##, you are forgetting that a depends on x. When ##a = c/x##, you get ##e^c## as ##x \to \infty##.

Jason Bennett said:
Also having some trouble understanding the limit of N to infinity in eq. 9.13 of the included picture. In my mind the limit of $$(1+a)^x$$ as x goes to infinity, is infinity... Can someone help me grasp this limit in the case of going from infinitesimal variations with Taylor expansions, to finite variations?
Look at 9.13 a bit more carefully. It's not the limit of ##(1+a)^N## as ##N \to \infty##; it's the limit of ##(1+\delta a)^N## where ##\delta a = a/N##.

## What is a Taylor expansion?

A Taylor expansion is a mathematical technique used to approximate a function by representing it as an infinite series of polynomials. It is named after the English mathematician Brook Taylor.

## How is a Taylor expansion calculated?

The Taylor expansion of a function can be calculated by taking the function's derivatives at a specific point and plugging them into the Taylor series formula. This formula involves the use of the function's value, its derivatives, and the point of expansion.

## What is the significance of Taylor expansions in physics?

Taylor expansions are used in physics to approximate complex functions and make them easier to solve. They are especially useful in situations where the exact solution is unknown or difficult to calculate. They are also used to model real-world phenomena in physics, such as motion and energy.

## What is the difference between a Taylor expansion and a Maclaurin expansion?

A Taylor expansion is a generalization of the Maclaurin expansion, which is a special case where the point of expansion is 0. This means that all the derivatives of the function at 0 are used in the Maclaurin series formula. In contrast, a Taylor expansion can be centered at any point.

## Can Taylor expansions be used to find exact solutions?

No, Taylor expansions are only used to approximate functions. As the number of terms in the series increases, the approximation becomes more accurate, but it will never be an exact solution. However, in some cases, a Taylor series may converge to the exact solution for certain values of the independent variable.

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