Help with infinitesimal transformation to finite transform

[TimeRip496]

<Moderation note: edited LaTex code>

E.g. A rotation by a finite angle θ is constructed as n consecutive rotations by θ/n each and taking the limit n→∞.
$$
\begin{pmatrix}
x' \\
y' \\
\end{pmatrix} =\lim_{x \to \infty} (I + \frac{\theta}{n} L_z )^n
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
$$ where I is the identity matrix and
$$L_z=
\begin{pmatrix}
0 & -1 \\
1 & 0 \\
\end{pmatrix}
$$

Instead of doing the above, why can't I do this instead?
$$
\begin{pmatrix}
x' \\
y' \\
\end{pmatrix} = (I + n*\frac{\theta}{n} L_z)
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
$$

This makes more sense as the small transform is constant unlike the above whereby exponential it will cause the transform to increase. Whats the rationale behind exponential the transform instead of just doing mine step?

I know only by doing the exponential can I get the rotational matrix but I can't see it intuitively.

 

[haushofer]

Your limit should have n--> oo.

Well, e.g. the Taylor expansion of sin(x) reads, up to first order, f(x) = x. So why shouldn't I take just the angle itself if i take the sine of it?

 

[haushofer]

Hint: the matrix components of your transformation are in general not linear functions of x and y ;)

 

[davidge]

I asked a pretty similar question just yesterday.

 

[robphy]

Does your proposed transformation preserve the magnitude of your vector?

 

[TimeRip496]

Hint: the matrix components of your transformation are in general not linear functions of x and y ;)

Sorry I don't really understand what you say. But is it ok to say that as I multiply
$$(I + \frac{\theta}{n} L_z )^2=(I + \frac{2\theta}{n} L_z+\frac{\theta^2}{n^2} L_z^2 )$$
As n goes to infinity, the higher order of (θ/n) can be ignored since they are negligible. Since we only care about the 1st order of (θ/n), as we multiply (I + (θ/n) L_z ) by itself, we say that we are increasing its transformation by 1
$$(I + \frac{\theta}{n} L_z )^2≈(I + \frac{2\theta}{n} L_z)$$
$$(I + \frac{\theta}{n} L_z )^3≈(I + \frac{3\theta}{n} L_z)$$ and so on.
Am I right to say how the above work?

My question here is why not save all the trouble and simply just sum up the infinitesimal transformation? Something like this
$$
\begin{pmatrix}
x' \\
y' \\
\end{pmatrix} = (I + k*\frac{\theta}{n} L_z)
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
$$ where k is no. of transformation and n goes to infinity

 

[TimeRip496]

I asked a pretty similar question just yesterday.

What a coincidence. And the timing is almost perfect
However I still can't find the answer I am looking for on your page

 

[pixel]

I don't understand the question here. If you want to do a 45 degree rotation, you can do it in one step. Why do it in an infinite number of infinitesimal rotations?

 

[TimeRip496]

I don't understand the question here. If you want to do a 45 degree rotation, you can do it in one step. Why do it in an infinite number of infinitesimal rotations?

Sorry if there is any confusion here. I am trying to understand how does the exponential of infinitesimal transformation gives a finite transformation

 

[stevendaryl]

I'm not sure that I understand the question.

Forgetting about matrices for a second, we can do a rotation of coordinate system $(x,y)$ through an angle of $\theta$ to get:

$x' = x cos(\theta) - y sin(\theta)$
$y' = y cos(\theta) + x sin(\theta)$

That can be written in matrix form as:

$\left( \begin{array} \\ x' \\ y' \end{array} \right) = \left( \begin{array} \\ cos(\theta) & -sin(\theta) \\ +sin(\theta) & cos(\theta) \end{array} \right) \left( \begin{array} \\ x \\ y \end{array} \right)$

The mathematical fact is that the square matrix can be written as an exponential:

$\left( \begin{array} \\ cos(\theta) & -sin(\theta) \\ +sin(\theta) & cos(\theta) \end{array} \right) = e^{\theta L_z}$

You can interpret the exponential in two different, equivalent ways:

1. $e^{\theta L_z} = lim_{n \rightarrow \infty} (1+ (\theta L_z)/n)^n$
2. $e^{\theta L_z} = \sum_{n=0}^\infty(\theta L_z)^n/n!$

I think it's easier to see the second one. Let's compute the first few terms:

$e^{\theta L_z} = I + (\theta L_z) + (\theta L_z)^2/2! + (\theta L_z)^3/3! + ...$

Note: $(L_z)^2 = -I$, $(L_z)^3 = - L_z$, $(L_z)^4 = -I$, etc. and so every other term is a multiple of the 2x2 identity matrix. So we can rearrange the terms to get:

$e^{\theta L_z} = I[1 - (\theta)^2/2! + (\theta)^4/4! - ...] + L_z(\theta - (\theta)^3/3! + (\theta)^5/5! ...)$

Those are just the taylor series for cosine and sine:
$e^{\theta L_z} = I cos(\theta) + L_z sin(\theta)$

I think it's a little harder with the first form of the exponential, but not too hard.

 

[TimeRip496]

I'm not sure that I understand the question.

Forgetting about matrices for a second, we can do a rotation of coordinate system $(x,y)$ through an angle of $\theta$ to get:

$x' = x cos(\theta) - y sin(\theta)$
$y' = y cos(\theta) + x sin(\theta)$

That can be written in matrix form as:

$\left( \begin{array} \\ x' \\ y' \end{array} \right) = \left( \begin{array} \\ cos(\theta) & -sin(\theta) \\ +sin(\theta) & cos(\theta) \end{array} \right) \left( \begin{array} \\ x \\ y \end{array} \right)$

The mathematical fact is that the square matrix can be written as an exponential:

$\left( \begin{array} \\ cos(\theta) & -sin(\theta) \\ +sin(\theta) & cos(\theta) \end{array} \right) = e^{\theta L_z}$

You can interpret the exponential in two different, equivalent ways:

1. $e^{\theta L_z} = lim_{n \rightarrow \infty} (1+ (\theta L_z)/n)^n$
2. $e^{\theta L_z} = \sum_{n=0}^\infty(\theta L_z)^n/n!$

I think it's easier to see the second one. Let's compute the first few terms:

$e^{\theta L_z} = I + (\theta L_z) + (\theta L_z)^2/2! + (\theta L_z)^3/3! + ...$

Note: $(L_z)^2 = -I$, $(L_z)^3 = - L_z$, $(L_z)^4 = -I$, etc. and so every other term is a multiple of the 2x2 identity matrix. So we can rearrange the terms to get:

$e^{\theta L_z} = I[1 - (\theta)^2/2! + (\theta)^4/4! - ...] + L_z(\theta - (\theta)^3/3! + (\theta)^5/5! ...)$

Those are just the taylor series for cosine and sine:
$e^{\theta L_z} = I cos(\theta) + L_z sin(\theta)$

I think it's a little harder with the first form of the exponential, but not too hard.

Thanks for the second one. However, I find the first one more intuitive in coming up with the exponential function that brings the infinitesimal transformation to a finite one.
______________________________________________________________________________________________
Lets say,
An infinitesimal rotation around the z-axis by δθ can be written as
$$\begin{pmatrix}
x' \\
y' \\
\end{pmatrix} =
\begin{pmatrix}
x-δ\theta y \\
y+δ\theta x \\
\end{pmatrix} =
\begin{pmatrix}
1 & -δ\theta \\
δ\theta & 1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}=(I+δ\theta L_z)
\begin{pmatrix}
x \\
y \\
\end{pmatrix}$$ ...(1)
with
$$L_z=
\begin{pmatrix}
0 & -1 \\
1 & 0 \\
\end{pmatrix}
$$
A rotation by a finite angle θ is constructed as n consecutive rotations by θ/n each and taking the limit n→∞. Using (1), it can be written be
$$
\begin{pmatrix}
x' \\
y' \\
\end{pmatrix} =\lim_{x \to \infty} (I + \frac{\theta}{n} L_z )^n
\begin{pmatrix}
x \\
y \\
\end{pmatrix} = e^{\theta L_z}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
$$...(2)
______________________________________________________________________________________________
The above makes sense to me as it shows me how to get to the exponential(which will subsequently bring me to the finite transformation) from the infinitesimal transformation.

However, how to get from (1) to (2) is a bit fuzzy to me.

How I rationalize this is that multiplying this (I + (θ/n) L_z ) to (x,y) to give it 1 unit of infinitesimal rotation. Multiply this (I + (θ/n) L_z ) twice gives (x,y) 2 units of infinitesimal rotation. Multiply this (I + (θ/n) L_z ) thrice gives (x,y) 3 units of infinitesimal rotation and so on.
In terms of matrix,
$$(I + \frac{\theta}{n} L_z )^2=(I + \frac{2\theta}{n} L_z+...)≈(I + \frac{2\theta}{n} L_z)$$
$$(I + \frac{\theta}{n} L_z )^3=(I + \frac{3\theta}{n} L_z+...)≈(I + \frac{3\theta}{n} L_z)$$ and so on.
I ignore the other power of (θ/n) cause they become negligible.
Hence, $$(I + \frac{\theta}{n} L_z )^n=(I + \frac{n\theta}{n} L_z+...)≈(I + \frac{n\theta}{n} L_z)$$ will means that there is n infinitesimal rotation which is equivalent to θ rotation.Here is the part where I am not very sure. Can I instead of doing step (2), I will just
$$
\begin{pmatrix}
x' \\
y' \\
\end{pmatrix} = (I + k*\frac{\theta}{n} L_z)
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
$$ where k is no. of transformation and n goes to infinity
So is something like this(n goes to infinity),
1 unit of infinitesimal rotation ⇒ (I + (θ/n) L_z ) , where k=1
2 units of infinitesimal rotation ⇒ (I + (2θ/n) L_z ) , where k=2
3 units of infinitesimal rotation ⇒ (I + (3θ/n) L_z ) , where k=3
and so on.

So is basically instead of multiplying (I + (θ/n) L_z ) to itself by n times, I will just sum (θ/n) by n times.

 

[stevendaryl]

Thanks for the second one. However, I find the first one more intuitive in coming up with the exponential function that brings the infinitesimal transformation to a finite one.

So you really want to know why $lim_{n \rightarrow \infty} (1+\frac{\theta}{n} L_z)^n$ is $e^{\theta L_z}$.

Your reasoning, based on small values of $n$, doesn't give you the right answer when you take $n \rightarrow \infty$.

In general, you can use the binomial theorem to calculate:

$(1+x)^n = 1 + n x + \frac{n (n-1)}{2!} x^2 + \frac{n (n-1) (n-2)}{3!} x^3 + ...$

So letting $x = \frac{\theta}{n} L_z$ gives us:

$(1+\frac{\theta}{n} L_z)^n = 1 + \theta L_z + \frac{n (n-1)}{n^2} \frac{1}{2!} (\theta L_z)^2 + \frac{n (n-1) (n-2)}{n^3} \frac{1}{3!} (\theta L_z)^3 + ...$

So term number $j$ is:

$\frac{n (n-1) (n-2) ... (n+1 -j)}{n^j} \frac{1}{j!} (\theta L_z)^j$

Then the only final step is to realize that $\frac{n (n-1) (n-2) ... (n+1-j)}{n^j} = \frac{n}{n} \frac{n-1}{n} ... \frac{n+1-j}{n}$

If $n \gg j$, then every term in that product is close to 1. So in the limit as $n \rightarrow \infty$, term number j is

$\frac{1}{j!} (\theta L_z)^j$

So we get back the same as the first expression : $e^{\theta L_z} = \sum_{j=0}^\infty \frac{1}{j!} (\theta L_z)^j$

 

[robphy]

It might be worth noting that the essence of this question can be phrased in terms of complex numbers
(which is neater and hopefully more intuitive than working with 2-by-2 real matrices).

$z'=e^{(i\theta)}z=\lim_{n\to\infty}\left( 1+ \frac{(i\theta)}{n} \right)^n\ z$
Recall that, geometrically, multiplication by $e^{i\theta}$ is a rotation operation in the complex plane
... and rotations preserve the magnitude of the complex number (which prompted my earlier comment).
In addition, rotations satisfy a group property... that the product of two rotations (by angles A and B) in the plane is equal to a rotation (of angle A+B).
You may wish to check if your proposed alternative satisfies these properties.

( https://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function )
( http://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-eit-cos-t-i-sin-t - has a neat animation that may be of interest)

(https://books.google.com/books?id=ogz5FjmiqlQC&printsec=frontcover&dq="geometry+of+the+mapping"+needham+complex - look at pg 80 from Needham's book )

 

[PeroK]

How I rationalize this is that multiplying this (I + (θ/n) L_z ) to (x,y) to give it 1 unit of infinitesimal rotation. Multiply this (I + (θ/n) L_z ) twice gives (x,y) 2 units of infinitesimal rotation. Multiply this (I + (θ/n) L_z ) thrice gives (x,y) 3 units of infinitesimal rotation and so on.
In terms of matrix,
$$(I + \frac{\theta}{n} L_z )^2=(I + \frac{2\theta}{n} L_z+...)≈(I + \frac{2\theta}{n} L_z)$$
$$(I + \frac{\theta}{n} L_z )^3=(I + \frac{3\theta}{n} L_z+...)≈(I + \frac{3\theta}{n} L_z)$$ and so on.
I ignore the other power of (θ/n) cause they become negligible.

Unfortunately, if you have $n$ in the $\frac{\theta}{n}$, you need $n$ in the exponent:

$(1 + \frac{\theta}{n})^n = 1 + n\frac{\theta}{n} + \dots = 1 + \theta + \dots$

 

[TimeRip496]

So you really want to know why $lim_{n \rightarrow \infty} (1+\frac{\theta}{n} L_z)^n$ is $e^{\theta L_z}$.

Your reasoning, based on small values of $n$, doesn't give you the right answer when you take $n \rightarrow \infty$.

In general, you can use the binomial theorem to calculate:

$(1+x)^n = 1 + n x + \frac{n (n-1)}{2!} x^2 + \frac{n (n-1) (n-2)}{3!} x^3 + ...$

So letting $x = \frac{\theta}{n} L_z$ gives us:

$(1+\frac{\theta}{n} L_z)^n = 1 + \theta L_z + \frac{n (n-1)}{n^2} \frac{1}{2!} (\theta L_z)^2 + \frac{n (n-1) (n-2)}{n^3} \frac{1}{3!} (\theta L_z)^3 + ...$

So term number $j$ is:

$\frac{n (n-1) (n-2) ... (n+1 -j)}{n^j} \frac{1}{j!} (\theta L_z)^j$

Then the only final step is to realize that $\frac{n (n-1) (n-2) ... (n+1-j)}{n^j} = \frac{n}{n} \frac{n-1}{n} ... \frac{n+1-j}{n}$

If $n \gg j$, then every term in that product is close to 1. So in the limit as $n \rightarrow \infty$, term number j is

$\frac{1}{j!} (\theta L_z)^j$

So we get back the same as the first expression : $e^{\theta L_z} = \sum_{j=0}^\infty \frac{1}{j!} (\theta L_z)^j$

Thanks for the response but my question is not so much about this part. It is about
$$\begin{pmatrix}
x' \\
y' \\
\end{pmatrix} =
\begin{pmatrix}
1 & -δ\theta \\
δ\theta & 1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}=(I+δ\theta L_z)
\begin{pmatrix}
x \\
y \\
\end{pmatrix}→?→
\lim_{x \to \infty} (I + \frac{\theta}{n} L_z )^n
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
$$
Basically what is the reason behind this conversion and hence this thread.

 

[stevendaryl]

Thanks for the response but my question is not so much about this part. It is about
$$\begin{pmatrix}
x' \\
y' \\
\end{pmatrix} =
\begin{pmatrix}
1 & -δ\theta \\
δ\theta & 1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}=(I+δ\theta L_z)
\begin{pmatrix}
x \\
y \\
\end{pmatrix}→?→
\lim_{x \to \infty} (I + \frac{\theta}{n} L_z )^n
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
$$
Basically what is the reason behind this conversion and hence this thread.

I really don't understand what it is that you are asking. You don't understand how to go from

1. $\left( \begin{array} \\ x' \\ y' \end{array} \right) = (I + \delta \theta L_z) \left( \begin{array} \\ x \\ y \end{array} \right)$

to

2. $\left( \begin{array} \\ x' \\ y' \end{array} \right) = (I + \frac{\theta}{n} L_z)^n \left( \begin{array} \\ x' \\ y' \end{array} \right)$
?

You just apply 1 over and over, using $\delta \theta = \frac{\theta}{n}$

 

[haushofer]

I guess the confusion of the TS is why every infinitesimal transformation one applies to obtain the finite transformation should contain the identity 1.

 

[TimeRip496]

I guess the confusion of the TS is why every infinitesimal transformation one applies to obtain the finite transformation should contain the identity 1.

Yes this is what I am looking for!

 

[haushofer]

Do you now see why every infinitesimal change by definition contains the identity?

 

[TimeRip496]

Do you now see why every infinitesimal change by definition contains the identity?

I know why it need to contain the identity but I don't quite get why every infinitesimal transformation(after the first one) one applies to obtain the finite transformation should contain the identity 1.
_____________________________________________________________________________________________
First transform: $$(I+δ\theta L_z)
\begin{pmatrix}
x \\
y \\
\end{pmatrix}$$

Second transform:
$$(I+δ\theta L_z)^2
\begin{pmatrix}
x \\
y \\
\end{pmatrix}$$

Third transform:
$$(I+δ\theta L_z)^3
\begin{pmatrix}
x \\
y \\
\end{pmatrix}$$
and so on.
___________________________________________________________________________________________Is there any reason why I can't do this instead
_____________________________________________________________________________________________
First transform: $$(I+δ\theta L_z)
\begin{pmatrix}
x \\
y \\
\end{pmatrix}$$

Second transform:
$$(I+2δ\theta L_z)
\begin{pmatrix}
x \\
y \\
\end{pmatrix}$$

Third transform:
$$(I+3δ\theta L_z)
\begin{pmatrix}
x \\
y \\
\end{pmatrix}$$
and so on.
___________________________________________________________________________________________

 

[haushofer]

Because your second transformation does not rotate your first transform; you simply add something to it. That is not how a rotation is defined. If it were, a rotation would only contain functions which are linear in the angle.

 

[haushofer]

E.g. how would your second transform look like when the angle of rotation is zero? Or your third?

With your definition one infinitesimal transfo with zero angle would put the whole finite transfo to zero, no?

 

[haushofer]

Maybe the following example enligthens. Imagine you work at a bank and have to calculate the amount an annual interest r makes an amount of money x grow. The amount at year t+1 is given by the following recursive formula:

$$x_{t+1} = (1+r)x_t $$

where x_0 is given. How would you calculate the solution of this formula: would it be
$$ x_t = x_0 (1+tr)$$

or would it be

$$ x_t = x_0 (1+r)^t? $$

The continuous limit of this is given by n payments of interest of r/n every year and sending n to infinity.

I hope you agree with me it should be the last formula. It is non-linear, which is to be expected, since the continuous limit gives an exponential solution.

 

[TimeRip496]

Maybe the following example enligthens. Imagine you work at a bank and have to calculate the amount an annual interest r makes an amount of money x grow. The amount at year t+1 is given by the following recursive formula:

$$x_{t+1} = (1+r)x_t $$

where x_0 is given. How would you calculate the solution of this formula: would it be
$$ x_t = x_0 (1+tr)$$

or would it be

$$ x_t = x_0 (1+r)^t? $$

The continuous limit of this is given by n payments of interest of r/n every year and sending n to infinity.

I hope you agree with me it should be the last formula. It is non-linear, which is to be expected, since the continuous limit gives an exponential solution.

Thanks! Your emphasis that it is non-linear really guide me to see why I need to include the Identity for all the subsequent transformations.