A Taking limits in discrete form

1. Mar 14, 2016

friend

I'm trying to calculate this limit to answer a question in Quantum Mechanics:
$$\mathop {\lim }\limits_{{t_1} \to 0} \,\,{\left( {\frac{m}{{2\pi \hbar i{t_1}}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{im{{(x' - x)}^2}/2\hbar {t_1}}}\,\,\,\,\, + \,\,\,\,\,\mathop {\lim }\limits_{{t_2} \to 0} \,\,{\left( {\frac{m}{{2\pi \hbar ( - i){t_2}}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{ - im{{(x - x')}^2}/2\hbar {t_2}}}$$
It seems as t → 0 in an arbitrary way, the complex exponentials circles wildly from +1 to i to -1 to -i to +1 again. And the two square roots approach ∞ in magnitude and are 90° out of phase with each other. So it seems the limit does not approach any particular value, not even to plus or minus ∞; the limit seem undefined.

However, I wonder if t1 and t2 could approach 0 in some controlled way that allows the two terms to cancel out.

Let
$$A = {\left( {\frac{m}{{2\pi \hbar i}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}$$
And let,
$$B = m{(x' - x)^2}/2\hbar$$
Then the above limit can be written,
$$A\left( {\mathop {\lim }\limits_{{t_1} \to 0} \frac{{{e^{iB/{t_1}}}}}{{{t_1}^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}} + i\mathop {\lim }\limits_{{t_2} \to 0} \frac{{{e^{ - iB/{t_2}}}}}{{{t_2}^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}}} \right)$$
which equals,
$$A\left( {\mathop {\lim }\limits_{{t_1} \to 0} \frac{{{e^{iB/{t_1}}}}}{{{t_1}^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}} + \mathop {\lim }\limits_{{t_2} \to 0} \frac{{{e^{ - iB/{t_2} + i\pi /2}}}}{{{t_2}^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}}} \right)$$
If we restrict t2 such that,
$$\frac{{ - iB}}{{{t_2}}} + \frac{{i\pi }}{2} = \frac{{iB}}{{{t_1}}} + i\pi - i2\pi n$$
for n any integer, then the $i\pi - i2\pi n$ factor in the exponent will insure that the second term is always 180° out of phase with the first term so that the two terms will cancel out to zero. In this case, t1 could be any arbitrary number approaching zero. But t2 would have to be
$${t_2} = \frac{B}{{\frac{{ - B}}{{{t_1}}} + 2\pi (n - \frac{1}{4})}}$$
We can see from this that t2 gets arbitrarily close to zero from above or below as n increases to plus infinity or negative infinity, respectively. It seems that for any other way of letting t2 approach zero, the limit is completely undefined.

The question is whether it is allowed to let parameters be discrete in the limiting process. Or must they always be continuous?

Last edited: Mar 14, 2016
2. Mar 14, 2016

friend

I was concerned that the two terms would not be equal if ${t_2} \ne - {t_1}$. But the last equation above can be written as,
$${t_2} = \frac{{B{t_1}}}{{ - B + {t_1}2\pi (n - \frac{1}{4})}}$$
Then it is seen that as t1→0, the second term in the denominator approaches zero compared to B so that
$${t_2} \to \frac{{B{t_1}}}{{ - B}} = - {t_1}$$
And when this is put in the first equation in the previous post, then the two square roots become equal. And since they have opposite phases, the two terms cancel.

And as for the limit of t2 being made discrete with n, doesn't this just change the limit of a function into a limit of a sequence, but still a legitimate limit process? I'm still not sure you can make two limiting processes somewhat dependent on each other. I would think that it doesn't matter if t2 is a function of t1 as long as you can still put the limit of t2→0 in the ε, δ formulation, right?

Last edited: Mar 14, 2016
3. Mar 15, 2016

Samy_A

Mathematically you can't do this.

Consider the following limit: $\displaystyle \lim_{(x,y) \rightarrow (0,0)} \frac{xy}{x²+y²}$.

This limit doesn't exist.

The limit exists if you go to $(0,0)$ along a line.
Take the limit with $x=2y$, then the limit is 2/5.
Take the limit with $x=y$, then the limit is 1/2.

You get a "limit" that depends on the slope of the line you use to approach $(0,0)$. Use polar coordinates to easily establish this.

Last edited: Mar 15, 2016
4. Mar 15, 2016

friend

If I could (as you do in your example) manipulate the limit I pose to various values, then I would agree that my manipulation to zero has no meaning. But I don't see that. I see either 0 or ∞, either existence or non-existence.

I suppose I could manipulate the limits to get any phase angle I wanted between the two terms. But then the magnitudes of the two terms would not match, and I'd get infinity as the parameters approached zero. So I'm beginning to see my manipulation as a regulatory method. If I carefully regulate the manner in which the limit is taken, I can guarantee that the limit converges. They do this a lot in quantum theory; it's call renormalization, where if they use some trick or another, then they can get a reasonable value where before they were getting infinity.

5. Mar 15, 2016

Samy_A

That's why my answer started with a bold "Mathematically".
I don't know whether your manipulation is acceptable in some specific physical context. Maybe you can ask in the appropriate physics forum. There is more likelihood that someone knowledgeable will read it.

6. Mar 15, 2016

friend

Mathematically, I think we are talking about convergence and whether it is acceptable to regulate the limiting process to get convergence. Does it really matter HOW I approach the limit of zero? Can I change a continuous parameter to a discrete parameter and still get the same and valid limit? Does it really matter if the limiting parameters are related? I think I'm asking a rather technical question about analysis. The physicists would probably direct me to the math department.

One thing that differs from your example using x and y is that my t1 and t2 are not entirely dependent on one another. Each can approach zero independently from the other (except one parameter become discrete). So the question of the slope or direction of approaching (0,0) is avoided.

7. Mar 15, 2016

Samy_A

The mathematical answer is yes, it matters HOW the limit is taken. If the two parameters must be related to get a limit, then it is not a limit for a math department.

I understand the difference, but still, mathematically, the limit doesn't exist.

Let me give you a simple example: does $\displaystyle \lim_{x \rightarrow \infty} \sin x$ exist?
The answer of course is no.
But, if I set $x_n=n\pi$, where $n\in \mathbb N$, then the "discrete" limit $\displaystyle \lim_{n \rightarrow \infty} \sin x_n=0$.

8. Mar 15, 2016

friend

I appreciate your answers. I'm trying to narrow down the issues and find principles that can be applied.

What I think I have so far is that I can indeed manipulate the two limits of t1 and t2 independently. I can certainly manipulate t1 independently as I wish. But if I keep t1 constant, then I can change the value of n independently to get an independent discrete value of t2. So I think the issue comes down to whether I can change a continuous parameter to a discrete parameter. My intuition tells me that this is what we do in practice anyway. We get out the calculator and discretely try ever smaller values to see if it is converging on a particular value.

9. Mar 15, 2016

Samy_A

Your intuition is wrong. Look at how a limit is defined using $\epsilon, \delta$.
$\displaystyle \lim_{(x,y)\rightarrow (0,0)}f(x,y)=L$ is defined by:
$$\forall \epsilon>0 \ \exists \delta>0 \ \ : 0\lt \|(x,y)\| \lt \delta \ \Rightarrow |f(x,y)-L|\lt \epsilon$$
If you take a "discrete" limit, you may not have $|f(x,y)-L|\lt \epsilon$ for all $(x,y)$ satisfying $\|(x,y)\| \lt \delta$.

Last edited: Mar 16, 2016
10. Mar 15, 2016

friend

Well, now you're talking. Yes, I see what you're saying. There could be continuous functions that are sharply spiked between adjacent discrete values so that you can not find a discrete values in the sequence for which $|f(n) - L| < \varepsilon$ with $n > N$. So in general you cannot replace a continuous parameter with a discrete parameter.

But in my case it seems obvious that you can always find ${t_1} < \delta$ and $n > N$ such that $|f({t_1},{t_2}(n)) - L| < \varepsilon$.

Or more precisely, you can always find $\left\| {\left( {{t_1},{t_2}} \right)} \right\| < \delta$ such that $|f({t_1},{t_2}) - L| < \varepsilon$, because you can always find a ${t_1} < \tau$ and an $n > N$ such that $\left\| {\left( {{t_1},{t_2}} \right)} \right\| < \delta$.

Last edited: Mar 15, 2016
11. Mar 15, 2016

Samy_A

You are missing the point about what the definition of limit means.
You don't have to find "a" $(t_1,t_2)$ such that $\left\| {\left( {{t_1},{t_2}} \right)} \right\| < \delta$ and $|f({t_1},{t_2}) - L| < \epsilon$, it must be true for all $(t_1,t_2)$ such that $\left\| {\left( {{t_1},{t_2}} \right)} \right\| < \delta$.

Try in on the following "limit": $\displaystyle \lim_{x \rightarrow 0} \sin \frac{1}{x}$. For each $\epsilon >0$ you can find an $x$ such that $|x|<\epsilon$ and $|\sin \frac{1}{x}|<\epsilon$.
But of course the limit doesn't exist.

Conclusion: mathematically your limit doesn't exist. Whether your manipulation to make it converge is acceptable is not a mathematical question.

12. Mar 15, 2016

friend

So let me fix the wording a little bit and see if it works for you.

My limit is such that, for all $\left\| {\left( {{t_1},{t_2}} \right)} \right\| < \delta$ there exists $|f({t_1},{t_2}) - L| < \varepsilon$, because for all ${t_1} < \tau$ and all $n > N$ it is true that $\left\| {\left( {{t_1},{t_2}} \right)} \right\| < \delta$. I don't see anything in the definition that prohibits t2 from being a function of t1 and n. I don't see anything that prohibit t2 from being discrete.

Last edited: Mar 15, 2016
13. Mar 15, 2016

Samy_A

Let me write the definition of limit (for a $\mathbb R² \to \mathbb R$ function) again, and add the elements that were implicitly assumed (as they seem obvious, but ok, let's be precise).

$\displaystyle \lim_{(x,y)\rightarrow (0,0)}f(x,y)=L$, where $L \in \mathbb R$, is defined by:
$$\forall \epsilon>0 \ \exists \delta>0 \ \ \ \text{such that}\ \ \forall (x,y) \in \mathbb R² \ :\\ 0 \lt \|(x,y)\| \lt \delta \ \Rightarrow |f(x,y)-L|\lt \epsilon$$

The $\forall (x,y) \in \mathbb R²$ is the key that prohibits $t_2$ being a function of $t_1$ and $n$. All that is required of $(t_1,t_2)$ is that $\|(t_1,t_2)\|\lt \delta$. For any $(t_1,t_2)$ satisfying this condition $|f(x,y)-L|\lt \epsilon$ must hold.
If that is the case (and can be achieved for every $\epsilon>0$), the limit exists. If not, the limit doesn't exist.

Again, I'm not even trying to say that what you do is wrong in your context. I just don't know about that.
But given how limits are defined, your expression doesn't have a limit.

Last edited: Mar 16, 2016
14. Mar 15, 2016

andrewkirk

No they won't. For that to work you would require that
$$\lim_{t_1,t_2\to 0}\left[\textrm{1st term}+\textrm{2nd term}\right]= \lim_{t_1,t_2\to 0}[\textrm{1st term}]+\lim_{t_1,t_2\to 0}[\textrm{2nd term}]$$
and that is only true if both limits on the right exist, which they don't.

If you replace $t_1$ in the formula by an expression involving only $t_2$ and you then replace the two limit operators by a single limit operator, outside the whole expression, that takes the limit as $t_2\to 0$ then, if your calcs are correct (which I haven't checked) you will get a valid limit. But the new expression will be a different expression from the original one. The two do not represent the same thing and one cannot be converted to, or substituted for, the other.

15. Mar 15, 2016

friend

Thanks for your input, guys. Let me see if I can reduce it to a more intuitive formula. I think what I have in essence is:

$$\mathop {\lim }\limits_{({t_1},{t_2}) \to 0} \left[ {\left( {\frac{1}{{{t_1}}}} \right) - \left( {\frac{1}{{{t_2}}}} \right)} \right]$$

And then put

$${t_2} = \frac{{{t_1}}}{{1 + {t_1}n}}$$

which gives the limit as

$$\mathop {\lim }\limits_{({t_1},{t_2}) \to 0} \left[ {\left( {\frac{1}{{{t_1}}}} \right) - \left( {\frac{1}{{\frac{{{t_1}}}{{1 + {t_1}n}}}}} \right)} \right]$$

And I think it is clear that for any value of n the two terms cancel out.

Although, I think Samy_A might disagree with you, andrewkirk, as to whether I can make t2 a function of t1.

16. Mar 15, 2016

friend

I take issue with the requirement that $\forall (x,y) \in \mathbb R²$. It seems that this prohibits discrete values of (x,y) and prohibits the convergence of sequences, since ALL values of (x,y) must be considered.

17. Mar 16, 2016

andrewkirk

As Samy has already pointed out (in the unsimplified case) the second expression is not the same as the first. In the second, you are taking the limit of the unlimited expression as $(t_1,t_2)$ approaches $(0,0)$ along a specific path (line). Along that path, there is a limit, and it is

$$\mathop {\lim }\limits_{({t_1},{t_2}) \to 0} \left[ {\left( {\frac{1}{{{t_1}}}} \right) - \left( {\frac{1}{{\frac{{{t_1}}}{{1 + {t_1}n}}}}} \right)} \right]= \mathop {\lim }\limits_{({t_1},{t_2}) \to 0} \left[ {\left( {\frac{1}{{{t_1}}}} \right) - \left(\frac{1 + {t_1}n}{t_1} \right)} \right]= \mathop {\lim }\limits_{({t_1},{t_2}) \to 0} \left[ \frac{1}{t_1}-\frac{1}{t_1}-n\right]=-n$$

So for every positive integer $n$ there is a path through the point $(0,0)$ and every path will give a different limit. There are other paths through $(0,0)$ for which there is no limit, such as the path on which $t_2=t_1{}^2$. So the limit as $(t_1,t_2)\to 0$ does not exist, because whether there is a limit, and what the limit is when there is one, both depend on the path taken.

This is directly similar to the example Samy gave in post 3, where he presented a function that has limits at a point if you approach it along specific paths, and those limits differ between paths, so that there is no simple (ie general) limit.

You felt that Samy's example didn't apply to your case, as you observed:
My calc above demonstrates that you can manipulate your limit, just by choosing different paths - ie by choosing different values of $n$. So there is no limit as $(t_1,t_2)\to 0$, as that notation requires that the limit exists on every path that approaches $(0,0)$ and that all such limits are the same.

18. Mar 16, 2016

Samy_A

No need to repeat what @andrewkirk wrote.

No, I do agree with andrewkirk. Of course you can make $t_2$ a function of $t_1$, or take a discrete path, or do whatever manipulation you want.
However, the result will not be the limit of the original expression.
Whether that result will be significant depends on the (physical) context, not on the mathematics.

19. Mar 16, 2016

Samy_A

It shows that a "continuous" limit is not the same as a "discrete" limit. In the first post you asked: "The question is whether it is allowed to let parameters be discrete in the limiting process. Or must they always be continuous?"
The answer you can deduce from the definition of limit is: the two are not the same.

See how $\displaystyle \lim_{x \rightarrow 0}\sin \frac{1}{x}$ doesn't exist while some "discrete" variants behave nicely: $\displaystyle \lim_{n \in \mathbb N \rightarrow \infty}\sin (\frac{1}{\frac{1}{n\pi}})=0$.

Last edited: Mar 16, 2016
20. Mar 16, 2016

friend

I think you need to change that to "may not be the limit of the original expression." For obviously,
$$\mathop {\lim }\limits_{n \to 0} \frac{1}{n} = 0$$
whether n is continuous or discrete. This is a counterexample to your use of "will not be".

And the reason I take issue with $\forall (x,y) \in \mathbb R²$ is because there are functions not even defined on some of R2 such as the probability of obtaining less than zero samples or the density of negative volume. It also contradicts the taking of a limit of a sequence.

Last edited: Mar 16, 2016