A Can Discrete Parameters Be Used in Limit Calculations?

[friend]

Because $|f(x,y)-L| \lt \epsilon$ must hold for all $(x,y) \in \mathbb R²$ satisfying $0<\|(x,y)\|<\delta$.
If the limit only exists along a specific path, or is different according to the path one uses, it is by definition not a limit.

If we must include all of R², then that would prohibit the convergence of a sequence since they consider only discrete values of R² (and not all values of R²). And it would prohibit functions not defined for some values of R², since the $|f(x,y)-L| \lt \epsilon$ would not hold. And I would contend that a specific path is no more restrictive on R² then using only discrete values of R₂ or using only the positive quadrant of R².

Well, basically you have $\displaystyle \lim_{t \rightarrow 0} \frac{1}{\sqrt t}e^{if(t)}$, and that doesn't converge.

OK, suppose I grant that. The question remains whether my limit in the first post converges or not. Each term by itself may not converge, but it is not clear that when added that the sum does not converge. For there may be some values of the parameters that put them out of phase so they do converge to zero. You claim that specific paths are not allowed by the definition of a limit. But I don't see that yet.

 

[Samy_A]

If we must include all of R², then that would prohibit the convergence of a sequence since they consider only discrete values of R² (and not all values of R²). And it would prohibit functions not defined for some values of R², since the $|f(x,y)-L| \lt \epsilon$ would not hold. And I would contend that a specific path is no more restrictive on R² then using only discrete values of R₂ or using only the positive quadrant of R².

All this is irrelevant as far as your expression in post #1 is concerned.

I gave a general definition of the limit of a function of two variables. If the domain is not the whole of $\mathbb R²$, the definition is adapted accordingly.
Still, that does not mean that you can pick a convenient path to the limit point, and claim convergence because the function converges along that convenient path.
We have seen examples in this thread, I won't repeat them.

OK, suppose I grant that. The question remains whether my limit in the first post converges or not. Each term by itself may not converge, but it is not clear that when added that the sum does not converge. For there may be some values of the parameters that put them out of phase so they do converge to zero. You claim that specific paths are not allowed by the definition of a limit. But I don't see that yet.

The question of whether your limit in the first post converges or not doesn't remain: it has been answered, your limit doesn't converge.

What you have is $\displaystyle \lim_{t_1 \rightarrow 0} \frac{c_1}{\sqrt t_1}e^{if_1(t_1)} + \lim_{t_2 \rightarrow 0} \frac{c_2}{\sqrt t_2}e^{if_2(t_2)}$, where $c_1, c_2$ are non zero constants and $f_1, f_2$ real functions.
None of these two limits converges.

You want to consider the following expression:
$\displaystyle \lim_{t_1 \rightarrow 0}( \frac{c_1}{\sqrt t_1}e^{if_1(t_1)} + \frac{c_2}{\sqrt t_2}e^{if_2(t_2)})$, where $t_2$ is now some specific function of $t_1$. Even if this limit converges, it will not be equal to the original expression, because we know that the original expression doesn't converge.

I'll conclude with a simple example:
Define $f(t)=\frac{1}{t}$ and $g(t)=-\frac{1}{t}$.
What is $\displaystyle \lim_{t_1 \rightarrow 0} f(t_1) + \lim_{t_2 \rightarrow 0} g(t_2) \ \ \ \ (1)$?
Clearly not convergent, as we get $+\infty - \infty$, an indeterminate form.
But what about $\displaystyle \lim_{t_1 \rightarrow 0}( f(t_1) + g(t_1))$?
Clearly $\displaystyle \lim_{t_1 \rightarrow 0}( f(t_1) + g(t_1))=0 \ \ \ \ (2)$

However, no mathematician will consider that the convergence of (2) somehow implies the convergence of (1).

 

[andrewkirk]

I may have oversimplified. Let me think about it a bit.
Since this question was conceived in the context of quantum mechanics, it may turn out to be a virtue instead of a vice.

It sounds like you are confident that your construction has some valid physical use because of that context. For all we know, you may be right. Since we know nothing of the context, we cannot have any informed opinion on that. All we can say is that the expression you arrive at after performing your manipulations is not mathematically equivalent to the one you started with. Whether the final expression is applicable to the physical situation you are envisaging depends on what that physical situation is.

So let me reiterate Samy's request that you explain the physical context. With that context provided, somewhere here may be able to help you. Without it, we can do no more than observe that your end result does not follow mathematically from your starting point.

 

[friend]

Thanks for the help, guys (which includes possibly gals). But my enthusiasm may have been premature. I did a little more algebra to see if my limit was path dependent as andrewkirk may have suggested. It turns out that, sure, I can get the two phase factors to cancel out, but I pick up a negative square root in the second term. So it seems I cannot get rid of the phase and make them cancel as I hoped. Sorry for putting you through this for nothing.

As for the physical context that originally got me excited, the first term is the transition amplitude for a (unspecified) particle to go from x to x'. The second term is for an antiparticle to transition from the same place. I was hoping that in the limit as the parameters go to zero, that I could get them to cancel out. This would be consistent with virtual particles that pop in and out of existence without any permanent wave function. Maybe there's still a way to do this, but I can't see it. But if I was able to get these conjugates to cancel out at least in the limit, then I would be able to say that they exist everywhere since there is no permanent effect (wave function). And I have some intuition and notes on how all of spacetime, matter, and energy could be described with these "virtual particles". And I wanted to get a firm mathematical foundation before getting too far. That dream has yet to be realized. Thanks, but I'm suspending my campaign until further notice.

 

[zinq]

Another point is that in the original question, where a factor of i is pulled out of the second term (in the 4th displayed equation), it would be necessary to take its square root first. I don't know if the 1/2 powers are well-defined in your physics situation. A simple way to do that is — if permitted by the physics — is: If angle is always chosen to be in the interval (-π, π) then the square root has an angle just half whatever the angle is. Since the terms taken to the 1/2 power are both on the imaginary axis, this would mean the square roots have angles -π/4 (first term), π/4 (second term), respectively.

 

[friend]

this would mean the square roots have angles -π/4 (first term), π/4 (second term), respectively.

Thanks for the hint, zinq. So let me start over with your suggestion.
I want to find the limit
\mathop {\lim }\limits_{{t_1} \to 0} \,\,{\left( {\frac{m}{{2\pi \hbar i{t_1}}}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}{e^{im{{(x&#039; - 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{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}{e^{ - im{{(x - x&#039;)}^2}/2\hbar {t_2}}}
because it represents a particle in superposition with its antiparticle. If I can get this limit to equal zero, then it can represent virtual particle/antiparticle cancellations that have no physical effects on their own. To simplify a bit, I will make
A = {\left( {\frac{m}{{2\pi \hbar i}}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}
and
B = m{(x&#039; - x)^2}/2\hbar
Then the above limit becomes
A\left( {\mathop {\lim }\limits_{{t_1} \to 0} \frac{{{e^{iB/{t_1}}}}}{{{t_1}^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}} + i\mathop {\lim }\limits_{{t_2} \to 0} \frac{{{e^{ - iB/{t_2}}}}}{{{t_2}^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}}} \right)
And since i = {e^{\frac{{i\pi }}{2}}}, this becomes
A\left( {\mathop {\lim }\limits_{{t_1} \to 0} \frac{{{e^{iB/{t_1}}}}}{{{t_1}^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \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{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}}} \right)
If we are talking about virtual particles that pop into existence at the same place at the same time and pop out of existence at the same place at the same time, we must have {t_1} = {t_2} = t, and the limit becomes
A\mathop {\lim }\limits_{t \to 0} \left( {\frac{{{e^{iB/t}}}}{{{t^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}}} + \frac{{{e^{ - iB/t + i\pi /2}}}}{{{t^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}}}} \right) = A\mathop {\lim }\limits_{t \to 0} \frac{1}{{{t^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}}}\left( {{e^{iB/t}} + {e^{ - iB/t + i\pi /2}}} \right)
Now, here's the new part that I just thought up, thanks to zinq. Let
\frac{{iB}}{t} = \frac{{i\pi }}{4} + \frac{{i\pi }}{2} + i2\pi n
with n any integer. Then t \to 0 means that n \to \infty. And we also have
t = \frac{B}{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}
Putting this into the limit gives
A\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}{B}} \right)^{\frac{1}{2}}}\left( {{e^{\frac{{i\pi }}{4} + \frac{{i\pi }}{2} + i2\pi n}} + {e^{ - \frac{{i\pi }}{4} - \frac{{i\pi }}{2} - i2\pi n + \frac{{i\pi }}{2}}}} \right)
The first and last terms in the second exponent add to give \frac{{i\pi }}{4} so that we have
A\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}{B}} \right)^{\frac{1}{2}}}\left( {{e^{\frac{{i\pi }}{4} + \frac{{i\pi }}{2} + i2\pi n}} + {e^{\frac{{i\pi }}{4} - \frac{{i\pi }}{2} - i2\pi n}}} \right)
The factor {e^{\frac{{i\pi }}{4}}} in both exponents can be pulled out to give
A\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}{B}} \right)^{\frac{1}{2}}}{e^{\frac{{i\pi }}{4}}}\left( {{e^{\frac{{i\pi }}{2} + i2\pi n}} + {e^{ - \frac{{i\pi }}{2} - i2\pi n}}} \right)
And since it is true that {e^{ix}} = {e^{ix \pm i2\pi n}}, we have
A\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}{B}} \right)^{\frac{1}{2}}}{e^{\frac{{i\pi }}{4}}}\left( {{e^{\frac{{i\pi }}{2}}} + {e^{ - \frac{{i\pi }}{2}}}} \right)
And with {e^{\frac{{i\pi }}{2}}} = i and {e^{ - \frac{{i\pi }}{2}}} = - i, the limit reduces to
A\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}{B}} \right)^{\frac{1}{2}}}{e^{\frac{{i\pi }}{4}}}\left( {i - i} \right) = 0
for any n whatsoever.

I got convergence by specifying a path of {t_1} = {t_2} = t and making t discrete, t = \frac{B}{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}. So the last equation is not equal to the first equation. But I still have to wonder if they are equal in the limit.

 

[Samy_A]

I got convergence by specifying a path of {t_1} = {t_2} = t and making t discrete, t = \frac{B}{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}. So the last equation is not equal to the first equation. But I still have to wonder if they are equal in the limit.

As both diverge, they can't be "equal in the limit".

 

[zinq]

Never mind, friend, I had overlooked that in your definition of A, you had included i in the denominator.

 

[friend]

It's possible my concerns about convergence of the limit may be rendered mute since it is identically 0 under certain circumstance. Maybe all I need do is specify under what conditions these virtual particle pairs can exist. I still don't know, however, if my condition for t is the only one that results in cancellation.

 

[friend]

Never mind, friend, I had overlooked that in your definition of A, you had included i in the denominator.

Still, it was your mention of iπ/4 that got me thinking. Thanks.