Hi,

That's an interesting question and that is probably why they need more peer review.
However, sometimes a sanity check helps to convince even if not prove. This is probably what they did.
For example, if we had a prime number generator and generated a set of primes and chose one HUGE number that a classical method would have a hard time doing, we could test just that one using classical methods. IF that huge number failed it would mean the new method failed, but if it was large enough and it proved to be a real prime, then the new method would have merit. Not a proof of course, but some indication that the method might work. On the contrary, in the 1980's i knew of one prime number generator that could generate a ton of primes but failed on just ONE because it generated ONE prime that was not a true prime.

For large factoring problems like cracking 2048 RSA it seems easy to verify the answer but very few general problems with have this trap-door factoring property.

 

How do you know that the quantum computer got the answer right in the first place? DO we just trust the answer with empirical proofs if classical computing the problem is impossible?

We'll have to wait for the paper to see, but my guess is that the toy problem is in NP (hard to solve, easy to verify).

 

Not sure what you are talking about when it comes to the sinusoid. You dont seem to have every analysed a circuit for a constant frequency sinusoid because you would have understood the point of view then.

You were trying to compare QC speedup with time vs frequency domain analysis, and I was trying to show you how they are not the same kind of thing. Quantum supremacy means that, for certain types of problems, quantum computers have more computational power than classical computers. In complexity terms, this means that the set of problems that can be efficiently solved by classical computers is a strict subset of the class of problems that can be solved by quantum computers (BPP ⊂ BQP).

In the case of Fourier duals, however, both domains contain precisely the same amount of information. Neither domain requires more computational power than the other, but even if we pretend that one does, the Fourier transform (and its inverse) is a polynomial-time algorithm. In other words, the set of problems that can be efficiently solved in time is equivalent to the the set of problems that can be efficiently solved in frequency.

The simple fact that we can easily go back and forth between the time and frequency domains, while there can be no such transform for classical vs quantum computing, should be a clear indication that you've made a category error.

 

You were trying to compare QC speedup with time vs frequency domain analysis, and I was trying to show you how they are not the same kind of thing. Quantum supremacy means that, for certain types of problems, quantum computers have more computational power than classical computers. In complexity terms, this means that the set of problems that can be efficiently solved by classical computers is a strict subset of the class of problems that can be solved by quantum computers (BPP ⊂ BQP).

In the case of Fourier duals, however, both domains contain precisely the same amount of information. Neither domain requires more computational power than the other, but even if we pretend that one does, the Fourier transform (and its inverse) is a polynomial-time algorithm. In other words, the set of problems that can be efficiently solved in time is equivalent to the the set of problems that can be efficiently solved in frequency.

The simple fact that we can easily go back and forth between the time and frequency domains, while there can be no such transform for classical vs quantum computing, should be a clear indication that you've made a category error.

Now i remember why i didnt reply to your other post. You assume too much and extract very little from what is being said. You missed my point entirely again.

 

For large factoring problems like cracking 2048 RSA it seems easy to verify the answer but very few general problems with have this trap-door factoring property.

Well i would think we would have the choice of what problem we picked to test with. So it doesnt have to be a general problem.

 

Now i remember why i didnt reply to your other post. You assume too much and extract very little from what is being said. You missed my point entirely again.

Ok, then what was your point?

 

Ok, then what was your point?

If you really want to understand my point then you would first have to look at a function like this:
y=A*sin(wt)

Now in the time domain we have to solve for every time point to see the result, so t=1 second, t=2 sections, etc., and everything between t=1.000001, t=1.000002, etc., and ideally with infinitesimal resolution.
In the frequency domain we only have to specify the 'A' and the 'w', and we have all of those solutions implied.

As for an example, in the time domain if we want to solve for the output of a simple RC low poss filter we have to calculate each time point so for input A*sin(wt) we get out B*sin(wt+ph) after steady state is reached, but that implies that we have to plot every point for every value of 't'.
In the frequency domain if we want to solve for the output we only have to solve for B and ph and everything else is implied.
So the spec for the output in time would have to be B*sin(wt+ph) but in the frequency domain just (B,ph).
The time spec implies that we have to run over all time from t=0 to t=infinity
the frequency spec tells us that the output is sinusoidal with amplitude B and phase ph but we may not even need the phase part of it.

That example is probably too simple to explain this let me see if i can show a better example.

[LATER]
Actually maybe that is enough but i'll show another perhaps anyawy.
The reason that is enough is because it does show that (with little thought) the time solution has time 't' in it, while the frequency solution does not have time 't' in it, therefore the solution in frequency is solved without requiring the knowledge of time or maybe more important, not requiring any time so it takes no real time to have all the solutions for all time in one package.
The solution with time in it implies that it takes real time to solve for all the time values.
Yes they both contain the same info, but they are presented in different ways so they are not EXACTLY the same in all senses of that word therefore we could look for differences and in fact we must find differences even though the actual solutions themselves may be the same. But saving time is what it is all about, and once the QC comes up with a complicated solution, we can 'say' that we have the solution but it's still in the box until we extract it, but extracting it is not considered to be part ot the solution time even though it is in real life, because extracting may still be faster than a classical method in it's entirety.

Indeed anyway arent all the inputs to the quantum computer frequencies anyway?
I may be implying something here that shouldnt be implied though so this statement really does require more review.

Also note that the time solution shown here is not the true time solution it's simplified to show only the steady state solution. That means in real life we'd have to wait for the exponentials to die down, and in theory we might be able to say that we never have the true solution in the time domain for what happens near t=inf because in theory the exponentials dont truely die until we reach infinity.

 

As for an example, in the time domain if we want to solve for the output of a simple RC low poss filter we have to calculate each time point so for input A*sin(wt) we get out B*sin(wt+ph) after steady state is reached, but that implies that we have to plot every point for every value of 't'.
In the frequency domain if we want to solve for the output we only have to solve for B and ph and everything else is implied.
So the spec for the output in time would have to be B*sin(wt+ph) but in the frequency domain just (B,ph).
The time spec implies that we have to run over all time from t=0 to t=infinity
the frequency spec tells us that the output is sinusoidal with amplitude B and phase ph but we may not even need the phase part of it.

This doesn't make sense. If you want to solve for the output magnitude B by leveraging the frequency domain, you need to know the frequency response of the filter, which means knowing the response at every frequency ω. This is no different than having to plot every value of t in the time domain.

Let's be explicit about this, since you seem to be missing the point. If we want to solve for the output of the filter strictly in the time domain, we use Kirchhoff's rules to set up a first-order differential equation relating the output \(\small{v_o(t)}\) to the input \(\small{v_i(t)}\) as functions of time:

\(\small{\frac{d}{dt} v_o(t) \, + \, \frac{1}{RC} v_o(t) \, = \, \frac{1}{RC} v_i(t)}\)

Solving the DE gives us a general solution in the form of a sum between a decaying exponential in time (the transient response) and a complex exponential in time (the steady-state response):

\(\small{v_o(t) = Ke^{-t / RC} \, + \, \frac{e^{j \omega t}}{j \omega RC + 1}}\)

Fixing the various parameters of the input gives us the output for all time.

Now let's do the same thing in the frequency domain. Using the voltage divider equation, we can write the output (in time) of the RC low pass filter as a function of its input:

\(\small{v_o(t) = v_i(t) \left( \frac{Z_C}{Z_R + Z_C} \right) }\)

where Z_R and Z_C are the resistive and capacitive impedances. From this we define the transfer function of the filter (in frequency) as the Laplace transform of the ratio of the output to the input:

\(\small{H(s) = \frac{V_o(s)}{V_i(s)} = \frac{Z_C}{Z_R + Z_C} = \frac{\frac{1}{sC}}{R + \frac{1}{sC}} = \frac{\frac{1}{RC}}{s + \frac{1}{RC}}}\)

where s = σ + jω is a complex number. Let σ = 0, making H a function of jω, in which ω represents angular frequency. The magnitude of this function gives us the filter's frequency response:

\(\small{|H(j \omega)| = \frac{1 / RC}{\sqrt{\omega^2 \, + \, (1 / RC)^2}}\)

with 1 / (RC) the filter's cutoff frequency. The independent variable ω ranges over all frequencies, i.e., ω ∈ (-∞, ∞). The filter output, then, is simply the product of the input (in frequency) and the transfer function.

Note that both the time and the frequency solutions have the same basic form: a function that ranges over all time or all frequency. There is no sense in which one solution requires "less" information than the other. Re-read the previous sentence, and then consider that both solutions can be easily transformed into each other without loss of information. In other words, they are equivalent.

The reason that is enough is because it does show that (with little thought) the time solution has time 't' in it, while the frequency solution does not have time 't' in it, therefore the solution in frequency is solved without requiring the knowledge of time or maybe more important, not requiring any time so it takes no real time to have all the solutions for all time in one package.

You seem to be using time in two different ways here: (1) time as in t, the independent variable of a function; and (2) time as in effort, i.e., the number of steps necessary to do something. In either case, the result is the same. In the sense of (1), remember that frequency is simply the reciprocal of time; they are bundled together, as it were. But what you're forgetting in your argument is that a solution in frequency is defined for as many different points along the ω axis as the corresponding solution in time is defined along the t axis. You seem to think that because a sinusoid passes through a linear filter and comes out the other end as a sinusoid -- i.e., a single real frequency -- that somehow it wasn't necessary to calculate the response for all the other infinite frequencies. But if you want to know the response for some particular frequency, you first need to know the response for all frequencies.

As for the use of time in the sense (2), certain operations are indeed more easily solved in one domain or the other. But it's very important to note that there's no preferred basis. So, for example, though it's true that convolution in time is computationally more difficult than the equivalent operation of multiplication in frequency, it is also true that convolution in frequency is computationally more difficult than multiplication in time. In other words, the computational power implicit in both domains is identically equal, a fact that should not be surprising as they both contain precisely the same information.

Quantum supremacy, on the other hand, means that classical computation and quantum computation do not have the same power. There is no sense in which time vs frequency analysis is comparable to classical vs quantum computing.

Indeed anyway arent all the inputs to the quantum computer frequencies anyway?

No. Quantum computers use entangled qubits to do their processing, with each qubit representing a generalized 2-state quantum system (e.g., electron spin measured along a spatial axis). What makes them different than classical bits is that, before measurement, a qubit can be in a superposition of its two possible states.

Also note that the time solution shown here is not the true time solution it's simplified to show only the steady state solution. That means in real life we'd have to wait for the exponentials to die down, and in theory we might be able to say that we never have the true solution in the time domain for what happens near t=inf because in theory the exponentials dont truely die until we reach infinity.

Again, no. Once we know the exponential of the transient response -- which I showed above -- then you know its value for any time t. You don't have to "wait" for anything.

 

This doesn't make sense. If you want to solve for the output magnitude B by leveraging the frequency domain, you need to know the frequency response of the filter, which means knowing the response at every frequency ω. This is no different than having to plot every value of t in the time domain.

Let's be explicit about this, since you seem to be missing the point. If we want to solve for the output of the filter strictly in the time domain, we use Kirchhoff's rules to set up a first-order differential equation relating the output \(\small{v_o(t)}\) to the input \(\small{v_i(t)}\) as functions of time:

\(\small{\frac{d}{dt} v_o(t) \, + \, \frac{1}{RC} v_o(t) \, = \, \frac{1}{RC} v_i(t)}\)

Solving the DE gives us a general solution in the form of a sum between a decaying exponential in time (the transient response) and a complex exponential in time (the steady-state response):

\(\small{v_o(t) = Ke^{-t / RC} \, + \, \frac{e^{j \omega t}}{j \omega RC + 1}}\)

Fixing the various parameters of the input gives us the output for all time.

Now let's do the same thing in the frequency domain. Using the voltage divider equation, we can write the output (in time) of the RC low pass filter as a function of its input:

\(\small{v_o(t) = v_i(t) \left( \frac{Z_C}{Z_R + Z_C} \right) }\)

where Z_R and Z_C are the resistive and capacitive impedances. From this we define the transfer function of the filter (in frequency) as the Laplace transform of the ratio of the output to the input:

\(\small{H(s) = \frac{V_o(s)}{V_i(s)} = \frac{Z_C}{Z_R + Z_C} = \frac{\frac{1}{sC}}{R + \frac{1}{sC}} = \frac{\frac{1}{RC}}{s + \frac{1}{RC}}}\)

where s = σ + jω is a complex number. Let σ = 0, making H a function of jω, in which ω represents angular frequency. The magnitude of this function gives us the filter's frequency response:

\(\small{|H(j \omega)| = \frac{1 / RC}{\sqrt{\omega^2 \, + \, (1 / RC)^2}}\)

with 1 / (RC) the filter's cutoff frequency. The independent variable ω ranges over all frequencies, i.e., ω ∈ (-∞, ∞). The filter output, then, is simply the product of the input (in frequency) and the transfer function.

Note that both the time and the frequency solutions have the same basic form: a function that ranges over all time or all frequency. There is no sense in which one solution requires "less" information than the other. Re-read the previous sentence, and then consider that both solutions can be easily transformed into each other without loss of information. In other words, they are equivalent.


You seem to be using time in two different ways here: (1) time as in t, the independent variable of a function; and (2) time as in effort, i.e., the number of steps necessary to do something. In either case, the result is the same. In the sense of (1), remember that frequency is simply the reciprocal of time; they are bundled together, as it were. But what you're forgetting in your argument is that a solution in frequency is defined for as many different points along the ω axis as the corresponding solution in time is defined along the t axis. You seem to think that because a sinusoid passes through a linear filter and comes out the other end as a sinusoid -- i.e., a single real frequency -- that somehow it wasn't necessary to calculate the response for all the other infinite frequencies. But if you want to know the response for some particular frequency, you first need to know the response for all frequencies.

As for the use of time in the sense (2), certain operations are indeed more easily solved in one domain or the other. But it's very important to note that there's no preferred basis. So, for example, though it's true that convolution in time is computationally more difficult than the equivalent operation of multiplication in frequency, it is also true that convolution in frequency is computationally more difficult than multiplication in time. In other words, the computational power implicit in both domains is identically equal, a fact that should not be surprising as they both contain precisely the same information.

Quantum supremacy, on the other hand, means that classical computation and quantum computation do not have the same power. There is no sense in which time vs frequency analysis is comparable to classical vs quantum computing.


No. Quantum computers use entangled qubits to do their processing, with each qubit representing a generalized 2-state quantum system (e.g., electron spin measured along a spatial axis). What makes them different than classical bits is that, before measurement, a qubit can be in a superposition of its two possible states.


Again, no. Once we know the exponential of the transient response -- which I showed above -- then you know its value for any time t. You don't have to "wait" for anything.

You tell me i am missing the point, but i tell you that you are missing the point.
You need to be able to see analogies on different levels. The world is not limited to only what YOU think is an analogy. Analogies can come from many angles.

But if i digress i see a total error in your first paragraph which may explain why you dont get this ponit.
That is, we dont need all frequencies to analyze an RC network in the form of a low pass filter. We may be dealing with only one frequency, and that one frequency depicts what happens over ALL time.

But back to the point, if i say to you "sine wave" you know right away what i am talking about, and you know it for ALL time, even without lifting a finger or thinking about it for very long. That's because a sine wave is a certain class of signals that is quite a bit different than other signals. The sine wave can be described by two CONSTANT numbers A and w, and the solution to the RC filter maps to B and w.

Now maybe a better example is a mix in the form of an addition of two sine waves of different frequencies.. The two added together create solutions in time out to infinity but we dont need time to show the solution unless we resort to time solutions, and time solutions after all take REAL time.

I understand completely what you say about the equivalency of time and frequency domains, but here we are dealing with a specific frequency or a set of specific frequencies (and amplitudes), and real time is never involved UNTIL we go to look for the solution at a certain point where A*sin(w*t) implies we have to look ever a time from 0 to that solution time, but in the frequency domain the response is captured for ALL time in just one set of numbers A and w, and both of these are a constant.

Now let's go back to the RC filter for a second. This might help explain better.
The simple RC filter excited by a sine wave:
y=A/(s*C*R+1)
and with R=C=A=1 we have:
y=1/(s+1)
and to analyze in frequency we can replace s:
y=1/(j*w+1)
the amplitude is:
|y|=1/sqrt(w^2+1)
and at a frequency of w=1 we get:
|y|=1/sqrt(2)
and see how we end up with a constant.
The interesting point is that this constant represents the entire sime solution without having to involve time.
Now the time solution is:
y=(w*sin(t*w))/(w^2+1)+cos(t*w)/(w^2+1)-e^(-t)/(w^2+1)
Now the interesting point is that to get an actual solution here we have to choose a time 't' and if we choose say 0.2 seconds for 't' we get a single result (w is constant here also). BUT, that result is only godd for that one time and we dont even know if it repeats or not because we have not yet done other time values like t=0.3, t=0.5, etc.
Also a side issue is that you can see the time solution has an exponential part where the frequency domain solution did not need that either. I wont stress that too much however because we can imagine time periods so long that the exponential part because irrelevant, but keep in mind that then we have an approximation in a sense. The frequency domain solution starts at a time value near or at infinity so we never have to deal with the approximation (even though it is extremely good).
Keeping to the main point though, the solution of
y=1/sqrt(2)
depicts what happens over all time while
y=1/sqrt(2)*sin(t+ph) (derived with A=1 and w=1 and R=1 and C=1 again and unspecified ph for now)
(note the ph comes in when we combine the sin and cos parts in the original frequency solution but i dont consider that a problem unless of course we see that we have to introduce another constant).

Now briefly look at the two solutions:
y=1/sqrt(2)
y=1/sqrt(2)*sin(t+pi/4) (where ph is the pi/4)

Note how the first can explain the second (dont worry about the phase here) with just ONE number, and it IMPLIES a solution over ALL time without calculating EVERY SINGLE point in time.

 

You tell me i am missing the point, but i tell you that you are missing the point.
You need to be able to see analogies on different levels. The world is not limited to only what YOU think is an analogy. Analogies can come from many angles.

The world is full of analogies, some good, some bad, and some are completely off the mark. A good analogy conveys the basic idea of an unfamiliar concept using familiar notions. A bad analogy causes more confusion than enlightenment. Your analogy, however, is completely off the mark because it fails to capture the single most important aspect of quantum supremacy.

Your analogy seems to be based on the idea that computationally difficult operations in one domain can be easier in another. A valid example is multiplication and division, which are computationally more difficult than addition and subtraction. If we can transform a problem over the reals into a problem over the logs of reals, then products and quotients are reduced to sums and we get a computational speedup. Would you agree that this is what you were trying to get at with your analogy?

But if i digress i see a total error in your first paragraph which may explain why you dont get this ponit.
That is, we dont need all frequencies to analyze an RC network in the form of a low pass filter. We may be dealing with only one frequency, and that one frequency depicts what happens over ALL time.

By definition of the Fourier transform and its inverse, a function that is defined over all time is defined over all frequency. It doesn't matter if the spectrum of the function consists of a single frequency, an infinite number, or something in between. For any f(t), we need to look at all frequencies to know F(ω). Even if f(t) is periodic, its Fourier series will contain an infinite number of spectral coefficients. In the case of single-frequency sinusoids, the fact that almost all spectral components are identically zero is the defining characteristic of a sinusoid. That is to say, f(t) is a sinusoid if and only if F(ω) = 0 everywhere except for precisely two points, F(±Ω).

But back to the point, if i say to you "sine wave" you know right away what i am talking about, and you know it for ALL time, even without lifting a finger or thinking about it for very long. That's because a sine wave is a certain class of signals that is quite a bit different than other signals. The sine wave can be described by two CONSTANT numbers A and w, and the solution to the RC filter maps to B and w.

I certainly don't know the value of any sine wave for all time. I'm guessing that if I ask you what sin(21.7) is, you won't "know right away" either. You likely think I'm being stubborn here, but you seem to be forgetting that we're talking about computational complexity. Any pre-existing "knowledge" we bring to the table is completely irrelevant. It's not generally true that sinusoids can be described by two constants (actually three, including phase); it's only true if we assume we're talking solely about sine waves. But so what? As long as we start with the same assumptions, there are an infinite number of functions that can be completely described by two constants. We can use two constants to characterize any square wave or triangle wave, if we assume we're talking about square waves or triangle waves. Such "pre-knowledge" has no bearing on either the spectrum of the signal or the complexity of any resulting calculations.

Now maybe a better example is a mix in the form of an addition of two sine waves of different frequencies.. The two added together create solutions in time out to infinity but we dont need time to show the solution unless we resort to time solutions, and time solutions after all take REAL time.

What do you mean by "REAL time"? Are you talking about clock time, as in how long it takes to perform the calculation? Because frequency solutions also take REAL time.

Now let's go back to the RC filter for a second.

The simple RC filter excited by a sine wave:
y=A/(s*C*R+1)
and with R=C=A=1 we have:
y=1/(s+1)
and to analyze in frequency we can replace s:
y=1/(j*w+1)
the amplitude is:
|y|=1/sqrt(w^2+1)
and at a frequency of w=1 we get:
|y|=1/sqrt(2)
and see how we end up with a constant.
The interesting point is that this constant represents the entire sime solution without having to involve time.

Note that you haven't shown an RC filter excited by a sine wave; what you've shown is that the given filter has its -3 dB point at ω = 1. Also note that the expression |y| = 1 / sqrt(ω^2 + 1) for the filter magnitude response is a function of ω, defined for all positive frequencies. Choosing a particular value of ω is no different than choosing a particular value of t in a function over time: in both cases, the function is reduced to a single number. In either case, there's no speedup because you have to start with the function defined for all time or all frequency.

Furthermore, it doesn't matter that we "know" that sine can only have two possible values of ω, because that presupposes the solution to the problem. That's like arguing that it's easy to win the lottery if you already know what the numbers will be.

And if you're still not convinced, consider that the input to an RC filter can never be a sine wave, because such things are mathematical abstractions. A signal that is time-limited cannot also be band-limited, therefore any signal that goes into an RC filter will necessarily contain energy at all frequencies. As such, your filter's response |y| needs to be evaluated over some range of ω, just as its time response needs to be evaluated over some range of t.

Do you now understand? Can you now see that the time and frequency domains are equal brothers? Everything that's hard to do in time -- like convolution -- is equally hard to do in frequency. Quantum supremacy, on the other hands, implies the complete opposite: some things that are hard to do using classical computers are easy to do using quantum computers. That's the headline point, the one that newspaper articles will tout. The analogy of logarithms simplifying multiplications speaks to this aspect. The more subtle, though far more interesting point is that our universe supports two valid but fundamentally different forms of computation. This is so breathtakingly deep that a good analogy might not exist.

 

Yeah you are insisting that your view is the correct one and cant take a fresh look at this.
You have to understand the difference between a constant and a variable and how one could define things more economically than another. It DOES NOT MATTER IF THEY ARE BOTH THE SAME once all is said and done, because one view encapsulates an entire solution set while the other does not.
And you are treating 'w' as a variable when it's not in AC analysis when we happen to be dealing with one frequency. The frequency is implied (to be the same on the output for a linear circuit).

The frequency viewpoint assumes a solution right at t=0 and we only need to look at certain time values when we get around to doing that. Mix two different frequencies and we get results that could take years to show in the time domain.

Going back to one frequency, if we apply A*sin(w*t) to a simple RC network we only have to specify the amplitude A and the frequency w, and we end up with a output that can be described by B and w. We dont have to specify A*sin(w*t) and thus there is NO TIME value to think about, so in a sense it takes no time to encapsulate the entire result we only need time when we feel like looking at specific results. This is just the same way the QC works in that same sense because the results are encapsulated until we feel like extracting them.

I dont know how else to explain this except when we specify
A,w
with respective values:
5vpk, 628Hz
it implies a sinusoid that starts at t=0 and goes on to infinity.
If we specify
A*sin(w*t)
we also have the same constants:
5vpk, 629Hz
so we end up with:
5*sin(628*t)

Now it's interesting that the first form has NO variable but yet we know we have a sinusoid. How is that possible? Because the two constants are enough to describe the phenomenon UNTIL such time as we want to look at specific values.
The second form also implies that we are dealing with a sinusoid, but we MUST specify a time values also. So we dont have the choice of waiting until we feel like looking at the solution, we must represent ONE AND ONLY ONE solution point with a value for 't'.

Let me reiterate...
In the first form we have two constants 5 and 628.
In the second form we have two constants AND a variable.
They might contain the same information but when we say that we have a solution in the first form we just have something like 3, 628 but in the second form we have the solution 3*sin(628*t) and obviously these two are different textually. But the important point is that the first IS A SOLUTION for ALL TIME, while the second is a solution for just one point in time given t a point to look at. Otherwise we'd have to calculate using every single time value from 0 to infinity.
Let''s plot them both.
In the first form, we just have to plot the value 5 at a frequency of 628.
In the second, we have to run though all the values of time we want to plot.
So to actually GET the plot we have to do a lot more work in the second form and we MUST do it WHEN we want the solution, not after we have (after we have a different form for) the solution.

Now obviously the QC is a practical real world thing. That means we are dealing with nature and when we do that we use up time and energy. which plot therefore would take up more time and energy? Obviously the second form is more complex.
Now imagine that we have two boxes A and B both inside a bigger box C.
In A we have 5,628 and
in B we have 5*sin(628*t).
Which one is going to take more time and energy to get their entire solution out of their respective boxes A and B. Obviously B will take more time and energy.
Box C is the environment.

The main point though is that A,w is a solution but we are not looking at any time values yet. with Asin(wt) we have to look at each point in time because that's what kind of solution it is.

Another view here, from the aspect of the nature of measurement and application, is that when we encounter two very very very tiny objects our experience as humans makes us consider both of them as one entity due to our observational statistical brain function. As we explore more deeply we find two extremely tiny objects that form one object as viewed by the brain. The question then is do we consider them two of the same objects (but certainly not the composite object) or do we look deeper. The answer is that sometimes when we look deeper we find that they are not the same. The next question then is, is there an application for this new knowledge in the real world? Sometimes the answer is yes and sometimes the answer is no. Once an application for one and not the other is discovered, they must then be considered as two separate objects and anyone that still thinks that they are the same is either dumb or has an application where that view works really well.

[LATER]
Actually this is my argument for quantum supremacy.
I am sorry if you dont understand or like my argument/opinion but i think i have explained my view enough now. The idea was not mine but the original idea was just about the difference between the two not for QS. It was explained to me by the chief engineer i worked with at the time (could have been as early as 1975).

 

You have to understand the difference between a constant and a variable and how one could define things more economically than another. It DOES NOT MATTER IF THEY ARE BOTH THE SAME once all is said and done, because one view encapsulates an entire solution set while the other does not.
And you are treating 'w' as a variable when it's not in AC analysis when we happen to be dealing with one frequency. The frequency is implied (to be the same on the output for a linear circuit).

You have yourself so wrapped up in the notion of a single frequency sine wave that you can't see the forest before you. In your RC filter example, you derived the frequency response as 1 / sqrt(ω^2 + 1). This is a function of the independent variable ω that tells you the magnitude of the filter at any frequency ω. It should be trivially obvious that the ω in the expression is most certainly not a constant, because if it were, then the filter would simply be a constant gain.

The frequency viewpoint assumes a solution right at t=0 and we only need to look at certain time values when we get around to doing that. Mix two different frequencies and we get results that could take years to show in the time domain.

What are you talking about?

Going back to one frequency, if we apply A*sin(w*t) to a simple RC network we only have to specify the amplitude A and the frequency w, and we end up with a output that can be described by B and w. We dont have to specify A*sin(w*t) and thus there is NO TIME value to think about, so in a sense it takes no time to encapsulate the entire result we only need time when we feel like looking at specific results. This is just the same way the QC works in that same sense because the results are encapsulated until we feel like extracting them.

Again, you seem to be blinded by the idea of a single frequency spectrum; you're ignoring the fact that we can never apply A sin(ωt) to an RC filter because such a signal doesn't physically exist. The closest we can get is a "pulsed sine" -- something of the form A sin(ωt) rect(t/T) -- which is the product of a sine and a unit rectangular function (zero everywhere outside of a finite time interval T). The spectrum of such a thing is two mirrored sinc functions, with main their lobes centered over ±ω. Please consider how you would analyze this signal with an infinite number of frequencies. As you are making a general claim about time vs frequency, making up physically-impossible special cases doesn't cut it -- we must use the tools available to us.

To that end, if we want to know the output y(t) of some linear filter with impulse response h(t), for some input signal x(t), we take the convolution of the input with the IR:

y(t) = x(t) * h(t)

Alternatively, we can get the output's magnitude in the frequency domain by taking the pointwise product of the filter's transfer function H(ω) with the spectrum of the input X(ω):

Y(ω) = X(ω) H(ω)

Let x(t) be our pulsed sine, presumably the simplest real-world signal we can apply, and let h(t) be the impulse response corresponding to our RC filter. Which form is computationally more expensive? At first-sight, we might think that the frequency-domain version is easier, because multiplication is easier than convolution. But x(t) is the product of two time-domain functions, and multiplication in the time domain transforms to convolution in the frequency domain. So, in order to calculate X(ω) we're going to need to perform a convolution anyway.

In either case, we're stuck with doing calculations over a range of t or ω. There is no sense in which one "encapsulates" the other.

I dont know how else to explain this except when we specify
A,w
with respective values:
5vpk, 628Hz
it implies a sinusoid that starts at t=0 and goes on to infinity.

What?! How does "5vpk" and "628Hz" imply a sinusoid? That's only in your head. What if I said "5vpk", "628 Hz", and held up a sign that said "triangle wave"?

If we specify
A*sin(w*t)
we also have the same constants:
5vpk, 629Hz
so we end up with:
5*sin(628*t)

Now it's interesting that the first form has NO variable but yet we know we have a sinusoid. How is that possible? Because the two constants are enough to describe the phenomenon UNTIL such time as we want to look at specific values.
The second form also implies that we are dealing with a sinusoid, but we MUST specify a time values also. So we dont have the choice of waiting until we feel like looking at the solution, we must represent ONE AND ONLY ONE solution point with a value for 't'.

I'm sorry but this is nonsense. When we write an expression such as y(t) = 5 sin(628t), we are describing an entire function y:ℝ→ℝ, not a single value at some point t. The function is precisely the map of all points (-∞, ∞) to all values y(t). There is no difference in meaning between writing 5 sin(628t), where we've explicitly labeled the independent variable, and writing "sine wave of amplitude 5 and frequency 628 Hz", where we don't mention the variable. One is not a compressed form of the other; both imply the same "amount of time".

Your argument is not even wrong.

 

See this is what i mean. I say w is a constant you say w is not a constant.
BUT it;s not your choice it's my choice what i make it because i am giving the example not you.
If you want to understand the viewpoint then you have to accept what i tell you too. You obviously do not so there's no point in talking about it.

So i am going to give this topic a rest because i want to get back to my theoretical full wave rectifier solution with a capacitor, load resistor, and series resistor. If you would like to critique that it may be welcomed.
I'll post it when i am ready in another thread.

 

See this is what i mean. I say w is a constant you say w is not a constant.
BUT it;s not your choice it's my choice what i make it because i am giving the example not you.
If you want to understand the viewpoint then you have to accept what i tell you too. You obviously do not so there's no point in talking about it.

You're perfectly free to insist that ω is a constant of |y| = 1 / sqrt(ω^2 + 1). But then, by definition, you cannot associate that same y with the transfer function of an RC filter. And so your argument fails either way.

I sincerely hope our little exchange has helped erode the misconception that the frequency domain has some sort of privileged status or computational advantage over the time domain. There's a deep, more general connection between any two Fourier duals -- like time and frequency, or position and momentum -- that can only be appreciated once you see them as equal spaces. See Parseval's theorem for more.

 

We'll have to wait for the paper to see, but my guess is that the toy problem is in NP (hard to solve, easy to verify).

https://www.scottaaronson.com/blog/?p=4317

 

You're perfectly free to insist that ω is a constant of |y| = 1 / sqrt(ω^2 + 1). But then, by definition, you cannot associate that same y with the transfer function of an RC filter. And so your argument fails either way.

I sincerely hope our little exchange has helped erode the misconception that the frequency domain has some sort of privileged status or computational advantage over the time domain. There's a deep, more general connection between any two Fourier duals -- like time and frequency, or position and momentum -- that can only be appreciated once you see them as equal spaces. See Parseval's theorem for more.

Yes but you keep insisting that w has to be a variable so you miss the whole point.
The problem is that you dont want to understand the point you just want to insist on one view of the world.
It seems hard for me to believe however that you've never worked with something like line frequency components where the frequency is always taken to be a constant. In fact, at 60Hz the value for w is VERY OFTEN taken to be 377 an approximation. Now let's see you find something wrong with that

What i like to tell students and members that insist on being hard left field theorists is that i understand their point of view AND my point of view, that's two views that i understand, but they only can see one point of view that's only one viewpoint they can see. So my viewpoint subsumes their viewpoint.

 

https://www.scottaaronson.com/blog/?p=4317

So they used a statistical proof, ala quantum correlations in games between Alice and Bob violating Bell's inequality. The fact that Scott Aaronson is this excited is a pretty good indication that Google has actually demonstrated quantum supremacy.

 

Yes but you keep insisting that w has to be a variable so you miss the whole point.
The problem is that you dont want to understand the point you just want to insist on one view of the world.

Dude, you're the one who is trying to argue the special case over the general case. You're essentially saying: "If we treat the transfer function as a constant, then . . ." Not only is this an incredibly myopic line of argument -- "please ignore every possible signal type except for a single-frequency sine wave" -- it doesn't even support your position.

Maybe you haven't noticed, but your responses to my counterarguments lately have been of the form "You're missing my point" without actually addressing the counterarguments themselves.

It seems hard for me to believe however that you've never worked with something like line frequency components where the frequency is always taken to be a constant. In fact, at 60Hz the value for w is VERY OFTEN taken to be 377 an approximation. Now let's see you find something wrong with that

What kind of weird trolling is this? Seriously, I'm baffled. You seem genuinely proud that you can give an example where people VERY OFTEN fix the value of ω, as if you were making a surprise reveal. "Tada! We can fix the value of ω! Bet you didn't see that coming!"

What i like to tell students and members that insist on being hard left field theorists is that i understand their point of view AND my point of view, that's two views that i understand, but they only can see one point of view that's only one viewpoint they can see. So my viewpoint subsumes their viewpoint.

LOL. Good luck with that.

 

So they used a statistical proof, ala quantum correlations in games between Alice and Bob violating Bell's inequality. The fact that Scott Aaronson is this excited is a pretty good indication that Google has actually demonstrated quantum supremacy.

You just know that someone is looking to find a loophole in their proof.
https://gilkalai.wordpress.com/2019...t-probably-false-supremacy-claims-google/amp/

 

Dude, you're the one who is trying to argue the special case over the general case. You're essentially saying: "If we treat the transfer function as a constant, then . . ." Not only is this an incredibly myopic line of argument -- "please ignore every possible signal type except for a single-frequency sine wave" -- it doesn't even support your position.

Maybe you haven't noticed, but your responses to my counterarguments lately have been of the form "You're missing my point" without actually addressing the counterarguments themselves.


What kind of weird trolling is this? Seriously, I'm baffled. You seem genuinely proud that you can give an example where people VERY OFTEN fix the value of ω, as if you were making a surprise reveal. "Tada! We can fix the value of ω! Bet you didn't see that coming!"


LOL. Good luck with that.

Sorry "DUDE" but you dont have the reasoning ability to ever understand this, and i dont even think you are trying. So quit trolling. Just so you know, i barely even read your long and drawn out posts because you just keep repeating yourself.
Say what you will im done with this topic for now. I have better things to do then argue with you.
Also, i suggest you work on some 50 or 60Hz systems where the frequency is, what? It is constant? Gee what a surprise