Zompist Bboard Again

Richard W wrote: ↑Sat Oct 07, 2023 6:18 pmThere are serious problems with the concept of a^b for arbitrary complex a and b.

I wouldn't say there are problems per se, the actual calculation is actually relatively simple. It is, however, multi-valued for anything other than an integer exponent - but that is the case for exponentiation in general. Complex exponents simply offer more ways to be non-integral.

KathTheDragon wrote: ↑Sun Oct 08, 2023 12:40 am

Richard W wrote: ↑Sat Oct 07, 2023 6:18 pmThere are serious problems with the concept of a^b for arbitrary complex a and b.

I wouldn't say there are problems per se, the actual calculation is actually relatively simple. It is, however, multi-valued for anything other than an integer exponent - but that is the case for exponentiation in general. Complex exponents simply offer more ways to be non-integral.

Yes, the calculation is simple for non-zero a. The problem is not the execution, but the concept. Exponentiation can make good sense for positive base and real exponents, and other special cases.

Richard W wrote: ↑Sun Oct 08, 2023 8:00 am Yes, the calculation is simple for non-zero a. The problem is not the execution, but the concept. Exponentiation can make good sense for positive base and real exponents, and other special cases.

What is the problem? We've been doing exponentiation on complex numbers for three centuries, and it's the basis for a huge amount of physics.

This thread has taken a direction I wasn't expecting.

I'm with Zomp. What's the conceptual problem? That you can't visualise the function?

KathTheDragon wrote: ↑Sun Oct 08, 2023 3:29 pm I'm with Zomp. What's the conceptual problem? That you can't visualise the function?

I think you already know the problem. It's the infinitely many branches in the log function. One of its manifestations is the failure of the identity (a^b)^c = a^(bc). For example, with a=(-1+√3i)/2, b = 2 and c = i, the LHS is e^−4π/3 and the RHS is e^2π/3.

The real oscillatory solution of the unforced quadratic response is normally written Ae^−ζtcos ωt+ε or as a real part of something.

Why is this a problem? On the contrary, the identity is true, but only if you don't choose your branches beforehand. Look, I can do it without complex numbers. Just arbitrarily choose the negative branch for √2 and the positive branch for √8, and (2^1/2)³ ≠ 2^{1/2 * 3}. It's only necessarily true with a, b, c all integers, or if you let the exponential be multivalued for non-integral operands.

KathTheDragon wrote: ↑Mon Oct 09, 2023 7:50 pm Why is this a problem? On the contrary, the identity is true, but only if you don't choose your branches beforehand. Look, I can do it without complex numbers. Just arbitrarily choose the negative branch for √2 and the positive branch for √8, and (2^1/2)³ ≠ 2^{1/2 * 3}. It's only necessarily true with a, b, c all integers, or if you let the exponential be multivalued for non-integral operands.

It's also true with a positive and b and c real, though it may be more natural to just say with all three positive.

With multivalued exponentiation, what are you doing for equality? Are you saying that sets are '=' if they have non-empty overlap? But doesn't transitivity matter for notions akin to equality?

Basically, complex exponentiation is a somewhat flaky operation that sometimes does what you want, but generally in important special cases.

Richard W wrote: ↑Tue Oct 10, 2023 5:06 pm

KathTheDragon wrote: ↑Mon Oct 09, 2023 7:50 pm Why is this a problem? On the contrary, the identity is true, but only if you don't choose your branches beforehand. Look, I can do it without complex numbers. Just arbitrarily choose the negative branch for √2 and the positive branch for √8, and (2^1/2)³ ≠ 2^{1/2 * 3}. It's only necessarily true with a, b, c all integers, or if you let the exponential be multivalued for non-integral operands.

It's also true with a positive and b and c real, though it may be more natural to just say with all three positive.

Do you not agree my counterexample is a counterexample to your position?

With multivalued exponentiation, what are you doing for equality? Are you saying that sets are '=' if they have non-empty overlap? But doesn't transitivity matter for notions akin to equality?

No, I'm using normal set equality. As in the same exact elements.

KathTheDragon wrote: ↑Tue Oct 10, 2023 6:02 pm

Richard W wrote: ↑Tue Oct 10, 2023 5:06 pm

KathTheDragon wrote: ↑Mon Oct 09, 2023 7:50 pm Why is this a problem? On the contrary, the identity is true, but only if you don't choose your branches beforehand. Look, I can do it without complex numbers. Just arbitrarily choose the negative branch for √2 and the positive branch for √8, and (2^1/2)³ ≠ 2^{1/2 * 3}. It's only necessarily true with a, b, c all integers, or if you let the exponential be multivalued for non-integral operands.

It's also true with a positive and b and c real, though it may be more natural to just say with all three positive.

Do you not agree my counterexample is a counterexample to your position?

No, because the principal values (as calculated by Fortran for example), are always positive.

KathTheDragon wrote: ↑Tue Oct 10, 2023 6:02 pm

With multivalued exponentiation, what are you doing for equality? Are you saying that sets are '=' if they have non-empty overlap? But doesn't transitivity matter for notions akin to equality?

No, I'm using normal set equality. As in the same exact elements.

But, returning to arbitrary complex numbers with multi-valued exponentiation, pace Cantor, (a^b)^c is a doubly infinite set whereas a^bc is a singly infinite set, in general a proper subset of the former.

Richard W wrote: ↑Tue Oct 10, 2023 6:41 pm No, because the principal values (as calculated by Fortran for example), are always positive.

Yes because we like to make an arbitrary (yes, arbitrary!) choice of branch because single-valued functions are easier to work with. But that doesn't mean the other branches are somehow less valid.

But, returning to arbitrary complex numbers with multi-valued exponentiation, pace Cantor, (ab)c is a doubly infinite set whereas abc is a singly infinite set, in general a proper subset of the former.

Ah, well then, I guess the strict equality just isn't true, and we should in fact say that (a^b)^c is a (non-strict) superset of a^bc. Again, what's the problem?

KathTheDragon wrote: ↑Tue Oct 10, 2023 7:49 pm

Richard W wrote: ↑Tue Oct 10, 2023 6:41 pm No, because the principal values (as calculated by Fortran for example), are always positive.

Yes because we like to make an arbitrary (yes, arbitrary!) choice of branch because single-valued functions are easier to work with. But that doesn't mean the other branches are somehow less valid.

But if we can have a continuous function in the domain of interest, it makes sense to use that. That means that if we are considering reals (or discarding complexes), {positive reals} ✕ {reals} is the natural domain. And a single branch can cover the positive reals.

Yes, each branch covers the entire codomain by definition. But the thing about having infinitely many branches is that the divisions are arbitrary. You could have the branch cuts at τn + 1 radians, or τn + 2.674 radians, or anywhere you like. And, for that matter, restricting the function to one branch means that you are opening yourself up to sacrificing the "nice" properties of the full multivalued function. You just have to accept that fact. The function is incomplete.

KathTheDragon wrote: ↑Thu Oct 12, 2023 8:26 am You could have the branch cuts at τn + 1 radians, or τn + 2.674 radians, or anywhere you like.

I note that you yourself are preferring straight line cuts! That's not what you used for your real-domain 'counterexample'.

Richard W wrote: ↑Sat Oct 07, 2023 6:18 pm

malloc wrote: ↑Thu Oct 05, 2023 4:04 pm But can you solve expressions like (3.89+2.5i)^(4.8-2.73i) in your head? My laptop can solve that in less than one second.

I'm not sure that you would ever want to evaluate that in your head. There are serious problems with the concept of a^b for arbitrary complex a and b. I recall not appreciating the involved nature of exponentiation when Simon Norton threw us the off-topic question of which fields had it at a supervision in around 1976 or 1977.

Then what is my computer doing when it calculates complex exponents of complex numbers and would you consider its answer mistaken?

malloc wrote: ↑Fri Oct 13, 2023 4:45 pm

Richard W wrote: ↑Sat Oct 07, 2023 6:18 pm

malloc wrote: ↑Thu Oct 05, 2023 4:04 pm But can you solve expressions like (3.89+2.5i)^(4.8-2.73i) in your head? My laptop can solve that in less than one second.

I'm not sure that you would ever want to evaluate that in your head. There are serious problems with the concept of a^b for arbitrary complex a and b. I recall not appreciating the involved nature of exponentiation when Simon Norton threw us the off-topic question of which fields had it at a supervision in around 1976 or 1977.

Then what is my computer doing when it calculates complex exponents of complex numbers and would you consider its answer mistaken?

Ok but LLMs are notoriously worse at mathematics the better they are at regurgitating written language. A simple math problem that chatgpt used to get correct 98%+ of the time when it wasn't as good at mimicking human language, it now gets correct less than 2% of the time. Meanwhile humans can get better at writing and math at the same time.

And if you want computers that are good at math (and nothing else) we already have those: calculators.

Well sure, but Richard W is claiming that raising one complex number to another is problematic regardless of what kind of software or human skill used. Yet the calculator program on my computer spits out an answer to such expressions. My question was what the calculator is doing and whether it has something resembling a correct answer.

Richard W wrote: ↑Fri Oct 13, 2023 4:18 pm

KathTheDragon wrote: ↑Thu Oct 12, 2023 8:26 am You could have the branch cuts at τn + 1 radians, or τn + 2.674 radians, or anywhere you like.

I note that you yourself are preferring straight line cuts! That's not what you used for your real-domain 'counterexample'.

Oh lord this is so tiring. Can you please just use your imagination for one minute and actually cooperate with me so I can make my point in the fewest number of increasingly exasperated posts possible? It doesn't matter what my "counterexample" is, it doesn't matter what the go-to branch cuts are. My point is ultimately very simple: you shouldn't expect properties to generalise cleanly when you expand a function's domain, especially when it becomes infinitely multivalued, and branch cutting to make it single-valued again will not save you. It's not a matter of the function being "conceptually problematic", you're just wanting it to be something it isn't.

KathTheDragon wrote: ↑Fri Oct 13, 2023 7:25 pm It's not a matter of the function being "conceptually problematic", you're just wanting it to be something it isn't.

That's a good summary of the problem - the 'function' isn't what one wants it to be, especially with non-positive bases. In particular, it doesn't satisfy the 'Maths 101' definition of a function in terms of a type of relation on sets.

However, 'principal values' mostly provide the answer one wants - which is why they're the principal values! Not always, though - the expression (-8)^(1/3.0) will usually not give -2.0, the answer one probably wants. Cube-rooting reals can easily defeat some programmers.

Note that Marx supported capitalist development even though he believed it eventually makes companies less profitable and thereby decreases general quality of life. He believed that this would motivate the revolutionaries to overthrow the profit motive and use the erstwhile "capitalist" developments for the betterment of society.

While I don't support accelerationism that deliberately seeks to make our lives worse, progressives can't oppose genuine developments like AI even though they have unintended antisocial consequences. These are the factors that motivate the opposition to profitability as the ruling ideology.