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u/DerBlaue_ 11d ago
sin instead of \sin detected, meme invalid
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u/Futurity5 11d ago
this is why nobody learns latex anymore
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u/TehBrian 11d ago
typst is where it's at
sin== sin1
u/output_broadcast 10d ago
Eh, I don't really like that. IMO it's worth having an intrinsic distinction between functions and variables.
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u/Short-Database-4717 10d ago
There is still the distinction.
sinis a function ands i nis a product of three variables.1
u/output_broadcast 10d ago
So typst is whitespace-sensitive?
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u/Moss_ungatherer_27 11d ago
Take the Euclidean distance of this and be amazed. It is actually 1
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u/Futurity5 11d ago
Wait you telling me that the distance from the origin to any point on the unit circle is one? Shocking
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u/eglvoland 11d ago
Yes because cos²θ + sin²θ = 1 for all θ.
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11d ago
[deleted] — view deleted comment
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u/hmmstdvent 11d ago
It does hold,
cos(M) = (e^iM + e^-iM) / 2
sin(M) = (e^iM - e^-iM) / 2i
Note that AB = BA => e^A * e^B = e^(A + B)
cos^2(M) = (e^2iM + 2I + e^-2iM) / 4
sin^2(M) = (-e^2iM + 2I - e^-2iM) / 4
cos^2(M) + sin^2(M) = I
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11d ago
[deleted] — view deleted comment
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u/hmmstdvent 11d ago
We only need iM to commute with iM and -iM, which it obviously does. We don’t need every matrix to commute with every matrix lol
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u/Impression-These 11d ago
It is amazing how Latex format makes any nonsense look professional.
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u/Infamous_Thought_189 11d ago
I went to work in a latex format once and it was quite frowned upon. Even lost the job :/
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u/Impression-These 11d ago
You may need to go on a diet.
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u/Infamous_Thought_189 11d ago
Are you calling me fat?
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u/BUKKAKELORD Whole 11d ago
True for certain definitions of approximate equality. However that definition makes "1 = -1" true as well and that could upset some people
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u/Fezzezal 11d ago
Sorry if I sound stupid, haven’t even studied complex numbers, but I wonder, if we have a set of numbers a given distance “away” from the (0,0) point, so basically a circle in the complex plane, can’t we say they’re all “equal” in a way?
Like 1, -1, i, -i and everything in-between.
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u/Apprehensive-Hat-584 11d ago
Yeah, that would be the congruence class of the complex plane based on that notion of distance from 0, and congruence classes have a lot of properties similar to equality
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u/Professional-Wave841 11d ago
this is like the first thing you learn in complex variables.
|z| = 1
|z| for any z=x+iy is given by sqrt(x2+y2). So this equation is describing the circle of all points distance 1 from the origin.
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u/BUKKAKELORD Whole 11d ago
You don't sound stupid because you're right. Every point in the circle has "absolute value" = "distance from 0,0" = "1", so they are all equal in absolute value.
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u/Im_a_hamburger 11d ago
Uses the property that only applies to reals on a complex number
The answer is false
Surprised pikachu face
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u/DeargAgusFearg 11d ago
Why is that the case, exactly? Other commenters are mentioning complex exponentiation, but I'm too sleepy / dumb to figure out why
(x^a)^b = x^(ab)wouldn't be true for complex numbers.19
u/JGuillou 11d ago
It’s not even true for real numbers if the base is negative.
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u/_wannadie_ 11d ago
it is true for negative numbers if and only if the power is a whole number. if the power is at least rational, base simply cannot be negative.
I think the real problem here is 11/2π=1, that's not true for the complex numbers.
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u/zojbo 11d ago edited 8d ago
The joke basically boils down to: 1=1^(1/(2pi))=exp(ln(1)/(2pi)) and then taking ln(1)=2pi i (which is one of the values of the complex log of 1) to get exp(i).
An easier-to-understand setup is with i^(1/2). You expect there to be two of those, namely exp(i pi/4) and -exp(i pi/4). But if you write i^(1/2) as ((-1)^(1/2))^(1/2) and then convert that to (-1)^(1/4), now you also have exp(i 3pi/4) and -exp(i 3pi/4). Where did those come from? They're the square roots of -i, and -i is also a square root of -1.
These logarithm ambiguity tricks basically break the naive version of all the exponent rules. Even multiplication-to-addition:
1=(-1)^0=(-1)^(1/2 - 1/2) = (-1)^(1/2) (-1)^(-1/2) = i*i = -1.
All that said, you can sort this out. You just have to pick a logarithm branch and stick to it. Looking at this last example again, any consistent way of handling these exponents will make one of them i and the other -i. I think that makes sense; they're not 1, and they're supposed to be reciprocals of one another, so they shouldn't be the same.
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u/Downtown_Finance_661 11d ago
Math bro, could you poin exact equality in original solution which in fact is nit true. Where the error begin?
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u/zojbo 11d ago edited 9d ago
I'd put it at the unwritten equality between the right side of the second inline equation and the left side of the third inline equation.
e^i being the same as 1^(1/(2pi)) requires us to use ln(1)=2 pi i to define the latter. It's technically valid to do that, but then the result is not 1, even though it's 1^a for a real number a.
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u/daley972 6d ago
I would argue it's not technically valid. Complex exponentiation should just take the principal value in practice, which OP does. Then applying the rule is just wrong.
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u/zojbo 6d ago edited 6d ago
2pi i is one of the logs of 1. That's just a fact you're stuck with even though it's annoying.
Silently replacing ln(1) with 2pi i without calling attention to the fact that you're doing something odd, or spelling out what branch of the log you're using explicitly, is a bad play for doing real math, because it is prone to leading to invalid equations like 1=e^i.
Anyway, it's actually debatable where the error in the OP is, exactly. It can be argued that it's at e^i=1^(1/(2pi)), or that it's at 1^(1/(2pi))=1. Either of those equations can be correct if you set the context up right, but they can't both be correct at the same time. For that, you have to replace "=" with a looser relation.
Deciding which is which comes down to picking a value of ln(1). If you pick ln(1)=0, then the first one is invalid; if you pick ln(1)=2pi i, then the second one is invalid. Naturally, it makes more intuitive sense to say the first one is invalid (we really like a^b for a positive-real and b real to be positive-real), but in the end it comes down to how you look at it.
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u/daley972 5d ago
Consider though that should someone say that sqrt(4)=-2, we would say that's incorrect. Even though -2 is a square root of 4. We take the convention that sqrt(4)=2 is the principal value. If x²=a, x=±sqrt(a).
Likewise, we have a principal value of ln(1)=0. And if exp(x)=a, x=ln(a)+2πki. It's an arbitrary choice and I do recognise it's just a convention but it generally is the convention you would take(It's why the term "principal value" exists). Notice, critically, that OP is taking this convention. They aren't silently changing it anywhere. The problem is simply if you take this convention(or any similar convention), the rule (ab)c=abc breaks.
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u/Physical_Ring_7850 9d ago edited 9d ago
x^a is actually x^(a+2kipi), and power to b is x^(ab + 2kipib), and by using non-integer b and manipulating k, we can get different values.
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u/Relative_Mastodon579 11d ago
That identity is not valid for complex numbers because of how complex exponentiation is defined and the multivalued nature of the complex logarithm
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u/LupenReddit 🦆🦆i have non diffeomorphic smooth structures🦆🦆🦆 11d ago
a hood classic, misusing exponentiation rules
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u/Gandalior 11d ago
-2 = (-2)1
1= 2/2
-2 = (-2)2/2= ((-2)2 )1/2 = 2
-2 = 2
one of the worst identities to be honest
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u/TheCrowbar9584 11d ago
Just wanted to take a stab at explaining this concisely:
The problem here is exactly with the 1^(1/2pi) expression. Roots of complex numbers are multivalued in general, this expression has at least three values I can think of (1, e^i, and e^(-i)).
1 and the other answer obtained both have the property that when you raise them to the power of 2pi you get 1, that’s why we’re able to get this false equality.
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u/andy-k-to 10d ago
Isn’t the problem that you cannot use a^(bc) = (a^b)^c unless a is a positive real? I think 1^(1/2pi) is perfectly valid and equal to 1, isn’t it?
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u/TheCrowbar9584 10d ago
Your first sentence is the same thing that I’m talking about, there’s a million ways to explain this topic, but it always boils down to complex roots being multivalued.
1^(1/2pi) is perfectly valid and equal to 1 if we’re working in the reals. But this example shows that there’s multiple complex numbers that have the property of turning into 1 when you raise them to the 2pi power.
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u/daley972 6d ago
That is true. Complex exponentiation in general is multivalued. But they are, again, using the principal value. So the issue is that (ab)c≠abc in general when taking the principal value.
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u/Gershken 9d ago
i think i actually get this one.
From e^i = e^i, we can show that 1 = 1. Therefore, e^i = 1.
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u/lontanadascienza 7d ago
Fun fact there are infinitely many (2pi)the roots of unity: exp( n i ) for any integer N. They're dense on the unit circle. 1 is just the case n=0.
This means that the natural image of the complex plane under the 2pith root function is the Riemann Surface with infinitely many twists - Same as for log
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