this post was submitted on 01 Sep 2024
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[–] AmbiguousProps@lemmy.today 8 points 2 months ago (1 children)

Can't wait to see the new images.

[–] xilliah@beehaw.org 2 points 2 months ago* (last edited 2 months ago) (1 children)

Could you use this to take a picture of an exo planet?

Also, who will we send in first? How about the orange man so we can get some data on redshift.

[–] OmnipotentEntity@beehaw.org 5 points 2 months ago* (last edited 2 months ago) (1 children)

Well, 13 microarcseconds is the resolution they claim to be shooting for. The nearest star is 4.2 light-years away. 13 microarcseconds at 4.2 light-years is 2500km, the earth is about 12742 km in diameter. So we can theoretically take an approximately 5x5 pixel image of Proxima Centauri b.

[–] xilliah@beehaw.org 2 points 2 months ago (1 children)

Wait, are you from star trek?

[–] OmnipotentEntity@beehaw.org 3 points 2 months ago (1 children)

No, I just understand math. So yes.

[–] xilliah@beehaw.org 2 points 2 months ago (1 children)

If you'd entertain me, why is that god formula such a big deal?

[–] OmnipotentEntity@beehaw.org 3 points 2 months ago (1 children)
[–] xilliah@beehaw.org 3 points 2 months ago* (last edited 2 months ago) (1 children)

Apologies, I was referring to Euler's identity.

What do you make of it? I'm just curious :)

[–] OmnipotentEntity@beehaw.org 3 points 2 months ago* (last edited 2 months ago) (1 children)

Oh, I'll try to describe Euler's formula in a way that is intuitive, and maybe you could have come up with it too.

So one way to think about complex numbers, and perhaps an intuitive one, is as a generalization of "positiveness" and "negativeness" from a binary to a continuous thing. Notice that if we multiply -1 with -1 we get 1, so we might think that maybe we don't have a straight line of positiveness and negativeness, but perhaps it is periodic in some manner.

We can envision that perhaps the imaginary unit, i, is "halfway between" positive and negative, because if we think about what √(-1) could possibly be, the only thing that makes sense is it's some form of 1 where you have to use it twice to make something negative instead of just once. Then it stands to reason that √i is "halfway between" i and 1 in this scale of positive and negative.

If we figure out what number √i we get √2/2 + √2/2 i

(We can find this by saying (a + bi)^(2) = i, which gives us (a^(2) - b^(2) = 0 and 2ab = 1) we get a = b from the first, and a^(2) = 1/2)

The keen eyed observer might notice that this value is also equal to sin(45°) and we start to get some ideas about how all of the complex numbers with radius 1 might be somewhat special and carry their own amount of "positiveness" or "negativeness" that is somehow unique to it.

So let's represent these values with R ∠ θ where the θ represents the amount of positiveness or negativeness in some way.

Since we've observed that √i is located at the point 45° from the positive real axis, and i is on the imaginary axis, 90° from the positive real axis, and -1 is 180° from the positive real axis, and if we examine each of these we find that if we use cos to represent the real axis and sin to represent the imaginary axis. That's really neat. It means we can represent any complex number as R ∠ θ = cos θ + i sin θ.

What happens if we multiply two complex numbers in this form? Well, it turns out if you remember your trigonometry, you exactly get the angle addition formulas for sin and cos. So R ∠ θ * S ∠ φ = RS ∠ θ + φ. But wait a second. That's turning multiplication into an addition? Where have we seen something like this before? Exponent rules.

We have a^(n) * a^(m) = a^(n+m) what if, somehow, this angle formula is also an exponent in disguise?

Then you're learning calculus and you come across Taylor Series and you learn a funny thing, the Taylor series of e^x looks a lot like the Taylor series of sine and cosine.

And actually, if we look at the Taylor series for e^(ix) is exactly matches the Taylor series for cos x + i sin x. So our supposition was correct, it was an exponent in disguise. How wild. Finally we get:

R ∠ θ = Re^(iθ) = cos θ + i sin θ

[–] xilliah@beehaw.org 2 points 2 months ago