Star trek w/ some personal

Mar 13, 2023 4:41 AM

Merdock

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871

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Last summer my aunt and I got into an argument about math at a family reunion. She's just retired from her long career as a math professor at a university and I'm a newbie engineer. The core of the issue was the equation √(x^2)=|x|. This is because, as she put it, unless stated otherwise we just want the positive value from the square root function rather than both the positive and negative value.

Last night something triggered my OCD mind to rethink the conversation and realized that if the solution would be a complex number (combination of real and imaginary), then this presents a problem that did not occur to me previously.

I fucking love my aunt and love having these conversations with her.

As a thank you for putting up with that, here's some star trek memes

I recently started a new binge of star trek

How do these ships take so much damage and yet the bridge almost always looks pristine

I feel ya O'Brien

Cat tax: Quinn (left) and his sister Claudia. They are definitely siblings.

Doggo tax: Wolfrik enjoying some recent snow. He loves snow more than anything in the world. He had one surgery, attempting to get a second for something that was missed in the first. His spirits are up and he will be ok however things move forward from here.

star_trek

dogtax

cat_tax

math

Soooo, is the Star Trek Adventures RPG by Modiphius any good? Seems like this thread would be a good one to ask in...

3 years ago | Likes 3 Dislikes 0

Congratulations on the math, but mostly on the fur babies. Mostly b/c of the 2, my brain only produces the happy chemical for the one.

3 years ago | Likes 3 Dislikes 0

Given that all these are jokes, you may think #5 is as well. But no no. That’s all Garak, baby! ?

3 years ago | Likes 4 Dislikes 0

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I'm quite glad that complex numbers reside in the world of things I had to learn but haven't needed to remember.

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So what is it?

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#10 Imaginary fucks are real. In fact, that's about the only way I get to experience them.

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v

3 years ago | Likes 36 Dislikes 0

"I believe that's the first totally honest thing you've ever said to me."

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Fools errand it might have been, I'm glad all the same that you took it with me.

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#8 If I steal, I upvote. Take your damn +1.

3 years ago | Likes 3 Dislikes 0

Currently I'm high on shrooms and am just tickled pink by all of the post but I'll come back in the morning because legit wanna see the end

3 years ago | Likes 2 Dislikes 0

of the sqrt(i) discussion

3 years ago | Likes 2 Dislikes 0

Enjoy your evening. I'll gladly talk about math if you talk about shrooms. I've never had the pleasure of trying them before.

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A starship for ants is brilliant.

3 years ago | Likes 2 Dislikes 0

#7

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ﻼƆUᆿ ƎHT TAHW HƆTIઘ

3 years ago | Likes 1 Dislikes 0

M/AM reactor excites exotic plasma, vented through nacelles to energize coils that warp/bend spacetime as an Alcubierre drive.

3 years ago | Likes 1 Dislikes 0

My point. It's not really space magic, although IIRC, the warp field creates a null mass pocket that can be accelerated indefinitely past C.

3 years ago | Likes 1 Dislikes 0

Anyway, I know later treks were real bad about technobabble, but TNG did a good job expanding past moral/ethical into data driven ideas.

3 years ago | Likes 1 Dislikes 0

Mass Effect series creates zero/negative mass ships that can zoom, but I thought Trek just bends space like Alcubierre's proposed drive.

3 years ago | Likes 1 Dislikes 0

Though the Impulse engines are as described, recusing effective mass to allow fusion drives to have greater acceleration.

3 years ago | Likes 1 Dislikes 0

@op the square root of i is the quad root of -1. or (-1)^(1/4) no?

3 years ago | Likes 3 Dislikes 0

I also love to piss off my math major drinking buddy with shit like, (d/dx)*x = (d/d)*(x/x) = 1

3 years ago | Likes 2 Dislikes 1

If f(x) = x, then (d/dx)f(x) does in fact equal 1... You were very close to the insight: (d/dx)*x = (dx/dx) = 1

3 years ago | Likes 2 Dislikes 0

Assuming "the" (singular) is a mistake. All roots of i are fourth roots of -1. Roots of z^4=-1 are +-1/rt2 for both real & im in sum.

3 years ago | Likes 2 Dislikes 0

z^2=i has 2 roots (a+ai, -a-ai), z^4=-1 has 4 (a+ai, -a+ai, a-ai, -a-ai). a =1/root2.

3 years ago | Likes 1 Dislikes 0

It is...

3 years ago | Likes 2 Dislikes 0

no differentiation is needed the. calc has become useless. newton and mexwell, eat me lol.

3 years ago | Likes 2 Dislikes 0

The first one is questionable since the derivative of |x| is not continuous, while sqrt(x^2) is. So the equation is a tricky one ;)

3 years ago | Likes 1 Dislikes 1

#1 ugh. I really should just remember the answer to this. I have forgotten most of what I learned in trigonometry.

3 years ago | Likes 1 Dislikes 0

To be fair... the square root of x² is not |x| in the slightest, which your high school math class should have taught you.

3 years ago | Likes 1 Dislikes 1

And if "my high school math class never taught me that", then maybe you live in one of the disappointing countries.

3 years ago | Likes 1 Dislikes 1

Oh my God Wolfrik, that's a fuckin amazing name and they're adorable

3 years ago | Likes 2 Dislikes 0

Nerd! Hah! J/k. I love this so much.

3 years ago | Likes 2 Dislikes 0

The negative results may not always by physically possible, but I'd much rather include ± than be beaten to death with my own skull.

3 years ago | Likes 1 Dislikes 0

Your doggo image -- when I was scrolling, the screen stopped with his eyes and ears above. Looked like a bovine skull, and I'm thinking WTF?

3 years ago | Likes 2 Dislikes 0

As someone who likely has dyscalculia I'll take your word for it.

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v

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I'm so happy that I've never been in a situation where I had to understand anything of what you just wrote.

3 years ago | Likes 1 Dislikes 0

#10 I'm sure at least one of them has experienced a negative fuck in their life.

3 years ago | Likes 1 Dislikes 0

the Vulcan might have but only gets to have any once every 7 years so will say it was at least agreeable

3 years ago | Likes 2 Dislikes 0

√(x^2)=|x| is ONLY true when x is a real number. if X is complex then the its different in part because |x| means something different there

3 years ago | Likes 5 Dislikes 0

I understood some of those words

3 years ago | Likes 1 Dislikes 0

I feel like √ means something different too, right, because typically √ is an operation that is only defined on positive real numbers?

3 years ago | Likes 2 Dislikes 0

no, it IS defined for real NEGATIVE numbers also. i.e. ax+i, or a real part coupled with an imaginary part.

3 years ago | Likes 1 Dislikes 0

for negative real numbers x= -k; k>0 √ x = i√ k

3 years ago | Likes 1 Dislikes 0

Root as an operation takes on the context of its domain. There is ambiguity if the range is just the positive root.

3 years ago | Likes 1 Dislikes 0

pretty sure its √ c = c^1/2 for Complex c. and of course c=k(e^(ni)) for some real k and n so √ c = k(e^((ni)/2))

3 years ago | Likes 1 Dislikes 0

I believe it should be √c = √(k)(e^((ni)/2). Though I typically use c =k /_ n° so √c = √k /_ (n/2)°

3 years ago | Likes 1 Dislikes 0

you're right. id probably have seen that if I were writing it out or doing it in Latex or something, harder to spot here with all the parens

3 years ago | Likes 1 Dislikes 0

#1 Yes, it's an "imaginary numbers are special" scenario. Math relies heavily on context. Unless there's reason to assume otherwise, we…

3 years ago | Likes 8 Dislikes 0

…generally assume a domain of ℝ. (The presence of i in an equation—as in your second example—would be one reason to assume otherwise.)

3 years ago | Likes 7 Dislikes 0

Correct me if I'm wrong, but the equation sqrt(x^2)=|x| still works even in the case of sqrt(i) since in that case x=sqrt(i), aka -1.

3 years ago | Likes 2 Dislikes 0

No, i^2 =-1. Sqrt(i) = (1+i)/sqrt(2). If sqrt(x^2)=|x| for complex numbers, then sqrt(i)=1. But 1^2 does not equal i

3 years ago | Likes 1 Dislikes 0

Ah right, I had that backwards for some reason.

3 years ago | Likes 2 Dislikes 0

I think you're right, and the aunt should say √(x^2)=|x|, x ∈ ℝ. On the complex plane, the √ and |x| operators mean something different.

3 years ago | Likes 4 Dislikes 0

I agree this works on the real plane, but I think if you want something equivalent on the complex plane you would need the radial sqrt eq.

3 years ago | Likes 1 Dislikes 0

√(x^2 /_ 2y°) = |x| /_ y°

3 years ago | Likes 1 Dislikes 0

That's interesting, I had always thought of abs as strictly distance from 0. Thank you!

3 years ago | Likes 2 Dislikes 0

The problem is you're treating i as a number, not a dimension. As NineLongWall eluded to, |x| is actually shorthand for "the distance of x".

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The most common measure of distance is the Euclidean distance, which is the square root of the sum of the squares of each component value of

3 years ago | Likes 1 Dislikes 0

x. That is to say, if x is multidimensional it will have component values associated with each of its dimensions that you sum the squares of

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and take the square root of. So, for a complex number, we see |c| = sqrt(r² + i²), r being the value of the real component, and i being the

3 years ago | Likes 1 Dislikes 0

value of the imaginary component. In your discussion, you are asking about the complex number 0 + 1i. Plugging into the formula, |0 + 1i| =

3 years ago | Likes 1 Dislikes 0

sqrt(0² + 1²), which is simply 1. Note that it is 1², and not i². i is simply a label, the dimension, not an actual value. 1 is the value.

3 years ago | Likes 1 Dislikes 0

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3 years ago (deleted Apr 4, 2023 9:10 AM) | Likes 0 Dislikes 0

I know Frakes was tall, but I'd have imagine Word to be the "Absolute Unit" here....

3 years ago | Likes 13 Dislikes 0

Worf*

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Word.

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Reminds me of that Russian version of return to treasure island cartoon

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Yeah it's a reference to dr livesey's phonk walk meme

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if you search Livesy Walk combined with any fandom you can think of you'll probably get a result

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What are they looking at on the ceiling?

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They struttin'

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Sqrt(i) = (1+i)/Sqrt(2).

3 years ago | Likes 50 Dislikes 1

Indeed, whereas the solution to x^2=i could be that or -(1+i)/sqrt(2).

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Dont forget -(1+i)/√2

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Oo yeah Sqrt all over my Imgur, daddy

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This comment made me laugh entirely too much.

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there are always two kinds of people

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And one of them is always something like a dwarf in a thong in a cask in the water

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Exactly, and the absolute value of that answer would be 1, which would make no sense as 1^2 does not equal i

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1) The absolute value of that is |√i | = |(1+i)/√2| = 1 indeed. You can also see it as |√i | = √((√i)²). I did this by simply following your

3 years ago | Likes 2 Dislikes 0

2) equation √(x²) = |x|. Where x = √i. So yeah, 1² ≠ i but that is totally unrelated here. So, what you are looking for is 1² = |√i|² = 1,

3 years ago | Likes 2 Dislikes 0

3) Which is true. Well perhaps I'm not getting at all where is the problem. Please explain the entire reasoning behind the "paradox".

3 years ago | Likes 1 Dislikes 0

There is no doubt that |√(i)|=1, it's that √(i)≠1, showing a counter-example to √(x^2)=|x|.

3 years ago | Likes 2 Dislikes 0

Instead of sqrt(x^2) = | x |, which isn't always true in C. A general statement would be sqrt(|x^2|)=|x|.

3 years ago | Likes 3 Dislikes 0

What part of that do you think doesn't make sense?

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I think a central issue here might be that we're abusing the idea of absolute value. What you're talking about is the magnitude

3 years ago | Likes 4 Dislikes 0

It seems the abs(x) that your aunt is referring to is just |x| = {x, for x >= 0; -x, for x < 0}, which is only the same as taking the

3 years ago | Likes 3 Dislikes 0

Magnitude if x is real, since real numbers only lie on 1 axis, the number line. On the complex plane, things are different.

3 years ago | Likes 2 Dislikes 0

If we're to take sqrt(x²) = |x| as meaning that sqrt() outputs the positive root (not the magnitude of the root) as standard, then

3 years ago | Likes 2 Dislikes 0

The absolute value of a complex number is its distance from 0 on the complex plane, |(1+i)/sqrt(2)| = 1

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It seems lot of the issue is i is being equated to x and not x^2. If it were then x^2=i so x=sqrt(i) and sqrt(x^2)=sqrt(sqrt(i))

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Yes, so the equation given in my post would mean that √(i)=1, which obviously isn't true. Hence her response to my question, "It's a trap"

3 years ago | Likes 15 Dislikes 1

The “equation” in your post is incorrect. It should read either: |sqrt(x^2)| = |x| Or: sqrt(x^2) = +/- x.

3 years ago | Likes 11 Dislikes 2

You have to either apply absolute to both sides, or indicate that there are two possible values of x

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In my previous discussion with her, I had several problems with said equation, but she brushed them off. I changed her opinion on it today

3 years ago | Likes 12 Dislikes 0

sqrt[sqrt(i)^2] = |sqrt(i)| ; sqrt(i) = |sqrt(i)| ; (1+i)/sqrt(2) = |sqrt(i)| ; how do you get 1? what am I missing here?

3 years ago | Likes 1 Dislikes 0

Is the notation ||x|| as in the L2 norm? Or is it abs as I have interpreted it as?

3 years ago | Likes 1 Dislikes 0

Ah because the terminology is garbage and changes definition from real to complex. Ha

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