/sci/ - Science & Math » Thread #12620268

271KiB, 3000x2265, schizo_math.png

View Same Google iqdb SauceNAO

Complex Pitagoras

Anonymous Sun 24 Jan 00:18:21 2021 No.12620268 View Reply Original Report

Quoted By: >>12620287 >>12620293 >>12621507 >>12621529 >>12621532 >>12621614

if you extend the Pythagoras theorem to complex numbers you obtain bullshit result a triangle with zero-length hypotenuse, maths are developed by schizos

can /sci/ explain this shit?

Anonymous

Anonymous Sun 24 Jan 2021 00:22:22 No.12620287 Report

Quoted By:

>>12620268
Triangles do not have side numbers, triangles have side $\bold{lengths}$ . Lengths are non-negative reals only.

Anonymous

Anonymous Sun 24 Jan 2021 00:22:57 No.12620293 Report

Quoted By:

>>12620268
Incorrect metric. A metric defines how you measure lengths in a space and squaring complex numbers isn't how it's done. You multiply the number with it's conjugate (a+bi => a - bi) so i*(-i)=1, you get sqrt(2).

Anonymous

View Same Google iqdb SauceNAO 1603824455065.jpg, 71KiB, 620x675

Anonymous Sun 24 Jan 2021 01:28:11 No.12620492 Report

Quoted By: >>12621610 >>12622016

$a = i$
$b = 1$
$c = 0$
So:

$A = \frac{ab}{2} = \frac{i}{2}$
According to heron's formula, which is $\sqrt{s(s-a)(s-b)(s-c)}$ the area ends up $\frac{i}{2}$ again

$a = b\cos\gamma+\sqrt{c^2-b^2\sin^2\gamma}$
then
$i = 1 * 0 + \sqrt{0 - 1 * 1}$
$i = \sqrt{-1}$

$c = \sqrt{a^2 + b^2 - 2ab\cos\gamma}$
then
$0 = \sqrt{-1 + 1 - 0}$
$0 = 0$

by definition
$\tan\alpha = \frac{a}{b}$
$\tan\alpha = i$

but also
$\tan\alpha = \frac{a\sin\gamma}{b-a\cos\gamma}$
then
$\tan\alpha = \frac{i * 1}{1 - i * 0}$
$\tan\alpha = i$

There are a couple of divisions by zero but since this is schizo general they're not illegal just undefined.
Also the division by zero loses commutativity and associativity.

Anonymous

Anonymous Sun 24 Jan 2021 07:43:39 No.12621507 Report

Quoted By: >>12621519

>>12620268
This sort of triangle actually occurs in special relativity; the zero-length hypotenuse is what you call a "light-like interval" or "null interval" and it represents how something moving at the speed of light doesn't experience any time.

Anonymous

Anonymous Sun 24 Jan 2021 07:49:11 No.12621519 Report

Quoted By: >>12621527 >>12621535

>>12621507
god physicists are so fucking stupid

Anonymous

Anonymous Sun 24 Jan 2021 07:51:22 No.12621527 Report

Quoted By:

>>12621519
Its just a function, not a direct observable. Spacetime distance is not space distance.

Anonymous

Anonymous Sun 24 Jan 2021 07:52:31 No.12621529 Report

Quoted By:

>>12620268
Pythagorean theorem only applies on the euclidean metric, which is defined for real numbers. You can construct triangles with imaginary lengths, but Pythagorean theorem is no longer valid and now you need to use something like a Reimann metric to find the hypotenuse.

Anonymous

View Same Google iqdb SauceNAO fuck math.jpg, 126KiB, 1131x622

Anonymous Sun 24 Jan 2021 07:54:16 No.12621532 Report

Quoted By:

>>12620268

Anonymous

Anonymous Sun 24 Jan 2021 07:55:56 No.12621535 Report

Quoted By: >>12621546 >>12621617 >>12621620

>>12621519
Physicist here. That guy is an idiot. That triangle doesn't exist in SR. We define a Minkowski Metric. Then you can construct a triangle with legs of values t, and r. The spacetime interval is given by s^2 = t^2 - r^2. Diagrammatically, it's a meaningless triangle used by brainlets who don't understand math.

Anonymous

Anonymous Sun 24 Jan 2021 08:00:04 No.12621546 Report

Quoted By:

>>12621535
>That guy is an idiot.
>repeats exactly what I said in different words

Anonymous

Anonymous Sun 24 Jan 2021 08:35:31 No.12621610 Report

Quoted By:

>>12620492
$\tan(ix) \to i$ as $x \to \infty$

Anonymous

Anonymous Sun 24 Jan 2021 08:37:26 No.12621614 Report

Quoted By:

>>12620268
>can /sci/ explain this shit?
not every quadratic form is positive definite

Anonymous

Anonymous Sun 24 Jan 2021 08:39:53 No.12621617 Report

Quoted By:

>>12621535
Brainlet. Let (t, r) be (0, 0), (0, 1), and (1, 0) for three vertices of a triangle and compute s for all three sides.

Anonymous

Anonymous Sun 24 Jan 2021 08:41:19 No.12621620 Report

Quoted By:

>>12621535
what the fuck are you smoking, that guy is 100% right and you haven't said anything different than him

Anonymous

View Same Google iqdb SauceNAO imaginenotknowingcomplex.png, 19KiB, 800x536

Anonymous Sun 24 Jan 2021 11:40:10 No.12621894 Report

Quoted By:

I was writing an elaborate response when realized I was just being jebaited by sum dumbass.

Lemme just leave here a quate from Gauss about the subject and an image, both from Wikipedia.

"If one formerly contemplated this subject from a false point of view and therefore found a mysterious darkness, this is in large part attributable to clumsy terminology. Had one not called +1, ?1, ??1 positive, negative, or imaginary (or even impossible) units, but instead, say, direct, inverse, or lateral units, then there could scarcely have been talk of such darkness." ~ Gauss (1831)

Lastly fuck you OP.

Anonymous

View Same Google iqdb SauceNAO 1603321180172.jpg, 7KiB, 250x250

Anonymous Sun 24 Jan 2021 12:41:23 No.12622016 Report

Quoted By:

>>12620492
let $r$ be the radius of the inscribed circle, $r = \sqrt{\frac{1}{s}(s-a)(s-b)(s-c)}$ , where $r = \frac{1}{2}+\frac{i}{2}$
and $s$ the semi-perimeter $s = \frac{a+b+c}{2}$ , where $s = \frac{1}{2}+\frac{i}{2}$

From the law of cotangents we have
$\cot \frac{\alpha}{2} = \frac{s - a}{r}$

So we get
$\cot \frac{\alpha}{2} = \frac{\frac{1}{2}+\frac{i}{2} - i}{\frac{1}{2}+\frac{i}{2}}$
$\cot \frac{\alpha}{2} = -i$

Extending the law further:

$\cot \frac{\beta}{2} = \frac{\frac{1}{2}+\frac{i}{2} - 1}{\frac{1}{2}+\frac{i}{2}}$
$\cot \frac{\beta}{2} = i$

$\cot \frac{\gamma}{2} = \frac{\frac{1}{2}+\frac{i}{2} - 0}{\frac{1}{2}+\frac{i}{2}}$
$\cot \frac{\gamma}{2} = 1$

We can then conclude with the last identity which is
$\frac{\cot \frac{\alpha}{2}}{s - a} = \frac{\cot \frac{\beta}{2}}{s - b} = \frac{\cot \frac{\gamma}{2}}{s - c} = \frac{1}{r}$
which becomes
$\frac{-i}{s - a} = \frac{i}{s - b} = \frac{1}{s - c} = \frac{1}{\frac{1}{2}+\frac{i}{2}}$
which is proven to be true here

https://www.wolframalpha.com/input/?i=%28-i%2F%28s-a%29%29%3D%28i%2F%28s-b%29%29%3D%281%2F%28s-c%29%29%3D%281%2F%281%2F2%2Bi%2F2%29%29+where+a%3Di%2C+b%3D1%2C+c%3D0%2C+s%3D%281%2F2%2Bi%2F2%29

Anonymous

Anonymous Sun 24 Jan 2021 22:37:37 No.12624014 Report

Quoted By:

As you know
$\sin(ix) = i \sinh(x)$
$\cos(ix) = \cosh(x)$
$\tan(ix) = i \tanh(x)$
$\sin(x+iy) = \sin(x) \cosh(y) + i \cos(x) \sinh(y)$
$\cos(x+iy) = \cos(x) \cosh(y) - i \sin(x) \sinh(y)$
$\tan(x+iy) = \displaystyle \frac{\tan(x) + i \tanh(y)}{1 - i \tan(x) \tanh(y)}$

cos is finite for finite inputs, but it approaches complex infinity as the imaginary part of the angle approaches $\pm \infty$ , with which direction it goes in the complex plane dependent on the real part of the angle. Similarly for sin.

$\tan(x + iy) \to i$ as $y \to \infty$ and $\tan(x + iy) \to -i$ as $y \to -\infty$ .

We can't assign finite angle values to the two non-right angles of this triangle, but we do approach its trigonometric ratios when the angle opposite the side of length i has an imaginary part tending towards positive infinity and the angle opposite the side of length 1 has an imaginary part tending towards negative infinity.

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches