/sci/ - Science & Math

[Anonymous No.2553148]

Nature has been triforcing since the start of our universe

ITT: Nature and its awesomeness

[Anonymous No.2553155]

Close up of the first section

[Anonymous No.2553156]

people die when they stop living

[Anonymous No.2553169]

>>2553148

My head is full of billions and billions of fuck

[Anonymous No.2553187]

>>2553148

What is this supposed to be a picture of?

[Anonymous No.2553196]

>>2553155

Oh fuck fibonnacci all up in this thread

[Anonymous No.2553206]

Nature is beautiful

[Anonymous No.2553226]

>>2553206

[Anonymous No.2553237]

I see the dimensions of the squares follow the fibonacci sequence, but how are the color distributions determined?

[Anonymous No.2553244]

fibonacci is a hack, he didn't even come up with it

[Filenamefag !ZkcfZDEybM!!kn0zMgLJnqE No.2553254]

Just so everyone knows what it's called:
Sierpinski's Gasket

[Anonymous No.2553263]

>>2553237

You evidently take the previous square, stick it in the bottom right corner, "fill" above it with a new color, take the next previous square, stick it in, then the next previous square, stick it in, and on and on until you arrive at two blank squares.

Hey, look, with enough rearrangements and colorings, I can make something look cool and meaningful.

[Anonymous No.2553286]

>>2553254

Doesn't look like sierpinski's gasket to me

[Anonymous No.2553293]

>>2553263

Do it faggot

[Filenamefag !ZkcfZDEybM!!kn0zMgLJnqE No.2553320]

>>2553286
Yeah, nevermind, I wasn't paying attention to what it was actually doing in the picture. I had just made an assumption from the triforce reference in OP.

Ignore me.

[Anonymous No.2553331]

sierpinski gasket

[Anonymous No.2553351]

Op here

Did you know that if you take the sum of the squares of 2 consecutive fibonnaci numbers, you get another fibonnaci number?

25+64=89
64+169=233

This is what inspired me to make this. I wanted to find out why this was true.

[Anonymous No.2553372]

>>2553351

Woah,

Why is this true? WHY?

[Anonymous No.2553383]

>>2553372

I dunno, lol.

I made the picture to find out why and didn't find any answers.

[Anonymous No.2553409]

>>2553383
>>2553372
I think the answer is obvious.

Magnets.

[Anonymous No.2553421]

>>2553383 I dunno, lol.
I loled, such a build up for that.

[Anonymous No.2553476]

>>2553421

I try.

But seriously, can any /sci/entist explain why the sum of the squares of 2 consecutive fibonnaci numbers is another fibonnaci number?

[Mayor Jim Suttle !OMAHa/k25s No.2553481]

Nature: it's fuckin' hot

[Anonymous No.2553532]

Nested Fibonacci squares

[Anonymous No.2553550]

>>2553532

Colored

[Anonymous No.2553552]

>>2553476
Well, the nth Fibonnacci number is
<div class="math">F_n = {\phi^n - (-\phi)^{-n} \over \sqrt{5}}</div>
where
<div class="math">\phi = {\sqrt{5} + 1 \over 2}</div>
So
<div class="math">F_n^2 = {\phi^{2n} - 2(-1)^n + \phi^{-2n} \over 5}</div>
<div class="math">F_{n+1}^2 = {\phi^{2n+2} + 2(-1)^n + \phi^{-2n-2} \over 5}</div>
<div class="math">F_n^2 + F_{n+1}^2
= {\phi^{2n} + \phi^{2n+2} + \phi^{-2n-2} + \phi^{-2n} \over 5}
= \left({1 \over \phi} + \phi\right) {\phi^{2n+1} + \phi^{-2n-1} \over 5}
= \left({\sqrt{5} - 1 \over 2} + {\sqrt{5} - 1 \over 2}\right) {\phi^{2n+1} - (-\phi)^{-2n-1} \over 5}
= F_{2n+1}
</div>
I'm hoping there's a more intuitive explanation, though. Anyone know one?

[Anonymous No.2553565]

>>2553552
Sorry, last line should be
<div class="math">F_n^2 + F_{n+1}^2 = {\phi^{2n} + \phi^{2n+2} + \phi^{-2n-2} + \phi^{-2n} \over 5} = \left({1 \over \phi} + \phi\right) {\phi^{2n+1} + \phi^{-2n-1} \over 5} = \left({\sqrt{5} - 1 \over 2} + {\sqrt{5} + 1 \over 2}\right) {\phi^{2n+1} - (-\phi)^{-2n-1} \over 5} = F_{2n+1}</div>
(sign error in the <span class="math">{\sqrt{5} + 1 \over 2}[/spoiler])

[Anonymous No.2553561]

>>2553206
dont forget deadly.

very very deadly

thats a mask shes wearing underneath are poison claws that will kill a man in under 2 minuets.

[Mayor Jim Suttle !OMAHa/k25s No.2553597]

>>2553552
I think there's a slightly more elegant inductive proof. Since <span class="math">f_{2n+1}^2 = f_{n+1}^2 + f_{n}^2[/spoiler] is what we're looking for, you could probably do it that way. I'd prove it myself, but I'm at work :-/

[Anonymous No.2553626]

>>2553597

It's

f_{2n+1} = f_{n+1}^2 + f_{n}^2

only one side is squared

[Mayor Jim Suttle !OMAHa/k25s No.2553637]

>>2553626
Er, yes, you are correct. I don't know why I put that there.

Correct version of >>2553597: f_{2n+1} = f_{n+1}^2 + f_{n}^2

[Anonymous No.2553667]

>>2553187

It's a visual demonstration of how consecutive terms of the the Fibonacci Sequence approximate the golden ratio.

[Anonymous No.2553668]

>>2553481
can i get a high res of this, gonna make it a poster.

[Anonymous No.2553710]

OP here, I got some more Fibonacci madness

[Anonymous No.2553729]

>>2553552
>>2553565
Let's see if we can generalize this a bit. For any integers m and n:
<div class="math">F_m^2 = {\phi^{2m} - 2(-1)^m + \phi^{-2m} \over 5}</div>
<div class="math">(-1)^{m+n} F_n^2 = {(-1)^{m+n} \phi^{2n} - 2(-1)^m + \phi^{-2n} \over 5}</div>
<div class="math">F_m^2 - (-1)^{m+n} F_n^2
= {\phi^{2m} - (-1)^{m+n} \phi^{2n} - (-1)^{m+n} \phi^{-2n} + \phi^{-2m} \over 5}
= {\phi^{m-n} - (-1)^{m+n} \phi^{n-m} \over \sqrt{5}}
\cdot {\phi^{m+n} - (-1)^{m+n} \phi^{-m-n} \over \sqrt{5}}
= F_{m-n} F_{m+n}
</div>

[Anonymous No.2553731]

>>2553667

How do they do that?

[Anonymous No.2553747]

Full 15000x15000 pic of this one at

http://restorethewonders.com/cgi-bin/jesus1.jpg

Worth it when you look at the very center at the beginning of the fibonacci sequence.

[Anonymous No.2553749]

>everyone is refusing to call it God's ratio
lol butthurt scientists

[Anonymous No.2553765]

>>2553731
A linear recurrence relation like
<div class="math">F_{n+2} = F_{n+1} + F_n</div>
can be solved by substituting in
<div class="math">F_n = r^n</div>
and solving for r:
<div class="math">r^{n+2} = r^{n+1} + r^n</div>
<div class="math">r^2 = r + 1</div>
<div class="math">r = {1 \pm \sqrt{5} \over 2}</div>
that is, <span class="math">r = \phi[/spoiler] or <span class="math">r = {-1 \over \phi}[/spoiler].
The general solution depends on the starting terms and is a linear combination of the possible <span class="math">r^n[/spoiler] solutions. For the starting terms of the Fibonacci sequence, it's:
<div class="math">F_n = {\phi^n - (-\phi)^{-n} \over \sqrt{5}}</div>
For large n, the term proportional to <span class="math">(-\phi)^{-n}[/spoiler] dies out, and the terms approximate a geometric series with ratio <span class="math">\phi[/spoiler].

[Anonymous No.2553768]

[Anonymous No.2553774]

[Anonymous No.2553777]

>>2553731
Well it's a ratio, so you divide one by the other.

So 5/8 = .625 which is a goodish approximation for [ sqrt(5) - 1 ] / 2
And 8/5 = 1.6 which is a good approximation for [ sqrt(5) + 1 0 / 2

Of course the approximation gets better with larger fibonacci numbers.

This trick also works for converting between miles and kilometers; in this case, the lower number has to be miles and the higher number is kilometers. For example, 89 miles = 143.231616 kilometers according to the Google conversion.

[Anonymous No.2553784]

[Anonymous No.2553788]

And I suppose I should explain why
<div class="math">{1 \over \phi} = {\sqrt{5} - 1 \over 2}</div>
since I've used it several times without proof. It's fairly simple to show:
<div class="math">\phi^2 = \phi + 1</div>
<div class="math">\phi = 1 + {1 \over \phi}</div>
<div class="math">{1 \over \phi} = {\sqrt{5} + 1 \over 2} - 1</div>
<div class="math">{1 \over \phi} = {\sqrt{5} - 1 \over 2}</div>

[Anonymous No.2553848]

>>2553729
So, for example, the difference of any two next-to-consecutive Fibonacci numbers is a Fibonacci number, e.g.:
64 - 9 = 55
169 - 25 = 144
And the sum of any two next-to-next-to-consecutive Fibonacci numbers is twice a Fibonacci number:
4 + 64 = 68 = 2*34
9 + 169 = 178 = 2*89

[Anonymous No.2553881]

>>2553788
how does that one thing squared equal that one thing plus one?

[Anonymous No.2553900]

>>2553848

math is fucking crazy

[Anonymous No.2553910]

>>2553881
The golden ratio <span class="math">\phi[/spoiler] is the positive solution to <span class="math">x^2 = x + 1[/spoiler]. Work it out and you get <span class="math">1 \pm \sqrt{5} \over 2[/spoiler] of which the positive solution, <span class="math">1 + \sqrt{5} \over 2[/spoiler], is what we call the golden ratio.

[Anonymous No.2553943]

>>2553910
ah I see now

[Anonymous No.2553971]

Factoid: The sum of two consecutive Fibonacci numbers is another Fibonacci number.

>yfw

[Anonymous No.2553985]

Another similar identity is the difference of the squares of consecutive Fibonacci numbers:
<div class="math">F_{n+1}^2 - F_n^2</div>
In this case, it's easy to show that
<div class="math">F_{n+1}^2 - F_n^2
= (F_{n+1} - F_n) (F_{n+1} + F_n)
= F_{n-1} F_{n+2}</div>
It looks very similar to (but not a case of) the identity in >>2553729 which makes me suspect there's a more general identity encompassing them both.

[Anonymous No.2554368]

>>2553985
>>2553985


So what's this formula?

[Anonymous No.2555302]

Wat