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/sci/ - Science & Math

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1944457 No.1944457 [Reply] [Original]

I think I found a general formula for <span class="math">f'(p)[/spoiler] where p is the number of times x is raised to itself. For example f(3) would be <span class="math">x^{x^{x}}[/spoiler]. The derivative seems to follow a nice pattern.

Discuss.

>> No.1944463

It's called Tetration.

http://en.wikipedia.org/wiki/Tetration

>> No.1944465

Methinks you should write <span class="math">f_p(x)[/spoiler] for clarity.

But do show. Should be easy with induction and motherfucking chain rule.

>> No.1944471

why does anyone give a fuck about tetration? As far as I can tell it's in the same shitheap as numerology.

>> No.1944477

Am I herping when I'm derping? How will you derivate f(4) when f(4.1) does not exist?

>> No.1944482

>>1944463
I'm specifically speaking about their derivatives.

>>1944465
Actually, no. You need logarithmic differentiation.

working on it... will delivar in a sec

>> No.1944484

>>1944477
It is possible because I'm not differentiation with respect to p, but with respect to x, and f(x) is not non-differentiable.

>> No.1944492

>>1944484
Oh sure, that's easy. Try f(x)=x⇈x.

>> No.1944494 [DELETED] 

>>1944482
I would do something like this. Since by definition
<span class="math">f_{n+1}(x) = x^{f_n(x)} = e^{f_n(x)*\ln x)[/spoiler]
<span class="math">f_{n+1}'(x) = e^{f_n(x)*\ln x)\cdot \frac{d/dx} f_n(x)\ln x = f_{n+1}(x) + \frac{f_n(x)}{x} + f'_n(x)\ln x[/spoiler]
Something like that.

>> No.1944500

>>1944482
I would do something like this. Since by definition
<span class="math">f_{n+1}(x) = x^{f_n(x)} = e^{f_n(x)\ln x}[/spoiler]
<span class="math">f_{n+1}'(x) = e^{f_n(x)\ln x}\cdot \frac{d}{dx} f_n(x)\ln x = f_{n+1}(x) + \frac{f_n(x)}{x} + f'_n(x)\ln x[/spoiler]
Something like that.

Now without missing close braces.

>> No.1944495 [DELETED] 

>>1944482
I would do something like this. Since by definition
<span class="math">f_{n+1}(x) = x^{f_n(x)} = e^{f_n(x)*\ln x}[/spoiler]
<span class="math">f_{n+1}'(x) = e^{f_n(x)*\ln x}\cdot \frac{d/dx} f_n(x)\ln x = f_{n+1}(x) + \frac{f_n(x)}{x} + f'_n(x)\ln x[/spoiler]
Something like that.

Now without missing close braces.

>> No.1944502

>>1944495
Looks legit. Can you do non-inductive form?

>> No.1944514 [DELETED] 

<span class="math">
\frac{d}{dx}x = 1
\frac{d}{dx}x^x = x^x(lnx+1)
\frac{d}{dx}x^x^x = x^x^x(x^{x-1}+x^xlnx(lnx+1)
\frac{d}{dx}x^x^x^x = x^x^x^x(x^{x^x-1}+x^x^xlnx(x^{x-1}+x^xlnx(lnx+1))
[/spoiler]

Hmm.. the previous answer appears in the next one...

>> No.1944528

<span class="math">\frac{d}{dx}x = 1 \frac{d}{dx}x^x = x^x(lnx+1)[/spoiler]
<span class="math">\frac{d}{dx}x^{x^{x}} = x^{x^{x}}(x^{x-1}+x^{x}lnx(lnx+1) [/spoiler]
<span class="math">\frac{d}{dx}x^{x^{x^{x}}} = x^{x^{x^{x}}}(x^{x^{x-1}}+x^{x^{x}}lnx(x^{x-1}+x^{x}lnx(lnx+1))[/spoiler]

>> No.1944535

>>1944528
Well the pattern is clear... how to state it?

>> No.1944540

>>1944502
I was wrong
<span class="math">f_{n+1}(x) + \frac{f_n(x)}{x} + f'_n(x)\ln x[/spoiler]
should be
<span class="math">f_{n+1}(x) \cdot\left( \frac{f_n(x)}{x} + f'_n(x)\ln x
\right[/spoiler].

The only thing you need to expand is <span class="math">f'_n(x)\ln x[/spoiler] but I don't want to... It's going to be messy.

>> No.1944542

>>1944540
Ah yes, collab ftw

>> No.1944546

>>1944535

series

>> No.1944563

>>1944540
correcting your latex (hopefully):
<span class="math">f_{n+1}(x) \cdot (\frac{f_n(x)}{x} + f'_n(x)\ln x)[/spoiler]

>> No.1944570

>>1944563
<span class="math">f_{n+1}(x) \cdot\left( \frac{f_n(x)}{x} + f'_n(x)\ln x \right)[/spoiler]
Actually. \left and \right. This board really needs preview.

>> No.1944569

<div class="math"> f_{\infty}'(x) = f_{\infty}^2(x) \left( \frac{1}{x ( 1 - f_{\infty}(x) \ln{x} ) } \right) </div>

weird thing is that f_\infty converges for some x greater than 1.

>> No.1944574

>>1944569
are you a wizard.jpg

>> No.1944585

>>1944570
same thing innit?

>> No.1944594
File: 13 KB, 363x364, 1282238405767.jpg [View same] [iqdb] [saucenao] [google]
1944594

>>1944585
Mine's prettier.

>> No.1944595

>>1944569
>>1944569
wait wut?

>> No.1944607

i think >>1944570 means /thread ?

>> No.1944614

>>1944595
>>1944595

<div class="math"> f_{\infty}(\sqrt{2}) = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} =2 </div>

>> No.1944633

>>1944614
Are you a wizard.jpg

>> No.1944640

>>1944614

Why does it converge?

>> No.1944648

>>1944640
Why is 1+1 equal to 2?

>> No.1944657

>>1944633
weirder still, there's a range of values less than 1 for which it diverges:

<div class="math"> x \in \left[0, e^{-e} \right] </div>

>> No.1944663

>>1944614
>>1944614
>>1944614
FALSE!!
(((sqrt2)^sqrt2)^sqrt2)^sqrt2 is 2.665

>> No.1944666
File: 4 KB, 121x126, snip.jpg [View same] [iqdb] [saucenao] [google]
1944666

>>1944657
Oh come on, wtf is this, divide by zero 2.0? I gettin mad nigga

>> No.1944672

>>1944663
yeh, 4 = infinity

>> No.1944675

>>1944672
>>1944672
No, but after that it will keep on increasing. So the limit (if there is one) is not 2 definitely.

>> No.1944681

>>1944675
keep going, trust me on this one.

>> No.1944682
File: 146 KB, 1280x2500, gal.png [View same] [iqdb] [saucenao] [google]
1944682

>>1944663
I love how much this thread is improving in quality, having begun from OP pic.

>> No.1944687

>>1944681
lol so you've been trolling all this time??!!!

>> No.1944692

>>1944682
wait what?? I just keep going up powers of two every other iteration.... ????

>> No.1944694

sqrt(1.9) = A
A^A^A^A^A^A^A^A^... reaches 1.75805..

WHY

>> No.1944693

>>1944681
actually i retract that, just realised your doin it wrong.

<div class="math"> f_{n+1}(x) = x^{f_n(x)} </div>
NOT
<div class="math"> f_{n+1}(x) = f_n(x)^x </div>

>> No.1944701

>>1944687
no, check on a calculator, enter this:

root(2) ^root(2)
root(2)^(answer)
repeat

>> No.1944712

>>1944693
I don't get the difference.

If n = 3: <span class="math">f_3(x) = {x^x}^x[/spoiler]
Then <span class="math">f_4(x) = {{x^x}^x}^x = (f_3(x))^x[/spoiler]

Please halp, I'm OP. I would really like to get this.

>> No.1944715

For those interested I'm not trolling, the reason it converges is because

<div class="math"> f_{\infty}(x) = -\frac{W(-\ln{x})}{\ln{x}} </div>

where W is the Lambert W function, the solution of the functional equation:

<div class="math"> W(x)e^{(W(x)} = x </div>

If you look at this function you will see why the range of convergence for f_infty is what it is, I won't ruin it for you.

>> No.1944719

>>1944701
I've been doing Answer^root2. I don't get the difference.

>> No.1944736

>>1944712
Just try a couple of examples and you'll see why the order of operation is important:

2^(2^(2^(2)))) = 2^(2^4) = 2^(16)

whereas, doing it incorrectly, i.e "bottom to top" we get:

(((2^2)^2)^2) = (4)^2^2 = (16)^2 = 256

which is a shit ton less than 2^16

>> No.1944752

>>1944719
see
>>1944736

>> No.1944754

>>1944736
>>1944736
But the order of operation dictates the "wrong" way I did it. Since it's all exponents, you just go left to right, right?

>> No.1944768

>>1944754
No. Go to wolframalpha and try it yourself:

enter 2^2^2^2 un-parenthesised and see what answer it gives.

You "read" it left to right, as in you read it

" two to the power of (two to the power of (two to the power of two))"

But you don't evaluate from "left to right" as that is incorrect.
If that makes sense

>> No.1944769
File: 28 KB, 600x214, QQ.jpg [View same] [iqdb] [saucenao] [google]
1944769

>>1944715
Oh wow, that was cool Anon, thanks. So then, what real-world phenomenon does this kind of behavior correspond to?

>mfw he applies mathematics

>> No.1944780

There are some certain enzyme-substrate kinematics that obey a x^(1/x) ~ y^(1/y) uptake rate.
As you can see, the f_infty function comes in very handy here.

>> No.1944781

>>1944780
aimed at >>1944769 sorry

>> No.1944783

>>1944768
Oh, thanks.

>> No.1944786

>>1944769
http://xkcd.com/435/
BITCH!

Regards, OP

>> No.1944824

>>1944614
why doesn't this work for sqrt3??

>> No.1944839

>>1944769
Furthermore, the Lambert function is analytic for a range of complex values, so using the analytic continuation of the log function (working on the principal branch) we can get nice results like

<div class="math"> i^{i^{i^{i^{\ldots}}}} = \frac{2i}{\pi} W \left( -\frac{i \pi}{2} \right) </div>

>> No.1944842

>>1944824
>>1944614
It doesn't seem to work for anything else.

>> No.1944865

>>1944824
because square root of 3 is not in the range of convergence.

try the cube root of 3 (i.e 3^(1/3)), it is in the range of convergence so we can write:

<div class="math"> (3^(\frac{1}{3}))^{(3^(\frac{1}{3}))^{(3^(\frac{1}{3}))^ {\ldots}}} = x </div>

If we do 3^(1/3) to the power of each side we get:

<div class="math"> (3^(\frac{1}{3}))^{(3^(\frac{1}{3}))^{(3^(\frac{1}{3}))^ {\ldots}}} = (3^(1/3))^x </div>

The left hand side is the same as before as the tower of exponents is and was infinite, therfore

<div class="math"> x = (3^(1/3))^x </div>
<div class="math"> x = 3 </div>

>> No.1944869

>>1944786
I wonder what that says about me for knowing it was purity without having to go to the link.

>> No.1944876

>>1944865
excuse that horrible formatting, you can see what it was supposed to be right?

>>1944842

It actually works for every x in the range [e^(-e) , e^(1/e)]

>> No.1944887

>>1944865
What is the range of convergence?

>> No.1944891

>>1944887
see
>>1944876

>> No.1944913

>>1944876
By "works," you mean converges, right?

Also, are you sure that converges to 3? If it does, it's awfully slow.

>> No.1944915

>>1944891
yeah hadnt refreshed

>> No.1944926

>>1944913
This???

>> No.1944935

>>1944913
Yeh. I said "works" because you had said "It doesn't seem to work.."

It doesn't converge to three I made a wooopsie sorry, converges to around: 2.48 or something like that

>> No.1944950

>>1944935
Exactly! But your logic seems right. It should converge to 3!? Maybe you did a divide-by-zero-equivalent in your infinite series manipulation?

>> No.1944952

>>1944648
You have to show it converges before actually finding its value.
Let's consider:

S = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + ......
S = 1 - ( 1 - 1 + 1 - 1 + 1 - 1 + ....)
S = 1 - S
S = 1/2

But also:
S = (1 - 1) + (1 - 1) + (1 - 1) + ....
S = 0

So 0 = 1/2
This contradiction comes from the fact that we assumed that S converged, which is false.

>> No.1944953

>>1944935
I guess I should clarify where I made a woopsie, feel like I'm confusing people.

The function converges if x is in the range I gave, but also it converges to a value in the range [1/e , e].

>> No.1944957

>>1944950
see>>1944953

>> No.1944967

>>1944952
>>1944952
Interesting!

>> No.1944974

>>1944953
And how did you work out the domain and range where it converges?

>> No.1945001
File: 17 KB, 604x587, plot.jpg [View same] [iqdb] [saucenao] [google]
1945001

>>1944974
I've given a hint ITT already, you only need to consider the function W(x) and where it converges.

To clear up any mess I've made, here's a plot showing why my method for evaluating f_infty(3^(1/3)) didn't work.

the f infty function kinda winds back on itself after it gets to e^1/e. Of course to be a function we can't let it do that so the winding bit is excluded (the bit in red)

The problem arose because if you look at 3^(1/3), you can see it hits the plot twice, once in the black range and once in the red. We only consider the black range, so the answer is 2.48.. or whatever. The red range gave 3, but it's nonsense.

To see why it would be silly to include the red range, repeat what I done before with the value 4^(1/4).

If you do as I done a few posts above you arrive at...

f(4^(1/4)) = 4

this is nonsense because 4^(1/4) = (2^2)^(1/4) = root(2)

and as we know, we already have f(root(2)) = 2.

In summary. The technique I used before is valid only if the answer it spits out is in the range [1/e , e]

>> No.1945017

>>1944974
p.s

I also arrived at the same domain and range of convergence using a different technique.
You need to use a few auxiliary results and functions (g(x) = x^(1/x) was one)

Basically split the number line into 4 regions:

1) 0<x<e^-e
2) e^-e <= x <= 1
3) 1 <= x <= e^(1/e)
4) x > e^(1/e)

Then though various methods I showed whether f converged or diverged on each domain. A few sequences and a bit of analysis was needed.

IIRC the domain in 3) required the implicit function theorem.

>> No.1945033

This thread... nerdgasm

OP signing out!