/sci/ - Science & Math

I would learn numeric analysis.

log(y)=x*log(x)

bump

>>3396395
this

The answer not expressible in the elementary functions in general case.

basically that equation is what defines x in terms of y. you can't write it in terms of logx or sinx, but if you think about it why does this matter? logx is itself simply defined as the solution of y = e^x. there is nothing special about it.

>>3396473
because if I could solve for a function then given any y I could find the x which raised to itself gives that y, and that is unique.

x y
1 1
2 4
3 27
4 256
5 3125

I dunno, fit a curve to the data and find your point x

take the x-root of each side.

you would have to make x>0 first, i guess.

>>3396490
thats how I started. But it is kind of funky between 0 and 1 (it is not always increasing, and has some sort of weird dip) and i was trying to find a function wherein I could solve/approximate by hand (most of the solutions are irrational, so it would likely be continuous approximation)

>>3396492
x = y^(1/x)?
how does this help?

>>3396507
Solve <span class="math">x^x=c[/spoiler] via Newton.

>>3396507
So, what exactly do you want?
Calculate value of x for any given y? Use numerical methods, for example, Newton method.
Get an approximation in a realtively small domain? Get enough points and try simple polynomial approximation.
Get an exact solution in elementary functions? No can do.

By the way, keep in mind that because of non-monotony of x^x, you get two answers for y lying in interval in <span class="math">[e^{1/e},1][/spoiler]

y=4
X=2

solved

You need that: https://secure.wikimedia.org/wikipedia/en/wiki/Lambert_W_function
So: x = log(y)/W(log(y))

>>3396528
>Get an exact solution
this.

>>3396528
>Calculate value of x for any given y?
and this

>>3396561
you don't understand what people are telling you.
y = e^x is an 'exact solution' of y = e^x, expressing a value of x for every (positive) y. we just write this as x = logy. that doesn't mean we've done anything special.

in the same way, y = x^x expresses a value of x for some y. write x = blob(y) if you want. this is exactly the same as what we do with log.

>>3396569
yes and if y is 1.5 how do I find x?

>>3396607
<div class="math">x=\exp(W_k(\ln(3/2)))</div>using the Lambert W function with branch choice <span class="math">k\in\{-1,0,1\}[/spoiler]. (Only <span class="math">k=0[/spoiler] is real)

>>3396607
Numerical methods, obviously.

I've got a similar question for you. How do you find <span class="math">e^{1.5}[/spoiler]? Or, for that matter, <span class="math">\sqrt{2}[/spoiler]?

I think about it this way, someone just comes to me and says "something is something else that is multiplied by something else something else times, how much is something else?" and I say "fuck you, I don't care".

>>3396632
Triangles on the number line.

>>3396649
Elaborate.

>>3396632
ok.
there's no function or equation I could use for any given y?

>>3396656
Right triangle with two legs measuring 1 cm, vertexes A, B, C. Right-angled at ABC. ABC on 0. Hypotenuse will measure sqrt(2). Then compass.

I'd go with Taylor series

http://www.wolframalpha.com/input/?i=y%3Dx^x

>>3396658
Again, Newton's method.
You know how to wikipedia, right?

>>3396660
Yes, really. Although, I think you'll have troubles calculating even first 5 digits of <span class="math">\sqrt{2}[/spoiler] that way, not to say 20 (I think, 27 digits is about the top you can achieve in a human lifespan).

<div class="math">
\left(\frac{x}{2},c_1\in \mathbb{Z}\land c_1\leq -2\land y=\left(-c_1\right){}^{c_1}\land x=c_1\right)
</div>or<div class="math">
(y=-1\land x=-1)
</div>or<div class="math">
\left(\frac{x+1}{2},c_1\in \mathbb{Z}\land c_1\leq -2\land y=-\left(-c_1\right){}^{c_1}\land x=c_1\right)
</div>or<div class="math">
(y=1\land x=1)
</div>or<div class="math">
\left(e^{-1/e}\leq y<1\land \left(x=e^{W(\log (y))}\lor x=e^{W_{-1}(\log (y))}\right)\right)
</div>or<div class="math">
\left(y>1\land x=e^{W(\log (y))}\right)
</div>

how would you differentiate x^x?

>>3396672
yes I do, and I'm righting down everything right now, Im just confirming and inquiring for all possible methods to solve.

>>3396684
It's one of the easiest to implement and fastest to converge methods I know.
I doubt anyone would advise you something different.

>>3396681
y=x^x=e^(xln(x))
y'=e^(xln(x))*(1+ln(x))=x^x(1+ln(x))