Tricky differentiaton 1
 

:smile:

∞
. .
x^
x^
x^
If y = x^ (meaning x to the power of (ttpo) x ttpo x ttpo x ..... till infinity.

Prove that x*dy/dx = y^2/(1-logy)

Please dont 'spoilerise' your method
Mally :coffee:

Anyone notice how more and more of these puzzles are looking like homework? :surprised :whistle:

I don't think this is a homework, except if it is his grandson's.

[tex]\large y=x^y[/tex]

[tex]\large \log y = y \log \,x[/tex]

[tex]\large \frac {y'}{y} = y' \log x + \frac {y}{x}[/tex]

[tex]\large y'(\frac {1}{y} - \log \,x) = \frac {y}{x}[/tex]

[tex]\large xy' \frac {1 - y \log \,x}{y} = y[/tex]

[tex]\large xy' = \frac {y^2}{1 - y \log \,x}[/tex]

[tex]\large xy' = \frac {y^2}{1 - \log \,x^y}[/tex]

From the first equation:

[tex]\large xy' = \frac {y^2}{1 - \log \,y}[/tex]

as stated by Mally.

The only tricky part in this exercise is to understand why Mally thinks that this is tricky. :grin:

[QUOTE=alpertron]The only tricky part in this exercise is to understand why Mally thinks that this is tricky. :grin:[/QUOTE]

It is "tricky" because one must be careful about the domain in which
the answer is correct.

Hint: The infinite exponentiation does not converge everywhere.......

Bonus points: Find the domain for x where it does converge.

[QUOTE=R.D. Silverman]It is "tricky" because one must be careful about the domain in which
the answer is correct.[/QUOTE]
Well, this is another exercise...

[QUOTE=R.D. Silverman]Hint: The infinite exponentiation does not converge everywhere.......[/QUOTE]
Obviously, for x = 2 we have 2^2^2^2^2^... which does not converge.

[QUOTE=R.D. Silverman]Bonus points: Find the domain for x where it does converge.[/QUOTE]

We have to find where the derivative is infinite.

1 - log y = 0

y = e

x^y = y

y ln x = ln y = 1

e ln x = 1

ln x = 1/e

x = e[sup]1/e[/sup]

So for x < e[sup]1/e[/sup] the function should be well defined.

Obviously x>0, so the domain is: 0 < x < e[sup]1/e[/sup].

[QUOTE=alpertron]Obviously x>0, so the domain is: 0 < x < e[sup]1/e[/sup].[/QUOTE]

Try again.

Hint: it does not converge for x close to 0.......

From the differential equation I don't see another restriction.

Anyway, I've just written an UBASIC program, and it finds that for both 10[sup]6[/sup] and 10[sup]7[/sup] iterations, for x = 0.001 I get y = 0.9927645..., and for x=0.0001 I get y = 0.9990714... so it appears that for x->0, y->1.

But 0.001[sup]0.9927645...[/sup] is not 0.9927645..., so I will have to continue with this problem.

Since the sequence is oscillating between two values for small value of x, there is no limit.

This is my next attempt:

I have to find the value of x for which the sequence x^y tends to y. So we consider a value y+r ds which is near to the result. We will consider x^y = y. Applying the iteration x^(y+r dx) we will get a value y+s dx. If -1<s/r<1, then the sequence is convergent.

[tex]\large x^{y+r\, dx} = y+s\, dx[/tex]

[tex]\large x^{y} x^{r dx} = y+s\, dx[/tex]

[tex]\large x^y (1+r \,\log \,x \,dx) = y+s\, dx[/tex]

[tex]\large yr \,\log \,x = s[/tex]

[tex]\large s/r = y \,\log \,x[/tex]

[tex]\large s/r = \log \,y[/tex]

s/r = 1 implies [tex]y = e \rightarrow x = e^{1/e}[/tex] as shown above.

s/r = -1 implies [tex]y = 1/e \rightarrow x = e^{-e}[/tex] because [tex]x^y = (e^{-e})^{1/e} = e^{-1} = y[/tex]

So the domain is [tex]e^{-e} < x < e^{1/e}[/tex]