Tricky differentiaton 1

mfgoode · 2006-01-20, 16:02

∞
. .
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

Ken_g6 · 2006-01-20, 17:18

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

alpertron · 2006-01-20, 17:52

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

$\large y=x^y$

$\large \log y = y \log \,x$

$\large \frac {y'}{y} = y' \log x + \frac {y}{x}$

$\large y'(\frac {1}{y} - \log \,x) = \frac {y}{x}$

$\large xy' \frac {1 - y \log \,x}{y} = y$

$\large xy' = \frac {y^2}{1 - y \log \,x}$

$\large xy' = \frac {y^2}{1 - \log \,x^y}$

From the first equation:

$\large xy' = \frac {y^2}{1 - \log \,y}$

as stated by Mally.

alpertron · 2006-01-20, 18:05

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

R.D. Silverman · 2006-01-20, 18:40

Quote:

Originally Posted by alpertron

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

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.

alpertron · 2006-01-20, 19:01

Quote:

Originally Posted by R.D. Silverman

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

Well, this is another exercise...

Quote:

Originally Posted by R.D. Silverman

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

Obviously, for x = 2 we have 2^2^2^2^2^... which does not converge.

Quote:

Originally Posted by R.D. Silverman

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

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^1/e

So for x < e^1/e the function should be well defined.

alpertron · 2006-01-20, 21:21

Obviously x>0, so the domain is: 0 < x < e^1/e.

R.D. Silverman · 2006-01-20, 21:29

Quote:

Originally Posted by alpertron

Obviously x>0, so the domain is: 0 < x < e^1/e.

Try again.

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

alpertron · 2006-01-20, 21:45

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

Anyway, I've just written an UBASIC program, and it finds that for both 10⁶ and 10⁷ 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.

alpertron · 2006-01-20, 21:59

But 0.001^0.9927645... is not 0.9927645..., so I will have to continue with this problem.

alpertron · 2006-01-20, 23:40

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.

$\large x^{y+r\, dx} = y+s\, dx$

$\large x^{y} x^{r dx} = y+s\, dx$

$\large x^y (1+r \,\log \,x \,dx) = y+s\, dx$

$\large yr \,\log \,x = s$

$\large s/r = y \,\log \,x$

$\large s/r = \log \,y$

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

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

So the domain is $e^{-e} < x < e^{1/e}$

2006-01-20, 16:02	#1
mfgoode Bronze Medalist Jan 2004 Mumbai,India 2²×3³×19 Posts	Tricky differentiaton 1 ∞ . . 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

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Lucas-like sequence of polynomials: a tricky thing	XYYXF	Math	7	2010-08-27 11:52
Tricky constant acceleration question	lavalamp	Puzzles	29	2009-12-14 15:27
A tricky way to estimate pi(x)	XYYXF	Math	13	2006-09-01 15:44
Tricky Problem with AMD 3700+	pseudorandom	Hardware	11	2006-01-23 08:53

2006-01-20, 17:18	#2
Ken_g6 Jan 2005 Caught in a sieve 5·79 Posts	Anyone notice how more and more of these puzzles are looking like homework? :surprised

2006-01-20, 17:52	#3
alpertron Aug 2002 Buenos Aires, Argentina 1,523 Posts	I don't think this is a homework, except if it is his grandson's. $\large y=x^y$ $\large \log y = y \log \,x$ $\large \frac {y'}{y} = y' \log x + \frac {y}{x}$ $\large y'(\frac {1}{y} - \log \,x) = \frac {y}{x}$ $\large xy' \frac {1 - y \log \,x}{y} = y$ $\large xy' = \frac {y^2}{1 - y \log \,x}$ $\large xy' = \frac {y^2}{1 - \log \,x^y}$ From the first equation: $\large xy' = \frac {y^2}{1 - \log \,y}$ as stated by Mally.

2006-01-20, 18:05	#4
alpertron Aug 2002 Buenos Aires, Argentina 1,523 Posts	The only tricky part in this exercise is to understand why Mally thinks that this is tricky.

2006-01-20, 21:21	#7
alpertron Aug 2002 Buenos Aires, Argentina 1,523 Posts	Obviously x>0, so the domain is: 0 < x < e^1/e.

2006-01-20, 21:45	#9
alpertron Aug 2002 Buenos Aires, Argentina 1,523 Posts	From the differential equation I don't see another restriction. Anyway, I've just written an UBASIC program, and it finds that for both 10⁶ and 10⁷ 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.

2006-01-20, 21:59	#10
alpertron Aug 2002 Buenos Aires, Argentina 1,523 Posts	But 0.001^0.9927645... is not 0.9927645..., so I will have to continue with this problem.

2006-01-20, 23:40	#11
alpertron Aug 2002 Buenos Aires, Argentina 5F3₁₆ Posts	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. $\large x^{y+r\, dx} = y+s\, dx$ $\large x^{y} x^{r dx} = y+s\, dx$ $\large x^y (1+r \,\log \,x \,dx) = y+s\, dx$ $\large yr \,\log \,x = s$ $\large s/r = y \,\log \,x$ $\large s/r = \log \,y$ s/r = 1 implies $y = e \rightarrow x = e^{1/e}$ as shown above. s/r = -1 implies $y = 1/e \rightarrow x = e^{-e}$ because $x^y = (e^{-e})^{1/e} = e^{-1} = y$ So the domain is $e^{-e} < x < e^{1/e}$