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 mfgoode 2004-08-14 16:09

Unusual differentiation

How would you differentiate :rolleyes:

y = x^( x ) ^(x^)----------x----right up to infinity ?
In other words y =X to the power of x to the power of x---up to infinity ?

Mally :coffee: :smile:

 davar55 2004-08-14 16:20

The function is infinite for x > 1,
is not well defined for x <= 0,
and is constantly 1 for 0 < x <= 1.

Hence the derivative is only defined for 0 < x < 1,
and equals zero there !!

 JuanTutors 2004-08-14 22:11

[QUOTE=mfgoode]How would you differentiate :rolleyes:

y = x^( x ) ^(x^)----------x----right up to infinity ?
In other words y =X to the power of x to the power of x---up to infinity ?

Mally :coffee: :smile:[/QUOTE]
For 0<x<1, this can't converge to 1. notice that y=x^y. So if, say for x=1/2, we would have that 1=y=(1/2)^1=1/2.

I'm not much into complex analysis, so I don't know all the places that this function converges. However, here is a hint on taking the derivative:

Let f[x] := x^f[x], where x^f[x] is defined as x^f[x] := exp[f[x]*Log[x]], i.e. the number e=2.71828... to the power of f[x]*Log[x], where Log[x] is the natural log. Take the derivative, and solve for df/dx

This has a fairly nice answer, and is related to the Lambert W-function. (I really don't know anything about it, though :innocent: )

 davar55 2004-08-15 16:16

^^ Oops. That post was incorrect.

If x == sqrt(2) for example, then x^x^x^... == 2

Hence the range of convergence is wider than given there,
and the function is not constant in that range.

 jinydu 2004-08-16 03:08

I do know how to differentiate f(x) = x^x. Maybe that can help:

f(x) = x^x = (e^ln x)^x = e^(x*ln x)

f '(x) = x*ln x * e^(x*ln x) = x*ln x * x^x = (x^x) * x * ln x

 jinydu 2004-08-16 03:45

Sorry, I made a mistake with my last post. The key property is:

If y = e^(f[x]), then dy/dx = f '(x) * e^(f[x])

So here's the corrected derivation:

f(x) = x^x = (e^ln x)^x = e^(x*ln x)

Let g(x) = x*ln x

g '(x) = (x * 1/x) + (ln x) = 1 + ln x

f '(x) = (1 + ln x) * e^(x*ln x) = (1 + ln x) * x^x

Now, I'll try to differentiate f(x) = x^(x^x):

f(x) = x^(x^x) = (e^ln x)^(x^x) = e^(ln x * x^x)

Let g(x) = x^x * ln x

g '(x) = (x^x * 1/x) + (ln x) * (1 + ln x) * x^x
g '(x) = ([x^x]/x) + ([x^x * ln x] * [1 + ln x])

f '(x) = ([x^x]/x) + ([x^x * ln x] * [1 + ln x]) * e^(ln x * x^x)
f '(x) = ([x^x]/x) + ([x^x * ln x] * [1 + ln x]) * x^(x^x)

Someone may want to check this. Obviously x^(x^(x^x)) is likely to be even more complicated.

 mfgoode 2004-08-18 15:49

Unusual differentiation

[QUOTE=mfgoode]How would you differentiate :rolleyes:

y = x^( x ) ^(x^)----------x----right up to infinity ?
In other words y =X to the power of x , to the power of x---up to infinity ?

Mally :coffee: :smile:[/QUOTE]

Thank you one and all for the interest shown in this problem and the keen insight in tackling it.

The Solution: Let Y = x^x^x^X--------- to infinity. :smile:
Therefore y =x^y -------- 0 <x<=1
Taking logs logy =ylogx
Hence log y/y=logx
Hence dx/dy = [ y*(1/y) -logy*1]/ (y^2) Quotient rule y not=0

Therefore dy/dx= (y^2)/ 1-logy
Now y tends to 1 in this range as given above
Therefore dy/dx = 1 as log1 =0

This gives the slope as 1 (tan inverse 1=45*) i.e. m=1
Hence this is the eqn. of a straight line bisecting the angle (90*) in 1st. quadrant
therefore eqn of line is y=x [y=mx+c and c=0]
I remain open for further discussion on this problem :rolleyes:

Mally :coffee:

 jinydu 2004-08-19 14:40

I assume you're saying that f' (x) = x. If that is the case, then f(x) = 0.5x^2 + c, where c is some constant.

I learned in my math class that integration is the reverse of differentiation. Also, when you integrate a function, you get a whole family of functions, which differ from each other only by a constant.

Thus, if your reasoning is correct, x^x^x^x^x... = 0.5x^2 + c.

That would be a surprising result.

 mfgoode 2004-08-19 16:40

Unusual differentiation

:smile:
You are making the wrong assumption so naturally you get a conflicting answer!
PLease re-read my post. It is better to use leibniz's notation. Its less confusing.
The diff. coeff. is 1 and not x
Thus dy/dx =1
integrating y=x+c. put c=0 as the eqn is derived to be y=x
Therefore y=x
Now if we integrate once again we get integral ydx=(x^2)/2+c and put c=0
therefore Integral ydx =(x^2)/2. The limits we set at 0 to 1 now draw the line y=x
The integral denotes the area of a triangle formed by points (0,0) (1,0) and (1,1). The area of this triangle is 1/2(base * height) =(1*1)/2=1/2
Compare with the integral above viz. (x^2)/2. Here x =1, so area =1/2.
Therefore both results from (Calculus and co-ord geom) tally Q.E.D
:confused :question:
I am open for further questions

Mally :coffee:

 Zeta-Flux 2004-08-19 18:47

The function x^{x^{x^{x...}}} converges (for real numbers) on the interval [e^{-e}, e^{1/e}]. You might want to google the term "infinite exponential."

mfgoode, you seem to have differentiated log(x) wrong.

Best,
Zeta-Flux

 mfgoode 2004-08-21 16:11

Unusual diferentiation