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 2004-08-14, 16:09 #1 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 22·33·19 Posts Unusual differentiation How would you differentiate 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
 2004-08-14, 16:20 #2 davar55     May 2004 New York City 419810 Posts 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 !!
2004-08-14, 22:11   #3
JuanTutors

Mar 2004

499 Posts

Quote:
 Originally Posted by mfgoode How would you differentiate 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
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 )

 2004-08-15, 16:16 #4 davar55     May 2004 New York City 2×2,099 Posts ^^ 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.
 2004-08-16, 03:08 #5 jinydu     Dec 2003 Hopefully Near M48 2·3·293 Posts 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 Last fiddled with by jinydu on 2004-08-16 at 03:09
 2004-08-16, 03:45 #6 jinydu     Dec 2003 Hopefully Near M48 2·3·293 Posts 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.
2004-08-18, 15:49   #7
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22·33·19 Posts
Unusual differentiation

Quote:
 Originally Posted by mfgoode How would you differentiate 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
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.
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

Mally

 2004-08-19, 14:40 #8 jinydu     Dec 2003 Hopefully Near M48 2×3×293 Posts 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.
 2004-08-19, 16:40 #9 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 40048 Posts Unusual differentiation Thank you for your query. 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 I am open for further questions Mally
 2004-08-19, 18:47 #10 Zeta-Flux     May 2003 5F216 Posts 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
 2004-08-21, 16:11 #11 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 22·33·19 Posts Unusual diferentiation Thank you for your astute observation on my derivation. Yes I left an 'x' out of differentiation. Since x =1 it does not affect the final conclusion. I will be highly obliged if you could elucidate on the two exponential limilts you have mentioned. What would the curve eqn. be and how did you arrive at the limits? It will be very interesting and valuable to me All the best, Mally

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