math problem

[Indolent]

My physics teacher asked this question in lecture the other day and would not give us the answer. It's been bugging me ever since. No, this isn't a homework or some type of extra credit question. I'm just curious about what the answer is.


if 2^2=4
2^2^2=16
2^2^2^2=256

Going in this same pattern, if you have y=a^a^a^a^a^a... and keep raising it to the "a"th power an infinite amount of times...

What is the largest falue for "a" where y is still finite?

He did say that the answer is not the obvious answer everyone was thinking of 1. It is greater than 1.

*edit* yeah....should have been 2 squared =4

 

[CTho9305]

well, (e^i)^infinity = 0 according to my TI-89, but unfortunately |e^i| = 1
(besides, complex numbers aren't ordered anyway)

 

[dougtran]

How does 2^2=8? Am I reading the ^ symbol to be an exponent sign?

 

[Shalmanese]

It cant be anything more than 1 since its a monotonically increasing function. for things that are less than 1, it should asymtote to 1. Maybe there is an asymtote for values larger than 1 but I doubt it.

 

[Indolent]

Originally posted by: Shalmanese
It cant be anything more than 1 since its a monotonically increasing function. for things that are less than 1, it should asymtote to 1. Maybe there is an asymtote for values larger than 1 but I doubt it.

He said it is greater than 1 but I don't see how either. I'll have to see if he'll cave in and give us the answer sometime.

 

[rtb21]

OK, I'm not very good at explaining but think about this...

sqrt(2) ^ 2 = 2

so let a = sqrt(2), then a ^ 2 = 2, working back from infinity starting with the very end (i.e. the one 'at' infinity) a = 2, you'd get y = 2.

now the end a = sqrt(2) < 2, so surely y = a^a^a^a^ etc. would be less than 2?

hmmm, not very well explained, but hopefully you get the idea?

 

[Shalmanese]

Its impossible now Ive had time to think it out. The two conditions for a limit of infinity is that a. each iteration must be breater than the last, b. the DIFFERENCE must be monotonically non-decreasing.

Now, assume that a > 1. let n be the nth iteration. for n = 1, n > 1 obviously, so n^n > n.

for n = k, the kth iteration = n, the k+1th iteration = n^a, the difference = n^a-n.
for n = k+1, the k+1th iteration = n^a, the k+2th iteration = (n^a)^a, the difference = ((n^a)^a-n^a)

now, we must prove that ((n^a)^a-n^a) > n^a-n for all a > 1.

uh, my brain is too tired to proceed from here but thats essentially what you have to prove.

Anyway. QED proof ny mathematical Induction.

 

[Haircut]

The answer is e^(1/e) which equates to around 1.444667
You can try to work out the convergence range of the Lambert-W function yourself which will give you the answer, or just look here for more info

 

[CTho9305]

Originally posted by: Haircut
The answer is e^(1/e) which equates to around 1.444667
You can try to work out the convergence range of the Lambert-W function yourself which will give you the answer, or just look here for more info

my TI-89 claims that (e^(e^-1))^infinity = infinity... what am I missing?

 

[Haircut]

(e^(e^-1))^infinity is infinity, your calculator is right.
That's not what we are asking here though.

x^infinity is infinite for all x>1

x^(x^(x^(x^(x^.... is not infinite for all x>1

(e^(e^-1))^((e^(e^-1))^((e^(e^-1))^... to infinity actually converges to e.

 

[kyu614]

Originally posted by: CTho9305

Originally posted by: Haircut
The answer is e^(1/e) which equates to around 1.444667
You can try to work out the convergence range of the Lambert-W function yourself which will give you the answer, or just look here for more info

my TI-89 claims that (e^(e^-1))^infinity = infinity... what am I missing?

Haircut: e^(1/e) wont work because its not the right form of y=a^a^a..., yours is in the form of a^(1/a)

CTho9305: your function is wrong, you are calculating y=(a^a)*(a^a)*(a^a)...
so is a=2, you are calculating y=2*2*2*2*2=32 for the fifth iteration and not
y=(2^2)^2^2^2 which is 65536

btw, Indolent. does the number have to be real?

 

[sao123]

Indeed I agree.
I took the derivative of the base case which would be x^x^x.

.d..[....(x)..]......[......(x).................].........[.............(2)................................]
---.[...(x)...]..=..[...(.x...+...x...-1.)..]..(*)..[....x(ln(x))....+....x(ln(x))....+1.....]
dx.[..(x)....]......[.(x)......................].........[..................................................]


I performed a graphing analysis on both the function and its 1st derivitive...

The derivitive is never zero or negative, therefore no asympotote exists which indicates infinate increasing without bounds instead of a maximum.

The minimum of the derivitave is approx sprt(2) / 2. Which would be the value for which the function x^x^x grows slowest, but it is still infinate.

I am also going to performed a similar analysis of x^x^x^x & x^x^x^x^x.

 

[Haircut]

kyu614, in my answer a = e^(1/e)

 

[kyu614]

Originally posted by: Haircut
kyu614, in my answer a = e^(1/e)

oh sorry. i misread

 

[Mattnum25]

It's probably worth pointing out at this point that
x^x^x does not equal (x^x)^x
it equals x^(x^x)

that should resolve any concerns about monotonicity in haircut's (correct) answer.

 

[Indolent]

Originally posted by: kyu614

Originally posted by: CTho9305

Originally posted by: Haircut
The answer is e^(1/e) which equates to around 1.444667
You can try to work out the convergence range of the Lambert-W function yourself which will give you the answer, or just look here for more info

my TI-89 claims that (e^(e^-1))^infinity = infinity... what am I missing?

Haircut: e^(1/e) wont work because its not the right form of y=a^a^a..., yours is in the form of a^(1/a)

CTho9305: your function is wrong, you are calculating y=(a^a)*(a^a)*(a^a)...
so is a=2, you are calculating y=2*2*2*2*2=32 for the fifth iteration and not
y=(2^2)^2^2^2 which is 65536

btw, Indolent. does the number have to be real?

Yeah, has to be real. Haircut's answer looks good. I'll try finding out next time I have class.

 

[MainFramed]

whoa

 

[sao123]

I must apologize for my analysis, I fell into the pitfall of incorrect order of operations.

After further analysis, the series y=(a^(a^(a^(a^....))))...is convergent on the interval e^-e < x < e^(1/e).

 

[TheMasterGhost]

Originally posted by: Mattnum25
It's probably worth pointing out at this point that
x^x^x does not equal (x^x)^x
it equals x^(x^x)

that should resolve any concerns about monotonicity in haircut's (correct) answer.

based on Indolent definition of a^a^a^..... the correct format of x^x^x is not x^(x^x) but (x^x)^x
this can easily be shown by using his Indolent's example.

In his post he stated 2^2^2^2=256 and 2^(2^(2^2)) =65536 while ((2^2)^2)^2=256
therefore in this particular situation x^x^x is equal to (x^x)^x and not the way your calculator probably defines it as.

 

[glugglug]

y=a^a^a^a^a^a.....

so y^(1/a) = (a^a^a^a^a^a^a^a...)^1/a = y (^1/a just cancels out the last ^a)
y = y^a.

Assuming y is positive (ln is not defined for values <= 0)
ln(y) = ln(y^a) = a * ln(y)
a = ln(y) / ln(y) = 1.

 

[RaynorWolfcastle]

Originally posted by: glugglug
y=a^a^a^a^a^a.....

so y^(1/a) = (a^a^a^a^a^a^a^a...)^1/a = y (^1/a just cancels out the last ^a)
y = y^a.

Assuming y is positive (ln is not defined for values <= 0)
ln(y) = ln(y^a) = a * ln(y)
a = ln(y) / ln(y) = 1.

actually, ln is defined for values smaller than 0, it's just imaginary (and multivalued at that). Just an interesting fact I thought I'd put out there for those who didn't know, as a lot of teachers try to drill that into your head when you're younger

 

[glugglug]

good point....
e^(+/- i*pi) = -1, so ln(-1) = +/- i*pi
ln(-x) = ln(x) +/- i * pi
Math works out the same as for positive on this problem and you still get that a=1. (Unless y=0 which happens for a in [0;1) )

 

[kyu614]

Originally posted by: glugglug
y=a^a^a^a^a^a.....

so y^(1/a) = (a^a^a^a^a^a^a^a...)^1/a = y (^1/a just cancels out the last ^a)
y = y^a.

correct me if im wrong but that is not correct. when the ^1/a cancels out the last ^a, then you no longer have y unless you make the assumption that a=1 which is why you have the final answer being a=1. so you got your answer because you already set a=1

assuming y=a^a^a
y^(1/a) = (a^a^a)^1/a = a^a != y
so y^(1/a) != y unless a = 1.

note: != is does not equal

 

[sao123]

I am now thoroughly confused...

My calculator performs a^a^a^a... as a^(a^(a^(...^a))))))...
My calculus book defines a^a^a^a.... as ((((a^a)^a)^a)^a)^......

So which is the correct form of the order of operations?

 

[kyu614]

Originally posted by: sao123
I am now thoroughly confused...

My calculator performs a^a^a^a... as a^(a^(a^(...^a))))))...
My calculus book defines a^a^a^a.... as ((((a^a)^a)^a)^a)^......

So which is the correct form of the order of operations?

what calculator do you use? my TI-86 outputs 256 when i type in 2^2^2^2 which is consistent with your calculus def and the order of operations