math problem

[Haircut]

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?

For the purpose of this question then the order of operations is what your calculator performs it as.
If we have it the other way then (a^a)^a = a^(2a).
((a^a)^a)^a = (a^(2a))^a = a^(3a) and so on.

(((a^a)^a)^a... n times is equal to a^(na).
Thus taking the limit of n to infinity we have that this value is not finite for any a > 1 and the professor in the original question said that there was an answer greater than 1 where the sum was finite. This only exists when we use the first order of operations you posted.

 

[kyu614]

Originally posted by: Haircut

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?

For the purpose of this question then the order of operations is what your calculator performs it as.
If we have it the other way then (a^a)^a = a^(2a).
((a^a)^a)^a = (a^(2a))^a = a^(3a) and so on.

(((a^a)^a)^a... n times is equal to a^(na).
Thus taking the limit of n to infinity we have that this value is not finite for any a > 1 and the professor in the original question said that there was an answer greater than 1 where the sum was finite. This only exists when we use the first order of operations you posted.

umm thats not correct, (a^a)^a = a^(a*a) and not a^(a+a)
i.e. 3^3^3 = 19683 = 3^9 where as 3^(2*3)=729
the order of operation is what your calculus book says

 

[glugglug]

Another approach:

a = a^1 = a^(a^0) (obvious)

assume (((a^a)^a)^a)^.....^a = a^(a^n) where there are n a's listed total
(((a^a)^a)^a)^....^a [n+1 a's] = (a^(a^n))^a
= a^((a^n) * a)
= a^(a^(n+1))

by induction this works for any natural # n.

So a^a^a^a.... infinitely repeating = a^(a^∞ )

I assume you know if a>1, a^∞ = ∞
So a^(a^∞ ) = ∞

Edit: WTF.... ∞ ) gets replaced with a smiley if you don't have space in between.

 

[Haircut]

Originally posted by: kyu614

Originally posted by: Haircut

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?

For the purpose of this question then the order of operations is what your calculator performs it as.
If we have it the other way then (a^a)^a = a^(2a).
((a^a)^a)^a = (a^(2a))^a = a^(3a) and so on.

(((a^a)^a)^a... n times is equal to a^(na).
Thus taking the limit of n to infinity we have that this value is not finite for any a > 1 and the professor in the original question said that there was an answer greater than 1 where the sum was finite. This only exists when we use the first order of operations you posted.

umm thats not correct, (a^a)^a = a^(a*a) and not a^(a+a)
i.e. 3^3^3 = 19683 = 3^9 where as 3^(2*3)=729
the order of operation is what your calculus book says

OK, It was late and I was drunk when I posted that 🙂
It still makes no odds though, it would be a^(a^2) and a^(a^3) etc. which still tends to infinity for all a > 1

 

[kyu614]

Originally posted by: Haircut

Originally posted by: kyu614

Originally posted by: Haircut

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?

For the purpose of this question then the order of operations is what your calculator performs it as.
If we have it the other way then (a^a)^a = a^(2a).
((a^a)^a)^a = (a^(2a))^a = a^(3a) and so on.

(((a^a)^a)^a... n times is equal to a^(na).
Thus taking the limit of n to infinity we have that this value is not finite for any a > 1 and the professor in the original question said that there was an answer greater than 1 where the sum was finite. This only exists when we use the first order of operations you posted.

umm thats not correct, (a^a)^a = a^(a*a) and not a^(a+a)
i.e. 3^3^3 = 19683 = 3^9 where as 3^(2*3)=729
the order of operation is what your calculus book says

OK, It was late and I was drunk when I posted that 🙂
It still makes no odds though, it would be a^(a^2) and a^(a^3) etc. which still tends to infinity for all a > 1

it is not a^(a^2) and a^(a^3) etc... the exponent does not change, the base number is changing and with each operation it is being taken to the same power, which is a.
your notation suggests that the operation is a^(a^(a^a))), but the operation is ((a^a)^a)^a

assume (((a^a)^a)^a)^.....^a = a^(a^n) where there are n a's listed total

thats incorrect

 

[DerwenArtos12]

it si impossible because any number greater than one infinatly increases with each exponent. and as soon as you put any number under one ,other than one, it becomes smaller than one. it is impossible.

 

[glugglug]

Originally posted by: kyu614

assume (((a^a)^a)^a)^.....^a = a^(a^n) where there are n a's listed total

thats incorrect

Actually its not. Its an assumption thats part of an induction proof -- the case for n=1 is obvious, from that statement about n you can get it for n+1, so it works for any n.

Not completely clearly stated though, what I mean is
(((a^a)^a)^a)....^a = a^(a^n) where there are n a's on the left side of the =.

 

[Jeff7]

Originally posted by: 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

To answer that, wouldn't you need to know exactly at what point infinity begins? There's not really a defined line between finite quantities and infinity.
I'm thinking too that if a is anything greater than 1, then the infinite string of exponents would still yield an infinite answer.
Or maybe I should just try to figure out that "e^(1/e) which equates to around 1.444667" answer...I remember seeing that e thing in some math class, maybe Calculus. And if it was Calculus, that'd be why I didn't remember it. I was hardly able to pass that class; that stuff just made no sense to me.

 

[rchiu]

Originally posted by: 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

what about a=1.000000000......1, you didn't say anything about a is an int.

and number of decimal=definition of finite.....

Yeah, I am just guessing, this is not based on scientific fact