Math Question

[Lord and Master]

Does 3^3^3^3 = 3^(3*3)

No.

or is that something different?

Yes.

3^3^3^3= 3^(3*3*3)

No it doesn't.

3^3= 3*3*3= 3*9= 27
3^(3^3)= 3^27= 3*3*3*3*...*3 (such that there is 27 3's in the multiplication)= 3^27 = 7.62559748 × (10^12)
3^(3^(3^3))= 3^(3^27)= 3^ 3^27 =3^(7.62559748 × (10^12))= which is some ridiculously large number, and several (to say the least!) orders of magnitude larger...
This is confusing! Interesting though

[PLAYER57832]

Does 3^3^3^3 = 3^(3*3)

No.

or is that something different?

Yes.

3^3^3^3= 3^(3*3*3)

No it doesn't.

3^3= 3*3*3= 3*9= 27
3^(3^3)= 3^27= 3*3*3*3*...*3 (such that there is 27 3's in the multiplication)= 3^27 = 7.62559748 × (10^12)
3^(3^(3^3))= 3^(3^27)= 3^ 3^27 =3^(7.62559748 × (10^12))= which is some ridiculously large number, and several (to say the least!) orders of magnitude larger...
This is confusing! Interesting though

You made an error somewhere, but I am too tired out to find it.

You multiply exponents of each other.

so 3^3^3 is 3^ (3*3) and 3^3^3^3 = 3^(3*3*3)

the other rule is you add multiplied exponents... so 3^3 * 3^3 = 3^(3+3)
(it helps to have a kid in 3rd grade... lol)

[Metsfanmax]

Does 3^3^3^3 = 3^(3*3)

No.

or is that something different?

Yes.

3^3^3^3= 3^(3*3*3)

No it doesn't.

3^3= 3*3*3= 3*9= 27
3^(3^3)= 3^27= 3*3*3*3*...*3 (such that there is 27 3's in the multiplication)= 3^27 = 7.62559748 × (10^12)
3^(3^(3^3))= 3^(3^27)= 3^ 3^27 =3^(7.62559748 × (10^12))= which is some ridiculously large number, and several (to say the least!) orders of magnitude larger...
This is confusing! Interesting though

You made an error somewhere, but I am too tired out to find it.

You multiply exponents of each other.

so 3^3^3 is 3^ (3*3) and 3^3^3^3 = 3^(3*3*3)

the other rule is you add multiplied exponents... so 3^3 * 3^3 = 3^(3+3)
(it helps to have a kid in 3rd grade... lol)

There was no error in what he wrote. The confusion comes precisely from the ambiguity in the association. Most people would assume that 3^3^3^3 = (((3^3)^3)^3), in which case what you said is right, Player, but if the association goes the other way, with the parentheses as written in Lord Master's post, then you have to do those operations first (remember PEMDAS?).

[PLAYER57832]

but 3^3^3^3 IS 3 to the third, to the third, to the third. or (3*3*3) ^ 3 ^3 or [(3*3*3)*(3*3*3)*(3*3*3)]^3..etc.

[Metsfanmax]

but 3^3^3^3 IS 3 to the third, to the third, to the third. or (3*3*3) ^ 3 ^3 or [(3*3*3)*(3*3*3)*(3*3*3)]^3..etc.

If we choose to interpret 3^3^3^3 as what you said, then sure, that's correct. But Lord Master's point is that by addition of parentheses in various places, you can change the result of that exponentiation (and he was correct).

[maasman]

To add some clarification, assume it is right to left so 3^3^3^3 would equal (((3^3)^3)^3) so in my original post it would have been ((2^2)^2). I realized the difference this morning when I thought about it more, but all the examples I was thinking about earlier worked no matter which direction I went and didn't think of it at the time
Also, ironically, there was a problem in my calc book the other day that said to prove that a repeating decimal like .74999...=3/4 or something like that.

[PLAYER57832]

but 3^3^3^3 IS 3 to the third, to the third, to the third. or (3*3*3) ^ 3 ^3 or [(3*3*3)*(3*3*3)*(3*3*3)]^3..etc.

If we choose to interpret 3^3^3^3 as what you said, then sure, that's correct. But Lord Master's point is that by addition of parentheses in various places, you can change the result of that exponentiation (and he was correct).

Well, his point is that we are free to add parenthesis, that there was "ambiguity". In math,this is not true. The rules for precedence are set. If there are not parenthesis, there is only one answer. Parenthesis would change the equation, they are not an "optional item".

[Metsfanmax]

but 3^3^3^3 IS 3 to the third, to the third, to the third. or (3*3*3) ^ 3 ^3 or [(3*3*3)*(3*3*3)*(3*3*3)]^3..etc.

If we choose to interpret 3^3^3^3 as what you said, then sure, that's correct. But Lord Master's point is that by addition of parentheses in various places, you can change the result of that exponentiation (and he was correct).

Well, his point is that we are free to add parenthesis, that there was "ambiguity". In math,this is not true. The rules for precedence are set. If there are not parenthesis, there is only one answer. Parenthesis would change the equation, they are not an "optional item".

Your statement is fundamentally incorrect. There exists no rule that says that the association must go down to up (or left to right, if you prefer), as opposed to up to down (or right to left). See the operation of tetration, for an example of how there exists different meanings to a given exponent chain depending on the context.

[PLAYER57832]

but 3^3^3^3 IS 3 to the third, to the third, to the third. or (3*3*3) ^ 3 ^3 or [(3*3*3)*(3*3*3)*(3*3*3)]^3..etc.

If we choose to interpret 3^3^3^3 as what you said, then sure, that's correct. But Lord Master's point is that by addition of parentheses in various places, you can change the result of that exponentiation (and he was correct).

Well, his point is that we are free to add parenthesis, that there was "ambiguity". In math,this is not true. The rules for precedence are set. If there are not parenthesis, there is only one answer. Parenthesis would change the equation, they are not an "optional item".

Your statement is fundamentally incorrect. There exists no rule that says that the association must go down to up (or left to right, if you prefer), as opposed to up to down (or right to left). See the operation of tetration, for an example of how there exists different meanings to a given exponent chain depending on the context.

No, my statement is absolutely correct. Exponents are merely shorthand for multiplying by the same number. Multiple exponents are multiplied. There is no "option". Inthis case, you could get the same answer whether you read it right to left or left to right, but that is cooincidence.

Math is not like English where there are rules "made to be broken". Math is precise. Any other answer is just plain wrong.

[Lord and Master]

but 3^3^3^3 IS 3 to the third, to the third, to the third. or (3*3*3) ^ 3 ^3 or [(3*3*3)*(3*3*3)*(3*3*3)]^3..etc.

Parentheses are necessary when we're writing it in this keyboard form (ie using "^" to represent the operation of raising to the power of) 'cos in normal writing we'd use the superscript form (where the first number 3 is normal and the second number 3 is a smaller 3 placed up on the right corner of the first one and so on) which is completely unambiguous and clearly shows that you have to start with the uppermost right-hand part of the equation and systematically work your way back until you end up with something along the lines of 3(superscript)(7.62559748 × (10^12)), however, if we don't have parentheses and we leave it in this keyboard fashion of 3^3^3^3 then if we start at the left instead (which to be fair we would do with most other equations) we'd end up with a different answer.
ie 3 cubed is 27 ( 3^3^3^3)
27 cubed is 19,683 ( 27^3^3)
19,683 cubed is (7.62559748 × (10^12)) ( 19,683^3)
So our answer this time is (7.62559748 × (10^12)) cubed (or raised to the power 3), which is clearly completely fundamentally different to the other (correct!) answer we derived which was 3 raised to the power (7.62559748 × (10^12))

I think what I'm basically saying is that in real life with a pen and paper it's fairly clear but when we use "^" on a screen it gets confusing unless we know that we're supposed to start over on the right, which is kind of counter-intuitive yet obvious when using superscript form

[jonesthecurl]

Heinlein once suggested that "the number of the beast" was 6 to the sixth power, to the sixth power.

[Metsfanmax]

Inthis case, you could get the same answer whether you read it right to left or left to right, but that is cooincidence.

Math may be precise, but your assertion that "this is the way it works" doesn't hold much water. The fact is, when we write 3^3^3^3 (and when I say this, I mean as it is on this page: http://en.wikipedia.org/wiki/Tetration#Iterated_powers), there is not only one correct way to read it. Both of the interpretations given there are legitimate, because the lack of parentheses means it is unclear what is meant by that chained exponentiation. You might say that if there are no parentheses, then in general we assume that exponentiation is right-associative; and if you said that, you would be correct. But to say that the left-associative way of doing it is incorrect is plainly wrong; the fact is that the notation specified can fall under either convention, it simply depends on what you meant to represent when you wrote down that symbol.

Furthermore, your claim is incorrect anyway, because even if I were to agree that there were only one legitimate way of interpreting that tower, it would be the conventional way, i.e. (3^(3^(3^3))), which is different from what you claim it is: "3 to the third, to the third, to the third", or (((3^3)^3)^3). And no, you don't get the same answer if you read it left to right or right to left; try typing those into a calculator and you'll see.

[Doc_Brown]

Here's a fun one for you all:
N^2 = N * N = N + N + N + ... + N (written out N times).
Now take the derivative with respect to N:
2*N = 1 + 1 + 1 + ... + 1 (with N 1s)
2N = N
2 = 1
1 = 0

Fun with derivatives! (Can anyone spot the error here?)

[Lord and Master]

Here's a fun one for you all:
N^2 = N * N = N + N + N + ... + N (written out N times).
Now take the derivative with respect to N:
2*N = 1 + 1 + 1 + ... + 1 (with N 1s)
2N = N
2 = 1
1 = 0

Fun with derivatives! (Can anyone spot the error here?)

Where you've put 2*N = 1 + 1 +...+1 (with N 1's), 'cos there should be 2*N 1's, so you've said immediately that 2N=N. Which is only true for N=0.

Actually scratch that, the error is that you can't differentiate (take the derivative) when there's only one variable, you need to find (for example) dx/dy or dN/dx etc, so in fact all you've done was to act as if you were differentiating N^2, thus deriving 2N, but then you simply divided the string of N N's by N. Which is bollocks!

[Metsfanmax]

Here's a fun one for you all:
N^2 = N * N = N + N + N + ... + N (written out N times).
Now take the derivative with respect to N:
2*N = 1 + 1 + 1 + ... + 1 (with N 1s)
2N = N
2 = 1
1 = 0

Fun with derivatives! (Can anyone spot the error here?)

Where you've put 2*N = 1 + 1 +...+1 (with N 1's), 'cos there should be 2*N 1's, so you've said immediately that 2N=N. Which is only true for N=0.

Actually scratch that, the error is that you can't differentiate (take the derivative) when there's only one variable, you need to find (for example) dx/dy or dN/dx etc, so in fact all you've done was to act as if you were differentiating N^2, thus deriving 2N, but then you simply divided the string of N N's by N. Which is bollocks!

Actually, there's nothing wrong with operating with d/dN on both sides of the equation. If you like, you can call the left side f(N) and the right side g(N), and then taking the derivative is equivalent to setting df/dN = dg/dN.

[Lord and Master]

Actually, there's nothing wrong with operating with d/dN on both sides of the equation. If you like, you can call the left side f(N) and the right side g(N), and then taking the derivative is equivalent to setting df/dN = dg/dN.

I'm puzzled! Unless it's simply that you can't differentiate when it's in the extended form of (N+N+...+N) and has to be in the simplest power form? Which would make sense seeing as the derivative is telling us the gradient of the graph at point N...

[Metsfanmax]

Actually, there's nothing wrong with operating with d/dN on both sides of the equation. If you like, you can call the left side f(N) and the right side g(N), and then taking the derivative is equivalent to setting df/dN = dg/dN.

I'm puzzled! Unless it's simply that you can't differentiate when it's in the extended form of (N+N+...+N) and has to be in the simplest power form? Which would make sense seeing as the derivative is telling us the gradient of the graph at point N...

Well, the flaw in the "proof" is actually that it's misleading to write N^2 as N+N+N...+N, because we must remember that we're working with continuous (not discrete) functions in order to take a derivative, and in general you cannot write N^2 that way. For example, if N = 3/2, then N^2 = 9/4, and you're "summing" N 1.5 times. The reason this causes an error is because the derivative measures rate of change with respect to N, and the number of terms in the sum changes when N changes, which you don't take into account if you attempt to write it the way it is posed in the question.

[Doc_Brown]

Actually, there's nothing wrong with operating with d/dN on both sides of the equation. If you like, you can call the left side f(N) and the right side g(N), and then taking the derivative is equivalent to setting df/dN = dg/dN.

I'm puzzled! Unless it's simply that you can't differentiate when it's in the extended form of (N+N+...+N) and has to be in the simplest power form? Which would make sense seeing as the derivative is telling us the gradient of the graph at point N...

Well, the flaw in the "proof" is actually that it's misleading to write N^2 as N+N+N...+N, because we must remember that we're working with continuous (not discrete) functions in order to take a derivative, and in general you cannot write N^2 that way. For example, if N = 3/2, then N^2 = 9/4, and you're "summing" N 1.5 times. The reason this causes an error is because the derivative measures rate of change with respect to N, and the number of terms in the sum changes when N changes, which you don't take into account if you attempt to write it the way it is posed in the question.

This is it exactly. Another way to put it is that writing out N*N as N + N + N +... + N (N times) hides the fact that the "N times" is not a constant value. Derivatives (or differentials, since that is the more accurate term when we're only dealing with a single variable) involve change, and the "N times" tries to hold constant that which has to change to properly take the derivative.

Basically, it's equivalent to saying that for some N = N0, N^2 = N*N0. Therefore, taking the derivative, 2N = N0. You can't say that N is constant and that it's changing.

[Doc_Brown]

How about another one for you?

You may recall Euler's relation:
e^(i 2pi) = 1
Take the natural logarithm of both sides:
i 2pi = 0
This means that either i=0, 2=0, or pi=0. Which is it? Or is something wrong in the proof?

[Metsfanmax]

How about another one for you?

You may recall Euler's relation:
e^(i 2pi) = 1
Take the natural logarithm of both sides:
i 2pi = 0
This means that either i=0, 2=0, or pi=0. Which is it? Or is something wrong in the proof?

This one's easy... on the complex plane, ln(z) is multi-valued, but you've chosen the one value that would correspond to the way we take logarithms of real numbers. The principal value of ln(z) must be between -pi and pi, so the principal value of ln[ e^(i 2pi) ] is zero, which does satisfy the relation.

[Doc_Brown]

How about another one for you?

You may recall Euler's relation:
e^(i 2pi) = 1
Take the natural logarithm of both sides:
i 2pi = 0
This means that either i=0, 2=0, or pi=0. Which is it? Or is something wrong in the proof?

This one's easy... on the complex plane, ln(z) is multi-valued, but you've chosen the one value that would correspond to the way we take logarithms of real numbers. The principal value of ln(z) must be between -pi and pi, so the principal value of ln[ e^(i 2pi) ] is zero, which does satisfy the relation.

I guess anyone that has taken a complex analysis class should know that one immediately. For those that didn't quite follow that, just realize that what's going on is exactly the same as in the following example:
cos( 0 ) = 1 = cos( 2*pi )
Take the arccosine of both sides and get
0 = 2*pi

[Doc_Brown]

Hmmmm. I can't remember any other cute mathematical oddities (though I've got a few mathematician jokes). Anyone else?

[Army of GOD]

sinx=n

cancel ns

six=1


??????

[AndyDufresne]

The last time I understood math was Factor Trees. Everything went downhill from there...but I like mathematics nonetheless!


--Andy

[maasman]

Thanks for the fun so far guys. My thoughts on the N^2 thing are that if its N*N then N+N... N times, I figured it would equal infinity and 2*infinity=infinity, and the fact that 2N=N only works for N=0 otherwise.