Wait a second... is that right??

[alkatran]

I was messing with my calculator and I found some odd stuff:

0^0 = 1 but 0/0 = undefined


but MAINLY!!

2^3^2 = 81 !!!!!!!

But shouldn't this be:
(2^3)^2 = (2*2*2)^2 = (2*2*2)*(2*2*2) = 8*8 = 64

Instead of:
2^(3^2) = (3*3)^2 = (3*3)*(3*3) = 81


But wait! Isn't the order of operations suppsoed to go from right to left?? My calculator did it from left to right!!!??


Explain. Now.


Oh, and what's the symbol for x^x^x^x^x = x?5
(like x*x*x = x^3)

[DiGiTaIErRoR]

2^3^2 is 2^(3^2)

and this is left to right

If you put this in form you'll see why:

[Guv]

0⁰ = 1 is what my HP calculator gets, and it seems reasonable.

x⁰ = 1 for all nonzero values of x.

(.0000001)⁰ = 1

(-.0000001)⁰ = 1

x⁰ = 1 as x approaches zero from either direction.

It would be strange for it to have some other value at zero.

In the complex plane, z⁰ = 1 for all nonzero values of z. This makes even even weirder for it to be nonzero at zero. Zero is surrounded in all directions by values for which z⁰ = 1

0 / 0 is undefined. The value depends on how you got the expression. For example consider the sin function.

sin(x) = x - x³ / 3! + x⁵ / 5! - x⁷ / 7! + . . .

From the above, sin(0) = 0 and

sin(x) / x = 1 - x² / 3! + x⁴ / 5! - x⁶ / 7! + . . .

The above indicates that sin(x) / x = 1 for x = 0 and

2 * sin(x) / x = 2 for x = 0

The last two expressions are 0 / 0 if you merely substitute 0 for x.

I think you made a keying error in your exponential, or your calculator got confused due to ambiguity. My HP is a stack calculator, and does not evaluate algebraic expressions. I cannot key in your exponential and test it.

2^3^2 might be ambiguous. I am not sure which of the following it might be.

( 2³ )² = 64

( 2 )^{3^2} = 2⁹, which is 512

I see no way to interpret it so that it evaluates to a power of 3.

[Guv]

DiGiTaIErRoR: You posted while I was typing my post. Your interpretation did not occur to me. Perhaps it is correct, but I do not think anybody would use a^b^c without parentheses to clarify the expression. I would ask the author of the expression before I attempted to evaluate it.

BTW: Why the strange capitalization of your name?

[alkatran]

I confused right and left in my first post

Yes, I know that's how it displays it (graphic calc). But my point is that it SHOULDN'T be like that. Although I will admit the order of ^s is more important then /*s and +-s.

[RAEsquivelC]

My good old QuickBASIC agrees with you, Alkatran! When I ran the code:

PRINT 2^3^2

the answer came back very clear as 64. Yeb, doing the operations left-to-right, we get 2^3 = 8, and 8^2 = 64.

So, how did you ever get 81??? Will you please describe exactly how you used your HP to get the wrong answer?

[alkatran]

I think I did the math wrong when I explained it...

2^3^2 should be (2^3)^2 = 8^2 = 64
but instead my calc does
2^3^2=2^(3^2) = 2^9 = 512

There's a slight difference in the answers

Imagine my surprise whne 5^5^5 overflowed my calculator .


BTW: If I put the formula back into the box after I type it in (erm... nm that hehe) it puts the parenthese as if it were right-to-left.

[twanvl]

Originally posted by Guv
x⁰ = 1 for all nonzero values of x.

Yes, but
0^y = 0 for all nonzero values of y

So according to one rule 0⁰ should be 1, and according to another rule it should be 0. Use whatever value suits you best .

[siyan]

Originally posted by alkatran
I think I did the math wrong when I explained it...

2^3^2 should be (2^3)^2 = 8^2 = 64
but instead my calc does
2^3^2=2^(3^2) = 2^9 = 512

There's a slight difference in the answers

Imagine my surprise whne 5^5^5 overflowed my calculator .


BTW: If I put the formula back into the box after I type it in (erm... nm that hehe) it puts the parenthese as if it were right-to-left.

Hrm....My HP39 gives 512 for 2^3^2, which implies 2^(3^2) like you said. And it also overflows on 5^5^5. But if we use the same a^b^c = a^(b^c) logic, then 5^5^5 on the HP should be 5^(5^5), which is 298023223876953125, well below the 9.9999999999 * 10^499 limit on my calc. So it would seem that its deciding to do (5^5)^5 which is indeed a huge number.

Odd. Bugged?

[sql_lall]

I don't really get your reasoning.
It seems you are implying (5^5)^5 > 5^(5^5)
LHS=5^(5*5) = 5^25
RHS=5^(5^5) = 5^125
Obviously, LHS<RHS, or
(5^5)^5 < 5^(5^5)

However, 5^125<10^125<10^499, so it should be within the limits.
The only thing i can think of is maybe the calculator for some reason doesn't allow numbers to be raised to such high powers, just in case there is some large base (i.e. not 5, but bigger)
try putting in 5^125, and see if that works

[NotLKH]

Originally posted by sql_lall
I don't really get your reasoning.
It seems you are implying (5^5)^5 > 5^(5^5)
LHS=5^(5*5) = 5^25
RHS=5^(5^5) = 5^125

I Fail to see the 25 and the 125.

LHS: (5^5)^5==> (3125)^5==>298023223876953125
RHS: 5^(5^5)==>5^(3125)==>1.9110125979454775203564045597e+2184

(Windows Calculater in Scientific mode)


-Lou

[siyan]

Originally posted by sql_lall
I don't really get your reasoning.
It seems you are implying (5^5)^5 > 5^(5^5)
LHS=5^(5*5) = 5^25
RHS=5^(5^5) = 5^125
Obviously, LHS<RHS, or
(5^5)^5 < 5^(5^5)

However, 5^125<10^125<10^499, so it should be within the limits.
The only thing i can think of is maybe the calculator for some reason doesn't allow numbers to be raised to such high powers, just in case there is some large base (i.e. not 5, but bigger)
try putting in 5^125, and see if that works

I mean the calc interprets
2^3^2 as
2^(3^2) = 2^9
instead of
(2^3)^2 = 8^2

but interprets
5^5^5 as
5^(5^5) = 5^3125
instead of
(5^5)^5 = 3125^5

[alkatran]

5^(5^5) = 5^3125 = SORTOF above 10^999

(5^5)^5 = 3125^5 = 2.98 * 10^17 or so