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Jan 25th, 2004, 11:17 AM
#1
Thread Starter
Fanatic Member
Higher levels
Well I just read about ^^ (2^^2 = 2^2)
I was wondering, it was said that 3^^3 = 3^(3^3) BUT
^^ is a series of exponentials, so the answer should be 3^3^3, and PEDMA says you read from left to right, so it would be (3^3)^3
3^(3^3) = 3^27
(3^3)^3 = 3^(3*3) = 3^9
I'd say that's a pretty big difference, what do you think?
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Jan 25th, 2004, 12:44 PM
#2
transcendental analytic
where did you read about it? ^ is right associative, and you never see x^ab as (x^a)^b
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Jan 25th, 2004, 12:50 PM
#3
Thread Starter
Fanatic Member
x^ab = (x^a)*b
I was saying that if you write:
2^2^2 should it be (2^2)^2 or 2^(2^2)
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Jan 25th, 2004, 01:00 PM
#4
transcendental analytic
Originally posted by alkatran
x^ab = (x^a)*b
I was saying that if you write:
2^2^2 should it be (2^2)^2 or 2^(2^2)
lets type it out explicitely:
xab=(x^a)^b
^ is right associative, which means that you put parentesis around the rightmost ^ first so say:
2^2^2^2 = 2^(2^(2^2))
left associative operators work the other way around:
2-2-2-2=((2-2)-2)-2
Use
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Jan 25th, 2004, 01:11 PM
#5
Thread Starter
Fanatic Member
Ok thanks.
Anybody else find it sortof funny that left associative operators can be rearranged any which way and still get the same answer?
2*3/4 = 3/4*2 = (2*3)/4 = 2*(3/4)
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Jan 25th, 2004, 01:16 PM
#6
transcendental analytic
no it doesn't work that way every time. * is not associative only /. try doing that with two / operators and you'll see what i mean.
Use
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Jan 25th, 2004, 01:22 PM
#7
Thread Starter
Fanatic Member
3/2/4 =1.5/4 = .375
3/4/2 = .75/2 = .375
/4/2*3 = 1/4/2*3 = .125*3 = .375
Seems pretty consistant to me.
Don't pay attention to this signature, it's contradictory.
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Jan 25th, 2004, 03:55 PM
#8
transcendental analytic
errr.. you're treating them as factors.. I thought you were talking about commutation of left associative operators..
Use
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Jan 25th, 2004, 05:55 PM
#9
So Unbanned
3^3^3 = 3^(3^3) <> (3^3)^3
The rule is, the highest power gets calculated first.
(2^3+4^2)^2^4
If you think about this problem. You should do 2^4 first. Then (...)^16.
Last edited by DiGiTaIErRoR; Jan 25th, 2004 at 06:00 PM.
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