Math Anyone?

Nitro

Flashlight Enthusiast
Joined
Feb 27, 2004
Messages
1,347
Take the formula y = x ^ x, then solve for x.
 

Saaby

Flashaholic
Joined
Jun 17, 2002
Messages
7,447
Location
Utah
All 3 numbers must be the same. Has to be 1.

1 = 1 ^ 1

That's the only number it can work for. Well, and 0 I suppose.
 

turbodog

Flashaholic
Joined
Jun 23, 2003
Messages
6,425
Location
central time
All 3? There are only 2 variables here, and they (X and Y) do not have to be the same.

y=x^x

if x is 5 for example: y=5^5, y=3125

This is a deceptively simple equation. I worked on it for a while and came up with log(y)=x*log(x).

This is a problem in that both x and log(x) are in different "exponent" forms. Fixing one breaks the other.

Furthermore, dividing the right side ends up with tying up an "X" into the left side.

There are many algebraic equations that do NOT have an algebraic solution, and this is one of them.
 

McGizmo

Flashaholic
Joined
May 1, 2002
Messages
17,291
Location
Maui
[ QUOTE ]
Nitro said:
Take the formula y = x ^ x, then solve for x.

[/ QUOTE ]

Why? I believe I have seen you propose this else where and I was curious as to what you were up to there?!?!? For a short while, a long time ago, I was a math major. I would probably have been better up to the challenge then, maybe.

I have no more obligation to try to solve for X than you do in explaining why you want us to. I assume that there might be some point to your challenge but it certainly elludes me. /ubbthreads/images/graemlins/icon3.gif

Perhaps you could proove the existance of aleph aleph for me?
 

Pi_is_blue

Flashlight Enthusiast
Joined
May 13, 2003
Messages
1,193
Location
Utah
There are infinitely many solutions if you are looking for a number that equals x. If you had given a system of equations it may have been solveable, but with two variables, one equation is not enough.
 

Nitro

Flashlight Enthusiast
Joined
Feb 27, 2004
Messages
1,347
Kind of like trying to find a representation of the Square of Root 2 as a rational number.

Consider this.

Let '#' be such that:
x # 2 = x ^ x
x # 3 = x ^ (x ^ x)
x # 4 = x ^ (x ^ (x ^ x))
.
.
.
And let '|' be such that:
x = x # a | a

Now
y = x # a
y | a = x # a | a
y | a = x

So if
y = x # 2
then
x = y | 2
Therefore if
x # 2 = x ^ x
and
y = x ^ x
then
x = y | 2
 

Nitro

Flashlight Enthusiast
Joined
Feb 27, 2004
Messages
1,347
[ QUOTE ]
McGizmo said:
Perhaps you could proove the existance of aleph aleph for me?

[/ QUOTE ]

I'll prove the existence of Aleph Aleph if and only if you prove the existence of the Number One.
 

McGizmo

Flashaholic
Joined
May 1, 2002
Messages
17,291
Location
Maui
Nitro,
I bailed from being a math major when in a class called introduction to algebra, one of the mid term problems was to prove 1+1=2 . I chose to take 1+1=2 as a given and to go on and do something else with my life. /ubbthreads/images/graemlins/icon3.gif Set theory and linear algebra were pushing me past my limits and that was OK. To each his own. I'll approach my asymtote (sp) and you can approach yours! /ubbthreads/images/graemlins/nana.gif It is likely we are following different paths and won't be getting in eachothers way! /ubbthreads/images/graemlins/thumbsup.gif
 

Nitro

Flashlight Enthusiast
Joined
Feb 27, 2004
Messages
1,347
McGizmo, I'm curious to know if you're a Pythagorean or Cantorist.
 

UncleFester

Flashaholic*,
Joined
Apr 28, 2004
Messages
1,271
Location
Desert Hlls,AZ
[ QUOTE ]
Nitro said:
[ QUOTE ]
McGizmo said:
Perhaps you could proove the existance of aleph aleph for me?

[/ QUOTE ]

I'll prove the existence of Aleph Aleph if and only if you prove the existence of the Number One.

[/ QUOTE ]
Number One grew up to become Nurse Chapel.
 

Mednanu

Enlightened
Joined
Feb 16, 2003
Messages
333
Location
Earth...
x = the square root of y

Edit: oops - just realized you meant x, to the xth power, rather than x times x. That equation would be a rather tougher nut to crack - I'll have to think about it some more.
 

McGizmo

Flashaholic
Joined
May 1, 2002
Messages
17,291
Location
Maui
Nitro,
I suppose I would be curious about which label I would get to wear as well! Being totally ignorant of either of those labels you have mentioned may not be an excuse but in ignorance I blissfully remain. I am curious to whether you have a point here and/or plan to enlighten us with something cool and interesting or if you just enjoy throwing out things to ponder. I am so out of it, I don't know what # and the straight line represent! I do recall ! used for factorial if memory serves. I didn't even know the symbol^ reffered raising something to the power until I followed some of your posts in the *other* forum which seems like a dejavu (sp) here! In retrospect, I should have answered your post in the other forum probably and you maybe should have kept your math posts confined to one or the other?!?!? I do believe there is some rule to that effect and although benign in content, perhaps you are also enjoying pushing some limits here as well? /ubbthreads/images/graemlins/icon3.gif No worries. I am free to move on and will do just that! I hope and trust that you and others who will take this thread on its course find value and satisfaction in doing so! /ubbthreads/images/graemlins/thumbsup.gif /ubbthreads/images/graemlins/wink2.gif
 

paulr

Flashaholic
Joined
Mar 29, 2003
Messages
10,832
You can solve that equation approximately, e.g. if y=100 then x=3.5972850236. See a book about numerical analysis to learn how to do that. Basically given y, you take a guess at x, then if x^x is too big, make x smaller; and if x^x is too small, make x bigger. Using a computer, keep improving the guess til it's accurate enough for your purpose.

You could also write out an infinite series solution, but it would be messy. There will not be any simple formula like y=sqrt(sin(x)) or anything like that.
 

turbodog

Flashaholic
Joined
Jun 23, 2003
Messages
6,425
Location
central time
[ QUOTE ]
paulr said:
You can solve that equation approximately, e.g. if y=100 then x=3.5972850236. See a book about numerical analysis to learn how to do that. Basically given y, you take a guess at x, then if x^x is too big, make x smaller; and if x^x is too small, make x bigger. Using a computer, keep improving the guess til it's accurate enough for your purpose.

You could also write out an infinite series solution, but it would be messy. There will not be any simple formula like y=sqrt(sin(x)) or anything like that.

[/ QUOTE ]

Exactly. I have done some more looking around on this. Several authors cite this exact example as a problem that cannot be expressed that way.

One guy suggested making a new function for this exact dilemma. Then you could answer this in simple algebraic terms.

When you think of it, it's not that far out. We currently have cube roots/etc that are very useful functions (on a calculator/etc) that are very hard to do by hand. If this was a more common problem, I am sure my HP calculator would have a button for it.

Just for the record, to anyone who said "square root": that's not right.
 

Nitro

Flashlight Enthusiast
Joined
Feb 27, 2004
Messages
1,347
[ QUOTE ]
McGizmo said:
Nitro,
I suppose I would be curious about which label I would get to wear as well! Being totally ignorant of either of those labels you have mentioned may not be an excuse but in ignorance I blissfully remain. I am curious to whether you have a point here and/or plan to enlighten us with something cool and interesting or if you just enjoy throwing out things to ponder. I am so out of it, I don't know what # and the straight line represent! I do recall ! used for factorial if memory serves. I didn't even know the symbol^ reffered raising something to the power until I followed some of your posts in the *other* forum which seems like a dejavu (sp) here! In retrospect, I should have answered your post in the other forum probably and you maybe should have kept your math posts confined to one or the other?!?!? I do believe there is some rule to that effect and although benign in content, perhaps you are also enjoying pushing some limits here as well? /ubbthreads/images/graemlins/icon3.gif No worries. I am free to move on and will do just that! I hope and trust that you and others who will take this thread on its course find value and satisfaction in doing so! /ubbthreads/images/graemlins/thumbsup.gif /ubbthreads/images/graemlins/wink2.gif

[/ QUOTE ]

I find basic math interesting, so I thought I'd share.

Think of this.

Take
y = x + x
Now solve of x

The only way to do so is to move up one level and define '*' to be
x * 2 = x + x
x * 3 = x + (x + x)
x * 4 = x + (x + (x + x))
.
.
.
Then you define '/' as
x = x * a / a

Now it's easy
y = x + x
y = x * 2
y / 2 = x * 2 / 2
y / 2 = x

The problem with solving for x in y = x ^ x is, there's no higher level, that's commonly defined. So the only way to solve for x is to define the higher operands '#' and '|'.
 

Nitro

Flashlight Enthusiast
Joined
Feb 27, 2004
Messages
1,347
BTW, Pythagoreans don't believe there are infinite numbers and Cantorists do.
 

Minjin

Flashlight Enthusiast
Joined
Sep 21, 2002
Messages
1,237
Location
Central PA
Huh?

y = x + x
y = 2x
x = y/2

I don't know why you are using the rest of that gobbledygook.

Mark
 

turbodog

Flashaholic
Joined
Jun 23, 2003
Messages
6,425
Location
central time
[ QUOTE ]
Minjin said:
Huh?

y = x + x
y = 2x
x = y/2

I don't know why you are using the rest of that gobbledygook.

Mark

[/ QUOTE ]

He must have been a math major. Has to use it someplace as there are no jobs for him with that degree.

/ubbthreads/images/graemlins/wink.gif
 
Top