Click here to go to the

By Zhalefarin (Zhalefarin) on Friday, July 02, 2004 - 08:04 pm: Edit |

i was intrigued... can anyone figure this out or prove it can't be.... x^x=5

and if you go straight to your ti-89 you are wrong... there MUST be multiple solutions.

By Rbc13 (Rbc13) on Friday, July 02, 2004 - 09:18 pm: Edit |

x^2-5=0

(x+rad5)(x-rad5)=0

x=plus or minus rad 5

I just realized that you want x^x, so forget this post. I gotta go now, but maybe later i will try it.

By Cuspidor (Cuspidor) on Friday, July 02, 2004 - 09:45 pm: Edit |

I did:

x^x=5

xlnx=ln5

lnx=ln5/x

Let lnx=y

Plot:

y=ln5/x

and y=lnx

Find the intercept.. but I only get one solution. I can't figure out how to find more, if there are.

By Tongos (Tongos) on Friday, July 02, 2004 - 10:11 pm: Edit |

try seeing a correlation to the taylor series, try to do the derivative or integrating it. no calcs please

By Tongos (Tongos) on Friday, July 02, 2004 - 10:20 pm: Edit |

just by looking at a graph you can't see that there is only one intersection, just like on a parabola above the x- axis, there could be more zeros on the other side, the imaginary side. caution: there could be imaginary solutions.....

By Ubercollegeman (Ubercollegeman) on Saturday, July 03, 2004 - 02:16 pm: Edit |

If you are really interested, you can take a look at the Lambert W-Function which is related to the power tower (not just x^x, but ((x^x)^x)^x...).

Not for the mathematical faint of heart though :D.

Basically, equations of the form x^x = z have the solution

x=ln(z)/W(lnz).

By Tongos (Tongos) on Saturday, July 03, 2004 - 05:15 pm: Edit |

so how do you find 2.19?

By Thermodude (Thermodude) on Saturday, July 03, 2004 - 05:34 pm: Edit |

graph y = x^x and y = 5....and see where they intersect...i'm gonna go try it after i sign out....

By Tongos (Tongos) on Saturday, July 03, 2004 - 05:38 pm: Edit |

yes, i know that, with a graphing calculator, but this guy ubercollegeman, really knows what he's talking about, the labert function? I've been in number theory for months now, and never heard of such. i have heard of riemannian curvatures which i believe have some correlation to the zeta of this. that's it, im going to snag this problem with the labert. it doesnt matter if i can do it the long way, the binomial theory!

By Thermodude (Thermodude) on Saturday, July 03, 2004 - 05:43 pm: Edit |

sorry 'bout my earlier post...i see how it was redundant..and i was sorta just trying to answer the first guy who posted's question....still...shouldn't X^X = 5 only have 1 real number solution ??....i mean...if x > or < whatever the solution is...it should come out as something different. Graphing equations and stuff will only yield real number solutions anyways...

By Goodchocolate (Goodchocolate) on Saturday, July 03, 2004 - 06:46 pm: Edit |

Anybody try Newton's Method...?

By Thermodude (Thermodude) on Saturday, July 03, 2004 - 07:46 pm: Edit |

I guess u could use Newton's Method...but I really don't know why everyone thinks there are multiple real number solutions to x^x = 5. There is ONLY ONE real number solution to x^x = 5....one can use simple logic to figure this out (note that if x is < or > the solution...the result will be different than 5)..and graphing will find it easily enough (or you could look at ubercollegeman's post above).

By Zhalefarin (Zhalefarin) on Saturday, July 03, 2004 - 07:50 pm: Edit |

there IS multiple solutions... end of story

By Zhalefarin (Zhalefarin) on Saturday, July 03, 2004 - 07:51 pm: Edit |

IMADOUCHE.... there ARE*

By Tongos (Tongos) on Saturday, July 03, 2004 - 08:19 pm: Edit |

imaginary solutions, right, i mean how could there be negative solutions if the power is irrational. i'll stay with my binomial sequencing, and probably end up with a partial factorial with 1.5!

why do you think that Zhalefarin, i need an explanation for your reasoning.

By Tongos (Tongos) on Saturday, July 03, 2004 - 08:27 pm: Edit |

I mean, whats i^i, much more important whats

xi^xi=5

if we know the real solution, is it possible that by the information, can we find the imaginary solution, if so, how? this is not your ordinary problem. not for the faint hearted in math at all! this one's harsh!

X^X, (X+1)^(X+1), (X+2)^(X+2)............

find a formula to somehow expand each, find connections between the terms......

By Seleucus26 (Seleucus26) on Saturday, July 03, 2004 - 10:33 pm: Edit |

since this has annoyed me twice now i will post even tho i have not been able to get anywhere.

wat we have so far is

a^a = 5, a apx.= 2.19.... (found graphically, but not algebraically)

did i miss anything?

ill add

(a*i^4)^(a*i^4) = 5 because i^4 = 1

(-a*i^2)^(-a*i^2) = 5 because i^2 = -1

maybe someone could start with these 'imaginary' solutions and find the 'real' imaginary solutions.

ill work on it for real soon, i promise

By Thermodude (Thermodude) on Saturday, July 03, 2004 - 11:34 pm: Edit |

Seleucus26....if the solution were a * i^4....wouldn't that just simply be the real number a(assuming that a is a real number)? I think we can safely assume that (a*i^4)^(a*i^4)...is the same thing as a^a.

Also...

(-a* i^2)^(-a * i^2) = (-a * -1)^(-a * -1) = a^a

By Thermodude (Thermodude) on Saturday, July 03, 2004 - 11:42 pm: Edit |

Zhalefarin, what makes you so sure that there are multiple solutions?....I think we have it established that X^X = 5 has only ONE real number solution....so if there would be multiple solutions...the rest would have to be imaginary. But to find imaginary solutions...one would...as Tongo said...have to have a understanding of what i^i is.

By Buddnutz (Buddnutz) on Saturday, July 03, 2004 - 11:52 pm: Edit |

2.12935 yadda yadda?

By Socalnick (Socalnick) on Sunday, July 04, 2004 - 12:07 am: Edit |

if i rember correctly there were 5 solutions 2 real 3 imaginary

i forgot the exact numbers but the imaganary ones involved log product

By Seleucus26 (Seleucus26) on Sunday, July 04, 2004 - 12:24 am: Edit |

yes, thermodude, i^4 is 1 and thats why it isnt a 'real' imaginary solution, its only a real solution in imaginary form, but sometimes you can go from this form and find imaginary solutions that cant be expressed in real terms.

By Thermodude (Thermodude) on Sunday, July 04, 2004 - 02:27 am: Edit |

Sorry 'bout that earlier post, Seleucus26....didn't read your whole post. :-)

By Tongos (Tongos) on Sunday, July 04, 2004 - 11:28 am: Edit |

"no negative solutions" i wouldnt be so sure, i said that there will be imaginary, but i didnt say no negatives.

think about it (-2)^-2 will yield 1/4 right, now solve.

X^X=1/4 viola, it has a positive and a negative solution!

By Tongos (Tongos) on Sunday, July 04, 2004 - 11:53 am: Edit |

i'm sorry, this one doesnt yield a positive solution.

By Tongos (Tongos) on Sunday, July 04, 2004 - 02:27 pm: Edit |

the only way it can yield a positive and a negative solution is only when z is between these values 0.6922 and 1.445.

and i^i is equal to cos(-ipi)+isin(-ipi)=

plug into the taylor series to solve.

By Averagemathgeek (Averagemathgeek) on Sunday, July 04, 2004 - 05:35 pm: Edit |

Wouldn't it be easier to use that Euler identity:

e^(pi*i)+1=0

e^(pi*i)=-1

[e^(pi*i)]^(i/2)=(-1)^(i/2)

i^i=e^(-pi/2)

By Tongos (Tongos) on Sunday, July 04, 2004 - 05:59 pm: Edit |

there's many ways to the solution, but i do see the eulerian identity as something easier.

By Ubercollegeman (Ubercollegeman) on Monday, July 05, 2004 - 06:17 am: Edit |

Proof that there are infinitely many complex solutions? Keep reading.

(a+bi)^(a+bi) =

(a+bi)^a * ((a+bi)^b)^i =

(c+di) * (e+fi)^i

So, the question is, can (a+bi)(c+di)^i = 5 for real constants a, b, c, and d?

If (c+di)^i is a real number, then there is no solution in the case.

We thus assume it is imaginary and rewrite it as (e+fi). (a+bi)(e+fi) = (ae - bf) + (af + be)i. If af is not equal to -be, there are no solutions. If it is, we set up the equations

ae - bf = 5

af + be = 0

When these conditions hold (which they can), we have ourselves a solution. So while we haven't actually found the solutions, we have proven that there must exist INFINITELY many complex solutions.

I don't see a flaw in logic, but I am convinced that the above must be flawed somewhere if not in multiple places (I am pretty tired). How can there be infinitely many complex solutions? Oh well, maybe there are. I'll ask my prof on tuesday and see what he thinks.

By Tongos (Tongos) on Monday, July 05, 2004 - 12:07 pm: Edit |

do you think it has a negative solution too?

By Ubercollegeman (Ubercollegeman) on Monday, July 05, 2004 - 03:54 pm: Edit |

The CC board messes up greater than or less than signs that it thinks is html, so I'm using funny symbols. "?" will be less than. "*" will be greater than.

It is almost trivial to prove that there exists only one positive solution (For f(x) = x^x, 0?f(x)?1 for 0?x?1, and for x * 1/e, f(x) is monotonically increasing.)

For all x?-1, rewrite x^x as (1/x)^(-x). -1? 1/x ? 0, and -x is a positive number greater than 1, and it follows that x^x ? 1, so there are no solutions for x?-1.

For -1?x?0, we have [(-1)(-x)]^x, or (-1)^x[(-x)^x]. (-x)^x is a positive number to a negative exponent and is thus positive.

(-1)^x is either going to be imaginary or -1 since we restricted the domain. In either case, our expression (-1)^x[(-x)^x] is going to yield either an imaginary or negative number, which is in either case not 5.

It's obviously not true for x=-1 or x=0 (some say 0^0 is undefined, some define it as 1 since lim x->0 x^x = 1; it's all the same).

So we've exhausted all posibilities to show that x~= 2.19 is the only real solution.

By Tongos (Tongos) on Monday, July 05, 2004 - 05:37 pm: Edit |

yeah i think so to, isnt it true that if it would be a negative and a positive then it would have to be between the values 0.6922 and 1.445? and even then, wouldnt it have to deal with the process of multiplying it by an odd or even to make it an integer?

By Ubercollegeman (Ubercollegeman) on Monday, July 05, 2004 - 05:44 pm: Edit |

Can you post the rationale for those bounds? I'm thinking that in x^x = a, all a greater than 1 have only one real solution and infinitely many complex solutions.

By Tongos (Tongos) on Monday, July 05, 2004 - 05:59 pm: Edit |

its an odd way of looking at the problem, my math is kind of sketchy, but bare, point my flaws, thanks.

two

x^x=z, z's lowest value is 0.6922 at minimum. its greatest value, we can call w for it to meet the criteria of negative and positive

(-y)^(-y)=z

disregard the negative sign to prove that it has the possibility of existence

y^(-y)=z

y^y=1/z, 1/z would have to be greater than 0.6922, 1/z would have to be greater than 1/0.6922 but if it would be greater, z would have to be carful of getting to small, then a positive might not exist. so its limit of z would have to have the very limit of 0.6922. 1/0.6922=1.445 which is w.

By Ubercollegeman (Ubercollegeman) on Monday, July 05, 2004 - 06:52 pm: Edit |

I'm more comfortable with exact values, so instead of 0.6922, it might be more useful to say e^(-1/e).

There is a problem with your logic, though, because e^(-1/e) is clearly not the minimum value of x^x. It is only the minimum for positive values of x. I believe the actual minimum value is at x=-1. You thought that e^(-1/e) was the minimum value because your dy/dx only works for positive x (because lnx is not real for x<=0). x^x is not differentiable by elementary means for negative x because it is not continuous.

Also, "disregard the negative sign to prove that it has the possibility of existence "--not sure what that means.

By Tongos (Tongos) on Monday, July 05, 2004 - 06:58 pm: Edit |

no, (e^-1)^(e^-1) which IS 0.6922!! is the y value for the minimum. this is the bump value for the minimum.

and disregard the negative sign clearly means that the negative only changes the sign value of the answer.

By Tongos (Tongos) on Monday, July 05, 2004 - 07:01 pm: Edit |

the sign does have some importance because if the power could be multiplied by an even or odd to make an integer, thus pointing out that it could be a solution or not.

By Tongos (Tongos) on Monday, July 05, 2004 - 07:10 pm: Edit |

try one, like 0.8^0.8, you get 0.8365116421

do one over that value yadda yadda on calculator. find the intersection between x^x and that line. you get the answer 1.1655154. now multiply it by -1 and raise it to that power. viola, you get 0.8365116421!

By Thermodude (Thermodude) on Monday, July 05, 2004 - 07:37 pm: Edit |

There is only 1 real solution for this problem...as I have said earlier. Since it appears that everyone agrees that there is only 1 real positive solution....I'll set out my logic for why there are no negative real solutions to x^x = 5.

Just simply graph y = abs(x^x) on your calculator (or you can take the derivative)...and you'll realize that the maximum from (-infinty, zero) is somewhere between x = -0.5 and 0. Now, taken into account that even with the abs of (x^x)...no value can even surpass 1....it means that without the abs...there can be no value of y which surpasses 1 when x < 0. Thus, there are no negative real number solutions to x^x = 5. There would be if say x ^ x = .5

By Ubercollegeman (Ubercollegeman) on Monday, July 05, 2004 - 07:39 pm: Edit |

"and disregard the negative sign clearly means that the negative only changes the sign value of the answer."

Is y positive or negative? Because if it's positive, (-1)^(-y) can also be imaginary.

Sorry, this would be a lot better if we were sitting next to each other, because I'm finding it hard to follow the logic. It's a little weird, and I'm sure it makes sense to you because you thought of it, but the words aren't translating to ideas inside my head.

By Ubercollegeman (Ubercollegeman) on Monday, July 05, 2004 - 07:41 pm: Edit |

Thermodude, you cannot take the derivative of x^x over negative x. The function is not continuous and therefore not differentiable.

Of course it seems "clear" from a graphing calculator that there are no negative solutions, but we're pretending we don't have any.

By Thermodude (Thermodude) on Monday, July 05, 2004 - 07:48 pm: Edit |

ops!...nm my part about taking the derivative of y = abs(x^x)...still...looking at a graph of it explains enough

By Tongos (Tongos) on Monday, July 05, 2004 - 07:57 pm: Edit |

okay, it can be imaginary, but it has a chance of yielding a real answer. how do we know though that an irrational to an irrational would yield either positive, negative or imaginary?

let's try the reciprocal of 0.8

1/0.8 is equal to

1.25

x^x=1.25

x will be positive. x can also be negative if and only iff a negative irrational to the negative irrational power will yield a positive.

then the solution would also be x^x=1/1.25 which is x^x=0.8, which the solutions are see my prior post and just multiply that by -1. AND THATS WHY I SAID IT HAS A CHANCE, ANYTHING OUTSIDE THESE BOUNDS WILL NOT EVEN HAVE A CHANCE! write back.

By Thermodude (Thermodude) on Monday, July 05, 2004 - 08:16 pm: Edit |

When analyzing the real solutions of x^x = 5....

....WE MUST analyze the equation y = x^x.

Now, as I have said before...looking at the equation y = abs(x^x)...notice that as x approaches -infinity...y approaches 0. If we look at the graph...the maximum value of y in the interval (-infinity, 0) is somewhere around 1.45.

This THUS means that there ARE negative values of x which satisfy y = x^x WHEN y < 1.45.

However, in the equation we are presently analyzing, y is 5...which is greater than 1.45.

By Tongos (Tongos) on Monday, July 05, 2004 - 08:37 pm: Edit |

try out the z value ubercollegeman, 1.045639553

wouldnt it make sense that it would have the roots -1.2, (-1.2)^12 makes it positive, and the tenth root of this is a pos because a number to an even power will yield a positive. -1.2 is easy to work with. now it also has the solution 1.043688. thus the chances are slim, 1/3 but still very possible. only if we could find the last digit to pi..... now lets deal with the question at hand, x^x=5, knowing and confirming no negative solutions exist. that guy zhalefarin, he made us go through all this trouble. its much more interesting to figure out to find the positive of the solution. not hassling with imaginaries and negatives and crap. i really didnt anticipate that people would veer off the subject so much, i posted this first on calculating pi, and i guess zhalefarin kind of veered us off subject.

By Ubercollegeman (Ubercollegeman) on Monday, July 05, 2004 - 08:39 pm: Edit |

You cannot look at the graph of abs(x^x) and say it's the same thing as x^x. This is because, at least on my 89, the graph is telling me absurdly that around x = -0.28481, x^x = 1.43004. Whether that is +-1.43004 is irrelevant, since (-0.28481)^-0.28481 is approximately 0.894788 - 1.11551i. When you graph the abs, it shows sqrt(a^2+b^2) of the imaginary number a + bi.

Try graphing it without abs and you will see jaggedness. This is the calculator's way of telling you that the function is discontinuous.

By Ubercollegeman (Ubercollegeman) on Monday, July 05, 2004 - 08:46 pm: Edit |

Tongos, I totally didn't understand that last post you just made.

What's special about z = 1.045639553 ?

And on a complete tangent, pi has not only been proven irrational but transcendental as well. So if you thought you'd make your name in the history books by finding a repeating sequence in pi, you won't.

By Tongos (Tongos) on Monday, July 05, 2004 - 09:07 pm: Edit |

sorry this was akwardly written maybe all false, i know its discontinuous

0.8^0.8=0.8365116421 right?

(-0.8)^(0.8) also yields this solution right?

(-0.8)^(-0.8) yields one over 0.8365116421

therefor we know -0.8 is a root of 1.195440625. we graph, right and find that 1.1655159 is also a solution, thus this is and so is -0.8

By Tongos (Tongos) on Monday, July 05, 2004 - 09:08 pm: Edit |

the pattern of pi is a total joke anyways, i was totally kidding, gosh, it is so hard to do math over the internet. again, what is w?

By Tongos (Tongos) on Monday, July 05, 2004 - 10:57 pm: Edit |

now, let's twist this. a number is irrational by the function x^x, can it be expressed in fractional terms, heck no!. if not then

infinith root of (-irrational)^(infinity)*(irrational). we know that infinity is even because every number could go into it evenly technically speaking(I'm asking for a controversy now!). thus yielding only imaginary solutions for that of an irrational most likely.

By Ubercollegeman (Ubercollegeman) on Tuesday, July 06, 2004 - 12:41 am: Edit |

Yeah, I wasn't accusing you of thinking pi was finite . Just a random thing. I wrote it because as a kid I always thought I might become famous because I'd find a flaw in multiplication .

Hmm, I see the point now. Can someone tell me why Mathematica keeps telling me (-0.8)^(-0.8) = -0.676752 + 0.491689i ?

By Thermodude (Thermodude) on Tuesday, July 06, 2004 - 01:23 am: Edit |

A little off tangent here...but infinity can't be even since it isn't a number...its merely a concept. I suppose Tongos wasn't actually serious...but just wanted to spark up a response with his statement.

By Tongos (Tongos) on Tuesday, July 06, 2004 - 12:10 pm: Edit |

mathematica doesnt do its real algebra

it doesnt raise -0.8 to the fourth power and doesnt do the fifth root of that. and plus, i guess mathematica might be using d'movieres theorem to find the roots, or does the circle thingy on the imaginary plane.

infinity is a concept, i was joking again, i just want a little attention, and i don't know how else to get it. i just woke up, well, i am in sunny but windy right now california talking to people across the country.

i still don't know how to make use of lambert's function, i guess thats still a little to high for me.

By Tongos (Tongos) on Thursday, July 08, 2004 - 03:14 pm: Edit |

i know how to find the solution, look at cool ways to approximate x^x

Posting is currently disabled in this topic. Contact your discussion moderator for more information. |

Administrator's Control Panel -- Board Moderators Only Administer Page |