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December 4

$x^{4}=4^{x}$

How do you find the solutions? there are exactly three I think, i can't find the one that's not 2 or 4 169.229.75.140 (talk) 01:32, 4 December 2008 (UTC)[reply]

I think there are exactly 2 positive real solutions. Why do you think there are 3? Algebraist 02:26, 4 December 2008 (UTC)[reply]

There are indeed three (real) solutions. The negative one is a bit nasty though, and I can see why you would have problems. A quick plug-n-chug gives you [-0.8,-0.7] as the interval, and consideration of derivatives gives you just the one solution in the interval, but beyond that you need some fancy tools (there are no obvious elementary methods). —Kan8eDie (talk) 03:59, 4 December 2008 (UTC)[reply]

Well, it's not entirely clear it's a well-defined problem in the negative case, though.

The thing is, exponentiation is not a single function. It's several functions defined differently in different contexts. The standard ones that apply here are:

you have integer-to-integer exponentiation, which is non-problematic in almost any context except for zero to a negative number
you have real-to-integer exponentiation, also non-problematic except for zero to a negative
you have real-to-real exponentiation, which most sources define only when the base is positive ( $a^{x}=e^{x\log a}$ , for $a>0$ )
you have complex-to-complex exponentiation, which is definable everywhere except ~~for 0 to the 0 power or a negative real power~~ when the base is 0, but you have to choose a branch cut

Whoops, saved when I meant to preview. The punch line is: when you're trying to solve

4^{x}=x^{4}

for x a negative real number, you can't be using the same sort of exponentiation on both sides of the equation, unless they're both complex-to-complex, in which case it's ambiguous because of the branch-cut issue. --Trovatore (talk) 04:23, 4 December 2008 (UTC)[reply]

See Lambert W function. I think the solution is

-W(-Q)/Q

where

Q=\log(4)/4

(although I have not had my coffee yet). Note this gives all complex solutions too. Robinh (talk) 08:23, 4 December 2008 (UTC).[reply]

In addition to the above (which gives two positive integer solutions) the negative real solution is

-W(Q)/Q

, which is about -0.766664695962123 (the standard manipulation loses this solution because it takes a fourth root and extra solutions are forgotten about. There may well be other 'lost' solutions lurking out there as well). Sorry for any confusion caused by my earlier incomplete edit Robinh (talk)

That the two kinds of exponentiation are different is a source of confusion, but not a fundamental obstruction; when asked to solve

x^{2}=\sin x

, you wouldn't object that one side is a polynomial and the other side is a trigonometry function. Eric. 131.215.158.213 (talk) 11:03, 4 December 2008 (UTC)[reply]

Sure, it's not a bad obstacle, just a point to note. The equation

4^{x}=x^{4}

, interpreted reasonably in context, is shorthand for

e^{x\log 4}=x\cdot x\cdot x\cdot x

, and it's perfectly reasonable to ask for values of x making that equation true. But it is important to understand that the left-hand side does not mean the product of x copies of 4, and the right-hand side does not mean

e^{4\log x}

(unless you're working in the complex case, in which case you need to specify a branch of

\log

).

It's my view that this confusion is a big part of the reason people get so misty-eyed about Euler's identity, which is indeed a nice result, but doesn't deserve to be piled with nonsense about how we don't know what it means. --Trovatore (talk) 20:01, 4 December 2008 (UTC)[reply]

Both sides are power series - how else does a first course in analysis define sin? Trig. functions are normally defined in terms of exp. There is only one sort of standard exponentiation: without further specification, it can be assumed that all powers are defined in terms of the power series for exp, which converges nicely with everything arbitrary and complex. (The inverse functions, as noted, are more of a nuisance.) Analytically, there are five solutions here: three real and two complex. Ignore anything your CAS might tell you; if all you are after is the real ones, the you easily find the two positive ones, and bound the negative one. Elementary methods (differentiation) quickly proves that there are no more, and after that you have to stop (no closed-form solution for the two complex ones and the negative real). —Kan8eDie (talk) 14:26, 4 December 2008 (UTC)[reply]

Kan8eDie, There are infinitely many nonreal complex solutions - not merely two. Trovatore, positive-to-complex exponentiation and nonzero complex-to-integer exponentiation is well defined, while nonpositive complex-to-noninteger exponentiation is a problem, but that is not the case here. Bo Jacoby (talk) 11:44, 6 December 2008 (UTC).[reply]

Sorry, but there is a closed form solution, using Lambert W function, which I give above. This will give all solutions on the complex plane. Robinh (talk) 08:17, 5 December 2008 (UTC)[reply]

There's always a closed-form solution, if you add enough function symbols :-)

To Kan8Die: I don't agree that there is only one sort of standard exponentiation. Note that, if you define exponentiation in terms of exp, using your preferred definition of exp in terms of power series, you'd have a circularity. The reason it's not circular is that the exponentiation used to define the power series is of a different sort, namely iterated multiplication. Real-to-real exponentiation cannot meaningfully be understood as iterated multiplication.

That was my point about the Euler identity. It would indeed be really really weird if we had proved that, if you take the operation of multiplying copies of e, and iterate it iπ times, you get −1. But the identity doesn't say any such nonsensical thing. --Trovatore (talk) 08:33, 5 December 2008 (UTC)[reply]

Interestingly enough, my CAS only returned an approximation to the negative solution. J kasd 09:08, 4 December 2008 (UTC)[reply]

Using Excel's goal seek (A1=-0.8; A2=A1*A1*A1*A1; A3=4^A1; A4=A2-A3. Goalseek A4=0 by adjusting A1) finds -0.7667052. -- SGBailey (talk) 09:36, 4 December 2008 (UTC)[reply]

Using J (programming language): p.(4&^-^&4)t.i.12. Approximate the function: 4&^-^&4 with a taylor series to the 11'th degree: t. i. 12, and solve the algebraic equation: p.. The reported 11 approximate solutions are: (_7.3535943j3.26425784 _7.3535943j_3.26425784 _2.46094861j7.44016249 _2.46094861j_7.44016249 3.48488995j6.4621893 3.48488995j_6.4621893 4.03176186 1.99994452 _0.270279229j0.869544826 _0.270279229j_0.869544826 _0.766664694). So the real solutions are 2 and 4 and −0.76666. Bo Jacoby (talk) 10:43, 4 December 2008 (UTC).[reply]

Here is a Python script that enumerates N-digit values of all the solutions (that is, using all branches of the Lambert W function and including all four fourth roots for each branch -- I think this will not miss any solutions):

from mpmath import mp, mpf, log, lambertw
N = 20
mp.dps = N
tol = mpf(10)**(-N*0.75)
q = -log(2)/2
m = 0
while 1:
    for p in [q, -q, 1j*q, -1j*q]:
        for n in set([m, -m]):
            x = lambertw(p,n)/q
            assert abs(x**4-4**x) / abs(x**4) < tol
            print ("%4s"%n), x
    m += 1

The output is:

   0 2.0
   0 -0.76666469596212309311
   0 (-0.27027922920791281722 + 0.8695448161806651849j)
   0 (-0.27027922920791281722 - 0.8695448161806651849j)
   1 (9.090729860748042345 - 21.50795035053775222j)
  -1 4.0
   1 (7.6545064969779272738 - 11.953678912263155851j)
  -1 (7.6545064969779272738 + 11.953678912263155851j)
   1 (6.4827738174005611491 - 6.8854213557517001695j)
  -1 (8.4646204891232434013 + 16.781492798875483318j)
   1 (8.4646204891232434013 - 16.781492798875483318j)
  -1 (6.4827738174005611491 + 6.8854213557517001695j)
   2 (10.746729525944597901 - 40.034536281487172114j)
  -2 (9.090729860748042345 + 21.50795035053775222j)
   2 (10.036848572412826468 - 30.818063282987779427j)
  -2 (10.036848572412826468 + 30.818063282987779427j)
   2 (9.6028744824351605501 - 26.179752701528789044j)
  -2 (10.413650262048970191 + 35.434122018232146873j)
   2 (10.413650262048970191 - 35.434122018232146873j)
  -2 (9.6028744824351605501 + 26.179752701528789044j)

(...)

Fredrik Johansson 12:33, 6 December 2008 (UTC)[reply]

One can get a good understanding of the geometry of the roots by plotting the phase of the function ( $4^{x}-x^{4}$ ) in the complex plane:

The points where the full cycle of colors meet are the roots. So there are infinitely many roots corresponding to periods of the exponential term ( $4^{x}$ ), with imaginary part asymptotic to $\pm n2\pi /\log(4)$ and with real part growing roughly like $\log n$ , plus a set of five roots near zero (2, -0.77, -0.27+0.87i, -0.27-0.87i, 4) which may be thought of as perturbations of the four roots of $x^{4}-1$ , intermixed with one of the exponential roots. Fredrik Johansson 15:22, 6 December 2008 (UTC)[reply]

Well done, Fredrik! Please tell me, how does the roots of x⁴−ε4^x = 0 depend on ε for 0 ≤ ε ≤ 1? Small values of ε gives 4 small (polynomial) roots: (+ε^1/4,−ε^1/4,+iε^1/4,−i·ε^1/4) and an infinite number of (exponential) roots far away from zero. Which 4 roots for ε = 1 are 'polynomial'? Bo Jacoby (talk) 16:27, 8 December 2008 (UTC).[reply]

Intuitively, the smallest roots. One can test by listing the roots as ε increases (using your idea to find the roots with a high order Taylor approximant):

>>> f = lambda eps: polyroots(taylor(lambda x:eps*4**x-x**4,0,12)[::-1])
>>> for eps in linspace(0.1, 1, 10):
...     nprint([u for u in f(eps) if abs(u) < 5])
...
[-0.476704, 0.722299, (-9.99425e-2 + 0.533923j), (-9.99425e-2 - 0.533923j)]
[-0.55225, 0.919824, (-0.136645 + 0.622999j), (-0.136645 - 0.622999j)]
[-0.600939, 1.07372, (-0.163303 - 0.680027j), (-0.163303 + 0.680027j)]
[-0.637598, 1.20929, (-0.184867 - 0.722644j), (-0.184867 + 0.722644j)]
[-0.667278, 1.33614, (-0.203232 - 0.756896j), (-0.203232 + 0.756896j)]
[-0.692354, 1.45957, (-0.219359 + 0.785633j), (-0.219359 - 0.785633j)]
[-0.714143, 1.58347, 4.7925, (-0.233815 + 0.810439j), (-0.233815 - 0.810439j)]
[-0.733459, 1.71152, 4.52719, (-0.246964 - 0.832294j), (-0.246964 + 0.832294j)]
[-0.750841, 1.84812, 4.27075, (-0.259058 - 0.851845j), (-0.259058 + 0.851845j)]
[-0.766665, 1.99999, 4.0126, (-0.270279 + 0.869545j), (-0.270279 - 0.869545j)]

From the looks of it, the roots at -0.77, 2, -0.27±0.87i approach from smaller values while the root at 4 approaches from larger values, so arguably 4 is the "non-polynomial" root. Fredrik Johansson 20:33, 8 December 2008 (UTC)[reply]

A question in trivialogy.

Is zero a limit ordinal? That depends on the definition we choose, of course; I would say: yes; the question is if you see any particular reasons why it should be so or it shouldn't. Here we have a peculiar taste to these theoretic question about the nature of trivial objects...--PMajer (talk) 10:18, 4 December 2008 (UTC)[reply]

Well, the first sentence in Limit ordinal says, "A limit ordinal is an ordinal number which is neither zero nor a successor ordinal." So it would seem that it is defined not to be. The reasons for such definitions are usually because if you defined it the other way you would have to say "non-zero limit ordinal" more often than not. Mathematical definitions are just whatever is most useful. --Tango (talk) 11:29, 4 December 2008 (UTC)[reply]

Ah, reading the article for a few more sentences finds: "Some contention exists on whether or not 0 should be classified as a limit ordinal, as it does not have an immediate predecessor; some textbooks include 0 in the class of limit ordinals[1] while others exclude it[2]." So it's really just a matter of opinion - define it in whatever way works best for you. --Tango (talk) 11:30, 4 December 2008 (UTC)[reply]

Limit ordinals are called limit because they are topologically a limit of another ordinals. Zero does not have this property, so calling it limit just does not make sense. Even if zero is grouped in the same class with the limit ordinals, the class should be called something else, IMHO. — Emil J. 12:56, 4 December 2008 (UTC)[reply]

But 0 is the supremum of its predecessors, which is another definition of the limit ordinals, and (I think) one sufficiently limit-like to justify the name. Algebraist 13:39, 4 December 2008 (UTC)[reply]

Jech, in his well-known book, says explicitly that 0 is a limit ordinal. 67.150.255.90 (talk) 20:05, 4 December 2008 (UTC)[reply]

Let's put it this way. When you're doing a careful proof by transfinite induction, or a construction by transfinite recursion, ordinarily you should probably have three parts to the proof: stage 0, successor stages, and limit stages. This is so even though, much of the time, the proof for the limit stages will work just fine for stage 0. The problem is not one of formalities; it's that you may not have really checked whether you're assuming something that depends on the existence of earlier stages. So for these purposes it's probably better to think of 0 as a third class, neither limit nor successor.

Of course in practice many inductive proofs/recursive constructions are trivial at the limit step, and start with a well-understood object at the base, so you write down only the successor step. But this does take a bit of practice before you can do it reliably. --Trovatore (talk) 00:48, 5 December 2008 (UTC)[reply]

Thank you all guys for these intelligent remarks. As a result, about 0, I personally remain a limitist :) In fact, the definitions of an (infinite) limit ordinal x that I can produce are almost all of the form: x satisfies "something", and is not 0 (here's another: "omega_0 divides x"). But I found very valuable also the other party's arguments. As Emil suggests, this classification should reflect the way one can approach x from below (i.e., say, the topological local type). It is true that under this respect 0 is different from a limit ordinal x>0, because 0 is isolated and x is not. But this is not all, because successors are also isolated, and on the other side, all the other limit ordinals are not unius generis as well, e.g. omega_0 or omega_1 + omega_0 are reachable with increasing countable sequences whereas omega_1 is not. The local type enters very much in the proofs by induction, as Trovatore says (in my own experience, the stage 0 is more often that not a sub-case of the limit stage, even if sometimes in a weird way!). In fact there are proofs by induction where the natural distinction is: zero stage, successor stage, countable limit stage, uncontable limit stage; that is, whether x=0 or x mod omega_1 is 0, finite, or omega_0. --PMajer (talk) 12:04, 5 December 2008 (UTC)[reply]

The finite fields are completely known

It says so on the page Finite field, but what does "known" mean? —Bromskloss (talk) 22:45, 4 December 2008 (UTC)[reply]

It means there is an explicit classification of finite fields. There is (up to isomorphism) exactly one field of each prime power order (not counting 1), and we can give the addition and multiplication tables of these fields explicitly. Algebraist 22:55, 4 December 2008 (UTC)[reply]

OK, so that's what it refers to. Thank you. —Bromskloss (talk) 23:13, 4 December 2008 (UTC)[reply]

December 4

x 4 = 4 x {\displaystyle x^{4}=4^{x}}

A question in trivialogy.

The finite fields are completely known

$x^{4}=4^{x}$