From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: How do you solve ln z = z ?
Date: 1 Nov 1995 19:18:46 GMT
In article <1995Oct31.171620.22793@lamont.ldgo.columbia.edu>,
VADIUM FELDBAUM wrote:
>The subject line is self-explanatory, I hope. Of course z is complex.
...
>Also, another equation:
>
> 1+z
>ln --- = 8z
> 1-z
The poster then goes on to exponentiate these equations -- a good move,
since ln isn't single-valued for complex inputs, making the original
question a little unclear.
What remains then is an equation of the form exp(f(z))=g(z) for nice
functions f and g. Then general question to answer, then, is
>Is this in any way related to solvability in complex?
Of interest here is a collection of theorems on the range of analytic
functions. I will quote one useful results from classical complex
analysis.
"Little Picard Theorem": Every complex analytic function defined on the whole
complex plane is either
(a) onto C (i.e., for every w there is a z with f(z)=w)
(b) onto C-{one point} (i.e., f misses at most one value)
(c) or constant.
In fact, in all cases, the inverse image of every point is either empty or
infinite.
So for example if f is the function f(z)=exp(z)-z, then since (as
another poster has pointed out) there is one value of z with f(z)=0,
there must be infinitely many. Moreover, the equations exp(z)=z+c
are also all solvable, except possibly for one value of c.
Likewise the equation (1+z)/(1-z) = exp(8z) may be recast as asking for
the zeros of f(z)=(1-z)exp(8z)-(1+z). Without looking at f any more than
checking it is entire, we know it would be nearly miraculous that it not
have any zeros, and if it had one, it would have infinitely many.
Incidentally, the Great Picard Theorem assures that f attains all
values (with, possibly, one exception) infinitely often in any neighborhood
of an essential singularity, so functions which appear to be poor candidates
for the Little Picard Theorem are still easy functions for which to
demonstrate the existence of many zeros.
You will no doubt have noticed that I only "solved" the equation in
a mathematician's sense: solutions exist, but I don't know where they are :-)
If I had to, I suppose I would use Newton's method. Once I had one zero, I
might look for others by applying Newton's method, not again to f(z) but
rather to f(z)/(z-z0), so as to avoid converging to the same root again.
dave
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From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: How do you solve ln z = z ?
Date: 2 Nov 1995 21:45:53 GMT
In a recent article, VADIUM FELDBAUM asked for
clarifications of a response I had made to his initial post:
>> "Little Picard Theorem": Every complex analytic function defined on the whole
>> complex plane is either
>> (a) onto C (i.e., for every w there is a z with f(z)=w)
>> (b) onto C-{one point} (i.e., f misses at most one value)
>> (c) or constant.
>
>Do you mean to say that each and every equation would have a solution?
That is correct: all of the equations f(z)=0, f(z)=1, f(z)=pi, etc. are
solvable. The example f(z)=exp(z) shows that there can be just one
equation (in this case, f(z)=0) which has no solution, but this theorem
asserts that for each nonconstant entire function there is at most one such
unsolvable equation.
>> In fact, in all cases, the inverse image of every point is either empty or
>> infinite.
>
>Is this a part of the above theorem? Does it mean that each and every
>equation would have infinitely many solutions?
As presented in Conway's "Functions of One Complex Variable", the preceding
line is actually a corollary of the Great Picard Theorem. I don't know if
historically they were presented as a unit or not; you may think of them
as all one theorem, I suppose.
>> Incidentally, the Great Picard Theorem assures that f attains all
>> values (with, possibly, one exception) infinitely often in any neighborhood
>> of an essential singularity, so functions which appear to be poor candidates
>> for the Little Picard Theorem are still easy functions for which to
>> demonstrate the existence of many zeros.
>
>Well, if I undertood correctly what you said here, I'll have to change
>the way I understood the idea of singularities.
Careful -- the phrase "essential singularity" has a technical meaning.
There are "removeable singularities" at a point a (meaning that if
a [new] value is assigned at f(a) then f is analytic at a -- for
example, f(z)=(z^2-1)/(z-1) has a removeable singularity at 1 ) and
there are "poles" (meaning (z-a)^n f(z) has a removeable singularity
at a); it's the other singularities which are "essential". The usual
example is f(z)=exp(1/z), which has an essential singularity at 0.
> Say if you take 1/x then it does take on every value near zero.
This is a simple pole, not an essential singularity. It does not
attain the value 0 (nor indeed any value of small absolute value)
in a neighborhood of zero.
>If you take
>|1/x|=ABS(1/x) then it take on every positive value, but not negative
>ones. Are you saying that this cannot be achieved for complex? ABS
>never takes on negative values, reals or complex! But then ABS is
>not an analytic f-n. Still I can't quite beleive same thing cannot
>be achieved with analytical f-ns only.
This is very much a theorem of analytic functions. This function
f(z)=1/sqrt(z*zbar) is not analytic at any point. (At 0 it fails
on three counts, so to speak: f is built up from the complex
functions "conjugate", "sqrt", and "reciprocal", none of which is
analytic here. At nonzero points, it fails to be analytic only because
of the first of these three: the function f(x+iy)=x-iy is often
the example used to show the difference between complex-analytic and
real-analytic functions f: C-->C.) It is a fundamental feature of
complex analysis that the inoccuous definition of analyticity is
actually quite restrictive, so that functions which do meet this
condition can be counted on to be extremely well-behaved (e.g. they
are _infinitely_-differentiable, uniquely determined by values
on small sets, etc.)
Take a good course on complex analysis (not just "Complex Variables").
It is an experience not to be missed in a student's mathematical
training.
dave (not a complex analyst)
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From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: How do you solve ln z = z ?
Date: 2 Nov 1995 22:00:19 GMT
Sorry, forgot one little thing:
> "Little Picard Theorem": Every complex analytic function defined on the whole
> complex plane is either
> (a) onto C (i.e., for every w there is a z with f(z)=w)
> (b) onto C-{one point} (i.e., f misses at most one value)
> (c) or constant.
...
> In fact, in all cases, the inverse image of every point is either empty or
> infinite.
This last line only applies if f is not a polynomial. (The trick is just
to observe f(1/z) has an essential singularity at 0 iff f is not a
polynomial, so that the Great Picard Theorem can be used.)
Sorry about that.
dave
==============================================================================
From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: How do you solve ln z = z ?
Date: 3 Nov 1995 18:33:56 GMT
Oh, dear: I've had to correct my own followups before, but I can't recall
having to correct a correction! It seems I have long labored under a
misconception which several in this group have now called to my attention.
Let me try this one more time. This much, posted a while ago, is correct:
> "Little Picard Theorem": Every complex analytic function defined on the whole
> complex plane is either
> (a) onto C (i.e., for every w there is a z with f(z)=w)
> (b) onto C-{one point} (i.e., f misses at most one value)
> (c) or constant.
Here's where I blew it:
> In fact, in all cases, the inverse image of every point is either empty or
> infinite.
I forgot to add that f is to be non-polynomial, but I hadn't realized
that even with that condition, the statement is incorrect. Rather than risk
botching it again, I will simply quote passages from Conway's book.
"Great Picard Theorem. Suppose an analytic function f has an essential
singularity at z=a. Then in each neighborhood of a, f assumes each complex
number, with one possible exception, an infinite number of times.
...
In the preceding chapter it was shown that an entire function of order \lambda,
where \lambda is not an integer, assumes each value infinitely often.
Functions of the form e^g, for g a polynomial, assume each value
infinitely often, although there is one excepted value -- namely, zero.
The Great Picard Theorem yields a general result along these lines (although
an exceptional value is possible, so that the following result is not
comparable with [that result from the preceding chapter].
Corollary. If f is an entire function which is not a polynomial then
f assume every complex number, with one exception, an infinite number of
times.
Proof. Consider the function g(z) = f(1/z). Since f is not a polynomial,
g has an essential singularity at z=0. The result now follows from the
Great Picard Theorem. [qed] "
(The main result of the preceding chapter is Hadamard's factorization theorem.)
==============================================================================
From: Toke Lindegaard Knudsen
Newsgroups: sci.math
Subject: Complex analysis question.
Date: 3 Nov 1995 16:17:14 GMT
Hello.
I was wondering about the following:
Suppose f is a mapping from the set of complex numbers into the set of
complex numbers and f is entire and that f does not take the value 0.
Will there then exist an entire function g, so that
f(x) = exp(g(x)) for all complex x.
?
Can anybody help me? Thanks.
Toke.
[sig deleted]
==============================================================================
From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Complex analysis question.
Date: 3 Nov 1995 23:03:10 GMT
In article <47dfaa$nkv@danmat.math.ku.dk>,
Toke Lindegaard Knudsen wrote:
>
>Suppose f is a mapping from the set of complex numbers into the set of
>complex numbers and f is entire and that f does not take the value 0.
>
>Will there then exist an entire function g, so that
>
> f(x) = exp(g(x)) for all complex x.
Since another respondent has posted something which I think misses
the heart of the issue, I thought I'd follow up. Of course there ought
to be such a g, namely g(x)=ln(f(x)), but there are two problems:
how do we know that a continuous choice of branch of the logarithm may be
made, and how do we know the resulting g is analytic?
(This g will be well-defined on the whole plane since f is
defined everywhere and assumed nonzero.)
Actually the analyticity is immediate from the chain rule since the
logarithm -- any branch -- is analytic wherever defined, and since
f was assumed to be analytic.
The remaining issue is the ability to choose a single branch of the
logarithm over the entire plane. This requires nontrivial use of the
hypotheses. Indeed, if f(x)=1/(x-a), then all the conditions are met
except for a single point where f is not defined; and sure enough,
there is no continuous function g with the same domain as f such
that f = exp o g.
The resolution of this problem is topological. Since f is assumed
to be a continuous function from the _simply-connected_ domain C
to the non-simply-connected region C-{0}, f factors through the
universal cover of C-{0}. But the universal covering map
p: C -> C-{0} is, you guessed it, the exponential map. (The lifting
g: C-> C is uniquely defined by the condition f = p o g after
choice of a single value g(x_0), which may be selected arbitrarily
in the inverse image p^(-1)(f(x_0)).)
The question of the existence of continuous liftings of this sort
is often treated in first courses in algebraic topology. See, for
example, Sieradski's "Introduction to Topology and Homotopy", Chap.14,
although this specific example is somewhat more transparent.
dave
==============================================================================
From: ags@seaman.cc.purdue.edu (Dave Seaman)
Newsgroups: sci.math
Subject: Re: Complex analysis question.
Date: 4 Nov 1995 11:59:05 -0500
>In article <47dfaa$nkv@danmat.math.ku.dk>,
>Toke Lindegaard Knudsen wrote:
[A good question, and not the one I answered.]
I misread your question, but fortunately it has been addressed
elsewhere. I had the Little Picard Theorem on the brain, having just
read about it in another thread, and I was reminded of an old paradox.
Those who are familiar with the theorem may as well stop reading,
because this will be old hat. You have been warned.
The question I was attempting to address here, not the one you asked,
was the following:
Consider the exponential function exp: C -> C. This is an entire
function whose image is C - {0}. If we now form the composition g(z) =
exp(exp(z)) we have an entire function that is obviously non-constant.
By the Little Picard Theorem, the image of g must be the entire complex
plane, with the possible exception of a single point.
Obviously, 0 is not in the image of g. But since exp(0) = 1, and 0 is
not in the image of exp, we find that 1 is also not in the image of g.
Therefore, 0 = 1.
(groan)
Dave Seaman
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