J
JosephKK
Actually, i am more interested in the reasoning about why the value of
your method is correct. But this is far enough from C already.
.
Actually, i am more interested in the reasoning about why the value of
your method is correct. But this is far enough from C already.
.
D
Dik T. Winter
The definition of analytic is that it has a derivative over its domain.
So if the requirement is that f'(x) = f(x) over the domain, the function
f can not be non-analytic because if it non-analytic over the domain there
would be an x for wich the derivative does not exist.
Is it? What more is required?
Can you show a non-analytic function in your sense where the derivative
does not exist for some 'x' within the domain?
With regards to a previous remark by you:
where f(x) = log(x) = int{1..x} dx/x.
I need to show that f(x) is a bijection from R+ to R, that f'(x) does not
vanish, and that is all. What more do you want?
J
JosephKK
OK. I can accept that.
.
J
JosephKK
Spending the time and effort to state it clearly.
.
J
JosephKK
I think you're making the mistake of assuming that a statement is
equivalent to the converse of its contrapositive
A function f is analytic on R if there is a power series Sum(a_n x^n)
such that the series converges to f(x) for every x in R.
IIRC the fact that that a function is analytic is not necessarily
proven by the fact that it can be represented by a power series. Try
arcsin(x) which has an infinite amount of correct values.
Question (since i no longer remember) are Reimann integratable
functions analytic?
.
J
JosephKK
Yes, it is a defensable position. It has had applications a few
times.
.
J
JosephKK
Yeow. <Withdraws singed fingers> In less space, a far better
explanation than in any of the textbooks that i failed to this learn
from way back when.
I do remember ordinary roots and powers on complex Reimann surfaces,
but misremembered application to ln() and exp().
.
L
luser-ex-troll
Have you tried any of the C-to-gates compilers you could
comment on for the curious?
lxt
N
Nate Eldredge
Dik T. Winter said:
That's not the definition of "analytic" that's usually used in
mathematics. Antonius has given the correct definition: a function f is
analytic at a point c if there is a power series
a_0 + a_1 (x-c) + a_2 (x-c)^2 + a_3 (x-c)^3 + ...
which converges to f(x) for all x in some neighborhood of c. f is
analytic if it is analytic at every point in R. (We often use the term
"real analytic" when talking about functions on R; "analytic" by itself
usually refers to functions on C.)
It's a theorem that if f is analytic, then it is infinitely
differentiable (has derivatives of all orders), and the power series
which converges to f is actually the Taylor series of f. In the case of
functions on R, the converse of this statement is false. See below.
Yes, it is.
There are functions on R which are infinitely differentiable but for
which the power series at some point does not converge to the function
in a neighborhood of that point.
The classical example is
f(x) = exp(-1/x^2) for x > 0
f(x) = 0 for x <= 0
You can verify that f has derivatives of all orders at every point, and
in particular at x=0. Moreover, at x=0, the derivatives of all orders
are equal to 0. So if you expanded f in a Taylor series about x=0, you
would get the series 0 + 0 + 0 + 0.... The power series converges to 0
at every point, but that's not the value of f(x) for any x > 0.
So f is a function which is infinitely differentiable but not real
analytic.
The point is that any argument which implicitly assumes that the
function can be written as a power series will only apply to analytic
functions.
Er, that's easy: f(x) = |x| is not analytic and not differentiable at
x=0. With more work I can give you one that's not analytic and not
differentiable at any point. Was that really what you meant to ask?
Anyway, exp(x) is not the only function on R which is its own
derivative; c*exp(x) for any real constant c will also do. But those
are the only examples. Here's a simple proof that doesn't assume f is
analytic.
Theorem. Suppose that f is a once-differentiable function on R such
that f'(x) = f(x) for all x in R. Then there exists a constant c such
that f(x) = c * exp(x).
Proof. Let f be such a function, and let g(x) = f(x) * exp(-x). By the
product rule, g'(x) = f'(x) * exp(-x) - f(x) * exp(-x). Since f'(x) = f(x),
g'(x) = 0 for all x. By the mean value theorem, g is constant. QED.
HTH, etc.
N
Nate Eldredge
Nate Eldredge said:
Sorry, I just noticed Antoninus already gave this example. (Also, I
misspelled Antoninus. Apologees.)
N
Nate Eldredge
JosephKK said:
Nope. They need not even be continuous.
f(x) = 1, if -1 <= x <= 1
f(x) = 0, otherwise
is Riemann integrable over all of R, but not continuous, hence not
differentiable and certainly not analytic.
f(x) = exp(-1/(x^2-1)^2), if -1 < x < 1
f(x) = 0, otherwise
is infinitely differentiable, Riemann integrable on all of R, but not
analytic.
P
Phil Carmody
Nate Eldredge said:
It's *one* of the definitions that's used. In recent years the
concept Dik is referring to has also been called holomorphic
in preference, but one can't deny that it's also commonly called
analytic.
Phil
N
Nate Eldredge
Phil Carmody said:
I think we are at cross purposes.
For functions on C, the property of having a derivative (in the complex
sense!) over an open set is equivalent to being given by a power
series. The terms "analytic" and "holomorphic" are both used for this
property.
For functions on R, the property of having a derivative over an open set
(interval) is strictly weaker than being given by a power series. The
term "analytic" (or "real analytic") is commonly applied to the latter
but never, in my experience, to the former. I have not seen the word
"holomorphic" used to describe functions on R at all.
The discussion was regarding functions on R.
If you have examples of conflicting uses of these terms, I'd be
interested to see them.
J
James Kuyper
Nate said:
"Mathematical Analysis", 2nd Edition, Tom M. Apostol, Definition 16.1:
"Let f = u + iv be a complex-valued function defined on an open set S in
the complex plane C. Then f is said to be analytic on S if the
derivative f' exists and is continuous at every point of S."
I've several other math books that say the same thing in various ways,
but Apostol says it most clearly. All of of them treat convergence of
the power series as a property to be derived from that definition, not
as the definition of the term.
D
Dik T. Winter
The definition I gave for exp(x) was over the complex plane.
A
Antoninus Twink
Note the crucial words "complex-valued function defined on an open set S
in the complex plane". For complex functions, being differentiable on an
open disc is equivalent to being given by a convergent power series
there, so authors are free to use whichever definition they find most
convenient.
For (real valued) functions of a *real* variable, the two conditions are
no longer equivalent - a famous counterexample has been provided at
least twice already in this thread - and "analytic" is conventionally
reserved for the stronger condition.
A
Antoninus Twink
Neat! The only solution that came to my mind was invoking uniqueness
results for solutions of linear first order ODEs, and I was sure that
was overkill.
N
Nate Eldredge
Antoninus Twink said:
I thought of the same thing. Then I looked up a proof of the uniqueness
theorem, which led me to a proof of Gronwall's lemma, at which point I
could see how the whole thing would simplify in this case.
A
Antoninus Twink
What assumed properties?
How do you define a^x for irrational a and x, other than in terms of
exp()?
Nate is just relying on the standard definition of exp, i.e.
exp(x) = 1 + x + x^2/2! + ...
Of course, one must prove that this power series converges for all x,
and that one can differentiate a convergent power series term by term
inside its radius of convergence, and that when one does this in this
case one gets the same power series back.
Not at all - see above.
One must prove that power series converging to functions f and g can be
multiplied in the "obvious" way(*) to give a power series that converges
to fg in a suitable radius. Once again, this will be in any
undergraduate analysis text.
Then one must apply this procedure to the power series for exp(x) and
exp(-x): not a difficult exercise.
(*) i.e. (Sum a_n)(Sum b+n) = Sum c_n, where
c_n = a_0.b_n + a_1.b_(n-1) + ... + a_n.b_0.
N
Nate Eldredge
Kaz Kylheku said:
Okay, perhaps you should propose a definition for exp(x), and we can see
how the desired properties follow.
exp(-x) is a notational convenience. You can replace it with 1/exp(x)
and the proof will go through the same way, using the quotient rule
instead of the product rule. Does that make you happier?
I don't think that it does.
Here is another theorem, which doesn't mention exp(x) at all.
Lemma. Suppose g is a once-differentiable function on R such that
g'(x) = g(x) for all x in R. Then either g is the zero function, or
g(x) is nonzero for all x in R.
Proof. Suppose, in order to get a contradiction, that there exist x0,
x1 such that g(x0) = 0, g(x1) != 0.
Suppose first that x0 < x1 and g(x1) > 0. Let y = sup { x < x1 : g(x) = 0 }.
Because g is differentiable, g is continuous; it follows that g(y) = 0
and g > 0 on the (nonempty) interval (y, x1). Since g = g', we have
that g is strictly increasing on (y, x1). Choose y1 in (y, x1) such
that 0 < y-y1 < 1. By the mean value theorem, there exists x in (y, y1)
such that g'(x) = g(y1)/(y-y1) > g(y1). Since g(x)=g'(x) we have g(x) >
g(y1). But this is absurd since x < y1 and g is increasing on (y, x1).
Suppose next that x0 > x1 and g(x1) > 0. Let y = inf { x > x1 : g(x) = 0 }.
As before, g(y) = 0 and g > 0 on (x1, y). Thus g is increasing on (x1, y).
But g(x1) > 0 and g(y) = 0 so this is absurd.
The cases where g(x1) < 0 are similar and left as an exercise to the
reader. QED.
Theorem. Suppose f,g are two once-differentiable functions on R such
that f'(x) = f(x) and g'(x) = g(x) for all x in R. Suppose further that
g is not the zero function. Then there exists a constant c such that
f(x) = c g(x) for all x in R.
Proof. By the lemma, g(x) is never zero, so the function h(x)=f(x)/g(x)
is differentiable on R. By the quotient rule, for any x in R we have
h'(x) = (f'(x)g(x) - f(x)g'(x))/g(x)^2
= (f(x)g(x) - f(x)g(x))/g(x)^2
= 0
since f'(x)=f(x), g'(x)=g(x). By the mean value theorem h is constant,
i.e. there exists c such that h(x)=c for all x. Thus f(x) = c g(x) for
all x in R. QED.
With this theorem in hand, start with your favorite definition for
exp(x). Use it to prove that exp'(x) = exp(x) for all x, and that
exp(x) is not the zero function. Then take g(x)=exp(x) in the theorem;
it follows that if f is any other function with f(x)=f'(x), then
f(x) = c*exp(x) for all x.
If you can also prove from your definition that exp(x) is never zero,
you can dispense with the lemma, which is most of the work.