Entire Function That Agrees with Log on Natural Numbers?
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Entire Function That Agrees with Log on Natural Numbers?
Do you think this can be done? I don't know much complex analysis, I suspect probably not, but I kind of need it.
Work so far:
I've been trying to work up something so far by cutting out a region and filling it in with something else, but no success.
The function Log(x)*[Cos((2^x)*Pi)+1]/2 agrees with Log on natural numbers but has a limit of zero as x goes to zero instead of negative infinity, so that might help with the singularity at zero, but I'm not sure what to do with the branch cut.
Danke
Work so far:
I've been trying to work up something so far by cutting out a region and filling it in with something else, but no success.
The function Log(x)*[Cos((2^x)*Pi)+1]/2 agrees with Log on natural numbers but has a limit of zero as x goes to zero instead of negative infinity, so that might help with the singularity at zero, but I'm not sure what to do with the branch cut.
Danke
Re: Entire Function That Agrees with Log on Natural Numbers?
I'm pretty sure you can find an entire function with prescribed values on any set that lacks limit points, but I can't find the theorem that says that.
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Re: Entire Function That Agrees with Log on Natural Numbers?
I mean, isn't anything that you can draw on a graph that passes the vertical line test considered a function? I mean, you probably won't be able to define it in terms of functions we're used to, like polynomials, sine and cosine, exponentials etc., but its still follows all the rules of a function. And I can imagine very easily a function that lines up with natural log on natural numbers
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Re: Entire Function That Agrees with Log on Natural Numbers?
He's asking about an entire function.
Re: Entire Function That Agrees with Log on Natural Numbers?
Oh. I just thought he wanted the whole function. >.>
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Re: Entire Function That Agrees with Log on Natural Numbers?
I'd suggest log(x+1) but it's not everywhere differentiable. But it's close!
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Re: Entire Function That Agrees with Log on Natural Numbers?
Eastwinn wrote:I'd suggest log(x+1) but it's not everywhere differentiable. But it's close!
It has to be Cdifferentiable and not just Rdifferentiable to be entire, and this function isn't Cdifferentiable anywhere. (The absolute value function is not terribly nice from the point of view of complex analysis.)
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Re: Entire Function That Agrees with Log on Natural Numbers?
Suffusion of Yellow wrote:Work so far:
I've been trying to work up something so far by cutting out a region and filling it in with something else, but no success.
This is impossible. The values on any open set force the values everywhere else.
I think that starting with log is also a mistake. The essential singularity will be a real killer to get past.
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Re: Entire Function That Agrees with Log on Natural Numbers?
In spite of the misunderstanding of "entire", this is probably true.Kurushimi wrote: you probably won't be able to define it in terms of functions we're used to, like polynomials, sine and cosine, exponentials etc.
Is there a limit for the sequence of polynomials of degree n1 that match log for the first n positive integers?
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Re: Entire Function That Agrees with Log on Natural Numbers?
Sure  the empty function.
... Oh, you meant the limit as n goes to infinity. Probably not.
... Oh, you meant the limit as n goes to infinity. Probably not.
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Re: Entire Function That Agrees with Log on Natural Numbers?
gmalivuk wrote:In spite of the misunderstanding of "entire", this is probably true.Kurushimi wrote: you probably won't be able to define it in terms of functions we're used to, like polynomials, sine and cosine, exponentials etc.
Is there a limit for the sequence of polynomials of degree n1 that match log for the first n positive integers?
I think that'd result in some infinitely large coefficients, which just breaks things.
Though if we divide one polynomial by another, we can do it, it won't look anything like the log function off the naturals, but hey.
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Re: Entire Function That Agrees with Log on Natural Numbers?
My idea is, take an entire function that's 0 on the naturals (for example the Weirstrass sigma function on the lattice generated by {1,i}), then multiply it by k/(zn) to get an entire function that is 0 on the naturals except for n, and log(n) at n. Then add them up. No idea if this converges Probably not since you'll get something looking like the harmonic series... so take the Weirstrass sigma function squared and multiply by k/(zn)^2 for appropriate k... then it might converge (what do the k's look like?).. but does it converge to an entire function?
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Re: Entire Function That Agrees with Log on Natural Numbers?
skeptical scientist wrote:Eastwinn wrote:I'd suggest log(x+1) but it's not everywhere differentiable. But it's close!
It has to be Cdifferentiable and not just Rdifferentiable to be entire, and this function isn't Cdifferentiable anywhere. (The absolute value function is not terribly nice from the point of view of complex analysis.)
Oops, this makes the problem far more interesting.
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Re: Entire Function That Agrees with Log on Natural Numbers?
I'm a calculus newbie, but what about
[math]\log\left(x\right)\left[\cos^2\left(\pi x\right)\frac{1}{\Gamma\left(1x\right)}\right]^2[/math]
EDIT: Apparantly, I'm a latex newbie as well
[math]\log\left(x\right)\left[\cos^2\left(\pi x\right)\frac{1}{\Gamma\left(1x\right)}\right]^2[/math]
EDIT: Apparantly, I'm a latex newbie as well
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Re: Entire Function That Agrees with Log on Natural Numbers?
undecim wrote:I'm a calculus newbie, but what about
[math]\log\left(x\right)\left[\cos^2\left(\pi x\right)\frac{1}{\Gamma\left(1x\right)}\right]^2[/math]
Even if you plug the "holes" of that function at the natural numbers by log(x), the function isn't entire since you're multiplying by the complex logarithm.
Re: Entire Function That Agrees with Log on Natural Numbers?
NathanielJ wrote:undecim wrote:I'm a calculus newbie, but what about
[math]\log\left(x\right)\left[\cos^2\left(\pi x\right)\frac{1}{\Gamma\left(1x\right)}\right]^2[/math]
Even if you plug the "holes" of that function at the natural numbers by log(x), the function isn't entire since you're multiplying by the complex logarithm.
I thought that the only reason the complex logarithm was not an entire function is because of the pole at 0. With this equation, the limit as x approaches 0 is 0 (unless I screwed up somewhere, but the plot of that function appears to agree with me. Maybe I should stop doing math before my morning coffee)
Or did I completely misunderstand something?
EDIT: Yup, I did misunderstand something. I just looked at the surface plot of Im(ln(z)) in Genius.
EDIT: Wouldn't this fix it, or does absolute value break it in another way?
[math]\left\log\left(x\right)\right\left[\cos^2\left(\pi x\right)\frac{1}{\Gamma\left(1x\right)}\right]^2[/math]
P.S: I learn about math on Wikipedia and Mathworld, so if you can be explicit about anything wrong with it, it would help me learn
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Re: Entire Function That Agrees with Log on Natural Numbers?
Generally, any function defined using absolute value in a nontrivial way is going to fail to be complex analytic.
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Re: Entire Function That Agrees with Log on Natural Numbers?
gmalivuk wrote:In spite of the misunderstanding of "entire", this is probably true.Kurushimi wrote: you probably won't be able to define it in terms of functions we're used to, like polynomials, sine and cosine, exponentials etc.
Is there a limit for the sequence of polynomials of degree n1 that match log for the first n positive integers?
This EXACT question was in a highschoollevel math competition I did a week ago. (Well, log base 2.)
I didn't solve it. :/
And the website is terribly organized, they don't have the answer up yet.
Ah well.
(Btw  it didn't seem to lead to any infinite coefficients. In fact, the terms seemed to generally shrink.)
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Re: Entire Function That Agrees with Log on Natural Numbers?
Given a set [imath]\{z_n\}[/imath] of complex numbers with [imath]\lvert z_n \rvert \rightarrow \infty[/imath], it is always possible to find an entire function with a prescribed value at each [imath]z_n[/imath]. This is a consequence of the Weierstrass Factorization Theorem and the MittagLeffler Theorem (see Ahlfors or any other complex variables text). Weierstrass can give you an entire function [imath]f[/imath] with a simple zero at each [imath]z_n[/imath], and MittagLeffler can give you a holomorphic function [imath]g[/imath] with singular part [imath]\frac{a_n}{zz_n}[/imath] at each [imath]z_n[/imath]. Then [imath]fg[/imath] is an entire function with [imath](fg)(z_n)=a_n[/imath] for each [imath]n[/imath].
So your function exists. If you need an explicit expression, try something like:
[math]\sum_{n=1}^\infty \frac{\sin(\pi z)}{\pi \cos(\pi n)} \frac{e^{c_n (zn)}}{zn} \log(n)[/math]
where the [imath]c_n[/imath] are some constants chosen so that the sum converges.
So your function exists. If you need an explicit expression, try something like:
[math]\sum_{n=1}^\infty \frac{\sin(\pi z)}{\pi \cos(\pi n)} \frac{e^{c_n (zn)}}{zn} \log(n)[/math]
where the [imath]c_n[/imath] are some constants chosen so that the sum converges.
Re: Entire Function That Agrees with Log on Natural Numbers?
Right, I was trying to remember MittagLeffler, but only remembered Runge (which is used in proving it). Thanks!
A small correction, [imath](fg)(z_n) =k_na_n[/imath] where k depends only on [imath]f[/imath], so you can pick the [imath]a_n[/imath] appropriately when building [imath]g[/imath].
As an aside, I wonder why the conditions on the two theorems are stated as [imath]z_n \to \infty[/imath] for one and discrete for the other... these seem equivalent to me, as any discrete set can only have finitely many values in any compact subset, in particular for the set [imath]z \le M[/imath], and any sequence whose magnitude grows without bounds cannot have a limit point (what would it's magnitude be?) Am I missing some subtle (or not so subtle) difference? (I guess one has a prescribed order, but a discrete set must be countable, so it may as well be a sequence)
A small correction, [imath](fg)(z_n) =k_na_n[/imath] where k depends only on [imath]f[/imath], so you can pick the [imath]a_n[/imath] appropriately when building [imath]g[/imath].
As an aside, I wonder why the conditions on the two theorems are stated as [imath]z_n \to \infty[/imath] for one and discrete for the other... these seem equivalent to me, as any discrete set can only have finitely many values in any compact subset, in particular for the set [imath]z \le M[/imath], and any sequence whose magnitude grows without bounds cannot have a limit point (what would it's magnitude be?) Am I missing some subtle (or not so subtle) difference? (I guess one has a prescribed order, but a discrete set must be countable, so it may as well be a sequence)
Last edited by mikel on Mon Nov 22, 2010 2:26 pm UTC, edited 4 times in total.
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Re: Entire Function That Agrees with Log on Natural Numbers?
forgetful functor wrote:Given a set [imath]\{z_n\}[/imath] of complex numbers with [imath]\lvert z_n \rvert \rightarrow \infty[/imath], it is always possible to find an entire function with a prescribed value at each [imath]z_n[/imath]. This is a consequence of the Weierstrass Factorization Theorem and the MittagLeffler Theorem.
Ah thanks. Mikel guessed this was true, and I also thought it was probably true, but we didn't have the proof.
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Re: Entire Function That Agrees with Log on Natural Numbers?
mikel wrote:As an aside, I wonder why the conditions on the two theorems are stated as [imath]z_n \to \infty[/imath] for one and discrete for the other... these seem equivalent to me, as any discrete set can only have finitely many values in any compact subset, in particular for the set [imath]z \le M[/imath], and any sequence whose magnitude grows without bounds cannot have a limit point (what would it's magnitude be?) Am I missing some subtle (or not so subtle) difference? (I guess one has a prescribed order, but a discrete set must be countable, so it may as well be a sequence)
No difference, must just be the text you are looking in. If you have any set [imath]S \subset C[/imath] without a limit point in [imath]C[/imath], then you can find a holomorphic (or meromorphic) function with zeros (or poles) at the points of [imath]A[/imath].
mikel wrote:A small correction, [imath](fg)(z_n) =k_na_n[/imath] where k depends only on [imath]f[/imath], so you can pick the [imath]a_n[/imath] appropriately when building [imath]g[/imath].
Thanks, I knew what I had seemed to easy. The example I gave should still work, though.

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Re: Entire Function That Agrees with Log on Natural Numbers?
forgetful functor wrote:Given a set [imath]\{z_n\}[/imath] of complex numbers with [imath]\lvert z_n \rvert \rightarrow \infty[/imath], it is always possible to find an entire function with a prescribed value at each [imath]z_n[/imath]. This is a consequence of the Weierstrass Factorization Theorem and the MittagLeffler Theorem (see Ahlfors or any other complex variables text). Weierstrass can give you an entire function [imath]f[/imath] with a simple zero at each [imath]z_n[/imath], and MittagLeffler can give you a holomorphic function [imath]g[/imath] with singular part [imath]\frac{a_n}{zz_n}[/imath] at each [imath]z_n[/imath]. Then [imath]fg[/imath] is an entire function with [imath](fg)(z_n)=a_n[/imath] for each [imath]n[/imath].
So your function exists. If you need an explicit expression, try something like:
[math]\sum_{n=1}^\infty \frac{\sin(\pi z)}{\pi \cos(\pi n)} \frac{e^{c_n (zn)}}{zn} \log(n)[/math]
where the [imath]c_n[/imath] are some constants chosen so that the sum converges.
Cool, thank you very much. I do need an explicit form (or something close to it), i'll look into that.
I've been messing around with something that looks like c1*(ex)+c2*(ex)(e^2x)+c3*(ex)(e^2x)(e^3x) but I haven't shown convergence (yet).

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Re: Entire Function That Agrees with Log on Natural Numbers?
mikel wrote:I'm pretty sure you can find an entire function with prescribed values on any set that lacks limit points, but I can't find the theorem that says that.
In this context, is infinity a limit point of the natural numbers?
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Re: Entire Function That Agrees with Log on Natural Numbers?
No. Only limit points in the complex numbers count.
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Re: Entire Function That Agrees with Log on Natural Numbers?
Suffusion of Yellow wrote:mikel wrote:I'm pretty sure you can find an entire function with prescribed values on any set that lacks limit points, but I can't find the theorem that says that.
In this context, is infinity a limit point of the natural numbers?
If you're talking about the Riemann sphere, then infinity is a limit point, and you can't necessarily find a function which is analytic on the Riemann sphere with given values on the natural numbers. But for this question, we're only interested in the function being analytic on C (an entire function), and the only possible obstruction would be a limit point in C.
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