User anixx - MathOverflow

Can an uniformly picked real number be an integer?

2013-05-24T16:50:32Z

In this highly upvoted answer it is claimed that by picking uniformly a random number from interval [0,100] one can pick an integer, literally "zero probability does not never happen".

When I asked the answerer for an example of a randomly-picked integer or any number from [0,100], he gave me a long binary expansion like "0.10011001011001..." (the comment is now deleted). I objected that obviously this is not a number. I was invited into a chat where I was told that I am silly and I do not know that I am speaking with two professors of mathematics. The discussion ended up here.

It seems to me that by claiming that one can uniformly randomly pick a number from an interval, the both professors were assuming Axiom of Choice. As such I do not understand why they were so offensive against me defending something that is correct only if this axiom is postulated. Especially given that this axiom does not reflect what we can do on Turing-complete computers in real life (either on analog, digital, probabilistic, quantum or in our brain), at least this would take infinite memory and time. As such their categorical claim that one can choose a random number from an interval seems unjustified to me.

But suppose now we take Axiom of Choice as true. Will it be possible given this axiom to get an integer as a result of uniformly random pick from [1,100]?

To my impression the result of such uniformly random choice inevitably should be

undefinable
normal (that is the digits of expansion should be uniformly distributed)

As it correct?

New differintegral formula: how is it related to other differintegral formulas?

2013-05-17T00:33:50Z

Lets define new differintegral formula as

$$\mathbb{D}^s_xf(x)= \sum_{m=0}^{\infty} \binom {s}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{(k)}(x)$$

or, equivalently,

$$\mathbb{D}^s_xf(x)= \lim_{t\to x} \lim_{n\to\infty}\frac{\sum_{k=0}^{n} \frac{(-1)^k f^{(k)}(t)}{(s-k)k!(n-k)!}}{\sum_{k=0}^{n} \frac{(-1)^k }{(s-k) k!(n-k)!}}$$

with the conditions that the above expression converges, and $(\mathbb{D}^{-s}_xf(x))^{(s)}=f(x)$ for each x.

Will this formula where the conditions are met give the same results as Riemann–Liouville differintegral, Grunwald–Letnikov differintegral and Weyl differintegral?

Answer by Anixx for Rational functions with a common iterate

2013-04-26T01:41:43Z

Regarding the second question the algorithm is as follows:

Find the flows (superfunctions) of the both functions in closed form
See if they coincide at integer points.

For example, $f(x)=x^2$, $g(x)=x^4$.

The flows will be respectively, $C^{2^x}$, $C^{4^x}$.

Now we solve

$$C^{2^x} = C^{4^y}$$

and find $y=x/2$

This equation obviously has infinitely many integer solutions.

A more complicated case is when $f(x)=\frac{x+1}{x-1}$, $g(x)=\frac{x-1}{x+1}$

In this case the flows are:

$$f^*(x)=\frac{C \cos \left(\frac{3 \pi x}{4}\right)+\sin \left(\frac{3 \pi x}{4}\right)}{\cos \left(\frac{3 \pi x}{4}\right)-C \sin \left(\frac{3 \pi x}{4}\right)}$$

$$g^*(x)=\frac{\left(\left(\sqrt{2}-1\right) C-1\right) (-1)^x+\sqrt{2} C+C+1}{\left(-C+\sqrt{2}+1\right) (-1)^x+C+\sqrt{2}-1}$$

Solving equation f*(x) = g*(y) for integer x and y gives $x=4m, y=n$

Answer by Anixx for What's a natural candidate for an analytic function that interpolates the tower function?

2010-11-03T04:44:12Z

The function you want grows too fast to be interpolated by usual method, but there exists an iterative solution with Cauchy integrals by Dmitry Kouznetsov

If you relax the condition so to find a solution for $f(x+1)=a^{f(x)}$ such that $$a \le e^{1/e} $$ then there are multiple expressions for your function:

$$f(x)=\sum_{m=0}^{\infty} \binom xm \sum_{k=0}^m \binom mk (-1)^{m-k}\exp_a^{[k]}(1)$$

$$f(x)=\lim_{n\to\infty}\binom xn\sum_{k=0}^n\frac{x-n}{x-k}\binom nk(-1)^{n-k}\exp_a^{[k]}(1)$$

$$f(x)=\lim_{n\to\infty}\frac{\sum_{k=0}^{2n} \frac{(-1)^k \exp_a^{[k]}(1)}{(x-k)k!(2n-k)!}}{\sum_{k=0}^{2n} \frac{(-1)^k }{(x-k) k!(2n-k)!}}$$

$$f(x)=\lim_{n\to\infty} \log_a^{[n]}\left(\left(1-\left(\ln \left(\frac{W(-\ln a)}{-\ln a}\right)\right)^x\right)\frac{W(-\ln a)}{-\ln a}+\ln \left(\frac{W(-\ln a)}{-\ln a}\right)\exp_a^{[n]}(1)\right)$$

Always here the number in square brackets designates n-th iteration and $W(x)$ is the Lambert's function.

There is also expression for inverse function:

$$ f^{[-1]}(x)=\lim_{n\to\infty} \frac{\ln \left(\frac{\frac{W(-\ln a )}{\ln a}+\exp_a^{[n]}(x)}{\frac{W(-\ln a)}{\ln a}+\exp_a^{[n]}(1)}\right)}{\ln \ln \left(\frac{W(-\ln a)}{-\ln a}\right)}$$

Proof that derivative of Hurwitz Zeta by the first argument is not expressable in terms of Hurwitz Zeta

2012-04-07T11:22:01Z

The set of elementary functions is defined so that it to be closed against operation of differentiation. It is also evidently close against discrete differentiation.

In the discrete calculus there is a similar set of functions, but with one sufficient difference. It appears not to be closed against normal (non-discrete) differentiation.

But I need a proof.

So I am asking for a proof for the following statement regarding Hurwitz Zeta: $$\frac{d}{dq}\zeta(q,p)$$ cannot be expressed in terms of elementary functions and Hurwitz Zeta.

UPDATE

I found the following formula which connects the two functions, but still a question remains whether one of them can be expresses explicitly.

$ \zeta '\left(z,\frac{q}{2}\right)-2^z \zeta '(z,q)+\zeta '\left(z,\frac{q+1}{2}\right)=\zeta(z,q)2^{z}\ln 2$

Is exponent of discrete-analytic function also discrete-analytic?

2012-03-09T18:32:01Z

Lets define a discrete analytic function such a function that is equal to its Newton series:

$$f(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k f\left (0\right)$$

Is function $g(x)=e^{f(x)}$ also discrete-analytic?

This question arose from the following considerations.

As you know the difference equation

$$\Delta y(x) = F(x)$$

has multiple solutions that differ only by an arbitrary 1-periodic function $C(x)$:

$$y(x)=y_1(x)+C(x)$$

At the same time there can be no more than one (up to a constant term) discrete-analytic solution which we can consider to be the natural solution of the equation.

But when considering multiplicative-difference equation $\frac{y(x+1)}{y(x)}=F(x)$ we come to a similar situation, this equation has multiple solutions which differ by an arbitrary 1-periodic factor:

$$y(x)=C(x)y_1(x)$$

Of these solutions, similarly, no more than one (up to a constant factor) is discrete-analytic which allows us to define the distinguished solution.

But on the other hand the following rule holds for indefinite product and sum:

$$\prod_x f(x)= e^{\sum_x \ln f(x)}$$

This means that we can obtain the solution to the equation $\frac{y(x+1)}{y(x)}=F(x)$ in the following form:

$$y(x)=e^{\sum_x \ln F(x)}$$

This allows us to select the distinguished solution by another method, that is taking the natural solution to the sum and taking exponent of it. The result will have a constant factor, but it is unevident whether it will be discrete-analytic or not, and as such, whether the both distinguished solutions coincide.

UPDATE

Due to the answer by David Speyer it is evident now that counter-examples exist among complex-valued functions and also there are instances when function $f(x)$ is discrete-analytic while the Newton series of its exponent does not converge.

So the question should be formulated more precisely: we assume that $f(x)$ is real-valued and Newton series for its exponent converges.

I started a bounty for this question

ADDENDUM

It would be even more great if somebody could prove a more general theorem about a composition of monotonous discrete-analytic functions. Whether the composition is also discrete-analytic and under what conditions.

Answer by Anixx for Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) )

2012-11-01T17:19:43Z

And regarding real solutions to the question, Alex Gavrilov is completely correct. A Taylor expansion at fixed point $p$ gives us the real solution. Existence of this solution is proven in the paper which I already referenced from my another answer.

$$f(z)=\sum_{n=0}^\infty \frac{d_n (z-p)^n}{n!}$$

where $d_n$ is defined as follows:

$$d_0=p$$ $$d_{n+1}=\sum _{k=0}^n d_k \operatorname{B}_{n,k}(d_1,...,d_{n-k+1})$$

where $B_{n,k}$ are the Bell polynomials

This gives the following starting coefficients:

$$d_1=p^2$$ $$d_2=p^3+p^4$$ $$d_3=p^4 + 4 p^5 + p^6 + p^7$$ $$d_4=p^5 + 11 p^6 + 11 p^7 + 8 p^8 + 4 p^9 + p^{10} + p^{11}$$

etc.

The fixed point $p$ here serves as a parameter, which determines the family of solutions. According the linked theorem, the expansion should converge in the neighborhood of $p$ for $0 < |p| < 1 $ or $p$ being a Siegel number.

Answer by Anixx for Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) )

2012-11-01T03:06:48Z

There are two closed form solutions:

$$\displaystyle f_1(x) = e^{\frac{\pi}{3} (-1)^{1/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$$ $$\displaystyle f_2(x) = e^{\frac{\pi}{3} (-1)^{11/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$$

The solution technique can be found in this paper.

For a general case, solution of the equation

$$f'(z)=f^{[m]}(z)$$

has the form

$$f(z)=\beta z^\gamma$$

where $\beta$ and $\gamma$ should be obtained from the system

$$\gamma^m=\gamma-1$$ $$\beta^{\gamma^{m-1}+...+\gamma}=\gamma$$

In your case $m=2$.

Can Bernoulli polynomials be extended to fractional orders without losing elementarity?

2012-06-19T08:02:37Z

Can Bernoulli polynomials $B_s(x)$ be extended to fractional $s$ in such a way so that for any given $s$ the function $B_s(x)$ still could be expressed in elementary functions of $x$?

Do complex iterates of functions have any meaning?

2011-07-27T19:00:58Z

Using a method explained in this answer it is possible to calculate not only integer and real iterates of functions but also complex ones, for example, the $i$-th iterate, where $i=\sqrt{-1}$. Here are graphs of the $i$-th iterates of some common functions (the blue is the real part and the red curve is the imagine part):

$$\arctan^{[i]}(x)$$

$$\sin^{[i]}(x)$$

So the questin is whether there is any intuitive meaning in complex iterates, especially, say, $i$-th iterates of functions?

Connection between Bernoulli polynomials and polygamma function

2010-10-18T21:07:06Z

There is an intricate connection between Hurwitz Zeta and the (traditional) polygamma function:

$$\psi_n(z)=(-1)^{n+1}n!\zeta(n+1,z)$$

If to use a generalization for Bernoulli numbers, this can be considered a formula, connecting polygamma and Bernoulli polynomials of negative order:

$$\psi_n(z)=(-1)^{n+1}n!\frac{B_{-n}(x)}n$$ (1)

As much as this equality is impressing, it is limited as it only holds for natural n.

After a couple of unsuccessful attempts to find a more general formula, I encountered a more natural, "balanced" generalization of polygamma function explained in this paper. It turned out that while the old formula still holds for natural n (since the old polygamma and balanced polygamma coincide in integer positive orders), a completely new formula connecting this balanced polygamma with Zeta and Bernoulli numbers can be derived which holds for any z:

$$\zeta(z,q)=\frac{\Gamma (1-z) \left(2^{-z} \left(\psi \left(z-1,\frac{q}{2}+\frac{1}{2}\right)+\psi \left(z-1,\frac{q}{2}\right)\right)-\psi(z-1,q)\right)}{\ln(2)}$$

$$B_z(q) = -\frac{\Gamma (z+1) \left(2^{z-1} \left(\psi\left(-z,\frac{q}{2}+\frac{1}{2}\right)+\psi\left(-z,\frac{q}{2}\right)\right)-\psi(-z,q)\right)}{\ln (2)}$$

Both of them can be expressed completely in terms of balanced polygamma and elementary functions if to notice that $\Gamma(x)=e^{\psi(-1,x)+\frac 12 \ln(2\pi)}$, which allows to get rid of the Gamma function.

While the target was reached, these expressions still leave a bad impression. I cannot simplify it as no CAS system is capable of operations with the balanced polygamma.

Hence I am asking for help on how to simplify the expressions so they could be easier to manage and use. It is also not evident how the letter formulas become the former ones at positive real z.

Convergence of Newton series for sin ax

2012-06-09T09:34:07Z

Let's define half discrete-analytic function as a function whose Newton series converges to that function for each $x>0$:

$$f(x)=\sum_{k=0}^\infty \binom{x}k \Delta^k f\left (0\right)=\sum_{m=0}^{\infty} \binom {x}m \sum_{k=0}^m\binom mk(-1)^{m-k}f(k)=\lim_{n\to\infty}\frac{\sum_{k=0}^{n} \frac{(-1)^k f(k)}{(x-k)k!(n-k)!}}{\sum_{k=0}^{n} \frac{(-1)^k }{(x-k) k!(n-k)!}}$$

Let's define weak discrete-analytic a function a function whose bi-directional Newton expansion converges to that function at any real $x$:

$$f(x)=\lim_{n\to\infty}\frac{\sum _{k=-n}^n \frac{(-1)^k f(k)}{(x-k) (k+n)! (n-k)!}}{\sum _{k=-n}^n \frac{(-1)^k}{(x-k) (k+n)! (n-k)!}}$$

It seems that the function $\sin x$ is both half discrete-analytic and weak discrete-analytic. But the function $\sin \frac{\pi x}2$ is only weak discrete-analytic.

So what is the maximum $a$ such that $\sin ax$ is half discrete-analytic?

I also interested to know at which $a$ $\sin ax$ is weak discrete-analytic. For example, is $\sin 3x$ weak discrete-analytic or not?

This question is motivated by my old search for a natural fractional integration and integration constant (integral analog of Ramanjuan sum). I want to find a natural generalization of Newton series to functions whose Newton series normally diverges. Once that accomplished it would be possible to find Newton expansions for consecutive derivatives of a function and by analytically continuing them into negative domain, get natural integral. Unfortunately finding natural fractional integral of even such simple function as $\sin x$ requires building Newton expansion for $\sin \frac{\pi x}2$ which diverges.

Is the analysis as taught in universities in fact the analysis of definable numbers?

2010-10-29T10:47:27Z

Ten years ago when I studied in the university I had no idea about definable numbers, but I came to this concept myself. My thoughts were as follows:

All numbers are divided into two classes: those which can be unambiguously defined by a limited set of their properties (definable) and such that for any limited set of their properties there is at least one other number which also satisfies all these properties (undefinable).
It is evident that since the number of properties is countable, the set of definable numbers is countable. So the set of undefinable numbers forms a continuum.
It is impossible to give an example of an undefinable number and one researcher cannot communicate an undefinable number to the other. Whatever number of properties he communicates there is always another number which satisfies all these properties so the researchers cannot be confident whether they are speaking about the same number.
However there are probability based algorithms which give an undefinable number in a limit, for example, by throwing dice and writing consecutive numbers after the decimal point.

But the main question that bothered me was that the analysis course we received heavily relied on constructs such as 'let's $a$ to be a number that...", "for each $s$ in interval..." etc. These seemed to heavily exploit the properties of definable numbers and as such one can expect the theorems of analysis to be correct only on the set of definable numbers. Even the definitions of arithmetic operations over reals assumed the numbers are definable. Unfortunately one cannot take an undefinable number to bring a counter-example just because there is no example of undefinable number, but still how to know that all those theorems of analysis are true for the whole continuum and not just for a countable subset?

Is there a known formula for fractional derivative of cot x?

2012-04-16T16:37:20Z

I wonder if there any established formula for fractional derivative of a function $\pi \cot (\pi x)$.

I derived the following expression:

$(\pi \cot (\pi q))^{(p)}=-\frac{\zeta'(p+1,q)+(\psi(-p)+\gamma ) \zeta (p+1,q)}{\Gamma (-p)}-\Gamma (p+1) \zeta (p+1,1-q)$

where $\psi$ is digamma, $\zeta$ is Hurwitz zeta, $\zeta'$ is the derivative by first argument

and I want to compare it with the other expressions, and, possibly, equate them to derive further results.

My derivation is as follows.

First of all there is a known formula that is only valid for integer $n$:

$\psi_n(z)=(-1)^{(n+1)}n!\zeta(n+1,z)$ (can be seen here: http://mathworld.wolfram.com/PolygammaFunction.html, formula 12).

Espinoza and Moll mention in their article on the definition of balanced polygamma that they were unable to find a generalization for this formula despite any attempts.

If to try to directly expand the formula to non-integer values, the resulting function will be non-real. If to replace the $(-1)^n$ with a cosine, the resulting polygamma generalization will be undefined at negative integer values. Red line on this graph:

This is highly undesirable because many formulas for integrals and discrete integrals use negapolygamma at negative integer order. If we take a traditional or balanced polygamma, multiplying it by cosine or another simple fiunction will not give the zeta function. Even more the resulting function will not be discrete-analytic. So this approach is also undesirable.

Following this a thought came to my mind that the alternating sign naturally arises if we differentiate a function of negative argument.

Since $\psi(1-x)=\psi(x) + \pi\cot\pi x$ we can rewrite the previously cited formula in the following form:

$(\psi(x) + \pi\cot\pi x)^{(n)}=-n! \zeta(n+1,1-x)$

By taking different definitions of fractional derivative of cotangent one can arrive at different generalizations of polygamma.

The formula also works in the opposite way: by taking a given generalization of polygamma one can arrive at different expressions for fractional derivative of cotangent.

Indeed,

$(\pi\cot\pi x)^{(n)}=-n! \zeta(n+1,1-x)-\psi^{(n)}(x)$

If we assume the polygamma being the balanced generalization, the formula

$(\pi \cot (\pi q))^{(p)}=-\frac{\zeta'(p+1,q)+(\psi(-p)+\gamma ) \zeta (p+1,q)}{\Gamma (-p)}-\Gamma (p+1) \zeta (p+1,1-q)$

arises.

This gives the formula $(\cot (q))^{(p)}=-\frac{\zeta'(p+1,\frac q\pi)+(\psi(-p)+\gamma ) \zeta (p+1,\frac q\pi)}{\pi^{p+1}\Gamma (-p)}-\frac 1{\pi^{p+1}}\Gamma (p+1) \zeta (p+1,1-\frac q\pi)$ for fractional derivative of cotangent.

Below is the graphic of function $2\cot x \csc x^2$ which is the second derivative of cotangent and the 1.9999th derivative of cotangent following the above formula. The graphs seem to coincide.

I wonder whether there are known other generalizations of fractional derivative of cotangent and which corresponding generalizations of polygamma will arise if to insert those in the previous formula.

Is Zeta function discrete-analytic?

2012-04-15T22:34:15Z

Let's define discrete-analytic functions as functions that are equal to their Newton series expansion:

$$f(x) = \sum_{k=0}^\infty \binom{x-a}k \Delta^k f(a)$$

My question is whether $\zeta(s,q)$ ($q$=const) is discrete-analytic against $s$?

That is whether its Newton series converges and is equal to the function itself.

For comparison, in the following graphic there are four functions:

red is the function $\zeta(x,3)$
blue is $\frac{\cos (\pi x)\psi_b^{(x+1)}(3)}{\Gamma(x+2)}$ where $\psi_b$ is the balanced polygamma
yellow is $\frac{\cos (\pi x)\psi^{(x+1)}(3)}{\Gamma(x+2)}$ where $\psi$ is the polygamma as implemented in Mathematica
green is the partial Newton expansion of the above functions taken at first 20 terms.

The three first functions and the Newton expansion, if it converges, have the same values at non-negative integer arguments.

notation $\zeta(x,q)$ is the Hurwitz zeta function, LINK

Balanced polygamma LINK

Discrete-analytic functions

2011-07-25T09:55:37Z

I do not know if such concept already exists but lets consider functions which are equal to its Newton series.

We know that functions which are equal to their Taylor series are called analytic, so lets call functions that are equal to their Newton series "discrete analytic".

The formula is alalogious to Taylor series but uses finite differences instead of dirivatives, so for any discrete-analytic function:

$$f(x) = \sum_{k=0}^\infty \binom{x-a}k \Delta^k f\left (a\right)$$

It is known that for a functional equation $\Delta f=F\, $there are infinitely many solutions which differ by any 1-periodic function. But it appears that there is only one (up to a constant) discrete-analytic solution, i.e. all discrete-analytic solutions differ only by a constant term.

Thus I have the following questions:

Do discrete-analytic functions express special properties on the complex plane?
Is there a method to extend the notion of discrete analiticity to a range of functions for which Newton series does not converge (so to make it possible to choose the distinguished solution to the abovementioned equation)?

For the second part of the question, as I know there is at least one one similar attempt, the Mueller's formula:

If $$\lim_{x\to{+\infty}}\Delta f(x)=0$$ then $$f(x)=\sum_{n=0}^\infty\left(\Delta f(n)-\Delta f(n+x)\right)$$

although it seems not to be universal and I do not now whether it is always useful.

Are these two functions equal?

2011-07-30T08:28:38Z

The question here is sparked by the discussion inside this question about indefinite sum(antidifference) of tan(x).

A proposed solution was a function $$f_1(x)=ix-\psi _{e^{2 i}}^{(0)}\left(x+\frac{\pi }{2}\right)+C$$ The function involved is the q-digamma function. Symbolically it resolves to be antidifference of tan(x), one can check it by following this link to Wolfran Alpha's result: http://tiny.cc/60mmf

But it turned out that neither Wolfram Alpha, nor any other software is able to evaluate the q-polygamma function with $q=e^{2 i}$ numerically. Attempting to evaluate it manually also turned out to be too difficult.

A second solution seemed to be easier to evaluate. It was a series that converged absolutely and also has been proven to be an antidifference of tan(x):

$$f_2(x)=-\sum _{k=1}^{\infty } \left( \psi \left(k \pi -\frac{\pi }{2}+1-x\right)+\psi \left(k \pi -\frac{\pi }{2}+x\right) \right. $$ $$ \qquad \left. -\psi \left(k \pi -\frac{\pi }{2}+1\right)-\psi \left(k \pi -\frac{\pi }{2}\right)\right)+C$$

This function has a fancy graphic (see the initial discussion). Since it is known that a function can have several antidifferences which differ between each other by a 1-periodic function, it is interesting, whether the both functions $f_1(x)$ and $f_2(x)$ are equal or what is the difference between them (of course if the first function actually can be evaluated numerically).

Answer by Anixx for Multiplicative integral of $\Gamma(x)$

2010-10-18T10:37:43Z

Multiplicative integral of gamma function is

$$\int \Gamma(x)^{dx}=C e^{\psi^{(-2)}(x)}$$

if to use the popular generalization of polygamma function $\psi^{(p)}(z)$ put forward by Grossman in 1976.

If to use a more modern "balanced" generalization $\psi(p,x)$ by Espinoza and Moll of 2004, the integral will be as follows:

$$\int \Gamma(x)^{dx}=C e^{\psi(-2,x)+\frac {x}{2}\ln 2\pi}$$

Can roots of any polynomial be expressed using Eulerian function?

2012-02-21T22:47:49Z

I encountered an interesting function which is called "Eulerian" by the Wolfram's MathWorld:

$$\phi(q)=\prod_{k=1}^{\infty} (1-q^{k})$$

It is interesting because it is claimed that roots of any polynomial can be expressed in this function and elementary functions. Is this true and how the roots of arbitrary polynomial can be expressed?

P.S. In Mathematica this function is inplemented as QPochhammer[q]

Answer by Anixx for How to solve f(f(x)) = cos(x) ?

2010-11-03T20:58:44Z

This answer has been deleted for the first time, so this is a re-post. The first answer also had an error in the plot. Since the questioner asked the following question "Is there a general solution strategy to equations of this kind?" I'll try to respond.

The half-iterate of a function can be found by expressing its superfunction in a form of Newton series:

$$f^{[1/2]}(x)=\sum_{m=0}^{\infty} \binom {1/2}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{[k]}(x)$$

Where $f^{[k]}(x)$ means k-th iterate of $f(x)$ This series converges if two criteria are met:

1) The superfunction of f(x) grows not faster than an exponent

2) Runge phenomenon does not appear.

There is a number of strategies to combat Runge phenomenon which are outside of this answer's scope. It is worth noting though that trying to find a half iterate of the function $f(x)=\cos x$ leads to this Runge swamp and one needs to employ one of the mentioned techniques to acheve convergence.

Opposite case is with the function $f(x)=\sin x$. The superfunction is limited by $\pm 1$ and the series converges without any problem.

Below is a plot of half-iterate of $\sin x$, obtained with this formula. It is periodic with the same period as $\sin x$. The blue curve is the half-iterate, and the red curve is the half-iterate, repeated twice, and we can see that it is indeed very similar to sine function.

This plot is made from the first 50 terms of the above series.

This formula for the half-iterate can be used to find not only half-itertes but any real (or even complex!) iterate of a function by substituting the needed value instead of 1/2.

The formula can be also written in the following forms:

$$f^{[s]}(x)=\lim_{n\to\infty}\binom sn\sum_{k=0}^n\frac{s-n}{s-k}\binom nk(-1)^{n-k}f^{[k]}(x)$$

$$f^{[s]}(x)=\lim_{n\to\infty}\frac{\sum_{k=0}^{n} \frac{(-1)^k f^{[k]}(x)}{(s-k)k!(n-k)!}}{\sum_{k=0}^{n} \frac{(-1)^k }{(s-k) k!(n-k)!}}$$

There are also some other formulas giving the same result.

Answer by Anixx for What is convolution intuitively?

2011-06-13T18:13:37Z

Algebraic meaning of convolution in analysis can be seen from the following formula:

$$\int_{-\infty}^{+\infty} f(x)*g(x)\ dx=\left(\int_{-\infty}^{+\infty} f(x)\right)\left(\int_{-\infty}^{+\infty} g(x)\right)$$

So, in short, convolution is the product of integrals.

This is also true in discrete calculus where discrete convolution is used when multiplying sums, and in particular n-times repeated convolution is used in binomial theorem to express the power of a binomial.

Answer by Anixx for Computer Algebra Errors

2011-05-03T00:46:16Z

Wolfram Mathematica 7 routinely confuses sums with integrals.

Example 1:

DSolve[(-Log[Log[a]] f'[x] + f''[x])/(Log[a] f'[x]) == D[Sum[f[x], x], x], f[x], x]

g[x_] := f[x] /. s
g[x]

Checking the result by inserting it into the equation shows the result is incorrect:

(-Log[Log[a]] g'[x] + g''[x])/(Log[a] g'[x]) - D[Sum[g[x], x], x]

Example 2:

s=NDSolve[{0.9159460564995328*Derivative[1][f][x] == f[x]*Product[f[x], x], f[0] == 1}, f, {x, -1.9, 15}]

Plot[Evaluate[f[x] /. s], {x, -0.4, 1.5}, AspectRatio -> Automatic, AxesOrigin -> {0, 0}]

In Mathematica 8.0 this has been fixed (i.e. it will report inability to solve the equations.

Answer by Anixx for Computer Algebra Errors

2011-04-27T10:31:23Z

This error affects all versions of Mathematica from 6 to 8. The result of a function depends on what letter is chosen for argument when calling it. In the simplest case it can be illustrated as follows:

in:

$A[\text{x_}]\text{:=}\sum _{k=0}^{x-1} x $

$A[k]$

$A[z]$

out:

$1/2 (-1 + k) k$

$z^2$

The correct answer is evidently, the later. This behavior affects not only sums but also integrals, so one have to check so that the letter user for the argument not to coincide with the index variable used for definition. In the case of recursion this becomes very difficult. The following example shows that moving a factor not dependent on the index variable out of the sum sign changes the result:

in:

A1[0,x_]:=1
A2[0,x_]:=1

A1[n_,x_]:=Sum[A1[-1 - j + n, x]*Sum[A1[j, k], {k, 0, -1 + x}], {j, 0, -1 + n}]
A2[n_,x_]:=Sum[Sum[A2[j, k]*A2[-1 - j + n, x], {k, 0, -1 + x}], {j, 0, -1 + n}]

A1[1,x]/.x->2
A1[2,x]/.x->2
A1[3,x]/.x->2

A2[1,x]/.x->2
A2[2,x]/.x->2
A2[3,x]/.x->2

A2[1,2]
A2[2,2]
A2[3,2]

out:

Is there integral analog of Ramanujan sum?

2010-12-20T08:20:34Z

I already asked a similar question http://mathoverflow.net/questions/48742/is-there-natural-integration-constant-closed, but it seems it was not understood properly, so I am trying now to formulate it differently.

Is there an operator R[f] that plays in integral world the same role as Ramanujan sum plays in the world of series?

It is known that Ramanujan sum plays the role of natural integration constant for discrete integration, that is if F(x) is the discrete integral of f(x), usually postulated that

$$F(0)=\sum^{\Re}f(x)$$

This yields the functions which have some very useful properties, for example, Bernoulli polynomials (which are the results of discrete integration of power function), the Hurwitz Zeta function (which is generalization of Bernoulli polynomials), etc. Discrete integrals which normalized with Ramanujan sum (or equal method) called "balanced" (as opposed to some functions normalized without it, for example to be zero in zero such as Harmonic numbers).

One of the properties of balanced functions is that $$\int_0^1 F(x)\ dx=0$$

To outline the properties of such operator in integral world we should admit that such operator should be symmetric against zero (unlike the Ramanujan sum).

It is also highly desirable (if possible) that R[exp(x)]=1 thus making integral of exponent itself exponent and thus invariant against differintegral operator. This would allow to provide integration constants for trigonometric functions that would work in agreement with known expressions for differintegral:

$$D^q \sin x=\sin \left(x+\frac{q\pi}{2}\right)$$

P.S. See my own answer below.

Answer by Anixx for Is there integral analog of Ramanujan sum?

2011-04-02T04:26:34Z

I will try to answer this question myself.

My aim was to find the most natural and universal constant of integration which would allow to define an operator of "natural integration", unambiguously selecting one distinguished integral function for any given function.

So let's designate as $F(x)=\int_N f(x) dx$ the natural integral which we are trying to define and $R[f]=F(0)$ is the value of the natural integral in zero, the constant of integration we are trying to find.

First of all we know that operation of integration is symmetrical over choice of direction. This means that we should require that $R[f(x)]=-R[f(-x)]$. This follows from the fact that $(F(x))'|_0=-(f(-x))'|_0$ for any function F. It allows us to directly define the constant of integration for even functions: its natural integral should be odd function and $R[f]=0$ if $F[0]$ is defined.

Since all functions can be represented as a sum of even and odd function, we only have to define $R[f]$ for odd functions.

This is a more difficult task.

But we can spot one more property of natural integral. If we want it to be independent of any particular point on the real axis, any shift in the given function should lead to a corresponding shift in the integral without any other change. This means that if a function can be made even by shifting it along real axis, we can find its natural integral by applying the rule for even functions.

I.e. for continuous f, if $f(x_0-x)=f(x_0+x)$ for any $x$, then $R[f]=\int_{x_0}^0 f(t)dt$.

A function can have more than one axis of symmetry though, but if they are more than one, the function is periodic, and $R[f]$ is still unique.

This method allows us to find natural integrals for sine, cosine, hyperbolic sine and cosine as well as exponent as in the following table:

$\int_N \sin x dx = -\cos x$

$\int_N \cos x dx = \sin x$

$\int_N \sinh x dx = \cosh x$

$\int_N \cosh x dx = \sinh x$

$\int_N \exp x dx = \exp x$

But how can we define the natural integrals for other functions?

To cover all analytic functions we have to define the natural integral on polynomials.

First of all we spot that natural integral of hyperbolic sine has 1 in zero. This means that natural integration it term by term adds a sequence that sums up to 1. The most simple sequence of this kind is 1/2+1/4+1/8+1/16+... . Since integrating exponent gives the same result, it is logical that the terms which stay in odd position after integration contribute nothing. Similarly as integrating minus sine gives the same result in zero, we can conclude that all terms that stay on even positions but do not divide by 4 also contribute nothing.

Simplifying all said above and accounting for a factorial which exists in each term we can obtain a simple formula:

$$R[f]=\sum _{k=1}^{\infty} \frac{f^{(4k-1)}(0)}{2^k}$$

and thus the general formula for natural integral:

$$F(x)=\sum _{k=1}^{\infty} \frac{f^{(4k-1)}(0)}{2^k}+\int_0^x f(t) dt$$

This formula again confirm the listed above results, derived by another method, but also adds the following list for polynomials:

$\int_N 0 dx = 0$

$\int_N 1 dx = x$

$\int_N x dx =\frac{x^2}{2}$

$\int_N x^2 dx = \frac{x^3}3$

$\int_N x^3 dx = 3+\frac{x^4}4$

$\int_N x^4 dx = \frac{x^5}5$

$\int_N x^5 dx = \frac{x^6}6$

$\int_N x^6 dx = \frac{x^7}7$

$\int_N x^7 dx = 1260 + \frac{x^8}8$

$\int_N x^n dx =\frac{x^{n + 1}}{n + 1}+\begin{cases} & \frac{n!}{2^{\frac{n + 1}4}}, & \mbox{if n+1 divides by 4,} \ 0, & \mbox{otherwise} \end{cases}$

Answer by Anixx for What does the adjective "natural" actually mean?

2011-03-01T13:42:21Z

I perceive that a "natural" solution to a problem differs from other solutions in that in addition to all the properties of other solutions, it also satisfies some additional requirements which either particularly useful for the solution's purpose or greatly simplify the expressions.

Just for example, when one speaks about interpolation of a function over a number of points, the "natural" solution will differ from the infinitely many other solutions in that it will also satisfy certain functional equations of the original function not only on the domain of definition of the original function, but also in the intermediate points. Thus one can speak about natural generalization/interpolation of the function.

For example, Gamma function is a natural generalization of factorial (up to a shift) because in addition to satisfying the functional equations of factorial in integer points it also satisfies the same equations in other points and has other useful properties.

The exponential function with base e is also natural in that in addition to having all properties of exponential function it also has a unique property of being its own derivative.

Answer by Anixx for What if Current Foundations of Mathematics are Inconsistent?

2011-02-20T08:52:56Z

In either case this will not affect any practical applications of mathematics, because practical mathematics deals only with finite quantities, and finite arithmetics has been shown to be consistent. The paradoxes arise only when using abstract axioms, such as axiom of infinity, axiom of choice etc. That is the major body of analysis will survive in a form of constructivist analysis or a stricter approach (depending on where the inconsistency is discovered).

Answer by Anixx for Summation of an expression

2011-02-20T07:40:33Z

You can use Wolfram Alpha to get exact solutions for different k, although the results are expressed in not-so-good functions

This allows to build smooth plots:

(both plots are for k=5)

Answer by Anixx for Does any research mathematics involve solving functional equations?

2011-01-27T21:21:16Z

Since the question is about "really contrived functional equations" you should define what do you count as "really contrived". There are some important classes of functional equations.

Difference equations. They are discrete difference analogs of differential equations
Iterative equations. They usually can be reduced to difference equations
Delay differential equations. The combinations of difference and differential equations.

These classes are applied in many spheres. That's why they deserve to be researched. If you are speaking about equations that are outside of these classes, then indeed they are not frequently used in applied mathematics, but still may be interesting for research. Mathematicians research what they perceive as interesting, some research things that even in theory cannot be used in applied fields.

I also want to add some considerations about usefullness of solving completely random contrived equations. Such equations may lead to interesting insights into special functions and uncover interesting connections between them. Of course the mathematicians usually want to find general method which may be applicable not only to a particular equation but to a large class.

Answer by Anixx for Verifying a sequence that converges to pi

2010-12-26T15:10:56Z

Regarding the second question, no.

But you can write $\Delta a_x=\sin a_x$.

Comment by Anixx

2013-05-17T11:24:11Z

@Gerald Edgar do you have something to say here?

Comment by Anixx

2013-04-26T16:26:56Z

@Peter Mueller I am sure that is the flows cannot be found, the question cannot be solved.

Comment by Anixx

2013-04-26T04:30:32Z

why the downvote?

Comment by Anixx

2012-11-02T10:36:15Z

See also my answer regarding real solutions below.

Comment by Anixx

2012-11-02T10:25:32Z

@fedja yes, the proof in the linked paper requires p<1 or a Siegel number. I wiil add this to the answer.

Comment by Anixx

2012-11-01T18:06:44Z

See my other answer regarding real solution.

Comment by Anixx

2012-11-01T11:04:38Z

Yes. +1 And I will give the expansion in another answer.

Comment by Anixx

2012-11-01T11:03:50Z

@Vel Nias I think math analysis questions are not welcome here.

Comment by Anixx

2012-11-01T03:54:49Z

@fedja for the formulas look below. I do not know whether a formula exists if to apply your limitations (requiring the solution to be real)

Comment by Anixx

2012-11-01T03:34:12Z

@fedja I only see the supposed proofs of existence. Regarding the solution, you yourself wrote "I have no hope for an explicit elementary formula for it."

Comment by Anixx

2012-11-01T03:27:46Z

I see no solution following the link.

Comment by Anixx

2012-06-19T10:15:30Z

Well, the Bernoulli polynomials can be extended by meant of Hurwitz zeta function, but in that case the non-integer orders would not produce an elementary function. I really doubt that an extension that always gives elementary functions is possible but who knows?

Comment by Anixx

2012-06-09T15:19:35Z

@Asaf Karagila strange question, I always hope to receive answers from anybody.

Comment by Anixx

2012-06-09T14:48:36Z

Great! It seems you are the only man in this site that really can answer the questions in my area of interest. I can accept now or wait for the second part.

Comment by Anixx

2012-05-31T17:10:25Z

It is only piecewise continuous.