## warning messages – Some of the functions have zero differential order

I am trying to solve the following system of differential equations, but Mathematica (11.3) complains that

NDSolveValue::pdord Some of the functions have zero differential order

I do not understand this message since all of the functions appear with their derivatives in the equations.

Here is the code:

``````ClearAll("Global`*")
(Rho)0 = 1;
M = 1;
eq1 = D(m(t, r), r) - 4 Pi (Rho)(t, r) R(t, r)^2 D(R(t, r), r);
eq2 = D(m(t, r), t) + (4 Pi)/3 (Rho)(t, r) R(t, r)^2 D(R(t, r), t);
eq3 = D(R(t, r), r, t) - D(R(t, r), t) D((Nu)(t, r), r) -
D(R(t, r), r) D((Lambda)(t, r), t);
eq4 = D((Rho)(t, r), r) + 1/4 (Rho)(t, r) D((Nu)(t, r), r);
eq5 = Exp(-2 (Nu)(t, r)) (D(R(t, r), {t, 2}) -
D((Nu)(t, r), t) D(R(t, r), t)) -
Exp(-2 (Lambda)(t, r)) D((Nu)(t, r), r) D(R(t, r), r) + m(t, r)/
R(t, r)^2 + (4 Pi)/3 (Rho)(t, r) R(t, r);
{Ro, Ra, Ma, nu, la} =
NDSolveValue({eq1 == 0, eq2 == 0, eq3 == 0, eq4 == 0,
eq5 == 0, (Rho)(0, r) == (Rho)0, (Rho)(t, 0) == (Rho)0,
m(0, r) == M,
m(t, 0) ==
M, (Nu)(0, r) == -(1/4) Log((Rho)0), (Nu)(t,
0) == -(1/4) Log((Rho)0), (Lambda)(0, r) ==
0, (Lambda)(t, 0) == 0, R(0, r) == ((3 M)/(4 Pi (Rho)0))^(1/3),
R(t, 0) == ((3 M)/(4 Pi (Rho)0))^(1/3),
Derivative(1, 0)(R)(0, r) == 1}, {(Rho), R,
m, (Nu), (Lambda)}, {t, 0, 1}, {r, 0, 1})
``````

Note: another message appears saying that the initial and boundary conditions are inconsistent, but that is another problem I have to solve later, I am more concerned with the other message at the moment.

## Solving functional equations including \$fof(x)\$ functions

Find all $$a$$ $$epsilon$$ $$R$$ for which there exists a function $$f : R$$ $$longrightarrow$$ $$R$$ , such that

(i) $$f(f(x))=f(x)+x$$, for all $$x$$ $$epsilon$$ $$R$$,

(ii) $$f(f(x)–x)=f(x)+ax$$, for all $$x$$ $$epsilon$$ $$R$$

Normally in such functional equations, I’d put different values of $$x$$ (like $$x+1$$, $$x+2$$ etc) and try to get an equation for $$f(x)$$. But here I’m not getting anywhere by that method. I can’t use the coefficient comparison method either, because $$f(x)$$ may not be a polynomial.

## functions – wp_register_script() and wp_register_style() 4th parameter (version) not working for logged out users

I’m enqueuing scripts and styles from functions.php in my custom theme. The 4th parameter is the version number. It’s working only for logged in users.

``````wp_register_script( 'my-handle', get_stylesheet_directory_uri( __FILE__ ) . '/js/my-script.js', array( '' ), filemtime( get_stylesheet_directory( __FILE__ ) . '/js/my-script.js' ), true );
wp_enqueue_script( 'my-handle' );
``````

Logged in:

``````<script src="https://mysite.com.local/path/to/my-script.js?ver=1614637944" type="text/javascript"></script>
``````

Logged out:

``````<script src="https://mercury.photo/wp-content/plugins/instagram-feed/js/sbi-scripts.min.js" id="sb_instagram_scripts-js" type="text/javascript"></script>
``````

Caching plugin on the live site is LiteSpeed Cache. The local dev site uses no caching. Same output on both sites.

## contest math – Find all functions satisfying \$f(m^{2}+f(n))=f(m)^{2}+ntextrm{, for all }m,nin N\$

Find all functions $$f:mathbb{N}rightarrowmathbb{N}$$ such that
$$f(m^{2}+f(n))=f(m)^{2}+ntextrm{, for all }m,nin N$$

I was initially unable to solve this problem, so I referred to a hint. The hint is as follows:

Let $$f(1)=k$$. From the given equation, we get $$f(m^{2}+k)=1+f(m)^{2}$$ and $$1+k^{2}=f(1+k)$$. Use this to prove that
$$f(f(m))^{2}-f(k)^{2}=m^{2}-1 (*)$$
Taking $$m=2$$, conclude that $$f(k)=1 (*)$$.
Obtain $$f(f(m))=m$$ for all $$min mathbb{N}$$. Use this to prove that $$f(n+1)=f(n)+k^{2} (*)$$
By induction $$f(n+1)=f(1)+nk^{2}=k+nk^{2}$$. Conclude that $$k=1$$ and hence $$f(n)=n$$ for all $$nin mathbb{N}$$

This solution is very unclear to me. I have marked all unclear steps with $$(*)$$. Please breakdown this solution and provide well explained arguments for the marked steps. Please try to use the same notation I have provided in the solution for ease of understanding. Thanks for all help 🙂

## co.combinatorics – proof of result from Ian Macdonald’s paper “A New Class of Symmetric Functions”

I’m currently working my way through Ian MacDonald’s somewhat seminal 1988 paper entitled “A New Class of Symmetric Functions” from Actes 20e Seminaire Lotharingien, p. 131–171. I’m fine with the paper up until p149 where MacDonald claims in result 3.9 the following:

$$omega_{q,t}E_{q,t}omega_{q,t}^{-1}=E_{t^{-1},q^{-1}}$$

Unfortunately MacDonald provides neither a proof nor a reference for this result. Instead he resorts to borrowing a margin from Fermat’s book, claiming to have a proof while not actually giving the proof. Fermat told a better tale, though, claiming to have a “marvelous” proof whereas MacDonald admits to having only a “rather messy” proof.

Does anyone know where I can find MacDonald’s “rather messy” proof, or better yet a more elegant proof of this result than the one MacDonald apparently had? Some of the results from this paper seem to also be in MacDonald’s earlier book “Symmetric Functions and Hall Polynomials” but I can’t find this one in that book either.

I’m trying to gain a rigorous understanding of this paper, but that is difficult for me to do if I need to accept MacDonald’s apparent request that I simply gloss over this rather important intermediate step.

## equation solving – How to make FindRoot work with PolyLog functions

I have the following code:

``````q = 1.6*10^-19;
hbar := 6.63/(2*(Pi))*10^-34;
m0 = 9.1*10^-31;
kb = 1.38*10^-23;
Cap := 4.5*10^-3;
W := 50*10^-6;
mu := 4.3*10^-4;
n := 20;
vth = ((3*kb*300)/m0)^0.5;
vt := -3.5;
Rex := 15 ;
vg = 0;
Plot(id /.
FindRoot((id - (Cap*W*mu*(n*vth)^2)/(
7*10^-6)*(PolyLog(2, -Exp((vd - id - vg + vt)/(n*vth))) -
PolyLog(2, Exp((id - vg + vt)/(n*vth))))), {id,
10^-6}), {vd, -1, .872}, ImageSize -> Large,
AxesLabel -> {vd, id}, LabelStyle -> {15, Bold, Black})
``````

It shows this error:

How can I specify the correct AccuracyGoal and PrecisionGoal for my problem. Also I want to plot the $$id$$ as a function of $$vd$$ for different $$vg$$, is there a way to automate it?

## Composition of functions and operator forms of built-in functions

I am trying to replicate the behaviour of some of the built-in operator forms (for `Map`, `Apply` etc) but struggling to understand the way `Composition` works. For a concrete example:

``````list = {1 (UndirectedEdge) 10, 2 (UndirectedEdge) 11, 3 (UndirectedEdge) 11, 4 (UndirectedEdge) 10}

Map(Apply(Rule))@list

%: {1 -> 10, 2 -> 11, 3 -> 11, 4 -> 10}
``````

produces the result I was looking for in this case.

My question is how to replicate the composition of operator forms above. This naive attempt doesn’t work as intended and produces an error message:

``````Map@*Apply@*Rule@list
``````

Rule::argr: Rule called with 1 argument; 2 arguments are expected.

``````%: Map(Apply(Rule({1 (UndirectedEdge) 10, 2 (UndirectedEdge) 11, 3 (UndirectedEdge) 11, 4 (UndirectedEdge) 10})))
``````

I’ve tried various combinations of parentheses to control the evaluation to avail – the example I started with appears to build an operator form of the composition of `Apply` and `Rule` then maps it over the list input but that is not what the composition I attempted does.

So, how to produce the same with `Composition` or other constructs?

I’m asking really to try and get a fuller understanding of these types of construction work.

## c – Naming convention for functions that mutate arguments vs creating a new object

Take the following signature

`BigInt* addBigInt(BigInt* arg1, BigInt* arg2);`

Traditionally, the safest way to implement this function would be for it not mutate the arguments, and would necessarily need to create a new object to result for the operation. Alternatively, if the aim is to conserve on system memory, `arg1` could be the object that the resulting operation would return having mutated that object.

The question is, is there an established naming convention the would delineate the nuance between the two implementation choices? Or is this something that would need to be clearly laid out in the documentation where the developer is establishing the convention?

Let’s assume this is in C where function overloading is not possible.

## google sheets – IF/AND functions from multiple cells

You need to use IFS() with AND.

See the reference here:

How to Correctly Use AND, OR Functions with IFS in Google Sheets

Hi This is my code currently i.ibb.co/HTz1L0T/qsdf.png – Lloyd 14 mins
ago

As outlined by OlegS you are still using multiple = signs in your code with the wrong syntax.

His comment that you should amend it to the following seems right in my view.

``````=IFERROR(IFS(AND(SEARCH("Paypal",D676)>0,C676=14),"Groceries", AND(SEARCH("Paypal",D676)>0,C676=45),"Internet", AND(SEARCH("Paypal",D676)>0,C676=21),"Telephone"),"-")
``````

## reference request – Ratio limit results for restricted partition functions

This concerns difference/limit ratio results for special restricted partitions.

Let $$r,a, b$$ be nonnegative integers; define $$p(r,a,b)$$ to be the number of partitions of the integer $$r$$ using at most $$b$$ positive integers and each of them is at most $$a$$; if $$r = 0$$ and $$a,b >0$$, then $$p(0,a,b)$$ is taken to be one. Then, assuming $$ab > 1$$, $$p(1,a,b) = 1$$, and $$p(ab,a,b) = 1$$; moreover, $$p(r,a,b) neq 0$$ iff $$0 leq r leq ab$$. These are well known and closely related to Gauss polynomials (explicitly, the coefficient of $$z^r$$ in the $$(a+b,b)$$ Gaussian polynomial is $$p(r,a,b)$$) (notation varies considerably in the literature; I may have messed things up).

It is also well known that if $$a, b$$ are fixed, then the sequence $$(p(0,a,b), p(1,a,b), dots , p(ab,a,b))$$ is unimodal, and symmetric (about $$ab/2$$).

What I am interested in is the ratios of consecutive $$p(r,a,m-a)$$ with $$r$$ varying (note the change in notation, $$b mapsto m-a$$) over the interval $$2a leq r leq a(m-a)/2$$, with estimates uniform in $$m$$ and $$a$$, with $$a$$ not too close to $$0$$ or $$m$$ (for the latter, we can just restrict to $$a leq m/2$$).

To that end, define $$R_{m} = inf_{ 2a leq r leq a(m-a)/2; {rm and }3 leq a leq m-3} frac{p(r-1,a,m-a)}{p(r,a,m-a}.$$
It is important to keep track of the conditions below the infimum. Since $$p(r, a,m-a)$$ is increasing (as $$r$$ runs over $$0,1,2, dots, a(m-a)/2$$) (with $$a$$, $$m$$ fixed), obviously $$R_{m} < 1$$. It is relatively easy to show that $$R_m geq 1/2$$ for all but a few values of $$m$$.

Question Is it true that
$$R_{m} geq 1 – pmb oleft( 1right)$$
or something similar?

This might require a further restriction on $$a$$; small values (or those close to $$m$$) tend to screw it up; alternatively, impose more conditions on $$r$$, e.g., $$r geq 3a$$).

I imagine there are such results in the literature, but I am unfamiliar with it. For $$r$$ near the middle ($$a(m-a)/2$$), asymptotic estimates for $$p(r,a,m-a)$$ given in (RP Stanley & F Zanello, Some asymptotic results on q-binomial coefficients, Ann Comb 20 (2016) 623–34, Thm 2.6)) suggest that the problem probably reduces to relatively small $$r$$. An exact formula for $$p(r,a,b)$$ is given in (G Almkvist & GE Andrews, A Hardy-Ramanujan formula for restricted partitions, J Number Theory, 38 (1991) 135–144, Thm 4), but it is difficult to see how to use it. Without the $$b$$, asymptotic formulas were obtained by Szekeres (An asymptotic formula in the theory of partions I and II, Quart J Math 4 (1951) 85–108 and 4 (1953) 96–11).

Motivation I am looking at very simple random walks on the simplest torsion-free nonabelian group, given by $$hg = z gh$$ where $$z$$ is central. Then the coefficient of $$z^r g^a h^b$$ in $$(g + h)^{a+b}$$ is $$p(r,a,b)$$. I want to obtain the limit ratio result above (or some version of it), because it will almost certainly lead to a description of all extremal harmonic functions on the subcone starting at $$1$$ and proceeding by multiplication by $$g+h$$. Less cryptically, I have a bunch of such harmonic functions, and want to prove they constitute all of them.

In general, limit ratio results are easier to prove than establishing asymptotics (since the latter usually yields the former).