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    $\begingroup$ I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. $\endgroup$ – Asaf Karagila Mod Jul 18 '12 at 8:35
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    $\begingroup$ (+1) For thinking outside the (sand)box. $\endgroup$ – cardinal Jul 18 '12 at 19:40
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    $\begingroup$ At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! $\endgroup$ – Grace Note ModStaff Oct 5 '12 at 14:45
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    $\begingroup$ To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. $\endgroup$ – leo Dec 17 '12 at 18:03
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    $\begingroup$ PSA: Between the creation of this sandbox (in July 2012) and today (December 2015), technology has advanced. Something like StackEdit (or others, it's simply the only one I know) essentially solves all the limitations of this sandbox. You can have multiple concurrent drafts, you don't have to worry about polluting meta's front page, you can leave your draft untouched for days and expect it to still be there, you don't have to explicitly clear up your draft when you're done... Maybe someday we can get rid of this outdated crutch. $\endgroup$ – Najib Idrissi Dec 2 '15 at 14:07

17 Answers 17


First of all to follows we remember some elementary definitions and results about manifolds.


A function $f$ defined in a subset $S$ of $\Bbb R^k$ is said of class $C^r$ if it can be extended to a function $\phi$ (said $C^r$-extension) that is of class $C^r$ in a open neighborhood of $S$.


If $f$ is a function defined in a subset $S$ of $\Bbb R^n$ such that for any $x\in S$ there exist a function $f_x$ defined in a neighborhood of $x$ that is of class $C^r$ and compatible with $f$ on $U_x\cap S$ then $f$ is of class $C^r$.


If $U$ is an open set of $H^n_k:=\Bbb R^{n-k}\times[0,+\infty)^k$ for any $k\le n$ then the derivatives of two different extensions $\phi$ and $\varphi$ of a $C^r$-function $f$ agree in $U$.


A $k$-manifold with boundary/corners in $\Bbb R^n$ of calss $C^r$ is a subspace $M$ of $\Bbb R^n$ whose points have a neighborhood $V$ in $M$ that is the immage of a homeomorphism $\phi$ of calss $C^r$ defined an open set $U$ of $\Bbb R^k$ or of $H^k_1/H^k_m$ and whose derivative has rank $k$.

So now let be $M$ a $(n-1)$-manifold with boundary of class $C^r$ in $\Bbb R^n$ and thus let be $\gamma$ a injective curve defined in the unit interval $I:=[0,1]$ such that $\gamma(0)=0$ and such that the unit tangent vector not lies -for each $t\in I$- to the tangent space at any point of $M$. So we call cylindroid $C$ of trajectory $\gamma$ and of section $M$ the set $$ C:=\bigcup_{t\in I}\big(M+\gamma(t)\big) $$ that is obtained moving along $\gamma$ the points of $M$. Click here to see an example. So if $\xi\in C$ then there exist a coordinate patch $\alpha:U\rightarrow V$ and $t\in I$ such that $$ \xi=\alpha(x)+\gamma(t) $$ for any $x\in U$ and thus let be $\phi$ the function from $U\times I$ to $\Bbb R^n$ defined through the equation $$ \phi(x,t):=\alpha(x)+\gamma(t) $$ for any $(x,t)\in U\times I$ and then we prove that any restriction of this function is a coordinate patch about $\xi$. So we observe that the immage of $\phi$ is open in $C$: indeed the immage $V$ of $\alpha$ is an open set of $M$ so that there exist an open set $W$ of $\Bbb R^n$ whose intersection with $M$ is $V$ and so remembering that the translation is a homeomorphism we observe that $$ \phi[U\times I]=\bigcup_{t\in I}\big(V+\gamma(t)\big)=\bigcup_{t\in I}\Big((W\cap M)+\gamma(t)\Big)=\bigcup_{t\in I}\Big(\big(W+\gamma(t)\big)\cap\big(M+\gamma(t)\big)\Big)=\\ \Biggl(\bigcup_{t\in I}\big(W+\gamma(t)\big)\Biggl)\cap\Biggl(\bigcup_{t\in I}\big(M\cap\gamma(t)\big)\Biggl)=\Biggl(\bigcup_{t\in I}\big(W+\gamma(t)\big)\Biggl)\cap C $$ and thus we conclude that the immage of $\phi$ is open in $C$ because $W+\gamma(t)$ is open for each $t\in I$. Now if $\overset{°}M$ and $\partial M$ are the sets of the interior and boundary points of $M$ then we observe that $C$ is union of the sets $$ \bigcup_{t\in\text{int}\, I}\big(\overset{°}M+\gamma(t)\big)\,\,\,\text{and}\,\,\,\bigcup_{t\in\text{bd}\, I}\big(\overset{°}M+\gamma(t)\big)\,\,\,\text{and}\,\,\,\bigcup_{t\in I}\big(\partial M+\gamma(t)\big) $$ so that we analyse separately the case where $\xi$ is an element of the first set and the case where $\xi$ is an element of the second or either of the third: in particular this means to analyse separately the case where the set $ U $ is open in $ \Bbb R^{n-1} $ and $ t $ is an element of $ \text{int} \, I $ and the case where this is not and so just this is what we will do to follows -clearly this can be done independentely from the definition of the three mentioned sets nevertheless we decided to define them because we thought it makes more clear the following argumentetions, that's all. In particular in the example posted above these sets are respectively the invisible part, the red part, the green and balck parts. So first of all we observe that if $\xi$ is an element of the first set then $\alpha$ is a coordinate patch of $M$ defined in an open set $U$ of $\Bbb R^{n-1}$ and so the map $\phi$ defined above is a diffeomorphism in a neighborhood at any point of $U\times\text{int}\,I$ where is contained $\phi^{-1}(\xi)$, because if the unit tangent vector of $\gamma$ not lies to the tangent space at any point of $M$ then the derivative of $\phi$ is an isomorphism and so the statement follows directely from the inverse function theorem. So we conclude that the set $$ \bigcup_{t\,\in\,\text{int}\,I}\big(\overset{°}M+\gamma(t)\big) $$ is a $n$-manifold without boundary. In particular in this way we proved that the invisible part of the linked example is a manifold without boundary. Now if $\xi$ is a not an element of the first set then the previous argumentations hold only with some efforts that we show to follow. So the functions $\alpha$ and $\gamma$ can be extended to two $C^r$-functions $\beta$ and $\psi$ defined in a open neighborhood of $U$ and $I$ respectively so that the function $\phi$ can be extended to a function $\varphi:=\beta+\psi$ defined in a open neighborhood of $U\times I$ and in particular at any point $(x,t)$ of $U\times I$ this function has not singular derivative so that by the inverse function theorem there exist a (rectangular) open neighborhood $W_x\times W_t$ where $\varphi$ is a diffeomorphism. Now the set $\varphi[W_x\times W_t]$ is open in $\Bbb R^n$ and it is not disjoint from $C$ -indeed $W_x\times W_t$ is not disjoint from $U\times I$ and $\varphi$ is compatible with $\phi$ in $U\times I$ and the immage of $\phi$ is contained in $C$- so that by the continuity of $\phi$ the set $$ \phi^{-1}[\varphi[W_x\times W_t]] $$ is (not empty and) open in $U\times I$ and contains $W_x\times W_t\cap U\times I$ where $\phi$ is a diffeomorphism. So if we prove that the immage of $W_x\times W_t\cap U\times I$ through $\phi$ is open in $C$ then we conclude that the restriction of $\phi$ to this set is a coordinate patch defined in a open set of $\Bbb R^{n-1}\times[0,+\infty)$ or in a open set of $\Bbb R^{n-2}\times[0,+\infty)^2$ if $U$ is open in $\Bbb R^{n-1}$ and $t$ is not an element of $\text{int}\, I$ or if $U$ is not open in $\Bbb R^{n-1}$ and $t$ is an element of $\text{int}\, I$ or either if $U$ is not open in $\Bbb R^{n-1}$ and $t$ is not an element of $\text{int}\, I$ respectively. So how do this? Could someone help me, please?

To follows some observations that I tried to use: naturally you are not constrained to read it if you do not desidre.


If the function $\phi$ was injective in $U\times I$ the statement follows immediately because in particular the function $\varphi$ would be injective in $W_x\times W_t\cup U\times I$ so that $$ \phi[W_x\times W_t\cap U\times I]=\varphi[W_x\times W_t\cap U\times I]=\varphi[W_x\times W_t]\cap\varphi[U\times I]=\\ \varphi[W_x\times W_t]\cap\phi[U\times I] $$ having remembered that $\varphi$ and $\phi$ are compatible in $U\times I$. So in particular we observe that the injectivity of $\phi$ follows immediately if the set $M+\gamma(t_0)$ and $M+\gamma(t_1)$ was disjoint for any $t_0,t_1\in I$ such that $t_0\neq t_1$ and in particular I tried to prove this using the injectivity of $\gamma$ that above I did not use. Moreover since $\alpha$ is a homeomorphism and since $W_x$ is open in $\Bbb R^{n-1}$ then $\alpha[W_x\cap U]$ is open in $M$ and so exist a open set $W$ of $\Bbb R^n$ whose intersection with $M$ is $\alpha[W_x\cap U]$ and thus implementing the argumentations used above it follows that $$ \phi[W_x\times W_t\cap U\times I]=...=\Biggl[\bigcup_{t\in W_t\cap I}\big(W+\gamma(t)\big)\Biggl]\cap\Biggl[\bigcup_{t\in W_t\cap I}\big(M+\gamma(t)\big)\Biggl] $$ so that if the set $\Biggl[\bigcup_{t\in W_t\cap I}\big(M+\gamma(t)\big)\Biggl]$ and $\Biggl[\bigcup_{t\in I\setminus W_t}\big(M+\gamma(t)\big)\Biggl]$ was disjoint then $\phi[W_x\times W_t\cap U\times I]$ was open in $C$ and this surely happens if the set $M+\gamma(t_0)$ and $M+\gamma(t_1)$ was disjoint for any $t_0,t_1\in I$ such that $t_0\neq t_1$.


How can you solve for alternate trigonometric functions in general? Besides the regular circle functions.

I was inspired to return to an old problem I came up with after seeing

this question.

This was the problem of finding the analogues of other trigonometric functions which would parametrize a certain graph f(x,y)=1 or y=f(x) taking the polar curve angle $\theta=\theta_0$ and finding the length of the curve for one intersection of f(x,y)+0:

enter image description here

This just boils down to finding trigonometric functions which parametrize the curve f(x,y). This is easy, but the goal is have the distance from the intersection of the angle with the curve to the x,y-axis be the these functions $(x(\theta), y(\theta))$. Here is how to find such functions. For convention, let $\mathrm{x(\theta)=cos_{f(x,y)=0}(\theta)=cos_f(x),y(\theta)=sin_{f(x,y)=0}(\theta)=sin_f(x)}$. Sorry for the notation, but it is intuitive. Here is how to find such functions easily:

$$\mathrm{f(x,y)=0,y=x\,tan(\theta)\implies f(x,x\,tan(\theta))=0\implies x=cos_{f(x,y)=0}(\theta),y=sin_{f(x,y)=0}(\theta)}$$

There is not an easy inverse function for both arguments. However, when y is given explicitly you can solve for the alternate trigonometric functions which sometimes works through recursion.

$$\mathrm{\frac{f(x)}{x}=tan(\theta)\implies x=f^{-1}(x\,tan(\theta))= f^{-1}(f^{-1}(f^{-1}(…x…\,tan(\theta))\,tan(\theta))\,tan(\theta))=cos_f(\theta)\implies y=sin_f(\theta)=tan(\theta) f^{-1}(f^{-1}(f^{-1}(…x…\,tan(\theta))\,tan(\theta))\,tan(\theta))}$$

Unless you use an inversion theorem, then you cannot always solve for the inverse of y=f(x), unless if there is a closed form inverse. For example if you modeled the trigonometric equations with y=x tan(θ) and f(x)=cos(x), you would find that $\mathrm{x(\theta)=\mathrm{cos_{cos(x)}(θ)}=cos_f(x), y(\theta)=x(\theta)tan(\theta)}$ would be the parametrization for cos(x) for $(x(\theta),y(\theta))$. This uses the cardinal sine/ sinc function. Note $x_0$ here is just a parameter as a consequence of the nesting:

$$\mathrm{\frac{sin(x)}{x}=sinc(x)=tan(\theta)\implies cos_f(x)=sin^{-1}(sin^{-1}(…x_0tan(\theta))tan(\theta)…),\frac{cos(x)}x=tan(\theta)\implies cos_f(x)=cos^{-1}(cos^{-1}(…x_0tan(\theta))tan(\theta)…)}$$

Here are the results in this graph. Notice that lengths of the legs of the right triangle, like in the diagram, will give the derived modified cosine and sine functions. These new trigonometric functions are now based on sine and cosine graph instead of the circle.

Of course we can just use an inversion theorem, in the link above, to find the alternate trigonometric functions. Here is a formula that will not need to be used very much as we can usually find these alternate trigonometric functions. This is the general result for a function with explicit y=f(x) with the Lagrange Inversion Theorem. This uses the Bell polynomials and the rising factorial

$$\mathrm{\frac {f(x)}x=tan(\theta)\implies x(\theta)=cos_f(x)\left(Inverse\ of\ \frac yx\right)[tan(\theta)]=a+\sum_{n=1}^\infty\lim_{x\to a}\frac{d^{n-1}}{dx^{n-1}}\left[\left(\frac{x-a}{\frac{f(x)}{x}-\frac{f(a)}{a}}\right)^n\right]\frac{\left(tan(\theta)-\frac{f(a)}{a}\right)^n}{n!}=\sum_{k=0}^\infty c^{-n}_1\sum_{K=1}^{k-1}(-1)^K k^{(K)}B_{k-1,K}\left(\frac{c_2}{2c_1}, \frac{c_3}{3c_1},…, \frac{c_{n-k+1}}{(n-k+1)c_1}\right),\frac{f(x)}x=\sum_{n=0}^\infty c_n\frac{x^n}{n!},n\ge 2}$$

All I did to solve for x was use the linked Lagrange Inversion theorem formulas.

As a side note, I will leave you with the alternate trigonometric functions for $a|x|^m+b|x|^n=1$. Here is an interactive graph The analogue for these trigonometric functions is just $$\mathrm{x(\theta)=\pm\frac{1}{\sqrt[n]{a+b|tan(\theta)|^m}},y(\theta)= \pm\frac{tan(\theta)}{\sqrt[n]{a+b|tan(\theta)|^m}}}$$.

The secant and cosecant analogues are just $\frac{1}{x(\theta)}$ and $\frac{1}{y(\theta)}$. The tangent and cotangent analogues are almost always tan(θ) and cot(θ) because $\mathrm{y(\theta)=x(\theta)tan(\theta)}$. Please tell me if any part of this is confusing.

I tried the best I could to propose new notations for the trigonometric analogues. See the d-analogue and the q-analogue as ways to back up my $\cos_f(\theta)$ and $\sin_f(\theta)$ notations where f is the function we would like the alternate trigonometric functions based on as seen in the graphic. I also could have easily made a typo.

The question now is how to solve $$\mathrm{y=x\,tan(\theta),f(x,y)=0\implies f(x,x\,tan(\theta))=0}$$ for x in exact form, not necessarily closed form, using any method except approximation. Please correct me and give me feedback!


$$\frac 1x + \frac 1y + \frac 1z = \frac 56$$

$$\frac 1x + \frac 1y = \frac 56 - \frac 1z$$

$$\frac{x+y}{xy} = \frac {5z-6}{6z}$$

$$$\frac{x+y}{xy} ​= \frac {5z-6}{6z}$6



For the general case a multistep-transformation, be it a cycle or not, of $N$ steps and the sum-of-exponents $S=A_1+A_2+...+A_N$ where $a_{k+1}={ 3a_k+1\over 2^{A_k} }$ and $A_k=\nu_2(3a_k+1)$ might be written as $$ a_{N+1}=T(a_1;[A_1,A_2,...,A_N]) $$ or - a so to say "canonical form" -: $$ a_{N+1}={3^N \over 2^S} a_1 + T(0;[A_1,A_2,...,A_N]) $$ The latter form I've once named "canonical form" because here $T()$ is free from any variable $a_1$ and the constant $0$ is inserted for all cases, and the $T()$-notation hides your explicite $\sum()$-notation of powers of products of $3$ and of $2$.
For example we can take the example $(N,S)=(3,5)$ and $E_N:=[A_1,A_2,A_3]=[1,2,2]$ then a first solution is via $$T(0;E_N)= {9+3\cdot 2 + 2^3\over 2^5}={23\over 32} $$ $$ a_4 = { 27\over 32} a_1 + {23\over 32} \\ \implies \\ a_1= 43 + k\cdot 2 \cdot 2^S \\ a_4= 37 + k\cdot 2 \cdot 3^N $$ Note, that the $+k \cdot 2 \cdot 2^S$-term (and following the $ \cdot 3^N$ term) occur by the modularity condition such that the rhs must be integral.

That notation shows modularity to the modulus $2 \cdot 2^S$: a second solution $b_1 \to b_4$ to such a transformation is in the modulus to $2 \cdot 2^S$; such that, for instance, $$ b_1 = a_1 + 2 \cdot 2^S$$ and then $$ b_{N+1}={3^N \over 2^S} b_1 + T(0;E_N) $$ $$ b_{N+1}={3^N \over 2^S} (a_1+2 \cdot 2^S) + T(0;E_N) $$ Then follows $$ b_{N+1}=a_{N+1}+ {3^N \over 2^S} (2 \cdot 2^S) \\ b_{N+1}=a_{N+1}+ 2 \cdot 3^N $$ So $$b_1 - a_1 = 2 \cdot 2^S \\ b_{N+1} - a_{N+1} = 2 \cdot 3^N $$ Another view, which is immediately derivable, is that $$ 2^S \cdot a_{N+1} - 3^N a_1 = 2^S \cdot T(0;E_N) \\ 2^S \cdot b_{N+1} - 3^N b_1 = 2^S \cdot T(0;E_N)$$

Of course we can as well subtract $2 \cdot 2^S$ from $a_1$ to get $c_1=a_1-2 \cdot 2^S$ and then $$ c_{N+1}=a_{N+1}- {3^N \over 2^S} (2 \cdot 2^S) = a_{N+1}- 2 \cdot 3^N $$ such that


First of all, your example equation doesn't correspond to your problem: your $b$ in $\sin(ax+b)=\sin(x)$ is not non-zero - differently from what you have prescribed by your constraints.

The conditions of your point 1 ("$x$ is a rational multiple of $\pi$") can only be met if $b=0$, but your constraints prohibit zero $b$.
Considering that $\sin$ and $\arcsin$ have the exceptional set $\{0\}$ (see below), the conditions of your point 2 can only be met if $b=0$, but your constraints prohibit zero $b$.
Your point 3 is just an assumption from you.
Clearly, your thoughts are far from complete.
$\ $

Let me present a more systematic way.

"Closed form" means expressions of allowed functions. If an equation is solvable in closed form depends therefore on the functions you allow.
What you are looking for are solutions that are closed-form numbers.
Clearly, answers for closed-form solutions can be given only for single given classes of functions. Clearly, integers, rational numbers and algebraic numbers are usually allowed as closed-form numbers.

Let's consider an equation $\text{trig}(ax+b)=P(x)$, with $a, b \in \mathbb{Q}$, $a\neq 0$, where $\text{trig}$ is one of the trigonometric functions and $P\in\mathbb{Q}(x)$ is not constant. The trigonometric functions are the functions $\sin$, $\cos$, $\tan$, $\cot$, $\sec$ and $\csc$.

We can treat this kind of equations in the same way for all trigonometric functions because the function term of a trigonometric function can be represented as a non-conctant rational expression of $e^{ix}$ over the algebraic numbers, e.g.: $\sin(x)=R(e^{ix})=-\frac{1}{2}i\left(e^{ix}-e^{-ix}\right)$, with non-constant $R\in Quot(\overline{\mathbb{Q}}(x))$. Let's call this kind of presentation of the trigonometric functions their exponential representation.

1.) Algebraic numbers

Let's allow the algebraic numbers as closed form.
Let $x$ be algebraic. $P(x)$ is algebraic then.
For the trigonometric functions, Lindemann-Weierstrass theorem helps to determine the exceptional sets. See my answer at On the behavior of transcendental functions. Because the only algebraic sets of trigonometric functions are $\{\}$ or $\{0\}$, $\text{trig}(ax+b)$ can for algebraic $x$ be algebraic only if $ax+b=0$ $\ \ $ ($x=-\frac{b}{a}$).
Let $P(x)$ a polynomial as defined above that has $-\frac{b}{a}$ as a solution. The trigonometric functions have the following exceptional points. $\sin,\tan\colon (0,0)$; $\cos,\sec\colon (0,1)$. $\cot$ and $\csc$ don't have an exceptional point. Of the equations of the kind prescribed in the question, exactly the following have an algebraic solution therefore.





2.) Elementary inverses, Elementary numbers

Let's now allow the elementary functions as closed form.
The elementary functions are according to Liouville and Ritt those functions of one variable which are obtained in a finite number of steps by performing algebraic operations and taking exponentials and logarithms.

a) Solvability of equations by closed-form expressions by rearranging the equation is related to the question of existence of inverses of the functions which are contained in the equation. The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in [Ritt 1925], that is also proved in [Risch 1979], answers which kinds of elementary functions can have an inverse which is an elementary function. We can extend this theorem from the inverses of the bijective elementary functions to the partial inverses of arbitrary elementary functions by treating restrictions of the non-bijective elementary functions by restricting their domains.
In some cases, you can also take the method of [Rosenlicht 1969].
As mentioned above, for each trigonometric function $\text{trig}$, there is an exponential presentation of the equation $\text{trig}(ax+b)=P(x)$: $\ R(e^{i(ax+b)})=P(x)$, with non-constant $R\in Quot(\overline{\mathbb{Q}}(x))$. This equation can be transformed to an equation $P_{1}(e^{i(ax+b)},x)=0$, where $P_{1}$ is a non-constant binary polynomial over the algebraic numbers because $e^{(ax+b)i}$ and $x$ are algebraically independent. The left-hand side of this equation is an elementary function of $x$. Let's name it $f$. If the equation is solvable by applying only elementary functions to the equation that we can read from the equation, the inverse of $f$ has to be an elementary function. A conclusion of the theorem of Ritt in [Ritt 1925] is: If $f$ and its inverse are both elementary, a function term for $f$ in exp-ln representation must exist that doesn't contain an multiary algebraic function of algebraically independent elementary functions. But for each trigonometric function $\text{trig}$, $P_1$ is multiary. Therefore we cannot solve the equations by rearranging them by applying only elementary functions that we can read from the equation.

b) Solving an equation by closed-form expressions means to find solutions that are closed-form numbers. A closed-form number is a number that is generated by applying a finite number of closed-form functions to a rational number. [Chow 1999] asked this problem. One first and simple solution of the problem is Lin's theorem in [Lin 1983]:
"If Schanuel's conjecture is true and $f(X,Y)\in\overline{\mathbb{Q}}[X,y]$" is an irreducible polynomial involving both $X$ and $Y$ and $f(\alpha,exp(\alpha))=0$ for some non-zero $\alpha\in\mathbb{C}$, then $\alpha$ is not in EL."
$EL$ are the elementary numbers. [Chow 1999] treats the explicit elementary numbers.
For each trigonometric function $\text{trig}$, the equation $P_{1}(e^{(ax+b)i},x)=0$ (see section a) is irreducible and therefore of the kind of Lin's theorem. That means, the equations don't have a non-zero solution that is an elementary number.

[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448

[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50

[Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math. 101 (1979) (4) 743-759

[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90

[Rosenlicht 1969] Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22.

3.) Lambert W and Elementary functions

Let's now allow the algebraic functions and Lambert W as closed form.
We use the exp-ln representation of the trigonometric functions (See e.g. the Wikipedia articles for the single functions or [Abramowitz/Stegun 1970].):

$$\sin(x)=-\frac{1}{2}ie^{ix}+\frac{1}{2}ie^{-ix},\ \cos(x)=\frac{1}{2}e^{ix}+\frac{1}{2}e^{-ix},\ \tan(x)=-i\frac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}},$$ $$\cot(x)=i\frac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}},\ \sec(x)=\frac{2}{e^{ix}+e^{-ix}},\ \csc(x)=\frac{2i}{e^{ix}-e^{-ix}}.$$

For each trigonometric function $\text{trig}$, the equation $\text{trig}(ax+b)=P(x)$ can be transformed to a quadratic polynomial equation over $\overline{\mathbb{Q}}$ of $e^{iax}$.
These equations are therefore not in a form to be solved by elementary functions and Lambert W. See my answer at How to check if some equation can be solved using Lambert $\operatorname{W}$ function.
We need a generalization of Lambert W therefore. See [Mezö/Baricz 2017].
Will be continued next days.

[Abramowitz/Stegun 1970] Abramowitz, M.; Stegun, I.: Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. National Bureau of Standard 1970

[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934

4.) Special functions and Elementary functions?

You could allow algebraic expressions of known Standard functions as closed form:
If the trigonometric functions can be decomposed into compositions of algebraic functions and other known algebraically independent Standard functions than $\exp$ and $\ln$, an analog theorem to the theorem of Ritt of [Ritt 1925] could be applied. I hope such a generalization of Ritt's theorem could be proved someday for this class of functions. I made a suggestion at How to extend Ritt's theorem on elementary invertible bijective elementary functions?.


Question title: Can first order logic be extended to include infinite conjunctions?

Can first order logic (FOL) be extended in some way where infinite conjunctions are permissible? Specifically, can it be extended and still refute statements in a finite number of steps?

I would like this question to be answered using Tarskian semantics, where names refer to objects external to the logic.

Self answer:

Suppose that an infinite conjunction is legal syntax in a first order theory. In first order logic, we are free to negate any formula we can reason about, so a theory which can reason about infinite disjunctions will necessarily reason about their negation.

\begin{equation}\tag{1} \{P_a,P_b,\dots\} \end{equation}

\begin{equation}\tag{2} \lnot(P_a\lor P_b\lor\dots) \end{equation}

Consider a first order theory which contains the infinite set of formulae $(1)$ and the negation of an infinite disjunction $(2)$. Any interpretation which satisfies $(2)$ will necessarily make the set of formulae $(1)$ unsatisfiable. However, every finite subset of formulae will be satisfiable, which by compactness means that the whole thing is.

This is a contradiction, so it must not be the case that an infinite conjunction can be made legal syntax in a first order theory.

Question title: What did Hilary Putnam mean by this following quote of his?

In Putnam's paper "The logic of quantum mechanics", he states:

There is nothing really answering to the Copenhagen idea that two kinds of description (classical and quantum mechanical) must always be used for the description of physical reality (one kind for the ways to be used for the 'observer' and the other for the 'system'), nor to the idea that measurement changes what is measured in a indescribable way (or even brings into existence), nor to the 'quantum potential', 'pilot waves', ect. of the hidden variable theorists. These no more than Reichenbach's 'universal forces'.

Precisely what did he mean by this? In particular, what did he mean by the "idea that measurement changes what is measured in a indescribable way (or even brings into existence)"? How does one clearly define the the issue raised in the first point, and has it been resolved today?


From this you have bounds for your $r(a_1)$-parameter: $$ r(a_1) = { E \over N+E } = 1-{ 1 \over E/N+1 } \implies $$

$$E_{min}=1-\{N \gamma_1 \} + N \gamma_1 \\ E_{max}=N $$

$ \displaystyle \lim_{N \to \infty} r(a_1)_{E_{min}} = 1-{ 1 \over (1-\{N \gamma_1 \} + N \gamma_1 )/N+1 } = 1-\frac 1{\gamma_1+1} = 1- \log_3(2) = \log_3(1.5) $

$ \displaystyle \lim_{N \to \infty} r(a_1)_{E_{max}} = { E_{max} \over N + E_{max} } = { N \over 2N } = \frac 12 $


{\lceil N \cdot \gamma_1 \rceil \over N+E} \le \frac E{N+E} = r(a_1) \le \frac N{N+E} $$

Let us assume the "Syracuse"-style notation of the Collatz-iteration $$ a_{k+1}= {3a_k+1\over 2^{A_1}} \qquad a_k \small \text{ from the odd integers} $$ and for a $N$-fold iterated transformation the short, vectorial, notation: $$ a_{N+1}=T(a_1;[A_1,A_2,...,A_N]) $$ So let $N$ denote the (N)umber of steps $3x+1$ and $S$ denote the (S)um of the exponents $A_k$, which is also the number of $x/2$-steps.

Then, to convert this into the version of $(3x+1)/2$ and $x/2$ -stepping, we introduce $E$ the number of even steps without the $(3x+1)/2$ steps, so $E=S-N$.

With that, I understand your $r(a_1)$ as $r(a_1)=E/(N+E) = (S-N)/S = 1- N/S$.

We can observe,

  • that the trival cycle $1 = T(1;[2,2,2,...2])$ to any length $N$, has the values $S=2N$ and $E=N$ and $r(a_1)= 1-N/S = 1-N/(2N)=1/2 $
  • that the first cycle in the negative numbers $-1= T(-1;[1,1,1,1,...,1])$ to any length $N$, has $S=N$, $E=0$, and $r(a_1)= 0$
  • that the second cycle in the negative numbers $-5=T(-5;[1,2,1,2,1,2,...,1,2])$ to any even length $N=2n$, has $S=3n$, $E=n$ and $r(a_1) = E/(N+E)= n/(3n) = 1/3 $

Now to have a cycle of any length, and other than $T(a_1;[2,2,2,...,2])$ we can use the well known multiplication-formula for the $N$ members of an expected cycle $a_k$ ($k=1..N$) $$ a_2 \cdot a_3 \cdot ... \cdot a_N \cdot a_1 = \left({3a_1+1\over 2^{A_1}}\right) \left({3a_2+1\over 2^{A_2}}\right) \cdots \left({3a_N+1\over 2^{A_N}}\right)$$ This can be rearranged to $$ 2^S = 2^{A_1+A_2+...A_N} =\left(3+{1\over a_1}\right) \left(3+{1\over a_2}\right) \cdots \left(3+{1\over a_N}\right)$$ We see, that the rhs must be at least as large as the smallest perfect power of $2$ larger than $3^N$, but at most as $4^N = 2^{2N}$ so we get for the lhs (writing $\gamma=\log_2(3)$, and further below $\gamma_1=\log_2(3)-1$): $$ 2^{\lceil N \cdot \gamma \rceil} \le 2^S \le 2^{2N} $$ which in terms of $S$ means $$ \lceil N \cdot \gamma \rceil \le S \le 2N \qquad \text{where } S \in \mathbb N^+ $$ and in terms of $E$ instead $$ \lceil N \cdot \gamma \rceil -N =\lceil N \cdot \gamma_1 \rceil \le E \le N $$ From this you have bounds for your $r(a_1)$-parameter: $$ r(a_1) = { E \over N+E } \implies \\ {\lceil N \cdot \gamma_1 \rceil \over N+E} \le \frac E{N+E} = r(a_1) \le \frac N{N+E} $$ Well, this formula, in which $E$ must be evaluated after $N$ is given and might be checked for calculation of $r(a_1)$ between $\lceil N \gamma_1 \rceil$ and $N$, looks not very nice to me, so I'd reconsider the choice for the ratio-parameter $r()$.

FREE REAL ESTATE $$ {}{}{}{}{} $$


$$=\underbrace{4n^2(n-1)(n+1)}_{\equiv ~0~(\text{mod}~~ 48)}+\underbrace{(n-1)n(n+1)(n+2)}_{\equiv ~0~(\text{mod}~ 24)}$$


Feel free to improve my post:

(This post is an attempted answer to the question: Has research involving the generalization of the arithmetic mean been done for the following?)

Defining First Measure

If $S\subseteq A$:

  • $\ell$ is the length of an interval
  • $\left(I_{n}\right)_{j=1}^{n}$ and $\left(J_{k} \right)_{k=1}^{m}$, for $m\in\mathbb{N}$, are a sequence of open intervals where $\left(I_1\right)=\left(I_2\right)=...=\left(I_{n}\right)=g\in\mathbb{R}^{+}$, $\;\ell(J_1)=...=\ell(J_m)=g\in\mathbb{R}^{+}$ and the infimum in the equations below are taken over all possible $I_j$ and $J_k$

\begin{align} \mathcal{M}(g,S)=\begin{cases} g\cdot\inf\left\{m\in\mathbb{N}: {S^{\prime}\text{ is countable}, \; \left(S\setminus S^{\prime}\right)\subseteq\bigcup\limits_{k=1}^{n}I_{n}}\right\} & A \text{ is uncountable}\\ g\cdot\inf\left\{m\in\mathbb{N}: {S\subseteq\bigcup\limits_{k=1}^{n}I_{n}}\right\} & A \text{ is countable} \end{cases} \end{align}


$$\mathcal{N}(S,A)=\lim\limits_{g\to 0}\frac{\mathcal{M}\left(g,S\right)}{\mathcal{M}\left(g,A\right)}$$

then if

\begin{align} \mathcal{O}(g,S)= \small{g\cdot\inf\left\{m\in\mathbb{N}: {S_{j}\subseteq S, \; \mathcal{N}(S_{j},A)=0,\;\bigcup\limits_{j=1}^{\infty}{S}_{j}=S^{\prime},\;\left(S\setminus S^{\prime}\right)\subseteq\bigcup\limits_{k=1}^{m}J_{k}}\right\}} \end{align}

(I added $\mathcal{O}(g,S)$ so when $A$ is an intervals of finite lengths, $\mathcal{O}(g,S)$ gives the Lebesgue Measure of $S$ divided by the Lebesgue Measure of $A$.)

therefore the outer measure is

$$\mathcal{P}^{*}(S,A)=\lim\limits_{g\to 0}\frac{\mathcal{O}(g,S)}{\mathcal{O}(g,A)}$$

And the inner measure is $\mathcal{P}_{*}(S,A)=1-\mathcal{P}^{*}(A\setminus S,A)$, which means measure $\mathcal{P}(S,A)$ exists when $\mathcal{P}^{*}(S,A)=\mathcal{P}_{*}(S,A)$.

Now note $S$ and $A$, when $\mathcal{P}(S,A)$ is undefined, can also have a measure when $S$ and $A$ have a algebraic notation written in terms of subsets of $\mathbb{R}$.

Defining Algebraic Form

For arbitrary set $A$, if

  • $p_{1,1},...,p_{w,q_{w}}\subseteq\mathbb{R}$ and $z_{1,1},...,z_{w,v_w}\subseteq\mathbb{R}$ (For example the subsets can be $\mathbb{Z}$, the Cantor Set, or finite length intervals and:)
  • $w=[1,...,l\in\mathbb{N}]$
  • $T_{1,1},...,T_{l,w}:p_{1,1,1} \times...\times p_{l,w,q_{l,w}}\times z_{1,1},...,z_{w,v_w}\to\mathbb{R}$

then the algebraic form of $A$, or $\alpha$ is:

\begin{align} & \alpha=\bigcup_{w=1}^{l}\bigcap\limits_{u_{w,1}\in z_{w,1}}...\bigcap\limits_{u_{w,v_{w}}\in z_{w,v_{w}}}\bigcup\limits_{s_{w,1}\in p_{w,1}}...\bigcup\limits_{s_{w,q_{w}}\in p_{w,q_{w}}}\left\{T_{l,w}\left(s_{w,1},...,s_{w,q_{w}},u_{w,1},...,u_{w,v_{w}}\right)\right\} \end{align}

To illustrate, when set $A$ has an algebraic form $\alpha_1=\bigcup\limits_{m\in\mathbb{Z}}\bigcup\limits_{n\in\mathbb{Z}}\left\{\frac{m}{8n+4}\right\}$ where

  • $l\in \left(k_1=\left\{1\right\}\right)$
  • $w\in \left(k_2=\left\{1\right\}\right)$
  • $s_{1,1,1}=m$ and $p_{1,1,1}=\mathbb{Z}$
  • $s_{1,1,2}=n$ and $p_{1,1,2}=\mathbb{Z}$

other algebraic forms must be equivelant to $\alpha_{1}$. For example, another algebraic form of $A$ is $\alpha_2=\bigcup\limits_{m_1\in\mathbb{Z}}\bigcup\limits_{m_2\in\mathbb{Z}}\left\{\frac{2m_1+1}{8m_2+4}\right\}\cup\bigcup\limits_{m_3\in\mathbb{Z}}\bigcup\limits_{m_4\in\mathbb{Z}}\left\{\frac{m_{3}}{12m_4+6}\right\}$ where

  • $w\in k_1=\left\{1\right\}$
  • $l\in k_2=\left\{1,2\right\}$
  • $s_{1,1,1}=m_1$ and $p_{1,1,1}=\mathbb{Z}$
  • $s_{1,1,2}=m_2$ and $p_{1,1,2}=\mathbb{Z}$
  • $s_{1,2,1}=m_3$ and $p_{1,2,1}=\mathbb{Z}$
  • $s_{1,2,2}=m_4$ and $p_{1,2,2}=\mathbb{Z}$

Since $\alpha_1=\alpha_2$, $\alpha_2$ is an algebraic form of $A$.

Moreover, note we can define the set of all algebraic forms of $A$ as $\mathcal{A}(A)$ where $\alpha\in\mathcal{A}(A)$. For example:


And an $\alpha\in\mathcal{A}\left(\mathbb{Q}\right)$ is $\alpha=\left\{\frac{m}{n}:m,n\in\mathbb{Z}\right\}$

Therefore, we can now define a "Folner-like" Sequence inorder to define a measure

Defining New Measure

Going back to $\alpha$. If

  • $\Phi_{1,w}=\left\{1,...,q_{w}\in\mathbb{N}\right\}$
  • $\Phi_{2,w}=\left\{1,...,v_{w}\in\mathbb{N}\right\}$
  • $i_{1,w},r_{w}\in\Phi_{{1,w}}$
  • $i_{2,w}\in\Phi_{2,w}$
  • $t_{l,w}\in\Phi_{l,w}\setminus\{r_{l,w}\}$

then $\alpha$ becomes

\begin{alignat}{2} &\mathcal{F}(w,\alpha,\psi,r_{l,w},\epsilon,\omega)= \\ &\bigcup\limits_{i_{l,w}\in\Phi_{l,w}}\bigcap\limits_{\large{u_{i_{w}}\in z_{i_{w}}}}\bigcup\limits_{\large{s_{w,i_{w}}\in p_{i_{w}}}}&&\left\{T_{w}\left(s_{w,1},...,s_{w,q_{w}},u_{w,1},...,u_{w,v_{w}}\right):\left|s_{w,t_{w}}\right|\le \omega, |u_{w,i_{w}}|\le\eta\right. \\ & &&\left. \left|T_{w}\left(s_{w,1},...,s_{w,q_{w}},u_{w,1},...,u_{w,v_{w}}\right)-\psi \right|\le\epsilon \right\} \end{alignat}

Note if all $p_{w}$ are subsets of $\mathbb{Z}$ and $\omega\to\infty$ for every $\eta$ when $\eta\to\infty$, then each $s_{r_{w}}$ would give different Folner Sequences. However, this forces us to choose a particular $r_{w}$.

Also note we add $\left|T_{w}\left(s_{w,1},...,s_{w,q_{w}},u_{w,1},...,u_{w,v_{w}}\right)-\psi \right|\le\epsilon$ so when $A$ is dense in an interval and $S\subseteq A$ is dense almost nowhere, the density of $S$ using Folner-esque sequences is always zero.

To avoid choosing between different $r_{w}$, take the intersection of all of them and take the Folner Sequences of the $l$ and $w$'s which helps obtain a measure that matches $\mathcal{P}$ when $\mathcal{P}(S,A)$ is defined.

\begin{align} \alpha_{\mathcal{F}}\left(\alpha,\psi,\epsilon,\omega,\eta\right)=\bigcup\limits_{w=1}^{l}\bigcap\limits_{r_{w}\in\Phi_{w}}\mathcal{F}\left(w,\alpha,\psi,r_{w},\epsilon,\omega,\eta\right) \end{align}

(Note we take $k_1$ less than $\eta$ so when $A=\mathbb{Q}$ one can derive a unique Folner Sequence which gives a unique measure.)

However, there are multiple algebraic forms of $\alpha$ in $A$ which prevent us from obtaining such a measure. Therefore we add additional criteria:

If $\alpha_1,\alpha_2\in\mathcal{A}(A)$ we solve

$$\beta=\inf\left\{\lim\limits_{\epsilon\to 0}\lim\limits_{\eta\to\infty}\lim\limits_{\omega\to\infty}\mathcal{P}\left(\alpha_{\mathcal{F}}(\alpha_1,\psi,\epsilon,\omega,\eta),\alpha_{\mathcal{F}}(\alpha_1\cup\alpha_2,\psi,\epsilon,\omega,\eta)\right)\right\}$$

and take the set of all $\alpha_1$ that gives $\beta$.

$$\mathcal{A}_{\beta}(A)= \left\{\alpha_1:\left(\forall\alpha_{2}\right)\left(\lim\limits_{\epsilon\to 0}\lim\limits_{\eta\to\infty}\lim\limits_{\omega\to\infty}\mathcal{P}\left(\alpha_{\mathcal{F}}(\alpha_1,\psi,\epsilon,\omega,\eta),\alpha_{\mathcal{F}}(\alpha_1\cup\alpha_2,\psi,\epsilon,\omega,\eta)\right)=\beta\right)\right\}$$

For example, when $A=\mathbb{Q}$, the Folner Sequence

$$\lim\limits_{\epsilon\to 0}\lim\limits_{\eta\to\infty}\lim\limits_{\omega\to\infty}\bigcap_{k\in\mathbb{N},|k|\le\eta}\left(\bigcup\limits_{n\in\mathbb{Z},|n|\le\omega}\bigcup_{m\in\mathbb{Z}}\left\{\frac{m}{n^{k}}:\left|\frac{m}{n^k}-\psi\right|\le\epsilon\right\}\cup\bigcup\limits_{n\in\mathbb{Z}}\bigcup_{m\in\mathbb{Z},|m|\le\omega}\left\{\frac{m}{n^{k}}:\left|\frac{m}{n^k}-\psi\right|\le\epsilon\right\}\right)$$

Should be in $\mathcal{A}_{\beta}(A)$

(We can picture the triple-limit as follows: For every $\epsilon>0$ and $\eta>0$, take $\omega$ as $\omega\to\infty$ then for every $\epsilon>0$ take $\eta$ as $\eta\to\infty$. As $\epsilon$ is evaluated for smaller values, the limit should converge.

Then take the union of all $\alpha$ in $\mathcal{A}_{\beta}(A)$ to get:


(Presumably), we are not only dealing with countable sets. Hence instead of taking the density of $S\subseteq A$ using Folner Sequences:

\begin{align} \lim\limits_{\epsilon\to 0}\lim\limits_{\omega\to\infty}\frac{\left|S\cap A_{\mathcal{F}}\left(\psi,\epsilon,\omega\right)\right|}{\left|A_{\mathcal{F}}\left(\psi,\epsilon,\omega\right)\right|} \end{align}

We take

\begin{align} d(S,\psi)=\lim\limits_{\epsilon\to 0}\lim\limits_{\omega\to\infty}\mathcal{P}\left(S\cap A_{\mathcal{F}}\left(\psi,\epsilon,\omega\right),A_{\mathcal{F}}\left(\psi,\epsilon,\omega\right)\right) \end{align}

Now, even if $d(S,\psi)$ exists, it only takes the density over $[\psi-\epsilon,\psi+\epsilon]$. Therefore, we broaden the density to all $A$ by taking the weighted average of $d(S,\psi)$ using $\mathcal{P}$. In formal terms:

Divide $[0,1]$ ($0\le d(S,\psi)\le 1$) into partitions $0=c_1\le \cdot\cdot\cdot\le c_i \le \cdot\cdot\cdot\le c_u=1$ where if $x_i\in[c_{i-1},c_i]$ and $C_i=[c_{i-1},c_i]$, then our final outer measure is

\begin{align} & \mathcal{U}^{*}(S,A)= \lim\limits_{u\to\infty}\sum\limits_{i=1}^{u}x_i\cdot\mathcal{P}^{*}\left(\left\{\psi:d\left(S,\psi\right)=C_i\right\},A\right) \end{align}

And our final inner measure is

\begin{align} & \mathcal{U}_{*}(S,A)= 1-\lim\limits_{u\to\infty}\sum\limits_{c=1}^{u}x_i\cdot\mathcal{P}^{*}\left(\left\{\psi:d\left(A\setminus S,\psi\right)=C_i\right\},A\right) \end{align}

Therefore, $\mathcal{U}(S,A)$ exists when:

$$\mathcal{U}^{*}(S,A)=\mathcal{U}_{*}(S,A) $$

Generalizing the Measure

We can generalize our measure to $\mu$:

\begin{equation} \mu(S,A)=\begin{cases} \mathcal{P}(S,A) & \mathcal{P}^{*}(S,A)=\mathcal{P}_{*}(S,A) \\ \mathcal{U}(S,A) & \mathcal{P}^{*}(S,A)\neq\mathcal{P}_{*}(S,A), \; \mathcal{U}^{*}(S,A)=\mathcal{U}_{*}(S,A) \end{cases} \end{equation}

Defining The Average

To define the average, first split $[a,b]=[\min(A),\max(A)]$ into sub-intervals using partitions $x_i$

$$a= x_0 \le \dots \le x_i \le \dots \le x_r =b$$

Here, if $1\le i \le r$, then $[x_{i-1},x_i]$ are sub-intervals of $[a,b]$.

For every $i$, choose a $v_i\in A\cap[x_{i-1}, x_{i}]$ and define $A_i=A\cap[x_{i-1},x_i]$. Next we define set $P$, such that $i \in P\subseteq\left\{1,...,r\in\mathbb{N}\right\}$ when $A\cap[x_{i-1},x_i]\neq\emptyset$. This gives

\begin{align} & \lim_{r\to\infty}\sum_{i\in P} f(v_i) \times \mu(A_i,A) \end{align}

(This may look tedious but we can use shortcuts to simplify the sum. If I write it out the post will be too long.)


$$\begin{align}\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y} &=\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos (x+y-y)}{x+y} \\ &\,=\lim_{(x,y) \to (0,0)} \dfrac{\,1 - \cos (x+y)\cos y-\sin (x+y)\sin x\,}{x+y} \\ &=\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos (x+y)}{x+y}\\ &= 0.\end{align}$$


(Deutsch: MathJax: LaTeX Basic Tutorial und Referenz)

  1. To see how any formula was written in any question or answer, including this one, right-click on the expression and choose "Show Math As > TeX Commands". (When you do this, the '$' will not display. Make sure you add these. See the next point.

Or click the Edit link at the bottom o a post to view the source code.

  1. For inline formulas, enclose the formula in $...$. For displayed formulas, use $$...$$.
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  8. Different Fonts

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$$\ln, \log,\lg,\log_2,\log_{16}$$ Use subscripts to attach a notation to \lim: \lim_{x\to 0} $$\lim_{x\to 0}$$ Nonstandard function names can be set with \operatorname{foo}(x) $\operatorname{foo}(x)$.

  1. There are a very large number of special symbols and notations, too many to list here; see this shorter listing, or this exhaustive listing. Some of the most common include:
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  • $\nabla$, $\partial$ \Im \Re $\Im$, $\Re$
  • For modular equivalence, use \pmod like this: a\equiv b\pmod n $a\equiv b\pmod n$.
  • For the binary mod operator, use \bmod like this: a\bmod 17 $a\bmod 17$.
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Detexify lets you draw a symbol on a web page and then lists the $\TeX$ symbols that seem to resemble it. These are not guaranteed to work in MathJax but are a good place to start. To check that a command is supported, note that MathJax.org maintains a list of currently supported $\LaTeX$ commands, and one can also check Dr. Carol JVF Burns's page of $\TeX$ Commands Available in MathJax.

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  1. A Pythagorean Triple (PT) is a triple $(a,b,c)$ of positive integers for which $a^2+b^2 = c^2$. Pythagorean triples can be parameterized by positive integers $s$ and $t$ by $$(a,b,c) = (2st, s^2-t^2, s^2+t^2)$$

  2. An arithmetic progression of squares (APS) is a triple $(a^2, b^2, c^2)$ for which $b^2 - a^2 = c^2 - b^2$. This is equivalent to $a^2+c^2 = 2b^2$. The value of $b^2 - a^2 = c^2 - b^2$ is called the common difference.

  3. If $(a,b,c)$ is a PT, then $$((b-a)^2, c^2, (b+a)^2)$$ is an APS. The common difference is $2ab$.

  4. Since $(2st, s^2-t^2, s^2+t^2)$ is a PT, then $$((s^2-t^2 - 2st)^2, (s^2+t^2)^2, (s^2-t^2+2st)^2)$$ is an APS. The common difference is $4st(s^2-t^2)$

We are given the APS $(205^2, 425^2, 565^2)$ with a common difference of $138600$. The corresponding PT is $(385, 180, 425)$

The common difference is $425^2 -


Notice that removing any edge $e\in T\setminus T'$, from $T$ itself, will result in two connected components: $V,U$. Since $T'$ is a tree, it is connected. Therefore, there must be some edge $e'\in T'$ between the components $V$ and $U$.

I claim, the three following things:

  1. $e'\notin T$
  2. By removing $e$ from $T$ and adding instead $e'$, we get a tree $\hat T$
  3. The tree $\hat T$ we get is a minimum spanning tree

Lets prove those statements one by one:

  1. Assume towards contradiction that $e'\in T$. Then, since $e\notin T'$ but $e'\in T'$ we can conclude that $e\neq e'$. Therefore, removing $e$ from $T$ will still keep $e'$ in $T$, and thus $V$ and $U$ will still be connected - contradiction.

  2. both $V$ and $U$ must be connected, since $T$ was. Adding the edge $e'$ will connect $V$ and $U$, hence the resulting graph must be connected. Additionally, the number of edges in the resulting graph is $|T|-1+1=|T|=n-1$ where $n$ is the number of vertices. Thus, the resulting graph must be a tree.

  3. This is the trickiest one to prove. Denote by $V',U'$ the subgraph of $T'$ on the vertices of $V,U$. It can be easily shown that $w(V')=w(V)$ ($w(\cdot)$ is the weight of the subgraph), since otherwise either $T$ or $T'$ would not be an MST. The same can be applied for $w(U')=w(U)$. Now, notice that $T'$ is a tree, and thus there is only one single edge between $V'$ and $U'$ that is also in $T'$, and this edge is $e'$ (if there were two edges, the graph would contain a cycle). Clearly, \begin{equation}{w(V)+w(U)+w(e)=w(T)=w(T')=w(V')+w(U')+w(e')}\end{equation} Substitute into this equation the fact that $w(V)=w(V')$ and $w(U)=w(U')$, and subtruct from both sides of the equation them. We finally get that $w(e)=w(e')$. Now, by the definition of $\hat T$, we have $w(\hat T)=w(V)+w(U)+w(e')=w(V)+w(U)+w(e)=w(T)$. And since $w(T)$ is minimal, then also $w(\hat T)$ is. Thus $\hat T$ is a minimum spanning tree.


$j=5^{(5^2)}$ $j-Z(j!)=j-($ $(j-1)-Z((j-1)!)=(j-1)-Z((j-1)!)$

$(5^1-5^0) - Z((5-5^0)!) = 4- 0 = 4$ $5^1 - Z(5!) = 5- 1 = 4$

$5^5 - Z(5^5!) = 5^5- (5^4 + 5^3 + 5^2 + 5 + 1)$ $(5^5-5)- Z((5^5-5)!) = (5^5-5)- (5^4-1 + 5^3-1 + 5^2-1 + 5 -1+ 1-1)=5^5- (5^4 + 5^3 + 5^2 + 5 + 1)$ $(5^5-5^1-5^0)- Z((5^5-5^1-5^0)!) = (5^5-5-1)- (5^4-2 + 5^3-1 + 5^2-1 + 5 -1+ 1-1)=5^5- (5^4 + 5^3 + 5^2 + 5 + 1)$

$(5^{25}-5) - Z((5^{25}-5)!) = (5^{25}-5)- (5^{24}-1 + 5 ^{23}-1 + 5^{22}-1 + ... +(5^4-1) + (5^3-1) + (5^2-1) + (5-1) + (1-1))$ $(5^{25}-5^1-5^0) - Z((5^{25}-6)!) = (5^{25}-5 -1 )- (5^{24}-2 + 5 ^{23}-1 + 5^{22}-1 + ... +(5^4-1) + (5^3-1) + (5^2-1) + (5-1) + (1-1))$ $(5^{25}+5^2-5^1) - Z((5^{25}+24)!) = (5^{25}+24 )- (5^{24}+4 + 5 ^{23} + 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)$ $(5^{25}+5^2+5^1) - Z((5^{25}+26)!) = (5^{25}+26 )- (5^{24}+5 + 5 ^{23} + 1 + 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)$

2: $5^1-5^0, 5^1\mid\mid 5^1-5^0$ First time a number appearing two times: $5^1-5^0$ $(5^1-5^0+4)$

3: $5^5-5^1-5^0, 5^5-5^1, 5^5 \mid\mid 5^5- (5^4 + 5^3 + 5^2 + 5 + 1)$ First time a number appearing three times: Two numbers: $5^5- (5^4 + 5^3 + 5^2 + 5 + 1) + 0$ $5^5- (5^4 + 5^3 + 5^2 + 5 + 1) + 4$

4: $5^{25}-5^1-5^0, 5^{25}-5, 5^{25}+5^2+5^1,5^{25}+5^2+5^1\mid\mid(5^{25}+26 )- (5^{24}+5 + 5 ^{23} + 1 + 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)$ First time ??? a number appearing four times: Five numbers: $5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+0??????$ a mild speculation Is this a missing one? $5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+4$ $5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+8$ $5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+12$ $5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+16$ $5^{25}- (5^{24} + 5 ^{23} + 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+20$

Some number near $5^{125}-(5^{125}+ 5^{124}+ 5 ^{123}+ 5^{122} + ... +5^4 + 5^3 + 5^2 + 5 + 1)$ appears five times as generated by a number near $5^{125}-(5^{3}+ 5^{2}+ 5 ^{1} + 1)$


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