gm.general mathematics – Is the solution to this trig function known to be algebraic or transcendental?

This largest solution to this gorgeous equation is the first local extremum on a function related to the Fibonacci sequence:

$$x^2 cdot sin left(frac{2pi}{x+1} right) cdot left(3+2 cos left(frac{2pi}{x} right) right) = (x+1)^2 cdot sin left(frac{2pi}{x} right) cdot left(3+2 cos left(frac{2pi}{x+1} right) right)$$

This is as simplified as I could get it. The largest solution to this equation is around $x = 2.1392.$

It appears there is no closed-form solution for this; is there any way to prove if the solution is algebraic or transcendental?

numerics – Solve a transcendental equation with some parameter in Mathematica

I’m trying to solve a transcendental equation in Mathematica,

NSolve(a (1/
  2 (a Sqrt(1 - a^4/b^4 - a^2/b^2) + (
    b^2 ArcSin((a Sqrt(a^2 + b^2))/b^2))/Sqrt(a^2 + b^2)) + 
 1/2 (Y Sqrt(1 - (a^2 Y^2)/b^4 - Y^2/b^2) + (
    b^2 ArcSin((Sqrt(a^2 + b^2) Y)/b^2))/Sqrt(a^2 + b^2))) == mg/(2 B), Y)

But it didn’t work, NSolve::nsmet: This system cannot be solved with the methods available to NSolve.

Any suggestion for me to get a numerical solution or even an approximation?

Why $f(x)=pi$ a rational function? Is a constant function a polynomial even though the constant is a transcendental?

Wikipedia said

A constant function such as $f(x) = π$ is a rational function since
constants are polynomials. The function itself is rational, even
though the value of $f(x)$ is irrational for all x.

But the definition stated:
A function ${displaystyle f(x)}$ is called a rational function if and only if it can be written in the form
$${displaystyle f(x) = {frac {P(x)}{Q(x)}}}$$
where ${displaystyle P,}$ and ${displaystyle Q,}$ are polynomial functions of ${displaystyle x,}$ and ${displaystyle Q,}$ is not the zero function. The domain of ${displaystyle f,}$ is the set of all values of ${displaystyle x,}$ for which the denominator ${displaystyle Q(x),}$ is not zero.

We can rewrite $f(x)=pi$ as $f(x)=frac{pi}{1}$

I could agree if $1$ is a polynomial since it’s a non-zero constant function that is a polynomial of degree 0. But $pi$ is a trancendental number here. I mean, i never saw a polynomial with a constant of $pi$. Can you explain it to me?


How to get F from this transcendental complex equation?

I want to solve this transcendental equation, but I don’t know the step by step to get F in form equation and value of F

f(F) = (exp^(I*p*b)/(16*A^2*B*H) {((A + H)^2)((A + B)^2*exp^(2*I (k (a - b) - q*a)) - (A - B)^2*exp^(-2*I (k (a - b) - q*a))) + ((A - H)^2)((A - B)^2*exp^(-2*I (k (a - b) + q*a)) - (A + B)^2*exp^(2*I (k (a - b) + q*a))) + 2 (A^2 - B^2) (A^2 - H^2)(exp^(2*I*q*a) - exp^(-2*I*q*a))}) = 0

I have changed this equation into another form (trigonometry), which is as follows

tan(qa)=(A*H - i*B*H* tan⁡(k(a-b)))/(i*A*B - A^2*tan(⁡k(a-b)))


n = 2;
h = 1;
a = 0.01;
b = 0.02;
m = 1;
h = 1;
c = 137.036;
S1 = 0;
S2 = 0;
V1 = 0;
V2 = 50.000;
k = Sqrt(F^2 - m^2*c^4)/(h*c);
p = Sqrt((F + m*c^2 - V2 + S2) (F - m*c^2 - V2 - S2))/(h*c);
q = Sqrt((F + m*c^2 - V1 + S1) (F - m*c^2 - V1 - S1))/(h*c);
A = Sqrt((F - m*c^2)/(F + m*c^2));
B = Sqrt((F - m*c^2 - V2 - S2)/(F + m*c^2 - V2 + S2));
H = Sqrt((F - m*c^2 - V1 - S1)/(F + m*c^2 - V1 + S1));

I need solution F in form equation

nt.number theory – Reference: Asymptotic bit-complexity of algebraic operations and transcendental functions

This question is a reference request. Does anyone know of a reference that lists the asymptotic bit-complexity of algebraic operations and transcendental functions implemented on a Turing machine that includes the leading coefficients? For example: $x^n$, $x^{1/n}$, $e^x$, $sin x$, $cos x$, and $|x|$.

I intend to combine the results by Schonhage (print ref 2) with the results by Brent (web ref 1). I will use Brent for everything that is at least as complicated as multiplication, and Schohage for everything simpler than multiplication.

My goal is to make the argument that “one equation is more difficult/complex than the other” based on this information. There is precedent for this in the paper by Borwein and Borwein (print ref 4). They write, “(1) How much work (by various types of computational or approximation measures) is required to evaluate n digits of a given function or number? (2) How do analytic properties of a function relate to the efficacy with which it can be approximated? (3) To what extent are analytically simple numbers or functions also easy to compute?”

Print references I have skimmed/read:

  1. “Elementary Functions: Algorithms and Implementation”
    Muller, J. 2nd Ed. Birkauser, Boston, 2006
  2. “Fast Algorithms: A Multitape Turing Machine Implementation”
    Schonhage, A; Grotefeld, A F W; Vetter, E. Wissenschaftsverlag,
    Mannheim, 1994
  3. “Pi and the AGM : a study in analytic number theory and
    computational complexity” orwein, J; Borwein, P. John Wiley & Sons,
    New York, 1987
  4. “On the Complexity of Familiar Functions and Numbers” Borwein, J; Borwein, P. SIAM Review, v30, n4, 1988

I’m currently in the process of requesting Volume 2 of Knuth’s “The Art of Computer Programming” through my university library (Chapter 4 is on arithmetic).

Partial list of websites I’ve read:

  1. Multiple-precision zero-finding methods and the complexity of elementary function evaluation

  2. Algorithmic Complexity of Multiplication

  3. Did the 2019 discovery of O(N log(N)) multiplication have a practical outcome?

  4. What is a plain English explanation of “Big O” notation?

  5. What is O(…) and how do I calculate it?

  6. Is there a system behind the magic of algorithm analysis?

  7. How to come up with the runtime of algorithms?

  8. How to discuss coefficients in big-O notation

  9. Why are we allowed to ignore coefficients in Big-O notation?

  10. Why do we ignore co-efficients in Big O notation?

  11. Should questions on Big-Oh be on-topic here?

  12. Computational complexity of calculating the nth root of a real number

closed form expressions – Why some kinds of elementary transcendental equations cannot be solved by elementary functions – Conjecture and proof

Are my conjecture and proof below correct and well formulated?

How can they be improved?

It’s a new conjecture. It’s an extension of Lin’s theorem in (Lin 1983) which is used in (Chow 1999) – see both references at the bottom of this page.

The proof is very very simple because the preconditions of my conjecture contain the preconditions of Lin’s theorem except irreducibility. The irreducibility follows directly from very simple arguments from elementary algebra. But I need qualified confirmations.

$mathbb{L}$ are the elementary numbers, or Liouvillian numbers:,,, (Chow 1999).
$ $

Theorem (Lin 1983):
If Schanuel’s conjecture is true and $tilde{P}(X,Y)inoverline{mathbb{Q}}(X,Y)$ is an irreducible polynomial involving both $X$ and $Y$ and $tilde{P}left(alpha,e^{alpha}right)=0$ for some non-zero $alpha$ in $mathbb{C}$, then $alpha$ is not in $mathbb{L}$.
$ $

$f$ be an elementary function,
$ninmathbb{N}_{ge 1}$,
$P(x,y)inoverline{mathbb{Q}}(x,y)setminus(overline{mathbb{Q}}(x)cupoverline{mathbb{Q}}(y))$ irreducible over $mathbb{C}$,
$Q(x,y)inoverline{mathbb{Q}}(x,y)$ not zero,
$q(x)inoverline{mathbb{Q}}(x)$ not zero,
$R(x,y)=frac{P(x,y)}{Q(x,y)cdot q(x)^n}$,
$r(x)=frac{p(x)}{q(x)}$ not constant
so that
$P(x,y)$ and $Q(x,y)$ coprime over $mathbb{C}$,
$p(x)$ and $q(x)$ coprime over $mathbb{C}$.
Suppose Schanuel’s conjecture is true.
If a $tilde{P}(x,y)inoverline{mathbb{Q}}(x,y)setminus(overline{mathbb{Q}}(x) cup overline{mathbb{Q}}(y))$ of $degree_x=n$ with $frac{P(x,y)}{{q(x)}^n}=tilde{P}(r(x),y)$ exists and $R(f(z_0),e^{r(f(z_0))})=0$ for $z_0inmathbb{C}$ and $r(f(z_0))neq 0$, then $z_0$ is not an elementary number.
$ $

Proof draft:
We will prove our conjecture using Lin’s theorem. Lin’s theorem makes statements about equations of the form $tilde{P}(alpha,e^alpha)=0$, wherein $alphainmathbb{C}$ and $tilde{P}(x,y)$ is irreducible over $mathbb{C}$. We consider the equation $R(f(z_0),e^{r(f(z_0))})=0$ from our conjecture and will show that under the preconditions of our conjecture there exists an equation equivalent to it in the form from Lin’s theorem.

Setting $f(z_0)=x$ and $e^{r(f(z_0))}=y$ gives the equation $R(x, y)=0$. We consider it in the form $frac{P(x, y)}{Q(x, y)cdot q(x)^n}=0$ from the preconditions of our conjecture. Because $P(x,y)$ and $Q(x,y)$ are coprime over $mathbb{C}$ by precondition, we can multiply both sides of the equation by $Q(x,y)$ and get the equation $frac{P(x,y)}{q(x)^n}=0$ equivalent to it. By precondition, $frac{P(x,y)}{{q(x)}^n}=tilde{P}(r(x),y)$ holds. So our equation becomes $tilde{P}(r(x),y)=0$. Backsubstitution of $x$ and $y$ gives $tilde{P}(r(f(z_0)),e^r(f(z_0)))=0$.
We have $forall z_0inmathbb{C}colon r(f(z_0))inmathbb{C}$. We can therefore set $r(f(z_0))=alpha$, which gives the form $tilde{P}(alpha,e^alpha)=0$ from Lin’s theorem. To fulfill all preconditions from Lin’s theorem, we still have to show that $tilde{P}(x,y)$ is irreducible over $mathbb {C}$.
If $tilde{P}(x,y)$ is reducible over $mathbb{C}$, then
$n_1inmathbb{N}_{ge 1}$,
$b_0,…,b_{n_2}inmathbb{C}(y)$ not all constant
exist so that
$a_{n_1},b_{n_2}neq 0$,
$tilde{P}(x,y)=(a_0+a_1x^1+…+a_{n_1}x^{n_1}) cdot(b_0+b_1x^1+…+b_{n_2}x^{n_2})$.
The highest degree regarding $x$ of the polynomial product on the right-hand side of the equation is $n_1+n_2$. This must be equal to the highest degree regarding $x$ of the polynomial $tilde{P}(x,y)$ on the left-hand side of the equation. This was $n$ according to the preconditions of our conjecture. We have therefore $n_1+n_2=n$.
By substitution of $x$ by $r(x)$, we get


Because, according to the preconditions of our conjecture, $r(x)=frac{p(x)}{q(x)}$, and $p(x)$ and $q(x)$ are coprime over $mathbb{C}$, we have





$$=frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+…+a_{n_1}p(x)^{n_1})cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+…+b_{n_2}p(x)^{n_2})}{q(x)^{n_1+n_2}}.$$

Because, as stated above, $n_1+n_2=n$, we have

$$tilde{P}(r(x),y))=frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+…+a_{n_1}p(x)^{n_1})cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+…+b_{n_2}p(x)^{n_2})}{q(x)^n}.$$

According to the preconditions of our conjecture, $tilde{P}(r(x),y))=frac{P(x,y)}{q(x)^n}$. We have therefore

$$frac{P(x,y)}{q(x)^n}=frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+…+a_{n_1}p(x)^{n_1})cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+…+b_{n_2}p(x)^{n_2})}{q(x)^n}.$$

Multiplying by $q(x)^n$ yields

$$P(x,y)=(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+…+a_{n_1}p(x)^{n_1})cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+…+b_{n_2}p(x)^{n_2}).$$

If $tilde{P}(x,y)$ is reducible over $mathbb{C}$, then $P(x,y)$ is accordingly reducible over $mathbb{C}$. Because, according to the preconditions of our conjecture, $P(x,y)$ is irreduible over $mathbb{C}$, $tilde{P}(x,y)$ is irreducible over $mathbb{C}$.
Thereby all preconditions from Lin’s theorem are fulfilled and we can apply it to our equation. According to Lin’s theorem, $alpha$ is not in $mathbb{L}$. Because $alpha=r(f(z_0))$, $r(f(z_0))notinmathbb{L}$. Because $r$ and $f$ are elementary functions, $r(f(z))inmathbb{L}$ holds for all $zcolon zin dom(rcirc f) land zinmathbb{L}$. But because, as just stated, $r(f(z_0))notinmathbb{L}$, $z_0notinmathbb{L}$, so $z_0$ is not an elementary number.
$ $

(Lin 1983) Ferng-Ching Lin: Schanuel’s Conjecture Implies Ritt’s Conjectures. Chin. J. Math. 11 (1983) (1) 41-50
(Chow 1999) Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448

Help me to solve a Transcendental equation by using FindRoot function

I am solving the free vibration of a beam with partially elastic support. I have to solve the following equation

2B^3*Cosh(BL)*Cos(x)-4(K^2/B^3)*Sinh(BL)*Sin(BL)-4K*Sinh(BL)*Cos(BL)+4K*Cosh(BL)*Sin(BL)-2B^3 = 0 

K and L are constant and B is the variable. I need roots of B.

I used the FindRoot function,

FindRoot(2*B^3*Cosh(B*L)*Cos(B*L) - 4*(K^2/B^3)*Sinh(B*L)*Sin(B*L) - 4*K*Sinh(B*L)*Cos(B*L) + 4*K*Cosh(B*L)*Sin(B*L) - 2*B^3, {B,4 (Pi)}) 

But it dose not give solution and it shows,

FindRoot::nlnum: The function value {-3968.8+3968.8 Cos(12.5664 L) Cosh(12.5664 L)+4. K Cosh(12.5664 L) Sin(12.5664 L)-4. K Cos(12.5664 L) Sinh(12.5664 L)-0.00201572 K^2 Sin(12.5664 L) Sinh(12.5664 L)} is not a list of numbers with dimensions {1} at {B} = {12.5664}.

I am a beginner to Mathematica and please answer these it will be very helpful for me. Thank you in advance.

cv.complex variables – Image of transcendental meromorphic functions

Let $f$ be a trancendental meromorphic function such that $f'(z) ne 0$ for all $z in mathbb{C}$. Let $Pi$ be the stereoprojection map from the north pole on the unit sphere. My question is the following:

For any two points $P,Q in mathbb{C}$, can we find a curve $gamma$ connecting $P$ and $Q$, such that $Pi^{-1}(f(gamma))$ lies in a great circle on the unit sphere, and that $Pi^{-1}(f(gamma))$ cover the circle at most once as points go from $P$ to $Q$ along the curve $gamma$?

Any ideas or comments are really appreciated!

solving equations – how to solve a system of transcendental functions?

Having trouble solving the following instructions:

Solve the system of simultaneous transcendental equations (e ^ x + ln (y) = 2; sin (x) + cos (y) = 1). Tip: find a pair of solutions (x, y) that meet both conditions.

  1. How can I get this to give me a value for x and y?
  2. What do the error messages mean?

Here are my inputs and outputs:

In(32):= eq1 = Exp(x) + Ln(y) -2
eq2 = Sin(x) + Cos(y) - 1
system = Solve(eq1 == 0 & eq2 == 0, {x, y})

Out(32)= -2 + E^x + Ln(y)

Out(33)= -1 + Cos(y) + Sin(x)

During evaluation of In(32):= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.

During evaluation of In(32):= Solve::svars: Equations may not give solutions for all "solve" variables.

Out(34)= {{y -> -ArcCos(1 - Sin(x))}, {y -> ArcCos(1 - Sin(x))}}

nt.number theory – rank 1 evaluations which are not discrete on finite transcendental extensions of rationals

assume $ K = mathbb {Q} (X_1, dots, X_n) $ is a purely transcendental extension of rationals on an infinitely indeterminate number. Can anyone give an example of rank $ 1 $ evaluation on $ K $ who fails to be discreet?

If not, is there a theorem which shows that such a rank $ 1 $ should the assessment be discreet?