Alfred_F

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These are questions asked by Alfred_F

Prove:
If a is an irrational number, then the function
y(x) = cos(a*x) + cosx is not periodic.

Further question:
Since every irrational number can be represented as the limit of a sequence of rational numbers and all software "only" works with finite representations of numbers, it would be interesting to know at what number length for a this becomes in examples apparent in the plot for y(x).

In the attached file, the trigonometric term (2, term) is transformed into a term (3, term1) consisting of radicals. Is there a Maple procedure that can be used to reverse this process? Given an algebraic term (e.g., consisting of radicals, powers, etc.), under what conditions can it be transformed into a trigonometric form (not a Fourier series) in the sense of (3) according to (2)?test.mw

 interface(version);

`Standard Worksheet Interface, Maple 2024.2, Windows 11, October 29 2024 Build ID 1872373`

(1)

restart

term := 2*cos(5*arcsin((1/2)*x))

2*cos(5*arcsin((1/2)*x))

(2)

term1 := expand(term)

-3*(-x^2+4)^(1/2)*x^2+(-x^2+4)^(1/2)+(-x^2+4)^(1/2)*x^4

(3)

convert(term1, trig)

-3*(-x^2+4)^(1/2)*x^2+(-x^2+4)^(1/2)+(-x^2+4)^(1/2)*x^4

(4)

simplify(term1, trig)

(-x^2+4)^(1/2)*(x^4-3*x^2+1)

(5)

solve(term1 = sqrt(2), x)

(1/2)*(8-2*(10+2*5^(1/2))^(1/2))^(1/2), (1/2)*(8+2*(10-2*5^(1/2))^(1/2))^(1/2), (1/2)*(8+2*(10+2*5^(1/2))^(1/2))^(1/2), -(1/2)*(8-2*(10+2*5^(1/2))^(1/2))^(1/2), -(1/2)*(8+2*(10-2*5^(1/2))^(1/2))^(1/2), -(1/2)*(8+2*(10+2*5^(1/2))^(1/2))^(1/2)

(6)

evalf(solve(term1 = sqrt(2), x))

.3128689302, 1.782013048, 1.975376681, -.3128689302, -1.782013048, -1.975376681

(7)

plot(term, x = -2.5 .. 2.5)

 

plot(term1, x = -2.5 .. 2.5)

 

NULL

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...flies around in the spatial Cartesian coordinate system and impacts the coordinate planes xy, xz, and yz with exactly one impact. This creates circular impressions on the coordinate planes in this order, with the radii r1, r2, and r3, which are assumed to be known. From these, the sphere's radius r and the coordinates of its center can be determined?

For practice, I would like to solve the following problem. It is from "Steven Weinberg, Gravitation and Cosmology, p. 7" and was posed in a NG. I planned to define coordinates (x_i; y_i), apply the Pythagorean theorem, and enter the tediously long terms as term1 and term2, respectively. The final result should be term1-term2=0, or "is" should be used. I failed to enter the coordinates - the error message "error=Null" appears.

Question:
How are coordinates and their names of the type described meaningfully defined and entered as term1 and term2, respectively?

The four points P_1, P_2, P_3, P_4 lie in the Euclidean plane. Let
(ij) be the square of the distance between P_i and P_j. Then, it must be proven that

(12)(12)(34)+(13)(13)(24)+(14)(14)(23)+(23)(23)(14)
+(24)(24)(13)+(34)(34)(12)+(12)(23)(31)+(12)(24)(41)
+(13)(34)(41)+(23)(34)(42)
=
(12)(23)( 34)+(13)(32)(24)+(12)(24)(43)+(14)(42)(23)
+(13)(34)(42)+(14)(43)(32)+(23)(31)(14)+(21)(13)(34)
+(24)(41)(13)+(21)(14)(43)+(31)(12)(24)+(32)(21)(14)

In the attached file, (6) is stated as "false." However, it is possible to prove with pen and paper that term1 = term2. In (5), the limit function is sought but not determined.
What am I doing wrong?test1.mw
 

restart

term1 := (2*cos(2^n*x)+1)/(2*cos(x)+1)

(2*cos(2^n*x)+1)/(2*cos(x)+1)

(1)

term2 := product(2*cos(2^k*x)-1, k = 0 .. n-1)

product(2*cos(2^k*x)-1, k = 0 .. n-1)

(2)

term3 := term1*(2*cos(2^n*x)-1)/(2*cos(x)-1)

(2*cos(2^n*x)+1)*(2*cos(2^n*x)-1)/((2*cos(x)+1)*(2*cos(x)-1))

(3)

simplify(term3)

(1+2*cos(2^(1+n)*x))/(1+2*cos(2*x))

(4)

limit((1+2*cos(2^(1+n)*x))/(1+2*cos(2*x)), n = infinity)

limit((1+2*cos(2^(1+n)*x))/(1+2*cos(2*x)), n = infinity)

(5)

is(term1 = term2)

false

(6)

NULL


 

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