Alfred_F

330 Reputation

9 Badges

0 years, 330 days

MaplePrimes Activity


These are questions asked by Alfred_F

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


 

Download test1.mw

 

In the attached file, I would like to determine the real part of the complex term2. I'm asking for your help.test.mw

restart

term1 := exp(I*t/2^k)

exp(I*t/2^k)

(1)

term2 := product(term1, k = 1 .. n)

(cos(2*t*(1/2)^(n+1))-I*sin(2*t*(1/2)^(n+1)))/(cos(t)-I*sin(t))

(2)

``

Download test.mw

On my journey of discovery through the world of Maple, I would like to ask for help again:
How are the properties of variables and the indexing of sequences handled/determined? For this purpose, I chose an old problem from a challenging MO as an exercise and tried it in the attached file, which, of course, failed.

restart

a[i] := 2*cos(t/2^i)-1

2*cos(t/2^i)-1

(1)

b[n] := product(a[i], i = 1 .. n)

product(2*cos(t/2^i)-1, i = 1 .. n)

(2)

limit(b[n], n = infinity)

limit(product(2*cos(t/2^i)-1, i = 1 .. n), n = infinity)

(3)

simplify(limit(b[n], n = infinity))

limit(product(2*cos(t*2^(-i))-1, i = 1 .. n), n = infinity)

(4)

NULL

Download test.mw

The attached problem is from a 1988 MO. It can be solved using complete induction, paper, and pencil, and with some effort, yields a simple answer. It's quite challenging to do by hand, but with "derive", it only takes three lines and a fraction of a second. Mow test.mw

restart

"puzzle(n):=(∑)(cos(k*Pi/(2*n+1))^())^(4);"

Download test.mw

I can't do it with Maple because I'm doing something wrong again. Therefore, I'm asking for help.

1 2 3 4 5 6 7 Last Page 1 of 12