Aixleft math

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These are questions asked by Aixleft math

Hi! how to find posint solution of this equation?

Is factorial move fast so that cannot solve?

any advice is appreciated.

restart

NULL

eq := `assuming`([factorial(x)+1 = y^2], [x::posint, y::posint])

factorial(x)+1 = y^2

(1)

`assuming`([solve(eq, {x, y})], [x <= 100, y <= 100])

{x = RootOf(factorial(_Z)-y^2+1), y = y}

(2)
 

NULL

Download find_posint_.mw

Hello, an interesting issue about set equivalence.

Logically, the two sets are equivalent by derivation.

The first principal case, for set A, k=0, the element values ​​are Pi/6 and 5*Pi/6, which corresponds to the case of k=0 (element value is Pi/6) and k=1 (element value is 5*Pi/6) in set B. Obviously, the k value is not one-to-one correspondence, but just a letter representing a positive integer. As for the second general case, it is the same steps I thought.

So, how to verify that the two sets are equivalent? I know Maple cannot do it in one step, but I don't know how to do it?

A := solve(sin(x) = 1/2, allsolutions = true)

(1/6)*Pi+2*Pi*_Z5, (5/6)*Pi+2*Pi*_Z5

(1)

A := `assuming`([`union`({(1/6)*Pi+2*k*Pi}, {Pi-(1/6)*Pi+2*k*Pi})], [k::integer])

{(1/6)*Pi+2*k*Pi, (5/6)*Pi+2*k*Pi}

(2)

B := `assuming`([{k*Pi+(1/6)*(-1)^k*Pi}], [k::integer])

{k*Pi+(1/6)*(-1)^k*Pi}

(3)

is(A = B)

false

(4)

restart

alpha = 'alpha'

alpha = alpha

(5)

solve(sin(x) = alpha, x, allsolutions = true)

2*Pi*_Z1+arcsin(alpha), -arcsin(alpha)+Pi+2*Pi*_Z1

(6)

A := `assuming`([`union`({arcsin(alpha)+2*k*Pi}, {Pi-arcsin(alpha)+2*k*Pi})], [k::integer])

{arcsin(alpha)+2*k*Pi, Pi-arcsin(alpha)+2*k*Pi}

(7)

B := `assuming`([{k*Pi+(-1)^k*arcsin(alpha)}], [k::integer])

{k*Pi+(-1)^k*arcsin(alpha)}

(8)

is(A = B)

false

(9)
 

NULL

Download verify_set_A_and_set_B_is_equivalent.mw

Hi! An basic and interesting issue.

First, why I run worksheet again and again, I will get different output (even seems understandable)? I guess maybe the memory is not renewed?

Second, how to obtain the periodic solution without adding ```k*Pi``` manually?

`assuming`([solve(sqrt(2)*sin(2*x-(1/6)*Pi) = 1)], [`in`(x, real)])

5/24

(1)

x := solve(sqrt(2)*sin(2*x-(1/6)*Pi) = 1)+k*Pi

5/24+k*Pi

(2)
 

NULL

Download periodic_solution_of_a_simply_Sine_function.mw

Hi! A basic issue.

Why view=[-2 ..1, -2 ..5]  is not useful here? According to the output, only the green line meets the view settings. I want to extend the left side of these three lines appropriately (show the intersection)

with(plots)

l := 2*x+y+1 = 0; l1 := 4*x+2*y+2 = 0; l2 := 4*x+2*y-2 = 0; l3 := 4*x-2*y+6 = 0

2*x+y+1 = 0

 

4*x+2*y+2 = 0

 

4*x+2*y-2 = 0

 

4*x-2*y+6 = 0

(1)

 

display({implicitplot(l, color = black, legend = l, thickness = 5, view = [-2 .. 1, -2 .. 5])}, {implicitplot(l1, color = red, legend = l1, view = [-2 .. 1, -2 .. 5])}, {implicitplot(l2, color = blue, legend = l2, view = [-2 .. 1, -2 .. 5])}, {implicitplot(l3, color = green, legend = l3, view = [-2 .. 1, -2 .. 5])})

 
 

NULL

Download The_intersection_parallelism_and_coincidence_of_two_straight_lines.mw

Hi. How to plot the 2d output like this? I don't know how to declare "a" and "b" random constant?
The pair of equations x = a and y = b graphically represents lines which  are, a. parallel , b. intersecting at (b, a), c. coincident, d.  intersecting at (a, b)

a := (rand(3 .. 4))(); b := (rand(0 .. 3))(); plot(x = a); plot(y = b)

3

 

2

 

Error, (in plot) unexpected option: x = 3

 

Error, (in plot) unexpected option: y = 2

 
 

``

Download x=a_y=b_plot.mw

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