Kitonum

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12 years, 73 days

MaplePrimes Activity


These are replies submitted by Kitonum

According to Markiyan's idea:

l1:=[a1*t+x1, b1*t+y1, c1*t+z1]:

l2:=[a2*s+x2, b2*s+y2, c2*s+z2]:

minimize(add((l1[i]-l2[i])^2, i=1..3), s=-infinity..infinity, t=-infinity..infinity) assuming a1^2+b1^2+c1^2>0 and a2^2+b2^2+c2^2>0:

RealDomain[simplify](sqrt(%)); 

 

 

@Markiyan Hirnyk 

Let the questioner will decide himself which option is more suitable for him. Purely terminological distinction.

@Markiyan Hirnyk 

Let the questioner will decide himself which option is more suitable for him. Purely terminological distinction.

@Markiyan Hirnyk 
The questioner wrote "two positive solutions and two negative solutions". If  m=5/2  then we have one negative and two positive roots.

@Markiyan Hirnyk 
The questioner wrote "two positive solutions and two negative solutions". If  m=5/2  then we have one negative and two positive roots.

@Markiyan Hirnyk 

Of course, I understand that the field is not potential, so the path of integration should be specified.

@Markiyan Hirnyk 

Of course, I understand that the field is not potential, so the path of integration should be specified.

This is a simple task, it is easy to solve by a procedure. But I did not understand the law of getting your matrix for any m.  What is  i ? Can you give an example of such a matrix for a given m, for example, for m = 8 ?

I noticed an error in the text of the procedure. Instead of  if alpha = 0 then {k> 0}  should be  if alpha = 0 then print(`No solutions`) .

 

@Markiyan Hirnyk

Of course, the advantage of substantially purely aesthetic.
Also, I think this form of writing can be interesting for educational purposes in the study of the concept of absolute value.

PS. The advantage is the same as   |x|  over 

 

 

I did not notice the condition that  x>=0, y>=0, z>=0. Therefore, the solution exists.

I did not notice the condition that  x>=0, y>=0, z>=0. Therefore, the solution exists.

@one_man

Thank you for your interest! I was waiting for your question, because I know about your enthusiasm implicitly given curves. Setting the boundary of the figure is not provided in this way in my procedures, but still there is a simple solution to the problem. Here it is: 

P1 := plots[implicitplot]((x1+0.7e-1*sin(30*x2)^2)^2+(x2+0.7e-1*sin(30*x1)^2)^2-1 = 0, x1 = -2 .. 2, x2 = -2 .. 2, numpoints = 100000):

P2 := plottools[getdata](P1):

L := [convert(P2[3], listlist)]:

Perimeter(L);

Area(L);

plots[display](Picture(L, color = green, [color = black, thickness = 2]));

 

 

@Alejandro Jakubi 

Of course, my statement is not accurate from a mathematical point of view. Would be more accurate to write that there are examples which algsubs command decides better than applyrule command. Maybe you know the opposite examples?

Thanks for your comment!

@Alejandro Jakubi 

Of course, my statement is not accurate from a mathematical point of view. Would be more accurate to write that there are examples which algsubs command decides better than applyrule command. Maybe you know the opposite examples?

Thanks for your comment!

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