vv

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10 years, 43 days

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@Carl Love 
Yes, I was once the author of such an "infringement", but I cannot understant how posting the result of a Maple command (showstat in this case) could be interpreted as a copyright problem.

@Carl Love 

plots:-transform works if the surface is simple wrt one of the axes (y in this case).
It will not work e.g. for
S1:= plot3d([1, theta, phi], theta= 0..3*Pi/2, phi= 0..Pi, coords= spherical):
But in such cases, for general parametric plots a small normal perturbation will work
(see http://www.mapleprimes.com/posts/203796-Equidistant-Surface-)

CP:=(v,w)->[v[2]*w[3]-v[3]*w[2], v[3]*w[1]-v[1]*w[3], v[1]*w[2]-v[2]*w[1]]: #Cross Product
IP:=(v,w)->v[1]*w[1]+v[2]*w[2]+v[3]*w[3]: # Inner Product
UV:=v -> v/~sqrt( IP(v,v)): #Unit Vector

r:=[sin(phi)*cos(theta), sin(phi)*sin(theta), cos(phi)]:  # Example
Nr:=UV( CP( diff(r,theta), diff(r,phi) ) ):  # Unit Normal vector to r
Er:=simplify(r + 1/200*Nr):;           # Equidistant to r

plot3d( [r,Er], theta= 0..3*Pi/2,  phi= 0.001..Pi,color=[red,green] );

 

The procedure `hypergeom/check_parameters` checks whether the upper parameters which are integer <=0 are at least as many as the lower ones. [maybe this could be relaxed].

I have noticed that:
limit( hypergeom([1, -1, 1/2], [k,-3], 1),  k=-12 ); 
      71/72

@lyakhovda 
But my previous reply answers to your question for r=23 too; it is NO.

@John Fredsted 

The OP wants to choose himself the free parameters.
Actually for a random linear system, this can be done with probability 1, but not for each and every system.

@lyakhovda 

If the rank is 6, then there will be 29-6=23 free variables, so {Z20,...,Z28} are not enough.
And you cannot know whether e.g. Z24 can be a free variable because the system could contain the equation Z24 = 7.

Yes, Maple has big problems with oscillating integrals.

J := int(abs(cos(1/t)), t = 0 .. 1);
is wrong (undefined).
And it is a challenge to compute J with 10 (or 30) digits in Maple!

 

But the angle at x=a is not  arctan(|f'(a) - g'(a)|). It should be  | arctan(f'(a)) - arctan(g'(a)) |.

PS. I doubt that a beginner who writes cosx^2  will grasp map2, ~, @

@Carl Love 

@wyoum 

Try:

subs( Physics:-`*` = :-`.`, expr );

subs( Physics:-`.` = :-`.`, expr );

 

@wyoum 
If you are interested in FP groups, the Physics package will be anyway not enough.
With GroupTheory you will be able to simplify much better.
Example:

with(GroupTheory):
G := < a, b | a^2 = 1, b^4 = 1, (a.b)^2 = 1 >:
E:=Subgroup([[]],G);  # trivial subgroup
Factor( a.b.a.b^3 .a.b.a, E);
            
[1,b]

So, in the group G one has:    a.b.a.b^3 .a.b.a = b

 

I think I have made an error while writing my answer. Trying to detete it and reload the worksheet, I think I have deleted  Markiyan Hirnyk' s answer.
Sorry, Markiyan, please repost the answer.


 

 

Jn:=-(1/n)*Int( y^(a+n)*(1-y)^b,y=0..1);

-(Int(y^(a+n)*(1-y)^b, y = 0 .. 1))/n

(2)

jn:=value(Jn);

-GAMMA(b+1)*GAMMA(a+n+1)/(n*GAMMA(a+b+n+2))

(3)

simplify( sum(jn, n=1..infinity) ) assuming a>-1,b>-1;

-(Psi(a+b+2)*b^2-Psi(3+b)*b^2+3*b*Psi(a+b+2)-3*Psi(3+b)*b+2*Psi(a+b+2)-2*Psi(3+b)+2*b+3)*GAMMA(a+1)*GAMMA(b+1)/((b^2+3*b+2)*GAMMA(a+b+2))

(4)

 

### For the original integral

ans:= GAMMA(phi) / (GAMMA(mu*phi)*GAMMA((1-mu)*phi) )  * eval(%, [a = mu*phi-1, b = (1-mu)*phi-1]);

-GAMMA(phi)*(Psi(mu*phi+(1-mu)*phi)*((1-mu)*phi-1)^2-Psi(2+(1-mu)*phi)*((1-mu)*phi-1)^2+3*((1-mu)*phi-1)*Psi(mu*phi+(1-mu)*phi)-3*Psi(2+(1-mu)*phi)*((1-mu)*phi-1)+2*Psi(mu*phi+(1-mu)*phi)-2*Psi(2+(1-mu)*phi)+2*(1-mu)*phi+1)/((((1-mu)*phi-1)^2+3*(1-mu)*phi-1)*GAMMA(mu*phi+(1-mu)*phi))

(5)

ans:=simplify(ans) assuming mu>0,mu<1, phi>0;

-(Psi(phi)*(mu*phi-phi+1)^2-Psi(-mu*phi+phi+2)*(mu*phi-phi+1)^2+3*(-mu*phi+phi-1)*Psi(phi)-3*Psi(-mu*phi+phi+2)*(-mu*phi+phi-1)+2*Psi(phi)-2*Psi(-mu*phi+phi+2)-2*(-1+mu)*phi+1)/(mu^2*phi^2-2*mu*phi^2-mu*phi+phi^2+phi)

(6)

eval(ans, [mu = 1/3, phi = 2]);

2-(1/6)*Pi*3^(1/2)-(3/2)*ln(3)

(7)

evalf(%);

-.554818117

(8)

 

@Markiyan Hirnyk 

All the source code is in my post; it is enough a copy & paste + execute.

I had the impression that you have tried lately to adopt a minimal politeness.
It seems that I was wrong. I am sorry.

Edit. It seems also that there is still hope, because you have removed your reply.

 

@Markiyan Hirnyk 

a = mu*phi-1
b = (1-mu)*phi-1

(The factor not depending on y was removed).

@krismalo 
I don't think it could be simpler; the procedure has only a few lines.
If f has multiple roots, solve(f,x) will find all of them. If you want the minimal one, simply replace
a := solve(f,x);
    with
a := min(solve(f,x) );
 

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