Kitonum

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11 years, 97 days

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These are replies submitted by Kitonum

@mehdi jafari   Probably no.

@digerdiga 

f:=ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2)) + ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2)) + ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2)) + ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2));
subsindets(f, `function`, p->applyop(t->numer(t)/expand(denom(t)),1, p));
simplify(%);
expand(%);

@waseem 

with(plots):
A:=plot([sin(x), cos(x)], x=-Pi..2*Pi, linestyle=[dash,dot], color=[red,blue], thickness=2):
B:=textplot([[3.6,0.8,sin(x)], [-1.8,0.5,cos(x)]], font=[times,bold,
15]):
C:=plot([[t,0.8,t=2.6..3.3],[t,0.5,t=-2.6..-2.2]], color=[red,blue], linestyle=[dash,dot], thickness=2):
display(A, B, C, size=[950,200], scaling=constrained);

 

@waseem   Add  evalf  command:

evalf(1-cos(0.1*Pi)^2);
                                           
 0.0954915028

I can not confirm it (Maple 2018.2  64 bit on Windows 10 without Physics version 272):

restart;
pde := diff(u(x, t), t) + diff(u(x, t),x) =0;
sol:=pdsolve(pde,u(x,t));
pdetest(sol,pde);
  
bc:=u(0,t)=0;
ic:=u(x,0)=sin(x);
sol:=pdsolve([pde,ic,bc],u(x,t)) assuming x>0;
pdetest(sol,pde);
                                

 

@bliengme 

Replace the line  
proc GCD3:=proc(n1,n2)  
with the line  
GCD3:=proc(n1,n2)

@Earl  If a fraction makes sense (the denominator is not 0) and this fraction is 0, then this means that its numerator is 0:

a/b-c/d=0;
# It is the same as
numer(lhs(%))=0;
 

@vanzzy  "The new version" above.

This is the new version of my answer that was deleted a few minutes ago.

Try the following:

plotmp_new1.mw

 

@greenworld   Use  continuous  option for this. In this case, Maple calculates the integral using the Newton – Leibniz formula, without checking the continuity of the integrand for continuity in the interval of integration:

Expr:=sin(x)/((a*b+a^2*sin(x)^2-d^2*cos(x)^2)*(c+cos(x)));
int(Expr, x=0..x, continuous);

 

 

@digerdiga  You wrote  "Apparently it can be transformed into a polylog(3,I)."

But

evalf([polylog(3,1+I) + polylog(3,1-I), polylog(3,I)]);
                 
 [1.742317767+0.*I, -.1126928347+.9689461463*I]

@vanzzy  Here are the plotting of 3 points for B=1, k=2  and 3 values of  V=[0.5, 1.5, 2] :

restart;
gm:=V->1/sqrt(1-V^2): T:=w-k*V: S:=w*V-k:
f:=unapply(I*B*gm(V)^3*T^3+gm(V)^2*T^2-I*gm(V)^3*T*S^2-1/3*I*B*gm(V)^3*S^3*T-1/3*gm(V)^2*S^2, w,B,V,k);
F:=(B,V,k)->fsolve(f(w,B,V,k),w, complex);
B:=1: k:=2: V:=[0.5, 1.5, 2]:
plots:-pointplot([seq([Re,Im](F(B,v,k)[1]), v=V)], color=red, symbol=solidcircle, symbolsize=17);

 

@vanzzy 
Example for the first root:
restart;
gm:=V->1/sqrt(1-V^2): T:=w-k*V: S:=w*V-k:
f:=unapply(I*B*gm(V)^3*T^3+gm(V)^2*T^2-I*gm(V)^3*T*S^2-1/3*I*B*gm(V)^3*S^3*T-1/3*gm(V)^2*S^2, w,B,V,k);
F:=(B,V,k)->fsolve(f(w,B,V,k), w, complex);
plot(Im('F'(B,0.5,2)[1]), B=0..1);


Addition. Your equation has 4 degree for w. Here is the plots of all the roots:

restart;
gm:=V->1/sqrt(1-V^2): T:=w-k*V: S:=w*V-k:
f:=unapply(I*B*gm(V)^3*T^3+gm(V)^2*T^2-I*gm(V)^3*T*S^2-1/3*I*B*gm(V)^3*S^3*T-1/3*gm(V)^2*S^2, w,B,V,k);
F:=(B,V,k)->fsolve(f(w,B,V,k),w, complex);
plot([seq(Im('F'(B,0.5,2)[i]), i=1..4)], B=0..1);

 

@brian bovril  Hi Brian! My procedure easily solves this problem. Replace the fountain with a large enough bucket (I took 40 gallons), half full (so that was where to pour the water if required). The middle branch of all the solutions tree gives the desired solution. We see what is required at least 6 steps:

Pouring([40,5,3], [20,0,0]);
Pouring([40,5,3], [20,0,0], trails)[2][1..7];


                                       

At the same time, we get all possible cases of filling these vessels with a capacity of 5 and 3 gallons.

 

Happy Xmas to you and all Maple fans!


Edit.

@asa12  You have long been on this forum and do not understand the obvious things. You must copy your code here in text form (and not as a picture) or upload your worksheet.

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