Kitonum

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17 years, 182 days

MaplePrimes Activity


These are replies submitted by Kitonum

@minhthien2016  
simplify((1/2)*ln(1+tan(x)^2))  assuming cos(x)>0;
                             
  -ln(cos(x))

@minhthien2016 

restart; 
int(sin(x)^n, x) assuming n::posint:
eval(%, n = 7); 
expand(simplify(value(%)));
algsubs(cos(x)^2=1-sin(x)^2, %);
factor(%);

 

@student_md 
Tm2 is present in your new code, but it is not defined. From your previous version it seems that  Tm2  is the same as Tm. Correct this.

The first error in your code occurs in the following line:

coz:=fsolve(sys);
indets(sys, name);

Error, (in fsolve) number of equations, 16, does not match number of variables, 21
{P_nn3, c[1, 1], c[1, 2], c[1, 3], c[1, 4], c[2, 1], c[2, 2],  c[2, 3], c[2, 4], c[3, 1], c[3, 2], c[3, 3], c[3, 4], [4, 1],  c[4, 2], c[4, 3], c[4, 4], t[1], t[2], t[3], t[4]}
 

 

@Jalale  See the corrected file:

QuestionSimulation_new1.mw 

@gaurav_rs  Maple often does not take into account restrictions on variables imposed by  assume  command.

An example:

assume(x>Pi/2, x<Pi):
solve(sin(x)=1/2);
           
 Warning, solve may be ignoring assumptions on the input variables.
                              1   
                              - Pi
                              6   
 

@Magma  See help on  ?remember . There everything is explained in detail with examples.

@vv   ``( ... )  construction prevents the automatic opening of the parentheses.

@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.

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