vv

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These are replies submitted by vv

@Alex0099  1. The order of summation does not matter, the summand being symmetric.

2. The proof could be very tricky. Probably here the Residue Theorem should be useful.  

@Thomas Dean  In Windows, after a restart everything is OK. Sorry, I don't use Linux.

@emendes  procedure is a type, so, its use is covered.

AFAIK the Draghilev method is to find a natural parametric representation of a curve given as an intersection of n surfaces in (n+1) unknowns. That is what Rouben did, but using direct computations.

(I was never able to find the original article of Draghilev, but there are many places -- even in this site -- where the method is used).   

@lcz Replace:

first:= fsolve((f(t)-x0)^2+(g(t)-y0)^2=0.02^2, t=0..1);
last := fsolve((f(t)-x3)^2+(g(t)-y3)^2=0.02^2, t=0..1);

 

@Stretto I don't see how you could make ArgK to work. As it is, after correcting the Is problem (see acer's comment), it would work only when computing the values in an array, but not randomly (in any language!). I'd suggest to take the dsolve approach.

It is interesting to reformulate eq2  and see what happens.

 

 

restart;

eq2:=int(diff(y(x),x)*x,x)+int(y(x),x) +1 = 0;
 

int((diff(y(x), x))*x, x)+int(y(x), x)+1 = 0

(1)

sys2:={
u(x)+v(x)+1 = 0,
diff(u(x),x) = diff(y(x),x)*x,
diff(v(x),x) = y(x)
}:

sol:=dsolve(sys2);

{u(x) = _C2*ln(x)+_C1, v(x) = -_C2*ln(x)-_C1-1, y(x) = -_C2/x}

(2)

odetest(sol, sys2);

{0}

(3)

eval(eq2, sol); # due to integration constants!

1 = 0

(4)

 

@Carl Love  A and B having polynomial entries, A.B = Id ==> A invertible, that's all.
BTW, your potential objection is also valit for the provided answer, if A = <p,0;0,1/p>, where det(A) gives 1. But if we disect such a simple problem, everything becomes complicated (more or less artificially).

Actually, you solved the problem! For a)  it's enough to compute  simplify(A . B) , obtaining the unit matrix, without any assumption on p. (Of course it's very useful to know what a singular matrix is.)

@fhn2022 A smooth discontinuous function? Analytic? In mathematics such things do not exist
[f analytic ==> f smooth ==> f continuous].

@fhn2022 Why are you so sure that the function is continuous? You have many complex square roots! Denominators too!

@fhn2022 You may need to increase Digits but then even a longer time will be needed.
If the problem is important for you, try to optimize/simplify the computations.

@fhn2022 I did not change anything in your code. The first output is just a check with the value of T2 for the those values of alf, y and x. You may delete it.
Only the method of integration in int was changed to method=_CubaCuhre. As you see, it works for alf > 0.277.

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