vv

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Int(cos(x)/(x^2+1),x=-infinity..+infinity) = 2*Pi*I*residue(exp(I*x)/(x^2+1), x=I);

        

@janhardo Edwards's book contains several obsolete notations including some used originally by Riemann himself. It is not a good idea to try to learn Complex Analysis from such a source (even if the book is a valuable monograph).   

You are a beginner in Complex Analysis. So, you should start with standard exercises. In this one, the path must be exactly defined; probably here we actually have a limit of a complex integral (e.g. tha radius of the arc around the origin tends to 0). 
The branches could be a problem; maybe it is not the Maple principal branch.
Not to mention z; the integral could diverge for some z, depending on how the path is defined.

dx / x is just for a convenient way of writing the integral: ∫ f(x)/x dx =  ∫ f(x) dx/x.

BTW, the notation Int( ..., x = +oo .. +oo)  is very exotic. Where did you find it?

@John2020 The methods do not work when the parameters depend on t; actually in such a case the problem does not make much sense.

@John2020 The method in my answer can be adapted here too.

restart;
r := x*(diff(theta(t), t))^2+y*(diff(varphi(t), t))^2+z*(diff(theta(t), t$2))+w*diff(varphi(t), t$2)+p*m:
g := (4*(f+T))*(diff(theta(t), t))^2+u*(diff(varphi(t), t))^2+(f+9)*(diff(theta(t), t$2))+4*s*diff(varphi(t), t$2)+p*4*cos(varphi(t)):

eval(r-g, [theta=(t->1+t+t^3), varphi=(t->1+t+t^7)]): # or similar
[coeffs(convert(series(%, t), polynom), t)]:
solve(%, [x,y,z,w,p]):
solve(%[1], [x,y,z,w,p]);
#      [[x = 4 T + 4 f, y = u, z = f + 9, w = 4 s, p = 0]]
simplify(eval(r-g, %[1]));   #check
#                               0

Note that you cannot take m as parameter; it must be p (it appears in both r and g); for a fixed p there is no solution.

@ecterrab  

eq:=sin(x)+cos(x) = a*sin(d*x^2+c*x+b):
solve( identity(eq, x), {a,b,c,d} );  # works
PDEtools:-Solve(eq, {a,b,c,d}, independentof=x); # must be interrupted

 

@Kitonum Yes, in my search I encountered the solve bug mentioned here

Sys:={x*y*z + y*z + y = -1, x*y*z + x*z + z = 0, x*y*z + x*y + x = 0}:
solve(Sys, [x,y,z]);

       [[x = 0, y = -1, z = 0]]

Anyway, the system with a,b,c has generically 3 solutions. It is impossible for the system to have exactly 3 integer solutions.

@nm More likely to be generic or defined in the code read at the beginning.

BTW, the algorithm is implemented in modern Maple, see  ?DEtools,kovacicsols

@janhardo No, it does not. Maple ignores the properties of ithprime(i), e.g. the fact that  ithprime(i) ~ i*ln(i)  for i --> oo.

@janhardo You should write

@jud (a)   int(F,...) works after a simplify and gives 4 - Pi  because you have a square minus a circle, not a circle.
Unfortunately, changing to the correct "f<0" the simplify is not enough.

(b) MultiInt works only for explicit functions, just like int; you can use of course piecewise or Heaviside as above (but the resulting function will be discontinuous, so, more difficult for Maple to handle).

@janhardo It's of course a typo, should be b.

expr := I*( ln(-z*I + b) - ln(z*I + b) ):
expr = combine(expr, symbolic):
simplify(eval(%, [z=-1, b=-1]));

        -(3*Pi)/2 = Pi/2

@janhardo You should use the symbolic option only in the rare cases when you don't care about the branch of the log.
Check here e.g. z=-1, a=-1.

@Carl Love Thank you Carl for this. For me it was a surprise that the presence of the lexical variable `a` is not visible in the body of the procedure F:

showstat(F);

F := proc(x)
   1   (3.5)*x*(1-x)
end proc

The only suspect element being (3.5) instead of 3.5.
The presence of `a` is revealed only in dismantle(eval(F)); and in op(7,eval(F));

BTW, I would have defined  f := unapply(F(x), x);

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