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These are replies submitted by vv

@nm It's not in the standard form (or reducible to)  d/dx y(x) = F(x, y(x)).

The reason is obvious. Anyway, I see ODESteps as designed for standard simple ODEs. 

@nm No bug without Physics updates.

@minhthien2016 Insert f:=simplify(f) assuming real; before calling extrema

@Carl Love Very nice, vote up! Unfortunately the rose is not quite 3d: rotating the plot, several gaps are visible and the stem is just "separately added".

@Hullzie16  Adding the option numeric, dsolve produces a nrmerical solution,
Your system has a lot of floats, so, approximations. In such situations, an exact (analytic) solution is usually useless (even when it exists). 

@Ali Hassani 

seriestorec(s, A(n)) 
   ==> for a given series s, finds a "suspected" recurrence relation of its coefficients A(n)
rectodiffeq(rec, A(n), f(z))
   ==> for a recurrence rec, finds the ode satisfied by the "generating function"
           f(z) = Sum(A(n)*z^n, n = 0 .. infinity).

@dharr  When checking the minimal polynomial we must use a much higher precision.

evalf(expr) - fsolve(poly)[2]: evalf[15](%);
#                        1.37954446875327 e-38

(it should be near 1e-200).
So, the first poly is incorrect.

The second one (of degree 20) is ok.

@sursumCorda  Why not use e.g.

Matrix( [ [x || (1 .. 2)] , [y || (1 .. 2)] , [1 $ 2] ] );



ode:=x*y(x)*diff(y(x), x) = (x + 1)*(y(x) + 1):
ic:=y(1) = 1:


y(x) = -LambertW(-1, -2*exp(-x-1)/x)-1



x+ln(x)-y(x)+ln(y(x)+1)+_C1 = 0


eq:= eval(sol_no_IC,[y(x)=1,x=1]);

ln(2)+_C1 = 0



_C1 = -ln(2)



x+ln(x)-y(x)+ln(y(x)+1)-ln(2) = 0


S:=solve(solC, y(x)) assuming x>1;

-LambertW(-2*exp(-x-1)/x)-1, -LambertW(-1, -2*exp(-x-1)/x)-1


sol:=select(s -> (eval(s,x=1)=1), [S])[];

-LambertW(-1, -2*exp(-x-1)/x)-1


odetest(y(x)=sol, [ode,ic]);

[0, 0]



Download W.mw

@nm Maple decided to use the -1 branch because it knows the LambertW function. For example,  the restrictions  f₁:(-∞,-1]→[-1/e,0)  and  f₂: [-1,∞)→[-1/e,∞)  of  f(x)=x*exp(x)  are inverted as LambertW(-1,y) and LambertW(0,y). Given IC, Maple knows which one to use.

The same thing happens in other situations, e.g to invert  cos : [4*Pi, 5*Pi] -> [-1,1]. So it's nonsense to look for an algorithm, because it would imply to rewrite Maple.

If you do not want IC, just use the option implicit as I said; most cases will be solved this way.

@C_R Actually, in both situations, solve only finds a few surfaces contained in the interior of the torus (the main components of the solution are missing).

@Preben Alsholm The result of

sol2:=SolveTools:-SemiAlgebraic({torus1<0}, [x,y,z]);

seems to be correct but more complicated than necessary.

After completing my last post I was said that there are no tags; but they were introduced! I accepted a proposed tag (a stupid one) and the post was accepted.

@janhardo As seen, Maple knows the symbolic value of Zeta(2).
It is possible to deduce it e.g. from the infinite product expansion of sin(x) (see below), but it would be much more useful for you to read a few known mathematical proofs in the given references.


f:=Product(1-x^2/(k^2*Pi^2), k=1..infinity); # sin(x)/x

Product(1-x^2/(k^2*Pi^2), k = 1 .. infinity)


eval(diff(f,x,x)/2, x=0);

(1/2)*(Sum(-2*(Product(1, i1 = 1 .. k-1))*(Product(1, i1 = k+1 .. infinity))/(k^2*Pi^2), k = 1 .. infinity))


C2:=simplify(%); # coeff of x^2 in f

-(Sum(1/k^2, k = 1 .. infinity))/Pi^2


S2:=coeff(series(sin(x)/x, x), x^2);



isolate(C2=S2, indets(C2, function)[]);

Sum(1/k^2, k = 1 .. infinity) = (1/6)*Pi^2



Download zeta2.mw

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