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These are questions asked by Annonymouse

I have an expression involving a sqrt, and i'd like to simplify it, but simplify(expression,sqrt) doesn't seem to fully help.

Eq1 := [(1/2)*(R[b]*kh[a2]+R[m]*kh[a2]-Rh[m]*kh[a2]+sqrt(R[b]^2*kh[a2]^2+2*R[b]*R[m]*kh[a2]^2-2*R[b]*Rh[m]*kh[a2]^2+R[m]^2*kh[a2]^2-2*R[m]*Rh[m]*kh[a2]^2+Rh[m]^2*kh[a2]^2))/kh[a2] = 0, 0 = 0, -(1/2)*(R[b]*kh[a2]+R[m]*kh[a2]-Rh[m]*kh[a2]+sqrt(R[b]^2*kh[a2]^2+2*R[b]*R[m]*kh[a2]^2-2*R[b]*Rh[m]*kh[a2]^2+R[m]^2*kh[a2]^2-2*R[m]*Rh[m]*kh[a2]^2+Rh[m]^2*kh[a2]^2))/kh[a2] = 0]

simplify(Eq1, sqrt)


[(1/2)*(R[b]*kh[a2]+R[m]*kh[a2]-Rh[m]*kh[a2]+sqrt(kh[a2]^2*(R[b]+R[m]-Rh[m])^2))/kh[a2] = 0, 0 = 0, -(1/2)*(R[b]*kh[a2]+R[m]*kh[a2]-Rh[m]*kh[a2]+sqrt(kh[a2]^2*(R[b]+R[m]-Rh[m])^2))/kh[a2] = 0]

which on paper simplifies to:


is there a way to get maple to show this?

[In part I am trying to better understand how to manipulate sqrt expressions in maple]

I have a complicated expression which includes RootOf( a quadratic ) but holds for all x what i'd like to do is turn it into a polynomial in x[1], x[2], x[3] so i can start looking at the monomial coefficients.

k[a1]*((x[1]+x[3])*k[d1]+C[T]*k[m])*(R[b]-x[1]-2*x[2])/((R[b]+R[m]-x[1]-2*x[2]-x[3])*k[a1]+k[m])-k[d1]*x[1]-k[a2]*x[1]*(R[b]-x[1]-2*x[2])+2*k[d2]*x[2] = (-R[b]*k[a2]+2*k[a2]*x[1]+2*k[a2]*x[2])*(k[a1]*kh[m]*((x[1]+x[3])*k[d1]+C[T]*k[m])*(R[b]+R[m]-Rh[m]-x[1]-2*x[2])/(k[m]*((R[b]+R[m]-x[1]-2*x[2]-x[3])*k[a1]*kh[m]/k[m]+kh[m]))-k[d1]*x[1]-kh[a2]*x[1]*(R[b]+R[m]-Rh[m]-x[1]-2*x[2])+2*kh[d2]*x[2])/(2*kh[a2]*RootOf(kh[a2]*_Z^2+(-R[b]*kh[a2]-R[m]*kh[a2]+Rh[m]*kh[a2]+2*kh[a2]*x[2])*_Z-2*k[a2]*x[1]*x[2]-k[a2]*x[1]^2+k[a2]*x[1]*R[b]+2*kh[d2]*x[2]-2*k[d2]*x[2])-R[b]*kh[a2]-R[m]*kh[a2]+Rh[m]*kh[a2]+2*kh[a2]*x[2])+(-2*kh[a2]*RootOf(kh[a2]*_Z^2+(-R[b]*kh[a2]-R[m]*kh[a2]+Rh[m]*kh[a2]+2*kh[a2]*x[2])*_Z-2*k[a2]*x[1]*x[2]-k[a2]*x[1]^2+k[a2]*x[1]*R[b]+2*kh[d2]*x[2]-2*k[d2]*x[2])+2*k[a2]*x[1]+2*k[d2]-2*kh[d2])*(kh[a2]*x[1]*(R[b]+R[m]-Rh[m]-x[1]-2*x[2])-2*kh[d2]*x[2])/(2*kh[a2]*RootOf(kh[a2]*_Z^2+(-R[b]*kh[a2]-R[m]*kh[a2]+Rh[m]*kh[a2]+2*kh[a2]*x[2])*_Z-2*k[a2]*x[1]*x[2]-k[a2]*x[1]^2+k[a2]*x[1]*R[b]+2*kh[d2]*x[2]-2*k[d2]*x[2])-R[b]*kh[a2]-R[m]*kh[a2]+Rh[m]*kh[a2]+2*kh[a2]*x[2])

If this were something like q(x)=p1(x)/sqrt(p2(x)) where p1 and p2 are polynomials and q is a quotient- this would be as simple as making sqrt(p2(x)) the subject and squaring both sides, and then movinbg everything onto one and multiplying out denominators. However RootOf is something I'm not used to manipulating.

Is there anyway of converting this expression to a polynomial using maple commands?

I am working on a problem that involves finding a map lambda(x) which maple stores using RootOf:

LambdaMap := [lambda[1] = RootOf(kh[a2]*_Z^2+(-R[b]*kh[a2]-R[m]*kh[a2]+Rh[m]*kh[a2]+2*kh[a2]*x[2])*_Z-2*k[a2]*x[1]*x[2]-k[a2]*x[1]^2+k[a2]*x[1]*R[b]+2*kh[d2]*x[2]-2*k[d2]*x[2]), lambda[2] = x[2], lambda[3] = x[1]+x[3]-RootOf(kh[a2]*_Z^2+(-R[b]*kh[a2]-R[m]*kh[a2]+Rh[m]*kh[a2]+2*kh[a2]*x[2])*_Z-2*k[a2]*x[1]*x[2]-k[a2]*x[1]^2+k[a2]*x[1]*R[b]+2*kh[d2]*x[2]-2*k[d2]*x[2])]

(as an aside, after a few years of using maple and reading the help page I find RootOf confusing, and I think this is the root of this question, I'd love to get some recomended reading on it in the answers)

I want to use the restriction lambda(0,0,0)=(0,0,0) to find relationships between the parameters (kh,k,Rh,R etc)

I tried:

`~`[`=`](`~`[rhs](subs([x[1] = 0, x[2] = 0, x[3] = 0], LambdaMap)), [0, 0, 0]);

which returned:
[RootOf(kh[a2]*_Z^2+(-R[b]*kh[a2]-R[m]*kh[a2]+Rh[m]*kh[a2])*_Z) = 0, 0 = 0, -RootOf(kh[a2]*_Z^2+(-R[b]*kh[a2]-R[m]*kh[a2]+Rh[m]*kh[a2])*_Z) = 0]
{R[b] = R[b], R[m] = R[m], Rh[m] = Rh[m]}

which seems wrong (both the first and third equations are quadratics sharing roots, the first root is _Z=0, and the second is _Z=-R[b]-R[m]+Rh[m]- i'm not sure how either result in this solve)

Is this the right way to do this? i.e. do solve and subs, work intuitively with RootOf expressions.

I am doing some calculus on lambda next - is there anything i should be mindful of when calculating its Jacobean or using diff?

Given a list with sublists e.g.


is there a way of merging everything into a single list, for example



I have a Vector of polynomials in x[1] and x[2]; and I want to use coeffs to get the coefficients of bothe elements of the Vector; using the command:

Eq2Coeffs := `~`[coeffs](%, [x[1], x[2]])

bizarely the result i get is something in terms of x[1] and x[2].

Here is the worksheet this problem emerged in, the specific command is 3.8.

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