sbh

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19 years, 37 days

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These are questions asked by sbh

Hello, I have a somewhat math and Maple question I'm hoping some can help with.

I have this curve,

curve:=alpha^2*beta^10-alpha^2*beta^9+4*alpha*beta^7-2*alpha*beta^6-2*alpha*beta^5-alpha*beta^4+alpha*beta^3-2*beta+1;

and if I solve this system (numerically);

sol:= solve([curve=0,diff(curve,beta)=0],[alpha,beta]);

I get 6 special points (8 actually but two are critical). So I'll refer to them by subscript "i".

If I then do a coordinate transformation by;

alpha=exp(u)

beta=exp(v)

and convert the 6 coordinates and curve, I'll get everything in terms of (u,v) coordinates.

curve2:=subs([alpha=exp(u),beta=exp(v)],curve);

soluv:=map(ln,sol);  (this is just pseudo - I don't know how to do it this way)

So now the 6 points are referred to by (u_i,v_i).

Next, I want to expand this curve locally around these six points, using the following (where "z" is the local coordinate);

u -> u_i + z^2

v -> v_i + sum(a_j*z^j,j=1..n)

where n is reasonable, though around 15.

curve3(i):=subs(u=z^2 + cat(Ubp,i),curve2);

curve4(i,n):=subs(v=cat(Vbp,i)+sum(a[i,j]*z^j,j=1..n),curve3(i));

Here, I'm not sure of the pros/cons of cat() vs a[i,j]....

 

Anyway, I'll then have an equation in terms of only the local coordinate "z". If I then solve each coefficient of "z", at each order, I should then be able to determine the power series v(z). I reason that since the curve is initially equal to zero, that every non-zero power of z will have a coefficient/equation (in terms of unknowns a[i,j]) that should be equal to zero.

This is analagous to solving differential equations with power series...

 

However, I'm a little lost in implementing this,

I am currently trying, for instance,

l1:= series(curve4(1,10),z,0,9);

e1:= seq(coeff(l1,z,i),i=1..4):

s1:=seq(a[1,j],j=1..4):

sol:=solve(e1,s1);

I believe I am doing something wrong though bcause every odd power is zero.

 

Thank you a lot for any suggestions and/or help,

sbh

PS>

Ubp and Vbp are those 6 points - I just kept them as symbols initially because I was still getting odd-powered coefficients as zero, regardless of their actual values. Hence a little bit of the "math" side of the problem...

Hi, as the title says I am animating an implicitplot and want a different color for each curve left by the trace. I realize it may be easier to build plot structures using seq and display(..,insequence=false) but I thought I should learn more Maple ha ha. Plus  it seems easier to implement.

 Forgot to say I use Maple 15

For equation sol[2][1]=constant A,

 

 

> animate(implicitplot, [sol[2][1] = A, x = -6 .. 15, y = -15 .. 15, numpoints = 10^5], A = -10 .. 10, trace = 2, frames = 20);

 

 

I am trying to invert a function and, during the course, wind up with something like,

 

a_1*V^P*V + a_2*V^P*V^2 + a_3*V^P*V^3 + a_4*V^P*V^3 + a_5*V^P*V^3 + a_6*V^P*V^4 + \cdots

 

is there a way to get the poly into a form 

 

(a_1+a_2+...)*V^P + (b_1+b_2+...)*V^(P+1) + (c_1+c_2+...)*V^(P+2) + .....

 

?

Thanks. Attached is my Maple 16 sheet.

restart

with(LargeExpressions)

``

main := V^P*(V-1)^Q = c^(P-Q)*X^Q*(V-c^2)^Q

V^P*(V-1)^Q = c^(P-Q)*X^Q*(V-c^2)^Q

(1)

``

assume(Q, real); additionally(Q > 0); additionally(Q, integer)

assume(P, real); additionally(P > 0); additionally(P, integer)``

NULL

``

you := proc (s, n) V[1] := s; for i to n do V[i+1] := (`@`(simplify, expand))(c^((P-Q)/P)*X^(Q/P)*(-c^2+V[i])^(Q/P)/(V[i]-1)^(Q/P), symbolic) end do end proc

proc (s, n) local V, i; V[1] := s; for i to n do V[i+1] := (`@`(simplify, expand))(c^((P-Q)/P)*X^(Q/P)*(V[i]-c^2)^(Q/P)/(V[i]-1)^(Q/P), symbolic) end do end proc

(2)

``

``

START*HERE

START*HERE

(3)

``

Rearranging "main" leads to the following.

``

main2 := X^Q = V^P*(V-1)^Q/(c^(P-Q)*(V-c^2)^Q)

X^Q = V^P*(V-1)^Q/(c^(P-Q)*(V-c^2)^Q)

(4)

``

It can't take a series due to the series about zero, so I split it into

``

split := X^Q = V^P*(a[1]+a[2]*V+a[3]*V^2+O(V^3))

X^Q = V^P*(a[1]+a[2]*V+a[3]*V^2+O(V^3))

(5)

````

Then I'll get a polynomial expression, switch places of the two variables' leading terms, and substitute into the "original" polynomial (split).

``

````

  one := (V-1)^Q/(c^(P-Q)*(V-c^2)^Q)

(V-1)^Q/(c^(P-Q)*(V-c^2)^Q)

(6)

``

two := `assuming`([simplify(`assuming`([series(one, V, 3)], [V > c^2, c > 1]))], [real])

series(c^(-Q-P)-(c^(-Q-P-2)*Q*(c^2-1))*V+((1/2)*c^(-Q-P-4)*Q*(-c^4+c^4*Q+1+Q-2*Q*c^2))*V^2+O(V^3),V,3)

(7)

``

three := convert(two, polynom)

c^(-Q-P)-c^(-Q-P-2)*Q*(c^2-1)*V+(1/2)*c^(-Q-P-4)*Q*(-c^4+c^4*Q+1+Q-2*Q*c^2)*V^2

(8)

``

four := simplify(expand(V^P*three))

V^P*c^(-Q-P)-V^(P+1)*c^(-Q-P)*Q+V^(P+1)*c^(-Q-P-2)*Q-(1/2)*V^(P+2)*c^(-Q-P)*Q+(1/2)*V^(P+2)*c^(-Q-P)*Q^2+(1/2)*V^(P+2)*c^(-Q-P-4)*Q+(1/2)*V^(P+2)*c^(-Q-P-4)*Q^2-V^(P+2)*c^(-Q-P-2)*Q^2

(9)

``

five := collect(four, V, factor)

V^P*c^(-Q-P)-V^(P+1)*c^(-Q-P)*Q+V^(P+1)*c^(-Q-P-2)*Q-(1/2)*V^(P+2)*c^(-Q-P)*Q+(1/2)*V^(P+2)*c^(-Q-P)*Q^2+(1/2)*V^(P+2)*c^(-Q-P-4)*Q+(1/2)*V^(P+2)*c^(-Q-P-4)*Q^2-V^(P+2)*c^(-Q-P-2)*Q^2

(10)

collect(five, V)

V^P*c^(-Q-P)-V^(P+1)*c^(-Q-P)*Q+V^(P+1)*c^(-Q-P-2)*Q-(1/2)*V^(P+2)*c^(-Q-P)*Q+(1/2)*V^(P+2)*c^(-Q-P)*Q^2+(1/2)*V^(P+2)*c^(-Q-P-4)*Q+(1/2)*V^(P+2)*c^(-Q-P-4)*Q^2-V^(P+2)*c^(-Q-P-2)*Q^2

(11)

factor(five, V)

Error, (in factor) 2nd argument, V, is not a valid algebraic extension

 

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Download invert2.mw

Hello everyone,
Hello All, I read some other posts on "stack limits" and they were not too helpful. In a particular case there seemed to be a bug, as stated by a Maple kernel developer, but I can't imagine that being the problem here. I have a MacBook Pro (Intel) using Maple 11.02 with 2GB of RAM. The following code displays the stacklimit eror causing a failure with the kernel connection whenever I try to simplify the last line. Does anyone have any ideas? Thanks much, Steve. P.S. Here is the worksheet lines: restart one := 1/((x*Dd-x^2*Dcj)*sqrt(x^2-1))
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