Ronan

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14 years, 31 days
East Grinstead, United Kingdom

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These are questions asked by Ronan

In the attached, I had to find the required coefficient. Now this has a relatively straight forward set of equation (which apparently form a closed group), so it was easy enough tee see what to do. In a more complicated situation that may be very difficult to see. I was a wondering is there is a better approach to use here?
 

restart

Prove/show the coefficient of c[0] in c[1]^11 is 1/12*(1-1/5^10)NULL

eq1 := c[1]^2 = (1/5)*c[0]+(2/5)*c[1]+(2/5)*c[2]

c[1]^2 = (1/5)*c[0]+(2/5)*c[1]+(2/5)*c[2]

(1)

eq2 := c[1]*c[2] = (2/5)*c[1]+(2/5)*c[2]+(1/5)*c[3]

c[1]*c[2] = (2/5)*c[1]+(2/5)*c[2]+(1/5)*c[3]

(2)

eq3 := c[1]*c[3] = c[2]

c[1]*c[3] = c[2]

(3)

eq4 := c[2]*c[3] = c[1]

c[2]*c[3] = c[1]

(4)

eq5 := c[2]^2 = (1/5)*c[0]+(2/5)*c[1]+(2/5)*c[2]

c[2]^2 = (1/5)*c[0]+(2/5)*c[1]+(2/5)*c[2]

(5)

eq6 := c[3]^2 = c[0]

c[3]^2 = c[0]

(6)

 

expand(eq1^5)

c[1]^10 = (1/3125)*c[0]^5+(32/3125)*c[1]^5+(32/3125)*c[2]^5+(16/625)*c[0]*c[1]^4+(16/625)*c[0]*c[2]^4+(2/625)*c[0]^4*c[1]+(2/625)*c[0]^4*c[2]+(8/625)*c[0]^3*c[1]^2+(8/625)*c[0]^3*c[2]^2+(16/625)*c[0]^2*c[1]^3+(16/625)*c[0]^2*c[2]^3+(32/625)*c[1]*c[2]^4+(32/625)*c[1]^4*c[2]+(64/625)*c[1]^3*c[2]^2+(64/625)*c[1]^2*c[2]^3+(64/625)*c[0]*c[1]^3*c[2]+(96/625)*c[0]*c[1]^2*c[2]^2+(64/625)*c[0]*c[1]*c[2]^3+(16/625)*c[0]^3*c[1]*c[2]+(48/625)*c[0]^2*c[1]^2*c[2]+(48/625)*c[0]^2*c[1]*c[2]^2

(7)

C111 := lhs(%)*c[1] = expand(rhs(%)*c[2]*c[3])

c[1]^11 = (1/3125)*c[2]*c[3]*c[0]^5+(32/3125)*c[2]*c[3]*c[1]^5+(32/3125)*c[2]^6*c[3]+(16/625)*c[2]*c[3]*c[0]*c[1]^4+(16/625)*c[2]^5*c[3]*c[0]+(2/625)*c[2]*c[3]*c[0]^4*c[1]+(2/625)*c[2]^2*c[3]*c[0]^4+(8/625)*c[2]*c[3]*c[0]^3*c[1]^2+(8/625)*c[2]^3*c[3]*c[0]^3+(16/625)*c[2]*c[3]*c[0]^2*c[1]^3+(16/625)*c[2]^4*c[3]*c[0]^2+(32/625)*c[2]^5*c[3]*c[1]+(32/625)*c[2]^2*c[3]*c[1]^4+(64/625)*c[2]^3*c[3]*c[1]^3+(64/625)*c[2]^4*c[3]*c[1]^2+(64/625)*c[2]^2*c[3]*c[0]*c[1]^3+(96/625)*c[2]^3*c[3]*c[0]*c[1]^2+(64/625)*c[2]^4*c[3]*c[0]*c[1]+(16/625)*c[2]^2*c[3]*c[0]^3*c[1]+(48/625)*c[2]^2*c[3]*c[0]^2*c[1]^2+(48/625)*c[2]^3*c[3]*c[0]^2*c[1]

(8)

c[1]^11 = simplify(rhs(C111), [eq2, eq3, eq4, eq5, eq6])

c[1]^11 = (1/9765625)*(4070573*c[3]+4067448)*c[2]+(813802/9765625)*c[0]+(813802/9765625)*c[3]

(9)

coeff(rhs(%), c[0], 1)

813802/9765625

(10)

eval(1/12*(1-1/5^10))

813802/9765625

(11)

NULL


 

Download Graphs_steps.mw

Recently I have wanted to add some questions/answers to "Favourites" . I added some posts a few years ago. Now I can't figure out how do do it. A good example is this evenings question on setting up a library. How to install a library on Maple? - MaplePrimes

The answer has useful links. And this is a particular topic I tend to have difficulty with.

This is a just a general type question. Often reading questions posted to do with modules and packages I see "foo" used. I have put together some simple packages for routines I find useful. I am week in general programming knowledge. I would like to know is "foo" something special or a generic name used for testing? Is is an acronym? ...

I have a procedure where M:=[m2,m3,m4...] and I  assign the m's values from list L:=[0.1,2....]
The procedure will only run once. Rerunning will not reassign the m's.

What is the correct way to do this?

Assign_in_proc.mw

 

I have a polynomial in c & d.  I wish to extract the coefficients of lets say 3rd order i.e c^3+c^2d+c d^2+d^3. 

The polynomials can be of 2,3,4.. variables.  

coeffs(14*c^4 + 84*c^3*d + 180*c^2*d^2 + 165*c*d^3 + 55*d^4 + 5*c^3 + 21*c^2*d + 28*c*d^2 + 12*d^3 + 2*c^2 + 5*c*d + 3*d^2 + c + d + 1, [c, d], 'l');
     14, 84, 180, 165, 55, 5, 21, 28, 12, 2, 5, 3, 1, 1, 1

l[1];
                                4
                               c 

l[2];
                               3  
                              c  d

l;
 4   3     2  2     3   4   3     2     2   3   2        2        
c , c  d, c  d , c d , d , c , d c , c d , d , c , d c, d , c, d, 

  1

 

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