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Mathematica Algebra

Interpreting Output

Jamie M

Posted 12 years ago

Hello, I've used the Mathematica 10.1 for the first time today. My initial reasoning was to solve a polynomial with 5 roots. My issue is I don't know how to apply the output with a '#1' symbol. How would I visualize this as standard form? *IN:* Solve[((A (120 x^5 - 300 x^4 + 200 x^3) - 20 B) U)/( F^5 X^5 (1 - X)^5) + ((C (120 x^5 - 300 x^4 + 200 x^3) - 20 D) V)/( F^5 X^5 (1 - X)^5) == 0, x] *OUT:* {{x -> Root[-B U - D V + (10 A U + 10 C V) #1^3 + (-15 A U - 15 C V) #1^4 + (6 A U + 6 C V) #1^5 &, 1]}, {x -> Root[-B U - D V + (10 A U + 10 C V) #1^3 + (-15 A U - 15 C V) #1^4 + (6 A U + 6 C V) #1^5 &, 2]}, {x -> Root[-B U - D V + (10 A U + 10 C V) #1^3 + (-15 A U - 15 C V) #1^4 + (6 A U + 6 C V) #1^5 &, 3]}, {x -> Root[-B U - D V + (10 A U + 10 C V) #1^3 + (-15 A U - 15 C V) #1^4 + (6 A U + 6 C V) #1^5 &, 4]}, {x -> Root[-B U - D V + (10 A U + 10 C V) #1^3 + (-15 A U - 15 C V) #1^4 + (6 A U + 6 C V) #1^5 &, 5]}} I would like to select root 1: X = -B U - D V + (10 A U + 10 C V) #1^3 + (-15 A U - 15 C V) #1^4 + (6 A U + 6 C V) #1^5 Provided the unknowns { B, U, D, V, A, C } Thank you for your time.

Hello,

I've used the Mathematica 10.1 for the first time today. My initial reasoning was to solve a polynomial with 5 roots. My issue is I don't know how to apply the output with a '#1' symbol. How would I visualize this as standard form?

IN:

Solve[((A (120 x^5 - 300 x^4 + 200 x^3) - 20 B) U)/(
   F^5 X^5 (1 - X)^5) + ((C (120 x^5 - 300 x^4 + 200 x^3) - 20 D) V)/(
   F^5 X^5 (1 - X)^5) == 0, x]

OUT:

{{x -> Root[-B U - 
      D V + (10 A U + 10 C V) #1^3 + (-15 A U - 
         15 C V) #1^4 + (6 A U + 6 C V) #1^5 &, 1]}, {x -> 
   Root[-B U - 
      D V + (10 A U + 10 C V) #1^3 + (-15 A U - 
         15 C V) #1^4 + (6 A U + 6 C V) #1^5 &, 2]}, {x -> 
   Root[-B U - 
      D V + (10 A U + 10 C V) #1^3 + (-15 A U - 
         15 C V) #1^4 + (6 A U + 6 C V) #1^5 &, 3]}, {x -> 
   Root[-B U - 
      D V + (10 A U + 10 C V) #1^3 + (-15 A U - 
         15 C V) #1^4 + (6 A U + 6 C V) #1^5 &, 4]}, {x -> 
   Root[-B U - 
      D V + (10 A U + 10 C V) #1^3 + (-15 A U - 
         15 C V) #1^4 + (6 A U + 6 C V) #1^5 &, 5]}}

I would like to select root 1:

   X = -B U - D V + (10 A U + 10 C V) #1^3 + (-15 A U - 15 C V) #1^4 + (6 A U + 6 C V) #1^5

Provided the unknowns { B, U, D, V, A, C }

Thank you for your time.

POSTED BY: Jamie M

2 Replies

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Philip Flanner

Philip Flanner, Independent Consultant

Posted 12 years ago

Root 1 is given by applying the first rule in the output: In[1]:= x /. Solve[((A (120 x^5 - 300 x^4 + 200 x^3) - 20 B) U)/(F^5 X^5 (1 - X)^5) + ((C (120 x^5 - 300 x^4 + 200 x^3) - 20 D) V)/(F^5 X^5 (1 - X)^5) == 0, x][[1]] Out[1]= Root[-B U - D V + (10 A U + 10 C V) #1^3 + (-15 A U - 15 C V) #1^4 + (6 A U + 6 C V) #1^5 &, 1] Solve[] gives roots in the form of `Root[function,rootNumber]` when it cannot give them in symbolic form. Basically it has just reduced the function as much as possible and left it as "root of this simpler function". Quintic polynomials are not generally solvable in terms of radicals. You can evaluate the root numerically if you provide numeric values for the undefined variables: In[2]:= N[% /. {B -> 1, U -> 2, D -> 3, V -> 4, A -> 5, C -> 6}] Out[2]= 0.452659

POSTED BY: Philip Flanner

Udo Krause

Udo Krause, NOrganization

Posted 12 years ago

In[73]:= X = Solve[((A (120 x^5 - 300 x^4 + 200 x^3) - 20 B) U)/(F^5 X^5 (1 - X)^5) + ((C (120 x^5 - 300 x^4 + 200 x^3) - 20 D) V)/(F^5 X^5 (1 - X)^5) == 0, x][[1, 1, 2,1]]; In[74]:= X Out[74]= -B U - D V + (10 A U + 10 C V) #1^3 + (-15 A U - 15 C V) #1^4 + (6 A U + 6 C V) #1^5 & as such not yet really useful, but navigated to the first root.

 In[73]:= X =  Solve[((A (120 x^5 - 300 x^4 + 200 x^3) - 
                   20 B) U)/(F^5 X^5 (1 - 
                    X)^5) + ((C (120 x^5 - 300 x^4 + 200 x^3) - 
                   20 D) V)/(F^5 X^5 (1 - X)^5) == 0, x][[1, 1, 2,1]];

 In[74]:= X
 Out[74]= -B U - D V + (10 A U + 10 C V) #1^3 + (-15 A U - 15 C V) #1^4 + (6 A U + 6 C V) #1^5 &

as such not yet really useful, but navigated to the first root.

POSTED BY: Udo Krause

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