Group Abstract

Message Boards

WOLFRAM COMMUNITY

34 Views

2 Replies

3 Total Likes

View groups...

Follow this post

Share this post:

GROUPS:

Algebra

Are there any other good methods for performing cubic completion?

Bill Blair

Posted 1 day ago

2 x^3 - 3 a x^2 + 1 == 1 - a^3/2 - 3/2 a^2 (-(a/2) + x) + 2 (-(a/2) + x)^3 Write a Mathematica function that performs cubic completion. For a cubic polynomial in x, such as the equation above, the function should be able to complete the cube to achieve the form shown on the right-hand side. Here is one way to solve it—are there any better approaches? AutoCompleteCube[poly_, x_] := Module[{a3, a2, x0, refinedPoly}, a3 = Coefficient[poly, x, 3]; a2 = Coefficient[poly, x, 2]; x0 = -a2/(3a3); refinedPoly = Total@Table[(Derivative[i][Function[x, poly]][x0]/i!)(x - x0)^ i, {i, 0, 3}]; refinedPoly] AutoCompleteCube[2 x^3 - 3 a x^2 + 1, x]

2 x^3 - 3 a x^2 + 1 ==     
1 - a^3/2 - 3/2 a^2 (-(a/2) + x) + 2 (-(a/2) + x)^3

Write a Mathematica function that performs cubic completion. For a cubic polynomial in x, such as the equation above, the function should be able to complete the cube to achieve the form shown on the right-hand side.

Here is one way to solve it—are there any better approaches?

AutoCompleteCube[poly_, x_] := 
 Module[{a3, a2, x0, refinedPoly}, a3 = Coefficient[poly, x, 3];
  a2 = Coefficient[poly, x, 2];
  x0 = -a2/(3*a3);
  refinedPoly = 
   Total@Table[(Derivative[i][Function[x, poly]][x0]/i!)*(x - x0)^
       i, {i, 0, 3}];
  refinedPoly]
AutoCompleteCube[2 x^3 - 3 a x^2 + 1, x]

POSTED BY: Bill Blair

2 Replies

Sort By:

Michael Rogers

Michael Rogers, Emory University

Posted 8 hours ago

Another way: With[{x0 = First@SolveValues[D[2 x^3 - 3 a x^2 + 1, {x, 2}] == 0, x]}, Expand[2 x^3 - 3 a x^2 + 1 /. x -> (u + x0)] /. u -> (x - x0) ] (* 1 - a^3/2 - 3/2 a^2 (-(a/2) + x) + 2 (-(a/2) + x)^3 *) Or: With[{x0 = First@SolveValues[D[2 x^3 - 3 a x^2 + 1, {x, 2}] == 0, x]}, Normal@Series[2 x^3 - 3 a x^2 + 1, {x, x0, 3}] ]

Another way:

With[{x0 = First@SolveValues[D[2 x^3 - 3 a x^2 + 1, {x, 2}] == 0, x]},
 Expand[2 x^3 - 3 a x^2 + 1 /. x -> (u + x0)] /. u -> (x - x0)
 ]
(*  1 - a^3/2 - 3/2 a^2 (-(a/2) + x) + 2 (-(a/2) + x)^3  *)

Or:

With[{x0 = First@SolveValues[D[2 x^3 - 3 a x^2 + 1, {x, 2}] == 0, x]},
 Normal@Series[2 x^3 - 3 a x^2 + 1, {x, x0, 3}]
 ]

POSTED BY: Michael Rogers

Michael Rogers

Michael Rogers, Emory University

Posted 23 hours ago

form = AA (x - x0)^3 + BB (x - x0) + CC; form /. First@Solve[ CoefficientList[2 x^3 - 3 a x^2 + 1, x] == CoefficientList[form, x] , {AA, BB, CC, x0} ] (* 1/2 (2 - a^3) - 3/2 a^2 (-(a/2) + x) + 2 (-(a/2) + x)^3 *)

form = AA (x - x0)^3 + BB (x - x0) + CC;
form /. First@Solve[
   CoefficientList[2 x^3 - 3 a x^2 + 1, x] == CoefficientList[form, x]
   , {AA, BB, CC, x0}
   ]

(*  1/2 (2 - a^3) - 3/2 a^2 (-(a/2) + x) + 2 (-(a/2) + x)^3  *)

POSTED BY: Michael Rogers

Reply to this discussion

Reply Preview

Attachments

Remove Add a file to this post

Follow this discussion

or Discard

Feedback