Re: Quadratic form: symbolic transformation

*To*: mathgroup at smc.vnet.net*Subject*: [mg76849] Re: Quadratic form: symbolic transformation*From*: dimitris <dimmechan at yahoo.com>*Date*: Mon, 28 May 2007 01:10:40 -0400 (EDT)*References*: <f3bj37$40i$1@smc.vnet.net>

Hello Wolfgang. It is very easy with the following code to make your procedure much more compact. If you are working in Mathematica 5.2 the following will take you in a section of Mathematica Book that contains material to understand the code. FrontEndExecute[{HelpBrowserLookup["MainBook", "3.6.1"]}] If you are working in 6 see here http://reference.wolfram.com/mathematica/tutorial/SolvingEquationsInvolvingPowerSeries.html Here are your polynomials in x. In[2]:= q1 = R*x^2 + S*x + T ; In[3]:= q2 = u*(x+v)^2 + w ; Here are the procedure (*with comments*) In[11]:= q1 == q2 + O[x]^3 (*this gives an equation involving the power series*) LogicalExpand[%] (*LogicalExpand generates a sequence of equations for each power of x*) Solve[%, {u, v, w}] (*now the rest is obvious!*) q1 == q2 /. %[[1]] (*verification*) Expand /@ % (*no need for Simplify here*) Out[11]= T + S*x + R*x^2 == SeriesData[Global`x, 0, {Global`u*Global`v^2 + Global`w, 2*Global`u*Global`v, Global`u}, 0, 3, 1] Out[12]= -R + u == 0 && -S + 2*u*v == 0 && -T + u*v^2 + w == 0 Out[13]= {{w -> (-S^2 + 4*R*T)/(4*R), u -> R, v -> S/(2*R)}} Out[14]= T + S*x + R*x^2 == (-S^2 + 4*R*T)/(4*R) + R*(S/(2*R) + x)^2 Out[15]= True True? We are indeed right! Let's make now an one-liner that will collect the steps (apart from the verification!). Note that this one liner is not something special. It is a quick attempt to show how you could work! Various modifications can be done to be better. The code looks big because I have add some coments to explain how it works. Please feel free to make me any questions. In[16]:= quadform[f_(*put here the expanded form*), g_(*put here the quadratic form*), x_(*the variable please*)] := If[PolynomialQ[f, x] && PolynomialQ[g, x] && (Exponent[#1, x] & ) /@ {f, g} == {2, 2}(*check if the expressions are polynomials in x and if their order is two*), Solve[LogicalExpand[f == g + O[x]^3], Variables[CoefficientList[g, x]](*this gives a list of the parameters which you want to express in known parameters*) ], Print["sorry; try again! I am a quick attempt not something special!"]] Let's taste it! In[17]:= quadform[q1, q2, x] Out[17]= {{w -> (-S^2 + 4*R*T)/(4*R), u -> R, v -> S/(2*R)}} Good! In[27]:= quadform[x^2 + 2*x + 4, q2, x] Out[27]= {{w -> 3, u -> 1, v -> 1}} Better! In[28]:= quadform[x^3 + a*x^2 + b*x + c, q2, x] >From In[28]:= "sorry; try again! I am a quick attempt not something special!" The best! My code knows its level! Regards Dimitris Anagnostou / Dr. Wolfgang Hintze : > Hello, > > this is a simple question but perhaps I can get here some information > towards a more apropriate way of using Mathematica. > > I take a very simple example: I would like to write the quadratic form > > q1 = R*x^2 + R*x + T > > in the form > > q2 = u*(x+v)^2 + w > > How can I find u, v, and w from R, S, and T? > > I'm sure there must be some symbolic way (using a sufficient amount of > _'s) to answer this question. > > My (cumbersome) procedure compares coefficients and looks like this > > (* writing down lhs == rhs) > In[112]:= > q = R*x^2 + S*x + T == u*(x + v)^2 + w > Out[112]= > T + S*x + R*x^2 == w + u*(v + x)^2 > > (* as q must be an identiy in x, i.e. must hold for all x, I compare > coefficients at x=0 *) > In[113]:= > eq1 = q /. {x -> 0} > Out[113]= > T == u*v^2 + w > In[114]:= > eq2 = D[q, x] /. {x -> 0} > Out[114]= > S == 2*u*v > In[115]:= > eq3 = D[q, {x, 2}] /. {x -> 0} > Out[115]= > 2*R == 2*u > In[119]:= > t = First[Solve[{eq1, eq2, eq3}, {u, v, w}]] > Out[119]= > {w -> (-S^2 + 4*R*T)/(4*R), u -> R, v -> S/(2*R)} > > (* writing down the result explicitly *) > In[120]:= > q /. t > Out[120]= > T + S*x + R*x^2 == (-S^2 + 4*R*T)/(4*R) + R*(S/(2*R) + x)^2 > In[122]:= > Simplify[q /. t] > Out[122]= > True > > Thanks in advance for any hints. > Regards, > Wolfgang