Re: How to calculate the partial derivative?

*To*: mathgroup at smc.vnet.net*Subject*: [mg128746] Re: How to calculate the partial derivative?*From*: Tang Laoya <tanglaoya1989 at gmail.com>*Date*: Fri, 23 Nov 2012 03:26:13 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net*References*: <k8jss7$5nl$1@smc.vnet.net> <k8ks07$858$1@smc.vnet.net>

Dear Dr. Hanlon, Thank you very much for your kindly reply. Your method works for me, that's great. Now I have another question: I wish Mathematica give me an express which I can code by Fortran with smallest number of floating operations. For example, if I have: bb1 = -Det[{{1, 1, 1}, {y2, y3, y4}, {z2, z3, z4}}]; the result by Mathematica is: y3 z2 - y4 z2 - y2 z3 + y4 z3 + y2 z4 - y3 z4 I wish it output the result like this: z2 (y3-y4) - z3 (y2-y4) + z4 (y2-y3) What should I do? Thanks, Tang Laoya On Thursday, November 22, 2012 5:43:45 PM UTC+8, Tang Laoya wrote: > Dear all, > > > > Sorry for my unsuitable example. My final program is as follows, could anyone please help me to take a look at it? > > > > > > Thanks, > > Tang Laoya > > > > > > Clear["Global`*"]; > > L1 = (a1 + b1 x + c1 y + d1 z)/6/V; > > L2 = (a2 + b2 x + c2 y + d2 z)/6/V; > > L3 = (a3 + b3 x + c3 y + d3 z)/6/V; > > L4 = (a4 + b4 x + c4 y + d4 z)/6/V; > > > > nod = 10; > > NN = Table[0, {nod}]; > > NN[[1]] = L1 (2 L1 - 1); > > NN[[2]] = L2 (2 L2 - 1); > > NN[[3]] = L3 (2 L3 - 1); > > NN[[4]] = L4 (2 L4 - 1); > > NN[[5]] = 4 L1 L2; > > NN[[6]] = 4 L2 L3; > > NN[[7]] = 4 L3 L1; > > NN[[8]] = 4 L3 L4; > > NN[[9]] = 4 L4 L1; > > NN[[10]] = 4 L2 L4; > > > > NNT = Flatten[Table[NN[[i]], {i, 1, nod}]]; > > Print["Dimensions of NNT:", Dimensions[NNT]]; > > > > DNx = D[NNT, x] > > (* How to do to let the DNx have such express? For example, DNx[[1]] = b1/6/V*(2*L1-1) + L1*b1/3/V *) > > > > DNy = D[NNT, y] > > DNz = D[NNT, z] > > DNxx = Transpose[{DNx}].{DNx}; > > DNyy = Transpose[{DNy}].{DNy}; > > DNzz = Transpose[{DNz}].{DNz}; > > Print["Dimensions of DNxx:", Dimensions[DNxx]]; > > > > K = DNxx + DNyy + DNzz; > > > > (* I wish to replace all express "L1^aa*L2^bb*L3^cc*L4^dd" in matrix K by "aa!*bb!*cc!dd!/((aa+bb+cc+dd+3)!)*6*V", how to do? in the express, the aa, bb, cc or dd could be ZERO *) > > > > K = Expand[K]; > > Print["Simplifying K"]; > > K = Simplify[K]