Re: Can't reproduce a solution found in a paper using
- To: mathgroup at smc.vnet.net
- Subject: [mg102881] Re: [mg102845] Can't reproduce a solution found in a paper using
- From: "Elton Kurt TeKolste" <tekolste at fastmail.us>
- Date: Tue, 1 Sep 2009 03:52:38 -0400 (EDT)
- References: <200908301006.GAA20849@smc.vnet.net>
The solution in the book agrees with Mathematica's solution after you fix syntax, divide and conquer the problem and note that Cos[m Pi]^2 is always equal to one. What you had in your email J \!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(b\)]\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)] \(\*FractionBox[\(\[Rho]\ J\), \(2\ t\)]\)[x \((a - x)\) - FractionBox[8 \*SuperscriptBox[\(a\), \(2\)], SuperscriptBox[\[Pi], 3]] \( \*UnderoverscriptBox[\(\[Sum]\), \(m = 0\), \(\[Infinity]\)] \*SuperscriptBox[\((2 m + 1)\), \(-3\)]\ * \*FractionBox[\(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ y\), \(a\)]]\), \(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ b\), \(a\)]]\)]\ *\ Sin[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ x\), \(a\)]]\)] \[DifferentialD]x \[DifferentialD]y\ \)\) Incorrect syntax : Replace [ ...] with ( ...) J \!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(b\)]\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)] FractionBox[\(\[Rho]\ J\), \(2\ t\)] \((x \((a - x)\) - FractionBox[8 \*SuperscriptBox[\(a\), \(2\)], SuperscriptBox[\[Pi], 3]] \( \*UnderoverscriptBox[\(\[Sum]\), \(m = 0\), \(\[Infinity]\)] \*SuperscriptBox[\((2 m + 1)\), \(-3\)]\ * \*FractionBox[\(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ y\), \(a\)]]\), \(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ b\), \(a\)]]\)]\ *\ Sin[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ x\), \(a\)]]\))\) \[DifferentialD]x \ \[DifferentialD]y\)\) Still syntax errors : FractionBox and SuperscriptBox are not a fraction and an exponent. (An artifact of the copy to text.) J \!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(b\)]\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)] FractionBox[\(\[Rho]\ J\), \(2\ t\)] \((x \((a - x)\) - \((8 \*SuperscriptBox[\(a\), \(2\)]/\[Pi]^3)\) \( \*UnderoverscriptBox[\(\[Sum]\), \(m = 0\), \(\[Infinity]\)] \*SuperscriptBox[\((2 m + 1)\), \(-3\)]\ * \*FractionBox[\(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ y\), \(a\)]]\), \(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ b\), \(a\)]]\)]\ *\ Sin[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ x\), \(a\)]]\))\) \[DifferentialD]x \ \[DifferentialD]y\)\) Pull out the constant J ((\[Rho] J)/(2 t)) \!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(b\)]\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)]\((x \((a - x)\) - \((8 \*SuperscriptBox[\(a\), \(2\)]/\[Pi]^3)\) \( \*UnderoverscriptBox[\(\[Sum]\), \(m = 0\), \(\[Infinity]\)] \*SuperscriptBox[\((2 m + 1)\), \(-3\)]\ * \*FractionBox[\(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ y\), \(a\)]]\), \(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ b\), \(a\)]]\)]\ *\ Sin[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ x\), \(a\)]]\))\) \[DifferentialD]x \ \[DifferentialD]y\)\) Distribute the integration over the sums (you probably need to verify some condition like absolute convergence to pull the integral into the infinte sum, but I am doing quick and dirty math here). J ((\[Rho] J)/(2 t)) (\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(b\)]\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)]x \((a - x)\) \[DifferentialD]x \[DifferentialD]y\)\) - (8 a^2/\[Pi]^3) \!\( \*UnderoverscriptBox[\(\[Sum]\), \(m = 0\), \(\[Infinity]\)]\( \*SuperscriptBox[\((2 m + 1)\), \(-3\)]\ *\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(b\)]\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)] \*FractionBox[\(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ y\), \(a\)]]\), \(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ b\), \(a\)]]\)]\ *\ Sin[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ x\), \(a\)]] \[DifferentialD]x \ \[DifferentialD]y\)\)\)\)) Separate the summands and isolate the constants. J ((\[Rho] J)/(2 t)) (\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(b\)]\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)]x \((a - x)\) \[DifferentialD]x \[DifferentialD]y\)\)) - J ((\[Rho] J)/(2 t)) (8 a^2/\[Pi]^3) (\!\( \*UnderoverscriptBox[\(\[Sum]\), \(m = 0\), \(\[Infinity]\)]\( \*SuperscriptBox[\((2 m + 1)\), \(-3\)]\ *\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(b\)]\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)] \*FractionBox[\(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ y\), \(a\)]]\), \(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ b\), \(a\)]]\)]\ *\ Sin[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ x\), \(a\)]] \[DifferentialD]x \[DifferentialD]y\)\)\ \)\)) In[20]:= Simplify[{J ((\[Rho] J)/(2 t)), J ((\[Rho] J)/(2 t)) (8 a^2/\[Pi]^3)}] Out[20]= {(J^2 \[Rho])/(2 t), (4 a^2 J^2 \[Rho])/(\[Pi]^3 t)} (J^2 \[Rho])/(2 t) (\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(b\)]\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)]x \((a - x)\) \[DifferentialD]x \[DifferentialD]y\)\)) - ( 4 a^2 J^2 \[Rho])/(\[Pi]^3 t) (\!\( \*UnderoverscriptBox[\(\[Sum]\), \(m = 0\), \(\[Infinity]\)]\( \*SuperscriptBox[\((2 m + 1)\), \(-3\)]\ *\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(b\)]\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)] \*FractionBox[\(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ y\), \(a\)]]\), \(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ b\), \(a\)]]\)]\ *\ Sin[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ x\), \(a\)]] \[DifferentialD]x \[DifferentialD]y\)\)\ \)\)) One integral is easy In[21]:= \!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(b\)]\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)]x \((a - x)\) \[DifferentialD]x \[DifferentialD]y\)\) Out[21]= (a^3 b)/6 (J^2 \[Rho])/(2 t) ((a^3 b)/6) - (4 a^2 J^2 \[Rho])/(\[Pi]^3 t) (\!\( \*UnderoverscriptBox[\(\[Sum]\), \(m = 0\), \(\[Infinity]\)]\( \*SuperscriptBox[\((2 m + 1)\), \(-3\)]\ *\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(b\)]\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)] \*FractionBox[\(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ y\), \(a\)]]\), \(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ b\), \(a\)]]\)]\ *\ Sin[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ x\), \(a\)]] \[DifferentialD]x \[DifferentialD]y\)\)\ \)\)) The other solves easily but does not agree with your book answer In[22]:= \!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(b\)]\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)] \*FractionBox[\(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ y\), \(a\)]]\), \(Cosh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ b\), \(a\)]]\)]\ *\ Sin[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ x\), \(a\)]] \[DifferentialD]x \[DifferentialD]y\)\) Out[22]= (2 a^2 Cos[m \[Pi]]^2 Tanh[(b (1 + 2 m) \[Pi])/a])/(\[Pi] + 2 m \[Pi])^2 (J^2 \[Rho])/(2 t) ((a^3 b)/6) - (4 a^2 J^2 \[Rho])/(\[Pi]^3 t) (\!\( \*UnderoverscriptBox[\(\[Sum]\), \(m = 0\), \(\[Infinity]\)]\( \*SuperscriptBox[\((2 m + 1)\), \(-3\)]\ * \*FractionBox[\(2\ \*SuperscriptBox[\(a\), \(2\)]\ \*SuperscriptBox[\(Cos[m\ \[Pi]]\), \(2\)]\ Tanh[ \*FractionBox[\(b\ \((1 + 2\ m)\)\ \[Pi]\), \(a\)]]\), SuperscriptBox[\((\[Pi] + 2\ m\ \[Pi])\), \(2\)]]\)\)) A bit of prettification yields In[26]:= {(J^2 \[Rho])/(2 t) ((a^3 b)/6), ( 4 a^2 J^2 \[Rho])/(\[Pi]^3 t \[Pi]^2)*2 a^2 , (2 m + 1)^-3 (2 m + 1)^-2} Out[26]= {(a^3 b J^2 \[Rho])/(12 t), (8 a^4 J^2 \[Rho])/(\[Pi]^5 t), 1/(1 + 2 m)^5} (a^3 b J^2 \[Rho])/(12 t) - (8 a^4 J^2 \[Rho])/(\[Pi]^5 t) (\!\( \*UnderoverscriptBox[\(\[Sum]\), \(m = 0\), \(\[Infinity]\)]\( \*SuperscriptBox[\((2 m + 1)\), \(-5\)]\ * \*SuperscriptBox[\(Cos[m\ \[Pi]]\), \(2\)]\ Tanh[ \*FractionBox[\(b\ \((1 + 2\ m)\)\ \[Pi]\), \(a\)]]\)\)) Which is almost but not quite the answer that you wanted, since there is an extra Cos[m \[Pi]]^2 in the infinite sum. (\[Rho] J^2)/(2 t)*[(a^3 b)/6 - (16 a^4)/\[Pi]^5 \!\( \*UnderoverscriptBox[\(\[Sum]\), \(m = 0\), \(\[Infinity]\)]\( \*SuperscriptBox[\((2 m + 1)\), \(-5\)]\ Tanh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ b\), \(a\)]]\)\)] We fix the syntax (\[Rho] J^2)/(2 t)*((a^3 b)/6 - (16 a^4)/\[Pi]^5 \!\( \*UnderoverscriptBox[\(\[Sum]\), \(m = 0\), \(\[Infinity]\)]\( \*SuperscriptBox[\((2 m + 1)\), \(-5\)]\ Tanh[ \*FractionBox[\(\((2 m + 1)\) \[Pi]\ b\), \(a\)]]\)\)) However, we note that Cos[0]^2, Cos[\[Pi]]^2, ... are all equal to one, so the book solution is correct since In[39]:= {(a^3 b J^2 \[Rho])/(12 t) == (\[Rho] J^2)/(2 t)*(a^3 b)/6 , ( 8 a^4 J^2 \[Rho])/(\[Pi]^5 t) == (\[Rho] J^2)/(2 t)*(16 a^4)/\[Pi]^5 } Out[39]= {True, True} On Sun, 30 Aug 2009 06:06 -0400, "Neelsonn" <neelsonn at gmail.com> wrote: > Guys, > > This is what I want to solve: > > J \!\( > \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(b\)]\( > \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)] > \(\*FractionBox[\(\[Rho]\ J\), \(2\ t\)]\)[x \((a - x)\) - > FractionBox[\(8 > \*SuperscriptBox[\(a\), \(2\)]\), > SuperscriptBox[\(\[Pi]\), \(3\)]] \( > \*UnderoverscriptBox[\(\[Sum]\), \(m = 0\), \(\[Infinity]\)] > \*SuperscriptBox[\((2 m + 1)\), \(-3\)]\ *\ > \*FractionBox[\(Cosh[ > \*FractionBox[\(\((2 m + 1)\) \[Pi]\ y\), \(a\)]]\), \(Cosh[ > \*FractionBox[\(\((2 m + 1)\) \[Pi]\ b\), \(a\)]]\)]\ *\ Sin[ > \*FractionBox[\(\((2 m + > 1)\) \[Pi]\ x\), \(a\)]]\)] \[DifferentialD]x \ > \[DifferentialD]y\)\) > > > ...and this is the solution that I found in a publication: > > (\[Rho] J^2)/(2 t)*[(a^3 b)/6 - (16 a^4)/\[Pi]^5 \!\( > \*UnderoverscriptBox[\(\[Sum]\), \(m = 0\), \(\[Infinity]\)]\( > SuperscriptBox[\((2 m + 1)\), \(-5\)] Tanh[ > \*FractionBox[\(\((2 m + 1)\) \[Pi]\ b\), \(a\)]]\)\)] > > I am simply not able to reproduce that with Mathematica. The obvious > questions is: why? If someone would be willing to have a look at the > the paper I could sent it over. I may say that the paper dated back > from the 70's and at that time Mathematica wasn't available (people > were smart at that time!!!! lol) > > Thanks again, > N > Regards, Kurt Tekolste