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