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Re: DSolve Issues
*To*: mathgroup at smc.vnet.net
*Subject*: [mg88356] Re: DSolve Issues
*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
*Date*: Sat, 3 May 2008 06:13:47 -0400 (EDT)
*Organization*: The Open University, Milton Keynes, UK
*References*: <fveguc$5ed$1@smc.vnet.net>
donkorgi12 wrote:
> I am solving the following Differential Equation
>
> Phi''[r]+2*r^(-1)Phi'[r]+0.104479*Phi[r]==0 and Phi[0]==K (some
> constant) ; kinda has a cos/sin solution
>
> 2.71828^(-0.323232 \[ImaginaryI] r) ((0.+
> 0. \[ImaginaryI]) + (0.+ 0. \[ImaginaryI]) 2.71828^(
> 0.646465 \[ImaginaryI] r) + (0.+ 1.54687 \[ImaginaryI]) K - (0.+
> 1.54687 \[ImaginaryI]) 2.71828^(0.646465 \[ImaginaryI] r) K)
>
> all divided by r.
>
> My problem is that Mathematica is not treating those "zeros".... as
> well zeros. Thus, the solution cannot really be used. In fact, if I
> manually reproduce the solution and remove those "zeros", then the
> solution is fine.
Say your solution is called sol. You can get ride of the spurious zeros
yield by numerical impression thanks to the command *Chop[]*. For instance,
In[1]:= sol = 2.71828^(-0.323232 I r) ((0. +
0. I) + (0. + 0. I) 2.71828^(0.646465 I r) + (0. +
1.54687 I) K - (0. + 1.54687 I) 2.71828^(0.646465 I r) K) // Chop
Out[1]=
0.646465 I r
1.54687 I K - 1.54687 I 2.71828 K
---------------------------------------------
0.323232 I r
2.71828
It would be better to use exact arithmetic as early as possible, either
by entering exact numbers (e.g. 1.0 is not the same thing as 1 from a
computational point of view since the underlaying computer
representation is not the same at all) or using the built-in function
*Rationalize[]*.
Moreover, one can convert exponential function thanks to the function
*ExpToTrig[]*. For instance,
In[2]:= ode = Rationalize[Phi''[r] + 2*r^(-1) Phi'[r] + 0.104479*Phi[r], 0];
sol = DSolve[{ode == 0, Phi[0] == K}, Phi, r]
Phi[r] /. sol[[1]] // ExpToTrig
Out[3]=
1/500 I Sqrt[104479] r
500 I (-1 + E ) K
{{Phi -> Function[{r}, -(-----------------------------------------)]}}
(I Sqrt[104479] r)/1000
E (Sqrt[104479] r)
Out[4]=
1 Sqrt[104479] r Sqrt[104479] r
-(-------------- (500 I K (Cos[--------------] - I Sin[--------------])
Sqrt[104479] r 1000 1000
Sqrt[104479] r Sqrt[104479] r
(-1 + Cos[--------------] + I Sin[--------------])))
500 500
[snip]
> Yet, Mathematica treats those "zeros" as something else.
No. It treats them as *inexact* numbers, which is fine since you are
using inexact (aka *machine-size*) arithmetic.
The following multimedia demo illustrates strikingly the difference
between floating-point arithmetic and arbitrary/exact arithmetic.
http://www.wolfram.com/technology/guide/precisiontracking.html
Best regards,
-- Jean-Marc
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