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Re: Mathematica can't win against Tiger Woods

  • To: mathgroup at smc.vnet.net
  • Subject: [mg19765] Re: [mg19677] Mathematica can't win against Tiger Woods
  • From: Leszek Sczaniecki <leszek2 at home.com>
  • Date: Wed, 15 Sep 1999 03:53:11 -0400
  • Organization: @Home Network
  • References: <7red0n$474@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com


To many times I tried to replicate simple computations done by hand with
Mathematica and was not able to get results that would justify the use of an
expensive computer algebra system. Therefore, I understand very well the
frustration of Prof. MacDonald. Here is a recent example.

Consider a very simple problem from analytical geometry. There an ellipse
with semi-latus rectum p and eccentricity e. You want to find semi-major
axis a, and semi-minor axis b. This is clearly a high school problem.

First, let's do it manually. We have p > 0, 0 < e < 1, a > b > 0.
  p = b^2/a
  e = Sqrt[a^2 - b^2]/a}
from 0.

e = Sqrt[a^2 - b^2]/a}
   => e^2 = (a^2 - b^2)/a^2
   => b^2/a^2 = 1-e^2
p = b^2/a = (b^2/a^2) a = (1-e^2)a
   => a = p/(1-e^2)
p = b^2/a
   => b^2 = p a = p^2 /(1-e^2)
   => b = p /Sqrt[1-e^2] or b = -p /Sqrt[1-e^2]
Because b > 0, b = p /Sqrt[1-e^2].

Here is the result from Mathematica 4.0
In[1]:=
Solve[{p == b^2/a, e == Sqrt[a^2 - b^2]/a}, {a, b}] // InputForm

Out[2]=
{{b -> (-I*p)/Sqrt[-1 + e^2], a -> -(p/(-1 + e^2))},
 {b -> (I*p)/Sqrt[-1 + e^2], a -> -(p/(-1 + e^2))}}

There is no way to transfer the solution to the form obtained by hand. The form
returned by Solve is purely developer's decision. BTW, Mathematica 3.0 produces a
different form of the result. Imagine yourself giving a lecture to high school
students who are not familiar with complex numbers. How would you explain them the
solution?

Luckily, there is a way to get the expected result by using InequalitySolve (in
4.0).

In[3]:= << Algebra`InequalitySolve`

In[4]:=
InequalitySolve[{p > 0, 0 < e < 1, a > b > 0, p == b^2/a,
    e == Sqrt[a^2 - b^2]/a}, {e, p, a, b}]

Out[4]=
0 < e < 1 && p > 0 && a == -(p/(-1 + e^2)) && b == Sqrt[a^2 - a^2*e^2]

In[5]:=
Simplify[{a, b} //. ToRules[Drop[%, 2]], Take[%, 2]] // InputForm

Out[5]//InputForm=
{-(p/(-1 + e^2)), p/Sqrt[1 - e^2]}

I am positive, that an average Mathematica user would not figured out to use
InequalitySolve. Also, notice that human would rather write the first term in the
form:
  p/(1 - e^2).
LeafCount for this form is 11. The Mathematica expression has LeafCount of 10.
That clearly proves that LeafCount alone is not necessarily the best measure of
the simplicity (in human terms) of an expression.

Let's now solve the equations posted by Prof. MacDonald with some human help. We
will solve first the equations for velocities (denoted by u[t] and v[t]). BTW, I
use InputForms to avoid expressions hard to read in ascii form.

In[1]:=
solution1 = {u[t], v[t]} /.
      DSolve[{u'[t] == -(a u[t] + b v[t]),
              v'[t] == -g - (a v[t] - b u[t])},
                  {u[t], v[t]}, t] // InputForm

Out[1]//InputForm=
{{(a^2*C[1]*Cos[b*t] + b^2*C[1]*Cos[b*t] +
    b*E^(a*t)*g*Cos[b*t]^2 - a^2*C[2]*Sin[b*t] -
    b^2*C[2]*Sin[b*t] + b*E^(a*t)*g*Sin[b*t]^2)/
   ((a - I*b)*(a + I*b)*E^(a*t)),
  -((-(a^2*C[2]*Cos[b*t]) - b^2*C[2]*Cos[b*t] +
     a*E^(a*t)*g*Cos[b*t]^2 - a^2*C[1]*Sin[b*t] -
     b^2*C[1]*Sin[b*t] + a*E^(a*t)*g*Sin[b*t]^2)/
    ((a - I*b)*(a + I*b)*E^(a*t)))}}

At this point imagine yourself advocating Mathematica to golf players not familiar
with the concept of complex numbers.:-) Good luck in explaining what ComplexExpand
and TargetFunctions do!

In[2]:=
FullSimplify[ComplexExpand[solution1, TargetFunctions -> {Im, Re}]]

Out[2]//InputForm=
{{(b*g)/(a^2 + b^2) + (C[1]*Cos[b*t] - C[2]*Sin[b*t])/
    E^(a*t), -((a*g)/(a^2 + b^2)) +
   (C[2]*Cos[b*t] + C[1]*Sin[b*t])/E^(a*t)}}

Very good! This is a pretty simple form. Now we have to integrate both terms and
add a constant to each of them.

In[3]:=
solution2 = Integrate[%, t] + {{C[3], C[4]}} // InputForm

Out[3]//InputForm=
{{(b*g*t)/(a^2 + b^2) + C[3] +
   ((-(a*C[1]) + b*C[2])*Cos[b*t])/((-I*a + b)*(I*a + b)*
     E^(a*t)) + ((b*C[1] + a*C[2])*Sin[b*t])/
    ((-I*a + b)*(I*a + b)*E^(a*t)),
  -((a*g*t)/(a^2 + b^2)) + C[4] -
   ((b*C[1] + a*C[2])*Cos[b*t])/((a - I*b)*(a + I*b)*
     E^(a*t)) - ((a*C[1] - b*C[2])*Sin[b*t])/
    ((a - I*b)*(a + I*b)*E^(a*t))}}

Well, we got complex expressions again. Simplify and FullSimplify don't help much.

In[4]:=
FullSimplify[ComplexExpand[solution2, TargetFunctions -> {Im, Re}]]

Out[4]//InputForm=
{{(E^(a*t)*(b*g*t + (a^2 + b^2)*C[3]) +
    (-(a*C[1]) + b*C[2])*Cos[b*t] + (b*C[1] + a*C[2])*
     Sin[b*t])/((a^2 + b^2)*E^(a*t)),
  (E^(a*t)*(-(a*g*t) + (a^2 + b^2)*C[4]) -
    (b*C[1] + a*C[2])*Cos[b*t] + (-(a*C[1]) + b*C[2])*
     Sin[b*t])/((a^2 + b^2)*E^(a*t))}}

Take the first term.

In[5]:=
LeafCount[(E^(a*t)*(b*g*t + (a^2 + b^2)*C[3]) + (-(a*C[1]) + b*C[2])*
          Cos[b*t] + (b*C[1] + a*C[2])*Sin[b*t])/((a^2 + b^2)*E^(a*t))]

Out[5]=
67

Any person with decent high school education can momentarily simplify this
expression.

((b*g*t) + E^(-a*t)((-(a*C[1]) + b*C[2])*Cos[b*t] + (b*C[1] + a*C[2])*
Sin[b*t]))/((a^2 + b^2)) + C[3]

In[6]:=
LeafCount[((b*g*t) +
          E^(-a*t)((-(a*C[1]) + b*C[2])*Cos[b*t] + (b*C[1] + a*C[2])*
                  Sin[b*t]))/((a^2 + b^2)) + C[3]]

Out[6]=
55

In[7]:=
((b*g*t) + E^(-a*t)((-(a*C[1]) + b*C[2])*Cos[b*t] + (b*C[1] + a*C[2])*
                    Sin[b*t]))/((a^2 + b^2)) +
      C[3] == (E^(a*t)*(b*g*t + (a^2 + b^2)*C[3]) + (-(a*C[1]) + b*C[2])*
            Cos[b*t] + (b*C[1] + a*C[2])*Sin[b*t])/((a^2 + b^2)*
          E^(a*t)) // FullSimplify

Out[7]=
True

Additionally, a human can notice that appropriately choosing the constants, one
can further simplify the expression.

b*g*t/(a^2 + b^2) + E^(-a*t)(C[1]*Cos[b*t] + C[2]*Sin[b*t]) + C[3]

(LeafCount of 38) or

b*g*t/(a^2 + b^2) + E^(-a*t)*C[1]*Cos[b*t + C[2]] + C[3]

(LeafCount of 32). As you can see, there are simpler solutions than those produced
by Mathematica.

10 - 12 years ago an average mathematics, physics, or engineering student could
solve these equations by hand and in time much shorter I needed to get a solution
with help of Mathematica for just one variable. Ironically, there is well known
exact solution for the differential equation of the form

   d x
   --- = A x + B
   d t

in Banach space. From there one can get a solution for the case when A is a
matrix, and x, B are vectors.

Here is my point. Mathematica can certainly do plenty of problems much better than
human. But, it is very, very frustrating, that in trivial cases the system often
produces results worse then those delivered by a human. I see this as the
challenge for Mathematica developers. The system should always produce better
results than human. Presently, Mathematica is a tool for some kind of "scientific
lower middle class". It is way to weak for people, who do serious mathematics or
theoretical physics, and way to complicated for pedestrians. If Wolfram Research
Inc. truly intents to reach "masses", it has to be more sensitive to their needs.

--Leszek

Andrzej Kozlowski wrote:

> I don't think of myself as a "computer algebra nerd" and I don't play golf
> but it seems to me that Mathemaitca does this problem rather well:
>
> In[2]:=
> solution = {y[t], x[t]} /. DSolve[{x''[t] == - (a x'[t] + b y'[t]),
>      y''[t] == - g - (a y'[t] - b x'[t])}, {y[t], x[t]}, t];
>
> In[3]:=
> Simplify[ComplexExpand[solution, TargetFunctions -> {Im, Re}]]
>
> Out[3]=
>       1        a t   4         3
> {{---------- (E    (a  C[1] + a  (-g t + C[3]) +
>     2    2 2
>   (a  + b )
>
>              2                  2
>           a b  (-g t + C[3]) + b  (-g + b (b C[1] + C[4])) +
>
>            2
>           a  (g + b (2 b C[1] + C[4]))) -
>
>          2    2
>        (a  + b ) (a C[3] + b C[4]) Cos[b t] -
>
>          2    2                                  a t
>        (a  + b ) (-b C[3] + a C[4]) Sin[b t]) / E   ,
>
>        1        a t   4         3
>    ---------- (E    (a  C[2] + b  (g t + b C[2] - C[3]) +
>      2    2 2
>    (a  + b )
>
>            2                              3
>           a  b (g t + 2 b C[2] - C[3]) + a  C[4] +
>
>           a b (-2 g + b C[4])) -
>
>          2    2
>        (a  + b ) (-b C[3] + a C[4]) Cos[b t] +
>
>          2    2                                 a t
>        (a  + b ) (a C[3] + b C[4]) Sin[b t]) / E   }}
>
> --
> Andrzej Kozlowski
> Toyama International University
> JAPAN
> http://sigma.tuins.ac.jp
> http://eri2.tuins.ac.jp
>
> ----------
> >From: "William M. MacDonald" <wm2 at umail.umd.edu>
To: mathgroup at smc.vnet.net
> >To: mathgroup at smc.vnet.net
> >Subject: [mg19765] [mg19677] Mathematica can't win against Tiger Woods
> >Date: Thu, Sep 9, 1999, 3:19 PM
> >
>
> >
> > I want to use the study of golf drives in teaching theoretical methods.  An
> > approximate pair of equations to get insight assumes that the drag force is
> >  linearly  proportional to velocity, instead of the actual quadratic
> >  dependence.  The equations for a ball with backspin to provide lift are
> >      x''[t]== - (a x'[t]+b y'[t]),
> >      y''[t]== - g - (a y'[t]- b x'[t])
> >  Mathematica returns a very complicated and apparently complex expression in
> >  about 9 seconds on my 250 MHz G3 Powerbook.  Simplify takes 1min and 20
> >  seconds and still returns an apparently complex expression.  If I apply
> >  FullSimplify on the solution for say x[t], I get no answer in 6 minutes.
> >
> >      I have a PC version of another system that I can run on my Powerbook
> using
> >  Virtual PC.  It requires 6 seconds to deliver a lengthy but obviously real,
> >  no Exp[(a+ I b)t] terms or (a + I b)(a - I b) terms.
> >
> >      I have never been able to learn why Mathematica is so slow in solving
> >  coupled equations and returns (as USUAL unless you use Simplify) such
> >  inelegant results.  Is there any computer algebra NERD out
> >  there who knows the answer.  (Don't tell me to use AlgebraicManipulation; I
> >  am trying to sell Mathematica to users who don't want to spend time
> > learning
> >  fancy tricks.)
> >
> > --
> > William M. MacDonald
> > Professor of Physics
> > University of Maryland
> >
> > Internet: wm2 at umail.umd.edu
> >
> >



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