From johnabloodworth3 at gmail.com  Thu Nov 14 00:58:57 2013
From: johnabloodworth3 at gmail.com (Jay Bloodworth)
Date: Wed, 13 Nov 2013 18:58:57 -0500
Subject: [Edu-sig] Syllabus
Message-ID: <52841241.7010505@gmail.com>

Does anyone have a link to a syllabus that might be appropriate for a 9
week (give or take) introductory programming course?

Thanks,
Jay

From kirby.urner at gmail.com  Sat Nov 23 04:04:30 2013
From: kirby.urner at gmail.com (kirby urner)
Date: Fri, 22 Nov 2013 19:04:30 -0800
Subject: [Edu-sig] Fibonacci Numbers and Phi (again)
Message-ID: <CAPJgG3T4huR3UY7kxjhCmVLdh5QM2+XFPkR5Z4Z2t73UokutRg@mail.gmail.com>

Here's something mathematical from the Poly list [1], adapting
something by David Koski.

The code shows how one might use a program to express a
cool relationship, without offering a proof.

The unit-test shows use raising PHI from the -6 to the 29th
power and finding an equivalent expression built solely from
three consecutive Fibonacci numbers and the sqrt(5).

Something like:

LOOP:
   PHI ** E = (FIB[N] + FIB[N-2] + FIB[N-1]*RT5) / 2
   E += 1
   N += 1

Start with:

E = -6

FIB[N-2] = 13
FIB[N-1] = -8
FIB[N]    =  5

in . . . 13, -8, 5, -3, 2, -1,1, 0, 1, 2, 3, 5, 8,13 . . .

===


"""
Encapsulating a discovery -- not a proof.
By David Koski (Python by K. Urner)

Failure at around 37th power is due to floating point limitations.
"""

import unittest

def fibonacci(a=0, b=1):
    while True:
        yield a
        a, b = b, a + b


series1 = fibonacci(5, -3)  # 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8
series2 = fibonacci(-8, 5)  # -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8
series3 = fibonacci(13,-8)  # 13, -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8

RT5 = pow(5, .5)   # "square root" of five
PHI = (1 + RT5)/2  # golden proportion

def cool_formula():
    """
    Two fib numbers, two apart, plus the one in the middle * sqrt(5)
    all over 2, gives PHI to a power.  Advancing all three sequences
    gives successive powers.
    """
    while True:
        yield (next(series1) + next(series3) + next(series2) * RT5)/2.0

class TestPhi(unittest.TestCase):

    def test_loop(self):
        gem = cool_formula()
        for e in range(-6, 30):  # adjust range (fails about 37th power)
            answer = next( gem )
            self.assertAlmostEqual( PHI ** e, answer)

if __name__ == "__main__":
    unittest.main()

[1]  Polyhedron mailing list
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From kirby.urner at gmail.com  Sat Nov 23 05:57:34 2013
From: kirby.urner at gmail.com (kirby urner)
Date: Fri, 22 Nov 2013 20:57:34 -0800
Subject: [Edu-sig] Fibonacci Numbers and Phi (again)
In-Reply-To: <CAPJgG3T4huR3UY7kxjhCmVLdh5QM2+XFPkR5Z4Z2t73UokutRg@mail.gmail.com>
References: <CAPJgG3T4huR3UY7kxjhCmVLdh5QM2+XFPkR5Z4Z2t73UokutRg@mail.gmail.com>
Message-ID: <CAPJgG3SGm_JGviVKPdwSK8g18EWT1ccJtcc_ASPjQ1NAGR58mA@mail.gmail.com>

"""
Another version, using a generator of Fibonacci "triples"...

Encapsulating a discovery -- not a proof.
By David Koski (Python by K. Urner)

Failure at around 37th power is due to floating point limitations.
"""

import unittest

def fib_triple(a=0, b=1):
    """
    >>> from anyproject import fib_triple
    >>> gem = fib_triple()
    >>> next(gem)
    (0, 1, 1)
    >>> next(gem)
    (1, 1, 2)
    >>> next(gem)
    (1, 2, 3)
    >>> next(gem)
    (2, 3, 5)
    >>> next(gem)
    (3, 5, 8)
    >>> gem = fib_triple(13,-8)
    >>> next(gem)
    (13, -8, 5)
    >>> next(gem)
    (-8, 5, -3)
    >>> next(gem)
    (5, -3, 2)
    >>> next(gem)
    (-3, 2, -1)
    >>> next(gem)
    (2, -1, 1)
    >>> next(gem)
    (-1, 1, 0)
    """
    c = a + b
    while True:
        yield a, b, a + b
        a, b, c = b, a + b, b + c


series = fib_triple(13,-8)  # 13, -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8

RT5 = pow(5, .5)   # "square root" of five
PHI = (1 + RT5)/2  # golden proportion

def cool_formula():
    """
    Two fib numbers, two apart, plus the one in the middle * sqrt(5)
    all over 2, gives PHI to a power.  Advancing all three sequences
    gives successive powers.
    """
    while True:
        f0, f1, f2 = next(series)
        yield (f0 + f2 + f1 * RT5)/2.0

class TestPhi(unittest.TestCase):

    def test_loop(self):
        gem = cool_formula()
        for e in range(-6, 30):  # adjust range (fails about 37th power)
            answer = next( gem )
            self.assertAlmostEqual( PHI ** e, answer)

if __name__ == "__main__":
    unittest.main()
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From litvin at skylit.com  Sat Nov 23 06:49:45 2013
From: litvin at skylit.com (Litvin)
Date: Sat, 23 Nov 2013 00:49:45 -0500
Subject: [Edu-sig] Fibonacci Numbers and Phi (again)
In-Reply-To: <CAPJgG3T4huR3UY7kxjhCmVLdh5QM2+XFPkR5Z4Z2t73UokutRg@mail.g
 mail.com>
References: <CAPJgG3T4huR3UY7kxjhCmVLdh5QM2+XFPkR5Z4Z2t73UokutRg@mail.gmail.com>
Message-ID: <7.0.1.0.2.20131123004922.05cca650@skylit.com>

Kirby,

I am sorry to spoil all the fun, but this is a not a gem but a 
mathematical trick, and I think the way to deal with mathematical 
tricks is to explain them, not to verify with code.

There are two Fibonacci sequences that are also geometric sequences 
1, x, x^2, ..., x^n,...: when x = phi or x = -1/phi, because phi and 
-1/phi are the roots of the quadratic equation x^2 = x+1.  Any 
Fibonacci sequence is a linear combination of these two: F[n] = 
a*phi^n + b*(-1/phi)^n.  a and b can be determined from the first two 
terms of the sequence:

a + b = F[0]
a*phi - b/phi = F[1]

If you choose F[0] and F[1] (not necessarily integers) in a tricky 
way, so that b = 0, then F[n] = a*phi^n.  In addition, if a = phi^e, 
then F[n] = phi^(e+n).  Working in reverse, you can find F[0] and 
F[1] that satisfy these conditions and get the trick going. Here

F[0] = (13 + 5 - 8*sqrt(5))/2 = 9 - 4*sqrt(5)
F[1] =  (-8 - 3 + 5*sqrt(5))/2 = (-11+5sqrt(5))/2

Easy to verify that F0*phi = F1, so, indeed, b = 0 and F[n] is a 
geometric sequence.  Also, phi^6 = 9 + 4*sqrt(5), so F[0]*phi^6 = 1 
and a = F[0] = phi^(-6).  You get F[n] = phi^(n-6).

Gary Litvin
www.skylit.com

At 10:04 PM 11/22/2013, you wrote:

>Here's something mathematical from the Poly list [1], adapting
>something by David Koski.
>
>The code shows how one might use a program to express a
>cool relationship, without offering a proof.
>
>The unit-test shows use raising PHI from the -6 to the 29th
>power and finding an equivalent expression built solely from
>three consecutive Fibonacci numbers and the sqrt(5).
>
>Something like:
>
>LOOP:
>    PHI ** E = (FIB[N] + FIB[N-2] + FIB[N-1]*RT5) / 2
>    E += 1
>    N += 1
>
>Start with:
>
>E = -6
>
>FIB[N-2] = 13
>FIB[N-1] = -8
>FIB[N]    =  5
>
>in . . . 13, -8, 5, -3, 2, -1,1, 0, 1, 2, 3, 5, 8,13 . . .
>
>===
>
>
>"""
>Encapsulating a discovery -- not a proof.
>By David Koski (Python by K. Urner)
>
>Failure at around 37th power is due to floating point limitations.
>"""
>
>import unittest
>
>def fibonacci(a=0, b=1):
>     while True:
>         yield a
>         a, b = b, a + b
>
>
>series1 = fibonacci(5, -3)  # 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8
>series2 = fibonacci(-8, 5)  # -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8
>series3 = fibonacci(13,-8)  # 13, -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8
>
>RT5 = pow(5, .5)   # "square root" of five
>PHI = (1 + RT5)/2  # golden proportion
>
>def cool_formula():
>     """
>     Two fib numbers, two apart, plus the one in the middle * sqrt(5)
>     all over 2, gives PHI to a power.  Advancing all three sequences
>     gives successive powers.
>     """
>     while True:
>         yield (next(series1) + next(series3) + next(series2) * RT5)/2.0
>
>class TestPhi(unittest.TestCase):
>
>     def test_loop(self):
>         gem = cool_formula()
>         for e in range(-6, 30):  # adjust range (fails about 37th power)
>             answer = next( gem )
>             self.assertAlmostEqual( PHI ** e, answer)
>
>if __name__ == "__main__":
>     unittest.main()
>
>[1]  Polyhedron mailing list
>Contributions to 
><mailto:polyhedron at lists.mathconsult.ch>mailto:polyhedron at lists.mathconsult.ch
>Admin: 
><http://lists.mathconsult.ch/mailman/listinfo/polyhedron>http://lists.mathconsult.ch/mailman/listinfo/polyhedron
>Archive: 
><http://lists.mathconsult.ch/mailman/private/polyhedron/>http://lists.mathconsult.ch/mailman/private/polyhedron/
>
>_______________________________________________
>Edu-sig mailing list
>Edu-sig at python.org
>https://mail.python.org/mailman/listinfo/edu-sig

At 10:04 PM 11/22/2013, kirby urner wrote:


>Here's something mathematical from the Poly list [1], adapting
>something by David Koski.
>
>The code shows how one might use a program to express a
>cool relationship, without offering a proof.
>
>The unit-test shows use raising PHI from the -6 to the 29th
>power and finding an equivalent expression built solely from
>three consecutive Fibonacci numbers and the sqrt(5).
>
>Something like:
>
>LOOP:
>    PHI ** E = (FIB[N] + FIB[N-2] + FIB[N-1]*RT5) / 2
>    E += 1
>    N += 1
>
>Start with:
>
>E = -6
>
>FIB[N-2] = 13
>FIB[N-1] = -8
>FIB[N]    =  5
>
>in . . . 13, -8, 5, -3, 2, -1,1, 0, 1, 2, 3, 5, 8,13 . . .
>
>===
>
>
>"""
>Encapsulating a discovery -- not a proof.
>By David Koski (Python by K. Urner)
>
>Failure at around 37th power is due to floating point limitations.
>"""
>
>import unittest
>
>def fibonacci(a=0, b=1):
>     while True:
>         yield a
>         a, b = b, a + b
>
>
>series1 = fibonacci(5, -3)  # 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8
>series2 = fibonacci(-8, 5)  # -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8
>series3 = fibonacci(13,-8)  # 13, -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8
>
>RT5 = pow(5, .5)   # "square root" of five
>PHI = (1 + RT5)/2  # golden proportion
>
>def cool_formula():
>     """
>     Two fib numbers, two apart, plus the one in the middle * sqrt(5)
>     all over 2, gives PHI to a power.  Advancing all three sequences
>     gives successive powers.
>     """
>     while True:
>         yield (next(series1) + next(series3) + next(series2) * RT5)/2.0
>
>class TestPhi(unittest.TestCase):
>
>     def test_loop(self):
>         gem = cool_formula()
>         for e in range(-6, 30):  # adjust range (fails about 37th power)
>             answer = next( gem )
>             self.assertAlmostEqual( PHI ** e, answer)
>
>if __name__ == "__main__":
>     unittest.main()
>
>[1]  Polyhedron mailing list
>Contributions to 
><mailto:polyhedron at lists.mathconsult.ch>mailto:polyhedron at lists.mathconsult.ch
>Admin: 
><http://lists.mathconsult.ch/mailman/listinfo/polyhedron>http://lists.mathconsult.ch/mailman/listinfo/polyhedron
>Archive: 
><http://lists.mathconsult.ch/mailman/private/polyhedron/>http://lists.mathconsult.ch/mailman/private/polyhedron/
>
>_______________________________________________
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>Edu-sig at python.org
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From kirby.urner at gmail.com  Sat Nov 23 17:08:17 2013
From: kirby.urner at gmail.com (kirby urner)
Date: Sat, 23 Nov 2013 08:08:17 -0800
Subject: [Edu-sig] Fibonacci Numbers and Phi (again)
In-Reply-To: <7.0.1.0.2.20131123004922.05cca650@skylit.com>
References: <CAPJgG3T4huR3UY7kxjhCmVLdh5QM2+XFPkR5Z4Z2t73UokutRg@mail.gmail.com>
 <7.0.1.0.2.20131123004922.05cca650@skylit.com>
Message-ID: <CAPJgG3R17iqVS+z4fSTr7yafHx+3G31LnG=ygf8i97D_uL8dCg@mail.gmail.com>

> On Fri, Nov 22, 2013 at 9:49 PM, Litvin <litvin at skylit.com> wrote:

> Kirby,
>
> I am sorry to spoil all the fun, but this is a not a gem but a
> mathematical trick, and I think the way to deal with
> mathematical tricks is to explain them, not to verify with code.


Having an algebraic verification is more like the proof I did not provide.
Thanks for the reasoning Gary.

I do think there's a two phase understanding for theorems where phase one
is too oft neglected, which is understanding what the theorem is claiming
or asserting, before attempting an analysis of "why?".

Euler's Theorem by way of example, the one with the totient, a
generalization of Fermat's:  it's too easy to just stare at the assertion
and not think of examples, not know what it says.

This is where I think a Python program may come in handy, not as a proof of
a theorem but as a illustration or demonstration of its consequences.  See
the theorem in action somehow.

Seeing something in Python that's runnable may close some critical circuits
in the brain of the theorem's reader.  "Now I better know what it means"
may be the aha experience.

Most people who have read up on Fibonaccis know they can be extended in the
negative direction.  Then the idea of starting with any two numbers, not
necessarily integers may be introduced.

The limit of F[N+1]/F[N] and its relationship to PHI (equal as N ->
infinity) may then be discussed.

I've taken exactly this same approach with one of Ramanujan's for 1/pi
(just flip it for pi).  A little Python generator will do the trick.  In
this case the gem being verified has not to my knowledge been fully
explained or reasoned about (the proof has yet to be supplied):

http://worldgame.blogspot.com/2012/01/testing-math-ml.html

(this blogpost uses MathJax, a good resource for the JavaScript-enabled)

I'm still without an explanation for this pi digits generator from one of
Michel Paul's students, but I share it around in hopes of finding one --
and then I hope its one I can follow.

https://groups.google.com/forum/#!msg/mathfuture/LA0pMPC6-HE/MBGWxn4ENsUJ

Thanks again for clearing up any mysteries that may have attached to the
PHI ** N formula (I originally called it a formula, not a gem... some gems
are only semi-precious).

Kirby
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From litvin at skylit.com  Sat Nov 23 19:54:11 2013
From: litvin at skylit.com (Litvin)
Date: Sat, 23 Nov 2013 13:54:11 -0500
Subject: [Edu-sig] Fibonacci Numbers and Phi (again)
In-Reply-To: <CAPJgG3R17iqVS+z4fSTr7yafHx+3G31LnG=ygf8i97D_uL8dCg@mail.g
 mail.com>
References: <CAPJgG3T4huR3UY7kxjhCmVLdh5QM2+XFPkR5Z4Z2t73UokutRg@mail.gmail.com>
 <7.0.1.0.2.20131123004922.05cca650@skylit.com>
 <CAPJgG3R17iqVS+z4fSTr7yafHx+3G31LnG=ygf8i97D_uL8dCg@mail.gmail.com>
Message-ID: <7.0.1.0.2.20131123134822.05c54de0@skylit.com>

Kirby,

You are right, a little program does help to clarify the statement 
and the formulas involved.  But then, keep it really simple (no 
classes, no generators), so that understanding of the code doesn't 
get in the way.  For example:

from math import sqrt

rt5 = sqrt(5)
phi = (1 + rt5)/2

def neat_formula(f0, f1, f2):
     return (f0 + f2 + rt5*f1)/2

power_of_phi = phi**(-6)
f0, f1, f2 = 13, -8, 5
for k in range(20):
     print("{0:8.5f} {1:8.5f}".format(power_of_phi, neat_formula(f0, f1, f2)))
     f0, f1, f2 = f1, f2, f1+f2
     power_of_phi *= phi

There are many other occasions to bring in more advanced tools.

Gary Litvin
www.skylit.com

At 11:08 AM 11/23/2013, kirby urner wrote:
> > On Fri, Nov 22, 2013 at 9:49 PM, Litvin 
> <<mailto:litvin at skylit.com>litvin at skylit.com> wrote:
>
> > Kirby,
> >
> > I am sorry to spoil all the fun, but this is a not a gem but a
> > mathematical trick, and I think the way to deal with
> > mathematical tricks is to explain them, not to verify with code.
>
>
>Having an algebraic verification is more like the proof I did not 
>provide.  Thanks for the reasoning Gary.
>
>I do think there's a two phase understanding for theorems where 
>phase one is too oft neglected, which is understanding what the 
>theorem is claiming or asserting, before attempting an analysis of "why?".
>
>Euler's Theorem by way of example, the one with the totient, a 
>generalization of Fermat's:  it's too easy to just stare at the 
>assertion and not think of examples, not know what it says.
>
>This is where I think a Python program may come in handy, not as a 
>proof of a theorem but as a illustration or demonstration of its 
>consequences.  See the theorem in action somehow.
>
>Seeing something in Python that's runnable may close some critical 
>circuits in the brain of the theorem's reader.  "Now I better know 
>what it means" may be the aha experience.
>
>Most people who have read up on Fibonaccis know they can be extended 
>in the negative direction.  Then the idea of starting with any two 
>numbers, not necessarily integers may be introduced.
>
>The limit of F[N+1]/F[N] and its relationship to PHI (equal as N -> 
>infinity) may then be discussed.
>
>I've taken exactly this same approach with one of Ramanujan's for 
>1/pi (just flip it for pi).  A little Python generator will do the 
>trick.  In this case the gem being verified has not to my knowledge 
>been fully explained or reasoned about (the proof has yet to be supplied):
>
><http://worldgame.blogspot.com/2012/01/testing-math-ml.html>http://worldgame.blogspot.com/2012/01/testing-math-ml.html
>
>(this blogpost uses MathJax, a good resource for the JavaScript-enabled)
>
>I'm still without an explanation for this pi digits generator from 
>one of Michel Paul's students, but I share it around in hopes of 
>finding one -- and then I hope its one I can follow.
>
><https://groups.google.com/forum/#!msg/mathfuture/LA0pMPC6-HE/MBGWxn4ENsUJ>https://groups.google.com/forum/#!msg/mathfuture/LA0pMPC6-HE/MBGWxn4ENsUJ
>
>Thanks again for clearing up any mysteries that may have attached to 
>the PHI ** N formula (I originally called it a formula, not a gem... 
>some gems are only semi-precious).
>
>Kirby
>
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From kirby.urner at gmail.com  Sat Nov 23 20:11:39 2013
From: kirby.urner at gmail.com (kirby urner)
Date: Sat, 23 Nov 2013 11:11:39 -0800
Subject: [Edu-sig] Fibonacci Numbers and Phi (again)
In-Reply-To: <7.0.1.0.2.20131123134822.05c54de0@skylit.com>
References: <CAPJgG3T4huR3UY7kxjhCmVLdh5QM2+XFPkR5Z4Z2t73UokutRg@mail.gmail.com>
 <7.0.1.0.2.20131123004922.05cca650@skylit.com>
 <CAPJgG3R17iqVS+z4fSTr7yafHx+3G31LnG=ygf8i97D_uL8dCg@mail.gmail.com>
 <7.0.1.0.2.20131123134822.05c54de0@skylit.com>
Message-ID: <CAPJgG3Rbyu+rbpQEe0Qgx+53tCvPnd53_f8NgVn=hHEEVLgR3w@mail.gmail.com>

Yes, I would consider that an improvement, though I might go back to a
generator with:

from math import sqrt

rt5 = sqrt(5)
phi = (1 + rt5)/2

def neat_formula(f0, f1, f2):
    while True:
        yield (f0 + f2 + rt5*f1)/2
        f0, f1, f2 = f1, f2, f1+f2

power_of_phi = phi**(-6)
results = neat_formula(13, -8, 5)  # seed me
for k in range(20):
    print("{0:8.5f} {1:8.5f}".format(power_of_phi, next(results)))
    power_of_phi *= phi

Going back and forth between yours and mine is introducing a small delta,
so if the game were to introduce generators as a genre, I could see doing a
bunch of small deltas like this.

Kirby



On Sat, Nov 23, 2013 at 10:54 AM, Litvin <litvin at skylit.com> wrote:

>  Kirby,
>
> You are right, a little program does help to clarify the statement and the
> formulas involved.  But then, keep it really simple (no classes, no
> generators), so that understanding of the code doesn't get in the way.  For
> example:
>
> from math import sqrt
>
> rt5 = sqrt(5)
> phi = (1 + rt5)/2
>
> def neat_formula(f0, f1, f2):
>     return (f0 + f2 + rt5*f1)/2
>
> power_of_phi = phi**(-6)
> f0, f1, f2 = 13, -8, 5
> for k in range(20):
>     print("{0:8.5f} {1:8.5f}".format(power_of_phi, neat_formula(f0, f1,
> f2)))
>     f0, f1, f2 = f1, f2, f1+f2
>     power_of_phi *= phi
>
> There are many other occasions to bring in more advanced tools.
>
> Gary Litvin
> www.skylit.com
>
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