From meyer@imap.theory.csail.mit.edu  Wed Nov 16 13:12:07 2005
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From: Harrison King Hall <hkhall@MIT.EDU>
Subject: [David] Week 11 Comments
Date: Wed, 16 Nov 2005 13:11:48 -0500
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This is really a cool way to do counting problems, and really intuitive 
since you only have to worry about how to select each class of element 
for n elements independent of all the others.  My only issue is that I 
don't always see the generating function .  Like in the "Impossible" 
counting problem, I didn't see that the number of oranges was O(x) = 1 
+ x + x ^2 + x^3 + x^4 = (1 $B!](B x^5)/(1 $B!](B x) and I don't really know how 
you got there.  That is probably something that comes with time 
however.  This is really slick.
-Harrison


From meyer@imap.theory.csail.mit.edu  Wed Nov 16 14:07:07 2005
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Subject: [Sayan] Week 11 comments
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Section 3.2 Finding a Closed Form for fib gen func

I don't really understand why we split the denominators. Why does
solving for the alphas give us the nth Fibonacci number?

From meyer@imap.theory.csail.mit.edu  Wed Nov 16 16:46:57 2005
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On page 1, the lecture notes state "A generating function is a “formal” 
power series in the sense that we usually regard x as a placeholder
rather than a number. Only in rare cases will we actually evaluate a 
generating function by letting x take a real number value, so we 
generally ignore the issue of convergence."

In what cases might we actually evaluate a generating function?

Jason.

From meyer@imap.theory.csail.mit.edu  Wed Nov 16 18:30:20 2005
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From: "Chris Yang" <yangc@MIT.EDU>
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I don't really understand how the Convolution rule works.  Will we be
covering it in class on Friday?  Also, the apples/oranges thing at the end
is pretty cool.

 

Thanks,

Chris Yang


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<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
style=3D'font-size:10.0pt;
font-family:Arial'>I don&#8217;t really understand how the Convolution =
rule works.&nbsp;
Will we be covering it in class on Friday?&nbsp; Also, the =
apples/oranges thing at
the end is pretty cool.<o:p></o:p></span></font></p>

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<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
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font-family:Arial'>Chris Yang<o:p></o:p></span></font></p>

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From hsoumare@MIT.EDU Wed Nov 16 21:52:36 2005
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Subject: {Sayan} Weekly Email Comment
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Hi

I would like you to go over how the closed form of the nth  
coefficient of the fibonacci series was obtained on page 9. Thanks.

Hamidou

From xiaoranz@MIT.EDU Thu Nov 17 00:23:58 2005
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Date: Thu, 17 Nov 2005 00:23:56 -0500
From: "Xiaoran (Sharon) Zhang" <xiaoranz@MIT.EDU>
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I thought the generating functions are interesting in that they can help solve
counting problems in a whole new way. The "Impossible" counting problem
solution is truly awesome.

**************************************************
Xiaoran (Sharon) Zhang
Class of 2008
Department of Biology &
Department of Electrical Engineering and Computer Science
Massachusetts Institute of Technology
320 Memorial Drive,
Cambridge, MA 02139
E-mail: xiaoranz@mit.edu
**************************************************

From mdmurray@MIT.EDU Thu Nov 17 03:08:57 2005
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From: Michael Murray <mdmurray@MIT.EDU>
Subject: [Hanson] Generating Functions Notes
Date: Thu, 17 Nov 2005 03:07:43 -0500
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Having never seen generating functions before I think the notes and  
lecture where a great introduction and everything so far is pretty  
clear. I like the use of whiteboard in lecture today, I thought it  
was much easier to follow than the slides and the lecture was at a  
much nicer pace than usual.

I am however a little unclear in the explanation of the product rule.  
I don't really understand the table and I don't know why

c_n = a_0*b_n + a_1*b_n-1 + a_2*b_n-2...

Thanks,
Michael

From mdmurray@MIT.EDU Thu Nov 17 03:08:57 2005
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Subject: [Hanson] Generating Functions Notes
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Having never seen generating functions before I think the notes and  
lecture where a great introduction and everything so far is pretty  
clear. I like the use of whiteboard in lecture today, I thought it  
was much easier to follow than the slides and the lecture was at a  
much nicer pace than usual.

I am however a little unclear in the explanation of the product rule.  
I don't really understand the table and I don't know why

c_n = a_0*b_n + a_1*b_n-1 + a_2*b_n-2...

Thanks,
Michael

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Hi,

That last "impossible" counting problem was cool. Funny though how the
numbers seemed to cancel perfectly... It's still cool to know that the
procedure works like that, even if the answer doesn't turn out as nice ever=
y
time.
Since most of the first half of the notes has been done in lecture already,
I'm guessing the rest will be done tomorrow. The question I have regarding
the second section is in reference to the Convolution Rule (page 10). Is
there an easy way to compute the generating function if the two sets A and =
B
are not disjoint? It's hard for me to even think of a common scenario when
this is the case, but I'd be interested to see if there's an easy way
someone has though of to do it.

Thanks,
Shreyes

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Hi,<br>
<br>
That last &quot;impossible&quot; counting problem was cool.&nbsp; Funny tho=
ugh
how the numbers seemed to cancel perfectly... It's still cool to know
that the procedure works like that, even if the answer doesn't turn out
as nice every time.<br>
Since most of the first half of the notes has been done in lecture
already, I'm guessing the rest will be done tomorrow.&nbsp; The
question I have regarding the second section is in reference to the
Convolution Rule (page 10).&nbsp; Is there an easy way to compute the
generating function if the two sets A and B are not disjoint?&nbsp;
It's hard for me to even think of a common scenario when this is the
case, but I'd be interested to see if there's an easy way someone has
though of to do it.<br>
<br>
Thanks,<br>
Shreyes<br>

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From hmzhou@blackbird.csail.mit.edu Thu Nov 17 16:18:19 2005
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Date: Thu, 17 Nov 2005 16:18:18 -0500 (EST)
From: Hanson Zhou <hmzhou@theory.csail.mit.edu>
To: Michael Murray <mdmurray@MIT.EDU>
cc: 6042-probs@theory.csail.mit.edu
Subject: Re: [Hanson] Generating Functions Notes
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The table is merely a multiplication table to show you the result of
multiplying a term from each sum.  From it you should be able to see which
terms contribute to the coefficient of x^n in the final product.

Specifically, each pair that multiplies to give x^n contributes to the
coefficient of x^n in the final answer.  The pairs that contribute are
therefore: (0,n), (1, n-1), (2, n-2)... and hence the coefficient on x^n
in the final product is:

 c_n = a_0*b_n + a_1*b_n-1 + a_2*b_n-2...

See me if you are still confused.

-Hanson

On Thu, 17 Nov 2005, Michael Murray wrote:

> Having never seen generating functions before I think the notes and
> lecture where a great introduction and everything so far is pretty
> clear. I like the use of whiteboard in lecture today, I thought it
> was much easier to follow than the slides and the lecture was at a
> much nicer pace than usual.
>
> I am however a little unclear in the explanation of the product rule.
> I don't really understand the table and I don't know why
>
> c_n = a_0*b_n + a_1*b_n-1 + a_2*b_n-2...

>
> Thanks,
> Michael
>

From ozcan@MIT.EDU Thu Nov 17 16:41:13 2005
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Subject: [Hanson]reading comments
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Section 3.2 as a whole, i.e. the construction of a closed form for the 
Fibonacci numbers:
It is quite impressive to be able to code this much information into only 
one simple function, I am really amazed.


From fluff@MIT.EDU Thu Nov 17 17:24:19 2005
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To: 6042-probs@theory.lcs.mit.edu
From: Crystal Chao <fluff@MIT.EDU>
Subject: [Jelani] week 11 reading
Date: Thu, 17 Nov 2005 17:22:08 -0500
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"Amazing!" It's still not entirely intuitive to me why the  
Convolution Rule works, but overall the reading was interesting,  
flowed well, and made sense. I liked how the reading was shorter and  
all on one topic, instead of like previous readings that were on a  
million different things all glommed onto some 25-page monstrosity.

~Crystal

From mracich@MIT.EDU Thu Nov 17 18:25:56 2005
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Subject: [Sayan] Comments for Course Notes, Week 11 (Generating Functions)
From: Moira Racich <mracich@MIT.EDU>
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I found section 3.2, "Finding a Closed Form", (starting on page 8)
confusing, both in the notes and when it was covered in lecture.  I
would appreciate it if this was explained further. I think I would also
find a brief partial fractions review very helpful.  

Moira Racich


From ereid@MIT.EDU Thu Nov 17 18:36:30 2005
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To: 6042-probs@theory.lcs.mit.edu
From: Elizabeth Reid <ereid@MIT.EDU>
Subject: [Hanson] Reading Comments
Date: Thu, 17 Nov 2005 18:36:10 -0500
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The reading makes sense, but I feel like I wouldn't be able to do it  
myself. Is there anywhere I can go that would help me practice this  
stuff?

From dangut@MIT.EDU Thu Nov 17 19:13:52 2005
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Date: Thu, 17 Nov 2005 19:13:49 -0500
From: Daniel A Gutierrez <dangut@MIT.EDU>
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These all make logical sense, but I was wondering if by chance you have a list
or know where to find one so that I can review geometric summing?
Thanks,
Daniel

From petek@MIT.EDU Thu Nov 17 20:53:08 2005
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Date: Thu, 17 Nov 2005 20:53:17 -0500
From: Pete Kruskall <petek@MIT.EDU>
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I'm still shaky on why a generating function is called what it is...  
Also need some more practice on counting with these things.  It is kinda 
cool that you can solve some complex probs that way....

-P

-- 
Pete Kruskall
28 The Fenway
Boston, MA 02215

:::::::::::::::::::::::::::::
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From tylerw@MIT.EDU Thu Nov 17 21:10:04 2005
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Subject: [TA-name] Week 11 Comments
Date: Thu, 17 Nov 2005 21:09:48 -0500
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I thought the impossible counting problem was interesting. 

From bilodeau@MIT.EDU Thu Nov 17 21:32:57 2005
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From: "Peter Bilodeau" <bilodeau@MIT.EDU>
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In this week's reading, part of what was hard to understand was the way the
formula for the infinite sum was used to describe the series that generates
the coefficients.  Trying to think about what is being substituted for x and
what operations happen on each term is a little difficult for me to grasp.


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to
understand was the way the formula for the infinite sum was used to =
describe
the series that generates the coefficients.&nbsp; Trying to think about =
what is
being substituted for x and what operations happen on each term is a =
little
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From jehan@MIT.EDU Thu Nov 17 21:33:32 2005
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Date: Thu, 17 Nov 2005 21:31:46 -0500
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From: Jehan deFonseka <jehan@MIT.EDU>
Subject: [Sayan] Week 11 Comments
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I don't understand relating generating functions to counting.


From dnreshef@MIT.EDU Thu Nov 17 21:36:34 2005
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Date: Thu, 17 Nov 2005 21:36:26 -0500
From: David N Reshef <dnreshef@MIT.EDU>
To: 6042-probs@theory.lcs.MIT.EDU
Subject: [Hanson] Reading 11 Comments
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Of all of the course material we have covered thus far, this has been the most
interesting, and I would like to get into it a lot more in class.  I had a
little bit of trouble with the section on counting with generating functions on
page 9.  Can we go over how to build generating functions that count?
-Dave

From hectorb@MIT.EDU Thu Nov 17 22:07:12 2005
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This week's reading was very straight forward; there wasn't anything I 
had a hard time understanding. I found the last section the most 
interesting, being able to generate functions for nasty counting problems.

-Hector

From icharny@MIT.EDU Thu Nov 17 22:08:33 2005
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I don't understand counting with functions (section 4, page 9). Please 
clarify this in lecture.


From vixen@MIT.EDU Thu Nov 17 22:38:02 2005
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I found the part about counting with generating functions confusing.

From dowgun@MIT.EDU Thu Nov 17 22:58:36 2005
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Date: Thu, 17 Nov 2005 22:58:28 -0500
From: Neil M Dowgun <dowgun@MIT.EDU>
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Subject: [Hanson] reading question
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So I realized that the example where we found the coefficiants for the Fibonacci
sequence doesn't really provide a clean fomulaic way to find coefficiants given
a closed form. I guess there's just always going to be some dirty math to do.
Anyway, I was wondering if using the Taylor expansion to find the coefficiants
was always a viable option. I mean, I realize that it would be a real pain to
take the derivative of an expression say, thirty times, but it just seems like
a more surefire way to get the answer if it really is foolproof.
Thanks, Neil

From brevzin@MIT.EDU Thu Nov 17 23:30:43 2005
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Date: Thu, 17 Nov 2005 23:30:35 -0500
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From: Barry Revzin <brevzin@MIT.EDU>
Subject: [Hanson] Reading
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For the convolution rule, I think it's really confusing that it's called 
the convolution rule. It really has nothing to do with the idea of a 
convolution from analysis, does it ? I'm really used to the convolution of 
f and g (f*g) being the integral of f(y)g(x-y)dy. Why IS it called that 
then ? Why not like... disjoint union rule ? Or Countable Additivity, which 
would be consistent with measure theory...

Barry


From tonyng@MIT.EDU Thu Nov 17 23:56:45 2005
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Date: Thu, 17 Nov 2005 23:56:41 -0500
To: 6042-probs@theory.lcs.mit.edu
From: Tony Ng <tonyng@MIT.EDU>
Subject: [Jelani] Reading Comments: Week 11
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I found the fruit counting problem ("An 'Impossible' Counting Problem, 
pg12) to be very interesting. I didn't understand the need for generating 
functions during the last lecture and it seemed excessively complicated to 
do, however, now I understand why it is necessary because it may be the 
only way we know how to solve some problems.

- Tony


From ctsims@MIT.EDU Fri Nov 18 00:03:39 2005
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From: Clayton Sims <ctsims@MIT.EDU>
Subject: [6.042] Reading Comments Lecture Notes 8 - Clayton Sims
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I felt like the material on page 7 was the most challenging. Generalizing 
the method of turning a sequence of coefficients into different Function 
Generators that are easily converted into a representation of the function 
is something that I feel I would benefit from practicing. Additionally, a 
quick review (5 mins.) of partial fractions in class would be nice, 
although not entirely necessary.

-Clayton Sims  


From jacques@MIT.EDU Fri Nov 18 00:17:13 2005
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Page 11-12.  I never thought the Taylor Series, which I'd seen and used many
times to approximate functions, could be related combinatorics (aka. most
surprising passage).  I actually thought generating functions were a difficult
concept until I read this example, since it was a series I was so familiar
with.

From sergiob@MIT.EDU Fri Nov 18 00:25:26 2005
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Date: Thu, 17 Nov 2005 22:25:17 -0700
To: 6042-probs@theory.csail.mit.edu
From: Sergio Bacallado <sergiob@MIT.EDU>
Subject: [David] Week 11 Comments
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I thought it was amazing how useful generating functions can be to solve 
counting problems. I would like to know if there is always an easy way to 
find a closed form for the nth term of a sequence, if you know the value of 
the generating function. That is, other than differentiating the function 
to get the Taylor expansion term.

Sergio Bacallado
Group A


From jjmonzon@MIT.EDU Fri Nov 18 00:38:56 2005
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Subject: David - Reading Comments
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I'm a little confused on the convolution rule. Perhaps this would be 
explained a little more in class. Also, I didn't really understand 
how they obtained the answer in the last problem of the tutorial. 
Hope tomorrow's lecture clears this up.

Josh



Joshua Jen C. Monzon
Massachusetts Institute of Technology
Electrical Engineering with Computer Science
jjmonzon@mit.edu   617-803-7497


From aston@MIT.EDU Fri Nov 18 00:51:55 2005
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I was really impressed by the fruit problem (pages 12 and 13), but I don't
feel like the explanation for why the convolution is the right answer to
problems like these (page 10) was a very convincing one.
 
    - Aston

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From cwong08@MIT.EDU Fri Nov 18 01:34:01 2005
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The part about convolution (pg 10) is confusing to me. I don't really 
get what it's trying to say.


From scot@MIT.EDU Fri Nov 18 01:57:35 2005
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Hi,

The biggest question I had from this week of notes was the reason that, 
while Prof. Meyer was on the board, he choose to round up on the sum, 
after erasing the second term. I was confused, and could not understand 
the conflicting explanations.

Thanks,

Scot

From cbossard@MIT.EDU Fri Nov 18 03:03:47 2005
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I find this to be an interesting topic although I have trouble
reproducing and creating these functions on my own they generally
make sense to me when I see them explained.  In section 2.5 on
products I am not seeing/understanding how the terms of the diagonal
were collected so I don't see why it's not aobo + a1b1 instead of
aobn etc.  Again with although the convolution rule makes some sense
to me in section 4.2 I dont see how they are getting the numbers from
the formula that was created.  I can see that each is (1+x) and they
should be multiplied.  Intuively I understand that there's 1 way to
choose 0 elements, 2 ways to choose 1, etc but I do not see how this
information can be attained from the 1+2x+x^2 equation.

Cynthia

From kjhollen@MIT.EDU Fri Nov 18 03:05:48 2005
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From: Kate Hollenbach <kjhollen@MIT.EDU>
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Subject: [Hanson] counting example
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the counting example for generating functions (p 12-13) is too cool!! hope we do
some problems like it in class.

From mpapi@MIT.EDU Fri Nov 18 03:31:57 2005
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Subject: [Jelani] week 11 comments
From: Matt Papi <mpapi@MIT.EDU>
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Section 3.2, page 8. I could use a review or appendix or something
regarding partial fractions. The examples are good, but once in a while
there are functions where the numerators of the split terms are
polynomials - anything more complex than the given examples I don't
remember how to do.

- Matt


From veracarr@MIT.EDU Fri Nov 18 03:59:05 2005
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Genearting Functions pg 1

When trying to find the sequence for a generating function when you have 
written the function in closed form what's the best way to find the 
sequence? Do a taylor expansion on the closed form?

From bakster@MIT.EDU Fri Nov 18 04:02:09 2005
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What kind of situations would require us to use values for x? In the 
readings it said that they are unusual or rare, I believe.

Alexander Bakst

From lmccart@MIT.EDU Fri Nov 18 04:19:49 2005
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I did not find anything particularly confusing in this reading, but it was a
pretty good reading nonetheless

-Lauren

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I'm not sure that I completely understand the last part of the reading.  The
application of generating functions to the fruit problem is kind of confusing. 
I hope to go over it in class on friday.

Thanks,
Praveen Pamidimukkala

From kevin08@MIT.EDU Fri Nov 18 05:07:50 2005
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Hi,

I thought this reading was fairly straightforward and understood all 
of it. With that being said though, if Prof. Meyers could go over 
proving the bookkeeper rule (Section 4.3) in class, that would be instructive.

Thanks,
Kevin


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HI,

I had problems with coming up with the generating functions.

Is there anything else that can be used apart from the sums?

 

Anton.

 


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From arup@MIT.EDU Fri Nov 18 06:09:14 2005
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Section 4.2:
I think this should be covered in lecture since I wasn't sure about the 
choosing with repetition stuff.

|Arup|

From lkini@MIT.EDU Fri Nov 18 08:36:40 2005
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Hi Hanson,

The most surprising thing I read in the lecture notes was how to use 
generating functions to count. This is immensely helpful for counting 
the larger, difficult counting questions such as that in the last few 
pages of the lecture notes.

Lohith

From kromer@MIT.EDU Fri Nov 18 09:17:06 2005
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p. 10 "Informally, the only restrictions are that (1) the order in which items
are selected is disregarded and (2) restrictions on the selection of items from
sets A and B also apply in selecting items from A union B".

Is there any way to use generating functions to count when the order in which
items are selected matters?

Katherine

From ajshafer@MIT.EDU Fri Nov 18 09:21:56 2005
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Generating functions still seem really hard.  What does G(x) represent when 
we plug in a value for x?  Is it only useful here for trying to derive the 
f(n) function?

-Andrew
----------------------------
Illegitmitatum Non Carborundum Est
Andrew Shafer, MIT Blog
http://shaferandrew.blogspot.com
Si hoc legere scis numium eruditionis habes.
----------------------------


From jstritar@MIT.EDU Fri Nov 18 09:34:01 2005
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3.1 Finding a Generating Function

I thought this section was pretty interesting because I remember using a 
method involving matrices to obtain generating functions. I don't 
remember the exact procedure, but I am wondering if the two are related 
or simply different techniques for the same solution.


Jon Stritar

From meyer@imap.theory.csail.mit.edu  Fri Nov 18 09:37:49 2005
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take another look at week 10 Monday, class problem 3, which we thought 
spelled out the Bookkeeper rule pretty thoroughly.  Did you get a chance 
to do that problem in class?

regards, A.

Kevin Wang wrote:

> Hi,
>
> I thought this reading was fairly straightforward and understood all 
> of it. With that being said though, if Prof. Meyers could go over 
> proving the bookkeeper rule (Section 4.3) in class, that would be 
> instructive.
>
> Thanks,
> Kevin
>


From letrec@MIT.EDU Fri Nov 18 10:04:53 2005
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From: Alton Torregano <letrec@MIT.EDU>
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This reading packet is one of the best written and most comprehensive. I
have been introduced to the concept of generating functions before in
Combinatorics, and the subject was never explained as succinctly and
straightforward.

In particular, the section on derivatives and Taylor's theorem are
useful in gaining a slightly more in-depth and abstract understanding of
the inner workings of the Generating Functions at this level of survey. 
Are we to cover exponential generating functions at all in this course?

~ Alton

From kktyan@MIT.EDU Fri Nov 18 10:06:44 2005
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Date: Fri, 18 Nov 2005 10:06:40 -0500
From: Karena Tyan <kktyan@MIT.EDU>
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Subject: [Hanson] Reading Comments 11
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I mostly understood everything in this reading, so I'd just like to note that
the method to discover the Fibonacci sequence's generating function was quite
fun and exciting.  Overall, I'd like to submit the suggestion (for the reading,
as Professor Meyer explained it better in class) how exactly to find the complex
roots (alpha_1 and alpha_2, per se) and giving a review on partial fractions for
those of us who don't quite remember how to do them.

- Karena

-- 
410 Memorial Drive
Cambridge, MA 02139
(585)957-5923

From meyer@imap.theory.csail.mit.edu  Fri Nov 18 10:19:12 2005
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Date: Fri, 18 Nov 2005 10:19:14 -0500
From: "Prof. Albert R. Meyer" <meyer@csail.mit.edu>
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Actually Prof. Meyer was a little confused too :-) and when he thinks 
about it, he thinks /rounding/ (either way) is wrong.  The right thing 
to  do is choose the /nearest/ integer:

The Fib formula looks like Term1 - Term2 where the absolute value of 
Term2 < 1/2.  But since Term1 - Term2 is an integer (a Fib. number), it 
follows that the integer nearest to Term1 must be the final answer.  So 
we can ignore Term2 and just pick the closest integer to Term1.

Make sense now?

Regards, A.

Scot Frank wrote:

> Hi,
>
> The biggest question I had from this week of notes was the reason 
> that, while Prof. Meyer was on the board, he choose to round up on the 
> sum, after erasing the second term. I was confused, and could not 
> understand the conflicting explanations.
>
> Thanks,
>
> Scot



From meyer@imap.theory.csail.mit.edu  Fri Nov 18 10:29:11 2005
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Date: Fri, 18 Nov 2005 10:29:13 -0500
From: "Prof. Albert R. Meyer" <meyer@csail.mit.edu>
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Even before the "amazing" fruit problem :-) ,  we used generating 
functions to find a formula, first for Fib nums, and then for another 
recursively specified sequence in the class prob.  Didn't that 
illustrate the usefulness of gen funcs?  .. or did you already know of 
some alternative way to find closed formulas for such sequences?

regards, A.

Tony Ng wrote:

> I found the fruit counting problem ("An 'Impossible' Counting Problem, 
> pg12) to be very interesting. I didn't understand the need for 
> generating functions during the last lecture and it seemed excessively 
> complicated to do, however, now I understand why it is necessary 
> because it may be the only way we know how to solve some problems.
>
> - Tony
>


From meyer@imap.theory.csail.mit.edu  Fri Nov 18 10:32:44 2005
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Date: Fri, 18 Nov 2005 10:32:46 -0500
From: "Prof. Albert R. Meyer" <meyer@csail.mit.edu>
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Neil M Dowgun wrote:

>So I realized that the example where we found the coefficiants for the Fibonacci
>sequence doesn't really provide a clean fomulaic way to find coefficiants given
>a closed form. 
>
Actually, it does.  What makes you think not?

>I guess there's just always going to be some dirty math to do.
>Anyway, I was wondering if using the Taylor expansion to find the coefficiants
>was always a viable option. I mean, I realize that it would be a real pain to
>take the derivative of an expression say, thirty times, but it just seems like
>a more surefire way to get the answer if it really is foolproof.
>  
>
both Taylor expansion and the partial fraction method are useful, though 
there are problems where neither one works.

>Thanks, Neil
>  
>
regards, A.


From meyer@imap.theory.csail.mit.edu  Fri Nov 18 10:35:44 2005
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glad you liked the Notes.  No, we won't cover exponential gen funcs: 
ordinary ones are enough for a broad intro course like 6.042.
regards, A.

Alton Torregano wrote:

>This reading packet is one of the best written and most comprehensive. I
>have been introduced to the concept of generating functions before in
>Combinatorics, and the subject was never explained as succinctly and
>straightforward.
>
>In particular, the section on derivatives and Taylor's theorem are
>useful in gaining a slightly more in-depth and abstract understanding of
>the inner workings of the Generating Functions at this level of survey. 
>Are we to cover exponential generating functions at all in this course?
>
>~ Alton
>  
>


From meyer@imap.theory.csail.mit.edu  Fri Nov 18 00:03:39 2005
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To: 6042-probs@theory.lcs.mit.edu
From: Clayton Sims <ctsims@MIT.EDU>
Subject: [6.042] Reading Comments Lecture Notes 8 - Clayton Sims
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I felt like the material on page 7 was the most challenging. Generalizing 
the method of turning a sequence of coefficients into different Function 
Generators that are easily converted into a representation of the function 
is something that I feel I would benefit from practicing. Additionally, a 
quick review (5 mins.) of partial fractions in class would be nice, 
although not entirely necessary.

-Clayton Sims  


From nedzel@MIT.EDU Fri Nov 18 10:42:49 2005
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From: "David A. Nedzel" <nedzel@MIT.EDU>
To: <6042-probs@theory.csail.mit.edu>
Subject: [Sayan] Week 11 Comments
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I didn't understand the notes on using generating functions to count.

- David


From vbrobbey@MIT.EDU Fri Nov 18 10:49:00 2005
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From: "Valery Kwasi Brobbey" <vbrobbey@MIT.EDU>
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Generating functions look very much like Taylor series expansion. I =
actually used some of my knowledge of Taylor series to do some of the TP =
problems. The lecture should probably have discussed the relationship =
between the two. It's interesting how you can use generating functions =
to count.=20
I like the lecture on Wednesday. I found that more interactive than =
lectures with slides. However, I think the lecture was quite long. It =
would have helped if we have the problem solving session right in the =
middle.

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<DIV><FONT face=3DArial size=3D2>Generating functions look very much =
like Taylor=20
series expansion. I actually used some of my knowledge of Taylor series =
to do=20
some of the TP problems. The lecture should probably have discussed the=20
relationship between the two. It's interesting how you can use =
generating=20
functions to count. </FONT></DIV>
<DIV><FONT face=3DArial size=3D2>I like the lecture on Wednesday. I =
found that more=20
interactive than lectures with slides. However, I think the lecture was =
quite=20
long. It would have helped if we have the problem solving session right =
in the=20
middle.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV></BODY></HTML>

------=_NextPart_000_0009_01C5EC2D.ABC42890--


From hmzhou@blackbird.csail.mit.edu Fri Nov 18 11:58:28 2005
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From: Hanson Zhou <hmzhou@theory.csail.mit.edu>
To: Elizabeth Reid <ereid@MIT.EDU>
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Subject: Re: [Hanson] Reading Comments
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Most good texts on discrete mathematics should have sections on generating
functions and exercises.  Look around or ask Prof. Meyer for pointers.

-Hanson

On Thu, 17 Nov 2005, Elizabeth Reid wrote:

> The reading makes sense, but I feel like I wouldn't be able to do it
> myself. Is there anywhere I can go that would help me practice this
> stuff?
>

From hmzhou@blackbird.csail.mit.edu Fri Nov 18 12:01:20 2005
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Well there is a formula for the geometric sum:
1+x+x^+...+x^n = (x^(n+1)-1)/(x-1).  You can review from the course
lecture notes or from just about any discrete math text.

-Hanson

On Thu, 17 Nov 2005, Daniel A Gutierrez wrote:

> These all make logical sense, but I was wondering if by chance you have
> a list or know where to find one so that I can review geometric summing?
> Thanks, Daniel
>

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From: Hanson Zhou <hmzhou@theory.csail.mit.edu>
To: Barry Revzin <brevzin@MIT.EDU>
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Well...and I'm purely speculating here:

Integral f(y)g(n-y)dy, with f(y) = a_y x^y, and g(n-y) = b_(n-y) x^(n-y),
would seem to be the continuous analogue of the discrete convolution we
talk about here.  Ask Prof. Meyer to make sure.

-Hanson

On Thu, 17 Nov 2005, Barry Revzin wrote:

> For the convolution rule, I think it's really confusing that it's called
> the convolution rule. It really has nothing to do with the idea of a
> convolution from analysis, does it ? I'm really used to the convolution of
> f and g (f*g) being the integral of f(y)g(x-y)dy. Why IS it called that
> then ? Why not like... disjoint union rule ? Or Countable Additivity, which
> would be consistent with measure theory...
>
> Barry
>
>

From lana@MIT.EDU Fri Nov 18 12:23:57 2005
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I was wondering if it is possible to have a function that can not be expressed
through a generating function or if 2 generating functions can result in the
same expression
Lana

From ksindi@MIT.EDU Fri Nov 18 13:47:01 2005
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I found it conceptiually confusing of how one treats generating functions only
formally without worry of convergence. However Im getting the hang of it.

From dshin@MIT.EDU Fri Nov 18 17:16:28 2005
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Interesting question!  I spent a bit of time thinking about whether 
generating functions could be used in this type of situation, and I'm 
leaning towards an answer of "no".  The reason is that multiplication is 
commutative:  x^a*x^b = x^{ab}.  It doesn't seem possible to encode 
order-dependent information with this order-independent operation 
(multiplication).  Maybe there exists some crazy method using 
non-commutative algebraic structures, but I am highly doubtful. 

DS

Katherine A Romer wrote:

>p. 10 "Informally, the only restrictions are that (1) the order in which items
>are selected is disregarded and (2) restrictions on the selection of items from
>sets A and B also apply in selecting items from A union B".
>
>Is there any way to use generating functions to count when the order in which
>items are selected matters?
>
>Katherine
>  
>

From dshin@MIT.EDU Fri Nov 18 17:40:52 2005
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Alexander Bakst wrote:

> What kind of situations would require us to use values for x? In the 
> readings it said that they are unusual or rare, I believe.

There are a few really ingenious applications of generating functions 
that involve using values for x.  Typically, the plugging-in-for-x step 
is to reach some sort of contradiction:  you have two generating 
functions that are supposed to be equal, but they don't agree at a 
particular value of x. 

Unfortunately, most examples are fairly complex (pun somewhat 
intended).  I tried to think of a simple example, but this is the best I 
could come up with:

-----

*Problem*:  Suppose we have a collection of infinite arithmetic 
sequences (like {1,4,7,10,...} or {10,12,14,16,...}) where each sequence 
has difference term larger than 1.  Suppose that every natural number 
occurs EXACTLY once among the sequences.  Prove that two of the 
sequences have the same difference term. 

*Solution*:  Suppose not.  Let the sequences be S_i = {a_i + n*b_i} for 
i=1,2,...  Then, since all the difference terms are distinct, the values 
of b_i are all distinct.  Assume without loss of generality that b_1 is 
the smallest of the b_i. 

Sequence S_i can be represented by a sequence of 0's and 1's, which can 
be represented by the generating function x^{a_i}/(1-x^{b_i}).  The 
natural numbers can be represented by a sequence of all 1's, which can 
be represented by the generating function 1/(1-x).  Since each natural 
number occurs exactly once among the sequences, we have that

1/(1-x) = SUM_i (x^{a_i}/(1-x^{b_i}))

Let x be close to a (b_1)-th root of unity.  I won't give a rigorous 
proof, but it's true that the right hand side blows up while the left 
hand side doesn't.  A contradiction. 

-----

Hopefully you find that somewhat cool. 

DS

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Alexander Bakst wrote:
<blockquote cite="mid437D988C.8020808@mit.edu" type="cite">What kind of
situations would require us to use values for x? In the readings it
said that they are unusual or rare, I believe.
  <br>
</blockquote>
There are a few really ingenious applications of generating functions
that involve using values for x.&nbsp; Typically, the plugging-in-for-x step
is to reach some sort of contradiction:&nbsp; you have two generating
functions that are supposed to be equal, but they don't agree at a
particular value of x.&nbsp; <br>
<br>
Unfortunately, most examples are fairly complex (pun somewhat
intended).&nbsp; I tried to think of a simple example, but this is the best
I could come up with:<br>
<br>
-----<br>
<br>
<b>Problem</b>:&nbsp; Suppose we have a collection of infinite arithmetic
sequences (like {1,4,7,10,...} or {10,12,14,16,...}) where each
sequence has difference term larger than 1.&nbsp; Suppose that every natural
number occurs EXACTLY once among the sequences.&nbsp; Prove that two of the
sequences have the same difference term.&nbsp; <br>
<br>
<b>Solution</b>:&nbsp; Suppose not.&nbsp; Let the sequences be S_i = {a_i +
n*b_i} for i=1,2,...&nbsp; Then, since all the difference terms are
distinct, the values of b_i are all distinct.&nbsp; Assume without loss of
generality that b_1 is the smallest of the b_i.&nbsp; <br>
<br>
Sequence S_i can be represented by a sequence of 0's and 1's, which can
be represented by the generating function x^{a_i}/(1-x^{b_i}).&nbsp; The
natural numbers can be represented by a sequence of all 1's, which can
be represented by the generating function 1/(1-x).&nbsp; Since each natural
number occurs exactly once among the sequences, we have that<br>
<br>
1/(1-x) = SUM_i (x^{a_i}/(1-x^{b_i}))<br>
<br>
Let x be close to a (b_1)-th root of unity.&nbsp; I won't give a rigorous
proof, but it's true that the right hand side blows up while the left
hand side doesn't.&nbsp; A contradiction.&nbsp; <br>
<br>
-----<br>
<br>
Hopefully you find that somewhat cool.&nbsp; <br>
<br>
DS
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From dshin@MIT.EDU Fri Nov 18 17:43:52 2005
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Glad you're liking this stuff. Do you see why the oranges thing works now?

DS

Harrison King Hall wrote:

> This is really a cool way to do counting problems, and really
> intuitive since you only have to worry about how to select each class
> of element for n elements independent of all the others. My only issue
> is that I don't always see the generating function . Like in the
> "Impossible" counting problem, I didn't see that the number of oranges
> was O(x) = 1 + x + x ^2 + x^3 + x^4 = (1 $B!](B x^5)/(1 $B!](B x) and I don't
> really know how you got there. That is probably something that comes
> with time however. This is really slick.
> -Harrison
>

From dshin@MIT.EDU Fri Nov 18 17:47:59 2005
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Jason Juang wrote:

> In what cases might we actually evaluate a generating function?

There are a few really ingenious applications of generating functions 
that involve using values for x.  Typically, the plugging-in-for-x step 
is to reach some sort of contradiction:  you have two generating 
functions that are supposed to be equal, but they don't agree at a 
particular value of x. 

Unfortunately, most examples are fairly complex (pun somewhat 
intended).  I tried to think of a simple example, but this is the best I 
could come up with:

-----

*Problem*:  Suppose we have a collection of infinite arithmetic 
sequences (like {1,4,7,10,...} or {10,12,14,16,...}) where each sequence 
has difference term larger than 1.  Suppose that every natural number 
occurs EXACTLY once among the sequences.  Prove that two of the 
sequences have the same difference term. 

*Solution*:  Suppose not.  Let the sequences be S_i = {a_i + n*b_i} for 
i=1,2,...  Then, since all the difference terms are distinct, the values 
of b_i are all distinct.  Assume without loss of generality that b_1 is 
the smallest of the b_i. 

Sequence S_i can be represented by a sequence of 0's and 1's, which can 
be represented by the generating function x^{a_i}/(1-x^{b_i}).  The 
natural numbers can be represented by a sequence of all 1's, which can 
be represented by the generating function 1/(1-x).  Since each natural 
number occurs exactly once among the sequences, we have that

1/(1-x) = SUM_i (x^{a_i}/(1-x^{b_i}))

Let x be close to a (b_1)-th complex root of unity.  I won't give a 
rigorous proof, but it's true that the right hand side blows up while 
the left hand side doesn't.  (It relies on the fact that b_1 is distinct 
from all the other b_i...)  A contradiction. 

-----

Hopefully you find that somewhat cool. 

If you want a really cool challenge, try your hand at this:

Problem:  Suppose you have a tiling of an 8x8 chessboard with 21 
trinominos (3x1 rectangles) and one monomino (1x1 rectangle).  Find all 
possible positions for the monomino.

Believe it or not, it can be done with (finite) generating functions!  
It involves cubic roots of unity...

DS

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Jason Juang wrote:
<blockquote cite="mid437BA8B1.2090306@mit.edu" type="cite">In what
cases might we actually evaluate a generating function?
  <br>
</blockquote>
There are a few really ingenious applications of generating functions
that involve using values for x.  Typically, the plugging-in-for-x step
is to reach some sort of contradiction:  you have two generating
functions that are supposed to be equal, but they don't agree at a
particular value of x.  <br>
<br>
Unfortunately, most examples are fairly complex (pun somewhat
intended).  I tried to think of a simple example, but this is the best
I could come up with:<br>
<br>
-----<br>
<br>
<b>Problem</b>:  Suppose we have a collection of infinite arithmetic
sequences (like {1,4,7,10,...} or {10,12,14,16,...}) where each
sequence has difference term larger than 1.  Suppose that every natural
number occurs EXACTLY once among the sequences.  Prove that two of the
sequences have the same difference term.  <br>
<br>
<b>Solution</b>:  Suppose not.  Let the sequences be S_i = {a_i +
n*b_i} for i=1,2,...  Then, since all the difference terms are
distinct, the values of b_i are all distinct.  Assume without loss of
generality that b_1 is the smallest of the b_i.  <br>
<br>
Sequence S_i can be represented by a sequence of 0's and 1's, which can
be represented by the generating function x^{a_i}/(1-x^{b_i}).  The
natural numbers can be represented by a sequence of all 1's, which can
be represented by the generating function 1/(1-x).  Since each natural
number occurs exactly once among the sequences, we have that<br>
<br>
1/(1-x) = SUM_i (x^{a_i}/(1-x^{b_i}))<br>
<br>
Let x be close to a (b_1)-th complex root of unity.  I won't give a
rigorous
proof, but it's true that the right hand side blows up while the left
hand side doesn't.  (It relies on the fact that b_1 is distinct from
all the other b_i...)  A contradiction.  <br>
<br>
-----<br>
<br>
Hopefully you find that somewhat cool.  <br>
<br>
If you want a really cool challenge, try your hand at this:<br>
<br>
Problem:  Suppose you have a tiling of an 8x8 chessboard with 21
trinominos (3x1 rectangles) and one monomino (1x1 rectangle).  Find all
possible positions for the monomino. <br>
<br>
Believe it or not, it can be done with (finite) generating functions! 
It involves cubic roots of unity...<br>
<br>
DS
</body>
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Did class help with convolution?

DS

Chris Yang wrote:

> I don’t really understand how the Convolution rule works. Will we be 
> covering it in class on Friday? Also, the apples/oranges thing at the 
> end is pretty cool.
>
> Thanks,
>
> Chris Yang
>

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Taylor expansion is sure to work.  If your function is a quotient of 
polynomials, then partial fractions often can do the trick. 

DS

Sergio Bacallado wrote:

> I thought it was amazing how useful generating functions can be to 
> solve counting problems. I would like to know if there is always an 
> easy way to find a closed form for the nth term of a sequence, if you 
> know the value of the generating function. That is, other than 
> differentiating the function to get the Taylor expansion term.
>
> Sergio Bacallado
> Group A
>

From dshin@MIT.EDU Fri Nov 18 17:53:34 2005
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It would have been great if you had come to lecture today - we covered 
much of these topics in class.  I would recommend coming to office hours 
if you are still having difficulty. 

DS

cbossard@MIT.EDU wrote:

>I find this to be an interesting topic although I have trouble
>reproducing and creating these functions on my own they generally
>make sense to me when I see them explained.  In section 2.5 on
>products I am not seeing/understanding how the terms of the diagonal
>were collected so I don't see why it's not aobo + a1b1 instead of
>aobn etc.  Again with although the convolution rule makes some sense
>to me in section 4.2 I dont see how they are getting the numbers from
>the formula that was created.  I can see that each is (1+x) and they
>should be multiplied.  Intuively I understand that there's 1 way to
>choose 0 elements, 2 ways to choose 1, etc but I do not see how this
>information can be attained from the 1+2x+x^2 equation.
>
>Cynthia
>  
>

From dshin@MIT.EDU Fri Nov 18 17:54:01 2005
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did class help?

DS

Joshua Jen C. Monzon wrote:

> I'm a little confused on the convolution rule. Perhaps this would be 
> explained a little more in class. Also, I didn't really understand how 
> they obtained the answer in the last problem of the tutorial. Hope 
> tomorrow's lecture clears this up.
>
> Josh
>
>
>
> Joshua Jen C. Monzon
> Massachusetts Institute of Technology
> Electrical Engineering with Computer Science
> jjmonzon@mit.edu   617-803-7497
>

From meyer@imap.theory.csail.mit.edu  Fri Nov 18 17:40:53 2005
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Alexander Bakst wrote:

> What kind of situations would require us to use values for x? In the 
> readings it said that they are unusual or rare, I believe.

There are a few really ingenious applications of generating functions 
that involve using values for x.  Typically, the plugging-in-for-x step 
is to reach some sort of contradiction:  you have two generating 
functions that are supposed to be equal, but they don't agree at a 
particular value of x. 

Unfortunately, most examples are fairly complex (pun somewhat 
intended).  I tried to think of a simple example, but this is the best I 
could come up with:

-----

*Problem*:  Suppose we have a collection of infinite arithmetic 
sequences (like {1,4,7,10,...} or {10,12,14,16,...}) where each sequence 
has difference term larger than 1.  Suppose that every natural number 
occurs EXACTLY once among the sequences.  Prove that two of the 
sequences have the same difference term. 

*Solution*:  Suppose not.  Let the sequences be S_i = {a_i + n*b_i} for 
i=1,2,...  Then, since all the difference terms are distinct, the values 
of b_i are all distinct.  Assume without loss of generality that b_1 is 
the smallest of the b_i. 

Sequence S_i can be represented by a sequence of 0's and 1's, which can 
be represented by the generating function x^{a_i}/(1-x^{b_i}).  The 
natural numbers can be represented by a sequence of all 1's, which can 
be represented by the generating function 1/(1-x).  Since each natural 
number occurs exactly once among the sequences, we have that

1/(1-x) = SUM_i (x^{a_i}/(1-x^{b_i}))

Let x be close to a (b_1)-th root of unity.  I won't give a rigorous 
proof, but it's true that the right hand side blows up while the left 
hand side doesn't.  A contradiction. 

-----

Hopefully you find that somewhat cool. 

DS

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Alexander Bakst wrote:
<blockquote cite="mid437D988C.8020808@mit.edu" type="cite">What kind of
situations would require us to use values for x? In the readings it
said that they are unusual or rare, I believe.
  <br>
</blockquote>
There are a few really ingenious applications of generating functions
that involve using values for x.&nbsp; Typically, the plugging-in-for-x step
is to reach some sort of contradiction:&nbsp; you have two generating
functions that are supposed to be equal, but they don't agree at a
particular value of x.&nbsp; <br>
<br>
Unfortunately, most examples are fairly complex (pun somewhat
intended).&nbsp; I tried to think of a simple example, but this is the best
I could come up with:<br>
<br>
-----<br>
<br>
<b>Problem</b>:&nbsp; Suppose we have a collection of infinite arithmetic
sequences (like {1,4,7,10,...} or {10,12,14,16,...}) where each
sequence has difference term larger than 1.&nbsp; Suppose that every natural
number occurs EXACTLY once among the sequences.&nbsp; Prove that two of the
sequences have the same difference term.&nbsp; <br>
<br>
<b>Solution</b>:&nbsp; Suppose not.&nbsp; Let the sequences be S_i = {a_i +
n*b_i} for i=1,2,...&nbsp; Then, since all the difference terms are
distinct, the values of b_i are all distinct.&nbsp; Assume without loss of
generality that b_1 is the smallest of the b_i.&nbsp; <br>
<br>
Sequence S_i can be represented by a sequence of 0's and 1's, which can
be represented by the generating function x^{a_i}/(1-x^{b_i}).&nbsp; The
natural numbers can be represented by a sequence of all 1's, which can
be represented by the generating function 1/(1-x).&nbsp; Since each natural
number occurs exactly once among the sequences, we have that<br>
<br>
1/(1-x) = SUM_i (x^{a_i}/(1-x^{b_i}))<br>
<br>
Let x be close to a (b_1)-th root of unity.&nbsp; I won't give a rigorous
proof, but it's true that the right hand side blows up while the left
hand side doesn't.&nbsp; A contradiction.&nbsp; <br>
<br>
-----<br>
<br>
Hopefully you find that somewhat cool.&nbsp; <br>
<br>
DS
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Subject: Re: [David] Lecture Notes 11 Comments
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>> The natural numbers can be represented by a sequence of all 1's,
which can be
>> represented by the generating function 1/(1-x)

Aren't the natural numbers represented by x times the derivative of 1/(1-x)?

-Jelani


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With this application, I'm using using coefficients of 1 and 0 to represent a
subset of the natural numbers.  For example, the subset {0,2,3} corresponds to
the sequence <1,0,1,1,0,0,0...> and thus to the function 1+x^2+x^3.

Sorry if it wasn't clear.

DS

Quoting Jelani Nelson <minilek@MIT.EDU>:

>>> The natural numbers can be represented by a sequence of all 1's,
> which can be
>>> represented by the generating function 1/(1-x)
>
> Aren't the natural numbers represented by x times the derivative of 1/(1-x)?
>
> -Jelani
>



From meyer@imap.theory.csail.mit.edu  Sat Nov 19 17:23:43 2005
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Subject: Re: [David] Lecture Notes 11 Comments
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With this application, I'm using using coefficients of 1 and 0 to represent a
subset of the natural numbers.  For example, the subset {0,2,3} corresponds to
the sequence <1,0,1,1,0,0,0...> and thus to the function 1+x^2+x^3.

Sorry if it wasn't clear.

DS

Quoting Jelani Nelson <minilek@MIT.EDU>:

>>> The natural numbers can be represented by a sequence of all 1's,
> which can be
>>> represented by the generating function 1/(1-x)
>
> Aren't the natural numbers represented by x times the derivative of 1/(1-x)?
>
> -Jelani
>



From meyer@imap.theory.csail.mit.edu  Sat Nov 19 17:20:34 2005
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Subject: Re: [David] Lecture Notes 11 Comments
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>> The natural numbers can be represented by a sequence of all 1's,
which can be
>> represented by the generating function 1/(1-x)

Aren't the natural numbers represented by x times the derivative of 1/(1-x)?

-Jelani


