UCBMath115 Google Group

Jessica Simpson incredible

vlizard0...@gmail.com (vlizard0179) — Sat, 06 Sep 2008 05:00:26 UT

Jessica Simpson, Live performance, however it's not the music that's
the star here!
[link]

a weird summer program

kri...@gmail.com (Kenneth A. Ribet) — Mon, 27 Nov 2006 23:42:22 UT

Hi All,
I got the message that follows from Paul Mantica. Two of my Math 54
students from last fall applied to the program and got accepted. I
saw one of them recently on campus. He told me that it was a good
thing.
By the way, we will be doing course evaluations in Math 115 on
Friday, December 1. Be sure to be in class that day!!

Re: problem 5.3a

zenka...@gmail.com (zenkalia@gmail.com) — Mon, 06 Nov 2006 19:24:43 UT

disregard this... all of it... i don't know how to use my ti-83

Re: problem 5.3a

garyd...@gmail.com (John Brooks-Jung) — Mon, 06 Nov 2006 19:03:41 UT

Try plugging it back into the recurrence relation, or better yet, find
a recurence relation that only uses the previous term to define the
next. Hope that helps
John

Re: problem 5.3a

zenka...@gmail.com (zenkalia@gmail.com) — Mon, 06 Nov 2006 19:01:00 UT

oops i'm an idiot... 0 1 2 4 8 is... 2^n, ish... it breaks at zero...
i don't know what i was thinking with n^2
so it should be...
a_0 = 3
a_n = 2^2^2^(n-1)+1 for n>0

Re: problem 5.3a

zenka...@gmail.com (zenkalia) — Mon, 06 Nov 2006 18:12:10 UT

well, if you take it just one step further, then you won't be getting
the next fermat number but you will get a fermat number;
a_0 = 3 = 2^2^0+1
a_1 = 5 = 2^2^1+1
a_2 = 17 = 2^2^2+1
a_3 = 257 = 2^2^4+1
a_4 = 65537 = 2^2^8+1

Re: problem 5.3a

kri...@gmail.com (Kenneth A. Ribet) — Mon, 06 Nov 2006 15:22:42 UT

Seems OK to me. The first element of the sequence is 3. The next is
3+2 = 5. The next is 3.5 + 2 = 17. These are the first few Fermat
numbers. The task is to show that you keep getting the Fermat
numbers 2^{2^n}+1, and I think that you do.
It's always striking to me that students are totally stymied when

problem 5.3a

zenka...@gmail.com (zenkalia@gmail.com) — Mon, 06 Nov 2006 06:37:24 UT

so the problem is just blatantly wrong, right?
it should read a_n = 2^2^n^2+1.

Re: Problem 5.6

kri...@gmail.com (Kenneth A. Ribet) — Tue, 31 Oct 2006 05:31:26 UT

I think that the authors are asking you to show that the set of
numbers that they ask about is equal to the union of finitely many
congruence classes mod some M. This means that if x is in S and y is
congruent to x mod M, then y is also in S.
Ken R

Problem 5.6

garyd...@gmail.com (garydoho@gmail.com) — Tue, 31 Oct 2006 03:29:54 UT

So I've read this problem a few times and have decided that I must be
reading it incorrectly, otherwise it is unbelievably easy. The way I
read it, the problem boils down to asking whether there are finite
congruence classes modulo a finite number. What am I doing wrong?
Thanks,
John

RIPS

kri...@gmail.com (Kenneth A. Ribet) — Tue, 31 Oct 2006 02:32:19 UT

Hi All,
Please look at [link] and see
whether this program might be the right REU-like activity for you
this coming summer. It has to do with real-world problems that help
industrial partners. In a recent summer (2005, I think), students
modeled the flow of ketchup for an animated film. It's at UCLA,

Re: problem 2 and some wikipedia junk

kri...@gmail.com (Kenneth A. Ribet) — Thu, 26 Oct 2006 13:33:14 UT

It might be fruitful to think of the situation in terms of two
functions. Let Q be the set of rational numbers. Let D be the set
of decimal numbers that are eventually periodic. There's a function
g:D -> Q that takes each decimal to the rational number that it
represents. You calculate it by the formula for the sum of a

Re: problem 2 and some wikipedia junk

zenka...@gmail.com (zenkalia) — Thu, 26 Oct 2006 06:10:47 UT

of course, but do they all have fractional representations other than
.999...?

Re: problem 2 and some wikipedia junk

zenka...@gmail.com (zenkalia@gmail.com) — Thu, 26 Oct 2006 04:53:22 UT

are there other repeating decimals that don't have fractional
representations? this might be high school talking but i thought that
they all did...

Re: problem 2 and some wikipedia junk

kri...@gmail.com (Kenneth A. Ribet) — Thu, 26 Oct 2006 03:17:35 UT

Not true that 0.999.... is irrational. It's 1. The problem states
that you will not encounter the sequence 99999... in converting a
fraction into a decimal by long division.
Ken R