math 201

[CrystyB]

just so it's started. I don't have any ideas yet - and i just recently read the topic.

First of all, this will not be a one-man course if i can help it. I would want mith and araya to be able to run this. mith already can, and if araya agrees to that i would give him rights to edit my posts. Or we would use a common username if he doesn't want to know my pwd.

For starters, i could only think of posting a few non-trivial problems and then i'll answer any question that comes up.

(a little bit of Algebra for kiddies is here)
Find the GCD of x⁷+6x⁶+4x⁴+5x+9 and 8x²+16x+64.

Analysis
Compute Integral_from_0_to_a of 1/((1-x)^a). i.e.


Arithmetics
Find a relation R based on a commonly used mathematical entity over the natural numbers such as "1 R n" and "n R 0" for any natural number n. That basically means that 1 is the smallest nat and 0 is the greatest, with all other nat's in between them!!

Combinatorics
Find the sum of the finite series S(2^n-kC_n^k). (k goes from 0 to n)

Game Theory

Geometry
Prove that in a tetrahedron all the four altitudes (*) intersect. i.e. ($)H so that H is on each of the 4 altitudes.
(*) i wrote 'heights' and explained: "sorry but i don't know the english term for the segment that is on a line perpendicular to a given face and is bounded by (the point of intersection with) that face and the opposing vertex."

Probabilities/Statistics
An urn contains 5 red balls, 3 white ones and 2 black ones.
a) Extracting 2 balls at the same time, what is the probability that one is white and the other black?
b) Extracting 2 balls at the same time, what is the probability that they aren't both black?
c) Find all n's for which the probability of "out of n balls extracted at the same time, at least one is black" is at least 8/15.

Trigonometry
Knowing that cos²x+sin²x=1 and cos(x+y)=cosxcosy-sinxsiny, deduce any other basic-trigonometry formula. You are of course allowed to use the unit circle and everything that is obvious on it.

Btw what's with all the 101's?

[This message has been edited by CrystyB (edited 02-04-2003 09:10 PM).]

[Hitchhiker]

101 is traditionally the course number of a beginner's or introductory course at American universities. Second-year courses generally have a number starting with "2", etc.

[Coyote]

I'd really like to learn the basics of trignometry, if anyone is interested in teaching it.

[CrystyB]

ok i will detail the solving of the Trig problem here, so *** SPOILERS ALERT ***

Or... WAIT! Would you like me to explain the using of the trig. circle too?

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Naturally, the sinx and cosx being defined as shown in the picture, we have:

sin(-x)= − sinx, cos(-x)=cosx. (sin is 'odd', cos is 'even')
So cos(x-y)=cosxcosy+sinxsiny.

sin(p/2-x)=cos(x), cos(p/2-x)=sin(x).
Then sub'ing this we would get sin(x+y)=cos(p/2-(x+y))=cos((p/2-x)-y)=cos(p/2-x)cosy+sin(p/2-x)siny=sinxcosy+cosxsiny. And sin(x-y)=sinxcosy-cosxsiny.
Which in turn gives sin2x=sin(x+x)=2sinxcosx. cos2x could've been deduced from the relation given in the hypothesis: cos2x=cos(x+x)=cosxcosx-sinxsinx=cos²x-sin²x. Two alternates could be reached using cos²x+sin²x=1 : cos2x=(1-sin²x)-sin²x=1-2sin²x, cos2x=cos²x-(1-cos²x)=2cos²x - 1.

Summary:

code:

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cos(x+y)=cosxcosy-sinxsiny


cos(x-y)=cosxcosy+sinxsiny


sin(x+y)=sinxcosy+cosxsiny


sin(x-y)=sinxcosy-cosxsiny

---

Let's look at this the other way around.
Say x+y=a, x-y=b. Then x=(a+b)/2=s, y=(a-b)/2=d.

code:

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cosa=cosscosd-sinssind


cosb=cosscosd+sinssind


sina=sinscosd+cosssind


sinb=sinscosd-cosssind

---

Summing and substracting these, we get:
cosa+cosb=2cosscosd
cosa-cosb= − 2sinssind
sina+sinb=2sinscosd
(sina-sinb=sina+sin(-b)=2sindcoss)

Now for the (c)tg's. tgx=sinx/cosx. So it's 'odd': tg(-x)=sin(-x)/cos(-x)= − sin(x)/cos(x)= − tgx. ctg is just its inverse, cosx/sinx, so again it's 'odd'.
Just for the fun of it, 1+tg²=(cos²x+sin²x)/cos²x=1/cos²x.

tg(x+y)=sin(x+y)/cos(x+y)=(sinxcosy+cosxsiny)/(cosxcosy-sinxsiny). 'Simplifying' with cosxcosy, =(tgx+tgy)/(1-tgxtgy).
tg(2x)=2tgx/(1-tg²x), tg(x-y)=(tgx-tgy)/(1+tgxtgy).
ctg(x+y)=(cosxcosy-sinxsiny)/(sinxcosy+cosxsiny)=(ctgxctgy-1)/(ctgx+ctgy).
ctg(2x)=(ctg²x-1)/(2ctgx), ctg(x-y)= − (ctgxctgy+1)/(ctgx-ctgy).

What else was there? Although i doubt i missed anything.

[This message has been edited by CrystyB (edited 01-07-2003 04:54 AM).]

[jeep]

Coyote, I don't know where you are starting from but this site has a good primer for beginning trig.

-JEEP

[Coyote]

Thanks jeep! That link looks more my speed at the moment. Maybe I'll try Crysty's lessons after I've learned a little more. I don't even know what a cosine is yet. I always thought it was something you did to a bank loan.

[jeep]

No problem. If you have any specific questions about that material, let me know. If you get that down, I can search for a 'next level' site or possibly put one together.

-JEEP

[CrystyB]

Thanks jeep, that site is absolutely perfect for learning basics of trig.

[This message has been edited by CrystyB (edited 01-09-2003 12:54 PM).]

[Hitchhiker]

If I may, professor(s), I'd also like to recommend for Coyote "Triangles, Circles, and Waves (oh my!): An Overview of Trigonometry" at The Math and Physics Help Home Page.

[Bicho the Inhaler]

By the way, Crysty, regarding your geometry problem, the English term is "altitude."

[Beartalon]

(Ack! I have a math degree and I can't remember how to do any of this stuff)

[CrystyB]

(bump to get mith's attention)

[robichelli]

I'm taking a state math test on February 11th, and I'm so screwed because I love math, but this test is notoriously hard. But as my sig says...

------------------
I'm always right.......Except when I'm wrong

[CrystyB]

Absolutely - think positive, and try to disregard any nastiness of the test. If you're going to study just as much for it (which is the logical choice in the situation) , you might as well stop worrying - it ain't of any help!

[This message has been edited by CrystyB (edited 02-09-2003 10:18 PM).]

[borschevsky]

A couple of quick questions - when you say natural numbers in the Arithmetics question, is that {0,1,2,..}? I thought zero was not included in N?

In the combinatorics question, C_n^k means n choose k, as in n!/k!(n-k)!, right?

[CrystyB]

Yes, it is usually included in N (i have *always* seen it included). At least let (N,+) be a monoid!

Of course. Other than with tables, there's no way to do BOTH a subscript and a superscript...

And tables have the drawback of having implied linebreaks before and after (except when nested)! But i could make it look like this:

Find the sum of the finite series

n S k=0

( 2n-k C

k n

).

[This message has been edited by CrystyB (edited 02-11-2003 07:53 PM).]

[borschevsky]

These kind of notational differences always mess with my mind. I always learned it as:
N = {1,2,3,...}
W = {0,1,2,...}
Z = {..,-2,-1,0,1,2,...}

So fine, I went over here, where they recommend Z* for {0,1,2,...}, which would give me even more problems, because we always use Z* for non-zero integers, (and Q* for non-zero rationals, R* for non-zero reals). But whatever.

And combinatorics is the same way; I've always used _nC_k or "n over k" in the big brackets, but a book I'm reading throws ⁿC_k at me, and you use C_n^k. It's a madhouse!

Oh, and the answer for the combinatorics one is 3ⁿ, yeah? And it should work the same for any number as the base of the exponent, so replacing 2^n-k with a^n-k, the sum is (a+1)ⁿ.

[Bicho the Inhaler]

In my high school math classes, N was considered to be {1, 2, 3, ...}. In my college math classes, I'm finding that N = {0, 1, 2, ...}.

In my high school math class (the only one where we talked about set theory), A Ì B meant "A is a proper subset of B," and in my college math classes, it almost universally means "A is a subset of B," where the inclusion is not assumed to be proper.

[mith]

In my day, we had to make due with just a 3 and a 5 in our N.



(I always saw it without 0, for the record)

[VinnyQ]

Show us again why .9999999... = 1

[Chuck]

Back in MY day we didn't have an infinite number of nines, so .999999999... was less than 1.

[Bicho the Inhaler]

Originally posted by VinnyQ:

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Show us again why .9999999... = 1

---

Hope you don't mind, Crysty; I'm stealing this one. Sorry if it's a bit long-winded.

First, you need to decide what you want this symbol, ".9999999...", to mean. Here is one way (the usual way, and arguably the most natural way) to interpret it:

(1) ".9999999..." represents a number, which we'll call x, in the real number system.
(2) Each term in the sequence 0.9, 0.99, 0.999, ... is at least as close to x as the previous number of the sequence.
(3) We can get arbitrarily good approximations to x from the sequence 0.9, 0.99, 0.999, ....

[Condition (1) just tells us what type of thing we're looking at (a real number), and conditions (2) and (3) justify the notation "0.9999999..."]

It's pretty easy to prove that under those conditions, x must be 1, and that the number 1 does indeed satisfy those requirements. Here's the proof:

Suppose x > 1. Then x - 1 = E is positive, and 1 = x - E. Clearly, each element of the sequence 0.9, 0.99, 0.999, ... is less than 1, and 1 < x, and 1 is E units from x. But this means no number from that sequence is less than E units from x, violating condition (3), which says that we should be able to get arbitrarily good approximations to x from the sequence.

Suppose x < 1. Then 1 - x = F is positive, and 1 - F = x. Since F is positive and real, 1/F is also positive and real, so it must be the case that 1/F < 10^k for some whole number k (or else 1/F would be infinitely large, which it isn't, since it's a real number). Take the reciprocals, so F > 1/10^k. Now subtract both sides from 1: 1 - F < 1 - 1/10^k = 0.999...9 (with k nines). But 1 - F = x, so we have x < 0.999...9. But 0.999...9 (with k - 1 nines) < 0.999...99 (with k + 1 nines), so x < 0.999...9 < 0.999...99, and this violates condition (2), which implies that 0.999...99 (with k + 1 nines) must be at least as close to x as 0.999...9 (with k nines), which we just showed it isn't.

This shows that if x is anything, then x = 1. We aren't done yet, though; we have to show that the number 1 actually does fit those requirements.

(1) Is 1 a real number? Of course.
(2) Is each number of the sequence 0.9, 0.99, 0.999, ... at least as close to 1 as the previous number? I think it's pretty clear that the answer is "yes."
(3) Can we get arbitrarily good approximations to x from the sequence 0.9, 0.99, 0.999, ...?
Well, suppose we want an approximation within G units of 1, where G is a positive real number which could be really small. No matter how small G is, though, if G is positive and real, then 1/G is positive, real, and finite, so 1/G < 10^j for some whole number j. Taking the reciprocals, G > 10^j. Subtracting from 1, 1 - G < 1 - 1/10^j = 0.999...9 (with j nines). Rearranging things a bit, we get 1 - 0.999...9 < G, so 0.999...9 is a number in that sequence within G units of 1.

Therefore, 1 is the only choice of x that satisfies those three requirements, QED.


I should note, though, that while the interpretation of the symbol ".9999999..." that I gave is the most common and arguably the most natural, it isn't the only one, and in fact there are ways to subtlely change that interpretation so that 1 is no longer the only choice for x. But that's beyond the scope of this post.

[CrystyB]

no it really isn't. I'm sure some of us would like to hear/read your subtleties.

(and if anyone wants to know, i do have the 0.999999... thread saved - i'm just not reading it yet)

Other justifications: 0.9999999...=sum(9/10ⁿ)_n>0=1;
Using the density propriety of R, there's no other real number between 1 and 0.99999...;
0.99999999...=0.FFFFFFFFF...₍₁₆₎=1

As for the N problem, Peano used zero as a constructor of N, so i can't see the logic in not including it in! Plus, since Z^* means non-zero integers, and Q^* non-zero rationals, and R^* and C^* likewise, why not having an N and an N^* too?

[This message has been edited by CrystyB (edited 02-15-2003 09:16 PM).]

[Agamemnon]

I always lokk at 0.999999999999 etc = 1, as a transplant question.
If I needed a heart transplant and the doctors said;

"Hey Aga, we have one of two hearts to give you. One is 0.99999999999 etc heart and the other is a 1 heart. Which one do you want?"

No matter how much twaddle on math you give me, I'll always go for the 1 heart.

[robichelli]

.99999... can't equal 1, because the limit of 9/10ⁿ is 1 just like the lim .5 is 0

------------------
I'm always right.......Except when I'm wrong

[Bicho the Inhaler]

Crysty -

One way is to tweak condition (1) so that we don't restrict 0.9999... to be a real number, but a member of the more general class of surreal numbers. In the surreal numbers, there are numbers greater than all of the numbers 0.9, 0.99, 0.999, ..., yet still less than 1 (1 - 1/w is the simplest such number, where w is the simplest infinite number, usually viewed as corresponding to the set of natural numbers) (in fact, there are an infinite number of such numbers (in fact, there are so many of them that they can't all fit in any one set!)). In this system, it isn't quite so obvious what "0.9999999..." is supposed to mean. Does it mean 1, or does it refer to one of the in-between numbers? We might be tempted to say "the least upper bound" of the set {0.9, 0.99, ...}, but the surreals don't actually have the least upper bound property, and indeed, that set has no least upper bound in the surreals. So it seems like expanding our view to include the surreal numbers just complicates matters.

I think the surreal numbers have at least as much real-life application (much more, probably) as the whole "0.9999999... = 1?" problem does, so I don't find the introduction of infinite and infinitesimal surreal numbers here to be contrived at all. Ultimately, it doesn't really bother me that there could be multiple interpretations of the symbol "0.9999999..." because it really is just a symbol. It means what it means only because it is convenient for it to mean that. In some contexts we might have certain expectations of a symbol that looks like "0.9999999...", and we might throw those expectations out the window in other contexts.

[Mikko]

Could someone answer this for me:

If we have two arbitrary sets, is it always possible to create an injection from one to the other?

I've searched for this on the internet and in a couple of books, but the best I could find was just a statement that the answer is yes (or actually that the cardinal numbers of sets form a total order which is equivalent). There was, however, no mention of how to prove this.

[CrystyB]

intuitively, start by making a bijection. The one which falls short is the smaller one.
(But actually i never tried to prove that particular fact... )

Bicho, do you really mean "so many of them that they can't all fit in any one set"?? That might belong to a math 301 topic though!

[This message has been edited by CrystyB (edited 02-17-2003 08:39 PM).]

[Bicho the Inhaler]

Yep, I mean too many to fit in any set. There are so many that given an arbitrary set (of any size), you could create a bijection to some "subset" of those. (Thus, if they could all be put in a set, say A, then there would be a bijection between A and its power set, contradicting Cantor's Theorem.)

[CrystyB]

ok i understand what surreal numbers are, but what does the notation in the second paragraph represent. I mean what is 'simple' - to have something be the simplest such and such??

[borschevsky]

[CrystyB]

hmmm, so i guess some notations are meaningless. Maybe even {6|w}.

And i don't pretty much expect operations to be easy. Addition just might be a nice chap and do {a|b}+{c|d} = {a+c|b+d}, but the multiplication is bound to mean trouble!

[This message has been edited by CrystyB (edited 02-19-2003 07:41 PM).]

[Bicho the Inhaler]

Crysty -

Actually, {6|w} = 7. But you're right, many plausible-looking "numbers" aren't surreal numbers, like {1|0}. If A and B are sets of surreal numbers, {A|B} is itself a surreal number if and only if every member of A is less than every member of B. If either A or B contains something other than surreal numbers, then {A|B} is not a surreal number.

Addition and multiplication are slightly more complicated than that, but not overly so. They both have recursive definitions, as does the "less than" relation. In fact, the "equal to" relation is defined in terms of the "less than" relation. Fancy that. I don't remember any of the definitions offhand, though.

It's all explained very clearly in On Numbers and Games by John H. Conway, which is probably the best resource available if you want to learn about surreal numbers. I don't have it with me now, unfortunately.

[Bicho the Inhaler]

"Simplicity" is pretty easy, though: it's all about how easy it is to derive a number. As borschevsky said, you start with {{}|{}} and define that to be "0". That's the simplest number. Then, we can get two numbers we call "-1" and "1", once we have "0". 0 is simpler than 1 and -1 because it takes fewer steps to derive starting with nothing, but 1 and -1 are simpler than everything else. Basically, a number is simpler than another number if it takes fewer steps to derive starting with nothing. w is the simplest infinite number because it's the first one that gets derived. (Actually, -w is equally simple.)

[borschevsky]

Here is a nice paper on the construction and use of surreals. It has the definitions for surreal addition and multiplication.

[mikegoo]

I think I remember reading about surreal numbers not holding under integration which is why they have never really received much attention...can anyone verify or dispute this?

[borschevsky]

Yeah. From that page I linked to earlier:

Originally posted by the guy who wrote that page I linked to earlier:

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It is hardly surprising that it has been difficult to find a useful definition of differentiation and integration using surreal numbers. What exactly does dy/dx mean when x and y themselves can be a fraction of an infinitesimal?

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But this kind of thing happens all the time. Surreals are good for game theory; reals are good for calculus. Sometimes we'll use R and other times S, just like we sometimes use N and sometimes N* .

[CrystyB]

so {4|7}=5.5? And how do i make 1/3? Using its base2 form of 0.(01)? And 1/5 using his 0.(0011)?

[This message has been edited by CrystyB (edited 03-02-2003 12:41 AM).]

[borschevsky]

{4|7} = 5. If you have a surreal {A|B}, it's equal to the oldest/simplest surreal between A and B.

Using the construction of surreals starting from {|}, the 'generations' look like this:

0
-1, 0, 1
-2, -1, -1/2, 0, 1/2, 1, 2
-3, -2, -3/2, -1, -3/4, -1/2, -1/4, 0, 1/4, 1/2, 3/4, 1, 3/2, 2, 3

So each new generation introduces the next available positive and negative integer, and the number halfway between each consecutive pair of existing numbers.

All these numbers are of the form x/2^y, so we're never going to generate 1/3 this way. 1/3 is defined as {L|R} where

L = the set of all x/2^y where x/2^y < 1/3
R = the set of all x/2^y where x/2^y > 1/3

There's a bit of a problem here: 1/3 isn't a surreal yet, and we're using it as part of the definition of the surreal number 1/3. To really be accurate, we need to define a function f:R->S to move between reals and surreals, and use that to specify the sets L and R.

[Bicho the Inhaler]

borschevsky, you actually don't need to do all that. Define the sequence a(n) as

a(1) = 1
a(n + 1) = 4*a(n) + 1

for finite positive integers n. (No problems here because surreal addition and multiplication are easy.) Let L be the set { a(n)*4^-n such that n is a finite positive integer } and let R be the set { (a(n) + 1)*4^-n such that n is a finite positive integer }. {L | R} is a valid surreal number and has all of the properties we associate with 1/3.

Mikko, I guess nobody responded to your post. I don't even know if you still check this thread, or if you've solved it already. I don't have a complete proof right on hand, but I think using the Axiom of Choice is the way to prove that. Given sets A and B, maybe you could use the Axiom of Choice to define an "attempt" at an injection from A to B as a bijection between some subset A' of A and a subset B' of B. Call this attempt F(A', B') (F being the function that maps a pair of subsets (A', B') to a bijection between A' and B'). Make sure it has the property that F(A', B') is an extension of F(A'', B'') whenever A'' Í A' and B'' Í B'. You'll also have to restrict the domain of the attempt-defining function F to a strategically chosen subset of Ã(A) × Ã(B). Then, look at the union G of all attempts. Prove that G is also an attempt F(A', B') for some A' Í A and B' Í B and that either A' = A or B' = B (otherwise you could extend G = F(A', B') to a larger attempt, which is a contradiction since G is the union of all attempts and thus the largest possible attempt). The Axiom of Choice would play a key role in defining the operator F and its domain. I've seen this technique used in a couple different proofs in this area of mathematics.

[edit: replace "surjection" with "injection"]

[This message has been edited by Bicho the Inhaler (edited 03-09-2003 07:56 PM).]