Proofs in sequences and series

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Alkatran

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mathwonk said:
proving e is irrational
I'm close to the end of my first year and I still haven't seen an irrationality proof. Come to think of it, I haven't heard the word irrational at all!

However:
While learning linear approximation, I linked that up with the physics equation, and saw that:
f(x+a) = f(x) + f'(x)*(a-x) + f''(x)*(a-x)^2/2! + ...

Then when we did taylor series, and we were told that e = (...) I wondered if it was related. Lo and behold:

e^(0+n) => 1 + n + n^2/2! ... = sum(e*x^n/n!)

I assume that equation itself is proof that e isn't rational, since if it is rational, the integer being divided needs to have an infinite amount of digits.
(n->inf)! = inf
 

mathwonk

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it is not quite so obvious that e is not rational.

the usual proof, from that first homework set freshman year in 1960 is as follows;

e is defined by the sum of the series e = 1 + 1/2 + 1/(2)(3) + 1/(2)(3)(4) + .....


If e were rational then there would exist an integer n such that n!e is an integer, and also m!e is an integer for all m ≥ n. We will show by using the formula for the sum of a geometric sequence that this is not possible.

I.e. assume that n!e is an integer.

then en! = n!+ (3)(....)(n) + (4)(....)(n)+ ......+(n) +1 + n!/(n+1)! + n!/(n+2)!+.... is an integer.

Now lets try to show this is impossible by estimating the size of the sum of the fractional part. I.e. since obviously the terms n!+ (3)(....)(n) + (4)(....)(n)+ ......+(n) +1 , are all integers, this would imply that n!/(n+1)!+.... is an integer.

Now the key is that once this is true for one n, it is also true for all larger n, so we only need to show there is some k such that n!/(n+1)!+.... is not an integer for any integer n ≥ k.

Well just estimate the sum n!/(n+1)!+..... I.e. this is
1/(n+1) + 1/(n+1)(n+2)+..... which is less than

the geometric series 1/(n+1) + 1/(n+1)(n+1) +..... = a + ar + ar^2 + ar^3 +....

= a/(1-r) = [1/(n+1)]/[1- (1/(n+1)] = [1/(n+1)]/[n/(n+1)]

= 1/n. now this is not an integer if n ≥ 2. i hope we are done, but i am a little "tired". (I did not get this problem in 1960 by the way.)
 
You wouldn't happen to be a teacher at the University of Florida would you Mathwonk? You're dilemma sounds like the one my own Honors Calc 2 math teacher is facing. If you do happen to be my teacher I think the problem is everyone in the class is retarded.
 

matt grime

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bfd said:
I was just as surprised as you were when I read that. After I wondered "If that is true for a majority of math students in the u.s. how are so many able to get accepted into grad school and survive?" I just can't imagine going to grad school without somewhat of a good exposure to proofs. Wouldn't they be at a major disadvantage?
The actual number of graduating mathematicians in the states is very small, as is the number of american participants in graduate school in mathematics. this is one of the reasons for Vigre's funding of able students. and there is also quite a high drop out rate, or non-completion rate (taking the masters rather than the phd).

Here is an example: Princeton, home to the IAS, and whose chair is Katz, and has seen some of the most important mathematical research in the history of the world has 30 undergraduate mathematics majors, 55 graduate students (source, their home page).

Cambridge each *year* has 250+ undergraduarte mathematics students. Ie in one year they graduate more mathematicians from the degree (I will refrain from comparing the content of the courses - the interested reader can do that themselves) than princeton graduate in 15 years.
 

Alkatran

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mathwonk said:
it is not quite so obvious that e is not rational.

the usual proof, from that first homework set freshman year in 1960 is as follows;

e is defined by the sum of the series e = 1 + 1/2 + 1/(2)(3) + 1/(2)(3)(4) + .....


If e were rational then there would exist an integer n such that n!e is an integer, and also m!e is an integer for all m ≥ n. We will show by using the formula for the sum of a geometric sequence that this is not possible.

I.e. assume that n!e is an integer.

then en! = n!+ (3)(....)(n) + (4)(....)(n)+ ......+(n) +1 + n!/(n+1)! + n!/(n+2)!+.... is an integer.

Now lets try to show this is impossible by estimating the size of the sum of the fractional part. I.e. since obviously the terms n!+ (3)(....)(n) + (4)(....)(n)+ ......+(n) +1 , are all integers, this would imply that n!/(n+1)!+.... is an integer.

Now the key is that once this is true for one n, it is also true for all larger n, so we only need to show there is some k such that n!/(n+1)!+.... is not an integer for any integer n ≥ k.

Well just estimate the sum n!/(n+1)!+..... I.e. this is
1/(n+1) + 1/(n+1)(n+2)+..... which is less than

the geometric series 1/(n+1) + 1/(n+1)(n+1) +..... = a + ar + ar^2 + ar^3 +....

= a/(1-r) = [1/(n+1)]/[1- (1/(n+1)] = [1/(n+1)]/[n/(n+1)]

= 1/n. now this is not an integer if n ≥ 2. i hope we are done, but i am a little "tired". (I did not get this problem in 1960 by the way.)
hmm. I would do it like this:
1: n -> infinity
2: fundamental theorem of arithmatic (all number can be represented as prime factors)
3: Since we're covering all n, we are covering all primes
4: there are infinetely many primes (proved elsewhere)
5: Whatever integer q, in p/q, must have all prime numbers as factors, because it must be a multiple of n!, which has this property (only in the limit, of course!)
6: There is no integer, q, that meets this criteria because whatever q you pick, there is a prime number which is higher.
 

matt grime

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I cannot make any sense of that. "covering"? what with? how? why does this prove anything about the sum of terms?
 

Alkatran

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matt grime said:
I cannot make any sense of that. "covering"? what with? how? why does this prove anything about the sum of terms?
Sorry. I see what's wrong with the proof now. (I know the wording is bad, anyways.)
 

bfd

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matt grime said:
Princeton, home to the IAS, and whose chair is Katz, and has seen some of the most important mathematical research in the history of the world has 30 undergraduate mathematics majors, 55 graduate students (source, their home page)
I assumed that they had a small number of math students but never knew that small. I'd imagine they pick the best of the best.
 

matt grime

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The place to focus is the grad schools, where princeton, harvard, chicago, yale, berkley, many more i'm not mentioning, are leading the way. the uk doesnt have anything like that culture, and i think it will suffer because of it.

one could make many guesses about the make up of the student bodies at each place, and the relative merits of each place, all of which would be very subjective.

one opinion that i hope isn't too controversial is that there are a large number of very good US universities (for mathematics), where as in the UK there are only 4 or 5 that i could hand on heart advise someone to go to (to do maths). What is worrying, i think, is that it's not just that princeton only has 30 undergraduate math majors (I think, but cannot corroborate, that harvard has about 16 but that was about 6 years ago now that i was told that figure, by someone from harvard), but that other quite good institutions where they aren't as selective have almost none.
 

mathwonk

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i am not a prof. at florida but at a similar place.

we think we have very few math majors, really only a handful, but we recently learned that relative to other places, we actually do not have so much fewer than average.

the paucity of math students puts pressure on those of us who try to elevate standards, because the prevailing way to increase numbers is to lower standards.


i am willing to answer any question, at any level, but i believe it is my duty to try to raise the level of the discourse to something like what is current at good schools elsewhere, and to also let students know what that level is.

some people say however that even at top schools the level is way down from the 60's. of course this is not the view of current students at top schools. they think they are better, just ask them.


perhaps one reason harvard has fewer math majors is they do not offer a spivak style intro to calc course, the one for math majors. they think they have no audience for it.

i.e. everyone who goes to harvard has already had intro calc at the AP level, which sadly is usually far below the spivak level.

thus students get into harvard or stanford, etc,, with woefully inadequate preparation in calc and then are plunged into a course designed to follow a spivak course, even though they have not had the spivak course first.

stanford for example a few years ago offered a course for good entering students out of voloume 2 of apostol, but no course out of volume 1 of apostol. the attrition rate was amazing. why they did not seem to care i have no idea.

a very few top schools still offered a spivak type intro to calc course recently, like chicago. some state schools also offer the course but do not attract the students that go to top schools.


another current phenomenon in grad schools in the us is a preference for american students, even if they are less well prepared, an attempt to reverse the trend of few native scientists. this forces a lower standard on the program in order to maintain these stduents.

I.e. the response to a low participation by certain groups in the US is always the same: instead of investing the money needed to raise the level of qualifications for that group, we just lower the standards of admission for that group.

thus there are special scholarships available for american citizen math grad students that insure their admission and support, over better foreign students.

notice this is not the case in basketball, where better players from europe are welcome. it is embarrassing to have higher standards in sports than in mathematics but there it is.
 
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