Official TSR Mathematical Society

http://en.wikipedia.org/wiki/Gabriel%27s_Horn

Reply 1862

Re Mr. Muon: just to provide some information overload, I just compiled a little collection of pdfs that deals, in as elementary a way as possible, with the concept of modular arithmetic, right here.

Reply 1863

Elongar

Re Mr. Muon: just to provide some information overload, I just compiled a little collection of pdfs that deals, in as elementary a way as possible, with the concept of modular arithmetic, right here.

how big is your - ahem - pdf collection on other topics ?

Reply 1864

Depends entirely on the topic :P

Reply 1865

Elongar

Depends entirely on the topic :P

you wouldn't happen to have any analysis? :smile:

Reply 1866

Dadeyemi

My Alt

It is impossible to find

\displaystyle\int_{1}^{\infty}\frac{1}{x}dx

, as

\displaystyle\lim_{x\to\infty}lnx = \infty

. But, as soon as I rotate this

2\pi

around the x-axis, I can evaluate the integral / limit, the answer is pi. Can anyone explain why this is so?

I posted a problem with a similar idea, it was about the Koch Snowflake IIRC a shape with an infinite perimeter but finite area.

Reply 1867

DeanK22

you wouldn't happen to have any analysis? :smile:

Sorry to disappoint, but only odds and ends (partly because there's rather a lot of it around, and it's hard to differentiate (pun intended) between texts without reading through them all - slightly different for elementary/olympiad mathematics).

Just have a look at the lists here and here and you'll see what I mean. Perhaps see if something there piques your fancy.

There's also a smattering of topics here, and if you're interested in videos, and you don't mind the overly enthusiastic voice of the lecturer, have a look here too (good site for other topics as well).

Reply 1868

Sorry for the double post, but I ran into something interesting :p:

(and I apologize in advance for the size of the post).

Back in the day, Leibniz apparently used a construction analogous to Pascal's Triangle to compute various summations.

Pascal's triangle is based on the following rules:

Denote

\displaystyle \dbinom{n}{k}

as the value of the kth number on the nth row, then,

1.

\displaystyle \dbinom{n}{0} = 1

\displaystyle \dbinom{n}{k} = \dbinom{n-1}{k} + \dbinom{n-1}{k-1}

Using these rules, we can prove by induction that

\displaystyle \binom{n}{k} = \frac{n!}{k!(n-k)!}

.

Leibniz considered a triangle based on the following rules:

Denote

\displaystyle \bigg[{n \atop k}\bigg]

as the value of the kth number on the nth row, then,

1.

\displaystyle \bigg[{n \atop 0}\bigg] = \frac{1}{n+1}

\displaystyle \bigg[{n \atop k}\bigg] = \bigg[{n+1 \atop k}\bigg] + \bigg[{n+1 \atop k+1}\bigg]

resulting in the following, so-called "Harmonic" triangle (note the appearance of the Harmonic Series):

Again, by induction, one can show relatively easily that,

\displaystyle \bigg[{n \atop k}\bigg] = \frac{k!(n-k)!}{(n+1)!} = \frac{1}{(n+1)\binom{n}{k}}

and we have a relationship between Leibniz's Harmonic Triangle, and Pascal's Triangle.

The Problem:

For Pascal's Triangle, we have the "Hockeystick identity":

\displaystyle S_k(N) = \sum_{n=k}^N \binom{n}{k} = \binom{k}{k} + \binom{k+1}{k} + \binom{k+2}{k} + \dots + \binom{N}{k} = \binom{N+1}{k+1}

N.B. If you draw a line through the summands and the summation on Pascal's Triangle, the namesake of the identity will become clear.

By considering a similar

S_k(N)

for the Harmonic Triangle, show that the Harmonic Series is divergent.

Hint

If this has you interested, try this too:

For Pascal's Triangle, the following identity results from summing all numbers in the nth row:

\displaystyle \sum_{k=0}^n \binom{n}{k} = 2^n

Find a similar identity for the Harmonic Triangle.

Reply 1869

Let f n be the nth fibbonaci number defined recursively from the series;

f_0 = 0 , f_1 = 1 , f_n+2 = f_n + f_n-1

What is the value;

Unparseable latex formula:

\displaystyle \sum_{\n=1}^{\infty} arctan(\frac{1}{f_n})

Reply 1870

I think you mean:

\displaystyle \sum_{n=1}^{\infty} \arctan \left ( \frac{1}{F_{2n+1}} \right )

Reply 1871

SimonM

I think you mean:

\displaystyle \sum_{n=1}^{\infty} \arctan \left ( \frac{1}{F_{2n+1}} \right )

I was interested as to whether someone could do the summation pf the reciprocals of the fibbonaci numbers because I got the odd indexed ones but haven't managed to get anywhere with the even indesed ones - they seem a bit more troublesome.

Reply 1872

I have no reason to believe that there should be a nice solution.

It's like when you stick lots of random elementary functions together, one shouldn't assume there will be a nice anti-derivative.

Reply 1873

Dadeyemi

SimonM

I think you mean:

\displaystyle \sum_{n=1}^{\infty} \arctan \left ( \frac{1}{F_{2n+1}} \right )

Spoiler

Reply 1874

Evaluate

\displaystyle \sum_{n \in S} \frac{1}{n^2}

where

S = \{ x : x\text{ is squarefree and has an odd number of prime divisors}\}

Reply 1875

Zhen Lin

I can confidently say that it is finite and not more than

\displaystyle \frac{\pi^2}{6}

Reply 1876

Dadeyemi

SimonM

Evaluate

\displaystyle \sum_{n \in S} \frac{1}{n^2}

where

S = \{ x : x\text{ is squarefree and has an odd number of prime divisors}\}

Would this work?

Spoiler

I'm assuming there is a more low-tech way of solving this though.

Reply 1877

Well, I don't think there is a more low tech way, but I'd like to see some justification for

\displaystyle \sum_{n=1}^{\infty} \frac{|\mu(n)|}{n^2} = \frac{15}{\pi^2}

(it's hardly the most obvious thing in the world)

Ack, I just want to post it 'cause it's just so hot

Spoiler

Reply 1878

DoMakeSayThink

SimonM

Well, I don't think there is a more low tech way, but I'd like to see some justification for

\displaystyle \sum_{n=1}^{\infty} \frac{|\mu(n)|}{n^2} = \frac{15}{\pi^2}

(it's hardly the most obvious thing in the world)

Ack, I just want to post it 'cause it's just so hot

Spoiler

That is rather clever. How on earth you are supposed to spot that relationship, though, I do not know.

Reply 1879