Transcendental Numbers 1

[Editor: The following question has not been answered in this thread, but it is preserved as the ensuing conversation on transcendental numbers proved constructive.]

By Andrew Hodges on Saturday, June 23, 2001 - 05:03 pm:

How would you calculate the sum to infinity (or just to n terms)of the series:

(ln 1)/2 + (ln 2)/4 + (ln 3)/8 + ..+ (ln(n))/2ⁿ

By Arun Iyer on Friday, June 29, 2001 - 08:24 pm:

I am trying and i am trying!!

love arun

By Andrew Hodges on Sunday, July 01, 2001 - 09:52 am:

me too!! I think it converges around the 0.5078 mark, but i want to know why, and if it is either transcendental or can be expressed in exact form.

By Arun Iyer on Sunday, July 01, 2001 - 07:33 pm:

Well it is quite easy to understand that answer is between 0 and 1.

Applying little of my common sense i see that,
The terms are so arranged that the final answer is approximately 0.5 just because as $n \to \infty$ $\log n / 2^{n}$ tends towards zero.
However, i could not come up with any good proof or anything.

One more thing I have heard this word "transcendental" time and again after i have joined nrich but i never really understood its meaning.

Can you please tell me?

love arun

By Olof Sisask on Monday, July 02, 2001 - 12:40 am:

A transcendental number is a number that isn't the root of any polynomial equation with integer coefficients. Both $e$ and $π$ are transcendental, though I don't know how to prove this.
Olof

By Andrew Hodges on Monday, July 02, 2001 - 07:33 pm:

Another more elegant way of putting is it by saying transcendental numbers are numbers which 'transcend' our number system. Our decimal system is one particular one, we count 1, 2, 3 and so on. You could have another system say that counted in $π$ : $1 π$ , $2 π$ , $3 π$ ... and so on. They may never have heard of this number '1'. 1 transcends their number system, they can calculate in terms of $π$ , it would be approximately $0.3183 \times π$ , but they could never find an exact value for it. They would have to give it a special symbol, like we do $π$ .

It is almost as if $π$ and $e$ are from a higher 'plane' of numbers!

By Arun Iyer on Monday, July 02, 2001 - 07:53 pm:

Ya i got it.

These numbers are something which are alien to the world & customs of algebraic numbers.am i right?

love arun

By Andrew Hodges on Tuesday, July 03, 2001 - 04:37 pm:

correct:)

By Olof Sisask on Tuesday, July 03, 2001 - 05:12 pm:

Are proofs that $π$ and $e$ are transcendental very difficult?

Andrew - I was thinking about that very thing just the other day. Wouldn't their number system be identical to ours, but just with different symbols? As in $π \Rightarrow 1$ , $2 π \Rightarrow 2$ , etc. I'm not sure, but isn't there really only one 'counting system' (I can't think of the right phrase) when talking about real numbers? With our base, 1, we have that $2 = (1 + 1) \times 1$ . If we were using $π$ s as our base, then we'd have $2 π = (π + π) \times π$ , which is clearly not true, unless $π = 1$ , i.e. the same base as ours? It's all very confusing!

Regards,
Olof

By Arun Iyer on Tuesday, July 03, 2001 - 08:03 pm:

OLOF,
I can quite understand your problem.
I was just thinking,
What would be your problem converted into if we take it as,
$2 π = (1 + 1) \times π$

$2 π = (π + π) \times 1$
arguments anybody!!!

love arun

By Michael Doré on Tuesday, July 03, 2001 - 10:49 pm:

The proofs that $e$ and $π$ are transcendental are indeed quite hard. They both follow from Lindemann's theorem which states that if $n$ is natural and $a_{1}$ , ..., $a_{n}$ are algebraic numbers and $b_{1}$ , ..., $b_{n}$ are distinct algebraic numbers then:

$a_{1} \exp (b_{1}) + \dots + a_{n} \exp (b_{n})$

is non-zero.

When I get time over the next few days I'll write out the proof of Lindemann's theorem, perhaps in a new thread to stop it cluttering this thread - the proof is pretty complicated!

Anyway, how does this help with $e$ and $π$ ?

Well if $e$ is algebraic then if you set $n = 2$ , $a_{1} = e$ , $a_{2} = 1$ , $b_{1} = - 1$ , $b_{2} = 0$ then since $a_{1}$ , $a_{2}$ , $b_{1}$ , $b_{2}$ are all algebraic (with $b_{1} \neq b_{2}$ ) we deduce by Lindemann's theorem that the following is non-zero:

$e \times \exp (- 1) + (- 1) \times \exp (0)$

This is clearly nonsense, so we are forced to conclude that $e$ is transcendental.

To prove that $π$ is transcendental, assume it is algebraic. It easily follows that $i π$ is algebraic (can you see why?) Hence using Lindemann's theorem, with $n = 2$ , $a_{1} = 1$ , $a_{2} = 1$ , $b_{1} = i π$ , $b_{2} = 0$ we deduce that the following is non-zero:

$\exp (π i) + 1$

As we all know, the expression does indeed equal zero, so we have our contradiction proving that $π$ is transcendental.

By Arun Iyer on Wednesday, July 04, 2001 - 06:35 pm:

Take your own time michael.
I'll be waiting.

Boy,this is getting interesting!!
i am getting to learn something new everyday!!

love arun