@DHMO Sure, any proof is fine by me. If you want to tell me about it I'm ears right now.

@BalarkaSen you should sleep now

I'll try in a few hours again.

So, $\int_0^t e^{t-x} f(x) \ \mathrm dx = e^t f(0) - f(t) + \int_0^t e^{t-x} f'(x)$

using integration by part

Agreed.

If $f(x)$ is a polynomial with degree $m$, then $f^{(m+1)}(x) = 0$.

Mhm.

So, if you apply the formula again and again, you would get $\displaystyle \int_0^t e^{t-x} f(x) \ \mathrm dx = e^t \sum_{k=0}^m f^{k}(0) - \sum_{k=0}^m f^{k}(t)$

By IBP'ing m+1 times. Gotcha.

yes

Now let $I(t) = \displaystyle \int_0^t e^{t-x} f(x) \ \mathrm dx$

Let $J = \displaystyle \sum_{j=0}^m a_j I(j)$

where $a_n e^n + \cdots + a_1 e + a_0 = 0$

and $(a_n)$ is an integer sequence

Go on.

(So this is where we are assuming, for a contradiction, $e$ is algebraic)

yes

Now, $f(x) = x^{p-1}(x-1)^p(x-2)^p\cdots(x-n)^p$

where $p$ is a prime larger than $n$

what would the degree of $f(x)$ (which is $m$) be?

$(n+1)p - 1$, yeah?

yes

Right, yeah.

oops

$J$ should be $\displaystyle \sum_{j=0}^n a_j I(j)$ instead

Ah, ok, ok.

So $J = \displaystyle \sum_{j=0}^n a_j e^j \sum_{k=0}^m f^{k}(0) - \sum_{j=0}^n a_j \sum_{k=0}^m f^{k}(j)$

The catch is $\displaystyle \sum_{j=0}^n a_j e^j = 0$

Ahh. So the first term vanishes.

yes

So $J = - \displaystyle \sum_{j=0}^n a_j \sum_{k=0}^m f^{k}(j)$

minus that you mean. Yep.

yes

@DHMO do you have any idea about this question math.stackexchange.com/questions/2061109/… it's somewhat relevant here

Now, if $k < p-1$, then the factor $(x-j)$ would not vanish

because if it is false then all the numbers whose partial quotients are bounded by reasonable functions are transcendental

so $f^{(k)}(j) = 0$

@Sophie sorry, no idea

Bit confused on that part, wait a second.

You're claiming $f^k(j) = 0$ for $k < p - 1$ and all $j = 0, .., n$?

yes

Recall that $f(x) = x^{p-1}(x-1)^p(x-2)^p\cdots(x-n)^p$

Ahh, I get it.

Each term would consist of a $(x - j)$ factor if $k < p - 1$ and everything would vanish

Duh. Alright, I agree.

Now, if $k = p-1$, then $f^{(k)}(j)$ is still zero as long as $j \ne 0$.

Yup.

could you compute $f^{(k)}(0)$?

For $k = p-1$ you mean? It's $(-1)^{np} n!$.

I beg to differ

Oh sorry.

You have a (p-1)! up front.

something is still wrong

Er

The $n!$ is wrong

You're right. Meh.

So what is it?

$(n!)^p$, ain't it?

yes

So $f^{p-1}(0) = (-1)^{np}(p-1)!(n!)^p$.

So $f^{(k)}(0) = (-1)^{np}(p-1)!(n!)^p$ when $k = p-1$

Yeah, keep going. I'm bad at doing algebra without pen and paper.

so $(p-1)!$ can divide this but not $p!$

Makes sense.

Denote this as $(p-1)!N$

now, for all $k > p-1$, $f^{(k)}(j)$ is divisible by $p!$

Why can't $p$, like, divide $n!$ or something?

Oh we chose $p$ to be larger than $n$

@BalarkaSen because I can always make $p$ bigger

Make $p$ bigger than $n!$

Just bigger than $n$ suffices, really.

yes

@DHMO Because we get $p!$ terms up front each term, right.

yes

Recall that $J = - \displaystyle \sum_{j=0}^n a_j \sum_{k=0}^m f^{k}(j)$

and $a_j$ are integers

So $J=-(p-1)!(a_0N+pM)$

Yup.

now, the latter term cannot be zero

Mhm.

So $|J| \ge (p-1)!$

and we established a lower bound for $|J|$

True enough.

Now, we will establish an upper bound

Recall that $J = \displaystyle \sum_{j=0}^m a_j I(j) = \sum_{j=0}^m a_j \int_0^j e^{j-x} f(x) \ \mathrm dx$

mhm

Now, let $F(x)$ be $f(x)$ with all coefficients absoluted.

$F(x)$ is increasing there

okay

obviously, $|J| \le \displaystyle \sum_{j=0}^m |a_j| \int_0^j e^j F(j) \ \mathrm dx$

Sure. Agree with the thing below too.

$|J| \le \displaystyle \sum_{j=0}^m |a_j| j e^j F(j)$

Let $A$ be a constant bigger than $|a_j|$

$|J| \le \displaystyle A \sum_{j=0}^m j e^j F(j)$

now, $je^j F(j)$ is increasing also

and I found the same mistake again

$|J| \le \displaystyle A \sum_{j=0}^n j e^j F(j)$

oops lol

no worries though

for $j=0$, it vanishes

nod

$|J| \le \displaystyle A \sum_{j=1}^n n e^n F(n)$

because everything's increasing, sure

so $\leq A n^2 e^n F(n)$.

now, $An^2e^n$ is fixed and does not depend on $p$

while obviously $F(n) \le (2n)^{(n+1)p-1}$

right.

so our upper bound for $|J|$ is $K\times c^p$

where $K$ and $c$ are independent of $p$

combining our lower bound and upper bound, we get $(p-1)! \le |J| \le kc^p$

the contradiction is left to the reader as an exercise

Heh. Neat.

I have literally no clue why the proof worked but it did.

exactly

how am I supposed to tell you the intuition behind

well, we look for it!

This is a pretty fun proof.

I agree

What time is it now?

About 7 in the morning.

and you haven't slept?

Well, I tried. Does that count? :P

Where do you live?

yes

what's your first language?

I'm from Hong Kong

Ah. I'm from India, and the national language is Hindi but we speak in Bengali from where I come from.

is there any specific reason why your English is better than the other Indians that I know?

most people who live in non english speaking countries who are on the english part of the internet have good english, it's just selection bias

well i'm referring to those I know from MSE chat

Well, I like reading literature. I think my English improved talking to people in the chat, but who knows. Historically Bengal was a British colony but I am not sure that counts for anything.

I see

so you know three languages?

More or less, yep. I'd like to know more but it's too hard for me.

I starred our discussion (where we started) for future reference, BTW.

thanks

I found a (bookkeeping) result online

8. If $a$ is a non-zero algebraic number, then $\cos(a)$ and $\sin(a)$ are both transcendental

Just look at $e^{ia}$ I guess

yes

$\cos(a)$ and $\sin(a)$ are algebraically dependent

yep

can you prove that $z=x+iy$ is algebraic iff $x$ and $y$ are algebraic?

no because it's not true

$\pi + i(i\pi)$

x and y have to be real

This is an interesting question.

the backwards direction is easy-ish but the forwards is hard

I agree

I still don't have a guess for its veracity

I read somewhere that it's true

Interesting

But that's trivial so whatever

Is there a theorem that if $x+iy$ is a root to an integer-coefficient polynomial, then $x-iy$ also is?

I think that's true of any real coefficient polynomial

complex roots come in conjugates

why?

Well if it satisfies $z^n + a_n z^{n-1} + \cdots + a_0 = 0$ then just take conjugate

conjugating is linear, so you get that $\bar{z}$ satisfies this too

so yeah, you're done

nice

@DHMO So I guess the first step to understand what the intuition for that $e$ proof is, is to understand what's special about $e$ so that it all panned out

clearly we used $(e^x)' = e^x$ multiple times in the definition of $I(t)$

why was it important?

@BalarkaSen I agree

what could go wrong if we used $22$ instead of $e$?

@BalarkaSen it is used in the iteration formula

nah, the first term vanishes due to hypothesis doesn't it

and the iteration formula will still work but we'd get a bunch of $\log 22$ terms.

i think in the end it'd mess up the inequalities/upper and lower bounds but how I don't see yet

the lower bound is based on the first term vanishing

the first term is based on the iteration formula

the iteration formula is based on $(e^x)'=e^x$

I'm not convinced. the iteration formula reads $I(t, f) = \log 22 \cdot 22^t f(0) - \log 22 f(t) + I(t, f')$ or something like that

if we do it with $22$ instead, I mean

continue

so $I(t, f) = \log 22 (\sum_{j =0}^m f^j(0) - \sum_{j = 0}^m f^j(t))$ after iterating $m$ times.

yes

sorry, I meant $\log 22(e^t \sum f^j(0) - \sum f^j(t))$.

yes

anyways, assume $a_n$ are integers such that $a_n 22^n + \cdots + a_0 = 0$. $J = \sum_{k=0}^n a_k I(k) = \log 22 \sum_{k = 0}^n a_k (22^k \sum_{j = 0}^m f^j(0) - \sum_{j = 0}^m f^j(k))$. The first term is $(\sum a_k 22^k) \cdot \{stuff\}$ which vanishes by hypothesis

wait

So what is $I$?

$I(t, f) = \int_0^t 22^{t-x} f(x) dx$

right

$\displaystyle I(t, f) = \int_0^t 22^{t-x} f(x) dx I(t, f) = \log 22 \cdot 22^t f(0) - \log 22 f(t) + \log 22 I(t, f')$ or something like that

main point being you would have $(\log 22)^j$

@BalarkaSen I meant $\log 22 (22^t \sum f^j(0) - \sum f^j(t))$ here.

@DHMO ah, right, I messed up

good catch. so that's the troublesome thing

yep

I'm still going to see why that causes trouble. $I(t, f) = 22^t \sum_{j =0}^m (\log 22)^{j+1} f^j(0)-\sum_{j=0}^m (\log 22)^{j+1} f^j(t)$ by that logic, yes?

I think so...

In which case, the first term of $J$ should still vanish

I don't think so...

you still get a factor of $\sum_{k=0}^n a_k 22^k$, don't you?

I don't think so

because there's that $22^t$ popping out the first term of $I(t, f)$

why's my argument invalid? p:

I'm thinking

you're right

yeah. but I think it's the second term which causes trouble

yes

we can't establish a lower bound

I'm certain we can. but I think it'll grow slower than the upper bound we'd establish, making the contradiction not happening or something

I see

@DHMO strange though. $J = -\sum_{j = 0}^m (\log 22)^{j+1} f^j(t)$, so $|J|$ should actually be larger than the lower bound we did for $e$. cuz $\log 22 > 1$.

that means the upper bound has to be lot larger than the upper bound for $e$, doesn't it?

@BalarkaSen actually it should be $(\log 22)^{-j-1}$

oh cruds right. see, i'm horrible at algebra

so yeah, of course things break

the lower bound is a lot lot lot smaller

then what about $2$?

or log(something very small)

I think the upper bound does something crazy then. let's have a look

I see

I think we'd just have $|J| \leq A n^2 2^n (2n)^{(n+1)p -1}$

it's not immediately clear why we won't get the desired contradiction

should we try it on real polynomials

like $x^3-3x+2=0$

by that you mean try a root of that instead of $e$?

well, I mean try $2$ by using a specific $f(x)$

instead of in general

well they are already using a specific $f$.

right

I mean a specific $a_j$

and a specific $n$

perhaps. i am not sure if it makes it simpler. numerically you won't get a contradiction, but i want to understand - qualitatively - what goes wrong with $2$

@DHMO I have to go to sleep now but I am sure this can be reasoned like the previous bit too. If you take anything other than $e$, stuff starts popping up (because (a^x)' is a multiple of a^x and the constant factor is 1 only if a = e), which messes the upper/lower bounds up. That's the intuition I gather, at least.

the planetmath pdf you linked previously gives a good summary and refers to Baker's textbook. I think you should have a peek there

actually it refers to Cohn's article: arxiv.org/pdf/math/0601660v3.pdf

also, literally the same technique is applied in the proof of transcendence of $\pi$.

goodnight

@DHMO Now that I am finally awake; the lower bound does not work anymore if you try with anything else than $e$. You get $J = -\sum_{k = 0}^n a_k \sum_{j = 0}^m (\log a)^{-1-j} f^j(k)$.

First off, divisibility arguments do not work if $\log a$ are not integer, and most of the time they are not.

I think you shouldn't have any trouble when they are integers though. The same argument should prove $e^N$ where $N$ is an integer is transcendental.

Because, notice, that the upper bound of $J$ is independent of whatever you choose instead of $e$ in the definition of $I$.

Therefore the key intuition really is that $e$ satisfies $(e^x)' = e^x$ :)

@BalarkaSen I see, thank you

@DHMO I think our next project is to learn and understand how to prove the Lindemann-Weierstrass theorem. It seems like similar ideas are involved.