Mathematics

Secret

12:00 AM

Suppose $\gamma^{n-1} (a_{n-1} + a_n\gamma)$ is T. Then $p(\gamma)$ is $T + \cdots + T + T = A/T + T = A/T$

Secret

12:16 AM

The final step of this experiment requires this question:

0

Q: Is a linear combination of algebraic powers of a transcendental transcendental?

Lindlemann Weierstrass theorem showed that the set $\{e^{\alpha_n}\}$ is algebraically independent for algebraic numbers $\alpha$. i.e. $$\sum_{n}a_ne^{\alpha_n} \neq 0$$ However, while the transcendence of sums and products of arbitrary transcendentals is an open question, is the above nonzer...

transcendental-numbers

actually nvm, this is not useful cause there is no reference to $\alpha, \beta$

Eric Silva

@Balarka cool, did you work any problems out?

Secret

12:43 AM

$a_n (\alpha + \beta)^n = a_n (\alpha^n + \alpha^{n-1}\beta + \alpha^{n-2}\beta^2 + \cdots + \alpha\beta^{n-1}+\beta^n)$

Let $s=(\alpha+\beta), p=\alpha\beta$

Assume $(s,p)$ is A,T

For $n$ odd:

$a_n (\alpha + \beta)^n = a_n (\alpha^{n-1}(\alpha + \beta) + \alpha^{n-3}\beta^2 (\alpha + \beta) + \cdots \alpha\beta^{n-1}(\alpha+\beta))$

=A(TA+TA+...+TA)=A(T+T+...+T)=AT=T

For $n$ even

$a_n (\alpha + \beta)^n = a_n (\alpha^{n-1}(\alpha + \beta) + \alpha^{n-3}\beta^2 (\alpha + \beta) + \cdots \alpha\beta^{n-1}(\alpha+\beta) + \beta^n)$

=A(TA+TA+...+TA+T)=AT=T

Tidying it up...

Let $p(x)=\sum_{n=1}^ma_nx^n$

Let $s=(\alpha + \beta), k = \alpha\beta$ where $\alpha,\beta$ transcendental

Suppose $s$ is algebraic. Then $k$ is transcendental by math.stackexchange.com/questions/899097/… there exists $p$ such that $p(s)=0$ i.e.

$p(s) = \sum_{n} a_n s^n = 0$

Expanding:

$a_ns^n = a_n (\alpha^n + \alpha^{n-1}\beta + \alpha^{n-2}\beta^2 + \cdots + \alpha\beta^{n-1}+\beta^n)$

For $n$ odd:

$a_ns^n = a_n (\alpha^{n-1}(\alpha + \beta) + \alpha^{n-3}\beta^2 (\alpha + \beta) + \cdots \alpha\beta^{n-1}(\alpha+\beta))$

= A(TA+TA+...+TA)=A(T+T+...+T)=AT=T

Hence $\sum_n a_ns^n= \sum_n T = T$

but this is a contradiction since $0=A$

For $n$ even:

$a_n (\alpha + \beta)^n = a_n (\alpha^{n-1}(\alpha + \beta) + \alpha^{n-3}\beta^2 (\alpha + \beta) + \cdots \alpha\beta^{n-1}(\alpha+\beta) + \beta^n)$

=A(TA+TA+...+TA+T)=A(T+...+T)=AT=T

but this is a contradiction since $0=A$

Suppose $k$ is algebraic. Then $s$ is transcendental by math.stackexchange.com/questions/899097/…, there exists $p$ such that $p(k)=0$ i.e.

$p(s) = \sum_{n} a_n s^n = \sum_{n} a_n\alpha^k\beta^k = 0$

=$\sum_n AA^k=\sum_n A = A$

There is no contradiction

Suppose $s,k$ both transcendental. Then:

7 mins ago, by Secret

= A(TA+TA+...+TA)=A(T+T+...+T)=AT=T

Acutally, this step is wrong

= A(TA+TA+...+TA)=A(T+T+...+T)=AA=A

Secret

1:18 AM

Let $a,b$ be transcendental and let $s=a+b,t=ab$

Suppose $s$ is transcendental. Then $t$ is algebraic

$\forall n \in \Bbb{N},(ab)^n$ is algebraic

Base case:

$ab$ is algebraic

$(ab)^2 = (a+b)^2-a^2-b^2-ab $ is $A=T^2-T^2-T^2-A \implies A=T+T+T$

and so on by induction

Suppose $t$ is transcendental. Then $s$ is algebraic

$\forall n \in \Bbb{N}, (a+b)^n$ is algebraic

Base case:

$(a+b)$ is algebraic

$(a+b)^2 = a^2 + 2ab +b^2$ is $A=T+T+T$

Semiclassical

Here's an integral challenge, mostly because I can't remember how to do it myself

Apparently, $\int_{-1}^1 (1-x^2)^p\,dx=\dfrac{\sqrt{\pi} \Gamma(p+1)}{\Gamma(p+3/2)}$ for $p>-1$. (i.e. Mathematica says this)

When $p$ is an integer, one can do this by hand (albeit tediously if $p$ is large)

But for $p$ non-integer, the only approach that comes to mind is to transform this into the beta function

and that seems a bit of a cop-out

(though it is immediate upon using symmetry and subbing $u=x^2$...oh well)

this is cute. If $p$ is a nonnegative integer, then $\int_0^1 (1-x^2)^p\,dx=\frac{(2p)!!}{(2p+1)!!}$.

Secret

2:02 AM

Let $a,b$ be transcedentals. Then suppose a+b is algebraic

let a+b=c algebraic. Then a-b=c-2b is transcendental

$(a+b)^2-2b^2$ is A-T is transcendental

=$a^2+2ab-b^2$ is T+A-T. Thus $a^2-b^2$ is transcendental

Secret

2:29 AM

Let $a,b$ be transcendentals

Suppose $(a+ib)$ is algebraic. Then a+ib=c is algebraic. Then a-ib=c-2ib is transcendental

$(a+ib)(a-ib)=a^2+b^2$ is AT thus transcendental

$a^2+b^2=(a+b)^2-2ab$ is A-T is transcendental (since $iab$ is transcendental and i is algebraic, hence $ab$ is transcendental, hence $a+b$ is algebraic)