show that $\ln x$ is not an algebraic function

https://en.wikipedia.org/wiki/Algebraic_function

Question

show that $\ln x$ is not an algebraic function

https://en.wikipedia.org/wiki/Algebraic_function

zzr0ck3r · Answer

I thought @oldrin.bataku solved this

ganeshie8 · Answer

I think he showed that $\ln x$ cannot be expressed as ratio of two polynomials
is it the same as proving $\ln x$ is not algebraic ?

anonymous · Answer

lnx is undefined for anything below 0

anonymous · Answer

lnx is a transcendental function not a algebraic one

zzr0ck3r · Answer

I thought it was, but this is not my best area...

ganeshie8 · Answer

thats okay, we only consider where it is defined

zzr0ck3r · Answer

@saseal that is what we are asked to prove

anonymous · Answer

$$\ln |x| \rightarrow \log_{e}=y \rightarrow e^y=x$$

anonymous · Answer

thats like getting e^x which is transcedental

ganeshie8 · Answer

oldrin's proof showing that $P_1(x)\ln x-P_0(x) = 0$ has no solutions https://i.gyazo.com/4340a5835350d4d1d8cac35a1bb4eb53.png

ganeshie8 · Answer

@saseal are you saying, if a function is transcendental, then its inverse is also transcendental ? this can be a tricky proof too yeah :)

geerky42 · Answer

*

anonymous · Answer

looks like the first sentence of the wiki is the answer

anonymous · Answer

and also this https://en.wikipedia.org/wiki/Transcendental_function

freckles · Answer

maybe we can try to prove by contradiction that ln(x) can not be expressed as a radical expression... or a ratio of radial expressions...
I don't know how we can prove it is not algebraic function without going through the possibilities.

freckles · Answer

or maybe we can also look at a sum of radical expressions and so on...

anonymous · Answer

well it says algebraic function is a function that can be as the root of polynomial equations

ganeshie8 · Answer

^ yeah i think that is a good place to start 

to prove it is not an algebraic function, we need to show that $t = \ln x$ is not a solution to below polynomial equation : 
$$P_n(x)t^n + P_{n-1}t^{n-1}+\cdots + P_1(x) t + P_0(x)=0$$

where $P_i(x)$ are polynomials

ganeshie8 · Answer

earlier, oldrin showed that $t=\ln x$ is not a solution to first degree polynomial equation :
$$P_1(x)t+P_0(x) = 0$$

we need to show that it is not a solution to polynomial equation of EVERY degree

anonymous · Answer

we can show that it cannot be expressed in terms of a finite sequence of +/-/x/ ÷  and root extracting, which makes it invalid for algebraic function

anonymous · Answer

$$\ln(x) = \ln(\frac{ 1+y }{ 1-y }) = 2(\frac{ 1 }{ 1 }+\frac{ 1 }{ 3 }y^2+\frac{ 1 }{ 5 }y^4 ....)$$its infinite sequence

anonymous · Answer

^ the taylor series to solve for ln(x)

anonymous · Answer

http://www.phengkimving.com/calc_of_one_real_var/07_the_exp_and_log_func/07_08_transcendency_of_the_exp_and_log_func.htm

ganeshie8 · Answer

thanks @ospreytriple that is very useful, according to the method described in that link, showing that $t=\ln x$ is not a solution to $G(t,x):=P_n(x)t^n+P_{n-1}(x)t^{n-1}\cdots + P_1(x)t+P_0(x) = 0$ is an one liner : 


suppose $\ln x $ is algebraic and let $n$ be the degree of such$G(t,x)$.
Then differentiating $G(t,x)$ with respect to $t$, we get another polynomial which has a solution $\ln x$ : 
$$\begin{align}\dfrac{\partial}{\partial t}G(t,x) &:= P_n(x)nt^{n-1} + P_{n-1}(n-1)t^{n-2}+\cdots +P_1(x)=0\~\
&:= Q_n(x)t^{n-1}+Q_{n-1}(x)t^{n-2}+\cdots + Q_1(x)=0
\end{align}$$
We can continue this(keep differentiating to get a lesser degree polynomial which has a solution lnx) till we reach a degree $1$ polynomial.  But we showed earlier that $\ln x$ is not a rational function. $\blacksquare$

ganeshie8 · Answer

Omg! that proof is so wrong because that shows every algebraic function is rational.