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

- ganeshie8

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

- schrodinger

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- zzr0ck3r

I thought @oldrin.bataku solved this

- ganeshie8

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

lnx is undefined for anything below 0

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## More answers

- anonymous

lnx is a transcendental function not a algebraic one

- zzr0ck3r

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

- ganeshie8

thats okay, we only consider where it is defined

- zzr0ck3r

@saseal that is what we are asked to prove

- anonymous

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

- anonymous

thats like getting e^x which is transcedental

- ganeshie8

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

- ganeshie8

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

- geerky42

*

- anonymous

looks like the first sentence of the wiki is the answer

- anonymous

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

- freckles

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

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

- anonymous

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

- ganeshie8

^ 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

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

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

- anonymous

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

- anonymous

^ the taylor series to solve for ln(x)

- anonymous

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

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

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

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