## ganeshie8 one year ago show that $$\ln x$$ is not an algebraic function https://en.wikipedia.org/wiki/Algebraic_function

1. zzr0ck3r

I thought @oldrin.bataku solved this

2. 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 ?

3. anonymous

lnx is undefined for anything below 0

4. anonymous

lnx is a transcendental function not a algebraic one

5. zzr0ck3r

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

6. ganeshie8

thats okay, we only consider where it is defined

7. zzr0ck3r

@saseal that is what we are asked to prove

8. anonymous

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

9. anonymous

thats like getting e^x which is transcedental

10. ganeshie8

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

11. ganeshie8

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

12. geerky42

*

13. anonymous

looks like the first sentence of the wiki is the answer

14. anonymous
15. 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.

16. freckles

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

17. anonymous

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

18. 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

19. 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

20. anonymous

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

21. 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

22. anonymous

^ the taylor series to solve for ln(x)

23. anonymous
24. 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. $$\blacksquare$$

25. ganeshie8

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