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

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zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I thought @oldrin.bataku solved this

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.0lnx is undefined for anything below 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lnx is a transcendental function not a algebraic one

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I thought it was, but this is not my best area...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2thats okay, we only consider where it is defined

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0@saseal that is what we are asked to prove

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\ln x \rightarrow \log_{e}=y \rightarrow e^y=x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats like getting e^x which is transcedental

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2oldrin's proof showing that \(P_1(x)\ln xP_0(x) = 0\) has no solutions https://i.gyazo.com/4340a5835350d4d1d8cac35a1bb4eb53.png

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2@saseal are you saying, if a function is transcendental, then its inverse is also transcendental ? this can be a tricky proof too yeah :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0looks like the first sentence of the wiki is the answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and also this https://en.wikipedia.org/wiki/Transcendental_function

freckles
 one year ago
Best ResponseYou've already chosen the best response.0maybe 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
 one year ago
Best ResponseYou've already chosen the best response.0or maybe we can also look at a sum of radical expressions and so on...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well it says algebraic function is a function that can be as the root of polynomial equations

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2^ 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 : \[P_n(x)t^n + P_{n1}t^{n1}+\cdots + P_1(x) t + P_0(x)=0\] where \(P_i(x)\) are polynomials

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2earlier, 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
 one year ago
Best ResponseYou've already chosen the best response.0we 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\ln(x) = \ln(\frac{ 1+y }{ 1y }) = 2(\frac{ 1 }{ 1 }+\frac{ 1 }{ 3 }y^2+\frac{ 1 }{ 5 }y^4 ....)\]its infinite sequence

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0^ the taylor series to solve for ln(x)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2thanks @ospreytriple that is very useful, according to the method described in that link, showing that \(t=\ln x\) is not a solution to \(G(t,x):=P_n(x)t^n+P_{n1}(x)t^{n1}\cdots + P_1(x)t+P_0(x) = 0\) 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 \(\ln x\) : \[\begin{align}\dfrac{\partial}{\partial t}G(t,x) &:= P_n(x)nt^{n1} + P_{n1}(n1)t^{n2}+\cdots +P_1(x)=0\\~\\ &:= Q_n(x)t^{n1}+Q_{n1}(x)t^{n2}+\cdots + Q_1(x)=0 \end{align}\] 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
 one year ago
Best ResponseYou've already chosen the best response.2Omg! that proof is so wrong because that shows every algebraic function is rational.
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