ganeshie8
  • ganeshie8
show that \(\ln x\) is not an algebraic function https://en.wikipedia.org/wiki/Algebraic_function
Mathematics
schrodinger
  • schrodinger
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zzr0ck3r
  • zzr0ck3r
I thought @oldrin.bataku solved this
ganeshie8
  • 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
  • anonymous
lnx is undefined for anything below 0

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anonymous
  • anonymous
lnx is a transcendental function not a algebraic one
zzr0ck3r
  • zzr0ck3r
I thought it was, but this is not my best area...
ganeshie8
  • ganeshie8
thats okay, we only consider where it is defined
zzr0ck3r
  • zzr0ck3r
@saseal that is what we are asked to prove
anonymous
  • anonymous
\[\ln |x| \rightarrow \log_{e}=y \rightarrow e^y=x\]
anonymous
  • anonymous
thats like getting e^x which is transcedental
ganeshie8
  • 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
  • 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
  • geerky42
*
anonymous
  • anonymous
looks like the first sentence of the wiki is the answer
anonymous
  • anonymous
and also this https://en.wikipedia.org/wiki/Transcendental_function
freckles
  • 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
  • freckles
or maybe we can also look at a sum of radical expressions and so on...
anonymous
  • anonymous
well it says algebraic function is a function that can be as the root of polynomial equations
ganeshie8
  • 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
  • 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
  • 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
  • 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
  • anonymous
^ the taylor series to solve for ln(x)
anonymous
  • 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
  • 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
  • ganeshie8
Omg! that proof is so wrong because that shows every algebraic function is rational.

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