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ganeshie8
 one year ago
show that \(\log x\) cannot be expressed as ratio of finite polynomials
ganeshie8
 one year ago
show that \(\log x\) cannot be expressed as ratio of finite polynomials

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ali2x2
 one year ago
Best ResponseYou've already chosen the best response.0Idk? try the qualified helper thread, it might help ;)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so maybe we can somehow do this by contradiction

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\log(x)=\frac{f(x)}{g(x) } \text{ where } f \text{ and } g \text{ are polynomials }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[10^{\frac{f(x)}{g(x)}}=x\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I think contradiction works well...

freckles
 one year ago
Best ResponseYou've already chosen the best response.3just trying to get it to work :p

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1isn't this a trouble that we have taylor series ? do i need to modify the question to "cannot be expressed as ratio of finite polynomials" or something ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1yeah that sounds better, il modify..

freckles
 one year ago
Best ResponseYou've already chosen the best response.3just for fun let's assume \[\log(x)=\frac{a}{cx+d} \\ \text{ differentiating both sides gives } \\ \frac{1}{x \ln(x)} =\frac{ac}{(cx+d)^2} \text{ but this equality doesn't hold for all } x \\ \text{ for example at } x=1 \text{ we have one side exist and the other side could exist } \\ \text{ depending on values } a,c, \text{ and } d \\ \text{ don't ask me why I skipped } b \\ \text{ just realize that } \\ \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3like maybe we can do something like that for polynomials of greater degree

freckles
 one year ago
Best ResponseYou've already chosen the best response.3but differentiating this looks ugly... \[\log(x)=\frac{a_nx^{n}+a_{n1}x^{n1}+ \cdots +a_3x^3+a_2x^2+a_1x+a_0}{b_nx^{n}+b_{n1}x^{n1}+\cdots +b_3x^3+b_2x^2+b_1x+b_0}\] but we still have the same problem at x=1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1is this a typo \[\log(x)=\frac{a}{cx+d} \\ \text{ differentiating both sides gives } \\ \frac{1}{x \color{Red}{\ln(x)}} =\frac{ac}{(cx+d)^2} \text{ but this equality doesn't hold for all } x \\ \text{ for example at } x=1 \text{ we have one side exist and the other side could exist } \\ \text{ depending on values } a,c, \text{ and } d \\ \text{ don't ask me why I skipped } b \\ \text{ just realize that } \\ \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I think we may crossmultiply before differentiatiating..

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[y=\log(x) \\ 10^{y}=x \\ \ln(10^{y})=\ln(x) \\ y \ln(10)=\ln(x) \\ y' \ln(10)=\frac{1}{x}\] lol yes it is

freckles
 one year ago
Best ResponseYou've already chosen the best response.3fine we have a problem at x=0

freckles
 one year ago
Best ResponseYou've already chosen the best response.3unless the bottom polynomial is x

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{1}{x \ln(10)}=\frac{f(x)}{xg(x)}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3probably shouldn't have used f and g

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so you think cross multiplying before differentiating could lead to better things maybe?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1was thinking.. we might be able to compare coefficients both sides then..

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\log(x)(b_nx^{n}+b_{n1}x^{n1}+ \cdots +b_2x^2+b_1x+b_0)= \\ a_nx^{n}+a_{n1}x^{n1}+ \cdots +a_2x^2+a_1x+a_0\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\log x \sum\limits_{i}^na_i x^i = \sum\limits_{i}^m b_i x^i \] differentating gives \[\frac{1}{x} \sum\limits_{i}^na_i x^i + \log x \sum\limits_{i}^nia_i x^{i1} = \sum\limits_{i}^m ib_i x^{i1} \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3actually that might work solve for log and compare log expression from before

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1looks it is better to differentiate first, that way we wont have log in the equation

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1ahh we could do that too

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\log(x)=\frac{ \sum_{}^{} i b_i x^i \sum a_i x^i}{\sum i a_i x^i}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\text{ I think } \frac{ \sum i b_i x^i  \sum a_i x^i }{\sum i a_i x^i} \neq \frac{ \sum b_i x^i }{\sum a_i x^i}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1g(x) = logx*f(x) satisfies the equation logx = (g'(x)  f(x)/x)/f'(x) so i dont see a problem ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1your claim : \[\text{ I think } \frac{ \sum (i b_i  a_i)x^i }{\sum i a_i x^i} \neq \frac{ \sum b_i x^i }{\sum a_i x^i}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I think it might be better to look at the way you had it \[ \\ \log(x)=\frac{g'(x)\frac{f(x)}{x}}{f'(x)} \\ \\ \text{ so we have that } \frac{g'(x)\frac{f(x)}{x}}{f'(x)}=\frac{g(x)}{f(x)} \\ \frac{ x g'(x)f(x)}{f'(x)}=\frac{g(x)}{f(x)} \\ \frac{(x g'(x)f(x))f(x)}{f'(x) f(x)}=\frac{g(x)f'(x)}{f(x) f'(x)} \\ \text{ which means } \\ (x g'(x)f(x)) f(x)=g(x)f'(x) \\ x g'(x) f(x)f^2(x)=g(x)f'(x)\] what would solving for g(x) look like..\[g(x)=\frac{x g'(x) f(x)f^2(x)}{f'(x)}\] but g is a polynomail and I have just written it as a rational function

freckles
 one year ago
Best ResponseYou've already chosen the best response.3unless f'(x)=c which means f(x) would have to be xc+d

freckles
 one year ago
Best ResponseYou've already chosen the best response.3where c and d are constants

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[g(x)=\frac{x g'(x)(cx+d)(cx+d)^2}{c} \\ g(x) =\frac{x}{c} g'(x)(cx+d)\frac{1}{c}(cx+d)^2 \\ g(x)=\frac{cx+d}{c}[ xg'(x)cxd] \] but this wouldn't be our original g(x)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I think it might be better to look at the way you had it \[ \\ \log(x)=\frac{g'(x)\frac{f(x)}{x}}{f'(x)} \\ \\ \text{ so we have that } \frac{g'(x)\frac{f(x)}{x}}{f'(x)}=\frac{g(x)}{f(x)} \\ \frac{ x g'(x)f(x)}{\color{Red}{x}f'(x)}=\frac{g(x)}{f(x)} \\ \text{ which means } \\ (x g'(x)f(x)) f(x)=\color{Red}{x}g(x)f'(x) \\ x g'(x) f(x)f^2(x)=\color{Red}{x}g(x)f'(x)\] what would solving for g(x) look like..\[g(x)=\frac{x g'(x) f(x)f^2(x)}{\color{Red}{x}f'(x)}\] but g is a polynomail and I have just written it as a rational function

freckles
 one year ago
Best ResponseYou've already chosen the best response.3but we need xf'(x) to be a constant so g(x) can be a polynomial

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1it need not be a constant, we just want \(xf'(x)\) to divide evenly into the numerator, right

freckles
 one year ago
Best ResponseYou've already chosen the best response.3yeah I guess it is possibly that the bottom could divide the top

freckles
 one year ago
Best ResponseYou've already chosen the best response.3the numerator does have a greater degree

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I think we can simply let x > 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Also, from your earlier work, it seems we can conclude that both polynomials must have same degree, if such a rational function exists

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1we have : \( x g'(x) f(x)f^2(x)=\color{Red}{x}g(x)f'(x)\) clearly the leading coefficient on right hand side is \(n*a_nb_n\) the leading coefficient on left hand side is \(m*a_nb_n\) comparing them gives \(m=n\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if \(f(x)=\log x=\frac{g(x)}{h(x)}\) for polynomial \(g,h\) note that we know $$\lim_{x\to0} xf(x)=x\log x=0$$ since \(\lim_{x\to0}\frac{xg(x)}{h(x)}=0\), meaning \(x g(x)\) is of higher degree than \(g\). now, if we instead look at \(\lim_{x\to0}\frac{g(x)}{h(x)}\) we should either get \(0\), meaning \(g\) is still of higher dimension than \(h\), or a finite constant, telling us \(g,h\) are of equal degree. instead, we see \(\lim_{x\to0} f(x)=\infty\), which surely means \(g\) is of lesser degree than \(h\). if the degree of \(g\) is \(n\) and the degree of \(h\) is \(m\), the earlier result tells us \(n+1>m\), while the second result tells us that \(n<m\). but the largest integer possible for \(n<m\) is surely \(n=m1\), and \(m1+1=m\not> m\), so \(n\) cannot be an integer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but by definition \(n\) is the degree of a polynomial, and must therefore be an integer  contradiction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so in other words, we cannot have that \(\log x\) is a rational function

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Maybe another rout would be to use the fact that \(e\) is transcendental.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Ahh that limit argument looks neat! this proof is equivalent to showing that \(\log x\) is not algebraic is it ? @zzr0ck3r
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