show that \(\log x\) cannot be expressed as ratio of finite polynomials

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show that \(\log x\) cannot be expressed as ratio of finite polynomials

Mathematics
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Idk? try the qualified helper thread, it might help ;)
so maybe we can somehow do this by contradiction
\[\log(x)=\frac{f(x)}{g(x) } \text{ where } f \text{ and } g \text{ are polynomials }\]

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\[10^{\frac{f(x)}{g(x)}}=x\]
just thinking so far
.
I think contradiction works well...
just trying to get it to work :p
isn'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 ?
maybe so
yeah that sounds better, il modify..
just for fun let's assume \[\log(x)=\frac{a}{cx+d} \\ \text{ differentiating both sides gives } \\ \frac{1}{x \ln(x)} =\frac{-ac}{(cx+d)^2} \text{ but this equality doesn't hold for all } x \\ \text{ for example at } x=1 \text{ we have one side exist and the other side could exist } \\ \text{ depending on values } a,c, \text{ and } d \\ \text{ don't ask me why I skipped } b \\ \text{ just realize that } \\ \]
like maybe we can do something like that for polynomials of greater degree
on top and bottom
but differentiating this looks ugly... \[\log(x)=\frac{a_nx^{n}+a_{n-1}x^{n-1}+ \cdots +a_3x^3+a_2x^2+a_1x+a_0}{b_nx^{n}+b_{n-1}x^{n-1}+\cdots +b_3x^3+b_2x^2+b_1x+b_0}\] but we still have the same problem at x=1
is this a typo \[\log(x)=\frac{a}{cx+d} \\ \text{ differentiating both sides gives } \\ \frac{1}{x \color{Red}{\ln(x)}} =\frac{-ac}{(cx+d)^2} \text{ but this equality doesn't hold for all } x \\ \text{ for example at } x=1 \text{ we have one side exist and the other side could exist } \\ \text{ depending on values } a,c, \text{ and } d \\ \text{ don't ask me why I skipped } b \\ \text{ just realize that } \\ \]
I think we may cross-multiply before differentiatiating..
\[y=\log(x) \\ 10^{y}=x \\ \ln(10^{y})=\ln(x) \\ y \ln(10)=\ln(x) \\ y' \ln(10)=\frac{1}{x}\] lol yes it is
fine we have a problem at x=0
unless the bottom polynomial is x
or has a factor of x
\[\frac{1}{x \ln(10)}=\frac{f(x)}{xg(x)}\]
probably shouldn't have used f and g
so you think cross multiplying before differentiating could lead to better things maybe?
was thinking.. we might be able to compare coefficients both sides then..
\[\log(x)(b_nx^{n}+b_{n-1}x^{n-1}+ \cdots +b_2x^2+b_1x+b_0)= \\ a_nx^{n}+a_{n-1}x^{n-1}+ \cdots +a_2x^2+a_1x+a_0\]
\[\log x \sum\limits_{i}^na_i x^i = \sum\limits_{i}^m b_i x^i \] differentating gives \[\frac{1}{x} \sum\limits_{i}^na_i x^i + \log x \sum\limits_{i}^nia_i x^{i-1} = \sum\limits_{i}^m ib_i x^{i-1} \]
actually that might work solve for log and compare log expression from before
looks it is better to differentiate first, that way we wont have log in the equation
ahh we could do that too
\[\log(x)=\frac{ \sum_{}^{} i b_i x^i- \sum a_i x^i}{\sum i a_i x^i}\]
\[\text{ I think } \frac{ \sum i b_i x^i - \sum a_i x^i }{\sum i a_i x^i} \neq \frac{ \sum b_i x^i }{\sum a_i x^i}\]
g(x) = logx*f(x) satisfies the equation logx = (g'(x) - f(x)/x)/f'(x) so i dont see a problem ?
your claim : \[\text{ I think } \frac{ \sum (i b_i - a_i)x^i }{\sum i a_i x^i} \neq \frac{ \sum b_i x^i }{\sum a_i x^i}\]
I 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
unless f'(x)=c which means f(x) would have to be xc+d
where c and d are constants
\[g(x)=\frac{x g'(x)(cx+d)-(cx+d)^2}{c} \\ g(x) =\frac{x}{c} g'(x)(cx+d)-\frac{1}{c}(cx+d)^2 \\ g(x)=\frac{cx+d}{c}[ xg'(x)-cx-d] \] but this wouldn't be our original g(x)
I 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
oops
but we need xf'(x) to be a constant so g(x) can be a polynomial
it need not be a constant, we just want \(xf'(x)\) to divide evenly into the numerator, right
yeah I guess it is possibly that the bottom could divide the top
the numerator does have a greater degree
I think we can simply let x > 0
Also, from your earlier work, it seems we can conclude that both polynomials must have same degree, if such a rational function exists
we 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 \(m=n\)
if \(f(x)=\log x=\frac{g(x)}{h(x)}\) 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 \(n m\), so \(n\) cannot be an integer.
but by definition \(n\) is the degree of a polynomial, and must therefore be an integer -- contradiction
so in other words, we cannot have that \(\log x\) is a rational function
Maybe another rout would be to use the fact that \(e\) is transcendental.
Ahh that limit argument looks neat! this proof is equivalent to showing that \(\log x\) is not algebraic is it ? @zzr0ck3r

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