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

- ganeshie8

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

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- chestercat

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- ali2x2

Idk? try the qualified helper thread, it might help ;)

- freckles

so maybe we can somehow do this by contradiction

- freckles

\[\log(x)=\frac{f(x)}{g(x) } \text{ where } f \text{ and } g \text{ are polynomials }\]

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## More answers

- freckles

\[10^{\frac{f(x)}{g(x)}}=x\]

- freckles

just thinking so far

- Astrophysics

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- ganeshie8

I think contradiction works well...

- freckles

just trying to get it to work :p

- ganeshie8

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 ?

- freckles

maybe so

- ganeshie8

yeah that sounds better, il modify..

- freckles

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

- freckles

like maybe we can do something like that for polynomials of greater degree

- freckles

on top and bottom

- freckles

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

- ganeshie8

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

- ganeshie8

I think we may cross-multiply before differentiatiating..

- freckles

\[y=\log(x) \\ 10^{y}=x \\ \ln(10^{y})=\ln(x) \\ y \ln(10)=\ln(x) \\ y' \ln(10)=\frac{1}{x}\]
lol yes it is

- freckles

fine we have a problem at x=0

- freckles

unless the bottom polynomial is x

- freckles

or has a factor of x

- freckles

\[\frac{1}{x \ln(10)}=\frac{f(x)}{xg(x)}\]

- freckles

probably shouldn't have used f and g

- freckles

so you think cross multiplying before differentiating could lead to better things maybe?

- ganeshie8

was thinking.. we might be able to compare coefficients both sides then..

- freckles

\[\log(x)(b_nx^{n}+b_{n-1}x^{n-1}+ \cdots +b_2x^2+b_1x+b_0)= \\ a_nx^{n}+a_{n-1}x^{n-1}+ \cdots +a_2x^2+a_1x+a_0\]

- ganeshie8

\[\log x \sum\limits_{i}^na_i x^i = \sum\limits_{i}^m b_i x^i \]
differentating gives
\[\frac{1}{x} \sum\limits_{i}^na_i x^i + \log x \sum\limits_{i}^nia_i x^{i-1} = \sum\limits_{i}^m ib_i x^{i-1} \]

- freckles

actually that might work solve for log
and compare log expression from before

- ganeshie8

looks it is better to differentiate first, that way we wont have log in the equation

- ganeshie8

ahh we could do that too

- freckles

\[\log(x)=\frac{ \sum_{}^{} i b_i x^i- \sum a_i x^i}{\sum i a_i x^i}\]

- freckles

\[\text{ I think } \frac{ \sum i b_i x^i - \sum a_i x^i }{\sum i a_i x^i} \neq \frac{ \sum b_i x^i }{\sum a_i x^i}\]

- ganeshie8

g(x) = logx*f(x) satisfies the equation
logx = (g'(x) - f(x)/x)/f'(x)
so i dont see a problem ?

- ganeshie8

your claim :
\[\text{ I think } \frac{ \sum (i b_i - a_i)x^i }{\sum i a_i x^i} \neq \frac{ \sum b_i x^i }{\sum a_i x^i}\]

- freckles

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

- freckles

unless f'(x)=c
which means f(x) would have to be xc+d

- freckles

where c and d are constants

- freckles

\[g(x)=\frac{x g'(x)(cx+d)-(cx+d)^2}{c} \\ g(x) =\frac{x}{c} g'(x)(cx+d)-\frac{1}{c}(cx+d)^2 \\ g(x)=\frac{cx+d}{c}[ xg'(x)-cx-d] \]
but this wouldn't be our original g(x)

- ganeshie8

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

- freckles

oops

- freckles

but we need xf'(x) to be a constant so g(x) can be a polynomial

- ganeshie8

it need not be a constant, we just want \(xf'(x)\) to divide evenly into the numerator, right

- freckles

yeah I guess it is possibly that the bottom could divide the top

- freckles

the numerator does have a greater degree

- ganeshie8

I think we can simply let x > 0

- ganeshie8

Also, from your earlier work, it seems we can conclude that both polynomials must have same degree, if such a rational function exists

- ganeshie8

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 \(m=n\)

- anonymous

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.

- anonymous

but by definition \(n\) is the degree of a polynomial, and must therefore be an integer -- contradiction

- anonymous

so in other words, we cannot have that \(\log x\) is a rational function

- zzr0ck3r

Maybe another rout would be to use the fact that \(e\) is transcendental.

- ganeshie8

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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