Question

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

Answer 1

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

Answer 2

so maybe we can somehow do this by contradiction

Answer 3

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

Answer 4

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

Answer 5

just thinking so far

Answer 6

.

Answer 7

I think contradiction works well...

Answer 8

just trying to get it to work :p

Answer 9

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 ?

Answer 10

maybe so

Answer 11

yeah that sounds better, il modify..

Answer 12

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

Answer 13

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

Answer 14

on top and bottom

Answer 15

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

Answer 16

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

Answer 17

I think we may cross-multiply before differentiatiating..

Answer 18

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

Answer 19

fine we have a problem at x=0

Answer 20

unless the bottom polynomial is x

Answer 21

or has a factor of x

Answer 22

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

Answer 23

probably shouldn't have used f and g

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

\[\log x \sum\limits_{i}^na_i x^i = \sum\limits_{i}^m b_i x^i \]

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

Answer 28

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

Answer 29

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

Answer 30

ahh we could do that too

Answer 31

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

Answer 32

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

Answer 33

g(x) = logx*f(x) satisfies the equation


logx = (g'(x) - f(x)/x)/f'(x)


so i dont see a problem ?

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

where c and d are constants

Answer 38

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

Answer 39

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

Answer 40

oops

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

the numerator does have a greater degree

Answer 45

I think we can simply let x > 0

Answer 46

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

Answer 47

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 \(m=n\)

Answer 48

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 \(n<m\). but the largest integer possible for \(n<m\) is surely \(n=m-1\), and \(m-1+1=m\not> m\), so \(n\) cannot be an integer.

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

Ahh that limit argument looks neat!

this proof is equivalent to showing that \(\log x\) is not algebraic is it ? @zzr0ck3r