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

1. ali2x2

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

2. freckles

so maybe we can somehow do this by contradiction

3. freckles

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

4. freckles

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

5. freckles

just thinking so far

6. Astrophysics

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7. ganeshie8

8. freckles

just trying to get it to work :p

9. 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 ?

10. freckles

maybe so

11. ganeshie8

yeah that sounds better, il modify..

12. freckles

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

13. freckles

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

14. freckles

on top and bottom

15. 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

16. ganeshie8

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

17. ganeshie8

I think we may cross-multiply before differentiatiating..

18. 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

19. freckles

fine we have a problem at x=0

20. freckles

unless the bottom polynomial is x

21. freckles

or has a factor of x

22. freckles

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

23. freckles

probably shouldn't have used f and g

24. freckles

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

25. ganeshie8

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

26. freckles

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

27. ganeshie8

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

28. freckles

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

29. ganeshie8

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

30. ganeshie8

ahh we could do that too

31. freckles

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

32. freckles

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

33. ganeshie8

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

34. ganeshie8

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

35. 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

36. freckles

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

37. freckles

where c and d are constants

38. 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)

39. 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

40. freckles

oops

41. freckles

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

42. ganeshie8

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

43. freckles

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

44. freckles

the numerator does have a greater degree

45. ganeshie8

I think we can simply let x > 0

46. ganeshie8

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

47. 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 $$m=n$$

48. 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 $$n<m$$. 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.

49. anonymous

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

50. anonymous

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

51. zzr0ck3r

Maybe another rout would be to use the fact that $$e$$ is transcendental.

52. ganeshie8

Ahh that limit argument looks neat! this proof is equivalent to showing that $$\log x$$ is not algebraic is it ? @zzr0ck3r