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ganeshie8

  • one year ago

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

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  1. ali2x2
    • one year ago
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    Idk? try the qualified helper thread, it might help ;)

  2. freckles
    • one year ago
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    so maybe we can somehow do this by contradiction

  3. freckles
    • one year ago
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    \[\log(x)=\frac{f(x)}{g(x) } \text{ where } f \text{ and } g \text{ are polynomials }\]

  4. freckles
    • one year ago
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    \[10^{\frac{f(x)}{g(x)}}=x\]

  5. freckles
    • one year ago
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    just thinking so far

  6. Astrophysics
    • one year ago
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    .

  7. ganeshie8
    • one year ago
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    I think contradiction works well...

  8. freckles
    • one year ago
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    just trying to get it to work :p

  9. ganeshie8
    • one year ago
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    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
    • one year ago
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    maybe so

  11. ganeshie8
    • one year ago
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    yeah that sounds better, il modify..

  12. freckles
    • one year ago
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    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
    • one year ago
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    like maybe we can do something like that for polynomials of greater degree

  14. freckles
    • one year ago
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    on top and bottom

  15. freckles
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    I think we may cross-multiply before differentiatiating..

  18. freckles
    • one year ago
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    \[y=\log(x) \\ 10^{y}=x \\ \ln(10^{y})=\ln(x) \\ y \ln(10)=\ln(x) \\ y' \ln(10)=\frac{1}{x}\] lol yes it is

  19. freckles
    • one year ago
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    fine we have a problem at x=0

  20. freckles
    • one year ago
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    unless the bottom polynomial is x

  21. freckles
    • one year ago
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    or has a factor of x

  22. freckles
    • one year ago
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    \[\frac{1}{x \ln(10)}=\frac{f(x)}{xg(x)}\]

  23. freckles
    • one year ago
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    probably shouldn't have used f and g

  24. freckles
    • one year ago
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    so you think cross multiplying before differentiating could lead to better things maybe?

  25. ganeshie8
    • one year ago
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    was thinking.. we might be able to compare coefficients both sides then..

  26. freckles
    • one year ago
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    \[\log(x)(b_nx^{n}+b_{n-1}x^{n-1}+ \cdots +b_2x^2+b_1x+b_0)= \\ a_nx^{n}+a_{n-1}x^{n-1}+ \cdots +a_2x^2+a_1x+a_0\]

  27. ganeshie8
    • one year ago
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    \[\log x \sum\limits_{i}^na_i x^i = \sum\limits_{i}^m b_i x^i \] 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
    • one year ago
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    actually that might work solve for log and compare log expression from before

  29. ganeshie8
    • one year ago
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    looks it is better to differentiate first, that way we wont have log in the equation

  30. ganeshie8
    • one year ago
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    ahh we could do that too

  31. freckles
    • one year ago
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    \[\log(x)=\frac{ \sum_{}^{} i b_i x^i- \sum a_i x^i}{\sum i a_i x^i}\]

  32. freckles
    • one year ago
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    \[\text{ I think } \frac{ \sum i b_i x^i - \sum a_i x^i }{\sum i a_i x^i} \neq \frac{ \sum b_i x^i }{\sum a_i x^i}\]

  33. ganeshie8
    • one year ago
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    g(x) = logx*f(x) satisfies the equation logx = (g'(x) - f(x)/x)/f'(x) so i dont see a problem ?

  34. ganeshie8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    unless f'(x)=c which means f(x) would have to be xc+d

  37. freckles
    • one year ago
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    where c and d are constants

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

  39. ganeshie8
    • one year ago
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    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
    • one year ago
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    oops

  41. freckles
    • one year ago
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    but we need xf'(x) to be a constant so g(x) can be a polynomial

  42. ganeshie8
    • one year ago
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    it need not be a constant, we just want \(xf'(x)\) to divide evenly into the numerator, right

  43. freckles
    • one year ago
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    yeah I guess it is possibly that the bottom could divide the top

  44. freckles
    • one year ago
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    the numerator does have a greater degree

  45. ganeshie8
    • one year ago
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    I think we can simply let x > 0

  46. ganeshie8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    but by definition \(n\) is the degree of a polynomial, and must therefore be an integer -- contradiction

  50. anonymous
    • one year ago
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    so in other words, we cannot have that \(\log x\) is a rational function

  51. zzr0ck3r
    • one year ago
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    Maybe another rout would be to use the fact that \(e\) is transcendental.

  52. ganeshie8
    • one year ago
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    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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