ganeshie8
  • ganeshie8
show that \(\log x\) cannot be expressed as ratio of finite polynomials
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ali2x2
  • ali2x2
Idk? try the qualified helper thread, it might help ;)
freckles
  • freckles
so maybe we can somehow do this by contradiction
freckles
  • freckles
\[\log(x)=\frac{f(x)}{g(x) } \text{ where } f \text{ and } g \text{ are polynomials }\]

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freckles
  • freckles
\[10^{\frac{f(x)}{g(x)}}=x\]
freckles
  • freckles
just thinking so far
Astrophysics
  • Astrophysics
.
ganeshie8
  • ganeshie8
I think contradiction works well...
freckles
  • freckles
just trying to get it to work :p
ganeshie8
  • 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
  • freckles
maybe so
ganeshie8
  • ganeshie8
yeah that sounds better, il modify..
freckles
  • 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
  • freckles
like maybe we can do something like that for polynomials of greater degree
freckles
  • freckles
on top and bottom
freckles
  • 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
  • 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
  • ganeshie8
I think we may cross-multiply before differentiatiating..
freckles
  • 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
  • freckles
fine we have a problem at x=0
freckles
  • freckles
unless the bottom polynomial is x
freckles
  • freckles
or has a factor of x
freckles
  • freckles
\[\frac{1}{x \ln(10)}=\frac{f(x)}{xg(x)}\]
freckles
  • freckles
probably shouldn't have used f and g
freckles
  • freckles
so you think cross multiplying before differentiating could lead to better things maybe?
ganeshie8
  • ganeshie8
was thinking.. we might be able to compare coefficients both sides then..
freckles
  • 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
  • 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
  • freckles
actually that might work solve for log and compare log expression from before
ganeshie8
  • ganeshie8
looks it is better to differentiate first, that way we wont have log in the equation
ganeshie8
  • ganeshie8
ahh we could do that too
freckles
  • freckles
\[\log(x)=\frac{ \sum_{}^{} i b_i x^i- \sum a_i x^i}{\sum i a_i x^i}\]
freckles
  • 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
  • ganeshie8
g(x) = logx*f(x) satisfies the equation logx = (g'(x) - f(x)/x)/f'(x) so i dont see a problem ?
ganeshie8
  • 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
  • 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
  • freckles
unless f'(x)=c which means f(x) would have to be xc+d
freckles
  • freckles
where c and d are constants
freckles
  • 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
  • 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
  • freckles
oops
freckles
  • freckles
but we need xf'(x) to be a constant so g(x) can be a polynomial
ganeshie8
  • ganeshie8
it need not be a constant, we just want \(xf'(x)\) to divide evenly into the numerator, right
freckles
  • freckles
yeah I guess it is possibly that the bottom could divide the top
freckles
  • freckles
the numerator does have a greater degree
ganeshie8
  • ganeshie8
I think we can simply let x > 0
ganeshie8
  • 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
  • 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
  • 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
  • anonymous
but by definition \(n\) is the degree of a polynomial, and must therefore be an integer -- contradiction
anonymous
  • anonymous
so in other words, we cannot have that \(\log x\) is a rational function
zzr0ck3r
  • zzr0ck3r
Maybe another rout would be to use the fact that \(e\) is transcendental.
ganeshie8
  • 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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