62soumik you have to still prove taht {Q(x)}2-{xQ(x)P'(x)-xQ'(x)P(x)} is a polynomial of finite degree!
62
Lokesh Verma
·2009-10-07 23:33:31
Yup I also was thinking of derivatives but could not complete the proof to my satisfaction :)
In fact one more struck me when I was travelling (Dont call me a geek .. bcos I travel a lot and these days I feel bored .. so use that time for reading and thinking on some problems :D)
The other idea was to take the limit after x-> infinity
then the limit will be an/bn which will mean that the limit will be finite.. but ln(x) is infinite...
Which is another contradiction (Much simpler than the proof I gave earlier..)
Btw there was another proof given in 500 mathematical challenges (where I happened to find the question 3 days back.. the idea of which i dont remember right now...)
66Hmm...lots of action has taken place over here while I was unaware.
Good work Nishant sir.
As far I am concerned I got this one from a book called Irresistible Integrals by George Boros and Victor H. Moll.
The basic idea is obviously contradiction. But the work can be reduced by using derivatives.
Assuming that
\ln x=\dfrac{P(x)}{Q(x)} where P and Q have common factor, we get after differentiating both sides and cross multiplying:
Q^2 = x (QP^\prime-Q^\prime P) --------- (1)
So x divides Q(x), say Q(x)=xk Q1(x), with Q1(0)≠0 and k ≥ 1. Replacing it back in (1), we get
x^k Q_1^2 = xQ_1P^\prime -kPQ_1-xPQ_1^\prime
Letting x=0 shows that P(0)=0, so x must also divide P(x) --- a contradiction.
Note that the proof is purely algebraic. The only property of ln x used is that its derivative is 1/x.
24very true sir...these kind of books are never available at stores and even in state libraries....which forces students to download them from sites...which is a form of piracy.....[2][2]
really sad situation.......
maybe if mr sibal thinks on this situation a bit more htan abolishing boards or not giving autonomy to IITs...then maybe India can progress more[1]
341You know I think its just that I got to these books before you. Read these books and regularly surf mathlinks.ro and your opinion may change :D
I was and am still crazy about downloading these kind of books. I read them for a while and then they hibernate :D.
I truly wish these books could be made widely available in our country. What a world that would open up to so many students!
62
Lokesh Verma
·2009-10-05 07:01:47
Is there any single book in the world that you have not read?
:D
(until i met you i used to think very high of my knowledge... :D But now i feel really really stupid)
I would have PMed you had i known that there was someone as knowlegeable on goiit to make sure that you were here :D (unfortunately i very rarely visit them.. that too only if i have something competitive or business related :D)
The good thing is that now you are here ;)
341
Hari Shankar
·2009-10-05 06:56:47
This one I encountered in Five Hundred Mathematical Challenges by Barbeau and Klamkin. I suspect it is also there in a book on Polynomials by Barbeau.
Problem Solving Strategies is truly a great resource. Actually Titu's book Putnam and Beyond is very much on the same lines. But a nice beginning can be made with The Art and Craft of Problem Solving by Paul Zeitz. This one should be freely gifted to all school libraries, IMHO!
62
Lokesh Verma
·2009-10-05 06:45:54
Thanks :)
BTW where is this question from ?? And where could you peek for help.. I almost found myself strangled.. that is why i took all the pain to solve it and more importantly type it down here [1]
* I have downloaded and printed a couple of books that you suggested.. some of them are really good.. but then none of them beats "problem solving strategies".. that is the best I have liked so far...
341
Hari Shankar
·2009-10-05 06:44:05
Great Stuff! I say that since you have worked out the whole thing by yourself, whereas I remember when I first saw this question I immediately turned the pages to peek at the solution :D
62
Lokesh Verma
·2009-10-05 06:32:09
An attempt (This time without higher maths shortcuts [6] ):
Lets assume that ln x = P(x)/ Q(x)
where P and Q are Prime Polynomials (Have no common polynomial factors) (I dont know if something like this exists but am assuming that to create a contradiction... If one does not like my using this then we can use infinite descent to prove the same....)
Look at ln(1/x) = - ln (x) for all x..
let P(x) be a0 + a1x+a2x2 ..... anxn
let Q(x) be b0 + b1x+b2x2 ..... bmxm
Where, an and bm are both non zero...
look at the identity.. ln(x) + ln (1/x) = 0
ln (1/x) = (a0xn + ..... an)/xn(b0+b1xm+ ..... bmx)/xm
ln (1/x) = (a0xn + ..... an)xm-n(b0+b1xm+ ..... bmx)
ln(x) = a0 + a1x+a2x2 ..... anxnb0+b1x+b2x2 ..... bmxm
But for the above identity to hold, we can say that m=n ... (That is because on cross multiplication, the higest power will otherwise never get cancelled .. To prove that we will need two cases m>n or n>m)
We can rewrite ln(1/x) and ln(x)
ln (1/x) = (a0xn + ..... an)(b0xn+ ..... bn)
ln(x) = a0 + a1x+a2x2 ..... anxnb0+b1x+b2x2 ..... bnxn
Once we have reached that far, we can now compare the constant on addition of ln(x) and ln(1/x)
anb0+bna0 = 0 (since it is an identity, all coeff will be zero)
Now we notice that 0 is a root of Q(x) because limit x->0 , LHS is -infinity
So 0 is a root of Q(x) .. hence, b0 = 0
Hence, we get a0b1+bna0 = 0
so either a0=0 or bn =0
a0 =0 will mean that the numerator also has a factor x, which due to our assumption that there are no common factors will fail...
bn=0 will mean that Q is not of power n..
hence there is a contradiction
24
eureka123
·2009-09-13 11:14:11
this still hasnt been done ..
does anyone has complete soln ??
11i have doubt sir
ln e =1
Therefore it can be rational
Rite Sir?
If the above is rite than wat do we have to prove
62
Lokesh Verma
·2009-08-24 23:14:26
but then, if i remember correctly, there is a theorem which says that if a function can be expressed as a rational function in any closed interval the function is the same on the whole of the real axis...
66
kaymant
·2009-08-24 22:40:41
Well that proof is not exactly a proof. Your arguments mean that for x>0, we might have
ln x = P(x)Q(x).
62
Lokesh Verma
·2009-08-24 01:53:58
i should have used the phrase "if acceptable"
62
Lokesh Verma
·2009-08-24 01:51:29
one very simple proof (if correct!)
if log(x) = p(x)/q(x) then the domain will be the same.
The domain of the LHS is only +ve real axis!
The domain of the RHS is most of the number line except when q is zero!
341
Hari Shankar
·2009-08-21 23:24:33
this has nothing to do with rational values.
He is asking can we write \log x = \frac{p(x)}{q(x)}
for some polynomials p(x) and q(x)
Hint: log (xm) = m log x
Note: the base need not be e
24
eureka123
·2009-08-21 23:14:17
Let us assume that log x is rational
=> It can be expressed in form of p/q
=> logx= p/q
=>x=ep/q
=>xq=ep
But this is not true
Hence contradiction
=>log x is not rational function
66
kaymant
·2009-08-21 10:28:13
@aieeee
consider what happens for x = e.
1just a logical attempt,may be wrong
ln x cn be written as log10x / log10e ,
whatever may be the value of x, log10e is not gonna be rational.hence proved.
66
kaymant
·2009-08-21 03:17:24
You have to prove that the f(x)=ln x is not a rational function. A rational function is the ratio of two polynomials.