Prove that the function f(x) = ln x is not rational.
(Note: ln x means loge x)
62soumik you have to still prove taht {Q(x)}2-{xQ(x)P'(x)-xQ'(x)P(x)} is a polynomial of finite degree!
62Yup 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!
62Is 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 ;)
341This 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!
62Thanks :)
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...
341Great 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
62An 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
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
62but 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...
66Well that proof is not exactly a proof. Your arguments mean that for x>0, we might have
ln x = P(x)Q(x).
62one 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!
341this 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
24Let 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
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.
66You have to prove that the f(x)=ln x is not a rational function. A rational function is the ratio of two polynomials.