Hi everybody, For a while i have been trying to prove the following, until now without succes: The integral of (x-1)*(x-1/2)*(x-1/3)*...*(x-1/k) for x=0 to 1, is smaller than zero for all k>3, where k is an integer. I got this one of a fellow student, he didn't know where he got it anymore. I'm almost sure that it holds, I computed it for many values of k with help of Maple. Increasing k seems to let the value of the integral converge to zero, but it remains negative. I tried estimating the integral with other integrals, but in this way i only succeeded in proving that the integral is smaller than 1/k! . I hope someone can help.. Please Register or Log in to view the hidden image! Pim
Hi again, A friend of mine came up with the following step to proving this inequality. Let's call the function (x-1)*(x-1/2)*..*(x-1/k) to be integrated f(x). Then we can estimate: integral(f(x),x=0..1/2) <= integral( |f(x)|, x=0..1/2 ) <=1/2 * |f(0)|=1/2 * 1/k! . The piece of f(x) between 1/2 and 1 is always negative. If we can prove that integral( f(x), x=1/2..1) < -1/2 *1/k!, we have proved the inequality. But how to do this? Any ideas?
Mm, now that you mention it. I thought that |f(x)| always had its maximum in x=0 on the interval [0,1/2], that's why i estimated it like that but i now wonder is this true...
anyway, as far as actually constructing a proof, i didn t have much luck. all i could come up with with my unoriginal brain is to expand the damn thing. which didn t get me anywhere. but i ll ask a friend or two, if you like.
I cannot get it proved either...very frustrating. Yes ask them please, anyone who can help Please Register or Log in to view the hidden image!
Yes true...but you want to find a maximum of |f(x)|. If the maximum of this function is for example in the point 0<=m<=1/2, then we can say: integral( |f(x)|, x=0..1/2 ) <=1/2 * |f(m)| . Am i making sense???
yes. ok, a maximum then. actually, i suppose then that f(0) is a maximum on the (0,1/2) interval, and the inequality holds.
No, that is not true. I just checked it in Maple..|f(0)| is not always the maximum of |f(x)| on the interval [0,1/2] for all k..
wait. no, i m still not convinced that it s true. actually, i m looking at graphs of the polynomial on mathematica right now, and it looks true.... certainly f(0) has to be a maximum on the [0,1/k].
it is certainly true on [0,1/k]. here is my proof: suppose f(0) is not a maximum (substitute absolute values, or change maximum to extremum as you please). then there must be some maximum on the interior of this interval. 1/k is a root, so the polynomial rises monotonically to this maximum, at which point it should start falling monotonically, until it reaches 0. since it is falling, it must have a root at some x<0. but it doesn t. the polynomial only has roots at 1/k > 0. f(0) is a maximum. as i write this though, i realize there is a problem with my proof. it fails if there are multiple extrema in the interval. so this proof is not correct, and it might not apply in all cases, unless i can show a few things about the roots of the derivative of the polynomial. i ll think about that some more.
Nobody tried proving it? Please let me know how you tried it, even if it did not work. Please Register or Log in to view the hidden image!
