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.. :)

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?

Originally posted by 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?

i m not sure i agree with the last step of that inequality. can you justify it?

You mean integral( |f(x)|, x=0..1/2 ) <=1/2 * |f(0)| ?

Originally posted by Pim
You mean integral( |f(x)|, x=0..1/2 ) <=1/2 * |f(0)| ?

yes

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...

No it is not true...thanks for pointing me this

f(1/2)=0, right? so that should be a minimum.

No, 1/2 is a root of this polynomial, not a minimum..

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.

Originally posted by Pim
No, 1/2 is a root of this polynomial, not a minimum..

yes yes, but when you take the absolute value, like you did, all roots become minima.

I cannot get it proved either...very frustrating. Yes ask them please, anyone who can help :)

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???

Originally posted by Pim
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..

Originally posted by lethe
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.

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].

No, try k=10...., on [0,1/k] it could be true though! Maybe that's an idea!

Originally posted by Pim
No, try k=10...., on [0,1/k] it could be true though! Maybe that's an idea!

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. :)