What do these new symbols mean...?I can't start this without knowing

LCKurtz

What level of course are you taking? I'm getting the impression that you haven't given any actual thought to what I posted and you just post a silly question in response hoping to get more detailed steps.

You are trying to show that function is nonnegative on [0,1]. Don't you have any idea how my suggestion might be relevant to that? Again, please tell me what level course this is that you are taking.

flyingpig

It is a 2nd year course...

flyingpig

Oh I get it, Intermediate value theorem.

I just test the end points [0,1] and do the horizontal line test.

flyingpig

Wait never mind...they both give me 0

flyingpig

No wait it is intermediate value theorem. I guess I don't have to prove, but just do this

[tex]f = x^k (1 - x)^{n-k}[/tex]

Since

[tex]f(0) = 0[/tex] and [tex]f(1) = 0[/tex] so the end points are roots. Pick a value in [0,1], say [tex]\frac{1}{2}[/tex]

Then

[tex]f\left (\frac{1}{2} \right) = \left(\frac{1}{2} \right )^{k} \left(\frac{1}{2} \right)^{n-k} \geq 0[/tex]

So for all values of x in [0,1], f is positive

flyingpig

I am just wondering, do I need ii) to do iii)...?

I just realize my above post did nothing at all because I needed to show all values in [0,1] is true...

LCKurtz

Dear flyingpig: Given the length of this thread and the lack of progress displayed, my advice is that you need to sit down with your teacher and discuss this problem with him. I think you have too many issues to resolve in any reasonable way in this forum.

flyingpig

But I got part i) with micromass's help...

flyingpig

I don't know if this was what you hinted me about, but here is the inequality

[tex]0 \leq x \leq 1[/tex]

[tex]0 \leq x^k \leq 1[/tex]

[tex]x^k \geq 0[/tex] as needed

[tex]0 \leq x \leq 1[/tex]

[tex]-0 \geq -x \geq -1[/tex]

[tex]0 \geq -x \geq -1[/tex]

[tex]1 \geq 1 - x \geq 0[/tex]
[

[tex] (1 - x)^{n - k} \geq 0[/tex] as needed again

This was easier than I thought...

So now I can multiply both inequalities to show that

[tex](1 - x)^{n - k} x^k \geq 0[/tex]

Now by definition

[tex]\binom{n}{k} > 0[/tex] for all k and n

So I could multiply them all together

[tex]\binom{n}{k} (1 - x)^{n - k} x^k \geq 0[/tex]

Now the only problem I have is the summation sign. I don't think I am wrong (and I am pretty confident this time lol) with the inequalities.

vela

You've shown [itex]P_k^n(x) \ge 0[/itex]. Now what?

flyingpig

The goal was to show that it is positive so I could multiply both sides to [tex]f \leq g[/tex] without flipping the sign.

Now my problem is that I can't figure out how to get the summation sign into the inequality.

vela

Use the fact that if a>b and c>d then a+c>b+d.

flyingpig

I am thinking of some kinda of summation property that goes with that inequality...

Am I going in the wrong direction...?

vela

No, that's essentially it. That inequality generalizes to any number of terms. If a>b, c>d, and e>f, then a+c+e > b+d+f, and so on. This is probably one of those things you can just go ahead and assume is true.

flyingpig

Oh that's what you were implying.

Do you think I still need to show that P => 0 when I write it out? Because it looked trivial when I typed it here...

flyingpig

Also how do I do iii)? I was thinking of doing the same thing with multiplying things out for ii), but P => 0, it could be 0, so it will mess things up.

vela

Like micromass said, it depends on the level of rigor your professor wants. In any case, if you've already written it up, you might as well include it.

Part iii assumes f(x)=1. What do you get if you put that into the summation? (You should recognize the sum.)