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

 Quote by LCKurtz One could observe that xk(1-x)n-k is continuous, has zeroes only at 0 and 1, and is positive at x = 1/2.
 Quote by flyingpig So should I include a derivative test saying it would be positive everywhere? I am not sure if it is everywhere yet because i haven't started on the proof, but you mentioned x = 1/2 being positive, so there is a negative point?
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.
 It is a 2nd year course...
 Oh I get it, Intermediate value theorem. I just test the end points [0,1] and do the horizontal line test.
 Wait never mind...they both give me 0
 No wait it is intermediate value theorem. I guess I don't have to prove, but just do this $$f = x^k (1 - x)^{n-k}$$ Since $$f(0) = 0$$ and $$f(1) = 0$$ so the end points are roots. Pick a value in [0,1], say $$\frac{1}{2}$$ Then $$f\left (\frac{1}{2} \right) = \left(\frac{1}{2} \right )^{k} \left(\frac{1}{2} \right)^{n-k} \geq 0$$ So for all values of x in [0,1], f is positive
 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...
 Recognitions: Gold Member Homework Help 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.
 But I got part i) with micromass's help...
 I don't know if this was what you hinted me about, but here is the inequality $$0 \leq x \leq 1$$ $$0 \leq x^k \leq 1$$ $$x^k \geq 0$$ as needed $$0 \leq x \leq 1$$ $$-0 \geq -x \geq -1$$ $$0 \geq -x \geq -1$$ $$1 \geq 1 - x \geq 0$$ [ $$(1 - x)^{n - k} \geq 0$$ as needed again This was easier than I thought... So now I can multiply both inequalities to show that $$(1 - x)^{n - k} x^k \geq 0$$ Now by definition $$\binom{n}{k} > 0$$ for all k and n So I could multiply them all together $$\binom{n}{k} (1 - x)^{n - k} x^k \geq 0$$ 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.
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus You've shown $P_k^n(x) \ge 0$. Now what?

 Quote by vela You've shown $P_k^n(x) \ge 0$. Now what?
The goal was to show that it is positive so I could multiply both sides to $$f \leq g$$ without flipping the sign.

Now my problem is that I can't figure out how to get the summation sign into the inequality.
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus Use the fact that if a>b and c>d then a+c>b+d.

 Quote by vela Use the fact that if a>b and c>d then a+c>b+d.
I am thinking of some kinda of summation property that goes with that inequality...

Am I going in the wrong direction...?
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus 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.
 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...
 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.

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