
#37
Sep1811, 05:32 PM

P: 2,568

Oh I get it, Intermediate value theorem.
I just test the end points [0,1] and do the horizontal line test. 



#38
Sep1811, 05:41 PM

P: 2,568

Wait never mind...they both give me 0




#39
Sep1811, 06:11 PM

P: 2,568

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)^{nk}[/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)^{nk} \geq 0[/tex] So for all values of x in [0,1], f is positive 



#40
Sep1811, 09:19 PM

P: 2,568

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... 



#41
Sep1811, 10:59 PM

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PF Gold
P: 7,221

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.




#42
Sep1811, 11:20 PM

P: 2,568

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




#43
Sep1811, 11:52 PM

P: 2,568

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. 



#44
Sep1911, 02:13 PM

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P: 11,530

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




#45
Sep1911, 02:17 PM

P: 2,568

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



#46
Sep1911, 02:21 PM

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P: 11,530

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




#47
Sep1911, 02:25 PM

P: 2,568

Am I going in the wrong direction...? 



#48
Sep1911, 02:28 PM

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P: 11,530

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.




#49
Sep1911, 03:07 PM

P: 2,568

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... 



#50
Sep1911, 03:09 PM

P: 2,568

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.




#51
Sep1911, 03:17 PM

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P: 11,530





#52
Sep1911, 03:36 PM

P: 2,568

Oh if f = 1, then P = 1.
Can I jsut state that or should I write up more on nCr? 



#53
Sep1911, 03:37 PM

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P: 11,530

That can't be right. P(x) doesn't depend on f(x) at all. Try again.




#54
Sep1911, 04:15 PM

P: 2,568

In the summation I get
[tex]\sum_{k=0}^{n} \binom{n}{k} x^k (1  x)^{nk}[/tex] 


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