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


by flyingpig
Tags: knowing, meani, start, symbols
flyingpig
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#37
Sep18-11, 05:32 PM
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Oh I get it, Intermediate value theorem.

I just test the end points [0,1] and do the horizontal line test.
flyingpig
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#38
Sep18-11, 05:41 PM
P: 2,568
Wait never mind...they both give me 0
flyingpig
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#39
Sep18-11, 06:11 PM
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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
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#40
Sep18-11, 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...
LCKurtz
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#41
Sep18-11, 10:59 PM
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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
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#42
Sep18-11, 11:20 PM
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But I got part i) with micromass's help...
flyingpig
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#43
Sep18-11, 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.
vela
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#44
Sep19-11, 02:13 PM
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You've shown [itex]P_k^n(x) \ge 0[/itex]. Now what?
flyingpig
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#45
Sep19-11, 02:17 PM
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Quote Quote by vela View Post
You've shown [itex]P_k^n(x) \ge 0[/itex]. Now what?
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
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#46
Sep19-11, 02:21 PM
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Use the fact that if a>b and c>d then a+c>b+d.
flyingpig
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#47
Sep19-11, 02:25 PM
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Quote Quote by vela View Post
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...?
vela
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#48
Sep19-11, 02:28 PM
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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
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#49
Sep19-11, 03:07 PM
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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
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#50
Sep19-11, 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.
vela
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#51
Sep19-11, 03:17 PM
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Quote Quote by flyingpig View Post
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...
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.

Quote Quote by flyingpig View Post
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.
Part iii assumes f(x)=1. What do you get if you put that into the summation? (You should recognize the sum.)
flyingpig
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#52
Sep19-11, 03:36 PM
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Oh if f = 1, then P = 1.

Can I jsut state that or should I write up more on nCr?
vela
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#53
Sep19-11, 03:37 PM
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That can't be right. P(x) doesn't depend on f(x) at all. Try again.
flyingpig
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#54
Sep19-11, 04:15 PM
P: 2,568
In the summation I get

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


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