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

[flyingpig]

Homework Statement

[PLAIN]http://img88.imageshack.us/img88/4418/unledekp.png

The Attempt at a Solution

What is the P^n _{k} part thing?

Someone should probably go over the i) for me too...

 

[micromass]

The P_n^k is just a symbol. It is a name for a polynomial defined by

P_n^k(x)=\binom{n}{k}x^k(1-x)^{n-k}

So it's just like we define P(x)=x^2. But now our polynomial depends of n and k.

 

[flyingpig]

For the i)

\sum^{n}_{k=0} \left( \alpha f\left(\frac{k}{n}\right) \binom{n}{k} x^k(1-x)^{n-k} + \beta g\left(\frac{k}{n}\right) \binom{n}{k} x^k(1-x)^{n-k}\right) = \alpha \sum^{n}_{k=0} f\left(\frac{k}{n}\right) \binom{n}{k} x^k(1-x)^{n-k} + \beta \sum^{n}_{k=0} g\left(\frac{k}{n}\right) \binom{n}{k} x^k(1-x)^{n-k}

Summation property??

 

[flyingpig]

Something is wrong...do I need induction? This was from proof class

 

[micromass]

For (i), you need to start from

B_n(\alpha f+\beta g)=\sum_{k=0}^n{(\alpha f+\beta g)(\frac{k}{n})x^k(1-x)^{n-k}}

 

[flyingpig]

For (i), you need to start from

B_n(\alpha f+\beta g)=\sum_{k=0}^n{(\alpha f+\beta g)(\frac{k}{n})x^k(1-x)^{n-k}}

What's wrong with what I did...? GO on grill me!

 

[micromass]

What's wrong with what I did...? GO on grill me!

Nothing is wrong, it's all correct, but it's not finished yet.

You still need to show that the left-hand-side of your post equals

B_n(\alpha f+\beta g)

and that the right-hand-side equals

\alpha B_n(f)+\beta B_n(g)

 

[flyingpig]

B_n(\alpha f+\beta g)=\sum_{k=0}^n{(\alpha f+\beta g)(\frac{k}{n})x^k(1-x)^{n-k}}

= \sum_{k=0}^n{(\alpha f (\frac{k}{n})x^k(1-x)^{n-k}} + \beta g(\frac{k}{n})x^k(1-x)^{n-k} ) = \sum_{k=0}^n{\alpha f (\frac{k}{n})x^k(1-x)^{n-k}} + \sum_{k=0}^n \beta g(\frac{k}{n})x^k(1-x)^{n-k} ) = \alpha B_n(f)+\beta B_n(g)

okay done, got lazy with the long tex

 

[micromass]

Yeah, that looks good!

 

[flyingpig]

How do I start ii) then?

 

[micromass]

You need to prove

B_n(f)\leq B_n(g)

Start by writing these things out according to the definition of the B_n.

 

[flyingpig]

\sum^{n}_{k=0} f\left(\frac{k}{n}\right) \binom{n}{k} x^k(1-x)^{n-k} \leq \sum^{n}_{k=0} g\left(\frac{k}{n}\right) \binom{n}{k} x^k(1-x)^{n-k}/tex]<br /> <br /> SOme kinda of cancellations...?

 

[micromass]

Well, you know that

f(k/n)\leq g(k/n)

Now try to introduce the terms needed to conclude that B_nf\leq B_ng.

 

[flyingpig]

Can you multiply P to both sides...? I don't know if P is always positive

 

[micromass]

Can you multiply P to both sides...?

That's the idea.

I don't know if P is always positive

Try to prove it then. Prove that P_k^n(x) is positive if x\in [0,1]...

 

[flyingpig]

So I have to prove two things...

OKay since x \in [0, 1] \leq 0, then it doesn't matter what i put in right? Now how do do that in proper English...?

 

[micromass]

OKay since x \in [0, 1] \leq 0, then it doesn't matter what i put in right?

This makes no sense to me...

 

[flyingpig]

I meant to say x \in [0, 1] \geq 0

sorry

 

[micromass]

I meant to say x \in [0, 1] \geq 0

sorry

Yeah, that also makes no sense. How can [0,1]\geq 0?? [0,1] is a set.

 

[flyingpig]

OKay I wanted to say that numbers in [0,1] are alll positive, so we never had to worry about odd powers messing up with negative numbers

 

[micromass]

OK, you need to show that for all x\in [0,1] and all k that

x^k\geq 0

and

(1-x)^{n-k}\geq 0

That shouldn't be too difficult??

 

[flyingpig]

This is a long problem lol

I have to do two induction problems first??

 

[micromass]

You can do it by induction if you want to. You can also say that it's obvious. Isn't it obvious that the power of a positive number is positive.

I don't know what your professor wants. If he wants you to prove every little thing, then you might want to do induction.

 

[flyingpig]

Do I have to prove that x^k and (1-x)^{n-k} > 0 first?

 

[micromass]

Do I have to prove that x^k and (1-x)^{n-k} > 0 first?

You don't need to show >0, you need to show \geq 0. But yes, if that's how you want to start, try to prove that first...

 

[flyingpig]

Is there another way

 

[micromass]

Is there another way

Another way for what?

 

[flyingpig]

To do thisproblem without doing two inductions?

 

[micromass]

To do thisproblem without doing two inductions?

No, I don't think that's possible if you really want to prove everything rigorously.

 

[flyingpig]

I do the first one first

S(k) : x^k \geq 0

1)Base Case for x \in [0,1]

x^k \geq 0
x^0 = 1 \geq 0

Thus the base case is true

2) Inductive Step.

Inductive Hypothesis: Assume that x^k \geq 0 is true for all k, then S(k + 1) is

x^{k +1} \geq 0

First

x^k \geq 0

x^k x \geq 0
x^{k+1} \geq 0

Thus by Induction, x^k \geq 0 for all k and for x \in [0,1]

 

[micromass]

I do the first one first

S(k) : x^k \geq 0

1)Base Case for x \in [0,1]

x^k \geq 0
x^0 = 1 \geq 0

Thus the base case is true

2) Inductive Step.

Inductive Hypothesis: Assume that x^k \geq 0 is true for all k, then S(k + 1) is

x^{k +1} \geq 0

First

x^k \geq 0

x^k x \geq 0
x^{k+1} \geq 0

Thus by Induction, x^k \geq 0 for all k and for x \in [0,1]

That's ok! And the second case follows from this case.

 

[flyingpig]

For the other one, Is it S(n + 1) or S(k + 1)...?

 

[LCKurtz]

To do thisproblem without doing two inductions?

No, I don't think that's possible if you really want to prove everything rigorously.

One could observe that x^k(1-x)^n-k is continuous, has zeroes only at 0 and 1, and is positive at x = 1/2.

 

[flyingpig]

One could observe that x^k(1-x)^n-k is continuous, has zeroes only at 0 and 1, and is positive at x = 1/2.

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?

 

[LCKurtz]

One could observe that x^k(1-x)^n-k is continuous, has zeroes only at 0 and 1, and is positive at x = 1/2.

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.

 

[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

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

 

[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

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} &gt; 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.

 

[vela]

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

 

[flyingpig]

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.

 

[vela]

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

 

[flyingpig]

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]

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.