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]

[tex] \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]

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.