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

1. Sep 17, 2011

### flyingpig

1. The problem statement, all variables and given/known data

3. The attempt at a solution

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

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

2. Sep 17, 2011

### micromass

Staff Emeritus
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.

3. Sep 17, 2011

### 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??

4. Sep 17, 2011

### flyingpig

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

5. Sep 17, 2011

### micromass

Staff Emeritus
For (i), you need to start from

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

6. Sep 17, 2011

### flyingpig

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

7. Sep 17, 2011

### micromass

Staff Emeritus
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

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

8. Sep 17, 2011

### 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

9. Sep 17, 2011

### micromass

Staff Emeritus
Yeah, that looks good!!

10. Sep 17, 2011

### flyingpig

How do I start ii) then?

11. Sep 17, 2011

### micromass

Staff Emeritus
You need to prove

$$B_n(f)\leq B_n(g)$$

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

12. Sep 17, 2011

$$\sum^{n}_{k=0} f\left(\frac{k}{n}\right) \binom{n}{k} x^k(1-x)^{n-k} \leq \sum^{n}_{k=0} g\left(\frac{k}{n}\right) \binom{n}{k} x^k(1-x)^{n-k}/tex] SOme kinda of cancellations...? 13. Sep 17, 2011 ### micromass Staff Emeritus Well, you know that [tex]f(k/n)\leq g(k/n)$$

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

14. Sep 17, 2011

### flyingpig

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

15. Sep 17, 2011

### micromass

Staff Emeritus
That's the idea.

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

16. Sep 17, 2011

### 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...?

17. Sep 17, 2011

### micromass

Staff Emeritus
This makes no sense to me...

18. Sep 17, 2011

### flyingpig

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

sorry

19. Sep 17, 2011

### micromass

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

20. Sep 17, 2011

### 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