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

[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

$$\binom{n}{k} > 0$$ 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.

 

[vela]

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.

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]

Oh if f = 1, then P = 1.

Can I just state that or should I write up more on nCr?

 

[vela]

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

 

[flyingpig]

In the summation I get

$$\sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k}$$

 

[flyingpig]

Oh wait, there are n - k + 1 terms...shoot I forgot how to do this. I am guessing that means the sum sums to 1.

 

[vela]

So I must find

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

Why? Where did you get this from?

 

[flyingpig]

Why? Where did you get this from?

Yeah scratch that I aws still thinking of multiplying and stuff. Forgot to get rid of it.

 

[flyingpig]

In

$$\sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k}$$

It can only be 1 when k = 0, and n = 0...

Is that what I Have to show?

This is due tomorrow and I m still scratching my head

 

[flyingpig]

In the picture, f(x) was shown to be $$f(\frac{k}{n})$$ does this imply that k = n since f(x) = 1?

If so then

$$\frac{n!}{n!(0!)} x^n (1-x)^0 = x^n$$

Darn something is still off.

 

[vela]

Do you understand sigma notation?