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

[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?

 

[flyingpig]

Do you understand sigma notation?

It means sum

EDIT: oh wait, are you referring when I said that n = k? Oh okay, so the sum is $$\sum_{n=0}^{n} \{whatever is here} = 1$$?

EDITING..not actually too sure of the property above. Looking through my old calc text

 

[vela]

I'm asking you what is $$\sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k}$$ shorthand for? In other words, if you write the sum out, what do you get? I'm asking because you're making a bunch of guesses that make no sense if you understand what the notation means.

 

[flyingpig]

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

$$\sum_{k=0}^{n} P$$

I am sorry for being so slow.wEDIT:writing out the sum...

 

[flyingpig]

I wrote the first three terms
$$\sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k} = (1-x)^n + \frac{n!}{(n-1)!}x(1-x)^{n-1} + \frac{n!}{2!(n-2)!}x^2 (1-x)^{n-2}...$$

 

[vela]

Ok, good. So you see how k isn't a variable you can really mess with, right? It takes on the values 0 to n just to generate the terms in the sum. In fact, when you expand the sum out, there is no k appearing anymore because it was just a dummy variable. So you can't do stuff like assume restrict k to just one value to try get the result you want.

Similarly, you're asked to evaluate the sum for any value of n, so you can't set n to anyone value.

Now, are you familiar with the binomial theorem? That is, what is the expansion of (a+b)ⁿ?

 

[flyingpig]

$$(a + b)^n = \sum_{k=0}^{n}\binom{n}{k} a^{n-k} b^k$$
By observations I took

x = b

1 - x = a

So that a + b = 1

1^n = 1 forever, so that completes the problem! YES THANK YOU VELA, Kurt, and micro and all who helped

*blows kisses*