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

[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*