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

by flyingpig
Tags: knowing, meani, start, symbols
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flyingpig
#55
Sep19-11, 04:20 PM
P: 2,568
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
#56
Sep19-11, 04:21 PM
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Quote Quote by flyingpig View Post
So I must find

[tex]P = \binom{n}{k} x^k (1 - x)^{n-k} = 1[/tex]
Why? Where did you get this from?
flyingpig
#57
Sep19-11, 04:23 PM
P: 2,568
Quote Quote by vela View Post
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
#58
Sep19-11, 09:13 PM
P: 2,568
In

[tex]\sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k}[/tex]

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
#59
Sep19-11, 09:16 PM
P: 2,568
In the picture, f(x) was shown to be [tex]f(\frac{k}{n})[/tex] does this imply that k = n since f(x) = 1?

If so then

[tex]\frac{n!}{n!(0!)} x^n (1-x)^0 = x^n[/tex]

Darn something is still off.
vela
#60
Sep19-11, 09:24 PM
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Do you understand sigma notation?
flyingpig
#61
Sep19-11, 09:24 PM
P: 2,568
Quote Quote by vela View Post
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 [tex]\sum_{n=0}^{n} \{whatever is here} = 1[/tex]?

EDITING..not actually too sure of the property above. Looking through my old calc text
vela
#62
Sep19-11, 09:28 PM
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I'm asking you what is [tex]\sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k}[/tex] 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
#63
Sep19-11, 09:31 PM
P: 2,568
[tex]\binom{n}{k} x^k (1 - x)^{n-k} = P[/tex]

[tex]\sum_{k=0}^{n} P[/tex]

I am sorry for being so slow.


wEDIT:writing out the sum...
flyingpig
#64
Sep19-11, 09:35 PM
P: 2,568
I wrote the first three terms
[tex]\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}...[/tex]
vela
#65
Sep19-11, 09:41 PM
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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 any one value.

Now, are you familiar with the binomial theorem? That is, what is the expansion of (a+b)n?
flyingpig
#66
Sep19-11, 09:48 PM
P: 2,568
[tex](a + b)^n = \sum_{k=0}^{n}\binom{n}{k} a^{n-k} b^k[/tex]



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*


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