prove it (2 Viewers)

shuning · Jun 25, 2009

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samthebear said:
shame 'bout that >.< but at least you got to see a BIGGER dot with lil dots around it haha =P
i'm amazed no one has answered your question yet considering how much this conversation is bumping the thread to the top.

oh well at least 4 ppl said their are working on it.... xD

all that to just see some bigger dots T.T not worth it LOL... i could of stayed home and had some nice sleep no stinky socks and no 12 @ night out in the cold... and I can still looked @ the stars LOL

ianc · Jun 25, 2009

shuning said:
if $In=\int_{0}^{1}x(1-x)^ndx$
show that $In=\frac{n}{n+2}I_{n-1}$
hence show that $In=\frac{1}{2\cdot ^{n+2}C_{2}}$

n = 1,2,3 ........

Hi there

I don't know if there is a proper "4 unit" method to do these, but this is sort of how i remember doing them in yr 12

First, calculate the integral using partial fractions.
choose:
$u=x, dV=(1-x)^n dx$

You should get:
$I_n=\frac{1}{n+1}\int_{0}^{1}(1-x)^{n+1}dx$

Using this (replace n with n-1):
$I_{n-1}=\frac{1}{n}\int_{0}^{1}(1-x)^{n}dx$

Okay, lets fiddle with our result for $I_n$ so we can substitute in $I_{n-1}$

$I_n=\frac{1}{n+1}\int_{0}^{1}(1-x)(1-x)^{n}dx$
$I_n=\frac{1}{n+1}\left[\int_{0}^{1}(1-x)^{n}dx - \int_{0}^{1}x(1-x)^{n}dx\right]$
$I_n=\frac{1}{n+1}\left[nI_{n-1} - I_n\right]$
$I_n=\frac{n}{n+2}I_{n-1}$

i dont know about the last bit of the question sorry

if you look at what they want, we need to get rid of the $I_{n-1}$ from the formula we derived for $I_n$

$nI_{n-1}=n\frac{1}{n}\int_{0}^{1}(1-x)^{n}dx$
$nI_{n-1}=\int_{0}^{1}(1-x)^{n}dx$
$nI_{n-1}=\frac{1}{n+1}$
So
$I_n=\frac{1}{(n+1)(n+2)}$
sorry i can't think of what to do after this, but hopefully this is enough to put you on the right track!

hope this helps

edit: beaten! but my choice of u and v made the integration a bit easier

jet · Jun 25, 2009

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$I_n = \frac{n}{n+1}I_{n-1} \\ \therefore I_n = \frac{n \cdot (n-1) \cdot (n-2) \cdot ... \cdot 3 \cdot 2 \cdot 1}{(n + 2)\cdot (n+1) \cdot n \cdot ... \cdot 3 \cdot 2 \cdot 1} \\ = \frac{n!}{(n+2)!} \\ = \frac{1}{\frac{(n + 2)!}{n!}} \\ = \frac{1}{2\frac{(n + 2)!}{n!\cdot 2}} \\ = \frac{1}{2^{n + 2}C_2}$

samthebear · Jun 25, 2009

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shuning said:
oh well at least 4 ppl said their are working on it.... xD

all that to just see some bigger dots T.T not worth it LOL... i could of stayed home and had some nice sleep no sleepy socks and no 12 @ night... and still looked @ the stars LOL

lol suddenly the whole thread got flooded with solutions haha - shame our conversation has now been derailed XD hope whatever solutions been posted helps =P

ianc · Jun 25, 2009

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jetblack2007 said:
$\\ = \frac{1}{2\frac{(n + 2)!}{n!\cdot 2}} \\ = \frac{1}{2^{n + 2}C_2}$

just curious -- as an hsc student, how did you do this step? have you memorised some formula that does it for you?

jet · Jun 25, 2009

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ianc said:
just curious -- as an hsc student, how did you do this step? have you memorised some formula that does it for you?

Well, the definition of the notation:
${}^nC_r = \frac{n!}{(n - r)!r!}$
so you can just work with that. btw, I did 4u last year so im prob not the best person to ask.

dog on heat · Jun 25, 2009

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bump

shuning · Jun 25, 2009

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samthebear said:
lol suddenly the whole thread got flooded with solutions haha - shame our conversation has now been derailed XD hope whatever solutions been posted helps =P

yea... i had anna on msn commenting to me about the camp LOL....

btw JetBlack.... ur PRO LOL

thx so much ^^

jet · Jun 25, 2009

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shuning said:
yea... i had anna on msn commenting to me about the camp LOL....

btw JetBlack.... ur PRO LOL

thx so much ^^

No problems and Thanks!

ianc · Jun 26, 2009

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jetblack2007 said:
Well, the definition of the notation:
${}^nC_r = \frac{n!}{(n - r)!r!}$
so you can just work with that. btw, I did 4u last year so im prob not the best person to ask.

haha whoops i thought the question was asking to us for $2^{n+2}C_2$ where $C_2$ was just a constant.... not $2\binom {n+2} {2}$

so if you did 4u maths last year, this year must be a bit boring for you then?

jet · Jun 26, 2009

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ianc said:
haha whoops i thought the question was asking to us for $2^{n+2}C_2$ where $C_2$ was just a constant.... not $2\binom {n+2} {2}$

so if you did 4u maths last year, this year must be a bit boring for you then?

Not really, I'm doing the Cosmology Distinction course, which is really cool. Plus I have Music and English extension to keep me busy.

prove it (2 Viewers)

shuning

Member

ianc

physics is phun!

jet

Banned

samthebear

Member

ianc

physics is phun!

jet

Banned

dog on heat

check it out uh uh uh uh

shuning

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jet

Banned

ianc

physics is phun!

jet

Banned

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