prove it (1 Viewer)

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 (1 Viewer)

shuning

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ianc

physics is phun!

jet

Banned

samthebear

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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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