4U Revising Game (2 Viewers)

gurmies · May 8, 2009

Damn, just beat me to it =)

EDIT: Oh, it's a hyperbola not a circle -.-

azureus88 · May 8, 2009

jetblack2007 said:
x^2 - y^2
Its a minus.

Yeh, that was silly of me. It's been edited. Gurms, you ask a question.

gurmies · May 8, 2009

Alrighty:

$(a)Resolve \,\, \frac{8}{x(x-2)(x+2)} \,\, into \,\, partial \,\, fractions.$

$(b)Show \,\, that \,\, \sum_{r=3}^{2001}\frac{8}{r(r-2)(r+2)}<\frac{11}{12}$

harism · May 8, 2009

omg! now i notice the minus.
great.

btw nice sol azureus.

azureus88 · May 8, 2009

lol gurmies, its kinda hard to type out a solution for that question. If you write it out in columns, a lot of stuff cancels out and you get:

[maths]\sum_{r=3}^{2001}\frac{8}{r(r-2)(r+2)}\\=\sum_{r=3}^{2001}(-\frac{2}{r}+\frac{1}{r-2}+\frac{1}{r+2})\\=(\frac{-2}{3}+\frac{1}{1}-\frac{2}{4}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\\=\frac{11}{12}+(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003})\\> \frac{11}{12}[/maths]

btw are you sure its less than and not greater than?

gurmies · May 8, 2009

Question says < than...

azureus88 · May 8, 2009

gurmies said:
Question says < than...

What did you get as an answer?

lolokay · May 8, 2009

i get less than

$\sum_{3}^{2001}\frac{1}{r-2}-\frac{1}{r}-\sum_{3}^{2001}\frac{1}{r}-\frac{1}{r+2}\\ = \sum_{1}^{1999}\frac{1}{r}-\frac{1}{r+2}-\sum_{3}^{2001}\frac{1}{r}-\frac{1}{r+2}\\ = 1/1 - 1/3 + 1/2 - 1/4 - \sum_{2000}^{2001}\frac{1}{r}-\frac{1}{r+2}\\ = 11/12- \sum_{2000}^{2001}\frac{2}{r(r+2)}\\ < 11/12$

azureus88 · May 8, 2009

lolokay said:
i get less than

oh yeh, dam, i see my mistake now. Im missing two terms, -2/2001 and -2/2000. cbb editing.

azureus88 · May 8, 2009

wow, thats a good solution. Nice work.

harism · May 8, 2009

ha! trust lolokay for a solution like that.

Drongoski · May 8, 2009

lolokay said:
i get less than

$\sum_{3}^{2001}\frac{1}{r-2}-\frac{1}{r}-\sum_{3}^{2001}\frac{1}{r}-\frac{1}{r+2}\\ = \sum_{1}^{1999}\frac{1}{r}-\frac{1}{r+2}-\sum_{3}^{2001}\frac{1}{r}-\frac{1}{r+2}\\ = 1/1 - 1/3 + 1/2 - 1/4 - \sum_{2000}^{2001}\frac{1}{r}-\frac{1}{r+2}\\ = 11/12- \sum_{2000}^{2001}\frac{2}{r(r+2)}\\ < 11/12$

lolokay

Brilliant formulation of solution ! I salute you.

gurmies · May 8, 2009

Drongoski said:
lolokay

Brilliant formulation of solution ! I salute you.

Concurrence

harism · May 9, 2009

Why so silent? has everyone died?

lol okay.

you were meant to put a question up.

heres one.

$Given\,\,:\,\,I_{n}=\int_{0}^{1}\frac{x^{n}}{1+x^{2}}\,\,dx$

$Show\,\,:\,\,I_{n}+I_{n+2}=\frac{1}{n+1}$

gurmies · May 9, 2009

harism said:
Why so silent? has everyone died?

lol okay.

you were meant to put a question up.

heres one.

$Given\,\,:\,\,I_{n}=\int_{0}^{1}\frac{x^{n}}{1+x^{2}}\,\,dx$

$Show\,\,:\,\,I_{n}+I_{n+2}=\frac{1}{n+1}$

Is this Raj Mahal's question?

omniscience · May 9, 2009

x + 2 = 2x + 4

find x

harism · May 9, 2009

x = 10000 billion.

oh no wrong again.

harism · May 9, 2009

gurmies said:
Is this Raj Mahal's question?

it is very.

Drongoski · May 9, 2009

$\100dpi I_{n+2}\ =\ \int_{0}^{1}\frac{x^{n+2}}{1+x^{2}}\ dx\ \ =\ \ \int_{0}^{1}\frac{(x^{2}+1-1)x^n}{1+x^{2}}\ dx\\ \\ \ \ =\ \ \int_{0}^{1} \left \{x^{n}\ -\ \frac{x^{n}}{1+x^{2}}\right \}\ dx\ \ =\ \ \left [\frac{x^{n+1}}{n+1} \right ]^{1}_{0}\ -\ \ I_{n}\\ \\ \therefore \ \ I_{n+2}\ +\ I_{n}\ \ =\ \ \frac{1}{n+1}$

Edit:

Above solution amended; missed out 'x^n' in numerator

harism · May 9, 2009

Drongoski said:
$\100dpi I_{n+2}\ =\ \int_{0}^{1}\frac{x^{n+2}}{1+x^{2}}\ dx\ \ =\ \ \int_{0}^{1}\frac{(x^{2}+1-1)}{1+x^{2}}\ dx\\ \\ \ \ =\ \ \int_{0}^{1} \left \{x^{n}\ -\ \frac{x^{n}}{1+x^{2}}\right \}\ dx\ \ =\ \ \left [\frac{x^{n+1}}{n+1} \right ]^{1}_{0}\ -\ \ I_{n}\\ \\ \therefore \ \ I_{n+2}\ +\ I_{n}\ \ =\ \ \frac{1}{n+1}$

can you explain to me the first line please. where did the x^n go?

thanks.

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