lolmath

[Cypher]

Math teacher posed the class this question to see who could work it out the fastest because we had nothing better to do, thought I'd repost here and see if anyone could work it out. (It only looks hard, it's not hard if you think about it logically)

" Simplify and factorise:
conjugate( ( (lim(n ->inf) sum(k=1)^inf 9/(10^k)) + (sum(m=1)^inf 1/(2^m)))^2 - x^2) / ( lim(p -> inf) 1 - 1/(10^p))! "

(Syntax of the question has been changed slightly so you can't just dump it into maple. )

/me is very bored

First one with the correct answer wins.

[schlumpf]

conjugate(4-x^2)

[Fault]

i started calc 2 yesterday, GAY!

[Marlo]

4 + 4

sorry what was the question again?

[Cypher]

conjugate(4-x^2)

Closest so far.

Now find the conjugate of that (easy). Then factorise (easy or hard, depending on the level of math you know).

[Cypher]

i started calc 2 yesterday, GAY!

This has absolutely nothing to do with calc.

[schlumpf]

= 4-conjuate(x)^2 ( x:= a+bi )
= 4-(a-bi)^2
= 4-(a^2-2abi-b^2)
= 4-a^2+2abi+b^2

[Cypher]

conjugate(4-x^2)
= x^2 + 4
Given:
a^2 + b^2 = (a-bi)(a+bi)
Then:
x^2 + 4 = (x-2i)(x+2i)

[Fearsoldier]

And the answer is


oh sorry my brain exploded

[schlumpf]

if x := a+bi is element of C then
x^2 = (a+bi)(a+bi)=(a^2+2abi-b^2)
conjugate(4-x^2)=conjugate(4-(a^2-b^2+2abi))=4-(a^2-b^2-2abi)=
=4-(a-bi)(a-bi) BUT NOT =4+x^2!! because x is not a-bi !!
so simplifing the solution "conjugate(4-x^2)" is not really useful because the solution 4-(conjugate(x)^2) is not much more easy

[Enfeebleness]

This question is not googleable.

New question plz.

[Loveshock]

/Initiate brain explosion

[Mr. Moose]

The answer is: impossible?

[Epic Sheep]

This question is not googleable.

New question plz.

What he said...:confused:um im second set year 11 this is some hard Shit...i know the anser though....



PIE!...i mean Pi