Mathematics Extension 2 Challenge Marathon

[RuiAce]

[pikachu975]

It's too hard for me to type out since I can't use latex, but would it be right to expand the brackets using binomial expansion then integrate and simplify?

[RuiAce]

It's too hard for me to type out since I can't use latex, but would it be right to expand the brackets using binomial expansion then integrate and simplify?

I just tried doing that. Pretty much gave up after line 3. (Which means it may actually work, but it's going to be really unpleasant.)

Think about what you normally do when you see random n's appearing in integrals.

[pikachu975]

I just tried doing that. Pretty much gave up after line 3. (Which means it may actually work, but it's going to be really unpleasant.)

Think about what you normally do when you see random n's appearing in integrals.

Integration by parts then? I'll try it tomorrow I've got the CSSA 4u on thursday so this will be a cool question.

[RuiAce]

This question first appeared as a meme, ironically enough. At first glance, it looks ridiculous. But it's all about finding the right starting point...


I had a go, and I promise it's doable within MX2 boundaries.

[ellipse]

This question first appeared as a meme, ironically enough. At first glance, it looks ridiculous. But it's all about finding the right starting point...


I had a go, and I promise it's doable within MX2 boundaries.

its a telescoping series,
hence you'll end up with the first term (J_0/I_0) + the last term (J_n)/(I_n)
as n goes to infinity, the last term disappears (ill try to prove this later) so evaluating J_0 and I_0, we get pi^2/6

its interestingly the basel problem, and also appeared as the last q in 2010 mx2

[RuiAce]

its a telescoping series,
hence you'll end up with the first term (J_0/I_0) + the last term (J_n)/(I_n)
as n goes to infinity, the last term disappears (ill try to prove this later) so evaluating J_0 and I_0, we get pi^2/6

its interestingly the basel problem, and also appeared as the last q in 2010 mx2

Of course, the right idea.

Care to show some working out now? You wouldn't be able to say in an exam, by inspection, the last term vanishes.


Edit: Had a look, because I definitely remembered the π^2/6 in the 2010 exam. Didn't realise the final result was the same.
But pretty much as you alluded to there's a faster way than the 2010 exam method.

[Paradoxica]

*briefly tries bashing*

*realises polar co-ordinates is faster*

*facepalms*

[Shlomo314]

i guessing this is wrong because pythag can only be used on the real plane and the imaginary doesn't exist on the im plane?

[RuiAce]

i guessing this is wrong because pythag can only be used on the real plane and the imaginary doesn't exist on the im plane?

Lol I saw this as a meme.

The fallacy is in that Pythagoras theorem uses lengths, so you need to take absolute values (moduli). It's a good joke though

[RuiAce]

Not too hard:


Extra credit: Rinse and repeat using \( \ge \) instead.

[RuiAce]

[RuiAce]

[ellipse]

[RuiAce]