Mega A Level Maths Thread MK II

I'm gonna go eat, I'll sort that in a bit

Have fun eating some
How's your maclaurin series/ do you know the Basel problem?

Have fun eating some
How's your maclaurin series/ do you know the Basel problem?

Okays, I look forward to seeing what method you decide use

Anyone got a nice question? I'm in the mood for some maths.

I was thinking of going polar/parametrised, but upon closer inspection this does look rather complicated.

My maclaurin series is ok, Basel's $\dfrac{\pi^2}{6}$ isn't it?

Yup, For this problem it is advisable to introduce polar coordinates (r and ϕ)

What modules are you doing this yr again?
I thought you were doing C1, C2 and an applied. But some of the questions you pose (assuming you can solve them) are surprising for an AS maths student!
But then, you wish to do maths @ Uni.

I see, that's fair enough and I don't think I've ever actually rolled on the floor laughing before

Also I would try it out but I'm hunting out a few STEP questions to try instead


Posted from TSR Mobile

To keep with the trend... Show that:

$\sqrt{1+\sqrt{2+\sqrt{5}}}<\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+ \cdots }}}}<\sqrt{1+\sqrt{2+\sqrt{4+ \sqrt{5}}}}$

C1, C2 & S1, I can only solve 'em with the use of the internet Although there's a really simple way of solving the rabbit one, he chose to do it the mathematical way
Some of the other questions came because I tried to get extra practice with weird papers which ended up including c3 and c4 syllabus which is why I had to get some help from here!



Really? we just end up losing balance, ain't really got a choice (and we can't even remember why most of the time)
Ahh good luck



I may attempt this, am I heading in the right direction?

Also (this probably won't help):

Also (this probably won't help):

Unless Lord of the Flies meant otherwise, I think $\sqrt{1+\sqrt{2+\sqrt{5}}}$ terminates at $\sqrt{5}$.

Unless Lord of the Flies meant otherwise, I think $\sqrt{1+\sqrt{2+\sqrt{5}}}$ terminates at $\sqrt{5}$.

I see. Ignore me then Tigerz!

could someone help me with this question: I am asked to use the substitution $x=sin \theta$ to find the exact value of: (Question 4 June 2005)

now i know $\frac{dx}{d \theta} =cos\theta$ and $dx=cos \theta d \theta$ but why is the upper limit $\frac{\pi}{6}$?

Because when you use substitution for definite integration you need to substitute the old limits to find the new ones so if $x=sin\theta$ then $0.5=sin\theta$ so $\theta=arcsin(0.5)=\dfrac{\pi}{6}$

but $arcsin(0.5)$ is also 30 so why is the upper limit not 30?

If I'm right, all trig calculus has to be done in radians.