1. ## Sequences

I'm getting a bit puzzled with regards to this question:

Explain why $\displaystyle \frac{a_{2n}\sqrt{A_n}}{(A_{n}+a_2n)(\sqrt{A_n}+\s qrt{a_n})}$ is never larger than 0.4 given that:

$\displaystyle a_{2n}=\sqrt{a_nA_n}$

$\displaystyle A_{2n}=\frac{2A_na_{2n}}{A_n+a_{2n}}$.

Hence show that the error $\displaystyle (A_n-a_n)$ in calculating $\displaystyle \pi$ reduces by at least 0.4 when replacing n by 2n. Show that by calculating $\displaystyle A_{2^{10}}$ and $\displaystyle a_{2^{10}}$ we can estimate $\displaystyle \pi$ to within 0.0014.
I'm not really sure where to start with this except to state that 0.4 is an upper bound of the sequence. That isn't really enough for the start bit and I can't see what the next step would be.

Okay, this may help a lot but i've just worked out that $\displaystyle 2 \leq a_n \leq \pi \leq A_n$ where $\displaystyle A_n$ tends to $\displaystyle \infty$ from above. Therefore the first bit should go a little something like this:

$\displaystyle \frac{\pi \sqrt{\pi}}{(\pi+\pi)(\sqrt {\pi}+\sqrt {\pi})}=\frac{\pi \sqrt{\pi}}{4\pi \sqrt {\pi}}=\frac{1}{4}$

The big problem here is that it doesn't give the required 0.4.

I'm also not sure if thiis helps but $\displaystyle a_4=2$ and $\displaystyle A_4=4$.

2. I'm stuck on this too! Were you able to explain why a4, a8, a16, a32, ... is increasing? I can do it geometrically, since obviously a polygon that is inscribed in a circle with a diameter of 1 with 4 sides would have a smaller perimeter than one with 8 and in turn the perimeter of a polygon with 8 sides would be smaller than the one of a polygon with 16 sides. But mathematically?

And I can't for the life of me figure out a proof for the product rule, ie prove that if an tends to zero and bn tends to zero then anbn tends to zero.

3. lol, I had no idea for any of it!!!

I basically wrote what you said and I can't do any of it mathematically. I left the last assignment completely blank since I had no idea what it was talking about. I even showed the 2nd year maths student who lives next door to me and even he said "what are they talking about??".

It didn't exactly fill me with confidence...

I got that one!!

$\displaystyle 0<b_n<a_n$

$\displaystyle |a_n|<\epsilon$ add the "for all" and the other mathy stuff.

$\displaystyle -\epsilon<a_n<\epsilon$

Therefore:

$\displaystyle -\epsilon<0<b_n<a_n<\epsilon$

$\displaystyle -\epsilon<b_n<\epsilon$

$\displaystyle |b_n|<\epsilon$

So therefore it tends to 0! It's nice helping each other, you rock nmathies!

(I couldn't quite remember the inequality it gives you at the start so just change the inequalities accordingly).

What small class are you in?

4. questions 13 and 14 are terrible!

Thanks for the proof but I actually meant another one, the one on the product rule You know, that if (an)->0 and (bn)->0 then (an x bn)-> 0

and I don't remember what small class I'm in, just that the the teacher is Dan Thompson. Which group are you in?

5. Oh! I'm also tired and I thought "product" meant "add some stuff together"

I don't think i've done that one either. This is getting waaaaay too hard...

I'm in Joerg Enders class. He's pretty cool in a "he's teaching us maths!" type way.

lol, what time were you on last night? I replied to your message a bit before 10 and you were offline. I tried to go to sleep but my housemates insisted on making a ton of noise right outside my room. Not good =(

So i'm tired this morning. Ready for an analysis lecture?