hey is it me or is spec way more easier and interesting than methods
honestly methods is actually feeling more difficult than spec in year 11
I really want to drop a math but I know I can't drop methods...so rip spec
Spec is definitely more interesting, and it's pretty common to find Units 1 & 2 Spec easier than Units 1 & 2 Methods. A lot of that is due to the crowded curriculum in Methods, where you have to learn the content at a very fast pace, but in Spec you have more time to absorb the concepts and consolidate what you've learned. It's worth persevering with both, they complement each other very well in year 12, with quite a lot of overlapping content.
Hi guys, first time posting on a forum! I feel like an old man with new technology, so if I've formatted anything wrong...oops.
Anyway, I just wanted to know if anybody could help me with this question? I don't know what I'm doing wrong, but I can't seem to come up with an answer (I sort of get it, but I need to show working/formal proof):
Prove that the recurrence relation xn+1 = (xn)2 + 2 does not converge to a single value.
If anyone could take a look at it it'd be so helpful for me! Thanks.
Firstly, the recurrence relation is only well defined if we have a value for x0 (the initial value). But in this case, the proof of divergence can be done for any value of x0.
There are a couple of approaches. One is to simply prove that the sequence does not converge. We can do this by supposing that the sequence converges to L and deriving a contradiction. Given that xn+1 = (xn)^2 + 2, if we assume that the limit is L, then we'll have L = L^2 + 2 as n approaches infinity. This equation has no solution, hence the limit does not exist.
A stronger argument is to prove not only that the sequence does not converge, but that it is unbounded (ie. the terms approach infinity). This can be shown by noting that (xn)^2 > xn when xn > 1, but for this sequence xn > 1 for all n ≥ 1. Hence x(n+1) > xn + 2. So for any number L, we can find always find a term in the sequence that is larger than L.