My Riemann zeta query was in reference to your "do you know that the Riemann Zeta function WAS actually examined in the HSC in 1975" comment. Anyway, I had a look back at buchanan's old post (http://community.boredofstudies.org/238/extracurricular-topics/102519/arc-length.html#post2220688) about old topics that have been removed, and it says that the Riemann zeta function was in the Level 1 (1966-1980) course.
The infinite sum is a first-order linear approximation of the following integral:Yeah. Well that's hard. But jokes aside, do you know that the Riemann Zeta function WAS actually examined in the HSC in 1975? So here is a not-so-hard question - HSC 1975 4 unit Q9i:
See if you can do it.
Indeed you are right (but also wrong).
Well although convergence was in the syllabus from 1965-1980, explicit mention of the Riemann zeta function was in the 1973 syllabus, not the 1965 one.
Yeah. Well that's hard. But jokes aside, do you know that the Riemann Zeta function WAS actually examined in the HSC in 1975? So here is a not-so-hard question - HSC 1975 4 unit Q9i:
See if you can do it.
Your way works, just evaluate the integral:The infinite sum is a first-order linear approximation of the following integral:
Two approximations are constructed such that they are definitively greater and lesser than the true value of the integral. Since the integral only converges for s>1, it follows that the sum only converges for s>1. QED.
This isn't a formal proof, strictly speaking, but I cbbs rn. I'll come back to patch it up later.
Bump
Nice.^s =\sum_{n=1}^\infty \frac{2^n}{2^{ns}} = \sum_{n=1}^\infty 2^{n(1-s)}$ which converges for all $|2^{1-s}| > 1 \Leftarrow s > 1$ (geometric series). Hence our original series converges for $ s >1 $ by the Cauchy Condensation test.$ )
Let D(a,b) be the number of ways to make a sum of a with b numbers.
It is obvious none of the digits can be greater than 6.
Since the sum of the digits must be 6, we can write it as:
Although the idea is good, I think the answer is not that, since multiple bars can exist next to each otherSince the sum of the digits must be 6, we can write it as:
1_1_1_1_1_1_
Just use the stars and bars method, since we need to put in 3 bars to represent the digits. So we get
i.e. 1_1 | 1_1 | 1 | 1 = 2211
56 is also what I got.Since the sum of the digits must be 6, we can write it as:
1_1_1_1_1_1_
Just use the stars and bars method, since we need to put in 3 bars to represent the digits. So we get
i.e. 1_1 | 1_1 | 1 | 1 = 2211
tywebb's site has a proof of Fermat's Theorem. I don't quite get it. But it is still fascinating.
Quite a lot of background is needed to understand Wiles' proof at any greater depth than the absolute "top level overview".
My method allows for that (56 is most definitely correct).
OP is a Year 12 student Extension 1. These are not in the scope of the HSC syllabus.1. By considering the cross-sections of the figure, sketch
2. Ifshow that
Note I am using older notation here (edit here is the equivalent notation)
Well, it is clear z > 0. By setting z to 1 (or any other positive constant), it's clear that the horizontal cross-section is that of an ellipse, which is stretched with major axis being the y-axis. By considering y or x being constant, it is clear the shape traced is that of a parabola. Hence the shape is an "elliptic paraboloid" (is that the right term?). Basically it's like a 3-d parabola that's stretched on the y-axis.1. By considering the cross-sections of the figure, sketch
