Search results

Re: Several Variable Calculus |sin(x)/x-1|=|x|^2 ? Surely you mean something like: |sin(x)/x-1| = O(|x|^2).

zero likely.

Re: Several Variable Calculus Might not be easy to visualise the graph of the function on all of R^2, but you should certainly be able to visualise what it looks like on the slices x=const. or y=const which is all that matters for seeing/proving the nonexistence of the iterated limit. It is an...

Re: Several Variable Calculus I assume d(x,y) is supposed to be ||x-y|| and the set is some normed vector space V with norm ||.|| (e.g. R^d). (Please specify more if this is not the intended setting.) Then these two metrics are not (strongly) equivalent because V is bounded with the delta...

Re: Extracurricular Integration Marathon One way (probably pretty inefficient so am not going to bother texing it) is to just differentiate with respect to the parameters. I let x=a^2, y=b^2, and evaluated \partial_xI(x,y)=\frac{\pi}{2(x-y)}\left(1-\frac{\sqrt{y}}{\sqrt{x}}\right) using the...

Re: Linear Algebra Ps this isn't really linear algebra at all.

Re: Linear Algebra In fact this argument straightforwardly generalises to tell us that the subset of Z_n consisting of only the equivalence classes coprime to n form a group w.r.t. multiplication. This yields Euler's theorem just as the version originally posted will yield Fermat's little theorem.

Re: Linear Algebra Suppose the integer a is a representative of a nonzero equivalence class in Z_p. Then a is coprime to p and hence am+pn=1 for some integers m and n. Projecting to equivalence classes we get [a][m]=1 (mod p). ([z] denotes the equivalence class in Z_p of the integer z.)...

Re: Several Variable Calculus Eg, consider the lines (x,y)=(t,0), (x,y)=(2t,0). These lines coincide exactly, so every point on the x-axis is a point of intersection. Yet the only place where (t,0)=(2t,0) is at the origin.

Re: Several Variable Calculus You have two parametric curves. Their points of intersection don't necessarily have the same parameter as a point on the first curve as they do as a point on the second curve.

Re: HSC 2017 MX2 Marathon Yeah fair enough, they were a lot easier back in the day. Still should probably put questions like it in the advanced thread (Dan only created it today I believe, but for the last couple of years we have usually split the questions into two marathons, one for...

Re: HSC 2017 MX2 Marathon Haha that was my suspicion.

Re: HSC 2017 MX2 Marathon It's most instructive for you to prove these yourself so I will just provide some hints. Consider the forms of sin(X+Y), sin(X-Y), cos(X+Y), cos(X-Y) upon expansion. Is it possible to combine these to obtain products of the form sin(X)cos(Y), sin(X)sin(Y)...

Re: HSC 2017 MX2 Marathon $Partial fractions can be used to decompose the LHS into \\ \\ $7+\sum_{k=0}^6 \frac{\omega^k}{z-\omega^k}$\\ \\ where $\omega=\textrm{cis}(2\pi/7)$\\ \\ Now let $H(z)=(z^7-1)\sum_{k=0}^6 \frac{\omega^k}{z-\omega^k}.\\ \\$ We have $H(\omega^k)=\omega^k\prod_{i\neq...

P_N is always positive by definition, so we just need to bound it above by something that tends to zero. In the notation I introduced earlier, this upper bound is: C/exp(h_k-h_N) where h_n = 1 + 1/2 + ... + 1/n denotes the n-th partial sum of the harmonic series, and C and k are constants...

Re: International Baccalaureate Marathon Depends what you mean by suitable. a) Will the answer be the same? Yes, obviously. Summing a geometric series manually will give you the same result as applying the formula for the sum of a geometric series. b) Will markers treat the solution as being...

Re: International Baccalaureate Marathon 0.892 and 0.655536 are wrong. By doing it manually like this you are greatly increasing the number of points at which you can mess up the calculation. And of course this won't be feasible if 9 is replaced by 1000. The correct answer is the expression...

It's a common trick to convert infinite products into infinite sums which are in some ways nicer to work with. There were also two facts used that I didn't bother writing proofs of (you should try to make sure you understand why they are true): 1. log(1+x) =< x for all x > -1 where these...

By the degree condition nP(n)/Q(n) tends to some positive constant limit (the ratio of leading coefficients.) Let's call this 2c. Then for sufficiently large n (say n > K), we have nP(n)/Q(n) > c, and (1-c/n) > 0. Then for N > k: P_N=\prod_{j=1}^K (1-\frac{P(j)}{Q(j)})\cdot \prod_{j=k+1}^N...

It is always zero when the degree of the denominator exceeds the degree of the numerator by exactly one. (As in the original question in this thread, although to prove this more general statement you do not have telescoping at your disposal.) Your second product does not satisfy P(n) < Q(n) for...