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Well the zero polynomial is clearly in S, so it is non-empty.

To be more precise, I think we need to be at least three times differentiable to show that the symmetric expression for f'' is better than the asymmetric one (in order to use Taylors to give us a handle on the error). If we are three times continuously differentiable, then the error term...

Re: MX2 Integration Marathon The meaning of infinities in the limits of the Riemann integral (ie improper integrals) is defined by the the limit notation you speak of braintic. \int_a^\infty f(x)\, dx := \lim_{T\rightarrow\infty}\left(\int_a^T f(x)\, dx\right).

Actually, the "symmetric formula" for the second derivative wouldn't converge for |x|. \lim_{h\rightarrow 0}\frac{f(h)-2f(0)+f(-h)}{h^2}=\lim_{h\rightarrow 0} \frac{2|h|}{h^2} which does not exist. The formula still doesn't tell us if a function is second differentiable though. Any odd...

Yeah it doesn't imply second differentiability because for functions like |x|, your limit still exists as you approach the point (0,0). Nice to have such a formula though.

Cool, please post anything interesting you come up with. By the way, those formulae don't imply second/third differentiability if the limits exist right? They just tell us the value of the derivative if it does happen to exist.

Ah right. In that case, I can't really think of any analogous picture for n differentiations.

I am still not sure what you are looking for here (in particular I have no idea what you mean by understanding the product rule in a geometric sense), but another way of thinking about it: If you are differentiating the product (uv) n times, you are hitting it with n successive...

Re: HSC 2014 4U Marathon I think I have stuck fairly rigidly to these guidelines (difficulty aside). The biggest reason why these questions are intimidating is that a lot of them are things that you would normally be guided through in baby-steps. Very few of these questions require any...

In future, remember that most of the questions of this form (where they first get you to differentiate something, and then ask you to integrate something similar) follow a predictable pattern: You want to write the function you are trying to integrate as the derivative of something (which will...

\frac{d}{dx}(xe^x)=xe^x+e^x \\ \\ \Rightarrow \int xe^x\, dx = \int \frac{d}{dx}(xe^x) \, dx - \int e^x \, dx = (x-1)e^x + C.

1. Because we explicitly know ALL functions whose n-th derivatives are equal to themselves, and the expression isn't much more complicated. Why specify one object that satisfies a property when it is just as easy to specify all of them? Also, just stating e^x gives us no understanding of how...

I think it is definitely be within the scope of at least some first year linear algebra courses. They vary pretty widely in content and flavour.

Motivated by Sy's last question, here is a fun problem for those with some linear algebra knowledge: $1. If $T:X\rightarrow Y$ is a linear map between vector spaces, and for some $q\in Y$, $p\in X$ is a solution to the equation $Tx=q$, prove that the general solution is given by $x=p+r$ where...

No worries. You might find it a fun/interesting linear algebra exercise to think about why the kernel of such a linear operator (an n-th order linear constant coefficient ode defined on the vector space of C^\infty functions) should have dimension n. Otherwise there's no reason why there cannot...

Then sure, all solutions will in fact be linear combinations of: \displaystyle e^{x\cdot e^{\frac{2\pi i k}{n}}} for k=0,1,...,n-1. (Think about what you are multiplying your function by with every differentiation, and the involvement of the roots of unity becomes clear. This works in greater...

So are you just looking for a function with f^{(n)}=f for some fixed n > 2? Or for a function with f^{(n)}=f for all n > 2?

Re: HSC 2014 4U Marathon (And y = -x + k, k being anything at all).

Re: MX2 Integration Marathon Lol. It's on the easier side of the ones that I post. There is just one thing that might be slightly difficult to spot (*), but the rest is okay. Edit: I also take no credit for this proof, just happened to stumble across it the other day whilst reading things on...

Re: MX2 Integration Marathon Yep. One of those ones where the difficulty is more mechanical than conceptual. (Worthwhile for any 2014'ers to crunch through the working and compare their answers with wolfram alpha's). Dayum as in a cool proof or a hard question? I wasn't sure how much leading...