@usukidoll Is it so? I was under the impression that it is elliptic when $b^2 - ac <0$, hyperbolic when $b^2 - ac >0$ and parabolic when $b^2 - ac = 0$
Anyone would mind having a quick look and making suggestions? I do obtain a solution but it doesn't exactly make sense, but I cannot find an error in my derivation.
@usukidoll It is elliptic if the characteristic equation has no real roots; it's parabolic if it has a repeated root; and it's hyperbolic if it has two distinct roots. In this case, $4x^2+(x+2)^2=0$ has no real roots.
The operator is represented by multiplying the Fourier Transform by $5x^2+4xy+4y^2$ and if there are no real roots, this can be inverted to give good smoothness to the solutions.
@usukidoll The Fourier transform maps linear differential operators to polynomials, so that's how you get the characteristic polynomial for these equations in the first place
Hey everyone, I have a basic understanding of mathematics (A-Level) and want to keep learning. What resources would you suggest when wanting to solve problems such as those on project euler?
@MikeCon94 Just keep trying them! And read the solution pdfs they upload when you finish one. It's handy to know some basic number theory like how modular arithmetic works but I think most problems you shouldn't need to know anything beforehand just keep experimenting and looking for patterns
@robjohn, Please let me know how $$\sum_{r=0}^{m+n} \left( \sum_{k=0}^{r} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r} = \sum_{r=0}^{m+n} \sum_{k=0}^{m+n} \binom{n}{k} \binom{m}{r} \ x^{r+k}$$? I followed your answer, but can't go further than $$\sum_{r=0}^{m+n} \sum_{k=0}^{m+n} \binom{n}{k} \binom{m}{r} \ x^{r+k}=\sum_{r=0}^{n} \sum_{k=0}^{m} \binom{n}{k} \binom{m}{r} \ x^{r+k}.$$
Thanks @AlexJBest I really struggle to spot the patterns and in the pdfs they often mention things I have heard of before but I guess like everything practice, practice, practice!
@MikeCon94 Yeah, half the idea is you learn tricks just by doing them that you can apply to later problems. I just generally make a really stupid way of solving a problem that is really slow, make it output loads of information as it goes along and try and search for a pattern which can then be used to speed the bad program up.
Not a bad idea in the past I have just wrote brute force programs and left them running until output but want to learn how the maths to make my future attempts better :)
@robjohn To better see that, one needs to write that as follows $$\sum_{n=1}^{\infty} \frac{H_{2n}}{(2n)^2}=\frac{11}{16} \zeta(3)$$ The rest gets reduced to what you said.
@Jasper, I find certain of his books far better than others. His complex analysis and real analysis books are quite good. Algebra, I don't like so much.
I would recommend you only go back to the elementary stuff when you find a lacuna in your knowledge.
@robjohn, when I do $r\mapsto r+k$, I get $$\sum_{r=0}^{m+n} \left( \sum_{k=0}^{r} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r}=\sum_{r+k=0}^{m+n} \left( \sum_{k=0}^{r+k} \binom{n}{k} \binom{m}{r} \right) \ x^{r+k},$$ so still not getting $$\sum_{r=0}^{m+n} \sum_{k=0}^{m+n} \binom{n}{k} \binom{m}{r} \ x^{r+k}.$$
@PedroTamaroff HMM! @PedroTamaroff, Will you please show the complete explanation here? I have encountered such problem first time, so am very confused.
For a symmetric real matrix $A$, denote by $\rho(A)$ the spectral radius of $A$, and by $\sigma(A)$ the largest absolute row sum of $A$; that is, $\sigma(A)=\max_i \sum_j |a_{ij}|$, where $a_{ij}$ are the elements of $A$. We thus have $\rho(A)\le\sigma(A)$. I don't get the final statement (it is that trivial ?)
I think probability is particularly alluring, @Studentmath. Because we love to puzzle over how we can see things different ways, and some of them aren't right :P
@Sush I was using $r$ and $k$ both going from $-\infty$ to $\infty$. Of course, because of the binomial coefficients, these sums are finite. If you insist on using limits that are codependent, then you must put $r$ on the inside before subsituting...
@Ted I think it is it, yeah. You can solve things in many different ways in calculus, linear algebra, so on. But here it's completely different approaches, and sometimes they seem completely reasonable, yet they are just wrong.
@Sush People cannot always be at your beck and call on chat. It is not really nice to demand that they respond. I was away from my computer at the time, anyway.
@BalarkaSen I like number theory and I like applied math, and both groups seem to use that notation differently (one for big O and one for little o). So I can never decide if I want to use it or not :)
How did your instructor expect you to do this, @Hippa? You need to know that $A^2 X = \mu X \implies AX = \lambda X$ for some $\lambda$. I don't see an immediate way without some knowledge, @Hippa.
@TedShifrin math.stackexchange.com/questions/831860/… @PedroTamaroff 's comment, does that mean that the sequence $\dfrac{1}{u_n}$ converges towards $0$ ? It seems way too easy ...
I've always found multivariable a lot more interesting, and it's something that most mathematicians don't know quite as well. It's more geometric and that doesn't appeal to all mathematicians.
I simply did the general thing for general i, got a nice expression and added up i=2,4,6
The exercise now is:
Let there be n independent normal distributions, $X_1,..., X_n$, each with expectency 0 and variance $\sigma^2$. We define $Y=exp(\bar{X}_n)$. Find E[Y].
Which makes me think it's simply the average all of the n' RV together, and then I can reach that with the moments generating function @cc
But I am unsure, so I will ask for clarification..
