OK, but I've done this by hand. I think I want to eliminate $t$.
It shows to go how difficult it is to see from an equation what is actually going on. There's another exercise (perhaps in Chapter 6) to give the equation of a torus, and it can be well-disguised.
If I eliminate $t$, I get $y=x^2-s^2$, $z=(x-s)^2(x+2s)$. This looks better.
I'm not seeing it now. I'll get back to you tomorrow :)
Except for $x^4y$ and $x^6$, all of those terms end up being used. I suspect it has something to do with the fact that $x^6$ is the only one with degree six in $s$, and $x^4y$ is the only one with degree five in $s$
so they can't cancel out
@arctictern Tried that, didn't work as well as I'd hoped
Note that if you have $ax+by+cz=0$, then $x=-\frac1a(by+cz)$, so you can set $y=1$, $z=0$ and get $(-b/a,1,0)$ and you can set $z=1$, $y=0$ and get $(-c/a,0,1)$.
Those are two vectors that satisfy the equation with $d=0$.
@Daminark: I actually did that in my course and then proved to them that a (constant coefficient) linear $n$th order ODE had an $n$-dimensional space of solutions.
They're fine questions, but not under hard time pressure, @Daminark. Computations take too much time. That 2nd order equation gives you $e^{t\begin{bmatrix}0&1\\-1&0\end{bmatrix}}$, which I presume he did in class?
1 was linear algebra, part a was to prove that a matrix $A\in M_2(\mathbb{C})$ is triangularizable, to prove that this did not hold in general for $A\in M_2(\mathbb{R})$, and to prove it did for $A\in M_3(\mathbb{C})$.
Problem 1: Find the largest open set in the plane such that $\omega = \frac{(x-1)dy + ydx}{(x-1)^2 + y^2} + \frac{(x+1)dx + ydy}{(x+1)^2 + y^2}$ is a smooth 1-form
in my text book when equation are given like ax+by+cz=0 and dx + ey+fz = 0 then ratio is $$\dfrac {x}{bf - ec} = \dfrac{y}{at - dc} = \dfrac{z}{ae - bd}$$ So in general how they get all denominators?
@eurocoder so you say that a method has order of convergence $q$ if the error in sup norm goes to $0$ as $h\to 0$ as $O(h^q)$, where $h$ is the offset between the points you exstimate the solution at
More formally you need $\lim\limits_{h\to 0} ||y_n-y(t_n)||=O(h^q)$, where $t_i$ are the points you estimate the solution at, $y$ is the actual solution and $y_n$ is the estimate solution at $t_n$
Actually there's no sup norm involved, take the max over $n$ and it's just a normal modulus
So it is kinda a nested version of the above equation, which then after some algebra (details forgot) simplifies into something that looks like the annuity formula
@Secret so do you think this would be a fair description of the situation: $\frac{\mathrm{d} P}{\mathrm{d} t}-2000t$ i.e. change in P (principal) - monthly payment
@Secret the question just asks me to find the form of the first order ODE for this situation
If you are unsure, the monthly payment formula can always be looked up, and you can get the ODE straight away by differentiate P wrt t that way. However I think it is more important to understand where it came from and why the formula is that formula
Let $S_n = \sum_{i=0}^n a_i$.
Now define The Cesàro Sum as
$$ C = \lim_{n \to \infty} \frac{ \sum_{k=0}^n S_k}{n} $$
Is it always true that
$$ C = \lim_{n \to \infty} \frac{\sum_{k= n - L(n)}^n S_k}{ L(n)} $$
Where $L(n)$ is any strictly nondecreasing function such that $ 2 < L(n) < ln(n) $.
If anyone is interested, I'm hosting a big number contest. Code your number in any language (no experience needed), maximum of 256 characters, not including spaces, and try to reach the largest number you can. First submissions due Saturday. :-) chat.stackexchange.com/rooms/51337/…
Let $S_n = \sum_{i=0}^n a_i$.
Now define The Cesàro Sum as
$$ C = \lim_{n \to \infty} \frac{ \sum_{k=0}^n S_k}{n} $$
Is it always true that
$$ C = \lim_{n \to \infty} \frac{\sum_{k= n - L(n)}^n S_k}{ L(n)} $$
Where $L(n)$ is any strictly nondecreasing function such that $ 2 < L(n) < ln(n) $.
