@Ashwin, it went alright for my first time. I definitely got one problem. Should've gotten a second but I made a stupid mistake. And of course some partial answers here and there.
so we have $e^{i \xi u}$ if I have to write it in terms of v, yet I have $v=\frac{u}{a}$ then I need to have $va=u$ to have a good subsitution for this $e^{i \xi u}$, so in the end I have $e^{i \xi va}$
What trick am I missing? This problem is too nasty to solve the normal way
"Let $\textbf{n}$ be the outer unit normal of the elliptical half-shell $S: 4x^2 + 9y^2 + 36 z^2 = 36, z \geq 0$, and let $\textbf{F} = 5 y \textbf{i} + 5 x^2 \textbf{j} + 5 (x^2 + y^4)^{3/2} \sin e^{\sqrt{xyz}}\textbf{k}$. Find the value of $\mathop{\iint}_S \big(\nabla \times F) \cdot \textbf{n}~\textrm{d}\sigma$
Hint: One parametrization of the ellipse at the base of the shell is $x = 3 \cos t, y = 2 \sin t, 0 \leq t \leq 2 \pi$. But you can avoid a line integral altogether."
@kaj $\operatorname{sup} \{ \frac{1}{n} : n \in \mathbb{N} \}=1$ i guess, because $n=0\pm.0000000000000001=\inf$ but $n=0$ is undefined and i can only take whole numbers in $\mathbb{N}$
A real of complex-valued function defined on $(-\infty, \infty)$ is said to be absolutely integrable on $(-\infty, \infty)$ if $\int_{-R}^{R} \mid f(x) \mid dx$ exists for all $R>0$ and\\
balarka had me spend 30 min trying to prove that $\mathbb{R}^2$ isnt ordered yesterday and then worked out i cant solve it if i dont know what a field is @ted @kaj lol nooo
@TedShifrin i was trying to show $\Bbb{R}^2$ wasnt ordered and then after 30 min he said i couldnt do it without knowing what a field was
@TedShifrin i said you can't make it ordered because you cant for example order $(2,0),(0,2)$ as they are equally far from the origin in two positive directions
Do you think user @Bahaviour is truly 24 as listed, or it is part of his absorbed persona for the current name(I don't recognize the person in his picture).
Find the determinant. To make calculations easier, work modulo $2$! The diagonal is $1$'s, the rest are $0$'s. The determinant is odd, and therefore non-zero.
How to find determinant matrix using modulo 2? Could you enlighten me @DanielFischer? Thanks.
@Venus: Since the determinant is a polynomial in the matrix entries, reducing the entries modulo $n$ does not change the value of the determinant modulo $n$.
I have an algorithm which takes as input the series expansion of:
$$\frac{-(1 + ax(-2 + x + ax))}{-1 + ax} \tag 1$$
or expressed differently:
$$\left\{a^0,(-a)^1,a^1,a^2,a^3,a^4,a^5,a^6,a^7\right\}$$
and applies the convoluted divisor recurrence via this Mathematica program:
Clear[t, n, k, i...
Find the determinant. To make calculations easier, work modulo $2$! The diagonal is $1$'s, the rest are $0$'s. The determinant is odd, and therefore non-zero.
I have an algorithm which takes as input the series expansion of:
$$\frac{-(1 + ax(-2 + x + ax))}{-1 + ax} \tag 1$$
or expressed differently:
$$\left\{a^0,(-a)^1,a^1,a^2,a^3,a^4,a^5,a^6,a^7\right\}$$
and applies the convoluted divisor recurrence via this Mathematica program:
Clear[t, n, k, i...
