@Chris'ssis: I know you've used Cesàro summation in the past. Have you seen this? $$\sum_{k=1}^\infty(-1)^k\sqrt{k}=(2\sqrt2-1)\zeta\left(-\frac12\right)$$
Are you trying to find the area of the triangle enclosed by the three coordinates you've listed, or the area of one of the two smaller triangles formed by drawing a vertical line down from (3,3)?
@Alizter You know what your error is? If the triangle is ABC with A(-1,0) B(3,3) and C(4,0), then the line segments AB and BC aren't orthogonal. Hence the discrepancy in answers
The line segment AB corresponds to the vector $4i+3j$, and the line segment BC corresponds to the vector $i-3j$. In order for two vectors to be orthogonal, the dot product must be zero. However, the dot product between these two vectors turns out to be $4-9=-5$
A function $f:\mathbb{R}^n\to\mathbb{R}^m$ is Lebesgue-Lebesgue(-Borel) measurable iff inverse images of Lebesgue(-Borel) measurable sets are Lebesgue(-Borel) measurable.
It is known that continous functions are borel-measurable and so every homeomorphism is Borel both ways. Is it true to concl...
I'm really stuck trying to prove that the euclidean norm is measurable from the lebesgue sigma algebra to the lesbegue sigma algebra...
it's supposed to be obvious, but i don't have any theorems i can use. My strategy is proving that a positive measure set is mapped to a positive measure set, but i don't see a way to do it yet.
@user116457 Just be patient, someone will post an answer to that question eventually. Some questions can't be answered immediately and require some thinking to truly get anywhere on
the fact is i have no idea how to prove that something is lesbegue-lesbegue measurable without resorting to the outer-measure and inner measure definition and proving equivalence
Wow, I've been building with LEGO and K'nex my whole life and it just dawned on me to try and calculate the amount of possible permutations in putting the pieces together...
A function $f:\mathbb{R}^n\to\mathbb{R}^m$ is Lebesgue-Lebesgue(-Borel) measurable iff inverse images of Lebesgue(-Borel) measurable sets are Lebesgue(-Borel) measurable.
It is known that continous functions are borel-measurable and so every homeomorphism is Borel both ways. Is it true to concl...
