TITLE: Textbook Answer vs. My (and Symbolab's!) answer Solving the Derivative of a Polynomial
QUESTION [1 upvotes]: The book Calculus in Five Hours asks readers to solve the following problem:

Determine the derivative of the following polynomial:
5x7 + 16x3 - 9 at x=4.

I've done this problem a few times now and have come up with:
35 * 46 + 48 * 42 = 144,128.  Symbolab agrees with me. Wolfram Alpha also agrees with me.
But the book (link above) gives the answer 144,110, as does the (slightly buggy) calculator page to which it links.  Am I missing something? Can anyone see how the (incorrect) answer might have been determined?

REPLY [1 votes]: Your answer is correct.  (One can tell from the last two digits that $144128$ is a multiple of $4$, as it should be, whereas $144110$ is not.)
The calculator page to which the book links gave me $144100$; perhaps it gave that rather than $144128$ because it gave only a few significant figures.
As far as why the book gives $144110$, I don't have a very satisfying explanation; it could be that $144100$ was intended but a typographical error was made -- that happens sometimes.


TITLE: Properness of moduli space
QUESTION [3 upvotes]: Here is a question about construction of moduli space of Higgs bundles or Hitchin pairs. Let's say we are constructing the moduli space of stable Hitchin pairs or Higgs bundles (doesn't matter at this point), E-->E\otimes K over a smooth curve, a surface, etc with the condition that E is a stable sheaf. Is this construction done somewhere? In particular I am concerned about the properness of this moduli space (the stable E's cut an open subscheme inside their projective Quot scheme). So before even discussing the Higgs stability, we are dealing with an open scheme of E's. Is there a proof somewhere which uses valuative criteria etc, to prove the properness of this particular moduli space?---It would be great if someone else already did this (?). Thank you.

REPLY [1 votes]: Even if we restrict to Higgs bundles where $E$ is stable as a vector bundle, the moduli space is not proper. If you take $(E,\sigma)$ with $\sigma \neq 0$, then $$\lim_{t \to \infty} (E,t\sigma)$$ does not exist.
Another way to see this is that there are nonconstant functions on the moduli space of Higgs bundles (think of the Hitchin fibration), which would be impossible if the moduli space were proper.
One way to think about this failure of this properness is that the moduli space of stable Higgs bundles is isomorphic to a moduli space of stable torsion sheaves on the total space of the cotangent bundle of your base $X$. The moduli space is not proper, since the total space isn't proper, but there's a fairly natural compactification coming from taking torsion sheaves on the projective completion of the total space of the cotangent bundle. We can think of this as allowing the spectral varieties of the Higgs bundles to go to infinity.


TITLE: Let $A=\{(x,x) \in \mathbb{R}^2 : 0 \le x \le 1\}.$ Find the $2-$dimensional Lebesgue measure $m^\ast(A).$
QUESTION [0 upvotes]: Let $A=\{(x,x) \in \mathbb{R}^2 : 0 \le x \le 1\}.$ Find the $2-$dimensional Lebesgue measure $m^\ast(A).$

I found this question here without answers and I cannot solve it. I have a bit trouble finding a cover for $A$. I tried something like $I_k=(-\frac{\varepsilon}{k}, 1+\frac{\varepsilon}{k})$, but this doesn't work. Is there some general method to find covers for sets or how should I do this? I feel like the measure should be zero since the line that this set is doesn't have any area, but I need to find some cover with what I can use here.

REPLY [2 votes]: Let $n\in\mathbb{N}$. For each $1\leq i\leq n$ define $I_n^i=[\frac{i-1}{n}, \frac{i}{n}]^2$. Then your set is contained in the union $\cup_{i=1}^n I_n^i$, and so its measure is bounded by $\sum_{i=1}^n m(I_n^i)=\sum_{i=1}^n\frac{1}{n^2}=\frac{1}{n}$. Since $n$ is arbitrary, this proves that the measure is $0$.


TITLE: Is $A \rightarrow S^{-1} A$ epi?
QUESTION [0 upvotes]: this question must be the most stupid I have ever asked.
If $A$ is a commutative ring and $S$ a multiplicative subset the usual inclusion induces a homeomorphism onto the image $\text{Spec} (S^{-1}A) \rightarrow \text{Spec} (A)$ and, then, it's a mono.
By the anti-equivalence between affine schemes and commutative rings, the inclusion $A \rightarrow S^{-1} A$ must be an epi. However this would be equivalent to the statement that given $y \in A$ and $s \in S$ there exists $x \in A$ and $s' \in S$ such that $s'xs = s'y$ , which apparently is not true for any ring $A$.
So what's the problem? And if it's true, how to prove the above statement without the anti-equivalence?

REPLY [1 votes]: No for "this would be equivalent to ...". See MO/109 for some results about epimorphisms of commutative rings. (Don't confuse them with regular epimorphisms of commutative rings, which are surjective.)
Also, it is not true that a morphism of schemes which is a homeomorphism onto its image is a monomorphism. Certainly it becomes a monomorphism of topological spaces, but applying the forgetful functor is often not enough! It is true that $A \to S^{-1} A$ is an epimorphism of commutative rings (by the universal property) and hence $\mathrm{Spec}(S^{-1} A) \to \mathrm{Spec}(A)$ is a monomorphism of affine schemes. Actually, it is a monomorphism in the whole category of schemes because $\mathrm{Spec} : \mathsf{CRing}^{op} \to \mathsf{Sch}$ is right adjoint to $\Gamma$ and hence preserves monomorphisms.


TITLE: Placing a cone in a circle
QUESTION [2 upvotes]: I am trying to make a neat charting app, yet I am having dificulty with some of the math. That's why I'm here!
I have a "pie slice" at a specific angle. I don't want the slice to just be a simple little pie-slice, no, I want to make it a cone shape!
Where do I place the circular cap? If I draw a triangle around my "pie-slice", I can see that the circle at the end must be tangent to all sides of the triangle. Whats the radius of that circle? Where's the center point? What are the points which are tangent to the "triangle"?

The variables R1, @1 (highlighted in blue), and @2 (highlighted in green) are known, but how could I find R2, P1, P2, P3, and @3 (highlighted in red)?
Thanks everybody!

REPLY [1 votes]: You are trying to find the incircle of the equilateral triangle with side lengths $R_1$ and enclosed angle $\theta_2$.
Confer http://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle for some formulae.
The missing side will have length $l = 2\sin(\frac{\theta_2}2) R_1$.
For instance the radius $R_2$ is given by
$$s := R_1 + \frac l2 = R_1(2 + \sin(\frac{\theta_2}2))\\
R_2 = \sqrt{\frac{(s-R_1)^2(s-2\sin\frac{\theta_2}2 R_1)}{s}} = R_1 \sqrt{\frac{(1-\sin\frac{\theta_2}2)^3}{1+\sin\frac{\theta_2}2}}$$
If you label the vertices of the triangle with $M, Q_1$ and $Q_2$,
$$P_3 = \frac{R_1 (Q_1 + Q_2) + l M}{2R_1 + l}$$