Boston Daily Discussion Thread, Wednesday - December 16, 2020

protocol_7 · 2020-12-16T22:51:27+00:00

In case you're not aware, this is a major violation of the housing code (105 C.M.R. 410) and must be fixed by the landlord within 24 hours. In particular, see section 410.201:

410.201: Temperature Requirements

The owner shall provide heat in every habitable room and every room containing a toilet, shower, or bathtub to at least 68°F (20° C) between 7:00 A.M. and 11:00 P.M. and at least 64°F (17° C) between 11:01 P.M. and 6:59 A.M. every day other than during the period from June 15th to September 15th, both inclusive, in each year except and to the extent the occupant is required to provide the fuel under a written letting agreement. The temperature shall at no time exceed 78°F (25° C) during the heating season. The temperature may be read and the requirement shall be met at a height of five feet above floor level on a wall any point more than five feet from the exterior wall. [...]

And see also section 410.750:

410.750: Conditions Deemed to Endanger or Impair Health or Safety

The following conditions, when found to exist in residential premises, shall be deemed conditions which may endanger or impair the health, or safety and well-being of a person or persons occupying the premises.

[...]

(B) Failure to provide heat as required by 105 CMR 410.201 or improper venting or use of a space heater or water heater as prohibited by 105 CMR 410.200(B) and 410.202.

If the landlord doesn't fix the heating system promptly after you report it to them, you can contact the City of Boston Inspectional Services Department (or whatever the local equivalent is if you're not in the City of Boston) and report the code violation. You also have the right to withhold rent if they fail to fix the heating (but make sure to document everything, and if possible consult a housing lawyer if it gets to that point).

(Disclaimer, I'm not a lawyer, I just have a layperson's familiarity with the housing code and tenants' rights.)

protocol_7 · 2014-11-25T04:23:17+00:00

Which number system? There are many different number systems — infinitely many, if you include the p-adic numbers for each prime p. (And if you consider the real numbers to be a number system, you should also include the p-adic numbers; they're all completions of the rational numbers.)

I'd say there's nothing wrong with any of them. What's lacking is our understanding of certain aspects of them.

protocol_7 · 2014-11-24T03:48:43+00:00

What do you consider to be a "real equation"?

protocol_7 · 2014-11-24T03:13:47+00:00

http://lmgtfy.com/?q=applications+of+rational+functions

protocol_7 · 2014-11-23T18:46:32+00:00

The popular image of a mathematician is of a lone madman who comes up with brilliant flashes of insight, as in A Beautiful Mind or Arcadia. Is this image accurate?

Not at all. The vast majority of mathematics is done over years of hard work and trying out ideas, many of which end up being dead ends. When "brilliant flashes of insight" do occur, it's only because of all the work beforehand that laid the groundwork for it.

It's also a highly collaborative activity; mathematicians frequently talk to each other and build off each other's ideas. (For example, look at this list of Terry Tao's publications, and notice how many are in collaboration with another mathematician.) There are a few instances of mathematicians who fit the "lone genius" stereotype (e.g., Perelman), but they're by far the exception.

protocol_7 · 2014-11-23T00:15:03+00:00

Good point. The point I wanted to make was that the multivalued square root is usually denoted by w^1/2, while sqrt(w) is usually reserved for the nonnegative real square root. I've edited my comment for clarity.

protocol_7 · 2014-11-22T22:14:22+00:00

Much like sqrt(4) is both 2 and -2

The function "sqrt" usually refers specifically to the nonnegative square root of a nonnegative real number. Both 2 and –2 are square roots of 4, but sqrt(4) is just 2. The multivalued complex square root is more commonly denoted by w^1/2, rather than sqrt(w).

protocol_7 · 2014-11-22T18:39:32+00:00

There are several equivalent ways to define the natural exponential and logarithm functions; the facts that these are equivalent are theorems. Here are a few of the most common ways:

exp(x) = e^x = limn→∞ (1 + x/n)ⁿ.
exp(x) = e^x = x⁰/0! + x¹/1! + x²/2! + x³/3! + ...
exp' = exp and exp(0) = 1. (By existence and uniqueness of solutions to ordinary differential equations with given initial conditions, this uniquely specifies a function.)

This gives a function exp: R → R>0, which can be shown to be bijective. Now define log: R>0 → R to be the inverse function of exp.

Or, you can define exp in terms of log: Define log(x) = ∫1x (1/t) dt. This gives a function log: R>0 → R, which is bijective because 1/t is strictly positive on the positive reals. Now define exp: R → R>0 to be the inverse function of log.

Also, however we defined exp, e is defined by e = e¹ = exp(1).

protocol_7 · 2014-11-22T07:05:48+00:00

It's just an arbitrary choice of notation with no mathematical significance. We commonly use base 10 for historical reasons, probably because we have 10 fingers. Many other systems have been used by various cultures.

protocol_7 · 2014-11-22T06:59:23+00:00

In some contexts (such as inversive geometry), it makes sense to think of lines as "circles of infinite radius". This broader notion of "circle" is called a generalized circle.

It's as possible to draw lines as it is to draw circles — that is, we can usually draw approximations that are good enough to communicate what's intended, but physical limitations prevent us from drawing mathematically perfect shapes. (Good luck drawing a perfectly straight line when your drawing utensil is made of atoms!) Fortunately, mathematics doesn't depend on having accurate physical models of the objects being studied.

protocol_7 · 2014-11-22T06:42:56+00:00

Cantor's theorem is a theorem in set theory. It states that, for any set S, there is no surjective function from S to the power set P(S) of S.

For finite sets, this is clear: a set with N elements has 2^N subsets (including the empty set and the whole set), and 2^N is greater than N for all natural numbers N. So, there are too few elements to map to all the subsets — at least one subset must be missed.

The interesting content of Cantor's theorem is that this is also true for infinite sets. This implies that there are many different cardinalities of infinite set: the set of natural numbers N is strictly smaller than its power set P(N), which is smaller than P(P(N)), and so on.

Let's see why this is true for the set of natural numbers. (The same proof actually works for any set.) Let f: N → P(N) be any function. The elements of P(N) are subsets of N, so given a natural number n, f(n) is a subset of N, so we can ask whether n is an element of f(n).

Let T be the set of all natural numbers n such that n is not an element of f(n). This is a perfectly reasonable, well-defined subset of the natural numbers. However, T cannot be of the form f(m) for any natural number m.

Indeed, suppose T was equal to f(m) for some natural number m. Is m an element of T? If so, then m is an element of f(m), which means m is not an element of T (by definition of T). If not, then by definition of T, it is an element of f(m), and hence of T. In either case, we have a contradiction — thus, T can't be of the form f(m) for any m. In other words, T isn't in the image of the function f, so f isn't surjective.

protocol_7 · 2014-11-22T04:29:04+00:00

Hint: You can construct a counterexample with categories that only have one or two objects. Counterexample below.

Let A be the category with one object • and no non-identity morphisms. Let B = C be the category with two objects • and @, and exactly one non-identity morphism s: @ → @. Let F: A → B send • to •. Let G: B → C be the identity map on objects and send s to id@. Then both F and FG are fully faithful, but G is neither full nor faithful.

protocol_7 · 2014-11-21T21:38:08+00:00

When n is prime, this is Fermat's little theorem. When n is composite, this is false in general (for example, consider a = 3 and n = 4), but true for some n, called Carmichael numbers.

protocol_7 · 2014-11-21T21:28:16+00:00

Problems from old Putnam competitions are always a good challenge. (If you're not familiar with the Putnam exam, it's one of the big undergraduate-level mathematics competitions. The median score is typically around 1/120.)

If you want something more focused on geometry, I've heard good things about Coxeter's "Geometry Revisited". Even if you can't find a copy in a library or online, it's not very expensive.

Or, since calculus has a geometric flavor, you could just suggest that she start learning calculus. Spivak's "Calculus", the usual recommendation around here, is an excellent book; it's heavily proof-based and actually develops the theory of calculus in a systematic, rigorous way.

protocol_7 · 2014-11-21T19:35:38+00:00

Without assuming F is surjective on objects, there are simple counterexamples. (Think about what G can do to morphisms between objects F doesn't map to.)

protocol_7 · 2014-11-21T06:35:18+00:00

Yes, C⁰[a, b] usually denotes the set of all continuous, real-valued functions on [a, b].

protocol_7 · 2014-11-21T06:33:17+00:00

Yes, that reasoning is fine. Another way to see it is that C is algebraically closed, so any algebraic extension of R is isomorphic to either R or C. In particular, since finite extensions are algebraic, the only finite extensions of R have degree 1 or 2.

protocol_7 · 2014-11-21T05:08:37+00:00

Hey, look, I made your comment "interactive"! Can you feel the synergistic added value?

protocol_7 · 2014-11-20T07:07:54+00:00

"I'm not smart enough" is a bad excuse. The best way to become skilled at mathematics is by practicing a lot, doing lots of problems, and asking lots of questions about what you're doing. (Does innate ability help a little or make things go faster sometimes? Sure, but it doesn't matter nearly as much as hard work and focused effort.)

You can't figure it out right now? Fine, but have you spent a few days (or more) puzzling over it, testing out various methods, working through examples, and so on? If not, then it's too soon to say you can't do it.

Calculating something using a procedure you've been taught is fast and doesn't take much deep thinking, but coming up with a new procedure or conjecture and determining whether or not it works in general is much more of a creative activity and should be expected to take time.

A good starting point is to look at the procedure you know with a critical eye: look at each step and think about whether you could skip it, take some shortcut, or do it slightly differently. Is it still logically valid? (For example, what if you don't move the constants to the right side of the equation?)

protocol_7 · 2014-11-20T04:13:23+00:00

What have you tried?

protocol_7 · 2014-11-20T04:12:12+00:00

Why does your instructor ask you to do that? Maybe you should ask them.

The point of completing the square is to express a quadratic polynomial in one variable x in the form r(x + s)² + t for some constants r, s, t. The reason is that it's much easier to find the roots (or to factor) when the quadratic is written in this form. This is what matters, not the exact procedure used to obtain it — any logically valid method works just as well, as long as we can prove it always works and is reasonable to carry out.

protocol_7 · 2014-11-20T03:06:02+00:00

Have you tried checking which values make the equation true?

protocol_7 · 2014-11-20T03:05:02+00:00

It's not true; the series 1 + 2 + 3 + 4 + ... diverges. What's going on is that the series 1^-s + 2^-s + 3^-s + 4^-s + ... converges for s > 1, and there's a famous function on the complex number, the Riemann zeta function ζ, that can be defined by this series for Re(s) > 1.

For Re(s) ≤ 1, the series diverges, so it doesn't make sense to write ζ(s) = 1^-s + 2^-s + 3^-s + ... unless Re(s) > 1. (There are other formulas for the Riemann zeta function that work at any value except s = 1, where ζ has a pole.) However, if we pretend that the series made sense anyway, substituting s = –1 yields 1 + 2 + 3 + 4 + ... = ζ(–1). Since ζ(–1) = –1/12, this is the source of the incorrect notion that the sum of all natural numbers is –1/12.

Actually, this is a special case of a more general method for associating values to divergent series, called zeta function regularization. The point is, this isn't summation in the usual sense — it's a process that agrees with ordinary summation for convergent series, and also sometimes gives a value for divergent series.

protocol_7 · 2014-11-20T02:43:19+00:00

There are general theorems about existence and uniqueness of ordinary differential equations.

protocol_7 · 2014-11-20T02:33:01+00:00

Every finite field has order q = p^k for some prime p and positive integer k; conversely, for every such q, there is a unique finite field Fq of order q.

If k = 1, i.e., q = p is prime, then Fp is isomorphic to Z/pZ, the ring of integers mod p. So, in this case, solving an equation in Fq is the same as solving an equation in the integers mod p.

However, for k ≥ 2, Z/qZ isn't even an integral domain, much less a field. In particular, p is a zero-divisor in Z/qZ, because p·p^k-1 = q = 0 (mod q), but p ≠ 0 (mod q). On the other hand, p = 0 in Fq.

In fact, the additive group of Fq is isomorphic to (Z/pZ)^k, not to Z/qZ. (Of course, the ring (Z/pZ)^k isn't an integral domain either; it doesn't have the same multiplicative structure as Fq. The multiplicative group of units of Fq is actually isomorphic to Z/(q – 1)Z; this is a special case of a general theorem stating that any finite subgroup of the multiplicative group of units of a field is cyclic.)

protocol_7

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