Math question thread

[Eternal_Freedom]

Is there an equation for a graph that will produce a spiral from the origin? If so, what is it?

Also, for a circle on a graph r^2=x^2 + y^2, does this hold true for a sphere on a three-dimenisional graph?

[Surlethe]

The curve t \to (t, t) is a spiral in polar coordinates. In Cartesian coordinates, that's t(cos(t),sin(t)).

I'm not sure what your second question means. Can you clarify?

[I am thinking still about the Bayesian problem. I want to prove the following conjecture:

Given this procedure and appropriate hypotheses on a and b (e.g., neither = 0.5), I can produce a sequence of distributions d_i "approaching" \delta_c in the sense that for any e > 0 and any M > 0 there exists some N > 0 and a sequence of integers i_1 < i_2 < ... with for each j, each 0 < k < M, | d_{j+k} - \delta_c | < e.

That is, the sequence does not NEED to converge to the delta function above c. However, it gets arbitrarily close to converging. A similar sort of weak convergence might be the sequence c_k = 1 if k is prime, 0 else: if you go far enough, you get arbitrarily many arbitrarily long sequence of 0, but the sequence doesn't converge to 0 because there are infinitely many primes.]

[Eternal_Freedom]

My second question was poorly worded:

On a 2-D graph, a sircle is defined as r^2=x^2 + y^2 around the origin.

On a 3-D graph, what is the equation for a sphere? Is it r^3 = x^3 + y^3 + z^3? Or are all those values supposed to be squared?

[starslayer]

They are squared*. Additionally, in polar coordinates, a spiral does not need to be parameterized; a simple Archimedean spiral has the form r = a+bθ, where a and b are constants. There are several other kinds of spiral as well which can be written as functions in polar coordinates.

*Think of what a sphere or a circle is: the set of all points that are the same distance from some central point (the distance is just the radius). By the Pythagorean Theorem, then, the surface r2 = x2+y2+z2 is a sphere, and similarly the curve r2 = x2+y2 is a circle in Cartesian coordinates. In polar coordinates, this is obviously extremely simple, since you just have r = const.

[Psawhn]

In addition to r2=x2 + y2 + z2, this property applies to any number of dimensions. A practical example is the error of GPS positioning. A GPS position not only estimates your x, y, and z position, but also the error in your receiver clock. DOP stands for "Dilution of Precision," and GDOP (Geometric Dilution of Precision, if I recall correctly) is actually related to the square root of the sum of the squares of the x, y, z, and t uncertainties.
(DOP actually takes measurement accuracy out of the equation - it is a measure of geometric strength, and is computed by the relative positioning of the satellites and your receiver)

My question is more of a simple one to do with statistics. Suppose we have a number of values that are normally distributed. We then take the 'sequential difference' between all the points (Is there a proper name for it?). So, for example, if we have a set of points {4, 6, 6.5, 7, 8}, then we also have the 'sequential difference' set {2, 0.5, 0.5, 1}. What probability distribution will this second set of values tend towards?
(Also: feel free to correct my conventions.)

[Surlethe]

A fellow grad student and I have had some fun thinking about the Bayesian problem. He thinks he can get some sort of strong convergence of a test statistic in almost all probabilities of false negative/positives. Maybe I'll make his answer rigorous.

Okay, so Psawhn, I don't quite understand your question. Are you asking what probability distribution will the second set of values tend towards as the number of points in the sequence tends to infty?

[Psawhn]

Maybe a better way of phrasing the question would be: "what probability distribution would be best used to model or approximate the second set of values?"

[freker]

Not so much a formula question, but more a math rule question.
if you take the square root of -4, you would end up with an imaginary number.
However, if you write it as (-4)^(2/4), and then rewrite it to become the fourth root of (-4)^2 you'll end up with the fourth root of 16, which is 2 or -2.
I have read that it is not allowed to take a fractioned power of a negative number if the denominator is even, but is there a deeper rule behind that? Because I can't find the logical error in my assumption above, unless the mathematical order of calculation places the numerator before the denominator.

[Surlethe]

There is deeper math happening here. Let's work with complex numbers, because that's the natural field to use when we're talking about algebra. I'm going to ramble for a bit about square roots and power functions before I answer your question.

The function z --> z^n is an n-to-1 function. That means that every complex number (except 0) is the image of n complex numbers. (For example, (2i)^2 = (-2i)^2 = -4.) This is why the square root is so weird: every nonzero number in the image of of z --> z^2 gets hit twice, so square rooting, undoing the square function, is most naturally thought of as a "multiple-valued function." In this case, sqrt(-4) = {2i, -2i}.

If you want to make square root an honest-to-god smooth function, you have to come up with a way of consistently choosing a single preimage of each number that gets hit. In the real numbers, there's an easy way to do this: pick the positive preimage. That's how the square root function is defined for real numbers (and incidentally why it looks like half a parabola). In the complex numbers, it's not possible to consistently define a square root everywhere, for topological reasons related to Mobius bands. The same goes for cube roots, fourth roots, and nth roots: these are most conveniently thought of as multiple-valued functions. (You can always locally, except at zero(?), pick a consistent root, but you can't do it everywhere.)

What does a rational exponent mean? In the real numbers, x^(p/q) = q-root(x^p). You should prove that this is consistent with (q-root(x))^p. But it's possible to make sense of this as a function because we are talking real numbers. As per the previous paragraph, in the complex numbers, roots are more complicated, so it's not so clear what "z^(p/q)" means. Let's say, just for shits and giggles, that it means "q-root(z^p)"; now it's a multiple-valued function which takes in z and spits out a list of the qth roots of z^p.

Now to your question. What happens when we have (-4)^(1/2)? Is this the same as, say, (-4)^(2/4)? Well, (-4)^(1/2) = {2i,-2i}. Meanwhile, (-4)^(2/4) = (16)^(1/4) = {2, -2, 2i, -2i}. The square roots of -4 are also fourth roots of 16, but the process of squaring picked up two more roots! That's where the 2 and -2 came from.

Exercise: If instead we define z^(p/q) := (qthrt(z))^p, raising each qth root of z to the pth power, what happens to the solution set? Is this better behaved? Work it out for z = -4.

[Surlethe]

BTW, those "topological reasons related to Mobius strips" are like this: z --> z^2 is a double-covering map of the unit circle to itself. This can be regarded as the collapse map of the outer circle of the Mobius band to the core circle. Since the Mobius strip is not a trivial vector bundle, there is no nowhere-zero section, so it's not possible to (on the unit circle) pick a continuous square root function: it has to break somewhere. In general, one takes the square root of a nonzero complex number z = r exp(iT) by picking a square root of the unit-length part exp(iT) and then multiplying it by the principal square root of r, so the nonexistence of a continuous square root on the unit circle implies the nonexistence of a continuous square root function globally on the complex numbers.

[Feil]

There is deeper math happening here. Let's work with complex numbers, because that's the natural field to use when we're talking about algebra. I'm going to ramble for a bit about square roots and power functions before I answer your question.

The function z --> z^n is an n-to-1 function. That means that every complex number (except 0) is the image of n complex numbers. (For example, (2i)^2 = (-2i)^2 = -4.) This is why the square root is so weird: every nonzero number in the image of of z --> z^2 gets hit twice, so square rooting, undoing the square function, is most naturally thought of as a "multiple-valued function." In this case, sqrt(-4) = {2i, -2i}.

If you want to make square root an honest-to-god smooth function, you have to come up with a way of consistently choosing a single preimage of each number that gets hit. In the real numbers, there's an easy way to do this: pick the positive preimage. That's how the square root function is defined for real numbers (and incidentally why it looks like half a parabola). In the complex numbers, it's not possible to consistently define a square root everywhere, for topological reasons related to Mobius bands. The same goes for cube roots, fourth roots, and nth roots: these are most conveniently thought of as multiple-valued functions. (You can always locally, except at zero(?), pick a consistent root, but you can't do it everywhere.)

What does a rational exponent mean? In the real numbers, x^(p/q) = q-root(x^p). You should prove that this is consistent with (q-root(x))^p. But it's possible to make sense of this as a function because we are talking real numbers. As per the previous paragraph, in the complex numbers, roots are more complicated, so it's not so clear what "z^(p/q)" means. Let's say, just for shits and giggles, that it means "q-root(z^p)"; now it's a multiple-valued function which takes in z and spits out a list of the qth roots of z^p.

Now to your question. What happens when we have (-4)^(1/2)? Is this the same as, say, (-4)^(2/4)? Well, (-4)^(1/2) = {2i,-2i}. Meanwhile, (-4)^(2/4) = (16)^(1/4) = {2, -2, 2i, -2i}. The square roots of -4 are also fourth roots of 16, but the process of squaring picked up two more roots! That's where the 2 and -2 came from.

Exercise: If instead we define z^(p/q) := (qthrt(z))^p, raising each qth root of z to the pth power, what happens to the solution set? Is this better behaved? Work it out for z = -4.

That is really quite cool, straightforward, and informative. Thanks, Surlethe

[Molyneux]

I had a thought pop into my head that I thought might be interesting to figure out.
What is the cardinality of the set of all lines in an infinite plane not passing through a particular point?

Is that cardinality larger, smaller, or equivalent to that of the set of all lines that do pass through that given point?

[K. A. Pital]

Okay, I'm bored and putting off studying for my exams. Post questions you have about math, math problems, the philosophy of math, and I (or those of us who are math/science-literate, if y'all want) will try to answer. If I/we have time.

Tell me about math and metodological materialism. Can the fact that math has undecideable problems mean that materialism is false, as some insinuate?

[Grog]

I had a thought pop into my head that I thought might be interesting to figure out.
What is the cardinality of the set of all lines in an infinite plane not passing through a particular point?

Is that cardinality larger, smaller, or equivalent to that of the set of all lines that do pass through that given point?

The two cardinalities are the same. It is the same as the cardinality of R. If I understand your question correctly.

EDIT:
The cardinality of the lines trough a point is R, since it is parameterized by a real number (the slope). The cardinality of all lines are also parameterized by a finite list of real numbers (figure out how!). And the cardinality of such a set will also be that of R (figure out why!). The set of lines not trough a point is at least as big as the set of lines trough that point since we can get an injection by translating the line trough the point.

[Surlethe]

Here's a little more of a hint to Grog's answer The set of lines in the plane through a single point can be thought of as a circle. See it by noting that, for a circle about a point, each line through the point intersects the circle twice. Then identify each point with its opposite on the circle*. This is again a circle (think of taking a rubber band and twisting it onto itself).

Now let's get a handle on the set of lines not through a given point (say, 0). Pick a line that does go through 0, say, the x-axis. Lots of lines that don't intersect 0 do intersect the x-axis, so we can parametrize them by a circle around every nonzero point on the x-axis. Now we've gotten almost all of them; which ones are we missing?

At the end of the day, for the one we have a circle, and for the other we have (circle - 2 pts) x (R - {0}) x (R - {0}). (Why?) Prove that these have the same cardinality.

Related question: prove that the unit circle group U(1) is isomorphic to the group of nonzero complex numbers C*. (It's on Wikipedia)

These are highly counterintuitive because we naturally associate a topology to these objects, but the question strips the topology and only asks about the underlying sets. The sets are very infinite, so all our natural intuition about how they should behave goes right out the window and we're left with only the formalism.

* by the map z --> z^2; see above! God I love math.

[Surlethe]

Okay, I'm bored and putting off studying for my exams. Post questions you have about math, math problems, the philosophy of math, and I (or those of us who are math/science-literate, if y'all want) will try to answer. If I/we have time.

Tell me about math and metodological materialism. Can the fact that math has undecideable problems mean that materialism is false, as some insinuate?

Okay, sorry, this is kind of rambly.

I do not see any connection between the two. I don't know much about decision theory or Godel's tour-de-force, but they are claims about the limits of logical and computational systems. I don't think they're very relevant to empirical questions. If they are, they perhaps merely prove that there are statements we'll never have enough information or observation to settle: "Caesar had a bowel movement at 3 pm on 21st August 52 BC," that kind of thing.

Methodological naturalism is the process of making observations and discovering "best models" for those observations. (I do not know what "best" means in this case ... .) Undecidability is, I guess, a fact about some statements in axiomatic systems. Insofar as a "model" is an axiomatic system used to generate predictions about observed patterns, perhaps problems of undecidability are somewhat relevant. However, since (I guess) "X is undecidable in this system" means "There's no algorithm to prove X given axioms of this system," the only relevant statement I see is, "There's an observation for which your axioms generate no true/false predictions."

And of course, there's also, "if not materialism, then what?" I'd argue that every other system also has the problems of methodological materialism, and then some.

I do have a tad bit of experience with undecidability: my area of interest, low-dimensional topology, had a big hole shot in the field's goal in the '40s or '50s when topologists realized that a classification of manifolds of dimension higher than 3 up to homeomorphism implied a classification of finite presentations of groups, because, given any group G and topological dimension higher than 4, one can construct a manifold whose fundamental group is G. The problem? Classification of finite presentations of groups is an undecidable problem. So you can't classify, up to homeomorphism, manifolds of dimension 4 on up. (However, classification in dimension 2 is done and taught in advanced undergraduate courses, and the possibility of classification in dimension 3 is still unknown.)

Meanwhile, I'd bet that many or all people who "insinuate" that undecidable problems imply materialism is false couldn't even precisely state what undecidability means, let alone understand the theory or its applications

[Molyneux]

Here's a little more of a hint to Grog's answer The set of lines in the plane through a single point can be thought of as a circle. See it by noting that, for a circle about a point, each line through the point intersects the circle twice. Then identify each point with its opposite on the circle*. This is again a circle (think of taking a rubber band and twisting it onto itself).

Now let's get a handle on the set of lines not through a given point (say, 0). Pick a line that does go through 0, say, the x-axis. Lots of lines that don't intersect 0 do intersect the x-axis, so we can parametrize them by a circle around every nonzero point on the x-axis. Now we've gotten almost all of them; which ones are we missing?

At the end of the day, for the one we have a circle, and for the other we have (circle - 2 pts) x (R - {0}) x (R - {0}). (Why?) Prove that these have the same cardinality.

Related question: prove that the unit circle group U(1) is isomorphic to the group of nonzero complex numbers C*. (It's on Wikipedia)

These are highly counterintuitive because we naturally associate a topology to these objects, but the question strips the topology and only asks about the underlying sets. The sets are very infinite, so all our natural intuition about how they should behave goes right out the window and we're left with only the formalism.

* by the map z --> z^2; see above! God I love math.

...OK, this is going to take some effort for me to understand, but thanks to both of you.

[Surlethe]

Any more burning math questions?

[Surlethe]

Any more burning math questions?

[Feil]

Here's a question that has been bumping around in my head for a day or two.

Let x be an integer greater than 1.
Is the product of a series of positive integer addends which sum to x always maximized by using no addends other than 2, for even numbers, and no more than one instance of 3 for odd numbers?
Why/why not?


Or, put as mathy as I can put it,

given:
a ∈Z and 0<=a
b=0 or b=1
a>0 or b=1
x := 2a +3b
y := x_1*x_2*x_3*...*x_n such that x_1+x_2+x_3+...+x_n = x
z := (2^a)*(3^b)

Is z>=y for all a and b? Why or why not?

[Darth Holbytlan]

No. Because there is an easy counterexample: 6 = 3+3 = 2+2+2, but 3*3 = 9 > 8 = 2*2*2.