Why do we care about irrational (or transcendental, or definable...) numbers? I'm a physicist by training and I've never cared if the number I am working with is rational or not
I think there are some uses in physics where the periodicity of events is concerned (such as waveforms/harmonics, or orbital periods). Diophantine approximation and Fourier analysis come to mind. However, I don't know too much about what physicists do.
Warp wrote:
Prove that there are always two points on opposite sides of Earth with the exact same temperature.
There is actually a two-parameter version of the question (e.g. Prove that there exist two points on opposite sides of the Earth with the exact same temperature and elevation.) I'll leave it to others before I give my answer.
By the way, the parameters are continuous on the surface of the sphere/S2.
Prove that there are always two points on opposite sides of Earth with the exact same temperature.
There is actually a two-parameter version of the question (e.g. Prove that there exist two points on opposite sides of the Earth with the exact same temperature and elevation.)
Ooh I wasn't sure until you gave this hint.
Let T1 be the minimum temperature on the sphere and T2 be the maximum temperature. Now consider the function f(lat,long) of temperature at a point, minus the temperature at the opposite point on the globe. f is positive in a region around T2, and negative in a region around T1. Because f is a sum of two continuous functions, it is itself a continuous function, and so there must be a zero contour. Any point on this contour has the property that it has the same temperature as the opposite side of the sphere.
Now, for the two-variable version: the same argument shows that there exists a contour line of equal antipodean elevation. But I'm not sure how to demonstrate that these contours must cross. I have a feeling that you need to show at least one contour exists which connects two points on opposite sides of the earth for each variable; two such contours in different variables would have to cross. But I'm not sure how to show that.
EDIT: I think I got it. First: show that the equal temperature contour joins a point with its opposite: this can be done by a symmetry argument because f(point opposite x) = -f(x), so the contour will join two symmetrical halves of the sphere and each point opposite a point on the contour will itself be on the contour.
Then perform the same argument on elevation, but constrained to the contour line: form the function g(x) = elevation at x minus elevation at point opposite x. At maximum elevation, this function is positive, at minimum elevation, it's negative, therefore there must be a zero on the contour. This point has equal temperature and elevation to the point opposite.
EDIT: I think I got it. First: show that the equal temperature contour joins a point with its opposite: this can be done by a symmetry argument because f(point opposite x) = -f(x), so the contour will join two symmetrical halves of the sphere and each point opposite a point on the contour will itself be on the contour.
I think this doesn't cover the case where there are disjoint zero contours.
How I remember the solution given was by the function h(x)=(f(x)-f(-x),g(x)-g(-x)), where f and g are temperature and elevation (-x means point opposite x). If there exists x such that h(x)=(0,0), then there are two opposite points on the sphere with equal temperature and elevation. Otherwise, suppose that there is no such x.
Let b(θ, t) be the half-great-circle on the sphere going from latitude 90° (at t=0) to latitude -90° (at t=1) along the longitude angle θ. Then h(b(0°, t)) is a path on two-dimensional space from a point (r,s)≠(0,0) to the point (-r,-s), not through (0,0). Likewise, h(b(180°, t)) also goes from (r,s) to (-r,-s), but on the "opposite side" of h(b(θ, t)). The argument is then that, as θ ranges from 0° to 180°, h(b(θ,t)) undergoes a "continuous deformation", and that is not possible without going through (0,0). This can be formalized by considering the angular displacement of the path (if h(b(θ, t)) is converted to polar coordinates with angular function φ(t), then take φ(1)-φ(0)). Paths h(b(0°, t)) and h(b(180°, t)) have opposite nonzero angular displacements, and h(b(θ, t)) as θ ranges from 0° to 180° does not change angular displacement unless it goes through (0,0).
Edit: This is the case n=2 of the Borsuk-Ulam theorem.
Edit 2: Cleaned up notation.
The intuitive way of understanding why there are always points on opposite sides of the Earth with the exact same temperature (or pressure, or height) is to draw an imaginary circle (with a maximum radius) around the Earth and choose any two opposite points on that circle. One of them will (most probably) have a higher temperature than the other, ie. their difference will be non-zero. Now start moving one of the points along the circle, and the opposite point as well (so that they remain opposite) and look at the difference between the two temperatures.
At some point this difference must reach zero. That's because if you move 180 degrees along the circle, you have now reversed the temperatures of the two points, ie. their difference is now the same but with the opposite sign, and thus by necessity the difference was zero at some point. (We can rightly assume that the change in temperatures along the surface of the Earth is contiguous, without jumps.)
The curious thing is that you can choose any circle around the Earth, and you will find at least one point which has the same temperature as its opposite point (and probably a lot more than one.)
Could someone give an intuitive explanation of how there can be two points on opposite sides of the Earth with the same two parameters (such as temperature and height)?
EDIT: I think I got it. First: show that the equal temperature contour joins a point with its opposite: this can be done by a symmetry argument because f(point opposite x) = -f(x), so the contour will join two symmetrical halves of the sphere and each point opposite a point on the contour will itself be on the contour.
I think this doesn't cover the case where there are disjoint zero contours.
I never claimed that the contour was the only zero contour (there may well be others); but the symmetry argument demonstrates that there is a unique zero contour which splits the sphere into two equally-sized regions, and that this zero contour must be invariant under transforming each point to its opposite. Even if other zero contours exist, only one can have this property -- the other contours are transformed into different contours on the opposite side by this transformation. (If there were two contours with this property, they would necessarily cross)
Let b(θ, t) be the half-great-circle on the sphere going from latitude 90° (at t=0) to latitude -90° (at t=1) along the longitude angle θ. Then h(b(0°, t)) is a path on two-dimensional space from a point (r,s)≠(0,0) to the point (-r,-s), not through (0,0). Likewise, h(b(180°, t)) also goes from (r,s) to (-r,-s), but on the "opposite side" of h(b(θ, t)). The argument is then that, as θ ranges from 0° to 180°, h(b(θ,t)) undergoes a "continuous deformation", and that is not possible without going through (0,0).
This is good, too. I'm not sure I understand the formalization (what does φ(t) mean?) but the proof sketch is good enough for me.
Could someone give an intuitive explanation of how there can be two points on opposite sides of the Earth with the same two parameters (such as temperature and height)?
I think I have the general idea:
You already said that it worked for every circle you can take, so let's take all the circles on which the north and south pole lie. Each of these circles will have at least one point (for the sake of intuitiveness, let's say its around the equator) at which the parameters are equal on both sides.
Due to the fact that the function is continuous, you can draw a line through all of these points and get a new "circle"; it doesn't have to be round per se, but we are working with points on opposite sides of a sphere, so this line will divide the sphere into two equal halves.
Now it goes that for any other parameter, this circle-of-equal-points can also be made. As this one divides the sphere into two equal halves as well, the two circles must either lie on eachother or cross eachother. Either case, there's at least one point were both parameters are equal to those at the other side.
I haven't looked at the link FractalFusion posted, but I have a feeling this works for N parameters on any N+1 (or greater) dimensional sphere.
Thanks for the explanation. It really helped to visualize the situation.
I wonder why it's so hard to intuitively accept that there are always points on opposite sides of the Earth with the exact same temperature. Much less points that have the exact same temperature and eg. atmospheric pressure at the same time.
There's a similar result about the fact that the wind is always calm somewhere. No matter what the pattern of wind is globally, there is always a point at which there is no wind.
Probably it is unintuitive because, if you take a random point on the Earth, the probability that the opposite side is the exact same temperature, or similarly, the probability that the wind is nothing, is 0.
It's quite easy to get two opposite points not at the same temperature and just as easy to get a point with wind on it, so...
The idea with these unintuitive results is that the use of a continuous parameter like temperature or wind (itself not always intuitive) enforces the existence of some point on the sphere satisfying some condition(s). Also, these are existence (nonconstructive) theorems; existence theorems aren't exactly intuitive.
Also, these proofs rely on the assumption that temperature, height and wind direction and continuous. At least for height I'd just point you to the nearest cliff.
For winds, the center of a small vortex is technically windstill, but standing in there you'd claim otherwise. The hairy-ball theorem also wouldn't count updraft as wind - I would.
Also, these proofs rely on the assumption that temperature, height and wind direction and continuous. At least for height I'd just point you to the nearest cliff.
Temperature and atmospheric pressure might be better parameters. They are practically continuous (unless you want to be really nitpicky and go to the quantum level...)
Given the classical problem "what's the sum of all integers from 1 to 1000" (or whatever number), most people have memorized the formula that gives the answer without understanding it.
I, however, like to think of the solution as: "The average of all the numbers multiplied by their amount." It's relatively easy even intuitively to understand why that gives the correct answer.
The great advantage of thinking about it like that, rather than blindly memorizing a formula, is that it can be generalized to other situations as well. For example, what if the problem were "what's the sum of all even numbers from 2 to 1000?" Many people would not know how to resolve the problem, but it becomes easy when you understand that it's, again, their average multiplied by their amount. (Their average is (2+1000)/2 and their amount is 500.) The same principle can be used for "all numbers divisible by 5" or whatever.
Now the problem becomes calculating the average of such a sequence. For a linear sequence it's trivial, but what about non-linear sequences? For instance, what if the problem is: "What's the sum of all n2, where n goes from 1 to 1000?" There are 1000 integers there, but what's their average?
Is there a generic way of calculating the average of np (where n is a linear sequence and p is a constant)? How about other possible functions?
Given the classical problem "what's the sum of all integers from 1 to 1000" (or whatever number), most people have memorized the formula that gives the answer without understanding it.
I, however, like to think of the solution as: "The average of all the numbers multiplied by their amount." It's relatively easy even intuitively to understand why that gives the correct answer.
The great advantage of thinking about it like that, rather than blindly memorizing a formula, is that it can be generalized to other situations as well. For example, what if the problem were "what's the sum of all even numbers from 2 to 1000?" Many people would not know how to resolve the problem, but it becomes easy when you understand that it's, again, their average multiplied by their amount. (Their average is (2+1000)/2 and their amount is 500.) The same principle can be used for "all numbers divisible by 5" or whatever.
Now the problem becomes calculating the average of such a sequence. For a linear sequence it's trivial, but what about non-linear sequences? For instance, what if the problem is: "What's the sum of all n2, where n goes from 1 to 1000?" There are 1000 integers there, but what's their average?
Is there a generic way of calculating the average of np (where n is a linear sequence and p is a constant)? How about other possible functions?
To expand on RGamma's response, I like to solve these kinds of problems using a little induction.
First, note that the finite sum of consecutive integers to the mth power is closely approximated by taking the integral of xm. Therefore, we expect it to be a polynomial of degree m+1. If it had a higher or lower degree, this approximation would surely diverge for large upper bounds.
(Incidentally, the coefficient you get by integrating matches the coefficient in the finite sum. For example, the sum of integers (m=1) up to n is equal to 1/2 n2 + 1/2 n. Likewise, the integral from x=0 to x=n of xdx is 1/2 n2. This fact isn't necessary for the remainder of the proof, but will cut down on your work in later steps.)
The rest of the process proceeds by induction. Assume a polynomial solution of the form
p(n) = a0n0 + a1n1 + ... + am+1nm+1
For the base case, we surely know the value when n=0. All that remains is to write the k+1th term in terms of the kth term and match all of the coefficients. You'll end up with m+1 equations and m+1 unknowns (the a coefficients). I'll leave that up to you to try out.
The utility of my method is that it allows for series such as the sum of squares of even numbers.
Edit: Changed my xs to ns.
So is there a simple formula for calculating the average of all squares between 12 and n2 (inclusive)?
Yes, I calculated it while I was writing up my previous post.
Assume it takes the form of the following polynomial
p(n) = a1n + a2n2 + a3n3.
We will show it does indeed take this form, but you should be led to try that solution because the integral of x2dx is 1/3 x3.
This formula works for n=0, the base case. (I altered your statement to all squares from 0 to n2. Clearly, this makes no difference.) Now we need to use induction to show that the formula works for n=k+1, assuming it works for n=k.
We require that p(k+1) = p(k) + (k+1)2 (the sum up to k, plus the k+1th term). This is the condition that will pin down the coefficients for us.
We start with...
p(k+1) = a1(k+1) + a2(k+1)2 + a3(k+1)3
and after a little expanding of terms (left to you), we find
p(k+1) = p(k) + (a1 + a2 + a3) + (2a2 + 3a3)k + 3a3k2.
Now apply the condition that p(k+1) = p(k) + (k+1)2 and we see that the polynomial coefficients of (k+1)2 must match up with the coefficients of the "leftover" powers of k from the induction step. This means
a1 + a2 + a3 = 1
2a2 + 3a3 = 2
3a3 = 1
which is just a linear system of equations. Solve this system of equations any way you like to obtain
a1 = 1/6
a2 = 1/2
a3 = 1/3
(or just verify that those are the correct values needed to solve the system of equations).
Putting this all together, the sum from 0 to n of i2 is equal to
1/3 n3 + 1/2 n2 + 1/6 n = (2n3 + 3n2 + n)/6.
Try it out!
Edit: As RGamma points out below (and I see from reviewing your question), you wanted to know the average, not the sum. Just as RGamma suggests, divide the above result by n to obtain
(2n2 + 3n + 1)/6.
How interesting! The average is always an integer! Who'd've thunk it!?
So is there a simple formula for calculating the average of all squares between 12 and n2 (inclusive)?
Assuming by 'average' you mean 'arithmetic mean', then the closed form formula for 1/n * sum(k=1,n)(k^2) is 1/n * 1/6 * n * (n + 1) * (2n + 1) = 1/6 * (n+1) * (2n+1) = 1/3 * n^2 + 1/2 * n + 1/6
All syllogisms have three parts, therefore this is not a syllogism.
(2n2 + 3n + 1)/6.
How interesting! The average is always an integer! Who'd've thunk it!?
What? No.
The sum is always an integer. Doesn't mean the average is.
Argh, you're right. It's been a rough day and I'm juggling four or five things at once. I'm actually a little surprised I didn't screw up the broader proof.
Anyway, I do think it's interesting that the sum is always an integer, despite being divided by six. There is nothing in the numerator of that expression that is a clear indication that it will be divisible by six, yet it is regardless. Once upon a time, I was really curious about that property and wanted to find other polynomials that, with integer arguments, were sure to be divisible by a specific number (in this case, 6). I seem to remember finding the general property at some point, but I can't remember what it is. It's somewhere in the study of modular arithmetic, rings, and ideals. Can anyone derive it? How can one quickly determine what polynomials up to order n-1 are always divisible by n?
Just as RGamma suggests, divide the above result by n to obtain
(2n2 + 3n + 1)/6.
If you need to know the actual sum of all the squares in order to get their average, it pretty much defeats the original purpose for the average in the first place, which was to calculate the sum... :P
It's just that for the sum of all integers that form a linear sequence (eg. all consecutive integers, or all even integers, etc.) you don't need to know the sum in order to know their average. You just take the first and the last integer in the sequence and divide their sum by 2. This, in turn, helps you calculate the sum of the entire sequence (by multiplying this average by the amount of integers in the sequence.) I was just wondering if there would be an equivalently easy way of calculating the average of a power series without having to know what the sum of the power series is (which is what we are trying to find out in the first place.)
There is nothing in the numerator of that expression that is a clear indication that it will be divisible by six, yet it is regardless.
2n3 + 3n2 + n =
(2n2 + n)(n + 1) =
n(2n + 1)(n + 1)
I think it's obvious that either n or n + 1 will be divisable by 2. Then remains to see that at least one term is divisable by 3. There are three cases:
1) n = 3k
2) n = 3k + 1
3) n = 3k + 2
In the first case the first term (n) will be divisable by three, in the last case the third term (n + 1) will be divisable by three. In the second case, we can show that
2n + 1 = 6k + 2 + 1 = 3(2k + 1)
which is again, divisable by 3. As at least one term is divisable by 2, and another by 3, the result will be divisable by 6. The proof is rather long, but I find the result to be at least somewhat intuitive, looking at the numerator.
There is nothing in the numerator of that expression that is a clear indication that it will be divisible by six, yet it is regardless.
2n3 + 3n2 + n =
(2n2 + n)(n + 1) =
n(2n + 1)(n + 1)
I think it's obvious that either n or n + 1 will be divisable by 2. Then remains to see that at least one term is divisable by 3. There are three cases:
1) n = 3k
2) n = 3k + 1
3) n = 3k + 2
In the first case the first term (n) will be divisable by three, in the last case the third term (n + 1) will be divisable by three. In the second case, we can show that
2n + 1 = 6k + 2 + 1 = 3(2k + 1)
which is again, divisable by 3. As at least one term is divisable by 2, and another by 3, the result will be divisable by 6. The proof is rather long, but I find the result to be at least somewhat intuitive, looking at the numerator.
Well when you put it that way, it seems so obvious! But I swear there was a nontrivial question in there somewhere. There are other integer-coefficient polynomials that are not factorable (I think...) but are divisible by six when you plug in an integer. There are others for all other integers. I think I proved something many years ago about the polynomials divisible by prime numbers using Fermat's little theorem, but I got stuck on nonprime numbers.
Basically, I wanted a way to produce the set of all polynomials that are always divisible by some integer, n. I quickly deduced the following:
• If some polynomial in x is always divisible by n, then that same polynomial plus n*xk, for any k, is also divisible by n. Therefore, the coefficients only needed to range from zero to n-1, inclusive.
• If a polynomial is always divisible by n, any integer times that polynomial is also divisible by n. The coefficients should have no common factors.
• The sum of two polynomials divisible by n is also divisible by n. These last two criteria taken together mean the set of polynomials always divisible by n form a basis set of a vector space.
• If a polynomial is always divisible by n, then it will still be divisible by n if it is multiplied by x.
• Since we are free to multiply by x, we have a strong incentive to look only for the polynomials up to order n-1. Our vector space will then have dimensionality n and if we can find n linearly independent polynomials that are all divisible by n, we should be able to reproduce all others by multiplying by x, multiplying by integers, adding the polynomials together, and/or taking the coefficients mod n.
I wrote up a Matlab program that would find integer coefficient polynomials. I believe I also had suspicions that non-integer coefficients could work as well. Would anyone like to pick up where I left off? I think a good start would be listing all third-degree polynomials divisible by 4. If they're all factorable, then I clearly overlooked the obvious.
There is nothing in the numerator of that expression that is a clear indication that it will be divisible by six, yet it is regardless.
2n3 + 3n2 + n =
(2n2 + n)(n + 1) =
n(2n + 1)(n + 1)
I think it's obvious that either n or n + 1 will be divisable by 2. Then remains to see that at least one term is divisable by 3. There are three cases:
1) n = 3k
2) n = 3k + 1
3) n = 3k + 2
In the first case the first term (n) will be divisable by three, in the last case the third term (n + 1) will be divisable by three. In the second case, we can show that
2n + 1 = 6k + 2 + 1 = 3(2k + 1)
which is again, divisable by 3. As at least one term is divisable by 2, and another by 3, the result will be divisable by 6. The proof is rather long, but I find the result to be at least somewhat intuitive, looking at the numerator.
A more straightforward solution would be using induction.
For the base case, n=1, we have n*(n+1)*(2n+1)/6 = 6/6 = 1.
For the step of induction, if for n = k, then k (2k +1) (k+1) = (k+1) (2k^2 + k) (1)
But for n = k + 1, the expression in the numerator is simply (k+1) (2(k+1)+1) (k+1+1) = (k+1) (2k^2+7k+6) = (k+1)(2k^2 + k) + (k+1)(6k+6)= (k+1)(2k^2+k) +6(k+1)^2 (2)
We see that the first term in (2) is simply (1), and thus divisable by 6. The second term in (2) is of the form 6x, and thus also divisable by 6. Their sum is thus also divisable by 6.
Hence, n = k+1 is also divisable by 6.
