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#### Ozle

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« on: May 08, 2014, 04:04:21 pm »
+3

Yeah so let's talk about maths....
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#### Awaclus

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« Reply #1 on: May 08, 2014, 04:05:47 pm »
+2

Shouldn't this be on the Innovation forum?
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#### sudgy

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« Reply #2 on: May 08, 2014, 04:27:06 pm »
+2

Actually, this might be a good idea.  It takes over the random stuff thread a lot.

Can anyone show me why the solution to y'' = -y is A*sin(x) + B*cos(x)?  It might be too complicated for me, but I'm still curious.  I can see that it needs to be sinusoidal, and that that equation is a solution, but how do you derive it?
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### DStu

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« Reply #3 on: May 08, 2014, 04:31:06 pm »
0

Actually, this might be a good idea.  It takes over the random stuff thread a lot.

Can anyone show me why the solution to y'' = -y is A*sin(x) + B*cos(x)?  It might be too complicated for me, but I'm still curious.  I can see that it needs to be sinusoidal, and that that equation is a solution, but how do you derive it?
In my analysis course, we proved that the solution is unique (given starting point and derivative), and defined sin/cos as the solution...
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#### Axxle

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« Reply #4 on: May 08, 2014, 04:33:43 pm »
+5

Is maths anything like math?
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#### Ozle

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« Reply #5 on: May 08, 2014, 04:44:56 pm »
+2

Is maths anything like math?

yes it's the proper short form of mathematics
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#### qmech

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« Reply #6 on: May 08, 2014, 04:53:44 pm »
+3

Actually, this might be a good idea.  It takes over the random stuff thread a lot.

Can anyone show me why the solution to y'' = -y is A*sin(x) + B*cos(x)?  It might be too complicated for me, but I'm still curious.  I can see that it needs to be sinusoidal, and that that equation is a solution, but how do you derive it?

The best way to see it is if you already know a little bit about exponentials.  The derivative of exp(kx) is k.exp(kx), for any k.  So the second derivative of exp(kx) is k^2.exp(kx).  So you get a solution if k^2 = 1, i.e. if k = +-i.  So for any constant A and B, A.exp(ix) + B.exp(-ix) is a solution.  This happens to be equivalent to your expression with sines and cosines (with a different A and B).

The reason why these are all the solutions is that the solution space is two-dimensional.  This is perhaps believable because you get one arbitrary constant from each of the implicit integrations, but showing it formally might be a bit beyond you right now.  (I can't remember how to do it offhand.)

[Aside that might be better ignored: you've almost certainly only be told that it means for a function from R to R to be differentiable, so my example that brings in complex numbers is a bit of a cheat.  It does go to the heart of the matter though: it's not so much of an exaggeration to say that making that kind of argument work is the main reason physicists and engineers care about complex numbers.]
« Last Edit: May 08, 2014, 05:02:18 pm by qmech »
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#### SirPeebles

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« Reply #7 on: May 08, 2014, 04:54:30 pm »
+2

Actually, this might be a good idea.  It takes over the random stuff thread a lot.

Can anyone show me why the solution to y'' = -y is A*sin(x) + B*cos(x)?  It might be too complicated for me, but I'm still curious.  I can see that it needs to be sinusoidal, and that that equation is a solution, but how do you derive it?

One way to derive it is to use Euler's formula!

Suppose we already know how to solve y' = k y to get y = e^(kt).  Then e^(kt) is also a solution to the equation y'' = k^2 y.  But of course so is e^(-kt).  It turns out that if y1 and y2 are both solutions to y'' = k^2 y, then so is A*y1 + B*y2 for any constants A and B (a differential equation with this property is called a homogeneous linear equation).  So A*e^(kt) + B*e^(-kt) is a solution to y'' = k^2 y.  To get your equation, just choose k = i.  So the solutions are A*e^(it) + B*e^(-it).  If A=B=1/2, you get cos(t).  If A= 1/(2i) and B = -1/(2i), you get sin(t).

edit:  ninja'd by qmech.  one way to get some intution for why that gives all solutions is that having these two parameters A and B provides precisely enough space to have a unique solution matching any pair of initial conditions of y(0) and y'(0).  Physically, y'' = - y models the motion of a ball hanging on a spring, and for any initial position and initial velocity there is "obviously" a unique motion which unfolds.
« Last Edit: May 08, 2014, 04:58:35 pm by SirPeebles »
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#### heron

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« Reply #8 on: May 08, 2014, 04:59:48 pm »
+6

Maths competitions is pretty much my biggest hobby, although currently I'm a little burned out after the USA Junior Maths Olympiad.

My favorite subject is combinatorics, and least favorite is geometry. The only way that I can solve geometry problems is plotting the figure into the complex plane.

Here is a fun maths problem: Let n be a natural number. 2n distinct points are plotted in the plane. Half of them are colored red, and the other half are colored orange. The points are then divided into n pairs, each with 1 red and 1 orange point. Then, a line segment is drawn between the 2 points in each pair. Prove that there is a way to pair the points such that after the line segments are drawn, none of the line segments intersect each other.
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#### sudgy

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« Reply #9 on: May 08, 2014, 05:13:34 pm »
+1

One way to derive it is to use Euler's formula!

The reason I was asking was because the derivation of Euler's formula that I saw used it at a part...
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### LastFootnote

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« Reply #10 on: May 08, 2014, 05:14:38 pm »
+2

Is maths anything like math?

yes it's the proper short form of mathematics

I disagree!

*drives off*
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#### SirPeebles

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« Reply #11 on: May 08, 2014, 05:27:41 pm »
+1

One way to derive it is to use Euler's formula!

The reason I was asking was because the derivation of Euler's formula that I saw used it at a part...

Yeah.  This is how I proved Euler's Formula in my differential equations class a few months ago.

In general, there is no such thing as "the definition" of a mathematical object like sine.  Rather, there is a web of equivalent statements, any one of which can be used as "the definition".  So to prove it in an entirely satisfactory way, I would need to know the definition you are starting with.

Another way to solve the equation y'' = -y is to search for a power series solution.  Suppose we want the solution which matches y(0) = 0, y'(0) = 1.  Then
y = 0 + x + a2 x^2 + a3 x^3 + ...
y'' = 2 a2 + 2*3 a3 x + 3*4 a4 x^2 + ...

Since we want y'' = -y, we match up coefficients and solve.  You'll arrive at the Taylor series for sin(x).
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#### qmech

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« Reply #12 on: May 08, 2014, 05:34:14 pm »
+1

Here is a fun maths problem:

I assume the points should be in general position?

Here's another: is it always possible to cover 10 marks on a white table cloth with 10 plates that cannot overlap?
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#### sudgy

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« Reply #13 on: May 08, 2014, 05:36:57 pm »
+1

One way to derive it is to use Euler's formula!

The reason I was asking was because the derivation of Euler's formula that I saw used it at a part...

Yeah.  This is how I proved Euler's Formula in my differential equations class a few months ago.

In general, there is no such thing as "the definition" of a mathematical object like sine.  Rather, there is a web of equivalent statements, any one of which can be used as "the definition".  So to prove it in an entirely satisfactory way, I would need to know the definition you are starting with.

Another way to solve the equation y'' = -y is to search for a power series solution.  Suppose we want the solution which matches y(0) = 0, y'(0) = 1.  Then
y = 0 + x + a2 x^2 + a3 x^3 + ...
y'' = 2 a2 + 2*3 a3 x + 3*4 a4 x^2 + ...

Since we want y'' = -y, we match up coefficients and solve.  You'll arrive at the Taylor series for sin(x).

Aaaaaaaaand I don't understand it.  I guess I'll wait until I know more to think about it.
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### scott_pilgrim

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« Reply #14 on: May 08, 2014, 05:58:39 pm »
+3

Here is a fun maths problem: Let n be a natural number. 2n distinct points are plotted in the plane. Half of them are colored red, and the other half are colored orange. The points are then divided into n pairs, each with 1 red and 1 orange point. Then, a line segment is drawn between the 2 points in each pair. Prove that there is a way to pair the points such that after the line segments are drawn, none of the line segments intersect each other.

Can we please color them red and blue?  Red and orange will look so similar...
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#### SirPeebles

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« Reply #15 on: May 08, 2014, 06:06:05 pm »
+2

One way to derive it is to use Euler's formula!

The reason I was asking was because the derivation of Euler's formula that I saw used it at a part...

Yeah.  This is how I proved Euler's Formula in my differential equations class a few months ago.

In general, there is no such thing as "the definition" of a mathematical object like sine.  Rather, there is a web of equivalent statements, any one of which can be used as "the definition".  So to prove it in an entirely satisfactory way, I would need to know the definition you are starting with.

Another way to solve the equation y'' = -y is to search for a power series solution.  Suppose we want the solution which matches y(0) = 0, y'(0) = 1.  Then
y = 0 + x + a2 x^2 + a3 x^3 + ...
y'' = 2 a2 + 2*3 a3 x + 3*4 a4 x^2 + ...

Since we want y'' = -y, we match up coefficients and solve.  You'll arrive at the Taylor series for sin(x).

Aaaaaaaaand I don't understand it.  I guess I'll wait until I know more to think about it.

Let me try one more.  I think the definition of sine you have in mind is geometric.  Here's a derivation that is based on the unit circle.

It is easier for me to think of if the independent variable is t for time.  In fact, let's rephrase the problem as looking for the function x(t) for which x'' = - x.

A common technique is to remove the second derivative by introducing a new function, y(t) = x'(t).  Then our differential equation just says y'(t) = -x(t).  In other words, we now have two equations with two unknown functions (but no second derivatives!)

x' = y
y' = -x

At each time t, let P(t) be the point in the plane with x coordinate x(t) and y coordinate y(t).  So P(t) = ( x(t), y(t) ).  As time proceeds, in which direction will this point move?  For that we need to see what the rate of change of P is.  That's the derivative dP/dt, which in this case is P'(t) = ( x'(t), y'(t) ) = (y,-x) from our pair of differential equations.

Now, the direction from the origin to (x,y) is perpendicular with to the direction from the origin to (-y,x),   So P(t) moves around the origin in a circle with a constant speed.

Now suppose our initial conditions are x(0) = 1, x'(0) = 0.  Well, y(0) = x'(0) = 0, so the point starts at (1,0).  The speed is given by v^2 = (x')^2 + (y')^2 = (y)^2 + (-x)^2 = y^2 + x^2.  At t=0 this gives v(0)=1, and since the speed is constant we have v(t) = 1.  Thus the distance traveled in time t is just v*t = 1*t = t, and so t is equal to the angle swept out as measured in radians.  The x-coordinate is therefore x(t) = cos(t).

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#### Polk5440

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« Reply #16 on: May 08, 2014, 06:08:44 pm »
+1

One way to derive it is to use Euler's formula!

The reason I was asking was because the derivation of Euler's formula that I saw used it at a part...

Yeah.  This is how I proved Euler's Formula in my differential equations class a few months ago.

In general, there is no such thing as "the definition" of a mathematical object like sine.  Rather, there is a web of equivalent statements, any one of which can be used as "the definition".  So to prove it in an entirely satisfactory way, I would need to know the definition you are starting with.

Another way to solve the equation y'' = -y is to search for a power series solution.  Suppose we want the solution which matches y(0) = 0, y'(0) = 1.  Then
y = 0 + x + a2 x^2 + a3 x^3 + ...
y'' = 2 a2 + 2*3 a3 x + 3*4 a4 x^2 + ...

Since we want y'' = -y, we match up coefficients and solve.  You'll arrive at the Taylor series for sin(x).

Aaaaaaaaand I don't understand it.  I guess I'll wait until I know more to think about it.

(Edit: I said something kind of stupid here. )

It's short, simple, and spawned a whole subfield in economics (matching theory). And it insults the general population at the very end (see page 15). What entertainment!
« Last Edit: May 08, 2014, 06:29:35 pm by Polk5440 »
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#### heron

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« Reply #17 on: May 08, 2014, 06:23:57 pm »
+2

Here is a fun maths problem: Let n be a natural number. 2n distinct points are plotted in the plane. Half of them are colored red, and the other half are colored orange. The points are then divided into n pairs, each with 1 red and 1 orange point. Then, a line segment is drawn between the 2 points in each pair. Prove that there is a way to pair the points such that after the line segments are drawn, none of the line segments intersect each other.

Can we please color them red and blue?  Red and orange will look so similar...

Actually, I think we might be best with black and white to minimize the effects of color blindness.
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#### Watno

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« Reply #18 on: May 08, 2014, 06:25:06 pm »
+3

But I guess the background is white, so you won't be able to see the white points.
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#### sudgy

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« Reply #19 on: May 08, 2014, 06:37:36 pm »
+1

(Edit: I said something kind of stupid here. )

I was about to respond to the thing you said was stupid, but it was gone!  :O :O :O

I'll respond to it anyway.  I've actually not taken a single bit of calculus officially.  I've just studied it in my free time, because I know that will help me in the future (I want to do physics in college, and from what I've seen, it requires a lot of calculus).  It's all for fun anyway, and I was just curious about something, so I thought I would ask here (I'd already looked at a few other places and couldn't find anything).
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### navical

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« Reply #20 on: May 08, 2014, 08:00:07 pm »
+5

Here is a fun maths problem: Let n be a natural number. 2n distinct points are plotted in the plane. Half of them are colored red, and the other half are colored orange. The points are then divided into n pairs, each with 1 red and 1 orange point. Then, a line segment is drawn between the 2 points in each pair. Prove that there is a way to pair the points such that after the line segments are drawn, none of the line segments intersect each other.

Assume that no three points lie on the same straight line, since otherwise it fails for e.g. four points on a straight line coloured RROO in that order.

Call the set of points A. We will do this by induction on n. Clearly, when n=1 it's possible.

So for n>1, consider the convex hull of A, which is a convex polygon with vertices at points of A such that no point of A is outside the polygon. There are two cases: all the vertices of the convex hull are the same colour, or some are different. If some are different, then two adjacent vertices F and G will be different, and we can safely draw a line between them; no line between two points will ever leave the convex hull, and F and G are both outside the convex hull of the remaining points. Then the number of remaining points is less than n, so by induction we're done.

If all the points on the convex hull have the same colour - red, without loss of generality - then pick two, call them X and Y, and consider a straight line L perpendicular to the line between X and Y. If the line L is very close to X, then there are more red points than orange points on the side of L closer to X. If the line L is very close to Y, then there are more orange points than red points on the side of L closer to X. The aim is to move L from close to X to close to Y, and at some point, there will be equal numbers of orange points and red points on the side of L closer to X.

The only way this can fail to happen is if there are two orange points P and Q which lie on the same line K which is perpendicular to the line between X and Y, and there is one more red point than orange point on the side of K closer to X. (So if L is just closer to X than K, there is one more red than orange on X's side, and if L is just closer to Y than K, there is one more orange than red on X's side). In this situation, however, putting L in the same place as K and then twisting it very slightly around the midpoint of the line segment PQ will give us a line L with equal numbers of orange points and red points on the side of L closer to X, as wanted.

Then we can apply induction to the points on the side of L closer to X, and separately to the points on the side of L closer to Y, to get a set of lines that works for all the points.

Remark: I think this proof only actually requires no four points to lie on the same line, but it was easier to write up with three.
« Last Edit: May 08, 2014, 08:02:33 pm by navical »
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#### AndrewisFTTW

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« Reply #21 on: May 08, 2014, 08:23:52 pm »
+1

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#### heron

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« Reply #22 on: May 08, 2014, 08:26:53 pm »
+7

I think that proof works. Nice. Also, yes, I probably should have specified that no three points are collinear.

Here is mine:

Consider 4 points which are the endpoints of two intersecting line segments. Those points form a convex quadrilateral, and the segments are the diagonals of that quadrilateral. We can re-pair those 4 points so that those two segments are two opposite sides of the quadrilateral and do not intersect. From the triangle inequalities, we know that the sum of the lengths of two opposite sides of a convex quadrilateral is strictly less than the sum of the lengths of the diagonals.

Now, consider the pairing of points which leads to the least total length of the line segments. We know that such a pairing exists since there is a finite number of points. Assume that two of the segments in this pairing intersect. Then, we can re-pair them as described above, and the total length of the line segments will decrease. But that is a contradiction, since we started with the pairing with the least total length.

Therefore the pairing with the least total length of line segments will not have any intersecting segments.
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#### SirPeebles

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« Reply #23 on: May 08, 2014, 08:36:04 pm »
+3

I didn't want to post a problem while heron's was still on the table, but now I'll share it.  This is my favorite math problem.  It is from a fairly well-known math contest, so don't spoil it if you've seen it.

Each point on the xy-plane is colored white.  You have a paintbrush tool with the peculiar property that when you click on a point, each point an irrational distance from where you clicked will be colored black.  For instance, if you click on the origin (0,0), then (1,1) will turn black since sqrt(2) is irrational, but (0,1) will not turn black.

Is it possible to color the entire plane black with a finite number of clicks?  And if so, what is the fewest number of clicks required?
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#### sudgy

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« Reply #24 on: May 08, 2014, 09:11:30 pm »
+1

I didn't want to post a problem while heron's was still on the table, but now I'll share it.  This is my favorite math problem.  It is from a fairly well-known math contest, so don't spoil it if you've seen it.

Each point on the xy-plane is colored white.  You have a paintbrush tool with the peculiar property that when you click on a point, each point an irrational distance from where you clicked will be colored black.  For instance, if you click on the origin (0,0), then (1,1) will turn black since sqrt(2) is irrational, but (0,1) will not turn black.

Is it possible to color the entire plane black with a finite number of clicks?  And if so, what is the fewest number of clicks required?

Ooh!  The lower bound is 2! (not 2!, but two with an actual exclamation mark (although 2! = 2...))
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Quote from: sudgy on June 31, 2011, 11:47:46 pm
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