Rawr! Sine System of Equations Problem

[Nootropic]

okay, so I attempted to solve the following problem:

 

sin(x) + 2sin(x +y +z) = 0

 

sin(y) + 3sin(x +y +z) = 0

 

sin(z) + 4sin(x +y +z) = 0

 

At first, I had values that would work for the two equations, but broke down in the third equation (they were, x = 2/3, y = 1, z = -2). Make sure it's in degrees. But I rechecked I was realtively sure I made a mistake in solving earlier on (which would lead to a mistake in the entire solution, since my values were all dependent on each other) and then redid it and then ended up with six linear functions and their inverses. Although I have another way of looking at it, I am not going to attempt now, as finding the answer will not bring joy or destroy the reality I have four mid-terms, which will most likely be easy, but I would like to be at least conscious for them. Anyhow, I'll post what I did earlier later, it's time to sleep!

[CraigD]

okay, so I attempted to solve the following problem:

sin(x) + 2sin(x +y +z) = 0

sin(y) + 3sin(x +y +z) = 0

sin(z) + 4sin(x +y +z) = 0

A trivial solution is x=y=z=0.

 

A general solution is, at least for me, challenging. So is a solution with non-zero values. I'm curious to see how you are approaching it, Nootropic.

[houserichichi]

Since it's all a bunch of sines and we know [math]sin(k\pi)=0[/math] so as long as x, y, and z are integer multiples of [math]\pi[/math] then the first argument will be zero as above but the second argument will be zero as well since the sum of any integer multiple of [math]\pi[/math] is also an integer multiple of [math]\pi[/math].

 

I don't believe that's the full solution set though because you're also trying to solve

 

[math]sin(x)=2sin(x+y+z)[/math]

[math]sin(y)=3sin(x+y+z)[/math]

[math]sin(z)=4sin(x+y+z)[/math]

 

so

 

[math]x=sin^{-1}(2sin(x+y+z)[/math]

[math]y=sin^{-1}(3sin(x+y+z)[/math]

[math]z=sin^{-1}(4sin(x+y+z)[/math]

 

You take it from there :D

[Nootropic]

 

[math]sin(x)=2sin(x+y+z)[/math]

[math]sin(y)=3sin(x+y+z)[/math]

[math]sin(z)=4sin(x+y+z)[/math]

 

so

 

[math]x=sin^{-1}(2sin(x+y+z)[/math]

[math]y=sin^{-1}(3sin(x+y+z)[/math]

[math]z=sin^{-1}(4sin(x+y+z)[/math]

 

You take it from there :D

 

 

Actually, that is slightly incorrect, the sin(x) = -2sin(x+y+z), since the two must be added together to produce zero. Though, I'm now doubting my methods, as I cannot, for the life of me, remember if sin^-1(2sin x) is just 2x or something different. If it is something different, then my method is completely wrong.

[CraigD]

… I cannot, for the life of me, remember if sin^-1(2sin x) is just 2x or something different.

It’s something different.

 

Consider the case of [math]x = \frac{\pi}6[/math]. [math]2 \sin \frac{\pi}6 = 1[/math], but [math]\sin^{-1} 1 = \frac{\pi}2 \ne 2 \frac{\pi}6[/math].

[houserichichi]

Nootropic...oops, how perfectly careless of me! At any rate I ran the problem through Mathematica just to see what came up (I won't spoil the surprise in case you're actually trying to work this out by hand) and it's plainly obvious that the integral multiples of [math]\pi[/math] aren't the only solutions. Unfortunately for us, however, the other ones aren't as pretty as the initial bunch!! :eek_big:

[Nootropic]

It’s something different.

 

Consider the case of [math]x = \frac{\pi}6[/math]. [math]2 \sin \frac{\pi}6 = 1[/math], but [math]\sin^{-1} 1 = \frac{\pi}2 \ne 2 \frac{\pi}6[/math].

 

Yes, I was relatively aware of that after I posted. But double checking is always a nice thing! I really don't have any experience in the inverse of anything like that, so that's what I'm going to attempt to find, unless someone explains it sooner. Although oddly enough, I used that (incorrect) method of solving and the first two equations had the correct solutions. It's quite an interesting system, I must admit.

[Qfwfq]

It's only a good first-order approximation around x = 0.

[Nootropic]

Yeah, so I have absolutely no idea how to solve this damn thing. I've tried different things, but I can't do it. However, I ran across an interesting identity over at the Dr. Math Forums...

 

Arcsin[sin(x)] = (-1)Floor[1/2 - x/] (x + Floor[1/2 - x/])

 

However, I have absolutely no idea what this is, nor have I ever heard of it and I couldn't find anything about what it was. Anyone?

[Qfwfq]

arcsin(sin(x)) = x, unless x is outside the range between -pi/2 and pi/2.

 

I don't know what the x/ are meant to be, are you sure you copied right?

 

P. S. I suspect this set of equations can only be solved iteratively, apart from the trivial case.

[Nootropic]

arcsin(asin(x)) is not equal to asin(x), see Craig's example. I have another idea, but I'm not quite sure how to solve it from there.

[Qfwfq]

That's not what I said, it was:

 

arcsin(sin(x)) = x

 

which is an identity for -pi/2 < x < pi/2.

[max4236]

Hi, I tried looking at this using phasors. Thinking of adding sine waves of the same frequency but with different phases, then finding x,y,z so that the phase shifts overlap. Not much luck with that. Then I tried a Monte Carlo approach on the computer generating a bunch of random x,y,z and plotted near solutions within an error threshold. It helped to visualize the results with 3 parallel number lines, x,y,z, (each 0 to 360) with lines connecting approximate solutions. Each approximate solution formed a unique shape and when combined formed a kind of webbed mesh map. It's pretty cool looking. There's an interesting pattern that emerges but as I reduced the acceptable error threshold it becomes finer and finer until you just get criss-crosses that represent 0 or multiples of 180 degrees. The approximate solutions spread out most for z and then y and then a much narrower band for x. Makes sense with the coeficents, 2,3,4. The first equation sin(x)+2*sin(x+y+z), x is the dominant variable. The map appears to be a symmetric mirror image on either side of 90 degrees. It would be interesting to make a 3D map with all the near solutions and fly through it like traversing a starfield. Haven't tried that yet.

 

Anyway scratch that. Then I thought about a geometric solution, 4 triangles inscribed inside a circle. It came to me in a vision while in the shower and I was sketching forumals on the foggy mirror. One triangle attatched to the other so the angles add up and all the sides satisfy those sine equations. Four triangles, all with a radius of R, one has a side of length 4, another 3, 2, and 1. -1=Rsin(x+y+z)=-1/4*Rsin(z). (sinx)/2=(siny)/3=(sinz)/4.

The 1-sqr(A^2+15)-R triangle forms a diagonal through the origin up through the 4-A-R triangle and then intersects the circle at a height of 1=1/4*Rsin(z).

 

It comes close, if you use the z angle first and assume that it's more than 90 degrees. Try solving for A by going from one side of the circle and then from the other. Match up the points to find out what A should be. Then use A to find all the angles. It never quite fits together though. Except as A approaches infinity, which corresponds to the trivial multiples of 180 degrees solution. Going around one direction I found a point pointed to by a vector which I called Bc (B reference to point C):

 

Bc=1/R < -12-A(A^2+7)^1/2, 4(A^2+7)^1/2 - 3A >

R=(A^2+16)^1/2

 

And then the other way Ba:

Ba=1/R < -2-(A^2+12)^1/2(A^2+15)^1/2,-(A^2+12)^1/2+2(A^2+15)^1/2>

 

In order for there to be a non-trivial solution x,y,z Bc has to equal Ba. Plotting A versus the angle between Ba and Bc, via the dot product, yields something that looks like a hyperbolic curve with the angle approaching 0 as A approaches infinity, which gets you back to the trivial mulitiples of 180 degrees..

 

When I first drew a 3-4-5 triangle I thought I was on to something. With a pencil compass and ruler it looked pretty close, but mathematically the points diverged by 4.2 degrees. pfft

 

Then I tried another approach. Maybe x,y, and z are complex angles, angles with an imaginary component. Imagine those triangles are veering off into an imaginary hyperspace and then linking back up in the real number space. If x is a complex number, say x=a+bi (let a & b be real numbers). Then you can write sin(x) like this:

sin(x)=(e^ix-e^-ix)/2

( also cos(x)=(e^ix+e^-ix)/2 )

 

Splitting it up into the real and imaginary parts you get this:

sin(a+bi)=sin(a)*(e^b+e^-:naughty:/2 + i cos(a)*(e^b-e^-:shrug:/2

 

(e^c+e^-c)/2=cosh©=hyperbolic cosine of c

(e^c-e^-c)/2=sinh©=hyperbolic sine of c

so sin(a+bi)=sin(a)cosh(+icos(a)sinh(

 

I let the computer think about it for a while and so far the best answer it can come up with is

X= (21.55645 +.0541012 i)°

Y= (28.84290 +.1034086 i)°

Z= (139.0789 +.5681697 i)°

"error factor" of 6.4e-3

 

This thing is a beast. You could try approximating the functions with a multidimensional Fourier transform or something? Where did it come from?

sin(x)+2sin(x+y+z)=0

sin(y)+3sin(x+y+z)=0

sin(z)+4sin(x+y+z)=0?

 

Anyway, good luck. I need some rest from this. :naughty:

[Nootropic]

So during the summer I find myself with spare time to solve things like this. If anyone's wondering where it came from you can look here Olympiad Math Madness. There's a ton of fun and interesting problems to keep anyone occupied.

 

So I've taken a second look at this thing. I was taking a look some simple functional equations (ie. the equation f(x+y) = f(x)f(y) is satisfied by all exponential functions) and didn't even occur to me to think of when x, y, or z is a certain number. So I first began by subtracting the second equation from the first to obtain sin(x) - sin(y) = sin(x+y+z). Then I looked at the case y = 0, which gives sin(x) = sin(x+z). I then expanded the right side with the identity sin(A+:shrug: = sin A cos B + cos A sin B. After a bit of manipulation and a handful more of identities I obtained tan(z/2) = cot(x). I used this method two more times (the third time I subtracted the third equation from the second and used that) and obtained cot(z/2) = -cot(y) and tan(y/2) = cot(x). From this it is obvious that tan(z/2) = tan(y/2) and then reciprocating gives cot(z/2) = cot(y/2), but cot(z/2) = -cot(y), this implies that cot(y/2) = -cot(y). More expansion and identities produces the quotient (2cos(y)^2-cos(y)-1)/((sin(y)cos(y)) = 0. This will of course be true when 2cos(y)^2-cos(y)-1 = 0. Using the quadratic equation I obtain cos y = 1,-.5. Which implies that y = 0 or arccos(-1/2) = 2pi/3. Now the only thing that led me to believe this was remotely correct was that I had obtained the trivial solution y = 0.

 

EDIT

Well, I solved for y and completely forgot I had other equations that will yield the values of x and z. DUH! So, I'm off to verify if this is correct or not. I'm certainly doubting it may be, but at least I tried.

[Nootropic]

Hmm...well, that appears to be wrong. I had x = pi/6 and y = z = 3pi/2. Which doesn't give correct solutions, though I could have made a mistake, so I'm gonna check that before heading back to the drawing board.

[max4236]

Make a 3D scalar field eps1(x,y,z) by combining the system of equations. Then find the minimum points inside the field. I tried a guassian traversal method. Pick random points and fall towards the local minimums.

 

x,y,z looks like table salt with the nodes at multiples of pi. I don't think there's anything hidden in between the nodes other than fuzzy near solutions. I haven't yet tried extending x,y,z into complex numbers with this method though.

 

function eps1(x,y,z)
'----------------------
f0=sin(x+y+z)
f1=sin(x)+2*f0
f2=sin(y)+3*f0
f3=sin(z)+4*f0
'----------------------
eps1=f1^2+f2^2+f3^2
end function

cls:screen 12:color 2
em=100

while not(done)
10 p=2
randomize timer
x=rnd*30:y=rnd*30:z=rnd*30

'gaussian traversal thingy
f0=-1:while f0
f1=-1:while f1
incr itr
e0=eps1(x,y,z):t1=e0
for i=-.5 to .5:for j=-.5 to .5:for k=-.5 to .5
 x1=x+i*p:y1=y+j*p:z1=z+k*p
 e1=eps1(x1,y1,z1)
 if e1<e0 then x0=x1:y0=y1:z0=z1:e0=e1
next:next:next
x=x0:y=y0:z=z0
if e0=t1 then f1=0
wend
p=p*.5:if p<1e-5 then f0=0
if itr>500 then itr=0:goto 10
wend'----------------------

'scatter plot
m=3
pset(x*m,y*m),1
pset(x*m+100,z*m),2
pset(x*m+100,z*m +100),3

'best guess so far
if e0<em then em=e0:xm=x:ym=y:zm=z
incr itr2
locate 22,1:?"itr";itr2;
locate 23,1:?"x,y,z";xm;ym;zm;"            ";
locate 24,1:?"e";em;

if inkey$="q" then end
wend

:(

[Nootropic]

Well I talked to a professor over at MSU, and he attempted to solve it "briefly", but then he resorted to maple, and maple only gave him the trivial solution of integer multplies of pi. He suggested attempting to prove that integer multiples of pi are the solution. Proof by contradiction, perhaps?