very difficult math problem


by elfboy
Tags: difficult, math
elfboy
elfboy is offline
#1
May31-08, 08:09 PM
P: 89
The equation cos(ax)=0 has a closed form solution in terms of x

so does cos(ax)+cos(bx)=0

but I am sure that cos(ax)+cos(bx)+cos(cx)=0 does not

I first encountered this problem back in late 2003 and began working on it again
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DavidWhitbeck
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#2
May31-08, 10:04 PM
P: 352
Quote Quote by elfboy View Post
so does cos(ax)+cos(bx)=0
Those two cos's are exactly out of phase (to cancel like that) so

[tex]ax = bx +(2n+1)\frac{\pi}{2}[/tex]
elfboy
elfboy is offline
#3
Jun2-08, 12:36 AM
P: 89
Wikipedia doesn't seem to have any identities for triple-cosine-sum-to-product

eok20
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#4
Jun3-08, 09:38 PM
P: 201

very difficult math problem


Quote Quote by elfboy View Post
The equation cos(ax)=0 has a closed form solution in terms of x

so does cos(ax)+cos(bx)=0

but I am sure that cos(ax)+cos(bx)+cos(cx)=0 does not

I first encountered this problem back in late 2003 and began working on it again
To clarify, when you say it has a closed form solution in terms of x, it seems like you're asking if given any x we can find a,b,c such that the equation is satisfied. Surely, you do not mean that since that problem is simple. I'm guessing you're asking it it is possible to find a,b, and c such that the equation holds for all x?
JG89
JG89 is offline
#5
Jun3-08, 11:57 PM
P: 724
Your question seemed a bit ambiguous to me, so I hope this is what you are asking for:

We start off with:

Cos(ax) + Cos(bx) + Cos(cx) = 0

Cos(bx) + Cos(cx) = -Cos(ax)

ArcCos[Cos(bx) + Cos(cx)] = ArcCos[-Cos(ax)]

Looking at the right side of the equation after taking the inverse Cos, it will be 180 - ax. It will be 180 - ax because the Cosine there is negative. Now, we want to write the sum of the Cosines on the left side as a single Cosine so we can take the inverse of it and get rid of Cosine all together. We can rewrite:

Cos(bx) + Cos(cx) as Cos(ax)

We can do this because the right side of the equation, before taking the inverse Cosine, is -Cos(ax). The left side must be the additive inverse of this, and so must be Cos(ax). However, the left side is Cos(bx) + Cos(cx), which means that Cos(bx) + Cos(cx) must be the additive inverse of -Cos(ax), which is Cos(ax).

So now, we have

ArcCos[Cos(ax)] = 180 - ax

ax = 180 - ax
2ax = 180
a = 90/x

So, we now know a can be written as 90/x.

Let's substitute that into the original equation:

Cos(ax) + Cos(bx) + Cos(cx) = 0
Cos( 90/x * x) + Cos(bx) Cos(cx) = 0
Cos(90) + Cos(bx) + Cos(cx) = 0

But Cos90 = 0, so:

Cos(bx) + Cos(cx) = 0
Cos(bx) = -Cos(cx)
ArcCos[Cos(bx)] = ArcCos[-Cos(cx)]
bx = 180 - cx
bx + cx = 180
x(b+c) = 180
b+c = 180/x

So, we now have two equations:

a = 90/x

and

b + c = 180/x

So, pick any value of x, substitute it into the first equation, you will get a value for a. Substitute a value for x into the second equation, then pick values of b and c to satisfy that equation and you will your values for a, b, c, and x that will satisfy your original equation (as long as x does not equal 0).

I hope that is what you were asking for, and sorry if the format is hard to follow. I don't know how to use math symbols on the computer.
quadraphonics
quadraphonics is offline
#6
Jun4-08, 12:05 AM
P: 270
I think what elfboy is asking for is a closed-form expression for the set of all x that satisfy the equation, given some a, b and c. For a single cosine, this is trivial: it's just the set of all integer multiples of [itex]\frac{\pi}{a}[/itex]. For a sum of two cosines, you can apply a trig identity to get a product of two cosines, and then apply the same reasoning as before to each one. This gives the solution as all integer multiples of either [itex]\frac{2\pi}{a+b}[/itex] or [itex]\frac{2\pi}{a-b}[/itex]. For the case of three cosines, it's not so obvious, at least when [itex]c \neq 0 [/itex]...
Kurret
Kurret is offline
#7
Jun4-08, 03:23 AM
P: 124
Quote Quote by JG89 View Post
Looking at the right side of the equation after taking the inverse Cos, it will be 180 - ax. It will be 180 - ax because the Cosine there is negative. Now, we want to write the sum of the Cosines on the left side as a single Cosine so we can take the inverse of it and get rid of Cosine all together. We can rewrite:

Cos(bx) + Cos(cx) as Cos(ax)
No, you cant substitute a value from one equation into itself. You will get:
Cosbx+Coscx=-cosax
-cosax=-cosax
0=0
Im not sure though that understand what you mean, but anyway what you have done is not the answer the the thread starters question. I think hes question was:
Given real numbers a,b and c, find the solution of
Cosax+Cosbx+Coscx=0
in terms of a,b and c.
JG89
JG89 is offline
#8
Jun4-08, 09:46 AM
P: 724
Quote Quote by Kurret View Post
No, you cant substitute a value from one equation into itself. You will get:
Cosbx+Coscx=-cosax
-cosax=-cosax
0=0
Im not sure though that understand what you mean, but anyway what you have done is not the answer the the thread starters question. I think hes question was:
Given real numbers a,b and c, find the solution of
Cosax+Cosbx+Coscx=0
in terms of a,b and c.

In that part I'm not substituting a value from one equation into itself. I am saying that the left side must be the additive inverse of the right side in this situation, so we can rewrite Cos(ax) + Cos(bx) as Cos(ax). The two equations I came up with work in any situation, except where x = 0.

But I'm not 100% sure I properly answered the threadmakers question, I agree with you there. Any input from the thread maker?
elfboy
elfboy is offline
#9
Jun5-08, 01:46 AM
P: 89
[QUOTE=quadraphonics;1754556]I think what elfboy is asking for is a closed-form expression for the set of all x that satisfy the equation, given some a, b and c. ..[QUOTE]

thats right

The cos(ax)+cos(bx)=0 roots can be derived by adding cos(ax+bx) and cos(ax-bx)

but the expanded form of cos(ax+bx+cx) doesn't yield a sum-to-product identity as the case above does.

the closest I have gotten is:

cos(a+b+c)+cos(a)+cos(b)+cos(c)=4*cos((b+c)/2)*cos((a+b)/2)*cos((a+c)/2)


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