How do I avoid losing solutions from a non-linear system like this one? What should I do?
I already solved it, but I almost miss it, I'm afraid I'll get more problems like this one in the future
Would appreciate some book recommendations about this topic.
khan academy
you also get a free colour blindness test
Most of the time you just reason it out.
In this case, it's the line x=y.
You have to use non-linear algebra. Scmoodleberg and Bloch is the classic reference.
Scmoodleberg?
Linearize it via Taylor expansion
I didn't now I could do that with the Taylor expansión, thanks I'll learn it.
It is not line x=y.
When you replace system1 with system2 always prove they have the same set of solutions.
{cos(x)+cos(x+y)=0,cos(y)+cos(x+y)=0} =>
{cos(x)+cos(x)cos(y)-sin(x)sin(y)=0,cos(y)+cos(x)cos(y)-sin(x)sin(y)=0}=>
{cos(x)(1+cos(y))=sin(x)sin(y),cos(y)(1+cos(x))=sin(x)sin(y)}
From this follows cos(x)(1+cos(y))=cos(y)(1+cos(x)) => cos(x)=cos(y). So we can add and apply this condition to the original system without changing solution set:
{cos(x)=cos(y),cos(x)(1+cos(x))=sin(x)sin(y),cos(x)(1+cos(x))=sin(x)sin(y)}=>
{cos(x)=cos(y),cos(x)+cos(x)^2=sin(x)sin(y)}=>
{cos(x)=cos(y),cos(x)+cos(x)cos(y)-sin(x)sin(y)=0}=>
{cos(x)=cos(y),cos(x)+cos(x+y)=0}
So here we are. cos(x)=cos(y) condition has solutions x+y=2pi*N, cos(x)=cos(x+y+pi) has solutions 2x+y+pi=2pi*M. With the first condition this means x+pi=2pi*M. So the final set of solutions is x=-pi+2pi M, y = 2pi*N - x. for any integers N and M.
>all that manipulation
pic related
also
>cos(x)=cos(y) condition has solutions x+y=2pi*N
is wrong, should be x +/- y, similarly for the next statement.
It isn't wrong for obvious reasons.
-pi is -2pi + pi.
I'm not overcomplicating I'm just being extra in detail
Ah, you mean it should be x-y too. Yes, my bad. :(
And can you show me a better solution if you think I'm overcomplicating?
The same thing you wrote but remove all the obviously superfluous steps where you rewrite cos(x+y) in various ways only to end up with cos(x+y) at the end.
You can't always. Not every system has closed-form, analytic solutions.
In cases like this though you need to pay attention to the domains and ranges of the functions you use, that's it.
There's no general method though.
Mathematica
That's only a partial set of solutions.
Cosx - siny = 0
Implies x = y
This should be obvious.
Take pi/3 for example:
.5 - .5 = 0
Satisfies the equations, but is not possible in your solution set.
Applying arbitrary transformations can give you tunnel vision. You're better off visualizing the solutions in polynomial space.
I mean, I understand making those manipulations when you're exploring the problem, but clean up before you post.
Don't forget function parity mate
I made a typo.
Cos (x) + cos (x+y) = cos (y) + cos (x+y) implies cos (x) = cos (y)
Implies x = y
To add to this: removing superfluous steps reduces your likelihood of error. You can check your work without your eyes glazing over.
>Every function is a one to one function
You're either in highschool or an engineer.
It obviously isn't, and I never said it was.
Either way, that other post was wrong because it won't provide all solutions.
The only thing he did was show that cos is an even function.
Cos(x)=Cos(y) => x=y absolutely assumes Cos is one-to-one. That's what a one to one function is.
What you said is akin to
x^2 + a = y^2 + a implies x^2 = y^2 implies x=y except more egregious.
Ohhh, I see.
You think I said that the function maps x to y.
That's not what I said or implied. Keep trying.
Am I missing something? Serious question. As far as I can see you literally said
> cos (x) = cos (y)
> Implies x = y
It is not set of solutions nor partial set of solutions.
take x=y=pi/2
cos(x)+cos(x+y)=cos(pi/2)+cos(pi/2 + pi/2)=0-1=-1. But it should've been zero.
>not calling a ""one to one"" function injective
how's middle school?
>You're better off visualizing the solutions in polynomial space.
What do you mean?
...
What I should have said is that the solutions will look like this:
let L be the set of all ordered pairs (x,y) where x,y belong to the reals R
let K be the set of all k such that -1
Eh, it's as much of a regional and field conventional thing I think, but I'm not sure because I don't really remember middle school You can certainly find academics still using the term though scholar.google.com
I'll agree it's not as nice of terminology, I've spent this summer teaching the intro to proofs class at my uni and I fell into the terminology we're used to using I guess.
Also, I apologize for this.
I wrote quickly and didn't think of the exact values that would work.
m.wolframalpha.com
My whole point was that the solutions to this systems of equations are either going to fall on the line y = x, or y = -x.
The solutions plotted out show that.
Was also close, but did not finish the problem either.
I prefer going about it my way and finding an intersection between the line, and the solutions, vs. showing that the solution sets are the same, because I'm more prone to miss solutions than I am to delete non solutions.
It's really looking at two sides of the same coin.
How does
cos(x)(1+cos(y))=cos(y)(1+cos(x))
immediately imply that cos(x)=cos(y)?