X^n +y^n

  • Thread starter phymatter
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  • #1
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Main Question or Discussion Point

What is the expansion of xn +yn , when is even ??/
 

Answers and Replies

  • #2
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I don't see anything that can be expanded.
 
  • #3
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I don't see anything that can be expanded.
i mean that xn - yn can be written as (x-y)(xn-1 +xn-2y ....+yn-1 )
similarly what can xn +yn be written as ????????
 
  • #4
jav
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Try alternating signs, and it becomes straightforward.
 
  • #5
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I think you need to think about zeros
[tex]x^n+y^n=0[/tex]
[tex]x^n=-y^n[/tex]
[tex]x=y\cdot\exp(i\pi k/n)[/tex]
[tex]\therefore x^n+y^n=\prod_k (x-\exp(i\pi k/n)y)[/tex]

Occationally combining a subset of these factors together will give you a real solution.

Now you need to think when... :)
 
Last edited:
  • #6
Mentallic
Homework Helper
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What is the expansion of xn +yn , when is even ??/
i mean that xn - yn can be written as (x-y)(xn-1 +xn-2y ....+yn-1 )
similarly what can xn +yn be written as ????????
Then you mean what are the factors :smile:

If n is odd, you can factor it as so:

[tex]x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^2-x^{n-4}y^3+...-xy^{n-2}+y^{n-1})[/tex]

However, if n is even, then [tex]x^n+y^n\neq 0[/tex] except for in the trivial case of x,y=0. This means you can't factor it over the reals. You'll need to use complex numbers. You could convert it into a few different ways, such as [tex]x^n-i^2y^n[/tex] and take difference of two squares, or, if you want to follow the same factorizing process as above, take [tex]x^n+(iy)^n[/tex] and take two cases, when [itexi^n[/itex] is equal to 1, and when equal to -1.
 

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