Prove x^n - y^n = (x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})

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In summary: It's Spivak's Calculus. He said so in the title.Hey, It's Spivak's Calculus. He said so in the title.
  • #1
rocomath
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Prove the following:

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

Hmm ... I factored x then -y and came out with:

[tex]x^{n}+...x^{2}y^{n-2}-x^{n-2}y^{2}...-y^{n}[/tex]

Argh. What's up with the middle part? I'm not sure where to go from here.
 
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  • #2
Direct expansion will verify it.
 
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  • #3
Oops, I mistyped something.
 
  • #4
Putting the right hand side in sigma form

[tex](x-y) \sum_{i=1}^{n} x^{n-i}y^{i-1}[/tex]

[tex]\sum_{i=1}^{n} x^{n-i+1}y^{i-1} - x^{n-i}y^{i}[/tex]

[tex]\sum_{i=0}^{n-1} x^{n-i}y^{i} - \sum_{i=1}^{n}x^{n-i}y^{i}[/tex]

[tex]x^n - y^n + \sum_{i=1}^{n-1} x^{n-i}y^{i} - \sum_{i=1}^{n-1}x^{n-i}y^{i}[/tex]

[tex]x^n - y^n[/tex]
 
  • #5
Too bad I have no idea what that means :-X I guess I'll just have to wait till I get to those types of methods. Thanks tho.
 
  • #6
Just expand the right side as mathboy suggested. chickendude just did it in abbreviated form.
 
  • #7
What does it mean to "expand" it. I think that's in the next chapter, so I'll just go back to it.
 
  • #8
It means to multiply the thing out, e.g. (x-y)(x+y) = x^2 + xy - yx - y^2. Have you done basic algebra? If not, I don't think Spivak is right for you just yet.
 
  • #9
morphism said:
It means to multiply the thing out, e.g. (x-y)(x+y) = x^2 + xy - yx - y^2. Have you done basic algebra? If not, I don't think Spivak is right for you just yet.
Uh. That is exactly what I did smart ***. And what I got in the middle makes no sense to me.
 
  • #10
No reason to get snappy. :tongue2: Everything between x^n and -y^n will cancel off, e.g. we're going to get x^(n-1) y and -y x^(n-1), etc. Try it out for n=3 to get a feel for it.
 
  • #11
Hey,

rocophysics said:
Prove the following:

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

Hmm ... I factored x then -y and came out with:

[tex]x^{n}+...x^{2}y^{n-2}-x^{n-2}y^{2}...-y^{n}[/tex]

Argh. What's up with the middle part? I'm not sure where to go from here.

Sorry, for posting a little late. This problem is interesting out of what textbook did you get this problem from?

Thanks,

-PFStudent
 
  • #12
It's Spivak's Calculus. He said so in the title.
 
  • #13
Hey,

Defennnder said:
It's Spivak's Calculus. He said so in the title.

Ahem. Spivak (that is, Michael Spivak) is the author of several calculus titles,

Calculus
Calculus on Manifolds: A Modern Approach to Classical Theorems of Advanced Calculus
A Comprehensive Introduction to Differential Geometry, Volumes 1-4: 3rd Edition
The Hitchhiker's Guide to Calculus
Calculus: Calculus of Infinitesimals

And no, he did not mention it was specifically from the text, Calculus.

What I wanted to know was which one of his texts had the problem. It was already obvious that it was from one of his several calculus texts, however which one was not.

Thanks,

-PFStudent
 
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1. What is the formula for x^n - y^n?

The formula for x^n - y^n is (x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1}), where n is a positive integer and x and y are real numbers.

2. How do you prove x^n - y^n = (x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})?

This can be proved using mathematical induction, starting with the base case of n=1 and then using the inductive hypothesis to prove the statement for n+1. Alternatively, it can also be proved using the binomial theorem.

3. What is the significance of the formula x^n - y^n?

The formula x^n - y^n is significant because it is a useful expression for factoring polynomials and solving equations involving exponents. It is also a key component in the binomial theorem and has applications in fields such as physics and engineering.

4. Can x and y be complex numbers in the formula x^n - y^n = (x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})?

Yes, x and y can be complex numbers in this formula. The formula holds true for all real and complex numbers.

5. Are there any special cases for the formula x^n - y^n = (x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})?

Yes, there are a few special cases for this formula. When n=0, the formula becomes x^0 - y^0 = 1-1 = 0. When n=2, the formula simplifies to x^2 - y^2 = (x-y)(x+y), which is a commonly used identity. Additionally, when x=y, the formula becomes 0=0, which is always true.

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