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Rudin Theorem 1.21. How does he get The identity ?

  1. Nov 8, 2013 #1
    Rudin Theorem 1.21. How does he get "The identity"?

    In Theorem 1.21, Rudin says:

    The identity [tex] b^n-a^n=(b-a)(b^{n-1}+b^{n-2}a+....+a^{n-1}) [/tex]yields etc etc.

    What is this "identity", and do we need to prove it first? If not, what assumption is Rudin making?
     
  2. jcsd
  3. Nov 8, 2013 #2

    lavinia

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    Just multiply it out
     
  4. Nov 8, 2013 #3
    So if we have [tex]b^3-a^3[/tex], we get
    [tex] (b.b.b)-(a.a.a)[/tex]

    I'm probably missing something obvious, but I'm not sure what to do from here on.
     
  5. Nov 8, 2013 #4
    ##(b-a)(b^{n-1}+b^{n-2}a+....+a^{n-1}) = b(b^{n-1}+b^{n-2}a+....+a^{n-1})-a(b^{n-1}+b^{n-2}a+....+a^{n-1}) = (b^{n}+b^{n-1}a+....+ba^{n-1}) - (b^{n-1}a+b^{n-2}a^2+....+a^{n}) = b^n-a^n##
     
  6. Nov 8, 2013 #5
    Thanks that works!

    It's curious though. We've proved it in one direction, but I wonder how someone made the formula in the first place.
     
  7. Nov 8, 2013 #6
    I suppose one might start by noticing that ##(b-a)## factors ##b^n-a^n##, since ##a## is clearly a root of the polynomial ##x^n-a^n##. So ##b^n-a^n=(b-a)p(a,b)## for some polynomial ##p(a,b)##. Then, to determine ##p(a,b)##, one could start doing long division to see the pattern. Alternatively, we can construct ##p(a,b)## by noticing that ##p(a,b)=p(b,a)##, and that ##p(a,b)## must have ##b^{n-1}## as a term.
     
  8. Nov 9, 2013 #7
    I linked to your reply from my blog post: http://www.bhagwad.com/blog/2013/un...ins-principles-of-mathematical-analysis.html/

    Thanks again!
     
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