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Extension of BinomialTheorem

  1. May 10, 2012 #1
    Thank you in advance, I need help proving or disproving this. In the binomial theorem, with a power (a+b)^n, I need to prove that a^n + b^n is greater than the rest, or in other words, (a+b)^n - (a^n + b^n).
     
  2. jcsd
  3. May 10, 2012 #2


    Uh? I think you forgot to add some info and/or to write some symbols, as it seems to the question doesn't make sense as it is.

    DonAntonio
     
  4. May 10, 2012 #3
    It depends on what you use for a,b, and n. In some cases, a^n + b^n will be greater than "the rest", in others, it won't.
     
  5. May 11, 2012 #4

    HallsofIvy

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    I think you mean that you want to prove that [itex](a+ b)^n- (a^n+ b^n)[/itex] is positive. As Matt Benesi said, that depends upon what a and b are.
     
  6. May 11, 2012 #5
    If the statement is suspicious, the first thing is to try a few numerical examples, to try to find a counterexample. Here is a hint: 1+1=2. :)

    P.S.: My understanding is that the OP tries to prove or disprove[tex]a^n + b^n > (a+b)^n - (a^n + b^n)[/tex]
     
  7. May 11, 2012 #6


    Too many assumptions: shall we let the OP to tell us what he meant, please?

    DonAntonio
     
  8. May 11, 2012 #7
    That is correct. It can also be assumed that a and b do not equal 1.
     
  9. May 11, 2012 #8


    Since the OP already wrote a post saying this is correct, this is the same as [tex]2(a^n+b^n)>(a+b)^n[/tex] which is greatly false, for example: for [itex]\,\,a=1\,,\,b=2\,,\,n=3\,\,,\,\,or\,\,a=2\,,\,b=3\,,\,n=4\,\,[/itex] , and infinite counterexamples more.

    DonAntonio
     
  10. May 11, 2012 #9
    Question: can a and b be less than 1? Can they be negative?

    Just some interesting cases: in all of these cases, x^n+y^n = "all the rest".

    n=2 and a=b.

    For n=3 we have [itex]x=-y[/itex], [itex]x=y\times\left[2-\sqrt{3}\right][/itex], [itex]x=y\times\left[2+\sqrt{3}\right][/itex].

    Getting more complicated for n=4, just one example (out of 4):
    [tex]x=\sqrt{2\,\sqrt{3}+3}\,y+\sqrt{3}\,y+y[/tex]
    or to preserve the format used above:
    [tex]x=y\times\left[\sqrt{2\,\sqrt{3}+3}\,+\sqrt{3}\,+1\right][/tex]
     
    Last edited: May 11, 2012
  11. May 12, 2012 #10
    a and b both must be integral and greater than 1. (a^n + b^n) must be greater than (a+b)^n - (a^n + b^n) by obvious reasoning. First of all, the two terms, a^n and b^n, contains the highest power of the binomial expansion. The next highest power,n-1, should be greater than a^n or b^n only when n is greater than the following coefficient
     
  12. May 12, 2012 #11


    Either you don't understand mathematically what is going on here or else you're misunderstanding and/or misreading big time

    the inequality you want/must prove.

    You wrote above "(a^n + b^n) must be greater than (a+b)^n - (a^n + b^n) by obvious reasoning", which means Gauss knows what, but

    this inequality is [itex]\,\,a^n+b^n>(a+b)^n-(a^n+b^n)\Longleftrightarrow 2(a^n+b^n)>(a+b)^n\,\,[/itex] , which already was show to be

    false for lots and lots of options...Please do read and write carefully what you exactly want to achieve.

    DonAntonio
     
  13. May 12, 2012 #12

    micromass

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    Please state exactly what you want to prove.
     
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