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Prove that |suba| < |a| for all a!=0

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that |sina| < |a| for all x≠0 and use the result to show that the only solution to the equation sinx=x is x=0. What happens if you try to find all intersections with a graphing calculator?


    2. Relevant equations



    3. The attempt at a solution
    I know for the first part you have to use MVT, but I've always been bad with the MVT, hence me being here :).
     
  2. jcsd
  3. Dec 4, 2011 #2

    Char. Limit

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    Well, first thing's first, it's good to remember that the absolute value of an odd function is an even function. So we can essentially ignore a<0, because if |sin(a)|≥|a| for any a<0, it'll also be true for some a>0. So it'll help to restrict the domain a bit. After that... well, I'm still working on after that.
     
  4. Dec 4, 2011 #3

    Char. Limit

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    Actually, after looking at this, I don't think the Mean Value Theorem is entirely necessary. (Feel free to correct me if I'm wrong, people who are better than me at this!) First, since we restricted our domain to non-negative a, we can throw the absolute value signs off the right side:

    |sin(a)|≤a for a≥0

    Then subtract |sin(a)| from both sides and prove that the resulting function's (a-|sin(a)|, to be exact) derivative is never negative. That, along with the fact that at a=0, the function is 0, and the fact that the original function is even, should prove what you're trying to prove.
     
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