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Finding the nontrivial zeros of Tan x = x

  1. Nov 19, 2011 #1
    2dj9zxz.png

    I can do the first part no problem.

    I then drew the graph, am I right in saying there is an infinite sequence because the lines intersect an infinite amount of times, because tan is periodic and has vertical asymptotes?

    I have no idea about showing why the first non rivial zero is bounded like that. I would have thought the trivial zero was at the origin, and the first non trivial was in the range of 0<lambda<pi/2.

    What am I not understanding here?

    Thanks

    edit:

    oh I think I'm suppose to be looking for solutions in the range of pi and 3pi/2?
     
  2. jcsd
  3. Nov 19, 2011 #2

    dynamicsolo

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    ...because [itex]\lambda[/itex] is a positive number, so the first non-trivial zero of the equation will be in the next quadrant where [itex]\tan \lambda[/itex] is positive, and you're after [itex]\lambda^{2}[/itex].

    [EDIT: On thinking about this a little more, we could also go in the negative direction, since [itex]\tan \lambda [/itex] and [itex] \lambda [/itex] both have odd symmetry and we're looking for solutions for [itex]\lambda^{2}[/itex]... But it is easier to think about going in the positive direction.]

    BTW, I think the problem-poser means the first non-trivial zero of the function [itex](\tan \lambda ) - \lambda [/itex] ; that last statement reads a little strangely...
     
    Last edited: Nov 19, 2011
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