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Finding the sum of inverse trigonometric expression

  1. Apr 20, 2015 #1
    1. The problem statement, all variables and given/known data

    Find ## \sum_1^{23} tan^{-1}(\frac{1}{n^2+n+1}) ##
    2. Relevant equations
    ## tan^{-1}x + tan^{-1}y = tan^{-1}(\frac{x+y}{1-xy} )##


    3. The attempt at a solution
    I think we have to split the question in a form of relevant equation given above.
    First thing what should I do?
     
  2. jcsd
  3. Apr 20, 2015 #2

    phyzguy

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    In order to use the identity you have been given, you want to get the term [itex]\frac{x+y}{1-xy}[/itex] into the form of [itex]\frac{1}{n^2+n+1}[/itex]. This requires that you find x and y in terms of n. Do you see a way to do this?
     
  4. Apr 20, 2015 #3
    Yes but it is by trial and error method.
    1/(n2+n+1) can be written as (n+1-n)/(1+n(n+1))
    Now it is seperatable.
     
  5. Apr 20, 2015 #4

    Ray Vickson

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    One possibility is to apply the equation in (2) recursively: If ##t_k = 1/(k^2+k+1)##, let ##T_n = \sum_{k=1}^n \arctan(t_k) \equiv \arctan(S_n)##. Then ##S_1 = t_1## and $$S_k = \frac{S_{k-1}+t_k}{1-S_{k-1} t_k}$$
    for ##k = 2, \ldots, 23##.
     
  6. Apr 20, 2015 #5

    phyzguy

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    An easier way is to see that in order for the forms to be the same, we must have x+y=1. Then solve for y in terms of x and substitute into 1-xy = n^2+n+1. Then solve for x, then y.
     
  7. Apr 20, 2015 #6

    SammyS

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    @Raghav Gupta ,
    Use this method suggested by phyzguy together with the difference identity:
    ##\displaystyle\ \tan^{-1}x - \tan^{-1}y = \tan^{-1}(\frac{x-y}{1+xy} )\ ## .​
     
  8. Apr 20, 2015 #7
    It's getting quadratic.
    I am getting ## x = \frac{1 ± \sqrt{1+4n^2 + 4n}}{2} ##
    following both of you @phyzguy and @SammyS
     
  9. Apr 20, 2015 #8

    phyzguy

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    OK, keep going. 1+4n^2+4n is a perfect square, so you can take the square root and simplify it further. There are two solutions for x and y, but you will see they are really the same solution, with x and y interchanged.
     
  10. Apr 20, 2015 #9
    Thanks, got it.
    Sorry @Ray Vickson for not following your method, as I am not so acquainted with it.
     
  11. Apr 20, 2015 #10

    Ray Vickson

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    No problem: whatever works for you is fine by me.

    However, just to clarify (in case you did not understand): one way of computing ##T_{23} = \sum_{k=1}^{23} \arctan(t_k)## is to compute the successive sums. Let ##T_2 = \arctan(t_1) + \arctan(t_2) = \arctan(S_2)## (using the equation in (2) to find ##S_2##), then ##T_3 = \sum_1^3 \arctan(t_k) = T_2 + \arctan(t_3) = \arctan(S_2) + \arctan(t_3) = \arctan(S_3)##, where we find ##S_3## in terms of ##S_2## and ##t_3## from equation (2). If you keep going like that you can find ##T_4, T_5, T_6, \ldots, T_{23}##.
     
  12. Apr 20, 2015 #11
    The mood is not coming for solving by this method.
    Will see it later if required. Although thanks.
     
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