Solving 4 π arctan ⁡ ( 1 + x ) cosh ⁡ x = 1

Question

Solving       4    π    arctan  ⁡      (    1    +    x    )    cosh  ⁡      x    =  1

Hector Hamilton · Accepted Answer

Set   f  (  x  )  :=  arctan  ⁡  (  1  +  x  )  cosh  ⁡  (  x  ) for   x  ∈      R  If   x  ≤  −  1, then f(x)&amp;lt;0, so x is not solution.On   [  0  ,  +  ∞  ), f is increasing function as the product of two non-negative increasing functions. We can check by derivation that f is also increasing on   [  −  1  ,  0  ]Indeed, for every   x  ∈      R        f    ′    (  x  )  =      1          1      +      (      1      +      x              )        2              cosh  ⁡  (  x  )  +  arctan  ⁡  (  1  +  x  )  sinh  ⁡  (  x  )  ,so f′(x) a the same sign as      f    ′    (  x  )  =      1          1      +      (      1      +      x              )        2              +  arctan  ⁡  (  1  +  x  )  tanh  ⁡  (  x  )  .Now assume   x  ∈  [  −  1  ,  0  ], so we set x=−t with   t  ∈  [  0  ,  1  ]. Then f′(x) a the same sign as      1          1      +      (      1      −      t              )        2              −  arctan  ⁡  (  1  −  t  )  tanh  ⁡  (  t  )  .In the one hand,   1  ≤  1  +  (  1  −  t      )    2    ≤  2, so      1          1      +      (      1      −      t              )        2              ≥      1    2  In the other hand, convexity inequalities   arctan  ⁡  (  1  −  t  )  ≤  1  −  t and   tanh  ⁡  (  t  )  ≤  t yield  arctan  ⁡  (  1  −  t  )  tanh  ⁡  (  t  )  ≤  (  1  −  t  )  t  ≤      1    4    .Hence      1          1      +      (      1      −      t              )        2              −  arctan  ⁡  (  1  −  t  )  tanh  ⁡  (  t  )  ≥      1    4    .so       f    ′    (  x  )  &amp;gt;  0Therefore, since   f  (  0  )  =  π      /    4, 0 is the only solution of your equation.

Solving (4)/(pi) arctan(1+x)cosh x=1

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