5
$\begingroup$
FindRoot[
  {Derivative[2][SinIntegral[x]], 0 < x < (Pi)},
  x
]

plot of the function of interest

I hoped that my code would return a value for x=0 but it is returning an error: "search specification should be a list with 1 to 5 elements"

edit: I want to iterate this multiple times so it will be helpful to be able to find the root within a given range for x

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2
  • $\begingroup$ To see the shape: Plot[SinIntegral''[x], {x, 0, 6 \[Pi]}] and then choose a suitable initial guess: FindRoot[SinIntegral''[x], {x, 5}] $\endgroup$
    – Syed
    Jan 18 at 5:26
  • $\begingroup$ For the interval 0 <= x <= 6 Pi use sol = FindRoot[SinIntegral''[x], {x, #}] & /@ {1/20, 5, 8, 11, 14, 17} $\endgroup$
    – Bob Hanlon
    Jan 18 at 6:16

1 Answer 1

4
$\begingroup$

Edit

FindRoot[Derivative[2][SinIntegral][x], {x, 1/2}]

{x -> 1.15898*10^-10}

sol=NSolve[{Derivative[2][SinIntegral][x], 0 < x < 6 Pi}, x]

{{x -> 4.49341}, {x -> 7.72525}, {x -> 10.9041}, {x -> 14.0662}, {x -> 17.2208}}

The concave and convex parts as below.

Plot[SinIntegral[x], {x, 0, 6 Pi}, Mesh -> {sol}, MeshStyle -> Red, 
 MeshShading -> {Cyan, Blue}]

enter image description here

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2
  • $\begingroup$ Limit[SinIntegral''[x], x -> 0] also evaluates to 0 $\endgroup$
    – Bob Hanlon
    Jan 18 at 6:23
  • $\begingroup$ The important thing for the OP to keep in mind here is that FindRoot (used in their original code) is designed to find one root and one root only, while NSolve can (sometimes) find all of the roots in a given range. $\endgroup$ Jan 18 at 14:49

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