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I need to solve an equation of this form:

NSolve[a + x^b + PolyLog[c, x] + PolyLog[d, 1/x] == 0, x]

Where $a, b, c, d$ are constants.

For the sake of simplicity we can make $a = b = c = d$. Ok I know that with no chosen variables, the question is basically impossible but besides that, what I would like to get is a sort of plot in function of those constant, in a range let's say from $-3$ to $+5$ each one.

But then another problem would arise: having the plot, let's say for $a = b = c = d = 2$, how can I ask Mathematica to give me the numerical solution of the above equation? Solve seems not working!

(WM 10.0.2.0)

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    $\begingroup$ Try with FindRoot. $\endgroup$ – corey979 Sep 12 '16 at 14:48
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    $\begingroup$ This is close. Since it is a transcendental equation you also need to restrict the range of the solution, so could do as follows. In[433]:= NSolve[ 2 + x^2 + PolyLog[2, x] + PolyLog[2, 1/x] == 0 && -100 < x < 100, x] Out[433]= {{x -> -0.370672770423}} $\endgroup$ – Daniel Lichtblau Sep 12 '16 at 14:49
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    $\begingroup$ @DanielLichtblau That excludes Complex solutions though $\endgroup$ – Feyre Sep 12 '16 at 14:50
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    $\begingroup$ One can give different restrictions, viable elsewhere in C. Example: In[438]:= NSolve[ 2 + x^2 + PolyLog[2, x] + PolyLog[2, 1/x] == 0 && -100 <= Re[x] <= 100 && 1 <= Im[x] <= 200, x] Out[438]= {{x -> -0.1147574412 + 1.19969565036 I}} $\endgroup$ – Daniel Lichtblau Sep 12 '16 at 14:52
  • $\begingroup$ @DanielLichtblau That is a cool method! Thank you! $\endgroup$ – Henry Sep 12 '16 at 15:09
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I would like to provide an answer slightly complementary to the comments. In fact, given equation has analytic solutions in special cases. We know that for negative integer orders PolyLog reduces to Bernoulli polynomials. Therefore we have exact roots for the orders $\ldots,-8,-6,-4,-2,-1$:

e = -4;
f = Simplify[
  a + x^b + PolyLog[c, x] + PolyLog[d, 1/x] /.{a -> e, b -> e, c -> e, d -> e}]
Solve[f == 0, x]

$x\to -\frac{1}{\sqrt{2}},\\x\to -\frac{i}{\sqrt{2}},\\x\to \frac{i}{\sqrt{2}},\\x\to \frac{1}{\sqrt{2}}$

e = -2;
f = Simplify[
  a + x^b + PolyLog[c, x] + PolyLog[d, 1/x] /. {a -> e, b -> e, c -> e, d -> e}]
Solve[f == 0, x]

$x\to -\frac{1}{\sqrt{2}},\\x\to \frac{1}{\sqrt{2}}$

e = -1;
f = Simplify[
  a + x^b + PolyLog[c, x] + PolyLog[d, 1/x] /. {a -> e, b -> e, c -> e, d -> e}]
Solve[f == 0, x]

$x\to \frac{1}{3} \left(\sqrt[3]{71+3 \sqrt{105}}+\sqrt[3]{71-3 \sqrt{105}}+5\right),\\x\to -\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{71+3 \sqrt{105}}-\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{71-3 \sqrt{105}}+\frac{5}{3},\\x\to -\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{71+3 \sqrt{105}}-\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{71-3 \sqrt{105}}+\frac{5}{3}$

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