#2351 Imaginary unit to the power of 4/3

None
closed
nobody
None
5
2012-12-10
2012-02-03
Anonymous
No

(%i1) (%i)^(4/3)
(%o1) 1

WxMaixima
System info
wxWidgets: 2.8.12
Unicode Support: no
Maxima version: 5.26.0
Lisp: GNU Common Lisp (GCL) GCL 2.6.8 (a.k.a. GCL)

Discussion

• Aleksas - 2012-02-04

After replace %i to polarform(%i) or domain:complex; we get correct result:

(%i1) polarform(%i)^(4/3);
(%o1) (sqrt(3)*%i)/2-1/2
(%i2) rectform(%);
(%o2) (sqrt(3)*%i)/2-1/2

or
(%i3) domain:complex;
(%o3) complex
(%i4) (%i)^(4/3);
(%o4) (-1)^(2/3)
(%i5) rectform(%);
(%o5) (sqrt(3)*%i)/2-1/2

Other example: integrate(exp(x^5),x,0,1)
Wrong:
(%i6) domain:complex;
(%o6) complex
(%i7) integrate(exp(x^5),x,0,1);
(%o7) (%e^((2*%i*%pi)/5)*(gamma_incomplete(1/5,-1)-gamma(1/5)))/5
(%i8) float(rectform(%)),expand;
(%o8) 0.37851290892278-1.164942948399964*%i

Correct:
(%i9) assume(k>1)\$ declare(k,odd)\$
(%i11) sol:integrate(exp(x^k),x,0,1);
"Is "(k-1)/k" an "integer"?"n;
(%o11) (gamma(1/k)/(-1)^(1/k)-gamma_incomplete(1/k,-1)/(-1)^(1/k))/k
(%i12) subst(k=5,sol);
(%o12) (gamma(1/5)/(-1)^(1/5)-gamma_incomplete(1/5,-1)/(-1)^(1/5))/5
(%i13) float(rectform(%)),expand;
(%o13) 1.1102230246251565*10^-16*%i+1.224893503635311
(%i14) realpart(%);
(%o14) 1.224893503635311
(%o15) [1.224893503635311,5.5812865751276883*10^-11,21,0]

Aleksas D

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Anonymous - 2012-02-13

Thank you kindly for the detailed answer.

I seem to recall reading that maxima uses complex numbers by default. As a general comment I find it odd that polarform(%i) would not give the same result as just %i.

However I've learned something. Much appreciated.

• Robert Dodier - 2012-12-10
• Description has changed:

Diff:

```--- old
+++ new
@@ -1,4 +1,3 @@
-
\(%i1\)  \(%i\)^\(4/3\)
\(%o1\) 1
```
• status: open --> closed
• milestone: --> None

• Robert Dodier - 2012-12-10

Observed behavior is to be expected given the current assumptions about simplification of complex numbers. Marking this report "closed" accordingly.