Summary

You searched for: sol=-6272

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1

New Number: 8.66 |  AESZ:  |  Superseeker: 4 12332  |  Hash: d941d8e5d41f2e7285be47b4fbc81023  

Degree: 8

\(\theta^4-2^{2} x\left(12\theta^4-24\theta^3-23\theta^2-11\theta-2\right)-2^{7} x^{2}\left(32\theta^4+392\theta^3+484\theta^2+223\theta+41\right)+2^{12} x^{3}\left(31\theta^4-30\theta^3-872\theta^2-801\theta-217\right)-2^{16} 3 x^{4}\left(140\theta^4+60\theta^3-1332\theta^2-971\theta-231\right)-2^{20} x^{5}\left(772\theta^4+7960\theta^3+7483\theta^2+1509\theta-266\right)+2^{26} x^{6}\left(46\theta^4+2766\theta^3+2333\theta^2+672\theta+19\right)-2^{30} 5 x^{7}\left(477\theta^4+930\theta^3+697\theta^2+232\theta+28\right)-2^{36} 5^{2} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, -8, 424, -6272, 859816, ...
--> OEIS
Normalized instanton numbers (n0=1): 4, 500, 12332, 358180, 15491360, ... ; Common denominator:...

Discriminant

\(-(64z+1)(4096z^3+6144z^2+48z-1)(1-32z+2560z^2)^2\)

Local exponents

≈\(-1.492036\) ≈\(-0.017379\)\(-\frac{ 1}{ 64}\)\(0\)\(\frac{ 1}{ 160}-\frac{ 3}{ 160}I\)\(\frac{ 1}{ 160}+\frac{ 3}{ 160}I\) ≈\(0.009415\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)
\(1\)\(1\)\(1\)\(0\)\(3\)\(3\)\(1\)\(1\)
\(2\)\(2\)\(2\)\(0\)\(4\)\(4\)\(2\)\(1\)

Note:

This is operator "8.66" from ...

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