Summary

You searched for: sol=128

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1

New Number: 8.69 |  AESZ:  |  Superseeker: 4 52  |  Hash: e303d10e77a367612be2fb706f37b895  

Degree: 8

\(\theta^4-2^{2} x\left(20\theta^4+34\theta^3+29\theta^2+12\theta+2\right)+2^{4} x^{2}\left(125\theta^4+362\theta^3+471\theta^2+284\theta+66\right)-2^{7} x^{3}\left(191\theta^4+606\theta^3+855\theta^2+588\theta+154\right)+2^{10} x^{4}\left(192\theta^4+552\theta^3+562\theta^2+268\theta+49\right)-2^{13} x^{5}\left(134\theta^4+380\theta^3+373\theta^2+124\theta+3\right)+2^{16} x^{6}\left(61\theta^4+150\theta^3+173\theta^2+93\theta+19\right)-2^{19} x^{7}\left(19\theta^4+50\theta^3+56\theta^2+31\theta+7\right)+2^{23} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 8, 128, 2816, 74896, ...
--> OEIS
Normalized instanton numbers (n0=1): 4, 15/2, 52, 1563/2, 7276, ... ; Common denominator:...

Discriminant

\((16z-1)(8z-1)(64z^2-48z+1)(1-4z+32z^2)^2\)

Local exponents

\(0\)\(\frac{ 3}{ 8}-\frac{ 1}{ 4}\sqrt{ 2}\)\(\frac{ 1}{ 16}-\frac{ 1}{ 16}\sqrt{ 7}I\)\(\frac{ 1}{ 16}\)\(\frac{ 1}{ 16}+\frac{ 1}{ 16}\sqrt{ 7}I\)\(\frac{ 1}{ 8}\)\(\frac{ 3}{ 8}+\frac{ 1}{ 4}\sqrt{ 2}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(0\)\(1\)\(1\)\(1\)\(1\)\(1\)\(1\)\(1\)
\(0\)\(1\)\(3\)\(1\)\(3\)\(1\)\(1\)\(1\)
\(0\)\(2\)\(4\)\(2\)\(4\)\(2\)\(2\)\(1\)

Note:

This is operator "8.69" from ...

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