solve wrt additive generator

[smichr]

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fixes #19509

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Release Notes

• solvers
  • equations with an additive generator are now solved more quickly

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• solvers
  • equations with an additive generator are now solved more quickly (#19524 by @smichr)

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fixes #19509

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* solvers
  * equations with an additive generator are now solved more quickly
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Update

The release notes on the wiki have been updated.

[codecov]

Codecov Report

Merging #19524 into master will increase coverage by 0.004%.
The diff coverage is 100.000%.

@@              Coverage Diff              @@
##            master    #19524       +/-   ##
=============================================
+ Coverage   75.690%   75.694%   +0.004%     
=============================================
  Files          653       654        +1     
  Lines       169873    169950       +77     
  Branches     40059     40073       +14     
=============================================
+ Hits        128577    128643       +66     
- Misses       35686     35693        +7     
- Partials      5610      5614        +4     

[oscarbenjamin]

This looks good. +1 to merge if this is done.

In general I wonder if a more principled way of doing this is to work one step at a time applying invertible operations to try and isolate the symbol:

1/(x - 1)**5 - 1  =  0
1/(x - 1)**5  =  1
(x - 1)**5  =  1

Then when an invertible operation isn't possible any more we can look for the substitution e.g. z = x-1 and solve sequentially:

1. z**5 = 1
2. z = x - 1

Does that not already happen in univariate solve?

[smichr]

Does that not already happen in univariate solve?

It goes for the polynomial first (after checking for linear cases). And yes, this could be done -- and _invert is used in the multi-equation solving -- but you would still want to check for the additive generator:

>>> _invert((x-1)**5-y*(x-1) - 3,x)
(3, -y*(x - 1) + (x - 1)**5)

So what is presented here covers this case, too.