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1 | 1 | # Frequently Asked Questions
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2 | 2 |
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3 |
| -Ask more questions. |
| 3 | +## How is the performance of Julia's NonlinearSolve.jl vs MATLAB's fzero? |
| 4 | + |
| 5 | +This is addressed in a [Twitter thread with the author of the improved fzero](https://twitter.com/ChrisRackauckas/status/1544743542094020615). |
| 6 | +On the test example: |
| 7 | + |
| 8 | +```julia |
| 9 | +using NonlinearSolve, BenchmarkTools |
| 10 | + |
| 11 | +N = 100_000; |
| 12 | +levels = 1.5 .* rand(N); |
| 13 | +out = zeros(N); |
| 14 | +myfun(x, lv) = x * sin(x) - lv |
| 15 | + |
| 16 | +function f(out, levels, u0) |
| 17 | + for i in 1:N |
| 18 | + out[i] = solve(NonlinearProblem{false}(NonlinearFunction{false}(myfun), |
| 19 | + u0, levels[i]), Falsi()).u |
| 20 | + end |
| 21 | +end |
| 22 | + |
| 23 | +function f2(out, levels, u0) |
| 24 | + for i in 1:N |
| 25 | + out[i] = solve(NonlinearProblem{false}(NonlinearFunction{false}(myfun), |
| 26 | + u0, levels[i]), NewtonRaphson()).u |
| 27 | + end |
| 28 | +end |
| 29 | + |
| 30 | +@btime f(out, levels, (0.0, 2.0)) |
| 31 | +@btime f2(out, levels, 1.0) |
| 32 | +``` |
| 33 | + |
| 34 | +MATLAB 2022a achieves 1.66s. Try this code yourself: we receive 0.06 seconds, or a 28x speedup. |
| 35 | +This example is still not optimized in the Julia code and we expect an improvement in a near |
| 36 | +future version. |
| 37 | + |
| 38 | +For more information on performance of SciML, see the [SciMLBenchmarks](https://github.com/SciML/SciMLBenchmarks.jl) |
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