Recently I had a small problem with Julia that confused me. The code is shown below: Julia> x=2.0e-10 2.0e-10 Julia> -1.0*(-0.5*x^3 - 3.0*x^2 - 7.5*x - 7.5)*exp(-x)/x^6 + 1.0*(0.5*x^3 - 3.0*x^2 + 7.5*x - 7.5)*exp(x)/x^6 0.0 Julia> x=2.1e-10 2.1e-10 Julia> -1.0*(-0.5*x^3 - 3.0*x^2 - 7.5*x - 7.5)…

https://0.30000000000000004.com/

[image] PSA: floating-point arithmetic Offtopic Sometimes people are surprised by the results of floating-point calculations such as julia> 5/6 0.8333333333333334 # shouldn't the last digit be 3? julia> 2.6 - 0.7 - 1.9 2.2204460492503…

In particular, you are seeing a catastrophic cancellation .

Results will behave by using big floats.

Only kind of. The errors will be smaller, but still non-zero.

[image] rafael.guerra: Results will behave by using big floats. @F-YF posted an ill-conditioned sum whose terms (in exact arithmetic) should exactly cancel to give zero. Any roundoff error can give a nonzero answer, which is an infinite relative error compared to 0.0 (no digits are correct)…

The general guidance is to mathematically rewrite your expressions to avoid catastrophic cancellation. In this case Wolfram Alpha simplifies this to (x(x^2 + 15)\cosh(x) - 3(2x^2 + 5)\sinh(x))/x^6 This doesn’t exhibit the same cancellation issue: julia> f(x) = (x*(x^2 + 15)*cosh(x) - 3(2x^2 + 5)…

I tried herbie for fun, and it suggested: 1.2025012025012024 \cdot 10^{-5} \cdot {x}^{5} + \left(0.0005291005291005291 \cdot {x}^{3} + x \cdot 0.009523809523809525\right) Not exact, but the error is small julia> function f(x) (1.2025012025012024e-5 * ^(x, 5.0)) + ((0.0005291005291005291 …

[image] Oscar_Smith: Only kind of. The errors will be smaller, but still non-zero. Humble question: using big floats the error is close to zero (~2e-12) and a 5% perturbation of the input doesn’t catastrophically explode the output from near 0 to ~ -1e43. This gain in stability should be imp…

Numerical precision

General Usage

lmiq March 30, 2022, 2:44pm 3

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Numerical precision

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