Generator

complex-symbol

One function in the convection library, called 4 times across 3 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 6 claims it put to the test while drawing them, and where it stands against the rule this site is named for.

At its defaults it draws the damping of each fourier mode, ε = 0.005, ω = 0.6667. The modulus of one weighted Jacobi sweep's effect on each grid mode, against the frequency. On the operator with no convection in it the effect is a real number and the oscillatory half is damped to ⅓ at ω = 2/3. With convection the symbol is complex, its imaginary part does not depend on ω, and the worst damping over the oscillatory half is 1.0937.

complex-symbol is one function in lib/figures/convection.js — convection — the stencil that is not symmetric, and the fourier arguments that assumed it was. Everything below came out of it during this build, at arguments taken from the essays rather than invented for this page. A figure here is the figure a reader meets in an essay, and if the generator changes, this page changes with it.

At its defaults

Drawn even though every essay passes arguments — which on this site is every essay, at 100% of placements since the standard pass. A default nothing exercises is a trap for the next essay to call this with none, and this is the page where a default that has drifted from the figures around it becomes visible.

The damping of each Fourier mode, ε = 0.005, ω = 0.6667The modulus of one weighted Jacobi sweep's effect on each grid mode, against the frequency. On the operator with no convection in it the effect is a real number and the oscillatory half is damped to ⅓ at ω = 2/3. With convection the symbol is complex, its imaginary part does not depend on ω, and the worst damping over the oscillatory half is 1.0937.00.7853981.57082.356193.1415900.250.50.751frequency θdamping |g(θ)|the oscillatory half →⅓ — the symmetric optimumno convectionwith convectionits imaginary parta modulus, not a valuesmoothing factor at this ω1.1best over every ω0.84the symmetric operator's, at ω = 2/30.33the imaginary part does not depend on ωso no ω removes it

The modulus of one weighted Jacobi sweep's effect on each grid mode, against the frequency. On the operator with no convection in it the effect is a real number and the oscillatory half is damped to ⅓ at ω = 2/3. With convection the symbol is complex, its imaginary part does not depend on ω, and the worst damping over the oscillatory half is 1.0937.

omega: 0.6666666666666666

The arguments are the ones A direction the smoother cannot see passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The damping of each Fourier mode, ε = 0.005, ω = 0.6667The modulus of one weighted Jacobi sweep's effect on each grid mode, against the frequency. On the operator with no convection in it the effect is a real number and the oscillatory half is damped to ⅓ at ω = 2/3. With convection the symbol is complex, its imaginary part does not depend on ω, and the worst damping over the oscillatory half is 1.0937.00.7853981.57082.356193.1415900.250.50.751frequency θdamping |g(θ)|the oscillatory half →⅓ — the symmetric optimumno convectionwith convectionits imaginary parta modulus, not a valuesmoothing factor at this ω1.1best over every ω0.84the symmetric operator's, at ω = 2/30.33the imaginary part does not depend on ωso no ω removes it

The modulus of one weighted Jacobi sweep's effect on each grid mode, against the frequency. On the operator with no convection in it the effect is a real number and the oscillatory half is damped to ⅓ at ω = 2/3. With convection the symbol is complex, its imaginary part does not depend on ω, and the worst damping over the oscillatory half is 1.0937.

omega: 0.35

The arguments are the ones The stencil that is not symmetric passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The damping of each Fourier mode, ε = 0.005, ω = 0.35The modulus of one weighted Jacobi sweep's effect on each grid mode, against the frequency. On the operator with no convection in it the effect is a real number and the oscillatory half is damped to ⅓ at ω = 2/3. With convection the symbol is complex, its imaginary part does not depend on ω, and the worst damping over the oscillatory half is 0.8495.00.7853981.57082.356193.1415900.250.50.751frequency θdamping |g(θ)|the oscillatory half →⅓ — the symmetric optimumno convectionwith convectionits imaginary parta modulus, not a valuesmoothing factor at this ω0.85best over every ω0.84the symmetric operator's, at ω = 2/30.33the imaginary part does not depend on ωso no ω removes it

The modulus of one weighted Jacobi sweep's effect on each grid mode, against the frequency. On the operator with no convection in it the effect is a real number and the oscillatory half is damped to ⅓ at ω = 2/3. With convection the symbol is complex, its imaginary part does not depend on ω, and the worst damping over the oscillatory half is 0.8495.

What it checked while drawing

Every figure above asserted its own claims on the way to being drawn, and a claim that failed would have failed the build rather than drawn a wrong picture. Those assertions used to leave no trace at all: a passing one returned true and the only evidence the figure had checked anything was that nothing crashed. The list below is what they actually said, collected by running this generator with an observer installed — not a description of what it is believed to check.

6 distinct claims across 3 sets of arguments, grouped below by shape — because most of them are one sentence with a different number in it, and how many separate times that sentence was put to the test is the informative part.

a diffusion coefficient inside the range drawn

a relaxation parameter inside the range

because the operator is far from symmetric

no relaxation parameter reaches the symmetric operator's smoothing factor

the symbol has an imaginary part

the symmetric smoothing factor is exactly one third at ω = 2/3

Against the rule

The rule does not apply to it. It factorises nothing, so there is no residual it could be withholding. That is worth stating rather than leaving blank: a site that reported the rule as satisfied by every generator would be counting mostly generators the rule never reached.

Across the library: the rule bites on 52 of 99 generators — 37 print a residual and 15 are exempt with a published reason; 47 factorise nothing. Read from lib/residual-rule.js, which is the same body the gate enforces from, and the gate's last check fails the build if this page and it disagree about any generator.

Where it is called

Changing this generator changes every figure on this list. That is what makes the list worth publishing rather than keeping in a check script.

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