Heaviside step function: Difference between revisions

Revision as of 08:43, 6 August 2025

Latest revision as of 08:49, 6 August 2025

(2 intermediate revisions by the same user not shown)

Line 37:

Approximations to the Heaviside step function are of use in [[biochemistry]] and [[neuroscience]], where [[logistic function|logistic]] approximations of step functions (such as the [[Hill equation (biochemistry)|Hill]] and the [[Michaelis–Menten kinetics|Michaelis–Menten equations]]) may be used to approximate binary cellular switches in response to chemical signals.

For a [[File:StepSmooth function|smooth]] approximation.png|alt=A setto ofthe functionsstep thatfunction, successivelyone approachcan use the step[[logistic function|thumb|500x500px|]]:<math display="block">H(x) \approx \tfrac{1}{2} + \tfrac{1}{2} \tanh( kx) = \frac{1}{1+e^{-2kx}},</math><br>approacheswhere thea steplarger function{{mvar|k}} corresponds to a sharper transition asat {{math|''kx'' →{{=}} ∞0}}.]]

For a [[Smooth function|smooth]] approximation to the step function, one can use the [[logistic function]]:<math display="block">H(x) \approx \tfrac{1}{2} + \tfrac{1}{2}\tanh kx = \frac{1}{1+e^{-2kx}},</math>where a larger {{mvar|k}} corresponds to a sharper transition at {{math|''x'' {{=}} 0}}. If we take {{math|''H''(0) {{=}} {{sfrac|1|2}}}}, equality holds in the limit:<math display="block">H(x)=\lim_{k \to \infty}\tfrac{1}{2}(1+\tanh kx)=\lim_{k \to \infty}\frac{1}{1+e^{-2kx}}.</math>There are [[Sigmoid function#Examples|many other smooth, analytic approximations]] to the step function.<ref>{{MathWorld | urlname=HeavisideStepFunction | title=Heaviside Step Function}}</ref> Among the possibilities are:

If we take {{math|''H''(0) {{=}} {{sfrac|1|2}}}}, equality holds in the limit:<math display="block">H(x)=\lim_{k \to \infty}\tfrac{1}{2}(1+\tanh kx)=\lim_{k \to \infty}\frac{1}{1+e^{-2kx}}.</math>

<math display="block">\begin{align}

For a [[SmoothFile:Step function|smooth]] approximation.png|alt=A toset theof stepfunctions function,that onesuccessively can useapproach the [[logisticstep function]]:|thumb|500x500px|<math display="block">H(x) \approx \tfrac{1}{2} + \tfrac{1}{2} \tanh (kx) = \frac{1}{1+e^{-2kx}},</math>where<br>approaches athe largerstep {{mvar|k}}function corresponds to a sharper transition atas {{math|''xk'' {{=}}→ 0∞}}. If we take {{math|''H''(0) {{=}} {{sfrac|1|2}}}}, equality holds in the limit:<math display="block">H(x)=\lim_{k \to \infty}\tfrac{1}{2}(1+\tanh kx)=\lim_{k \to \infty}\frac{1}{1+e^{-2kx}}.</math>none]]There are [[Sigmoid function#Examples|many other smooth, analytic approximations]] to the step function.<ref>{{MathWorld | urlname=HeavisideStepFunction | title=Heaviside Step Function}}</ref> Among the possibilities are:<math display="block">\begin{align}

H(x) &= \lim_{k \to \infty} \left(\tfrac{1}{2} + \tfrac{1}{\pi}\arctan kx\right)\\

H(x) &= \lim_{k \to \infty}\left(\tfrac{1}{2} + \tfrac12\operatorname{erf} kx\right)