Peaksigmoid

Purpose

Outputs a sigmoid function.

Synopsis

[y,y1,y2] = peaksigmoid(x,ax)

Inputs

• x = 3 element vector where

• x(1) = coefficient
• x(2) = offset
• x(3) = decay constant

Outputs

${\sqrt {1-e^{2}}}$

$f\left({{a}_{i}},\mathbf {x} \right)={{x}_{1}}\left[{{x}_{4}}{{\operatorname {e} }^{\frac {-4\ln \left(2\right){{\left({{a}_{i}}-{{x}_{2}}\right)}^{2}}}{x_{3}^{2}}}}+\left(1-{{x}_{4}}\right)\left[{\frac {x_{3}^{2}}{{{\left({{a}_{i}}-{{x}_{2}}\right)}^{2}}+x_{3}^{2}}}\right]\right]$

$f\left({{a}_{i}},\mathbf {x} \right)={{x}_{1}}\left[{{x}_{4}}{{\operatorname {e} }^{\frac {-4\ln \left(2\right){{\left({{a}_{i}}-{{x}_{2}}\right)}^{2}}}{x_{3}^{2}}}}+\left(1-{{x}_{4}}\right)\left[{\frac {1-x_{3}^{2}}{{{\left({1}+{{x}_{2}}\right)}^{2}}+x_{3}^{2}}}\right]\right]$

$f\left({{a}_{i}},\mathbf {x} \right)={{x}_{1}}\left[{{x}_{4}}{{\operatorname {e} }^{\frac {-4\ln \left(2\right){{\left({{a}_{i}}-{{x}_{2}}\right)}^{2}}}{x_{3}^{2}}}}+\left(1-{{x}_{4}}\right)\left[{\frac {x_{3}^{2}}{{{\left({{a}_{i}}-{{x}_{2}}\right)}^{2}}+1}}\right]\right]$

Failed to parse (syntax error): {\displaystyle f\left( {{a}_{i}},\mathbf{x} \right)={{x}_{1}}\left[ {{x}_{4}}{{\operatorname{e}}^{\frac{-4\ln \left( 2 \right){{\left( {{a}_{i}}-{{x}_{2}} \right)}^{2}}}{x_{3}^{2}}}}+\left( 1-{{x}_{4}} \right)\left[\frac{{{\operatorname{e}}^{-z}}}}{{{\left( {{a}_{i}}-{{x}_{2}} \right)}^{2}}+{{\operatorname{e}}^{-z}}} \right] \right]}

$f\left({{a}_{i}},\mathbf {x} \right)={{x}_{1}}\left[{{x}_{4}}{{\operatorname {e} }^{\frac {-4\ln \left(2\right){{\left({{a}_{i}}-{{x}_{2}}\right)}^{2}}}{x_{3}^{2}}}}+\left(1-{{x}_{4}}\right)\left[{\frac {1}{{{\left({{a}_{i}}-{{x}_{2}}\right)}^{2}}+{{\operatorname {e} }^{-z}}}}\right]\right]$

$f\left({{a}_{i}},\mathbf {x} \right)={{x}_{1}}\left[{{x}_{4}}{{\operatorname {e} }^{\frac {-4\ln \left(2\right){{\left({{a}_{i}}-{{x}_{2}}\right)}^{2}}}{x_{3}^{2}}}}+\left(1-{{x}_{4}}\right)\left[{\frac {1}{{{\left({{a}_{i}}-{{x}_{2}}\right)}^{2}}+{{\operatorname {e} }^{-z}}}}\right]\right]$

• y(1) =

• y(2) = ${\operatorname {d} \!y \over \operatorname {d} \!{x}_{i}}$

• y(3) = ${\operatorname {d^{2}} \!y \over \operatorname {d} \!{{x}_{i}}^{2}}$