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3 arguments transit integrand for modelling

Scalar Integral model curve:

This document is Edu. Introduction to integrate.

For a fast visit, integral accuracy 100 .

3 equivalent scalar representations

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u F 11 x 2^+ / :

Transit the system, plot/interpolate z Int u F u x - x int :

Indefinite integral z z int u F u x - x int :

3 equivalent forms

∫(z) => function symbol z ∫ u F u x - x int :

x ∫ z ∫ :

X 4 :

X ∫ 0.470588

x F x Int 2 1 sys x F x x int 2 1 sys x F x ∫ X X ∫ + 15 black 1 5 mat 3 1 sys

2 resulting models z z int z Int u x - x int : x Int2 z z int : 21 sys

x X 0.75 :

x ∫ x X - x solve 0.4514 2.2153 21 mat

X Int2 3.7647

x Int2 x ∫ x F 3 1 sys

11 x 2^+ / x int maple z z int maple 21 mat x atan z 2^2 / 21 mat

z ∫ u F u x - x int :

u F 11 x 2^+ / :

Create an integral model:

1. Copy/paste in blank sheet. 2. Plug your experimental model in F(u). 3. Hope for some result. 4..... Experiment 1/[1+x]

2 resulting models z z int z ∫ u 1 - x int : x Int2 z z int : 21 sys

x Int2 z ∫ x F 31 sys

x Int2 z ∫ x F 3 1 sys

transit/recast x ∫ z ∫ :

1. ½ amplitude
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2. govern assymmetry # # 0 x int # # # ± 0 ≠ x int 21 sys

5 - ∫ 5 ∫ 21 mat 5 - Int2 5 Int2 21 mat 21 mat 0.3846 - 0.3846 21 mat 1.5385 2.3077 21 mat 21 mat

z ∫ u F u x - x int :

u F 11 x 3^+ / :

Interesting reaction rate model .. copy/paste in blank sheet ..

Explain . . .

Modelling data set is not easy, even assisted from

a generous library. Sufficiently clean data sets can

be smoothed. Noisy ones should be fitted .

Assume a data set could be accepted red curve.

At this point, you don't have an explicit model

but you may populate a clean experimental data,

that can be further tabulated/interpolated.

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2 resulting models z z int z ∫ u 0 x int : x Int2 z z int : 21 sys

x c 10 x ≤ (x 10 ≤ (& cases :

x Int2 x c * z ∫ x c * x F x c * 3 1 sys

transit/recast x ∫ z ∫ :

1. ½ amplitude
--------------
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2. govern assymmetry # # 0 x int # # # ± 0 ≠ x int 21 sys

5 - ∫ 5 ∫ 21 mat 5 - Int2 5 Int2 21 mat 21 mat 0.3846 - 0.3846 21 mat 1.9231 1.9231 21 mat 21 mat

data 00 0.1 0.0011 0.2 0.0043 0.3 0.0097 0.4 0.0173 0.5 0.0269 0.6 0.0386 0.7 0.0523 0.8 0.0679 0.9 0.0854 1 0.1047 112 mat :

Exercise with Voltera IntegralEquation

data char size clr plot k 1 data rows range r3 k el char : r4 k el size : r5 k el clr : 31 line for data r3 r4 r5 augment 21 line :

u Φ 1 u mwbesselj u 1015 -^+ / :

x c 10 x ≤ (x 1 ≤ (& cases :

t0 1 time :

x abs Φ 1 1 sys

pilot with λ [easier] λ 0.9 :

μ 1 λ 2^- sqrt : 0.4359

Voltera 2nd kind x f μ x t - (* sin t Φ * t 0 x int :

V data . 12 black plot :

V x f x c * 2 1 sys

10 f 0.692 -

1 time t0 - min 1.1

About this example . . .

It may be tempting to argue about other fitting methods,

polynomials, ExpGrow ... It may be valid/invalid, just

a matter of trying. Polynomial must be rejected as it does not

highlight physical parameters. Even if ExpGrow looks adequate

on small range, it may fail on longer range simply because it

does not represent the entire physical underlying phenomenon.

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https://nvlpubs.nist.gov/nistpubs/jres/73B/jresv73Bn3p191_A1b.pdf

Sub-integrate from spline

Voltera x 01 0.05 range : i 1 x rows range F i el x i el f eval : for x F augment 31 line :

x q x 2^:

Voltera Voltera :

X Voltera 1 col : Y Voltera 2 col : x S X Y x cinterp : 31 sys

integrates nicely x R 0 x t - S t q * t 0 x int + :

x R x c *