da084a54-8a21-446c-85e8-dfd6401c3bdb 37 4 5 decimal

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1. Inverse/Iterate a scalar cumulative integration

algo style "iterative inverser" on range [L, H] ...N discrete points f L H N Inverse dx H L - N / : i 1 N range ω i el L dx - (dx i * + : for j 1 ω rows range sol j el x f ω j el - 0 ≡ x 05 solve eval : for ω sol augment 41 line :

Setup the companion system

Note:

Set integration accuracy 50 ... 5 min

Inversing a function is solving for 'x' given 'y' ...

a most common task in Engineering/Scientific works.

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5953 / 11 x 2 / - (2^- (* 0 ≡ x 35 solve 4

Hi 4 :

linear cumulative
scalar on canvas,
... point evaluate z ∫ u F u 0 x int : x ∫ z ∫ : 21 sys

x F 5953 / 11 x 2 / - (2^- (* 0 x ≤ (x Hi ≤ (& cases :

1 ∫ 2 ∫ 4 ∫ 31 mat 82.639 264.444 528.889 31 mat

x Σcumul x ∫ 0 x ≤ (x Hi ≤ (& cases :

X 256 : Y x Σcumul X - 0 ≡ x 05 solve : 21 line

X Y 21 mat 256 1.957 21 mat

recast the Σcumul(x) => for f(x) in the solver x f x Σcumul :

tabulate the iterative solver Inv x Σcumul 0 528.88888 50 Inverse :

solve for intersection Φ x F x Σcumul - 0 ≡ x 1.5 2 solve : 1.6277

1. X => the seeked value of cumul Σ 2. Y => 'x' at which Σcumul occurs

Inv X Y + 250 red 1 5 mat 2 1 sys

1. blue => the function 2. red => its cumulative integration

x F x Σcumul Φ Φ F + 30 black 1 5 mat 3 1 sys

Inv 00 10.578 0.336 21.156 0.482 31.733 0.596 42.311 0.695 52.889 0.783 63.467 0.865 74.044 0.941 84.622 1.013 95.2 1.082 102 mat

2. Maple classical/direct analytical inverser

x a b q 1 x b / - exp (a^- :

y x a b q ≡ x solve maple 1 y - ln b * a / -

y a b sol 1 y - ln b * a / - :

a 2 : b 1 : 12 mat

x middle b x * :

x a b q x a b sol x middle 3 1 sys

An inverse function:

1. may exist simply, this example: q(x,a,b)

2. may exist via "horrible maths" !!!.

3. may not exist at all ... inverse ∫(x)

4. inverse ∫(x) is evaluated via the iterative "solve bloc".

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x a b q x a b sol - 0 ≡ x 0.5 1 solve 0.797

γ 1.4 : (Air isentropic ≡ k 0.88128485 : (Pitot constant ≡ p static pressure ≡ qc dynamic pressure ≡ x Pitot dynamic ratio qc/P ≡ x qc p / ≡ (supersonic Pitot ratio ≡ 61 line

3. Mach => scalar iterate on canvas

x Subsonic 2 γ 1 - / (x 1 + (γ 1 - γ /^1 - (* sqrt 0 x ≤ (x k ≤ (& Supersonic cases :

x Supersonic U x 1 + (117 U 2^* / - (2.5^* sqrt / k - 0 ≡ U 1 roots k x ≤ Subsonic cases :

x Subsonic x Supersonic k 1 + 20 black 1 5 mat 3 1 sys

Pitot 3 :

Mach 1 :

supersonic Mach vs Pitot ratio Pitot Supersonic Mach 1.65

Observe the canvas solutions:

as 'x' in Supersonic(x) runs on the plot canvas,

the 'roots' solves and plots each canvas pixel result ...

each result on the basis of 96 ppi [pixel per inch].

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4. Golden-Ratio

1. All modules/functions in "Symbolic" optimiz. 2. Once in target: shrink the range 3. respect the search y1≥y2≡"max in range" y1≤y2≡"min in range"

x f x 4^2 x 3^* - 7 x 2^* - 2 - : a b G 32 / b * 12 / a * - 12 / 5 sqrt * a b - (* + : 21 line

Code adapted from Valery Ochov Mathcad 11 f a b test TOL Extrema y1 y2 ≥ max in range ≡ y1 y2 ≤ min in range ≡ x1 a b G eval : y1 x1 f eval : x2 b a G eval : y2 x2 f eval : 41 line iter 0 : iter 1000 < a b - abs TOL ≤ break test b x2 : x2 x1 : y2 y1 : x1 a b G eval : y1 x1 f eval : 51 line a x1 : x1 x2 : y1 y2 : x2 b a G eval : y2 x2 f eval : 51 line if iter iter 1 + : 21 line if while a b + 2 / iter 21 mat 61 line :

X f 1.5 - 1 - y1 y2 ≤ 1012 -^Extrema 1 el f 0.5 - 0.5 y1 y2 ≥ 1012 -^Extrema 1 el f 2.5 3 y1 y2 ≤ 1012 -^Extrema 1 el 31 mat :

x f X 1 el X 1 el f + 15 red X 2 el X 2 el f + 15 red X 3 el X 3 el f + 15 red 3 5 mat 2 1 sys

X 1.265564 - 2.816018 108 -^* - 2.765564 31 mat

x f x 4^2 x 3^* - 7 x 2^* - 2 - : x g x : 21 line

solve root(s) X x f x 3 - 5 solve : 1.9225 - 3.8522 21 mat

solve intersection(s) C x f x g - 0 ≡ x 3 - 5 solve : 1.837 - 3.8963 21 mat

x f x g X 1 el X 1 el f . 15 red 1 5 mat X 2 el X 2 el f . 15 red 1 5 mat C 1 el C 1 el f o 10 green 1 5 mat C 2 el C 2 el f o 10 green 1 5 mat 6 1 sys

1. roots
2. intersections X C 21 mat 1.9225 - 3.8522 21 mat 1.837 - 3.8963 21 mat 21 mat

5. Miscellaneous

Berechnen Sie den Durchmesser d und die Höhe h eines Zylinders mit gegebenem Volumen V und gegebener Oberfläche A

Compute the diameter 'd' and the height 'h' of a cylinder of given volume 'V' and given surface area 'A' Source: Re-published from Martin Handbook.

Define the cylinder A 10 m 2^* : Surface area ≡ V 1 m 3^* : Volume ≡ 21 line

Transport informations data surface area A volume V 21 mat :

System of equations f A 2 π 4 / d 2^* (* π d * h * + ≡ V π 4 / d 2^* h * ≡ 21 mat :

(embedded assignment) f d 1 m * ≡ h 1 m * ≡ 21 mat FindRoot 0.4109 m * 7.5411 m * 21 mat

Startwerte mit passenden Maßeinheiten vorgeben und Gleichungssystem auflösen.

Provide initial guesses with appropriate units and solve for the unknowns.

Sind die Anfangswerte in Gleichungsform gegeben, dann werden die Variablen mit den Lösungswerten definiert.

If the initial guesses are given as equalities name=value, then the solution is assigned to these names (embedded assignment).

solved variables d h 21 mat 0.4109 m * 7.5411 m * 21 mat

Sanity check of the "Given/Find Cylinder" data d h 21 mat 2 π 4 / d 2^* (* π d * h * + π 4 / d 2^* h * 21 mat 13 mat surface area A volume V 21 mat 0.4109 m * 7.5411 m * 21 mat 10 m 2^* 1 m 3^* 21 mat 13 mat

Circle/line intersection

C 6 : Circle ≡

L 0 : Line ≡

Φ vector equalities Φ C x 2^y 2^+ ≡ L x y + ≡ 21 mat :

init x 1 ≡ y 1 ≡ 21 mat :

Collect solutions at kernel level x y sol Φ init FindRoot : 1.732052 - 1.732052 21 mat

Assign the solutions sol x y sol vectorize : 1.732052 - 1.732052 21 mat

C x 2 ^ - sqrt C x 2 ^ - sqrt - L x - sol 1 el sol 2 el . 14 black sol 2 el sol 1 el . 14 black 2 5 mat 4 1 sys

Solving numeric a user system ======================

b 1 : c π : 21 line

f a0 0 ≡ a1 c ≡ a1 a2 + a3 + 0 ≡ a1 2 a2 * + 3 a3 * + b ≡ 41 mat :

init a0 1 ≡ a1 1 ≡ a2 1 ≡ a3 1 ≡ 41 mat :

a0 a1 a2 a3 sol f init FindRoot :

solutions in lexicographic order of "init" solutions a0 a1 a2 a3 sol vectorize : 0 3.1416 8.2832 - 5.1416 41 mat

Solving quaternion ======================================

α 0.5 :

β 1.5 :

The Quaternion system f p q + r - 1012 -^≡ p 2^q 2^+ r 2^+ 1 ≡ β p * α r * + 1012 -^≡ 31 mat :

init p 1012 -^≡ q 1012 -^≡ r 1012 -^≡ 31 mat :

Sanity check vs Mathcad 11 f init FindRoot 0.1961 - 0.7845 0.5883 31 mat

p q r 31 mat 0.1961 - 0.7845 0.5883 31 mat

Quaternion p α - 2 α 2^* 2 α * β * + 2 β 2^* + sqrt / α 2 α 2^* 2 α * β * + 2 β 2^* + sqrt / q 12 α 2^* 2 α * β * + 2 β 2^* + sqrt / α β + (* 1 - 2 α 2^* 2 α * β * + 2 β 2^* + sqrt / α β + (* r 12 α 2^* 2 α * β * + 2 β 2^* + sqrt / β * 1 - 2 α 2^* 2 α * β * + 2 β 2^* + sqrt / β * 33 mat :

Next: evaluate numeric

modulo 2 β 2^* 2 β * α * + 2 α 2^* + sqrt :

α 0.5 :

β 1.5 :

Quaternion 2 col 0.1961 - 0.7845 0.5883 31 mat

V α - modulo / α modulo / α β + (modulo / α β + (- modulo / α modulo / α - modulo / 32 mat :

V 0.1961 - 0.1961 0.7845 0.7845 - 0.1961 0.1961 - 32 mat

Next: stack a 'p' value

p 0.6 :

q Quaternion 2 col p stack :

q 0.1961 - 0.7845 0.5883 0.6 41 mat

Nex: the unit vector

q 0.1961 - 0.7845 0.5883 0.6 41 mat :

q q q norme / : 0.1682 - 0.6727 0.5045 0.5145 41 mat

solutions are in lexicographic order

Solving a chemical balance ===============================

k1 0.5 : k2 0.25 : k3 0.1 : 31 line

init γ1 1012 -^≡ γ2 1012 -^≡ 21 mat :

f γ1 γ2 + k1 k2 + k3 + ≡ γ1 γ2 * k1 k2 * k1 k3 * + ≡ 21 mat :

f init FindRoot 0.5006 0.3494 21 mat

γ1 γ2 21 mat 0.501 0.349 21 mat

Clear "initial" to renew the solver

γ1 Clear 1

γ2 Clear 1

k1 10.052 : k2 14.836 : k3 4.982 : 13 mat

init γ1 1012 -^≡ γ2 1012 -^≡ 21 mat :

f γ1 γ2 + k1 k2 + k3 + ≡ γ1 γ2 * k1 k2 * k1 k3 * + ≡ 21 mat :

f init FindRoot 19.818 10.052 21 mat

γ1 γ2 21 mat 19.818 10.052 21 mat