Saturday, December 23, 2023

Swiss Micros DM32: Solving Integral Equations

Swiss Micros DM32:  Solving Integral Equations



Introduction


The programs presented today solves the following equation for X:


X

∫   F(T)  dT = C

0


Since we are not able to use an integration command in solving a program on the DM32 (and the HP 32S and 32SII), we will have to use a manual method, mainly Newton's Method:


X_n+1 = X_n - [ ∫( F(T) dT from T = 0 to T = x_n) - C ] ÷ f(x_n) 


with in tolerance D.


To see a similar program for the HP Prime, posted on April 3, 2020, please click here:

https://edspi31415.blogspot.com/2020/04/hp-prime-solving-integral-equations.html



DM32 Code:  Integral Equations


LBL H:  Help File


 LBL H

  SF 10

  EQN: L B L _ F - F ( T )

  EQN: L B L _ S - S O L V E

  EQN: S _ H A S _ R C L _ T

  EQN: C = C O N S T

  EQN: D = T O L E R

  EQN: G U E S S _ X E Q _ S

  CF 10

  RTN


Note:  Underscore is the space key.   Press R/S after each message.   


LBL F

  RCL T

  enter f(T), the integrand, here

  RTN


Note:  The variable used is T.   If you want to test out the function, store a value in the variable T first.  


 LBL S

  RAD

  STO X

  LBL A

  FN= F

  0

  RCL X

  ∫ FN d T

  RCL- C

  STO Y

  RCL X

  STO T

  XEQ F

  STO÷ Y

  RCL X

  RCL- Y

  STO Y

  RCL Y

  RCL- X

  ABS

  RCL D

  x<y?

  GTO B

  RCL Y

  RTN

  LBL B

  RCL Y

  STO X

  GTO A


Note:  This is the main program.  Enter a guess, and then key in XEQ S.  



Variables Used:


T = independent variable

C = constant

D  = tolerance (i.e. 10^-4, 10^-5, 10^-6, etc)

X = x_n

Y = x_n+1, final approximation



Download the state file here:  ntegralequ.dm32



The state file includes a sample integrand:


e^(-T ÷ 4):


LBL F

RCL T

x^2

+/-

4

÷

e^x

RTN



Examples



In the following examples, the tolerance is 10^-5  (5 +/- 10^x STO D) and FIX 5 mode is set.  


1.   ∫( e^(-T^2 ÷ 4) dT for T = 0 to X) = 1


C = 1


Guess = 2,   Result:  1.10208

Guess = 1,   Result:  1.10208

Guess = 3,   Result:  Division by 0 error


Note that initial guesses are important.  



2.  ∫( e^-sin(T + 1) dT for T = 0 to X) = 10


C = 10


LBL F

RCL T

1

+

SIN

+/-

e^x

RTN


Guess = 5;  Result:  9.10014



3.  ∫( T^3 - 2 × T dT for T = 0 to X) = 40


C = 40


LBL F

RCL T

3

y^x

RCL T

×

-

RTN


Guess = 5; Result:  3.84789



4.  ∫( sin^2 T dT for T = 0 to X) = 1.4897


C = 1.4897


LBL F

RCL T

SIN

x^2

RTN


Guess = 2;  Result:  2.49991




Eddie


All original content copyright, © 2011-2023.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 


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