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


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