RPN with HP 15C & DM32: Solving Simple Systems

RPN with HP 15C & DM32: Solving Simple Systems

Welcome to another edition of RPN with HP 15C & DM32.

Many Approaches to a Solving Problems

This blog covers two ways to solve the simple linear system of two equations:

x + y = a

x – y = b

In matrix form:

[ [ 1, 1 ] [ 1, - 1] ] * [ [ x ] [ y ] ] = [ [ a ] [ b ]

The inverse of the coefficient matrix [ [ 1, 1 ] [ 1, - 1] ]:

[ [ 1, 1 ] [ 1, - 1] ] ^-1 = [ [ 0.5, 0.5 ] [ -0.5, 0.5 ] ]

[ [ x ] [ y ] ] = [ [ 0.5, 0.5 ] [ -0.5, 0.5 ] ] * [ [ a ] [ b ] ]

The solution to the system is:

x = (a + b) / 2

y = (a – b) / 2

As far as keying the solutions in the calculator, we can address this two ways. The first, and probably the easier method, is to use memory registers. The second is to use stack operations such as swap (x<>y), roll up (R↑), roll down (R↓), enter (ENTER) as duplication, and the Last X function (LST x). Both the HP 15C and DM32 use a four-level stack.

The Memory Register Approach

Start with the stack as follows:

Y: a

X: b

The results are presented as:

Y: y

X: x

Memory Registers used:

HP 15C: R1 = a, R2 = b

DM32: A, B

HP 15C Code:

001	42, 21, 11	LBL A
002	44, 2	STO 2
003	34	X<>Y
004	44, 1	STO 1
005	30	-
006	16	CHS
007	2	2
008	10	÷
009	45, 2	RCL 2
010	45, 40, 1	RCL + 1
011	2	2
012	10	÷
013	43, 32	RTN

DM32 Code (31 bytes):

T01 LBL T

T02 STO B

T03 x<>y

T04 STO A

T05 -

T06 +/-

T07 2

T08 ÷

T09 RCL B

T10 RCL + A

T11 2

T12 ÷

T13 RTN

The Stack Approach

Start with the stack as follows:

Y: a

X: b

The results are presented as:

T : original contents of the z stack

Z: original contents of the z stack

Y: y

X: x

This time no memory registers are used

HP 15C Code:

001	42, 21, 12	LBL B
002	40	+
003	43, 36	LST x
004	2	2
005	20	×
006	34	X<>Y
007	36	ENTER
008	33	R↓
009	30	-
010	16	CHS
011	2	2
012	10	÷
013	43, 33	R↑
014	2	2
015	10	÷
016	43, 32	RTN

DM32 Code (24 bytes):

S01 LBL S

S02 +

S03 LAST x

S04 2

S05 ×

S06 x<>y

S07 ENTER

S08 R↓

S09 -

S10 +/-

S11 2

S12 ÷

S13 R↑

S14 2

S15 ÷

S16 RTN

Examples

Example 1:

x + y = 32

x – y = 16

32 ENTER 16 (run program)

Results: y = 8, x = 24

Example 2:

x + y = 9

x – y = -8

Results: y = 8.5, x = 0.5

Until next time,

Eddie

All original content copyright, © 2011-2025. 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.

Casio fx-CG 50: Pseudorandom Number Generator (PRNG) Stat Plot

Casio fx-CG 50: Pseudorandom Number Generator (PRNG) Stat Plot

Introduction

This program is an inspiration from a HHC 2024 talk given by Kuba Tatarkiewicz. Tatarkiewicz’s talk is about testing RNG (random number generators). You can see it here: https://www.youtube.com/watch?v=vSDfqCK-ENk

Premise of RANDGRPH: Generate a recursive sequence

r_n = frac( (A * r_n-1 + B) ^ C)

The program builds two lists and develops a scatter plot. I have the program set up to plot 60 points, but we can up to 999 points. The list starts with the initial point (0, seed). The program asks you whether to have the calculator provide the seed or you provide a seed (between 0 and 1).

Casio fx-CG 50 Code: RANDGRPH (252 bytes)

“RAN # GRAPH”

“(A×R+B)^C” ◢

“A”? → A

“B”? → B

“C”? → C

Menu “SEED?”, “RANDOM”, 1, “YOUR OWN”, 2

Lbl 1

Ran# → R

Goto 0

Lbl 2

“0≤R<1, SEED”? → R

Lbl 0

{0} → List 1

{R} → List 2

For 1 → I To 75

Augment(List 1, {R}) → List 1

Frac((A×R+B)^C) → R

Augment(List 2,{R}) → List 2

Next

S-Gph1 DrawOn, Scatter, List 1, List 2, 1, Dot, ColorLinkOff, Black, AxesOn

DrawStat

Examples

Example 1: r_n = frac(991 * r_n-1)

Example 2: r_n = frac( (0.3 * r_n-1 + 1)^2 )

Example 3: r_n = frac( (r_n-1 + π)^5 )

Enjoy! Happy New Year, be safe, sane, strong, and take care. Forever grateful,

Eddie

All original content copyright, © 2011-2025. 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.

TI-30Xa Algorithms: RLC Series Circuit

TI-30Xa Algorithms: RLC Series Circuit

The task is to calculate the total impedance and phase angle for an RLC circuit in a series. An RLC circuit contains a resistor ( R ), an inductor ( L ), and a capacitor ( C ) , all powered by a voltage supply. The resistor, inductor, and capacitor are all in a single path.

The three elements are measured in the following units:

Resistor ( R ): Ohms ( Ω )

Inductor ( L ): Henry ( H )

Capacitor ( C ): Farads ( F )

The impedance for the RLC circuit is calculated by:

Z = √( R^2 + ( (2 * π * f * L) - 1 / (2 * π * f * C) )^2 )

The phase angle is calculated by:

θ = arctan( ( (2 * π * f * L) - 1 / (2 * π * f * C) ) / R )

where: f = frequency in Hertz (Hz)

We can use the rectangular to polar conversion function (R>P) to calculate both the impedance and phase angle.

x = R

y = (2 * π * f * L) - 1 / (2 * π * f * C)

If we have an RL circuit, with no capacitor, then C = 0.

If we have an RC circuit, with no inductor, then L = 0.

If we have an LC circuit, with no resistor, then R = 0.

TI-30Xa Algorithm – RLC Circuit

Step 1: Store resistance in Memory 1.

R [ STO ] 1

Step 2: Calculate angular frequency, ω = 2 * π * f, store in memory 3.

2 [ × ] [ π ] [ × ] f [ = ] [ STO ] 3

Step 3: Calculate (2 * π * f * L) - 1 / (2 * π * f * C)

[ RCL ] 3 [ × ] L [ - ] [ ( ] [ RCL ] 3 [ × ] C [ ) ] [ 1/x ] [ = ] [ STO ] 2

If we have an RL circuit, Steps 2 and 3 can be shortened to:

2 [ × ] [ π ] [ × ] f [ × ] L [ = ] [ STO ] 2

If we have an RC circuit, Steps 2 and 3 can be shortened to:

[ ( ] 2 [ × ] [ π ] [ × ] f [ × ] C [ ) ] [ 1/x ] [ +/- ] [ = ] [ STO ] 2

Step 4: Calculate Impedance and Phase Angle:

[ RCL ] 1 [ 2^nd ] [ π ] (x<>y) [ RCL ] 2 [ 2^nd ] [ - ] (R>P)

What is shown: (r) Impedance

Press [ 2^nd ] [ π ] (x<>y) for phase angle (θ)

Examples

Example 1: RLC Series Circuit

R = 50 Ω

L = 3.8 H

C = 0.7 F

f = 40 Hz

Step 1: Store resistance in Memory 1.

50 [ STO ] 1

Step 2: Calculate angular frequency, ω = 2 * π * f, store in memory 3.

2 [ × ] [ π ] [ × ] 40 [ = ] [ STO ] 3 (251.3274123)

Step 3: Calculate (2 * π * f * L) - 1 / (2 * π * f * C)

[ RCL ] 3 [ × ] 3.8 [ - ] [ ( ] [ RCL ] 3 [ × ] 0.0007 [ ) ] [ 1/x ] [ = ] [ STO ] 2

(949.36000616)

Step 4: Calculate Impedance and Phase Angle:

[ RCL ] 1 [ 2^nd ] [ π ] (x<>y) [ RCL ] 2 [ 2^nd ] [ - ] (R>P)

Impedance: 950.6758262

Phase Angle: 86.9851854°

Example 2: RLC Series Circuit

R = 450 Ω

L = 1.15 H

C = 3.5 μF = 3.5 * 10^-6 F

f = 60 Hz

Step 1: Store resistance in Memory 1.

450 [ STO ] 1

Step 2: Calculate angular frequency, ω = 2 * π * f, store in memory 3.

2 [ × ] [ π ] [ × ] 60 [ = ] [ STO ] 3 (376.9911184)

Step 3: Calculate (2 * π * f * L) - 1 / (2 * π * f * C)

[ RCL ] 3 [ × ] 1.15 [ - ] [ ( ] [ RCL ] 3 [ × ] 3.5 [ EE ] 6 [ +/- ] [ ) ] [ 1/x ] [ = ] [ STO ] 2

(-324.3408952)

Step 4: Calculate Impedance and Phase Angle:

[ RCL ] 1 [ 2^nd ] [ π ] (x<>y) [ RCL ] 2 [ 2^nd ] [ - ] (R>P)

Impedance: 554.7044405

Phase Angle: -35.78246242°

Source

“Impedance and Complex Impedance” Electronics Tutorials. AspenCore, Inc. 2024. Retrieved October 7, 2024. https://www.electronics-tutorials.ws/accircuits/impedance.html

This wraps up the TI-30Xa Algorithm Series. Next up will be a series on the Casio fx-991CW.

I want to wish you all a Happy New Year and a prosperous, sane, and happy 2025! Be safe, everyone, it’s a very crazy world we live in.

Eddie

All original content copyright, © 2011-2024. 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.

TI 84 Plus CE: Consolidated Debts

TI 84 Plus CE: Consolidated Debts  

Disclaimer: This blog is for informational and academic purposes only. Financial decisions are your own.

Introduction

The program CONSOL consolidates a number of debts into a single loan, typically of a lower interest rate than the rate paid of the original debts. The loan can include additional cash and fees charged.

TI-84 Plus CE Program: CONSOL

Notes:

The percent sign is comprised of three symbols: a degree sign ( [ 2^nd ], [ apps ] (angle), 1, °), the division slash, and a decimal point. The percent symbol will look something like this: ° / .

All fees are assumed to be financed. At the fees question, if you enter an amount less than 100, it will be treated as a percent of the desired cash out. If you enter an amount 100 or above, it will be treated as an amount. For no fees, enter 0.

Monthly payments at the end of the month are assumed.

The cash sign convention is used. If the interest difference is positive, you are saving interest costs. If the interest difference is negative, you have an additional interest cost.

The list of original debts are stored to list L₄.

The list of associated rates are stored to list L₅.

The list of associated number of payments renaming to list L₆.

The payment function, tvm_Pmt is from the Finance app ( [ apps ], 1. Finance…, 2: tvm_Pmt). Syntax: tvm_Pmt( number of payments, annual interest rate, present value, future value, payments per year, compound periods per year).

Code (520 bytes):

ClrHome

Disp “CONSOLIDATED ANALYSIS”

Input “LIST OF DEBTS? “, L₄

Input “LIST OF RATES (°/.)? “, L₅

Input “LIST OF PMTS REMAINING? “, L₆

If dim(L₄)≠dim(L₅) or dim(L₅)≠dim(L₆) or dim(L₄)≠dim(L₆)

Then

Disp “LISTS NOT EQUAL SIZE”

Stop

End

0 → J

For(I, 1, dim(L₄))

L₆(I) * tvm_Pmt(L₆(I), L₅(I), L₄(I), 0, 12, 12) + L₄(I) + J → J

End

sum(L₄) → D

Disp “CONSOLIDATED DEBT:”, D

Input “DESIRED AMOUNT? “, C

Disp “FEES AND COSTS”, “<100 (°/.), ≥100 (AMT)”

Prompt F

If F<100

Then

C*(1+F/100) → F

Else

C+F → F

End

Input “NEW RATE (°/.)? “, R

Input “NO. OF PMTS? “, N

tvm_Pmt(N,R,F,0,12,12) → p

N*P+F → K

Disp “CASH OUT: “, C-D

Disp “NEW PMT: “, P

Disp “INTEREST DIFF: “,”(>0 SAVINGS)”, K-J

Examples

Example 1: A Simple Consolidation

Consolidate two debts, take no cash out, no fees.

Original Debt	Interest Rate	Number of Payments Left
48000	8%	240
17500	19.99%	24

Desired Amount: 65500. No fees. New rate: 6%. Term: 240 months

Result:

Cash Out: 0

New Pmt: -469.2623433 (469.26)

Interest Diff: 5109.118945 (save 5109.12)

Example 2: Consolidate Three Debts: 3% fees or a flat 600 fee.

Original Debt	Interest Rate	Number of Payments Left
36500	12.99%	180
6800	8%	80
1800	20.99%	12

Desired Amount: 52000. New Rate: 7.25%. Term: 240 months

Result with 3% fees:

Cash Out: 6900

New Pmt: -423.3253775 (-423.33)

Interest Diff: 752.12157 (752.12)

Result with 600 flat fee:

Cash Out: 6900

New Pmt: -415.7377681 (415.74)

Interest Diff: 1613.147839 (1613.15)

Example 3: Not every scenario will have interest savings.

Original Debt	Interest Rate	Number of Payments Left
40000	13%	90
4000	20.99%	12
1800	9.95%	12

Desired Amount: 50000. New Rate: 7%. Term: 180 months. Fees: 3%

Result with 3% fees:

Cash Out: 4200

New Pmt: -462.8965595 (462.90)

Interest Diff: -8434.001105 (-8434.00)

Download the file here: https://drive.google.com/file/d/171aj-kTKH0-YZCyeMDHZMxqpGERaaAhd/view?usp=sharing

Happy Holidays! Take care and be well,

Eddie

All original content copyright, © 2011-2024. 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.