fx-991CW: Sums with a Step Not Equal to One

fx-991CW: Sums with a Step Not Equal to One

Step ≠ 1? An Adjustment is Necessary

The summation function on the Casio fx-991 CW calculator, and a lot of other advanced scientific calculators, calculates the sum of f(x) from x = a to x = b, with an increment of 1.

But what if have a sum of following:

log(2) + log(4) + log(6) + log(8): sum(f(x) from x = 2 to x = 8, step = 2)

Clearly, the increment is not 1, but 2 instead. Believe it or not, we can still use the summation function. We will need to adjust both the function and the limits to fit the requirements of the summation function.

1. Adjust f(x) to f(step * x)

2. Adjust the limits to x = a ÷ step and x = b ÷ step

For this particular problem:

1. f(x) becomes log(2 * x)

2. The limits become x = 2 ÷ 2 = 1 to x = 8 ÷ 2 = 4 (this allows the step to be 1)

Which transforms:

sum(log(x) from x = 2 to x = 8, step = 2)

to:

Σ(log(2 * x) from x = 1 to x = 4)

Sum: 2.584331224

A Couple More Examples

Example 1:

sum(x^2 from x = 0.5 to x = 2, step = 0.5), f(x) = x^2

The key here is the step (also known as the increment). The adjustments needed are:

1. f(x) becomes (0.5 * x)^2 = (x / 2)^2

2. The limits become x = 0.5/0.5 = 1 to x = 2/0.5 = 4

Sum: 15/2 = 7.5

Example 2:

sum(e^(x) from x = 1.5 to x = 2.1, step = 0.1), f(x) = e^(x)

The adjustments needed are:

1. f(x) becomes e^(0.1 * x) = e^(x / 10)

2. The limits become x = 1.5/0.1 = 15 to x = 2.1/0.1 = 21

Sum: 43.1994368

In Summary…

For the sum:

sum(f(x), x = a to x = b, step = Δx):

1. Adjust f(x) to f(Δx * x)

2. Adjust the limits: x = a/Δx to x = b/Δx

3. Summation function looks like this: Σ( f(Δx * x), x = a/Δx to x = b/Δx)

Hope you find this useful! Take care,

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.

The author does not use AI engines and never will.

Σ(1 / (a^n)) from n=1 to m

 Σ(1 / (a^n)) from n=1 to m

This blog entry covers the sum of the series:

Σ[1 / (a^n), n=1 to m] with n and m positive integers

Specific Cases:  a = 2 and a = 3

When a = 2:

m = 1:   1/2

m = 2:   1/2 + 1/4  = (2 + 1)/4 = 3/4

m = 3:   1/2 + 1/4 + 1/8 = (4 + 2 + 1)/8 = 7/8

m = 4:   1/2 + 1/4 + 1/8 + 1/16 = (8 + 4 + 2 + 1)/16 = 15/16

Going from the pattern,

Σ[1 / (2^n), n=1 to m] = 1/(2^m) * Σ[(2^n), n=0 to m-1] = (2^m - 1) / 2^m 

When a = 3:

m = 1:  1/3

m = 2:  1/3 + 1/9  = (3 + 1)/9 = 4/9

m = 3:  1/3 + 1/9 + 1/27 = (9 + 3 + 1)/27 = 13/27

m = 4:  1/3 + 1/9 + 1/27 + 1/81 = (27 + 9 + 3 + 1)/81 = 40/81

Going from the pattern,

Σ[1 / (3^n), n=1 to m] = 1/(3^m) * Σ[(3^n), n=0 to m-1]

Finding the General Formula and Proof

Let's presume that, for any a:

Σ[1 / (a^n), n=1 to m-1] = 1/(a^(m-1) * Σ[(a^n), n=0 to m-2]

Let's add 1/(a^m) to the series:  

Σ[1 / (a^n), n=1 to m-1]  + 1/(a^m)

= (1 + a + a^2 + ... + a^(m-3) + a^(m-2)) / (a^(m-1)) + 1 / (a^m)

= (a * (1 + a + a^2 + ... + a^(m-3) + a^(m-2)) + 1) / (a^m)

= (a + a^2 + a^3 + ... + a^(m-2) + a^(m-1) + 1) / (a^m)

= 1/(a^(m)) * Σ[(a^n), n=0 to m-1]

The general formula is now: 

Σ[1 / (a^n), n=1 to m] = 1/(a^(m)) * Σ[(a^n), n=0 to m-1]

Until next time,

Eddie

All original content copyright, © 2011-2021.  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. 

DM41 and DM42: Self Timing Programs

 DM41 and DM42: Self Timing Programs

Timing a Program's Performance

The Swiss Micros calculators DM41 (including the DM41L, DM41X, Hewlett Packard HP 41CX, HP41C with a Time Module installed) and DM42 have an internal clock which allows us to set the date and time on the calculator.  With the use of the TIME function we can determine the how long a program takes to execute.

One structure that can be used:

[ preliminary storage]

TIME

HR (DM 41), →HR (DM 42)

STO Rxx  (register xx)

[ loop/main program starts here ]

TIME

HR (DM 41), →HR (DM 42)

RCL Rxx

-

ABS 

3600

*

The routine TIME, HR, RCL xx, -, ABS, 3600, * determines the amount of time in term of seconds.  

DM42S:  Accessing the TIME function:  [SHIFT], [ + ] (CATALOG), [down arrow], ( TIME )  ("F1")

Let's demonstrate this technique on three examples.  The time record is the calculator ran on batteries.  

Example 1:  40th Fibonacci Numbers 

Registers:  

R01:  counter

R02:  sum

R03:  time

(counter of 39 is set because the first Fibonacci number is loaded at the beginning)

DM41 Program TFIB

01  LBL ^T TFIB

02  39

03  STO 01

04  0

05  STO 02

06  TIME

07  HR

08  STO 03

09  1

10  0

11  +

12  LBL 00

13  LASTX

14  X<>Y

15  +

16  DSE 01

17  GTO 00

18  TIME

19  HR

20  RCL 03

21  -

22  ABS

23  3600

24  *

25  STO 03

26  END

Y:  Results:  102334155.0

X: Seconds:  0.5100

DM42 Program TFIB

00  {45-Byte Prgm}

01  LBL "TFIB"

02  39

03  STO 01

04  0

05  STO 02

06  TIME

07  →HR

08  STO 03

09  1

10  0

11  +

12  LBL 00

13  LASTX

14  X<>Y

15  +

16  DSE 01

17  GTO 00

18  TIME

19  →HR

20  RCL 03

21  -

22  ABS

23  3600

24  ×

25  STO 03

26  END

Y:  Results:  102334155.0

X: Seconds:  0.0200

Example 2:  The Sum of Cubes Using a Loop from 1 to 250

Registers:  

R01:  counter

R02:  sum

R03:  time

DM41 Program TSUM

01  LBL ^T TSUM

02  250

03  STO 01

04  0

05  STO 02

06  TIME

07  HR

08  STO 03

09  LBL 00

10  RCL 01

11  3

12  Y↑X

13  ST+ 02

14  DSE 01

15  GTO 00

16  RCL 02

17  TIME

18  HR

19  RCL 03

20  -

21  ABS

22  3600

23  *

24  STO 03

25  END

Y:  Results:  984390625.0

X: Seconds:  12.07

DM42 Program TSUM

00  {45-Byte Prgm}

01  LBL "TSUM"

02  250

03  STO 01

04  0

05  STO 02

06  TIME

07  →HR

08  STO 03

09  LBL 00

10  RCL 01

11  3

12  Y↑X

13  STO+ 02

14  DSE 01

15  GTO 00

16  RCL 02

17  TIME

18  →HR

19  RCL 03

20  -

21  ABS

22  3600

23  ×

24  STO 03

25  END

Y:  Results:  984390625.0

X: Seconds:  0.24

Example 3:  Savage Test

Let A = 1, then for 2499 loops:

A = tan(atan(exp(ln(sqrt(A^2))))) + 1   (radians mode is used)

Registers:  

R01:  counter

R02:  sum

R03:  time

DM41 Program SVGE

01  LBL ^T SVGE

02  2499

03  STO 01

04  1

05  STO 02

06  TIME

07  HR

08  STO 03

09  LBL 00

10  RCL 02

11  X↑2

12  SQRT

13  LN

14  E↑X

15  ATAN

16  TAN

17  1

18  +

19  STO 02

20  DSE 01

21  GTO 00

22  RCL 02

23  TIME

24  HR

25  RCL 03

26  -

27  ABS

28  3600

29  *

30  STO 03

31  END

Y:  Results:  2499.970322

X: Seconds:  246.4500240  (about 4 minutes, 6.45 seconds)

DM42 Program SVGE

00  {51-Byte Prgm}

01  LBL "SVGE"

02  2499

03  STO 01

04  1

05  STO 02

06  TIME

07  →HR

08  STO 03

09  LBL 00

10  RCL 02

11  X↑2

12  SQRT

13  LN

14  E↑X

15  ATAN

16  TAN

17  1

18  +

19  STO 02

20  DSE 01

21  GTO 00

22  RCL 02

23  TIME

24  →HR

25  RCL 03

26  -

27  ABS

28  3600

29  ×

30  STO 03

31  END

Y:  Results:  2500  (just over 2500)

X: Seconds:  13.24

Source:

Burkett, John.  "The Savage Benchmark"  TI-89 / TI-92 Plus Tip List 10.0   July 20, 2002.  http://www.technicalc.org/tiplist/en/files/pdf/tips/tip6_50.pdf   Retrieved November 28, 2020 

Eddie 

HP 12C: Last X: Sums and Products

HP 12C:   Last X:   Sums and Products

Introduction

It is possible to use the LST X feature on the HP 12C calculators assist us in quickly calculate sums or products  of a terms, especially when the list of terms are in a series. 

For today's blog, we are starting with a base amount, a, and then adding 1 to each term.   You can use a similar algorithm for a sequence where each term is doubled, tripled, 1 is subtracted from the previous term, and so on.  The key is to complete the adjustment, use storage arithmetic, and then use LST X.

Accessing LST X:

HP 12C (Classic):  [ g ] [ ENTER ]

HP 12C Platinum:  [ g ] [ + ]

These algorithms can be used in program. 


Sum:  a + (a+1) + (a+2)  + (a+3) + ....

Let n represent the storage register to be used.  On the HP 12C, only storage registers R0 through R4 (classic HP 12C).

Algorithm:

a 
STO n

Loop:  
LST X 
1     [ or 2, x  to double each term,  1, - to subtract 1, etc.]
+  
STO+ n    [stores what is in the X display to the Last X register]

Finish:  
RCL n


Example:  7 + 8 + 9 +10 = 34.   Use register 0 to store the sum.

7
STO 0

LST X
1
+
STO+ 0

LST X
1
+
STO+ 0

LST X
1
+
STO+ 0

RCL 0


Sum:  1/a + 1/(a+1) + 1/(a+2) + 1/(a+3) + ....

Algorithm:

a 
1/x
STO n

Loop:  
LST X 
1     
+  
1/x
STO+ n    

Finish:  
RCL n

Example:  1/7 + 1/8 + 1/9 + 1/10 ≈ 0.47897

7
1/x
STO 0

LST X
1
+
1/x
STO+ 0

LST X
1
+
1/x
STO+ 0

LST X
1
+
1/x
STO+ 0

RCL 0

Sum:  √a + √(a+1) + √(a+2) + √(a+3) + ...

Algorithm:

a 
√
STO n

Loop:  
LST X 
1     
+  
√
STO+ n    

Finish:  
RCL n

Try the algorithm on this example:  √7 + √8 + √9 + √10 ≈ 11.63646

Sum:  a^2 + (a+1)^2 + (a+2)^2 + (a+3)^2 + ...

Hint:  Use ENTER, x instead of 2, y^x

Algorithm:

a 
ENTER 
*
STO n

Loop:  
LST X 
1     
+  
ENTER
*
STO+ n    

Finish:  
RCL n

Try the algorithm on this example:  7^2 + 8^2 + 9^2 + 10^2 = 294

Let's move on to products.

Product:  a * (a+1) * (a+2) * (a+3) * ....

a 
STO n

Loop:  
LST X 
1     
+  
STOx n    

Finish:  
RCL n


Example:  7 * 8 * 9 * 10 = 5040.   Use register 0 to store the product.

7
STO 0

LST X
1
+
STOx 0

LST X
1
+
STOx 0

LST X
1
+
STOx 0

RCL 0

Product:  1/a * 1/(a+1) * 1/(a+2) * 1/(a+3) * ....

a 
1/x
STO n

Loop:  
LST X 
1     
+  
1/x
STOx n    

Finish:  
RCL n


Try the algorithm on this example 1/7 * 1/8 * 1/9 * 1/10 ≈ 0.0020

Product:  √a * √(a+1) * √(a+2) * √(a+3) * .......

a 
√
STO n

Loop:  
LST X 
1     
+  
√
STOx n    

Finish:  
RCL n


Try the algorithm on this example √7 * √8 * √9 * √10 ≈ 70.99296

Hopefully this will help your calculations on your RPN calculators faster and more effective,

Eddie

All original content copyright, © 2011-2020.  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.

HP 12C: Sums of x

HP 12C:  Sums of x

Introduction

The following are four programs to calculate:

*  ∑ x
*  ∑ x^2
*  ∑ x^3
*  ∑ ax + b

with limits x = 0 to n.

Each program uses a closed sum formula.  The second and fourth program uses Horner's Method. 

Note:  You can use the first three if your limits are x = 1 to n (because at x = 0, 0 is added to the sum).  However if you use the fourth program, to get the sum from 1 to n, subtract b. 

HP 12C Program:  ∑x (x=0 to n)

n
∑ x =  n * (n + 1) / 2
x=0

Enter n, execute program

STEP:  KEY CODE  KEY
01:  36  ENTER
02:  36  ENTER
03:  1  1
04:  40  +
05:  20  *
06:  2   2
07  10  ÷
08:  43,33,00  GTO 00

Keys Only:
ENTER
ENTER
1
+
2
*
÷
GTO 00

Example:  (n = 48)

48
∑ x  = 1176
x=0

HP 12C Program:  ∑x^2 (x=0 to n)

n
∑ x^2 =  n * ( ( n / 3 + 1 / 2 ) * n + 1 / 6 )
x=0

Enter n, execute program

STEP:  KEY CODE KEY
01:  36  ENTER
02:  36  ENTER
03:  36  ENTER
04:  3  3
05:  10  ÷
06:  2  2
07:  22  1/x
08:  40  +
09:  20  *
10:  6  6
11:  22  1/x
12:  40  +
13:  20  *
14:  43,33,00  GTO 00

Keys Only:
ENTER
ENTER
ENTER
3
÷
2
1/x
+
*
6
1/x
+
*
GTO 00

Example:  (n = 48)

48
∑ x^2  = 38,024
x=0

HP 12C Program:  ∑x^3 (x=0 to n)

n
∑ x^3 =  1 / 4 * (n^2 + n)^2
x=0

Enter n, execute the program

STEP:  KEY CODE  KEY
01:  36  ENTER
02:  36  ENTER
03:  2  2
04:  21  y^x
05:  40  +
06:  2:  2
07:  21  y^x
08:  4  4
09:  10  ÷
10:  43,33,00  GTO 00

Keys Only:
ENTER
ENTER
2
y^x
+
2
y^x
4
÷
GTO 00

Example:  (n = 32)

32
∑ x^3  = 278,784
x=0

HP 12C Program:  ∑(a*x + b)  (x=0 to n)

n
∑ (a*x + b) =  n * (a * n / 2 + (a / 2 + b) ) + b
x=0

Store a in R1, store b in R2, enter n on the x-stack, execute the program

STEP:  KEY CODE   KEY
01:  36   ENTER
02:  36   ENTER
03:  36   ENTER
04:  45, 1  RCL 1
05:  20   *
06:  2    2
07:  10   ÷
08:  45, 1  RCL 1
09:  2   2
10:  10   ÷
11:  45, 2   RCL 2
12:  40   +
13:  40   +
14:  20   *
15:  45, 2  RCL 2
16:  40  ÷
17:  43, 33, 00   GTO 00

Keys Only:
ENTER
ENTER
ENTER
RCL 1
*
2 
÷
RCL 1
2
÷
RCL 2
+
+
*
RCL 2
÷
GTO 00

Example:  (a = R1 = 3,  b = R2 = 4, n = 32)

3 STO 1,  4 STO 2,  32, R/S

32
∑ (3x + 4) = 1,716
x=0


I tried something new with my RPN keystroke programs, the second versions of each I just listed what keys to press without the step number or code.  I think this would be easier to read, but does leave out the key and key code.  Any comments on which style is preferred is appreciated. 


Eddie

All original content copyright, © 2011-2020.  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.

Adventures in Python: List Manipulation

Adventures in Python: List Manipulation

This program creates two lists, one from 0 to 11, the other from 12 to 23.  Lists are enclosed by square brackets ( [ ] ).  Here is a short demo script.

# Program 004:  List manipluation

# Start with a range of 0 to 11

list1 = list(range(12))

# Let a second list range from 12 to 23

list2 = list(range(12,24))

print(list1)

print(list2)

# reverse the order

# use sorted in Python 3

print("Reverse List Elements")

list3 = sorted(list1, reverse=True)

list4 = sorted(list2, reverse=True)

print(list3)

print(list4)

# combine two (or more) lists

print("Combine lists with + ")

list5 = list1 + list2

print(list5)

# print the first two elements of each elements

print("First two elements of each list")

for k in range (0,2):

    a = list1[k]

    b = list2[k]

    print([a,b])

# sums, minimum, and maximum of each list

print("Sum, minimum, and maximum of each list")

print("Sum : ",sum(list1),sum(list2))

print("Minimum : ",min(list1),min(list2))

print("Maximum : ",max(list1),max(list2))

Output:

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

[12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]

Reverse List Elements

[11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

[23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12]

Combine lists with +

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]

First two elements of each list

[0, 12]

[1, 13]

Sum, minimum, and maximum of each list

Sum :  66 210

Minimum :  0 12

Maximum :  11 23

Coming up, combinations and permutations, random numbers, and plotting. 

Eddie

This blog is property of Edward Shore, 2017

Fun with the Casio fx-3650p

Links to previous fx-3650p programs:

5/11/2014:

http://edspi31415.blogspot.com/2014/05/casio-fx-3650p-programming.html

Contents:
1. Circular Sectors
2. Stopping Sight Distance
3. Resistors in Parallel
4. Net Present Value
5. Rod Pendulum
6. Vectors: Dot and Cross Products

10/27/2015:

http://edspi31415.blogspot.com/2015/10/casio-fx-3650p-programming-sequel.html

Contents for this blog:

1.  Combination with Replacement

2.  Great Circle (Distance in km) 

3.  Orbital Speed and Period 

4.  Eccentricity and Area of an Ellipse

5.  Super Factorial

6.  Escape Velocity 

7.  Finance: Payment of a Monthly Mortgage

8.  Wind Chill Factor

9.  Speed of Sound in Dry Air 

-------

Contents for this blog entry (7/2/2017)

1.  Modulus Function

2.  Normal CDF

3.  Sum:  Σ (AX + B)^C,  from X = 0 to X = Y

4.  Sun Altitude and Azimuth Based on the Vernal Equinox

5.  Trapezoid: Midsegment, Height, and Area

6.  Solar Irradiance

7.  General a list of X Random Integers from 0 to Y

Modulus Function

Calculates A mod B for A > 0 and B > 0.   Since the fx-3650p has no integer or fraction part functions, a loop of repeated subtractions are needed.

Program (25 steps):

? → A : ? → B : Lbl 1 : A – B → A : A ≥ B ⇒ Goto 1 : A

Examples:

Input:  A = 77, B = 9.  Result: 5

Input:  A = 92.38, B = 2.38.  Result:  1.94 

Normal Distribution CDF

This calculates the area of a normal distribution curve between points A and B, given that mean = 0 and deviation = 1.  Radians mode is set.  The result is stored in C.

Program (34 steps):

? → A : ? → B : Rad : ∫ ( e (-X² ÷ 2 ), A, B → C : C ÷ √ (2 π → C

Note e is the exponential function (e^x).

Examples:

Input:  A = 0, B = 2.  Result:  0.47725066

Input:  A = -1, B = 1.  Result:  0.682709924

Sum:  Σ (AX + B)^C,  from X = 0 to X = Y

Program (51 steps):

? → A : ? → B : ? → C : ? → Y : 0 → X : 0 → M : Lbl 1 : (AX + B)^C M+ : 1 + X → X : Y ≥ X ⇒ Goto 1: M

Examples:

Input:  A = 2. B = 6, C = 2, Y = 4.  Result:  540

Input:  A = -3, B = 1, C = 3, Y =6.  Result:  -9632

Sun Altitude and Azimuth Based on the Vernal Equinox

Original Post:  http://edspi31415.blogspot.com/2013/06/hp-35s-sun-altitude-azimuth-solar-pond.html

Input:

Y =  days after the vernal equinox, usually March 21

A = latitude on Earth (north-south, -90° to 90°)

X = the time before solar noon, local time

For example, for 10 AM (10:00), enter 2.  For 3 PM (15:00), enter -3.  Hence:  12 – time.

Output:

D = approximate declination of the sun (-23.45° to 23.45°)

B = sun’s altitude

C = sun’s azimuth (from ground wise north)

Program (69 steps):

? → Y : ? → A : ? → X : Deg : 23.45 sin(.9856Y → D ◢  sin¯¹ (cos A cos D cos (15X) + sin A sin D → B ◢ cos¯¹ ( ( sin B sin A – sin D) ÷ (cos A cos B → C

Examples:

Input:  Y = 90 days, A = 25°, X = -3  (3 PM)

Results: 

D = 23.44400127° = 23°26’38.4”

B = 48.81756385° = 48°49’3.23”

C = 99.85903298° = 99°51’32.52”

Input:  Y = 68 days, A = 46°, X = 4 (8 AM)

Results:

D = 21.58916369° = 21°35’20.99”

B = 35.989914° = 35°59’23.69”

C = 84.40834691° = 84°24’30.05”

Source: Sun Altitude, Azimuth, Solar Pond Absorption, HP 67/97 Energy Conservation December 1978, Author: HP

Trapezoid: Midsegment, Height, and Area

Original Post:  http://edspi31415.blogspot.com/2017/06/hp-prime-and-casio-fx-cg50-trapezoid.html

Input:

A = length of top side

B = length of bottom side

C = length of left side

D = length of right side

Output:

X = Midsegment length

Y = Height

M = Area

Program (92 steps):

? → A : ? → B : ? → C : ? → D : .5(A + B → X ◢ (-A + B + C + D)(A – B + C + D)(A – B + C – D)(A – B- C- + D) → Y : √Y ÷ ( 2 √( (B – A)² ) → Y ◢ (A + B)Y ÷ 2 → M

Example:

Input:  A = 18, B = 16, C = 12, D = 11

Results:  X = 17, Y = 9.921567417, M = 168.6666461

Source:  “Trapezoid”  Wikipedia.  Edited July 7, 2014.  Retrieved July 8, 2014

Current page:  https://en.wikipedia.org/wiki/Trapezoid

Solar Irradiance

Original Post:  http://edspi31415.blogspot.com/2015/03/hp-prime-and-casio-prizm-solar.html

The program calculates:

1. The solar angle of incidence given the angular elevation and azimuth (from south going “counterclockwise”:  east-north-west) of both the sun and panel.

2. The irradiance given by the solar panel. 

Input:

X = elevation of the sun

A = azimuth of the sun

Y = elevation of the solar panel

Z = azimuth of the solar panel

M = the sun’s power or irradiance.  Often this is treated as a constant, which is approximately 1367 W/m^2 for extraterrestrial solar power, or approximately 1000 W/m^2 when we are dealing with the Earth’s surface (taking scattering of light into account)

Output:

C = incidence angle

D = solar irradiance

Program (46 steps):

? → X : ? → A : ? → Y : ? → B : ? → M : Deg : cos¯¹ ( cos Y sin X + sin Y cos X cos (A – B → C ◢ M cos Y → D

Example:

Input:

Sun:

X = 55°24’21”

A = 175°15’44”

Panel:

Y = 40°

B = 90°

Sun’s Irradiance:

M = 1000 W/m

Results:

C = 48.6431686°

D = 766.0444431 W/m

Sources (you might have to copy and past these links)  

Baldocchi, Dennis “Lecture 7, Solar Radiation, Part 3, Earth-Sun Geometry”  Biometeorogy, ESPM 129  University of California, Berkeley.

http://nature.berkeley.edu/biometlab/espm129/notes/Lecture%207%20Solar%20Radiation%20Part%203%20Earth%20Sun%20Geometry%20notes.pdf

Retrieved February 17, 2015. 

Mortimer, David  “Lambert’s Cosine Law”  30 January 2014.  The Solar Bucket.

http://thesolarbucket.blogspot.com/2014/01/lamberts-cosine-law.html

Retrieved March 18, 2015

University of Oregon Solar Radiation Monitoring Laboratory  “Solar Radiation Basics”  University of Oregon.  http://solardat.uoregon.edu/SolarRadiationBasics.html   Retrieved February 10, 2015

General a list of X Random Integers from 0 to Y

This program generates a list of random integers (X) from 0 to upper limit Y.  This program uses the Fix 0 mode and makes use of the Rnd (round the number in the display) command.  Note the integers as they appear.  The program finishes by setting the calculator back in Norm 1 mode.

Program (33 steps):

? → Y : ? → X : 1 → M : Fix 0 : Lbl 1 : Ran# Y : Rnd ◢ 1 M+ : X ≥ M ⇒ Goto 1 : Norm 1

Example:

Generate 5 random integers from 0 to 10.  (Y = 10, X  = 5)

Result:  6, 7, 6, 6, 10 (your results will vary)

Comments

As with the other fx-3650p I and others have posted, they can easily be adapted to the fx-50fH, fx-5800p, fx-6300g, fx-CG50, and (almost) any other Casio programming calculator. 

Recently I learned that Casio updated the fx-3650P and the fx-50F(H) with the fx-3650P II and fx-50F(H) II respectively.  The only difference I was able to spot is that the fx-3650P II now has 390 programming steps instead of 360. http://www.casio-intl.com/asia/en/calc/school/programmable/

And yes, I still wish Casio sold these models in stores in the United States.  Currently, for us U.S. residents, they can only be purchased online. 

Now if Casio made a solar version of the fx-6300g, with more memory.

Eddie

This blog is property of Edward Shore, 2017.