Casio fx-991CW: Second Derivative

Casio fx-991CW: Second Derivative

Calculating Higher Order Derivatives

A general formula to estimate derivatives of order n, where n is a positive integer (see Sources, Wikipedia):

d^n/dx^n = lim h→0 (1÷h^n * Σ((-1)^(k+n) * comb(n,k) * f(x+k*h), k=0, n)

The first, second, and third derivatives are derived from the above formula like so:

n = 1:

d/dx

= lim h→0 (1÷h * Σ((-1)^(k+1) * comb(1,k) * f(x+k*h), k=0 to 1)

= lim h→0 (1÷h * ((-1)^(0+1)*f(x) + (-1)^(1+1)*f(x+h))

= lim h→0 (1÷h * (-f(x) + f(x+h))

= lim h→0 (f(x+h) - f(x)) ÷ h

This is the famous forward difference formula.

n = 2:

d^2/dx^2

= lim h→0 (1÷h^2 * Σ((-1)^(k+2) * comb(2,k) * f(x+k*h), k=0 to 2)

= lim h→0 (1÷h^2 * ((-1)^2*comb(2,0)*f(x) + (-1)^3*comb(2,1)* f(x+h) + (-1)^4*comb(2,2)*f(x+2*h))

= lim h→0 (f(x) - 2*f(x+h) + f(x+2*h)) ÷ h^2

n = 3:

d^3/x^3

= lim h→0 (1÷h^3 * Σ((-1)^(k+3 * comb(3k) * f(x+k*h), k=0 to 3)

= lim h→0 (1÷h^3 * ((-1)^3*comb(3,0)*f(x) + (-1)^4*comb(3,1)*f(x+h) + (-1)^5*comb(3,2)*f(x+2*h)

+ (-1)^6*comb(3,3)*f(x+3*h))

= lim h→0 (-f(x) + 3*f(x+h) - 3*f(x+2*h) + f(x+3*h)) ÷ h^3

Using the fx-991CW

- - - - - - - - - -

Start with a note: Commentary and limitations: The functions f(x) and g(x), along with the calculus functions sum (Σ), integral (∫), and derivative (d/dx), has x as variable. It makes it a challenge that x is the only variable the functions f and g can take.

Example:

f(x)=x^2

g(x)=Σ(f(x),x=0 to 5)

Executing g(x) will take the values x=0 through x=5 no matter what value we put for g. The answer, in this example, will always return 0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55.

- - - - - - - - - -

We can still use f and g to set up specific derivatives in order n. For the second dervative, set up f and g:

f(x) = <function in terms of x>

g(x) = (f(x+2×A)-2×f(x+A)+f(x))÷A²

Store h in A. We can also choose an h and write in the formula directly.

Remember, this will calculate an approximation. With the most appropriate settings for h, we can get the best approximation.

Examples

For all the examples, A is set as 10^-7. The calculator is set to Radians. g(x) is the second derivative approximation.

Example 1:

f(x) = sin(x), f''(x) = -sin(x)

g(x) = (f(x+2×A)-2×f(x+A)+f(x))÷A²

x = 0.6; g(x): -0.564642551 (actual: -0.5646424734)

x = 2.8; g(x): -0.334988057 (actual: -0.3349881502)

Example 2:

f(x) = 2×e^(0.3×x), f''(x) = 0.18×e^(0.3×x)

g(x) = (f(x+2×A)-2×f(x+A)+f(x))÷A²

x = 0; g(x): 9/50 = 0.18 (actual: 0.18)

x = 1.2; g(x): 0.25799928 (actual: 0.2579992946)

Example 3:

f(x) = 1.1×x^3, f''(x) = 6.6×x

g(x) = (f(x+2×A)-2×f(x+A)+f(x))÷A²

x = 0; g(x): 6.6×10^-7 (actual: 0)

x = 2.2; g(x): 14.52 (actual: 14.52)

Sources

McCarty, George. Calculator Calculus. EduCALC Publications. E. & F.N. Spon: London. 1975, ISBN 0- 419-12910-3

Wikipedia "Numeric differentiation" Wikimedia Foundation, Inc. Last Edited June 17, 2025. Last Accessed July 7, 2025. https://en.wikipedia.org/wiki/Numerical_differentiation

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.

Basic (TI-81) vs. Python (TI-84 Plus CE Python): First and Second Derivative and Integral by Simpson’s Rule

Basic vs. Python: First and Second Derivative and Integral by Simpson’s Rule

Calculators Used

Basic: TI-81

Python: TI-84 Plus CE Python

Task

Estimate the numerical first derivative by the Nearby Secant Line Method:

f'(x) = lim (f(x + h) - f(x - h)) / (2 * h) as h → 0

Estimate the numerical second derivative by:

f''(x) = lim (f(x + h) - 2 * f(x) + f(x - h)) / h^2 as h → 0

Estimate the definite integral by Simpson’s rule:

∫( Y1 dx, x = a to x = b) ≈

(b - a)/(3 * n) + (f(a) + 4 * f1 + 2 * f2 + 4 * f3 + 2 * f4 + ... + 2 * f_n-2 + 4 * f_n-1 + f(b))

n must be an even integer.

fi = f(a + i * (b – a) / n)

BASIC: TI-81

First let’s look at the TI-Basic version with the classic TI-81:

First Order Derivative  (Nearby Secant Line Method)

f'(x) ≈ (f(x + h) - f(x - h)) / (2 * h)

DER1  (61 bytes)

Function

Rad

Disp "D/DX Y1"

1E-4 -> H

Disp "X?"

Input A

A+H -> X

Y1 -> D

A-H -> X

(D-Y1)/(2H)->D

Disp D

Second Order Derivative

f''(x) ≈ (f(x + h) - 2 * f(x) + f(x - h)) / h^2

DER2  (75 bytes)

Function

Rad

Disp "D²/DX² Y1"

1E-4 -> H

Disp "X?"

Input A

A+H -> X

Y1 -> D

A -> X

D-2*Y1 -> D

A-H -> X

(D+Y1)/(H^2) -> D

Disp D

INTGY1  (134 bytes)

Function 

Rad

Disp "SIMPSONS RULE"

Disp "A,B,N?"

Input A

Input B

Input N

A -> X

Y1 -> S

B -> X

S+Y1 -> S

1 -> I

Lbl 1

A+I*(B-A)/N -> X

S+2*Y1 -> S

If Fpart (I/2)≠0

S+2*Y1 -> S

IS>(I,N-1)

Goto 1

S*(B-A)/(3*N) -> S

Disp S

Examples

f(x) = 3*x^3 + 4*x – 1

Y1 = 3 * X³ + 4 * X – 1

f’(2.4): DER1 returns: 55.84

f’’(2.4): DER2 returns: 43.2

∫( 3 * X³ + 4 * X – 1 dX, 0.5, 1.5):

INTGY1 returns with N = 40 returns 6.75

Notes for BASIC

Before running the programs, store the function to the variable Y1. Calling Y1 evaluates the function of the value stored in variable X. Evaluate Y1 at X: [2nd], (Y-VARS), 1

The TI-81 MUST be in Function mode to evaluate Y1, unless an error occurs. The programs set the angle mode to Radians. The value of h is set to a "default" value of 10^-4. We want h to be small, but not too small. A value of h too small can run the risk of floating point errors.

In the integral program:

* I/2 - Ipart (I/2) is Fpart(I/2), but the long way. Python does not have a built-in fractional part function.

* IS>(I,N-1) translates to:

I=I+1

If I>N-1

Goto 1

IS> is the increment (by 1) and skip, a command in TI-Calculator Basic.

In general BASIC languages, evaluation Y1 may take a different approach, depending on the calculator, pocket computer, or computer:

General Basic:

### ...

### X = g(A)

### GOSUB ***

### ...

*** Y = <evaluate f(X)>

*** RETURN

Another approach on a calculator: Use another program

Prog <main>

...

g(A) -> X

Prog "FX" (see below, answer is stored in Y)

...

Prog "FX" (or any name of your choosing)

<evaulate f(X)> -> Y

<end program, an implied Return is executed here>

PYTHON: TI-84 Plus CE Python Edition

The code presented here uses the math module and built-in functions, and should work with any calculator with Python.

DERINTG.py

from math import *

# derivatives and integrals

# 2025-06-01

print("MATH MODULE IS IMPORTED")

# first derivative function

def der1(fx,x,h=1E-3):

  # fx is a string

  # h is optional (def 1E-3)

  # set up lambda

  f=eval("lambda x:"+fx)

  d=(f(x+h)-f(x-h))/(2*h)

  return d

# second derivative function

def der2(fx,x,h=1E-3):

  # fx is a string

  # h is optional (def 1E-3)

  # set up lambda

  f=eval("lambda x:"+fx)

  d=(f(x+h)-2*f(x)+f(x-h))/(h**2)

  return d

# integrals by simpsons rule

def intg(fx,a,b,n=64):

  # fx is a string

  # n is even, optional (def 64)

  # set up lambda

  f=eval("lambda x:"+fx)

  s=f(a)+f(b)

  for i in range(1,n):

    w=f(a+i*(b-a)/n)

    s+=(2*w) if (i/2-int(i/2)==0) else (4*w)

  s*=(b-a)/(3*n)

  return s

I decided to make a script of three functions.

Python uses binary arithmetic, so be prepared to possibly see answers such as 2.999999999999 and 5.00000000000001. The code leaves the answer in floating point form. Feel free to use the round function and/or format specifiers as you see fit.

f(x) = 3*x^3 + 4*x – 1

In Python:

f(x) = 3*x**3+4*x-1

Examples

Store the string to s:

s=”3*x**3+4*x-1”

der1(s,2.4) returns 55.84000299999303

der2(s,2.4) returns 43.19999999324864

intg(s,0.5,1.5) returns 6.75

I hope you find this useful. Enjoy,

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.

Plus42 Solver: Derivatives

Plus42 Solver: Derivatives 

Introduction

The Technical Applications book for the HP 27S and HP 19B (and can apply to the HP 17B outside of trigonometry) shows the numerical first and second derivative can be calculated by the formulas:

f ' (x) = (f(x+h) - f(x-h)) / (2 * h)

f ' ' (x) = (f(x + h) - 2 * f(x) + f(x + h)) / h^2 

where h is sufficiently small, like 10^-5 to 10^-12.

Legacy Formulas vs. Plus42 Formulas 

The formulas suggested by the Technical Applications Book are:

First Derivative:

Second Derivative:

Depending on the function FX, this above can turn the above into long equations.  With the ability of user functions, this allows us to use the original definitions.

FX(X): f(x) (insert f(x)

First Derivative:

F'X=(FX(X+H)-FX(X-H))÷(2×H)

Second Derivative:

F''X=(FX(X+H)-2×FX(X)+FX(X-H))÷SQ(H)

SQ:  press by the key sequence [(shift)] (x^2)

':  (ALPHA) [ ↓ ] (PUNC) [ ↓ ] ( ' )

Note:  Radians mode 

Examples 

FIX 5 mode is set.

Example 1:

f(x) = 0.5 * cos(3*x)

x = π/4

FX(X):0.5×COS(3×X)

f'(x) ≈ -1.06066

f''(x) ≈ 3.18198

Example 2:

f(x) = (x^2 + 3*x + 5) / (4*x - 1)

x = 2

FX(X):(X^2+3×X+5)÷(4×X-1)

f'(x) ≈ -0.22449

f''(x) ≈ 0.54227

Functions with Variable Constants

It is easy to expand the user function FX to include variable constants.  For example:

f(x) = -ln(cos(√(a*x)))

Calculate the value and first derivative at x = 0.11 and a = 0.46

Attach variable constants at the end of FX:

FX(X:A):-LN(COS(SQRT(A×X)))

F'X=(FX(X+H:A)-FX(X-H:A))÷(2×H)

f(x:a) ≈ 0.02552

f'(x:a) ≈ 0.23396

The user function makes the calculating numerical derivatives easier.  

Source:

Technical Applications: Step-by-Step Solutions for Your HP-27S or HP-19B Calculator Hewlett Packard.   Edition 2.  Corvallis, OR.   November 1988.  pg. 44

Have any Halloween plans?  Wishing you a great day,

Eddie 

All original content copyright, © 2011-2022.  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. 

Fun with the Infinite Series 1 + x + x^2 + x^3 + x^4 + x^5 + ...

Fun with the Infinite Series 1 + x + x^2 + x^3 + x^4 + x^5 + ...


The Series and Its Derivatives

Let F be the infinite series:

F = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ... = ∑ x^k from k = 0 to ∞.

Working with derivatives:

----

First Derivative of F:  (F' = dF/dx)

F' = 1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5 + 7*x^6 + ... 
= ∑ (k+1)*x^k from k = 0 to ∞

----

Second Derivative of F:  (F'' = d^2F/dx^2)

F'' = 2 + 6*x + 12*x^2 + 20*x^3 + 30*x^4 + 42*x^5 + 56*x^6 + ....

Factor out a 2:
= 2 * (1 + 3*x + 6*x^2 + 10*x^3 + 15*x^4 + 21*x^5 + 28*x^6 + .... )

Note the sequence 1, 3, 6, 10, 15, 21, 28...  These are triangle numbers, denoted as T_n.  

T_1 = 1
T_2 = 1 + 2 = 3
T_3 = 1 + 2 + 3 = 6
T_4 = 1 + 2 + 3 + 4 = 10 
and so on.

Using summation notation,  T_n = ∑ k from k = 1 to n

Going back to the series:

F'' = 2 + 6*x + 12*x^2 + 20*x^3 + 30*x^4 + 42*x^5 + 56*x^6 + ....
= 2 * (1 + 3*x + 6*x^2 + 10*x^3 + 15*x^4 + 21*x^5 + 28*x^6 + .... )
= 2 * (∑ x^k * T_k+1 from k = 0 to ∞)

In nested summation notation:

= 2 * (∑ x^k * (∑ m from m = 1 to k+1) from k = 0 to ∞)


Addition with F, F', and F"


F + F' = 2 + 3*x + 4*x^2 + 5*x^3 + 6*x^4 + 7*x^5 + 8*x^6 + ...
= ∑ ((k + 2) * x^k from k = 0 to ∞)

----

F + F' + F'' = 4 + 9*x + 16*x^2 + 25*x^3 + 36*x^4 + 49*x^5 + 64*x^6 + ...
= ∑ ((k + 2)^2 * x^k from k = 0 to ∞)

----

F' + F" = 3 + 8*x + 15*x^2 + 24*x^3 + 35*x^4 + 48*x^5 + 63*x^6 + ...

Note the sequence 3, 8, 15, 24, 35, 48, 63... where
3 = 4 - 1 = 2^2 -1
8 = 9 - 1 = 3^2 - 1
15 = 16 - 1 = 4^2 - 1
24 = 25 - 1 = 5^2 - 1
35 = 36 - 1 = 6^2 - 1
48 = 49 - 1 = 7^2 - 1
63 = 64 - 1 = 8^2 - 1
and so on...

This can be summarized as ∑( (k + 2)^2 - 1 from k = 0 to ∞)

Hence:
F' + F" = 3 + 8*x + 15*x^2 + 24*x^3 + 35*x^4 + 48*x^5 + 63*x^6 + ...
= ∑  ((k + 2)^2 - 1) * x^k from k = 0 to ∞)

Multiplying F and F' by x and x^2 

F = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ... = ∑ x^k from k = 0 to ∞.
x * F = x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + ... = ∑ x^(k+1) from k = 0 to ∞.
x^2 * F =  x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + ... = ∑ x^(k+2) from k = 0 to ∞.

-----

F + x * F = 1 + 2*x + 2*x^2 + 2*x^3 + 2*x^4 + 2*x^5 + 2*x^6 + ...
= 2 - 1 + 2*x + 2*x^2 + 2*x^3 + 2*x^4 + 2*x^5 + 2*x^6 + ...
= 2 + 2*x + 2*x^2 + 2*x^3 + 2*x^4 + 2*x^5 + 2*x^6 + ... - 1
= 2 * F - 1

This is one way to dervie the formula for the Infinite Geometric Series (for |x| < 1), to solve for F:

F + x * F  = 2 * F - 1
F + x * F - 2 * F = -1
F * (1 + x - 2) = -1
F * (x - 1) = -1
F = -1 / (x - 1)
F = 1 / (1 - x)   (keep this mind, this is true only when |x| < 1)

----

F - x * F = (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ... ) - (x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + ... )
= F * (1 - x)

For |x| < 1,

F * (1 - x) = 1 / (1 - x) * (1 - x) = 1

In general:
F - x * F = (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ... ) - (x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + ... )
= 1 + (x - x) + (x^2 - x^2) + (x^3 - x^3) + (x^4 - x^4) + (x^5 - x^5) + (x^6 - x^6) + ...
= 1

F - x * F = 1

----

F' = 1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5 + 7*x^6 + ... 
= ∑ ((k + 1) * x^k from k = 0 to ∞)

x * F' = x + 2*x^2 + 3*x^3 + 4*x^4 + 5*x^5 + 6*x^6 + 7*x^7 + ... 
= ∑ ((k + 1) * x^(k + 1) from k = 0 to ∞)


x^2 * F' = x^2 + 2*x^3+ 3*x^4 + 4*x^5 + 5*x^6 + 6*x^7 + 7*x^8 + ... 
= ∑ ((k + 1) * x^(k + 1) from k = 0 to ∞)

----

F' + x * F' = 1 + 3*x + 5*x^2 + 7*x^3 + 9*x^4  + 11*x^5 + 13*x^6 + ...

The sequence of 1, 3, 5, 7, 9, 11, 13, ... is the sequence of odd numbers which can be summarized as:

∑(2 * k + 1 from k = 0 to ∞)

Then:

F' + x * F'  =  F' * (1 + x) = ∑( (2*k + 1) * x^k from k = 0 to ∞)

----

F' + x * F' + x^2 * F' = 1 + 3*x + 6*x^2 + 9*x^3 + 12*x^4 + 15*x^5 + 18*x^6 + ...
= 1 + ( 3*x + 6*x^2 + 9*x^3 + 12*x^4 + 15*x^5 + 18*x^6 + ... )
= 1 + ∑(3 * k * x^k from k = 0 to ∞)

Eddie

All original content copyright, © 2011-2018.  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.  Please contact the author if you have questions.