## Saturday, February 6, 2016

### TI 84 Plus: Numerical Second Derivative of Two Variables and OS 5.1 Update

TI 84 Plus:  Numerical Second Derivative of Two Variables and OS 5.1 Update

TI 84 Plus OS 5.1 Update

Texas Instruments released an update for the TI-84 Plus CE calculator 5.1. The link to download this update is:

6
TI-84 Plus: Numeric Second Derivatives of Two Variables

The program DBLDERIV approximates the double second order partial derivatives of the two variable function f(x,y).  The program returns three results:

d^2/dx^2, stored in variable C.
d^2/dy^2, stored in variable D.
d^2/dxdy, stored in variable E.  For continuous functions, d^2/dxdy = d^2/dydx, and the program assumes that this is the case.

Formulas Used

For the function f(x,y) given Δx and Δy:

Let H = Δx and K = Δy.  Note that H and K do not have to be equal values. H and K are typically small (such as 0.0001 to 0.01).

Then:

d^2/dx^2 = 1/H^2 * ( f(x+H,y) – 2 * f(x,y) + f(x-H,y) )
d^2/dy^2 = 1/K^2 * ( f(x,y+K) – 2 * f(x,y) + f(x,y-K) )
d^2/dxdy = 1/(4*H*K) * ( f(x+H,y+K) – f(x+H,y-K) – f(x-H,y+K) + f(x-H,y-K) )

When prompted for f(x,y), enter the equation in quotes.  For example, enter f(x,y) = x – y^2 as “X-Y^2”.

Program DBLDERIV

Disp "DOUBLE PARTIAL"
Disp "DERIVATIVES"
Input "F(X,Y)=",Y₁
Input "X0=",A
Input "Y0=",B
Input "ΔX=",H
Input "ΔY=",K
A+H→X:B→Y:Y₁→C
A→X:C-2Y₁→C
A-H→X:(C+Y₁)/H²→C
A→X:B+K→Y:Y₁→D
B→Y:D-2Y₁→D
B-K→Y:(D+Y₁)/K²→D
A+H→X:B+K→Y:Y₁→E
B-K→Y:E-Y₁→E
A-H→X:B+K→Y:E-Y₁→E
B-K→Y:(E+Y₁)/(4HK)→E
Disp "C: D/DX²=",C
Disp "D: D/DY²=",D
Disp "E: D/DXDY=",E

Note:
Pressing [ VARS ], 1, 8 types the ΔX character.
Pressing [ VARS ], 1, 9 types the ΔY character.

Examples

Remember that this program gives approximate values!

Example 1:  f(x,y) = sin(x*y), Radians Mode, x0 = π/2, y0 = π, H = K = 0.001

Results:
d^2/dx^2 = C = 9.62648815    (actual is about 9.62649603)
d^2/dy^2 = D = 2.40662356    (actual is about 2.406624007)
d^2/dxdy = E = 5.033820798   (actual is about 5.03382056)

Example 2:  f(x,y) = x^2*y^2 – x^3,  x0 = 1, y0 = 2, H = K = 0.001

Results:
d^2/dx^2 = C = 2
d^2/dy^2 = D = 2
d^2/dxdy = E = 8
In this case, we get the exact answers.

Example 3:  f(x,y) = y^2*e^x, x0 = 0.1, y0 = 1.1, H = K = 0.001

Results:
d^2/dx^2 = C = 1.337257  (actual is about 1.337256811)
d^2/dy^2 = D = 2.203418  (actual is about 2.210341836)
d^2/dxdy = E = 2.4313764  (actual is about 2.43137602)

Source:

“Lecture 27: Numerical Differentiation”  This page is from the textbook:  Young, Todd & Mohelnkamp, Martin  “Introduction to Numerical Methods and Matlab Programming for Engineers”  Ohio University Math 3600 Course.   Retrieved January 30, 2016.  Link: https://www.math.ohiou.edu/courses/math3600/lecture27.pdf

Happy computing!

Eddie

This blog is property of Edward Shore. 2016