**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:

I recommend that you save your programs and downloaded apps
because installation will erase unarchived memory.

**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

Eddie

This blog is property
of Edward Shore. 2016

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