SERIES - TI 89 or TI 92

Introduction
This program evaluates Taylor series of the form $\sum\limits_{K=0}^N$ C_KX^K or $\sum\limits_{K=0}^N$ C_K(X-A)^K. It also compares that value to the corresponding function value, f(x).

If you have not used one of the programs posted on this website before, you should read through the information in the Intro to Programming section first.

If you have a TI Connectivity Cable, you can download the program main.series.89p

The Program

:series( ) {This will already appear if you named the program series}

:Prgm {This will already appear}

:Lbl p {Lbl is in the CATALOG menu}

:ClrIO {ClrIO is in the CATALOG menu}

:Input "x", x {Input is in the F3 menu}

:Input "n", n

:0 $\to$ s {The 0 is a zero} {The arrow is STO }

:For k,0,n {For EndFor is in the F2 menu} {The 0 is a zero}

:y1(x)+s $\to$ s {y1 is the Y followed by a 1}

:EndFor

:Disp "series value", s {Disp is in the F3 menu}

:Disp "f(x) value", y2(x)

:Pause {Pause is in the CATALOG menu}

:Goto p {Goto is in the CATALOG menu}

:EndPrgm

Running the Program

You will need to enter an expression and the corresponding function f(x) into y1 and y2, respectively. The program will ask for the x value and N (the number of terms to be evaluated).

To test the program try the following:

Enter $\hbox{$X^{\hbox\ {K}}$}\ \over \hbox{K!}$ into y1 and e^x into y2 . Use x = 0.5, n = 3.

Your answer will be

series value
1.6458333333
f(x) value
1.6487212707

:series( )	{This will already appear if you named the program series}
:Prgm	{This will already appear}
:Lbl p	{Lbl is in the CATALOG menu}
:ClrIO	{ClrIO is in the CATALOG menu}
:Input "x", x	{Input is in the F3 menu}
:Input "n", n
:0 $\to$ s	{The 0 is a zero} {The arrow is STO }
:For k,0,n	{For EndFor is in the F2 menu} {The 0 is a zero}
:y1(x)+s $\to$ s	{y1 is the Y followed by a 1}
:EndFor
:Disp "series value", s	{Disp is in the F3 menu}
:Disp "f(x) value", y2(x)
:Pause	{Pause is in the CATALOG menu}
:Goto p	{Goto is in the CATALOG menu}
:EndPrgm