As far as I know, the calculator uses *only* iterative methods for commands like solve( (solves an expression for a variable), fnInt (calculates the integral, the area below (or above) a function) or nDeriv (numerical derivation, calculates the gradient at X of an function).

So, for all these commands, the calculator only uses iterative methods and does not really calculate it.

For example the nDeriv command:

A example of finding the derivative at a location X in an equation y with Ti 84 Basic:

```
:Input “Equation“, Y1
:Input “X: “, C
:Disp “The gradient at the location
:Disp C
:Disp “of
:Disp Y1
:Disp “is equal to
:C-10^-4→X
:expr(Y1→A
:C+10^-4→X
:expr(Y1→B
:(B-A)/2*10^-4
:Disp Ans
```

As we all know, the iterative derivative (using lim) is defined as:

For precision you will have to minimize Deltax, so that lim DeltaX → 0

This is, why I chose 10^-4 ;)

Of course this was not iterations, but the

fnInt command uses iterations (it uses the method, where you part the function area into many rectangles)

And here an example for the iterative solving:

```
:Disp “Input the equation in such a way that it is equal to 0
:Input “Equation: 0=“, Strq
:Input “Guess“, G
:while inString(Str1, “X
:inString(Str1, “X
:sub(Str1, 1, Ans-1)+“L1“+sub(Str1,Ans+1,length(Str1)-Ans-1→Str1
:End
:0→D
:{G,G→L1
:while (D<100) and (10^-4<min(abs(expr(Str1
:D+0.1→D
:L1(1)-D→L1(1
:L1(2)+D→L1(2
:End
:If D=100
:Then
:Disp “No solution found
:Else
:Disp “Zero at
:10^-4<abs(expr(Str1
:If Ans(1):Then
:Disp G-D
:Else
:Disp G+D
:End:End
```

(Did not test it)

And here an example for the integral of a function being calculated iteratively using the rectangle method:

```
:Input “Equation: f(x)=“,Str1
:Input “Start: “,A
:Input “End: “,B
:Input “Iterations: “, M //this is the number of iterations. The greater it is, the greater also the precision is. Something like 100 should do a good job
:0→F
:(B-A)/M→D
:For(X,A,B,D
:expr(Str1
:F+Ans*D→F
:End
:Disp “The integral is equal to
:Disp F
Or here an optimized version:
:Input “Equation: f(x)=“,Str1
:Input “Start: “,A
:Input “End: “,B
:Input “Iterations: “, M //this is the number of iterations. The greater it is, the greater also the precision is. Something like 100 should do a good job
:(B-A)/M→D
:Disp “The integral is equal to
:Disp sum(seq(expr(Str1)*D,X,A,B,D
```

(I also did not test this one)

However I only wanted to show you the methods, how the calculator only iteritavely calculates things like this