Small Solver Program

[Thomas Klemm]

If you compare your algorithm with Newton's method:

\(x_{n+1}=x_{n}-\frac {f(x_{n})}{f'(x_{n})}\)

you may notice that the number in register 0 should in fact be \(f'(x_{n})\) or at least a good approximation thereof.

And with this in mind indeed the solution can now be found:

\(f'(x) = \frac{d}{dx}4 x (x + 1) = 8 x + 4\)

\(f'(-1) = -4\)

\(f'(0) = 4\)

Examples:

-1.5
ENTER
-4
A

-1.00000

---

0.5
ENTER
4
A

0.00000

---

Thus your 2nd parameter is an estimate of the slope at the root.

---

And then you don't really have to call the function twice with the same value:

Code:

LBL A
STO 0
Rv
STO 1
28
STO I  // Store Loop Count Limit
LBL 0
GSB B
RCL 0
÷
STO - 1
X=0  // End if Root is found
GTO 1
DSE  // Loop Limit Counter
GTO 0
CLx
FIX 9  // 0.000000000 indicate that Maximum Loops is use up.
PSE
PSE
FIX 4
LBL 1
RCL 1  // Answer 
RTN
LBL B
.  // f(x) equation start here
.
.  // X = Register 1 [R1]
.
.
RTN

So the question remains how to approximate the derivation at the roots.
If you keep record of the previous function evaluation you can get an estimate of the slope using the secant method.

Cheers
Thomas