Extremums on HP 42S

[lrdheat]

The equation sqrt(ABS(x^3 +2*x +4)) produces extremums where I expect at x~-2.59 and x~0, but does not find the extremum at x~-1.33. Anyone know why?

LBL “FX”
ENTER
ENTER
ENTER
3
Y^X
X<>Y
X^2
2
*
+
4
+
RTN
END

[rprosperi]

Presuming you are using SOLVE (you don't say what you are using, or how, so we can only guess) you need to provide 2 initial guesses which straddle the desired root. What guesses are you providing and what results to they lead to?

[Thomas Okken]

Keep in mind that SOLVE was designed to find roots, not extrema. It will report extrema if it gets stuck on them, but there is no guarantee it will find any, regardless of what starting guesses you provide.

If you are specifically looking for extrema, you should use SOLVE to find roots of the derivative.

[lrdheat]

I used the solve protocol, adding MVAR X, RCL X to the routine. Tried -1.35, -1.31 as the bound, and it finds the extremum at X~0.

[Pekis]

Hello,

Is it this function ? If it's the case, I don't understand your values ...

[Sylvain Cote]

(12-24-2021 09:13 AM)Pekis Wrote:  Is it this function ?

The formula is: \( \mathsf{y} = x^3 + 2(x^2) + 4 \)

[Albert Chan]

I think the function is y = sqrt(abs(x^3+2*x^2+4))

(x^3+2*x^2+4)' = 3*x^2 + 4*x = 3*x*(x+4/3)

To understand why extremum -4/3 is misssed, we need to understand how Secant method work.

Assuming plot is above x-axis:

From (x1,y1),(x2,y2), secant line locate the root at the x-axis, go up, to get y3.
Then, it uses (x2,y2),(x3,y3) to locate x4, go up, to get y4 ...

Extremum that shaped like a \(\bigcup\) may be located.
Extremum that shaped like \(\bigcap\) will not (new x's will move away from it)

[lrdheat]

Thanks! My equation did use 2*X^2, sorry for my typing. This is neat in that it demonstrates why this particular integration approach may not find all extremums! There is no perfect numerical approach to integration approximations, but some amazingly good ones!

[toml_12953]

(12-24-2021 09:13 AM)Pekis Wrote:  Hello,

Is it this function ? If it's the case, I don't understand your values ...

I wondered that, too. There's only one extremum AFAICS: -1.18 or so as your graph shows.

[C.Ret]

In the case the investigated function is \( f(x)=\sqrt{\left| x^3+2x^2+4 \right|} \), the extremums can be found solving \( \frac{\partial f(x)}{\partial x}=0 \) equation.

The only problem here is that \( f(x) \) isn't continuous all over the interval and the derivative function \( \frac{\partial f(x)}{\partial x} \) is not definite at the point \( x_d \) where \( x_d^3+2x_d^2+4=0 \) . Numerically, \( x_d\approx-2.5943 \).

This point is indicated by a cross on the following screen capture.

   

At the discontinuity point \( x_d \), the extremums or root can't be determinate using method based on function continuity.

This explain why the algorithm of the SOLVE instruction of the HP-42S (or related clone or simulators) can only be used inefficiently in vicinity of the discontinuity.

Numerically, if the investigated function is \( f(x)=\sqrt{\left| x^3+2x^2+4 \right|} \); one may determine three extremums :

• discontinued root at \( x_d\approx-2.5943 \) where \( f(x_d)=0 \)
• intermediate extremum at \( x_b=-\frac{4}{3} \) where \( f(x_b) = \frac{2.\sqrt{105}}{9}\approx2.2771 \)
• last extremum on y-axis at \( x_0=0 \) where \( f(x_0)=2 \)

[lrdheat]

What I find interesting is that the HP 42S does find the zero at ~-2.59!