HP 49g/ 50g: Wrong extremum

[rawi]

I put in (Rad Mode): APPS - 1 (Plot finction) - 1 (equation entry) -
Y1(X) = (e^(-X²)+X²-5.)/(SIN(X)+2.)
Plot X from -4 to 4, Y from -5 to 10 - DRAW
FCN - EXTR -
I get: Extrm: (0.6831, -1.4846)
which is not an extremum as can be seen easily from the function plot.
Is there anything I did wrong?
Thanks and best regards
Raimund

[carey]

(06-04-2024 08:01 PM)rawi Wrote:  I put in (Rad Mode): APPS - 1 (Plot finction) - 1 (equation entry) -
Y1(X) = (e^(-X²)+X²-5.)/(SIN(X)+2.)
Plot X from -4 to 4, Y from -5 to 10 - DRAW
FCN - EXTR -
I get: Extrm: (0.6831, -1.4846)
which is not an extremum as can be seen easily from the function plot.
Is there anything I did wrong?
Thanks and best regards
Raimund

I confirmed your finding on an HP49g+.

The point (0.6831, -1.4846) that the calculator misidentifies as an extrema is actually one of four inflection points (points where 2nd derivative = 0) in the x = [-4,4] interval. The four inflection points are: (-3.5395, 3.15313), (-1.9825, -0.969083), (-0.489831, -2.59779), (0.683099, -1.48459).

There are no extrema in the x = [-4,4] interval in the sense that there are no global or local maxima or minima. The maximum absolute value of this function in this interval occurs at the positive end (x=4) of the interval (where the function is still increasing) and is 8.84815169836. Perhaps the calculator’s extrema function only correctly identifies points if there is a local or global maxima and minima, but I’m not sure.

[roem76]

My HP50G shows the same behaviour.

I agree with almost all the remarks made by carey but not the remark that there is no local maxima or minima in the range x=-4..4. At x=-1.07957 is a local minima (y=-3.15025).

The HP50G actually finds that local minima but the cursor needs to be placed somewhat near to it prior to calling [FCN]-[EXTR] in order to find it or it will misidentify the inflection point mentioned as an extrema.

Anyway, good to know that it seems mandatory to check if an extrema reported by the HP49G/HP50G is actually an extrema.

Roman

[rawi]

Thanks carey and Roman for the reply.
Actually I am playing around with my HP 50g I own since many years but never got a real understanding of it. So I was not sure whether I did something wrong. Now I am sure.

[carey]

(06-05-2024 10:07 AM)roem76 Wrote:  My HP50G shows the same behaviour.

I agree with almost all the remarks made by carey but not the remark that there is no local maxima or minima in the range x=-4..4. At x=-1.07957 is a local minima (y=-3.15025).

The HP50G actually finds that local minima but the cursor needs to be placed somewhat near to it prior to calling [FCN]-[EXTR] in order to find it or it will misidentify the inflection point mentioned as an extrema.

Anyway, good to know that it seems mandatory to check if an extrema reported by the HP49G/HP50G is actually an extrema.

Roman

roem76, thanks for catching my error about missing the local minima and the good suggestion when calling [FCN]-[EXTR] on the HP49g/50g!

Just for curiosity I tried a couple of other graphing calculators to see how they compare to the 49g/50g in finding extrema for this function. Both the Casio CG50 and TI-84 correctly determine the local minimum -- the CG50 did so automatically while the TI-84 asked for left and right bounds and a guess, though succeeded no matter how far away the guess was within the bounded interval.

[Albert Chan]

HP50g likely use just enough points to make plot.
Examine the local few points, it found a flat spot, and consider it as relative extremum.

Modern calculator, with more available points, see finer details, thus less false positives.

[rawi]

Albert Chan wrote:

Quote:Examine the local few points, it found a flat spot, and consider it as relative extremum.

This does not sound plausible to me.
Why does the calculator find an inflection point instead of an extremum? The HP 50g has a powerfull solver routine. So I would assume that for finding an extremum he solves for something like (f(x)-f(x+delta))/delta = 0 with a small delta. OK, the curve is relatively flat at that point, but f'(0.6831) = 0.631, so it is not near to zero.

So in this case the calculator seems to search for an extremum of the derivate which would be a strange bug.