Someone said you can use where the relative min and max of a continuous function occur to approximate the inflection point's x value by finding the average of those extrema value. But for some reason I don't think this would be an appropriate approximation in all cases. Does anyone know a link I can look at proving this or verifying it in some way? I never actually seen this stated anywhere before.

Question

Someone said you can use where the relative min and max of a continuous function occur to approximate the inflection point's x value by finding the average of those extrema value.  But for some reason I don't think this would be an appropriate approximation in all cases. Does anyone know a link I can look at proving this or verifying it in some way? I never actually seen this stated anywhere before.

tkhunny · Answer

You can use them to APPROXIMATE some of the inflection points.  Don't expect wonderful results.  Don't expect no value at all.  As with all things, learn the value and go with it.

freckles · Answer

I have no problem finding the actual. I'm curious about the approximation thingy.

fibonaccichick666 · Answer

Thatworks for sine, cosine, but I doubt it works on let's say a modified x^3

freckles · Answer

I was trying to find somewhere to read up on it but not having any luck.

fibonaccichick666 · Answer

hmm, let me see.  I never saw that as a theorem, and I can tell you we didn't prove that in analysis

freckles · Answer

Well the thing is like @tkhunny said I think it would lead to bad approximations in some cases.

freckles · Answer

But someone must have said it it is a good approximation for some cases and which cases maybe.

freckles · Answer

like I can see the whole sin and cos thing you mentioned

fibonaccichick666 · Answer

Only on maybe uniformly continuous functions

tkhunny · Answer

What's to be curious?

Minima and Maxima must lie between zeros (of things that have potential to lead to zeros with slight adjustments.)

Inflection Points must lie between Minima and Maxima (or things that have potential to lead to minima and maxima with slight adjustments.)

In the approximation business, one should ALWAYS know what one is doing, rather than rely on any method or rumor.

freckles · Answer

because those things have symmetry

fibonaccichick666 · Answer

I think you are right, the more symmetric a graph is, the more accurate the approximation

fibonaccichick666 · Answer

like x^3, I think you could get it on that too

tkhunny · Answer

In other words, the better the approximation, the more accurate the approximation.  This is not a great discovery.  Tautology, maybe?

If you alter the cubic so that it has three zeroes, there will be better results.  y = x^3 doesn't do much.

fibonaccichick666 · Answer

ah, I mean the approximation of the inflection point, ie. the zeroes must be evenly spaced so that the inflection point lies exactly between the two

fibonaccichick666 · Answer

in order to get an accurate answer, the less symetric the less accurate the approx.

tkhunny · Answer

You're talking in circles.  Of course the approximation will be better with better-behaved functions.  So what?  If the point of the exercise is to find points of inflection, this is not a good way to go about it.  You may narrow down the location of a few such things with this gross approximation.  Don't expect any better than that unless you hand-pick your fucntions.

freckles · Answer

@tkhunny You wouldn't be curious if there is a pretty error function for the naughty functions?

tkhunny · Answer

You can't find an application where this method would be particularly helpful - not in this day and age.

fibonaccichick666 · Answer

But it is curious, he could create a whole new theory!

fibonaccichick666 · Answer

a function equivalent to the peterson graph... It could be very interesting to study

tkhunny · Answer

No.  It's been done and it's not that interesting.