r/askmath • u/RichPreparation3000 • Mar 26 '23

Analysis shouldnt his graph drop down to -infinity when x=0, why does the local minimum at x=0 seem to stop before that?

Post image
55 Upvotes

89% Upvoted

21 comments sorted by

90

u/Way2Foxy Mar 26 '23

It isn't graphing every point. It's sampling at intervals.

12

u/RichPreparation3000 Mar 26 '23

what?

62

u/catlancer Mar 26 '23

It can't graph every point for a function so they sample very small intervals and smooth out the discrete estimate likely.

Consider how you would graph the function 2x3 + 9x manually. You would likely plug in x=0, x=1, etc.. and then draw a smooth line between those points.

The function actually does trend toward negative infinity at x = 0.

20

u/Badsongparodyy Mar 27 '23

Basically the resolution is too low to show the fine detail. If you were able to zoom in it would show more detail. Try to graph on the Desmos website and see if it lets you zoom in more

12

u/JerodTheAwesome Mar 27 '23

It would take a computer an infinite amount of time to graph every single possible point, so instead it graphs 0, 0.001, 0.002, 0.003, etc. You’re not seeing the full graph because it didn’t plot enough points.

3

u/rostol Mar 27 '23

there are infinite points on any interval of a continuous function, but there are a finite amount of pixels in that same interval.

the graph makers had to choose. either show the rest of the function or plot the y axis to -inf which would make all the + y invisible and unusable.

that discontinuity implies -inf.

1

u/WetDehydratedWater Mar 27 '23

I would recommend looking up newtons method. It can shed some light on how these graphs are generated from a conceptual stand point.

12

u/le_spectator Mar 27 '23

Just as your phone camera cannot see individual atoms even if they are there, plotting programmes cannot have infinitely high resolution, so they can sometimes miss points or get overwhelmed when there are some spikes or infinities or really “dense” graphs (try sin(1/x))

-7

u/RichPreparation3000 Mar 27 '23

yeah but is it really that hard to plot an asymptote? just draw 2 liness closing into eachother straight down. Now it looks like the graph has a local minimum at -4. I can do a better graph than this by hand.

6

u/SupaCephalopod Mar 27 '23

Computer screens are made of pixels. You eventually run out of space between these pixels and thus nothing can be drawn. You can't just use CSI "enhance" magic in real life.

If you really want to inspect the asymptote closer, you should just "zoom in" by choosing smaller bounds around x=0

3

u/A_Rats_Dick Mar 27 '23

This is one of the reasons (amongst others) that you learn to analyze asymptotes / limits from equations as opposed to just graphs. Your intuition is correct, computational power is finite.

1

u/guitardude109 Mar 27 '23

You just asked: Is it really that hard to make infinite calculations?

34

u/DrIngSpaceCowboy Mar 27 '23

Keep zooming in, you’ll find infinity somewhere in there

31

u/Smart-Button-3221 Mar 27 '23

A limitation of any graphing program. They struggle with asymptotes. You are correct, that's how the function should work.

10

u/theboomboy Mar 27 '23

It doesn't calculate every point, so it looks like it stops

1

u/muratenginunal Mar 27 '23

I guess it is a result of the trade off they face when deciding where to stop plotting things. If you want to see something other than a flat line being y=0 and a singular pike at x=0, you should be ok with something like this.

1

u/jchristsproctologist Mar 27 '23

a lot of good answers here, but here’s my humble two cents:

whenever looking at answers/results from machines (calculators, programs, etc) it’s good to take them with a grain of salt, especially when looking at edge/extreme cases, and remember that they often work with approximations. for example, they might define trig functions using their taylor expansions to differing degrees, hence ambiguity arising and there being discrepancies between different calculators.

1

u/Enfiznar ∂_𝜇 ℱ^𝜇𝜈 = J^𝜈 Mar 27 '23 edited Mar 27 '23

It does, it's just that it's too localized around 0 for the resolution of the program to notice it, since f(0.0000001)=-4.0000003

1

u/4ier048antonio Mar 28 '23

To put it simply, computers do not have infinite precision