Understanding The Third Derivative Geometrically

[@FildasFilipi]

"the rate of increase of inflation was decreasing" - Richard Nixon 1972, the first use of the 3rd derivative in economics 😃

[@yunoewig3095]

omg

[@HiHello-dj8if]

Inflation isn’t a derivative of anything, so he used a second derivative.

[@yunoewig3095]

@@HiHello-dj8if what do you mean?? inflation is the rate of change of prices

[@canaDavid1]

Also, if you accept wages as the derivative of cash, then inflation is the second derivative, "rate of change of inflation" is 3rd, and your statement is a fourth derivative

[@MattM-oe6qs]

@@canaDavid1 no his statement is giving the direction of the change so it is the third derivative not the fourth. He is actually describing the change, not adding another level of change.

[@shlokdave6360]

Pretty much how we define and look at skewness. That one is the third moment around the central tendency, ie the mean. The fourth derivative mirrors the properties of kurtosis ie the spread around the tails. Its nice to see how everything ties up so nicely. Thank you for a nice video. Perhaps a segment on skewness would add to the understanding plus application of the concept.

[@DrBarker]

That's a good point. We could even try to understand higher order derivatives by looking at higher order moments, but I don't really have a clear image in mind for anything beyond the 4th moment.

[@shlokdave6360]

@@DrBarker The fifth moment measures the symmetry of the tails. It can be used as a measure of the symmetry/asymmetry of the two tail-risks in distribution of returns of assets, for example. The fifth derivative should, by logical extension, just be a measure of the symmetry of the extreme values about f(x = 0) as we move away from x = 0 in either direction. I have no clue as to what the sixth and higher derivatives will translate into. I also just now remembered my physics teacher defining the third derivative of position vector as the Jerk felt by the particle in question.

[@DrBarker]

@@shlokdave6360 I could see the 7th, 9th, and other odd moments as related to the symmetry, where higher order odd moments are more heavily weighted by the more extreme values than lower order odd moments. I'm not sure how much insight that actually gives me though, especially for derivatives. A higher order derivative should still presumably be a local property(?), so I'm not sure how the notion of extreme values extends there?

[@shlokdave6360]

@@DrBarker I agree with both points. I wrote down the point on odd and even derivatives twice before deleting since they were just observations and not concrete derivations- I was trying to avoid making a silly mistake, something I excel in. 😆 But it seems someone as learned as you also saw the patterns. Beyond the 4th and 5th things do get silly to be honest. I am sure we can find interpretations of the same but how much do the 6th 7th and higher terms really contribute in the Taylor’s series? One real application of higher order derivatives I have used are the option Greeks in applied economics. And over there, nothing useful is obtained beyond the third derivatives.

[@shlokdave6360]

@@DrBarker again at the risk of making a complete ass of myself, higher derivatives should not be a local property for translationally invariant situations. I have a very weird understanding of why I think this to be the case with zero confidence in saying so. How do we test for it, any ideas?

[@txikitofandango]

The way I always thought of the 3rd derivative: When I'm cruising at constant speed in my car, my back's position is at zero relative to the seat back. When I accelerate, the seat back squishes proportionally. So the seat squish and car acceleration differ by two derivatives. Therefore, the third derivative of car motion is related to the speed of seat squish. So 1 m/s³ of car jerk equals so many cm/s of seat squish. Roughly.

[@txikitofandango]

I like the geometric interpretation, the gradients of chords, it plays nicely with the physical one

[@DrBarker]

This is a really fun physical interpretation - thanks for sharing!

[@ianfowler9340]

The derivative of acceleration is sometimes referred to as "jerk" motion. When I was in university and someone was being an idiot we used to say: "Stop acting like the derivative of acceleration."

[@bobh6728]

@@ianfowler9340 That “jerk” measurement is what determines the comfort of a passenger in a vehicle or airplane. Commercial airplanes are actually programmed to make turns so that the acceleration is down into the seat and jerk is as close to zero as possible. It does this by banking the plane. So the passenger just feels pressure down into the seat, not a leaning to the side. That is why passengers have to fasten the seatbelts when landing. The jerk as the acceleration decreases from the braking throws the passengers forward.

[@mosquitobight]

To use a rocket analogy, if you're in a rocket in space with a positive second derivative in the direction of travel, you're accelerating, so you'll feel a G-force like gravity pulling you to the rear of the rocket. You can stand on a scale and watch your weight change. If the third derivative is zero, you'll feel a constant G-force and you'll have a constant weight on the scale. If the third derivative is positive, you'll feel the G-force increasing, and the scale will say you're getting heavier. If the third derivative is negative, you'll feel the G-force decreasing, and the scale will say you're getting lighter.

[@bradsword5822]

Before I watched, I was kind of like “no way!” Surely the third derivative is too deep to see or intuit or something. But it makes sense. If the second derivative is concavity, then the third derivative is like changing concavity. Since the second derivative is related to curvature, it’s basically changing curvature. A curve curves more or less, depending on the third derivative.

[@eliyahzayin5469]

This made a surprising amount of sense to me. A while ago I tried to find the "asymmetry" of a curve and found that the most sensible definition is the derivative of curvature with regards to arc-length. Since curvature is analogous to the second derivative, then this asymmetry would be analogous to the third derivative

[@ianfowler9340]

I'd like your take on the following: Third Derivative Test for Inflection Points. For y = f(x) at x = c: (c,f(c)) is an inflection point iff f ' '(c) = 0 and f ' ' '(c) 0 If f ' '(c) = 0 and f ' ' '(c) = 0 then no conclusion can be drawn. In this case you must go back to the 2nd derivative test and check the concavity around x = c. This is just anther way of saying that the slopes of the tangent reach a max or a min at the point of inflection.

[@DrBarker]

This works nicely, and if f''(c) = f'''(c) = 0, then we could also check the fourth derivative, fifth derivative, and so on. If the 4th derivative is non-zero, it's not a point of inflection, and if the 4th derivative is zero, we need to keep going. If the 5th derivative is non-zero, it is a point of inflection.

[@ianfowler9340]

@@DrBarker I never thought of the before! But when you think it through it makes perfect sense. Learn something new everyday.

[@MichaelRothwell1]

@@DrBarker Nice! This is basically saying that if our function has a Taylor series about the point of interest, we look at the first non-zero term after the (x-a)² term [x² term if we have translated the point to 0], then the local behaviour of f (in relation to the tangent at that point) mimics that of the (x-a)ⁿ term - point of inflection iff n is odd.

[@jens6076]

Your choice of topics and case studies are so tasteful and refreshing! Thank you❤

[@bosorot]

so Nike is a negative 3rd derivative.

[@bernardlemaitre4701]

Thanks ! I Always ask me what should be useful in the third derivative, now i have the answer !

[@francescocostanzo8225]

I did too and now I know why teachers don't go over it as much. A lot more involved than initially would expect.

[@General12th]

Very cool! Although I'm curious how often we'd have a situation where we're dealing with the third derivative but we don't want to reach for a full algebraic approach yet. Anyway, I wonder if this can be extended any further.

[@DrBarker]

The third derivative gives us a nice way to understand how the curvature of a curve changes as we move along it. I'm thinking of curve sketching (by hand) as a nice opportunity to apply this knowledge. We could extend this further if we want to understand the 4th derivative geometrically, etc. The higher order derivatives could help us to understand exactly why the graphs of y = x^2, y = x^4, and y = x^6 all look subtly different.

[@emanuellandeholm5657]

In mechanics, there's this entire hierarchy for time derivatives of displacement: position, velocity, acceleration, jerk, snap, crackle, pop. I'm not even kidding. :D

[@kindreon]

Wow that's a sick visual representation. I'm working with something related to this in cognitive science surprisingly. It's kind of like how fast you're pushing your foot down on the gas pedal.

[@alexismiller2349]

8:42 no assumption about Taylor series are needed; little-o notation gets the job done too btw is the eps'eye'lon pronunciation of epsilon common somewhere? It's the first time I hear it

[@tessleblanc8050]

The curves look like boats! Love it

[@shadow-ht5gk]

Great channel man, definitely underrated

[@DrBarker]

Thank you!

[@TheMrbigfresh]

Perhaps you could do a video on a geometrical understanding og fractional derivatives aswell? :D That topic boggles my mind.

[@DrBarker]

This is a really interesting question. After a brief search, I'm not sure if there even is a widely-accepted geometrical interpretation!

[@TheMrbigfresh]

@@DrBarker Yes, i have seen some other content exploring some of the interpretations, but i really enjoy your style and presentation, so i thought that perhaps you had some thoughts on this subject.

[@DrBarker]

@@TheMrbigfresh Thank you! This would be a fun topic to cover, if I can actually work out a satisfactory way of interpreting fractional derivatives... Even e.g. if it's just for the half-derivative it would still be nice.

[@codatheseus5060]

Pretty sure the word for change in acceleration over time is jerk

[@behzat8489]

So, the 3rd derivative of a circle arc at any point is 0. Am I missing something?

[@APaleDot]

Well, if you define a circle parametrically then its curvature is constant. Curvature is kind of like a second derivative of a parametric equation, so the change in curvature would be 0. However, he never even mentions parametric equations, so this is just me trying to find _some_ interpretation of his lecture that fits. For anyone who wants to see that the 3rd derivative of a circle is _not_ zero, here's an intuitive explanation: consider the top half of the circle given by x² + y² = 1. When x = -1, the tangent line is vertical and the derivative shoots off to infinity at that point. But for every other point within the range (-1, 1), the derivative is well-defined and finite. There's no way for the derivative to shoot off to infinity over a finite range if it was changing at a constant rate. Therefore, the second derivative cannot be constant, and the third derivative is not zero. So as far as I can tell, this geometric interpretation is just wrong. Don't trust everything you see on the internet, I guess.

[@APaleDot]

Actually, I've found a way to salvage a part of the intuition. At 4:58 he says that we can simplify the proof by rotating the function such that the tangent line is horizontal. Of course, you are not allowed to "rotate" a function, that could make it not a function anymore! This simplification is invalid, as shown by the circle example which is symmetric under rotation. However, there is a way to make the tangent flat while maintaining the function: a skew transformation. If we skew the function vertically, all x positions remain fixed, and the functionality is preserved. This is equivalent to adding an extra linear term to our function that perfectly cancels the first derivative at the point of interest. Since this linear term goes to zero on the second and third derivative, it doesn't effect our results. So what is the upshot? Well, unfortunately, the first part of the intuition about mid-points being to the left or right of the normal is dead in the water, it's unsalvageable. This is because when we do the vertical skew, the normal rotates, but the x positions of the midpoints don't change, they only move vertically. Therefore, it is entirely possible for the normal to rotate from one side of the midpoints to the other during the skew. *However,* the part about the slope of the chord after the skew is still valid. After we skew, we can look at the asymmetry of the function by measuring that slope and it tells us which way the 3rd derivative is going. This also works with the circle because a circle _is_ asymmetrical under a vertical skew.

[@kepler4192]

What would happen if the 100th derivative is positive

[@epimaths]

math is good.....

[@APaleDot]

Unfortunately, this intuition is wrong. You should take this video down, or issue a correction.

[@General12th]

Thank you for your thorough and well-reasoned analysis.

[@APaleDot]

​@@General12th I have a longer comment in response to another comment that I didn't want to reproduce here. The counterexample that commenter gave is the third derivative of a circle. This intuition would imply that a circle has a third derivative of 0, since the circle is symmetric. But circles (when written as functions) don't have a third derivative equal to 0. Ipso facto, this intuition is wrong.

[@donegal79]

what a jerk. Great video.

[@johnchristian5027]

I was going to make the exact same joke :(

[@wilderuhl3450]

Insert some quip about you being a jerk for not referring to the rate of change of acceleration as the jerk, but my brain can’t think in terms of puns right now.

[@guruone]

He said "epsilon" incorrectly SO MANY TIMES................... LOL......... Cheers from Down Unda............ ps. I found the "God" Formula........ News at 10 :P