a cubic fact you probably didn’t know

Рет қаралды 31,286

2 жыл бұрын

What is so magical about the cubic function a x^3 + b x^2 + c x + d? Watch this video and find out!
Parabola fact you probably didn't know: • a parabola fact you pr...
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@jacemandt 2 жыл бұрын

Here is a fact about cubics a lot of people don't know that I learned about a long time ago: Let f(X) be a cubic polynomial with the stipulation that it has three real roots-call them r1, r2, r3 in any order. Find the average of r1 and r2-call it r_a. Then draw the tangent line to the cubic at (r_a,f(r_a)). That tangent line will always pass through (r3,0). The proof with calculus isn't super hard. I have been told that there's also a group theoretical explanation for this, but I don't know enough group theory to understand it!

@drpeyam 2 жыл бұрын

So cool!!!

@fedem8229 2 жыл бұрын

Damn, that's such a nice property

@kobethebeefinmathworld953 2 жыл бұрын

If I'm not mistaken, it is related to the Galois group of the cubic equation which is isomorphic to the permutation group with 3 objects.

@goldfing5898 Жыл бұрын

That was new to me and sounds very interesting! I will exam and verify this, at least for some examples.

@aryanjoshi3342 Жыл бұрын

@@goldfing5898 does it work for complex roots too?

@MathFromAlphaToOmega 2 жыл бұрын

This is an interesting property! I think you could also prove it by shifting the cubic by replacing x with x-b/(3a). That will get rid of the x^2 term, so we'll be left with a cubic of the form px^3+qx+r. Then shifting the graph down r units, we get px^3+qx, which is symmetric around the origin.

@Unc4pped 2 жыл бұрын

Very nice!

@anshumanagrawal346 2 жыл бұрын

That's how I thought about it

@chitranghosal879 2 жыл бұрын

Absolutely

@MichaelRothwell1 2 жыл бұрын

Yes, that was my approach. You can simplify a cubic by translating so that the inflexion is at the origin. This is very much like translating a quadratic so its vertex is on the y-axis, which algebraically corresponds to completing the square.

@chitranghosal879 2 жыл бұрын

Yes. And once you shift a quadratic and drag it so that is symmetric about the y axis, the cubics you build by integrating it will necessarily see inflection only at x=0. Then you can shift it vertically till you put the inflection at (0,0)

@henryginn7490 2 жыл бұрын

I think this was in Mathologer's video on the cubic formula, I remember doing the exact same proof of this fact back then. It feels a lot more surprising than it should be

@GreenMeansGOF 2 жыл бұрын

6:07 Notice that this equation is equivalent to showing that the average of the two symmetric function values equals f(k). So perhaps this can be used to simplify the derivation?

@slavinojunepri7648 Жыл бұрын

Good to know the symetry of the cubic polynomial about the point of inflection. Thank you Dr. Peyam!

@MathSolvingChannel 2 жыл бұрын

So it is anti-symmetric respect to the inflection point. I think you can proof it by shifting the graph by that vector

@sharpnova2 2 жыл бұрын

yep this is the half turn symmetry about the inflection point that makes cardano's method work. it transforms a cubic into the depressed cubic. and from what I've heard was rather depressing for its discoverer as well due to publishing issues and squabbles

@konstantingukov4752 2 жыл бұрын

This immediately follows from the fact that the derivative (parabola) is always symmetric :)

@atil4 2 жыл бұрын

You make math so much fan, not only because of your interesting topics, but also the energy and passion have when explaining math :)

@xenorzy9331 2 жыл бұрын

Amazing!

@manyirons 2 жыл бұрын

Awesome finish!

@magnuswootton6181 2 жыл бұрын

thumbs up. pretty good, gave me lots of ideas!

@paulopanarella1839 2 жыл бұрын

Congratulations, Professor! I really enjoyed it.

@JohnRandomness105 2 жыл бұрын

This is related to the start of the method for solving cubic equations: first divide by a and get the form x³ + bx² + ... Then let x = y - b/3, putting the equation in the form y³ + cy + d = 0. That puts y=0 at the inflection point.

@douglasmagowan2709 2 жыл бұрын

f(x-b/(3a)) will transform a general cubic into a diminished cubic, which is an odd function plus a constant. This is clearly symmetric about a point.

@michaelz2270 2 жыл бұрын

Cardano's formula for solving ax^3 + bx^2 + cx + d = 0 does something like that... after shifting so that the x^2 term goes away, it becomes of the form x^3 + ex + f = 0, which scales to x^3 + x + g = 0 or x^3 - x + g = 0. So really Cardano's formula boils down to solving equations of those two forms.

@sharpnova2 2 жыл бұрын

yep and the next step is basically "completing the cube" thx mathologer

@gersantru 2 жыл бұрын

Nice!

@dgrandlapinblanc 2 жыл бұрын

Thank you very much.

@parshvpatel9644 2 жыл бұрын

I love your accent and smile 😄

@koenth2359 2 жыл бұрын

Starting with f(x)=ax^3+bx^2+cx+d, it is easy to show that f(-b/3a + x) - f(-b/3a) = (a)x^3 + (c - b^2/a) x This is an odd function of x, so that f(-b/3a - x) - f(-b/3a) = f(-b/3a) - f(-b/3a + x) follows directly ■

@paolopalmisano827 2 жыл бұрын

For the polynomio the simmetrc point=inflection point

@paolopalmisano827 2 жыл бұрын

Vert cool

@DoctrinaMathVideos 2 жыл бұрын

Great video! I want to come up with a new kind of Simpson's Technique for approximating integrals using a cubic polynomial instead of a quadratic polynomial. I can probably implement some of the ideas that you present here.

@mychessnotebook8653 2 жыл бұрын

That's already been done. I think it's called Simpson's 3/8 rule. I'm pretty sure generalizations to any degree polynomials have been done too. But I'm sure it would be beneficial to you to come up with your own formulas and then compare to what's been published.

@sonarbangla8711 2 жыл бұрын

Beautiful.

@VolkGreg 2 жыл бұрын

Easier solution. Substitute z = x - k, k = b/3a so the general cubic becomes g(z) = Az³ + Cz + D = f(x) = ax³ + bx² + cx + d where A = a, B = 0, C = c - kb, D = d + kc - 2k³a. Symmetry of g(z) requires g(z) + g(-z) = 2g(0) This is obviously true, since A and C are coefficients of odd powers. That is... g(z) + g(-z) = (Az³ + Cz + D) + (-Az³ - Cz + D) = 2D = 2g(0)

@seth333 2 жыл бұрын

Thanks!

@drpeyam 2 жыл бұрын

Thanks again! Your contributions are greatly appreciated 🤗

@CAG2 2 жыл бұрын

Funny thing is I had all the prerequisite knowledge to know this... I knew that any cubic could be reduced to a "depressed" cubic by using letting u = x - b/(3a) (thanks Veritasium) then considering f(u) (ie, moving the graph right by a units), which removes the even x^2 term. And of course then you can add or subtract a constant term to get rid of the even constant term. But I never thought to combine these!

@nnaammuuss 2 жыл бұрын

Something like _‘every origin-fixing orientation-preserving rigid motion in R³ must be a rotation’_ because every 3×3 real matrix must have a bloody eigen value?

@paolopalmisano827 2 жыл бұрын

The only inflection point=simmetric point for the polinomio

@philipoakley5498 2 жыл бұрын

Fun, fun, fun. Enjoyable!

@francoisseguin6692 2 жыл бұрын

For a polynomial f of degree 3 f(x+h)=f(h)+xf'(h)+x^2/2f"(h)+x^3/6f'''(h) (Taylor lagrange identity for polynomials). So if f"(h)=0 x->f(x+h)-f(h)=xf'(h)+x^3/6f'''(h) is a odd function.

@tomctutor 2 жыл бұрын

And the point on a quadratic that has its line of symmetry is x= -b/2a so is the quintic (order 5 polynomial) x=-b/5a because every quintic does have a point of inflexion? [Notice that quartic f(x)=x^4+x^3+x^2 has no point of inflexion ]

@gurkiratsingh7tha993 2 жыл бұрын

Please make a video on the riemann zeta function and it's analytic continuation

@drpeyam 2 жыл бұрын

🤮🤮

@cparks1000000 Жыл бұрын

I always think of symmetry about a point in terms of 180 degree rotation about that point.

@erfanmohagheghian707 2 жыл бұрын

It's surprising that you didn't notice it before! shifting x to x-b/3a is the first step of solving a cubic polynomial.

@chitranghosal879 2 жыл бұрын

The fact is that every cubic is just a shifted scaled and stretched equivalent of y=×^3. None of which changes the fact that it is symmetric about its inflection point. It is similar to the fact that every quadratic is symmetric by reflection about the line x=-b/2a. It seems strange only when you think about it purely from an algebraic standpoint

@drpeyam 2 жыл бұрын

I think it’s very strange, even that x^3 has this property

@person1082 2 жыл бұрын

how can you get a function like x³-x just from transformations of x³

@anshumanagrawal346 2 жыл бұрын

That's not quite true for Cubics

@chitranghosal879 2 жыл бұрын

@@person1082 just like you can get x(x-1) from x^2 but still have it reflect symmetrically about x=-b/2a.

@chitranghosal879 2 жыл бұрын

@@anshumanagrawal346 why?

@Joe_Payne 2 жыл бұрын

You can just use the fact that ax^3 is odd bx^2 is even and cx is odd. Adding d just moves it up/down. Odd and even functions are symmetrical so Adding them will be too

@bjornfeuerbacher5514 2 жыл бұрын

Actually, adding an odd function and an even function usually results in a function which is _not_ symmetric. Try x^5 + x^4.

@MrBeen992 2 жыл бұрын

I had to watch it two times to understand what was the purpose of doing all of this. I will do it for the quadratic with k= -b/2a and the fact that f(x)=x^2 is even

@danielantone6216 2 жыл бұрын

Would a translation of the origin to the centre of symmetry do the trick? After transformation the function will be in the form f(X)=X^3, where (X, f(X)) is measured from the new origin.

@bjornfeuerbacher5514 2 жыл бұрын

Actually, it will be in the form f(x) = ax³ + cx.

@jamiewalker329 2 жыл бұрын

f'(x) has even symmetry about the x = -b/3a (complete the square to see this). Hence the integral between -b/3a to x of f'(t) will have odd symmetry about x = -b/3a. Adding a constant of integration +d doesn't change this symmetry property. DONE.

@MrBeen992 2 жыл бұрын

Is there something similar for quintic polynomials ?

@gurkiratsingh7tha993 2 жыл бұрын

I wonder, is the zeta function odd or even or none of these???

@goldfing5898 Жыл бұрын

Well, I did know :-) but anyway the point symmetry of the cubic parabola is relatively little known in comparison to the axis symmetry of the quadratic parabola. And I appreciate your enthusiasm.

@nedmerrill5705 2 жыл бұрын

Algebra fun?!? You bet!! Very instructive! A crystallographic _p2_ space group.

@praveenb9048 2 жыл бұрын

If we take x^3 and add -10(x-5)^2 , we would expect that the parabolic term would totally skew the result and destroy the symmetry. To me, this is why this video is so astonishing. I still find it sort of hard to imagine.

@VeteranVandal 2 жыл бұрын

I guess it gets easier to see if you go to the complex plane. If you make everything symmetric, in a cubic equation, you can trivialise the solution based on simple geometry. I've seen other vids about this doing exactly that.

@paolopalmisano827 2 жыл бұрын

Grazie mille for the video

@carstenbode8251 2 жыл бұрын

Dear Doc, honestly, I knew this already!

@chessematics 2 жыл бұрын

Can we do the same for any odd polynomial? I suspect symmetry at the point where the n-1 derivative is 0.

@paolopalmisano827 2 жыл бұрын

I suspect symmetry at inflection points

@bjornfeuerbacher5514 2 жыл бұрын

No. Try the simple example x^5 + x^4. That has no symmetry at all.

@kyleneilson1457 2 жыл бұрын

A cool property of cubic curves: If two cubic curves intersect at 9 points, say p1,...,p9, then any cubic curve going through p1,...,p8 also goes through p9. This is the Cayley-Bacharach Theorem.

@drpeyam 2 жыл бұрын

Interesting

@dramwertz4833 2 жыл бұрын

i remember realizing that when starting with calculus in school. think i wrote a proof for it too

@dragonflycrashed5511 2 жыл бұрын

i`m a hopeless idiot when it comes to math, but your enthusiasm is contagious.....all those funny characters on your board really seem to make some sense; but allow me to ask: does that guy drawn on the left of the board at 0:27" have a mole on his shoulder?

@jesusthroughmary 2 жыл бұрын

I think I learned this or thought it instinctively

@nbooth 2 жыл бұрын

I don't consider myself a math genius, but I find this fact totally unsurprising, even obvious.

@lacasadeacero 2 жыл бұрын

all polynoms are symmetric or antisymmetric.

@brendanlawlor2214 2 жыл бұрын

nice symmetric shifting.....bet Dr TP would do even better when he gets another Rabbit ! 🐰🐰🐇

@drpeyam 2 жыл бұрын

Awwww probably

@eamonnsiocain6454 2 жыл бұрын

It is good to see someone who loves Maths as well as I. Thanks.

@dylanogden9337 2 жыл бұрын

I heard a rumor that Dr. Peyam is right-handed, he just writes backwards and flips the image to give himself something to do.

@drpeyam 2 жыл бұрын

😂😂😂

@citizencj3389 2 жыл бұрын

The cubic formula is just an extension of the pythagorean theorem!

@willie333b 2 жыл бұрын

We’re taught this way at school lol, minus calculus

@Dziaji 2 жыл бұрын

Nice dude. You must get all the ladies with this one. “Hey baby, wanna see how all cubics have 180 degree rotational symmetry? Come to my room, and I’ll show you.”

@Aramizyera23 2 жыл бұрын

It's symmetric 2 order derivative

@SimonPetrikovv 2 жыл бұрын

If the point of symmetry is (x0,f(x0)) (we show here that x0 = -b/3a, yeah) in the cubic, shouldn't that be the mid point of (x+x0,f(x+x0)) and (-x+x0,f(-x+x0)), thus 2f(x0) = f(x+x0)+f(-x+x0)? For the example you used, x0 = -b/3a = 2 and then f(x+2)+f(-x+2) = 2f(2) f(x-b/3a)+f(-x-b/3a) = 2f(-b/3a), isn't it?

@SimonPetrikovv 2 жыл бұрын

Oh, you considered f(-x+k)+f(x+k) - 2f(k) = 0, so it's not really an issue at the end, it was only a typo in the sign there but it was corrected after, ok, that's nice

@mininougaming757 2 жыл бұрын

thank god i knew u re channel

@drpeyam 2 жыл бұрын

Thank you!!

@mininougaming757 2 жыл бұрын

@@drpeyam u too i wish you can record a video about some logharithme limits

@Reza_Audio 2 жыл бұрын

Interesting

@adki231 2 жыл бұрын

Why do all these presentations always show a misleading diagram in -x- axis?

@indarajgochermaths5176 2 жыл бұрын

nice

@jlol4951 2 жыл бұрын

Cool

@8jhjhjh 2 жыл бұрын

Why does this guy kinda look like electroBOOM

@ff-ti7nj 2 жыл бұрын

I think both are the same race.

@uy-ge3dm 2 жыл бұрын

was this not obvious??

@KiwiSteveYT 2 жыл бұрын

Why does he say "minus x" when he means 'negative x'. I see this often, and think it reveals a basic misunderstanding of integers vs addition and subtraction.

@drpeyam 2 жыл бұрын

Why misunderstanding?

@KiwiSteveYT 2 жыл бұрын

@@drpeyam I always understood and taught that minus is, like plus, an instruction about what to do, whereas negative and positive are the sign of places on the number line, or equivalent with algebraic terms. So, negative 3 minus 5 equals negative 8 means start at negative 3 and 'go back' 5 places, and you'll arrive at negative 8. But minus 3 minus 5 equals minus 8... to me it has no meaning. If you look at the international Pisa studies, you'll see that the operations on integers remains a weak point in maths understanding. Perhaps this is one reason why ? I'm happy to hear your view on this :)

@jamiewalker329 2 жыл бұрын

@@KiwiSteveYT You are being pedantic in the extreme - there is no ambiguity here, and no one else seems to be having a misunderstanding.

@KiwiSteveYT 2 жыл бұрын

@@jamiewalker329 perhaps, but the international test data shows many students do have a misunderstanding. You, as an educated adult have long since resolved this, but young students do not learn by being presented with ambiguity and semantics. That is my point. Perhaps the others you refer to have never had to teach young people, or to clarify their misunderstanding brought on by the lack of ambiguity, such as presented above. If you were to take off your educated hat for a minute, you might see what I am trying to point out. I mean no disrespect, I just know where many students stumble and that is when they are not taught mathematics - better still, arithmetic - as a symbolic language. Shalom

@KiwiSteveYT 2 жыл бұрын

I said 'lack of ambiguity'. I meant 'ambiguity'