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.
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.
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)
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!
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
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?
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 ■
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
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
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.
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.
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)
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.
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
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?
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 ]
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.
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.
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.
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!
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.
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.
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?
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?
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
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.
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.
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.
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
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 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 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
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.”
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.
Very nice!
That's how I thought about it
Absolutely
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.
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)
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!
So cool!!!
Damn, that's such a nice property
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.
That was new to me and sounds very interesting! I will exam and verify this, at least for some examples.
@@goldfing5898 does it work for complex roots too?
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
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?
Thanks!
Thanks again! Your contributions are greatly appreciated 🤗
mistake at 4:15 it is f( -x - b/(3a)) instead of f( -x + b/(3a)) because it is f( -x + k) and k= - b/(3a)
Good to know the symetry of the cubic polynomial about the point of inflection. Thank you Dr. Peyam!
So it is anti-symmetric respect to the inflection point. I think you can proof it by shifting the graph by that vector
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
■
For the polynomio the simmetrc point=inflection point
Vert cool
This immediately follows from the fact that the derivative (parabola) is always symmetric :)
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
I think it’s very strange, even that x^3 has this property
how can you get a function like x³-x just from transformations of x³
That's not quite true for Cubics
@@person1082 just like you can get x(x-1) from x^2 but still have it reflect symmetrically about x=-b/2a.
@@anshumanagrawal346 why?
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
You make math so much fan, not only because of your interesting topics, but also the energy and passion have when explaining math :)
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.
yep and the next step is basically "completing the cube"
thx mathologer
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.
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.
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)
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.
Awesome finish!
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
Actually, adding an odd function and an even function usually results in a function which is _not_ symmetric.
Try x^5 + x^4.
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.
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?
I always think of symmetry about a point in terms of 180 degree rotation about that point.
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 ]
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.
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.
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.
Congratulations, Professor! I really enjoyed it.
The only inflection point=simmetric point for the polinomio
I wonder, is the zeta function odd or even or none of these???
Please make a video on the riemann zeta function and it's analytic continuation
🤮🤮
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!
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.
Actually, it will be in the form f(x) = ax³ + cx.
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.
Is there something similar for quintic polynomials ?
thumbs up. pretty good, gave me lots of ideas!
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?
Beautiful.
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?
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
I love your accent and smile 😄
Why do all these presentations always show a misleading diagram in -x- axis?
Thank you very much.
Fun, fun, fun. Enjoyable!
I heard a rumor that Dr. Peyam is right-handed, he just writes backwards and flips the image to give himself something to do.
😂😂😂
Can we do the same for any odd polynomial? I suspect symmetry at the point where the n-1 derivative is 0.
I suspect symmetry at inflection points
No. Try the simple example x^5 + x^4. That has no symmetry at all.
I don't consider myself a math genius, but I find this fact totally unsurprising, even obvious.
Dear Doc, honestly, I knew this already!
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.
Interesting
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.
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.
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
It's symmetric 2 order derivative
Nice!
Amazing!
all polynoms are symmetric or antisymmetric.
I think I learned this or thought it instinctively
It is good to see someone who loves Maths as well as I. Thanks.
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.
Why misunderstanding?
@@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 :)
@@KiwiSteveYT You are being pedantic in the extreme - there is no ambiguity here, and no one else seems to be having a misunderstanding.
@@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
I said 'lack of ambiguity'.
I meant 'ambiguity'
We’re taught this way at school lol, minus calculus
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.”
i remember realizing that when starting with calculus in school. think i wrote a proof for it too
Algebra fun?!? You bet!! Very instructive! A crystallographic _p2_ space group.
thank god i knew u re channel
Thank you!!
@@drpeyam u too i wish you can record a video about some logharithme limits
The cubic formula is just an extension of the pythagorean theorem!
nice symmetric shifting.....bet Dr TP would do even better when he gets another Rabbit ! 🐰🐰🐇
Awwww probably
was this not obvious??
Grazie mille for the video
Boom boom boom boom
I want you in my room
Let's do some math together
From now until the limit where _t_ goes to infinity
Why does this guy kinda look like electroBOOM
I think both are the same race.
This was surprising? All cubics are a scaled / shifted version of x^3+ax for some a controlling the slope at the origin. Learned this in Calc I.
Interesting
nice
3곱하기 변곡점
Cool